Física Para Ingenieria y Ciencias - Ohanian - 3ed Solucionario.pdf

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CHAPTER 1

SPACE, TIME, AND MASS

Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions Manual, available for purchase. Answers to all solutions below are underscored. 1-1.

Assume a height of 5 ft 10 in. Then, 5′10″ = 70 in = 70 in × 2.54 cm/in = 178 cm.

1-2.

There are approximately 300 actual pages in this text (vol. 1). The thickness is 2.5 cm. Therefore each page = 2.5 cm/300 pg = 8.3 × 10−3 cm.

1-3.

100 yd × 0.914 m/yd = 91.4 m; 53 1/3 yd × 0.914 m/yd = 48.7 m 1 step 1000 m steps N = × = 1.7 × 103 0.60 m 1 km km 1 pica 17 1 pica L = 11 in × = 66 picas. W = in × = 51 picas 1 1 2 in in 6 6 −8 Virus: 2 × 10 m × 10/0.3048 ft/m × 12 in/ft = 8 × 10−7 in

1-4. 1-5. 1-6.

Similarly: Atom: 1 × 10−10 m × 39.4 in/m = 4 × 10−9 in Fe Nucleus: 8 × 10−15 m × 39.4 in/m = 3 × 10−13 in Proton: 2 × 10−15 m × 39.4 in/m = 8 × 10−14 in †1-7.

1-8.

Let’s convert 1/2 inch to mm using conversion factors, then use proportional reasoning to do the others until the number of significant figures becomes large. 1 in × 25.4 mm/in = 12.7 mm 2 1 12.7 mm in = = 6.35 mm 4 2 1 6.35 mm in = = 3.175 mm = 3.18 mm (to three significant figures) 8 2 1 3.175 mm in = = 1.5875 mm = 1.59 mm (to three significant figures) 16 2 The number of digits is becoming large, so let’s do direct conversions for the rest of the problems. 1 in × 25.4 mm/in = 0.794 mm 32 1 in × 25.4 mm/in = 0.397 mm 64 10−3 in 2.54 cm 10−2 m 10−6 µm 1 mil × × × × = 25.4 µm. mil in cm m 1000 µm 1 mil 1 mm × × = 39.4 mil mm 25.4 µm

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CHAPTER 1-9.

1

(a) Grapefruit diameter ≈ 0.1 m Ratio of grapefruit/sun = 0.1 m/(1.4 × 109 m) = 7 × 10−11 Earth diameter ≈ 13 × 106 m Comparative size of Earth = 13 × 106 m × (7 × 10−11) = 9 × 10−4 m ≈ 1 mm. Nearest star distance = 4 × 1016 m Comparative distance = 4 × 1016 m × (7 × 10−11) = 2.8 × 106 m (b) Head diameter ≈ 0.2 m Earth diameter ≈ 13 × 106 m Earth/head ratio ≈ 7 × 107 Size of atom = 10−10 m Comparative size of atom = 10−10 m × (7 × 107) = 7 × 10−3 m = 7 mm Size of red blood cell ≈ 7.5 × 10−6 m Comparative size of cell = 7.5 × 10−6 m × (7 × 107) ≈ 500 m = 1/ 2 km

1-10.

Distance to Q1208 + 1011 = 12.4 × 109 × 9.47 × 1015 = 1.17 × 1026 m Distance on the diagram (PRELUDE, p. 6) 1.17 × 1026 = = 7.8 × 105 m 20 1.5 × 10

1-11.

Size (diameter) of the sun = 2 × 6.46 × 108 = 1.4 × 109 m distance on the diagram (PRELUDE, p. 1.4 × 109 6) = 10−3 m = 1 mm 1.5 × 1012

1-12.

†1-13.

10−9 m = 6.33 × 10−13 m. According to Table 1-1, the diameter of an nm atom is about 1 × 10−10 m, so this is 6.33 × 10−3 times the diameter of an atom, or roughly 1/100 the diameter of an atom. 1 turn = 360°, so 5° × 1 turn/360° = 0.0139 turn. For an English thread, 1in 0.0254 m 106 µm 0.0139 turn × × × = 4.41 µm. For a metric thread, 80 turns in m ∆l = 10−6 × 633 nm ×

0.5 mm 10−3 m 106 µm × × = 6.94 µm. turn mm m 1 nmi = 1852 m; Circumference of Earth = 4.00 × 107 m Circumference = (4.00 × 107 m)/1852 m/nmi = 21, 600 nmi 0.0139 turn ×

1-14.

Also 360° × 60 min/deg = 21, 600 min, so 1nmi ⇒ 1min


CHAPTER †1-15.

For one of the triangles, (R + 1.75 m)2 = R2 + (4700 m)2. Expand this to get R 2 + 2(1.75 m) R + (1.75 m) 2 = R 2 + (4700 m) 2 . We expect R to be much larger than 1.75 m, so we can ignore (1.75 m)2 relative to all the other terms. The R2 terms cancel, leaving (3.50 m)R = (4700 m)2, which gives R = 6.3 × 106 m.

1-16.

9400

R R + 1.75 m

22 yr, 5 mo, 23 days = (8035 + 153 + 23) = 8211 days (This excludes leap years and assumes average 30.5-day month.) 1 day = 1 day × 24 h/day × 60 min/h × 60 s/min = 86,400 s 8211 days = 8211 days × 86,400 s/day = 7.1 × 108 s

1-17. 1-18. 1-19.

1-20. †1-21.

1-22. 1-23. 1-24. 1-25. 1-26.

1 yr = 365.25 days. Therefore, 4.5 × 109 yr = 4.5 × 109 yr × 365.25 day/yr × 86,400 s/day = 1.4 × 1017 s 60 s 1 calculation = 3.6 × 1012 calculations/h. × min 10−9 s 3600 s 60 s 2 h 9 min 21s = 2 h × + 9 min × + 21s = 7761 s h min 3600 s 60 s 2 h 24 min 51s = 2 h × + 24 min × + 51s = 8692 s h min 365.25 solar days/year 24 h 1 sidereal day × × = 23.934 h/sidereal day. Using 1 h = 60 366.25 sidereal days/year solar day N =1h ×

60 min h

×

min to convert the 0.934 h to minutes gives 1 sidereal day = 23 h 56 min. s N = 4 ticks × 3.2 × 107 × 10 years = 1.2 × 109 ticks year 1h 1day 106 s × × = 11.6 days 3600 s 24h 7 days 24 h 3600 s 1week × × = 168 h. 168 h × = 6.048 × 105 s hr week day beats 1 min 3.2 × 107 s = 3.8 × 107 beats/year ticks × × min 60 s yr (a) June 24−25: (20 − 4) s/day = 16 s/24 h = 0.67 s/h N = 71

June 25−26: (34 − 20) s/day = 14 s/24 h = 0.58 s/h June 26−27: (51 − 34) s/day = 17 s/24 h = 0.71 s/h (b) Average rate = (51 − 4)/3 day = 47 s/(24 × 3)h = 0.65 s/h


1

CHAPTER

1

(c) 10h30m on June 30 is 70.5 h after noon June 27. By average loss, watch should have lost 70.5 h × 0.65 s/h = 46 s. Combined with loss of 51 s on June 27 gives total loss of 97 s. Therefore the correct WWV time is 10h 31m 37s . With the largest rate, loss is 70.5 h × 0.71 s/h = 50 s. Combined loss is 51 + 50 = 101 s. Then estimated WWV time is 10h 31m 41s . The wristwatch can be trusted to †1-27.

1-28.

about ±4 s on June 30. 1 day = 24 hr = 86,400 s. The earth rotates 360º per day, which corresponds to a rotation rate of 360° 60 min × = 0.250 min/s. A timing error of 1 s will result in an angular error of 0.250 86, 400 s degree min. According to Problem 1-14, 1 min = 1852 m, so the corresponding error in position is 0.250 min/s × 1852 m = 463 m = 0.463 km. 140 lb-mass = 140 lb-mass × 0.454 kg/lb-mass = 64 kg 140 lb-mass = 140 lb-mass × 0.454 kg/lb-mass × 1/14.6 slug/kg = 4.35slug 140 lb-mass = 140 lb-mass × 0.454 kg/lb-mass × 1 amu/(1.67 × 10−27 kg) = 3.8 × 10 28 amu

†1-29.

m planets = (0.33 + 4.9 + 5.98 + 0.64 + 1900 + 553 + 87.3 + 10.8 + 0.66) × 1024 kg = 2.56 × 1027 kg (to three significant figures). msun = 1.99 × 1030 kg, so the total mass is mtotal = msun + m planets = 1.99 × 1030 kg + 2.56 × 1027 kg = 1.99 × 1030 kg (to three significant figures). The fraction of the total mass included in the planets is m planets 2.56 × 1027 × 100% = × 100% = 0.134%. The fraction of the mass in the sun is 100% − mtotal 1.99 × 1030

1-30.

0.134% = 99.9%. largest length 1 × 1026 m 4 × 1017 s 40 longest time = = 5 × 10 . = = 4 × 1041. smallest length 2 × 10−15 m shortest time 1 × 10−24 s largest mass 1 × 1053 kg = = 1 × 1083. The first two ratios are within an order of magnitude of −31 smallest mass 9 × 10 kg

†1-31.

1-32.

1-33.

1-34.

each other, and the third is roughly equal to the square of the other two. (Note that the first two ratios will keep increasing because the universe is expanding and aging.) From the periodic table in the Appendix, we see that the uranium nucleus contains about 238 nucleons, each with the mass of a proton. Table 1.7 gives 1.7 × 10−27 kg for the mass of a proton. Then the total mass of the electrons is (92)( 9.1 × 10−31 kg) = 8.4 × 10−29 kg, and the total mass of the nucleus is (238)( 1.7 × 10−27 kg) = 4.0 × 10−25 kg. To two significant figures, the total mass of the atom is 4.0 × 10−25 kg. The fraction of the total mass in the electrons is 8.4 × 10−29/4.0 × 10−25 = 2.1 × 10−4 = 0.021%. The fraction of mass in the nucleus is 99.98%. Let m represent the mass of material and M represent the mass of one mole. Then atoms ⎛ 0.1 × 10−6 g ⎞ ⎛m⎞ 14 N = N A ⎜ ⎟ = 6.02204 × 1023 ×⎜ ⎟ = 3 × 10 atoms. mol ⎝M ⎠ ⎝ 197 g/mol ⎠ 1 lb avoirdupois = 0.453 59 kg = 435.59 g (given in the text) 0.453 59 kg 1 lb troy = 0.822 86 lb avoidupois × = 0.373 24 kg = 373.24 g lb avoidupois ⎞ atoms ⎛ 10−21 kg ⎛m⎞ 4 N = N A ⎜ ⎟ = 6.02204 × 1023 ×⎜ ⎟ = 1.1 × 10 atoms mol 0.055 85 kg/mol ⎝M ⎠ ⎝ ⎠


CHAPTER 1-35.

1

(a) Since the density of water is 1 g/cm3, 250 cm3 of water has a mass of 250 g, which is 250 g/(18 g/mol) = 14 mol. Therefore, the number of molecules is 14 mol × 6.02 × 1023 molecules/mol = 8.4 × 1024 molecules. (b) Mass of sea-water (density = 1030 kg/m3) is 1.3 × 1018 m3 × 1030 kg/m3 = 1.3 × 1021 kg = 1.3 × 1024 g Number of molecules of sea-water is 1.3 × 1024 g × 1 mol/18 g × 6.02 × 1023 molecules/mol = 4.3 × 1046 molecules. (c) Ratio of molecules of water from cup to molecules from sea is 8.4 × 1024/4.3 × 1046 = 2.0 × 10−22 . The probability of a single molecule drawn from the ocean being originally from the cup is therefore 2.0 × 10−22. Drawing a cup full then should result in 2.0 × 10−22 × 8.4 × 1024 = 1680 molecules.

1-36.

The mass of H2 and He, respectively, is 0.70 × 1.99 × 1030 kg = 1.39 × 1030 kg and 0.30 × 1.99 × 1030 kg = 0.60 × 1030 kg. The atomic mass of hydrogen and helium, respectively, is 1 g/mol and 4 g/mol. Therefore, the number of hydrogen atoms is NH = 1.39 × 1033 g × 1 mol/g × 6.02 × 1023 molecules/mol = 8.38 × 1056 molecules. The number of helium atoms is NHe = 0.60 × 1033 g × 1 mol/4 g × 6.02 × 1023 molecules/mol = 9.03 × 1055 atoms. Total number = 9.28 × 1056

†1-37.

1-38.

Molecular mass of N2 = 28 g/mol Molecular mass of O2 = 32 g/mol Molecular mass of Ar = 40 g/mol Therefore, 1000 g of air will contain: 755 g N2 = 755 g/(28 g/mol) = 27.0 mol 232 g O2 = (232/32) mol = 7.25 mol 13 g Ar = (13/40) mol = 0.325 mol The percentage by number of molecules of these substances is: N2: 27.0/(27.0 + 7.25 + 0.325) × 100% = 27.0/34.575 × 100% = 78.1% O2: (7.25/34.575) × 100% = 21% Ar: (0.325/34.575) × 100% = 0.9% Therefore, the “molecular mass” of air is (0.781 × 28) + (0.21 × 32) + (0.009 × 40) = 28.95 g/mol. The mass of elements is: Oxygen: 0.65 × 73 kg = 47.4 kg; atomic mass = 16 Carbon: 0.185 × 73 kg = 13.5 kg; atomic mass = 12 Hydrogen: 0.095 × 73 kg = 6.94 kg; atomic mass = 1.008 Nitrogen: 0.033 × 73 kg = 2.41 kg; atomic mass = 14 Calcium: 0.015 × 73 kg = 1.09 kg; atomic mass = 40.08 Phosphorous: 0.01 × 73 kg = 0.73 kg; atomic mass = 31


CHAPTER

†1-39.

1-40. 1-41. 1-42.

1-43.

1

The number of atoms of these substances is: O: 47.4 × 103 g × 1/16 mol/g × 6.02 × 1023/mol C: 13.5 × 103 g × 1/12 mol/g × 6.02 × 1023/mol H: 6.94 × 103 g × 1 mol/g × 6.02 × 1023/mol N: 2.41 × 103 g × 1/14 mol/g × 6.02 × 1023/mol Ca: 1.09 × 103 g × 1/40 mol/g × 6.02 × 1023/mol P: 0.73 × 103 g × 1/31 mol/g × 6.02 × 1023/mol Total number of atoms: 0.53° = 0.53/360 × 2π rad = 9.25 × 10−3 rad d = 9.25 × 10−3 rad × 1.5 × 1011 m = 1.4 × 109 m r = 6.9 × 108 m

= 1.78 × 1027 = 6.77 × 1026 = 4.17 × 1027 = 1.04 × 1026 = 1.64 × 1025 = 1.42 × 1025 = 6.76 × 1027

m⎞ days h s ⎛ 1 ly = ⎜ 3.00 × 108 ⎟ × 365.25 × 24 × 3600 = 9.47 × 1015 m s ⎠ year day h ⎝ 15 1 ly = 9.47 × 10 m 2.2 × 106 ly = 2.2 × 106 ly × 9.47 × 1015 m/ly = 2.1 × 1022 m 1 light-s (ls) = 3.00 × 108 m/s × 1 s = 3.00 × 108 m 1 light-min (lm) = 3.00 × 108 m/s × 60 s = 1.80 × 1010 m E-S distance = 1.50 × 1011 m × 1/(1.80 × 1010) lm/m = 8.3 lm E-M distance = 3.84 × 108 m × 1/(3.00 × 108) ls/m = 1.28 ls 1 2π 1 degree = 1 second of arc = × rad = 4.85 × 10−6 rad 3600 360 3600 1 AU = 4.85 × 10−6 (a) pc 1 pc = 1 × 1/(4.85 × 10−6)AU = 2.06 × 105 AU (b) 1 pc = 2.06 × 105 AU × 1.496 × 1011 m/AU = 3.08 × 1016 m 1 ly = 9.47 × 1015 m 1 pc = 3.08 × 1016 m × 1 ly/(9.47 × 1015 m) = 3.25 ly (c) 1 ly = 3.00 × 108 m/s × 1 yr = 3.00 × 108 m/s × 3.156 × 107 s = 9.47 × 1015 m 2

1-44.

⎛ 1 ft ⎞ 2 1m × ⎜ ⎟ = 10.76 ft . Note that the conversion factor must be squared. ⎝ 0.3048 m ⎠ 2

3

†1-45. 1-46. †1-47.

⎛ 1 ft ⎞ 3 1m3 × ⎜ ⎟ = 35.31 ft . Note that the conversion factor must be cubed. 0.3048 m ⎝ ⎠ 1 m2 Use the result from 1−44: A = (78 ft)(27 ft) × z = 196 m2. 10.76 ft 2 His height is measured to a precision of 0.1 inch, so we want to see how many significant digits this implies. 8 feet = 96 inches (exactly), so to the nearest 0.1 inch his height can be expressed as 107.1 inches, which contains four significant figures. Converting 11.1 inches to feet gives his height in feet: 8 ft + 11.1 in = 8.925 ft, which also contains four significant figures. Thus his m height in meters should be specified to four figures: 8.925 ft × 0.3048 = 2.720 m. ft


CHAPTER

1

2

1-48.

⎛ 1 m2 ⎞ ⎛ 3 ft ⎞ = 4.46 × 103 m2, using the result from 1−44. A = 100 yd × 53.33 yd × ⎜ × ⎜ ⎟ 2 ⎟ yd 10.76 ft ⎝ ⎠ ⎝ ⎠ 3

†1-49.

⎛ 1 kg ⎞ ⎛ 100 cm ⎞ 3 3 8.9 g/cm3 × ⎜ ⎟×⎜ ⎟ = 8.9 × 10 kg/m . Note that the cm to m conversion factor 1000 g m ⎝ ⎠ ⎝ ⎠ must be cubed! 1 ft = 0.3048 m, 1 lb = 0.454 kg. 3 ⎛ 1 lb ⎞ ⎛ 0.3048 m ⎞ 2 3 3 8.9 × 103 kg/m3 × ⎜ × ⎟ ⎜ ⎟ = 555 lb/ft , or 5.6 × 10 lb/ft to two significant 0.454 kg ft ⎝ ⎠ ⎝ ⎠ figures. Again, note that the m to ft conversion factor must be cubed. 1 ft = 12 in, so 3

1-50.

1-51. 1-52.

†1-53.

1-54. †1-55.

1-56.

†1-57.

1-58.

⎛ 1 ft ⎞ 3 555 lb/ft 3 × ⎜ ⎟ = 0.32 lb/in . ⎝ 12 in ⎠ Assume a mass of 73 kg and a density of 1000 kg/m3. mass V= = 73 kg × 1/1000 m3/kg = 0.073 m3 density cm3 3600 s 10−6 m3 = 7.9 m3/day. × 24 h × × 3 s h cm 5.2 × 103 cm3 1 liter = 103 cm3, so 5.2 liter = 5.2 × 103 cm3. t = = 57 s. 92 cm3 /s 10−4 m 2 1 cm 2 × cm 2 = 108 transistors. If they’re stacked, N = the number of transistors per N = m2 10−12 transistor layer (108) × number of layers. If the cube is 1 cm high and each layer is 10−7 m, or 10−5 cm, thick, then the cube holds 105 layers and the cube can hold 108 × 105 = 1013 transistors! 3 g 1 lb liter 3 cm −3 kg × 10 × 10 × × 3.875 1 gal = 3.785 liter. Density = 1.00 = 8.34 3 cm g liter 0.4536 kg gal 92

lb/gal (a) (3.6 × 104) × (2.049 × 10−2) = (3.6)(2.049) × 104−2 = 7.4 × 102 (b) (2.581 × 102) − (7.264 × 101) = (2.581 − 0.7264) × 102 = 1.855 × 102 7.9832 × 10−2 (c) 0.079832 ÷ 9.43 = = 0.847 × 10−2−0 = 8.47 × 10−3 0 9.43 × 10 3 3m 3(2.0 × 1030 kg) m m 3 g −6 m = = = = 1.5 g/cm3 × 10 × 10 3 8 3 3 4 π π V 4 R 4 (7.0 × 10 m) kg cm 3 πR 3 3 3m 3(2.0 × 1030 kg) 1 metric ton m m −6 m = = = Density = × × 10 3 4 V 4π (20 × 103 m)3 103 kg cm3 π R 3 4π R 3 = 6.0 × 107 metric tons/cm3 Oceans of the earth have 1.3 × 1018 m3 of water. The mass of the oceans is 1.3 × 1018 m3 × 1030 kg/m3 = 1.3 × 1021 kg. Mass of the earth is 5.98 × 1024 kg, so that the percentage of the mass of the earth that is water is: (1.3 × 1021 kg/5.98 × 1024 kg) × 100% = 0.02%.

Density =


CHAPTER

1

†1-59.

From Table 1.10, 1 liter = 10−3 m3. According to data given in the “Conversion of Units” section of the chapter, the density of water is 1000 kg/m3, so 10−3 m3 1 min = 5.00 × 10−3 m3 /s, and 300 liters/min × × liter 60 s 5.00 × 10−3 m3 /s × 1000 kg/m3 = 5.00 kg/s.

1-60.

1 in = 1 in × 2.54 cm/in × 1/100 m/cm = 0.0254 m. Therefore volume on 1 m2 is V = 0.0254 m × 1 m2 = 0.0254 m3 /m 2 .

†1-61.

Mass of this much water is 0.0254 m3/m2 × 1000 kg/m3 = 25.4 kg/m2. 3m m m This can be solved using proportional reasoning. Density = , which means = = 3 4 π 4 V R 3 πR 3 1/ 3

⎛ mlead mcopper mlead R R = , from which we get = ⎜ lead copper 3 3 ⎜m Rlead Rcopper ⎝ copper

⎞ ⎟⎟ ⎠

1/ 3

= (4.8 × 10

−15

⎛ 3.5 ⎞ m) ⎜ ⎟ ⎝ 1.06 ⎠

1/ 3

1/ 3 ⎛ moxygen ⎞ ⎛ 0.27 ⎞ −15 = 7.1 × 10 m. Likewise we get Roxygen = Rcopper ⎜ ⎟⎟ = (4.8 × 10 m) ⎜ ⎟ ⎜m ⎝ 1.06 ⎠ ⎝ copper ⎠ = 3.0 × 10−15 m. Note that it was not necessary to include the factor (× 10−25) when expressing the masses because it cancels in the ratio. mass 4 where V = πr3. Density is calculated as volume 3 Planet Mass (kg) Vol (m3) Density (kg/m3) 23 20 Pluto 6.6 × 10 ? 1.13 × 10 5800 ? 23 19 Mercury 3.3 × 10 5.94 × 10 5600 1.09 × 1021 5500 Earth 5.98 × 1024 24 20 9.51 × 10 5100 Venus 4.9 × 10 Mars 6.40 × 1023 1.62 × 1020 4000 26 22 4.58 × 10 2250 Neptune 1.03 × 10 5.65 × 1022 1550 Uranus 8.73 × 1025 27 24 Jupiter 1.90 × 10 1.52 × 10 1250 9.23 × 1023 600 Saturn 5.53 × 1026 One dimension of each roof segment must be divided by cos 45°, or multiplied by 2. Then total roof area = floor area × 2 = 250 m2 × 2 = 354 m2. Use the radius of the earth from Table 1-1. ∆s 15 m ∆θ = = = 2.3 × 10−6 radian. In ∆θ ∆s = 15 m R 6.4 × 106 m

−15

1-62.

1-63. 1-64.

R

degrees, this is (2.3 × 10−6 radian) × (180°/π radian) = 1.3 × 10−4 degree. †1-65.

The slope is the tangent of the required angle. For a slope of 1:5, θ = tan −1

1 = 11°. For a slope of 5

1 = 5.7°. For a slope angle of 0.1°, tan θ = 1.7 × 10−3, so the slope is 10 1:( 1.7 × 10−3)−1 = 1: 5.7 × 102, so the rise is about 1 atom per 570 atoms. 1:10, θ = tan −1


CHAPTER

1-66.

†1-67.

1-68.

1-69.

1-70.

h , where D = 75 m. Both D and θ are D specified to two significant figures, so the value calculated for h must be specified to two figures. Thus h = D tan θ = (75 m)( tan 78°) = 3.5 × 102 m.

1

tan θ =

h

θ D

In this calculation, assume that the calendar year is exactly 365 days long and a circle contains exactly 360°, so these are not interpreted as numbers with only three significant figures. The 365 × 360° = 359.76°. In four angle the earth moves through in one calendar year is θ = 365.24 years, including one leap year with exactly 366 days, the total number of days is 3 × 365 + 366 = 1461 × 360° = 1440.0°, 1461 days. The angle the earth moves through in four years is θ = 365.24 which is exactly four complete circles to five significant figures. (The angle is actually a little larger than 1440.0°, which is why there are some four-year intervals that do not include an extra day.) The sun will set at Marchena n-minutes after 8 PM where minutes in a day × distance between islands 24 × 60 × 60 × 103 = = 2.15 min n= circumference of the earth 4 × 107 Therefore, the sun sets at Marchena at 2.15 min after 8 PM. The diameter, d of the tree trunk is related to its length, L by d = AL3/2, where A is a constant 7.6 = A(81)3/2 if L = 90 m, then 3/ 2 ⎛ 90 ⎞ 3/2 d = A(90) = ⎜ ⎟ 7.6 = 8.9 m. ⎝ 81 ⎠ The mass, m of the tree trunk is related to d and L by 2 ⎛d⎞ m = ρπ ⎜ ⎟ L, where ρ is the mass density. ⎝2⎠ π 6100 = ρ (7.6) 2 81 4 for L = 90 m, 2 π ⎛ 8.9 ⎞ 90 = 9295 tons m = ρ (8.9) 2 90 = 6100 ⎜ ⎟ 4 ⎝ 7.6 ⎠ 81 Distance from pole to equator = 1/4 circumference. Therefore 1 π π × 6.37 × 106 m = 1.00 × 107 m. d = (2πr ) = r = 4 2 2 Straight line distance is, using the Pythagorean theorem: d=

r 2 + r 2 = 2r 2 =

2 × 6.37 × 106 m = 9.0 × 106 m.


CHAPTER

†1-71.

1-72.

1-73.

1

M 235 g/mol = 3.902 × 10−22 g = 3.902 × 10−25 kg. In atomic mass units, = NA 23 atoms 6.02204 × 10 mol 3.902 × 10−25 kg = 235.0 u. this is matom = kg 1.66054 × 10−27 u The molar mass M of water (H2O) is 18.0 g/mol. The mass m of 1 liter (1000 cm3) of water is m 1000 g molecules 1000 g. N = × NA = × 6.02204 × 1023 = 3.35 × 1025 molecules. Each M 18.0 g/mol mol matom =

molecule contains one oxygen atom and two hydrogen atoms, so there are 3.35 × 1025 oxygen atoms and 6.70 × 1025 hydrogen atoms. Molecular mass of N2 is 2 × 14 g/mol = 28 g/mol. The density of air is 1.3 kg/m3. The number of molecules in 1 cm3 of air is then number of molecules = 10−6 m3 × 1.3 kg/m3 × 1000 g/kg × 1 mol/28 g × 6.02 × 1023 molecules/mol = 2.8 × 1019 molecules. 3

1-74. 1-75.

3 cells ⎛ mm ⎞ −3 m N = 5.1 × 10 × 1000 × 10 × 5.2 liters = 2.7 × 1013 cells. ⎜ ⎟ 3 mm m ⎠ liter ⎝ Volume of paint = area × thickness. ⎛ volume m3 ⎞ −4 2 = (1 liter) × ⎜ 10−3 Thickness = ⎟ ÷ (8 m ) = 1.25 × 10 m (0.125 mm) area liter ⎠ ⎝ 6

2

1-76.

m ⎞ ⎛ 12 2 The area is 9.4 × 106 km 2 × ⎜ 103 ⎟ = 9.4 × 10 m . km ⎠ ⎝ mass kg ⎛ ⎞ 12 2 −4 2 Area density = = 8 × 106 metric tons × ⎜ 103 ⎟ ÷ (9.4 × 10 m ) = 9 × 10 kg/m area metric ton ⎠ ⎝

1-77.

55 mi/h = 55 mi/h × 1.609 km/mi = 88.5 km/h = 55 mi/h × 5280 ft/mi × 1/3600 h/s = 80.7 ft/s = 88.5 km/h × 1000 m/km × 1/3600 h/s = 24.6 m/s

1-78.

Density =

1-79.

kg 1 metric ton ⎛ 1 m ⎞ 8 3 cm this is 2.33 × 10 × ×⎜ 2 ⎟ = 2.3 × 10 metric tons/cm . 3 3 m 10 kg ⎝ 10 cm ⎠ h 3600 s 120 yr × 365 days/yr + 237 days = 44067 days. 44067 days × 24 × = 3.81 × 109 s day h 1 h = slope × 300 m = × 300 m = 33.3 m. d 9 h

m m 3m 3(9.5 × 10−26 kg) = 2.33 × 1017 kg/m3. In metric tons per = = = 3 3 −15 4 V 4 R 4 (4.6 × 10 m) π π 3 πR 3 2

3

1-80.

17

d = (33.3 m) 2 + (300 m) 2 = 302 m.

300 m


CHAPTER 1-81.

1

The distance traveled by the plane gliding: x = 5 × 103/tan 15° = 18.7 × 103 m = 18.7 km. Therefore the pilot can reach San Francisco.

1-82.

⎛ R ⎞

θ = cos−1 ⎜⎝ Rth ⎟⎠ ⎛ ⎞ 6.4 × 106 = cos−1 ⎜ 6 3 ⎟ ⎝ 6.4 × 10 + 2.3 × 10 ⎠ = 1° = 0.027 rad τ = Rθ = 6.4 × 106 × 0.027 = 1.71 × 105 m = 171 km

†1-83.

The arc AB has a length of 3900 km, so the half angle θ is 1950 km/RE, where RE is the radius of the earth. Using the value from Table 1-1 gives θ = (1950 km)/(6.4 × 103 km) = 0.305 radian = 17.5°. (a)The linear distance d from A to B is d = 2 RE sin θ = 2(6.4 × 103 km)(sin17.5°) = 3840 km. (b)The depth h at the midpoint is h = RE − RE cos θ = RE (1 − cos θ ), which gives 3

A

C h

φ

d/2 RE

θ θ

B

RE

O

h = (6.4 × 10 km)(1 − cos17.5°) = 296 km. (c) “Horizontal” means parallel to the surface of the earth, or tangent to the surface of the earth at that location. The tangent line is shown in the diagram, and the slope angle φ between the tangent and the line AB is shown. Since the radius OA is perpendicular to the tangent line and the radius OC is perpendicular to the line AB, the angle φ must be the same as θ. Thus φ = 17.5° and the slope is the tangent of that angle: slope = tan 17.5° = 0.315. Using the ratio form of slope gives slope = 1:(0.315)−1 = 1:3.2. (This is a quite steep slope!) Note that to observers standing at each end, the tunnel appears to be sloping down into the earth.


CHAPTER 2

MOTION ALONG A STRAIGHT LINE


2-2. †2-3.

distance 30 = = 0.3 s 100 speed distance 100 yd 1 mi 3600 s = Avg speed = = 23 mi/h × × time 9.0 s 1760 yd 1h 20 m 1 year = 3.156 × 107 sec, so 20 m/year = = 6.34 × 10−7 m/s (6.3 × 1 year × 3.156 × 107 s/year Time required, ∆t =

10−7 m/s to two significant figures). 1 day = 24 hr = 86,400 s. In cm/day the rate is 6.34 × 10−7 m/s × 8.64 × 104 s/day × 100 cm/m = 5.4 cm/day. 2-4.

Assume the butterfly’s speed is 0.5 m/sec. Then the travel time is d 3500 × 103 m 1 = × ≈ 81 days. t= v 0.5 m/s 24 hr/day × 3600 s/hr

2-5.

6 days 12 hrs = 156 hrs. dist. 5068 ν = = = 32.5 km/h time 156 d 1.4 × 109 ly t = = 7 2.16 × 10 m/s × 9.47 × 1015 m/ly × 1/(3.16 × 107 )yr/s ν

2-6.

t = 1.9 × 1010 yr 2-7.

2-8.

2-9. 2-10.

†2-11.

Estimated distance (by sea) between Java and England is 20,000 km. d 20,000 km ν = = ≈ 600 km/h t 32 h 4000 nmi (a) Average speed = = 8.3 nmi/hr. (b) He must cover the remaining 1720 20 days × 24 hr/day 1720 nmi = 10.2 nmi/hr. This is nmi in 7 days, which requires an average speed of 7 days × 24 hr/day about the same as his maximum possible speed. Since it’s unlikely that he can maintain the highest possible speed for the entire 7 days, he should probably conclude that he will not be able to complete the trip within the 20-day limit. d 35 km = 14 km/hr Average speed = = 2.5 hr t Average speed =

110 km 110 × 103 m = = 1.27 m/s. Burst speed = 32 km/hour = 1 day 24 h × 3600 sec

32 × 103 m = 8.9 m/s 3600 s d 1 km 1000 m t = ⇒ t1 = = = 2.5 × 104 years. Moving 1000 km will take v 4 cm/year 4 × 10−2 m/year 1000 times as long, or t2 = 2.5 × 107 years.


CHAPTER

2-12. 2-13. 2-14. †2-15. 2-16. 2-17. 2-18.

2

d 402 m = = 16.9 m/s t 23.8 s 500 m Average speed = = 12.8 m/s, or 12.8 × 10−3 km/m × 3600 s/hr = 46.0 km/hr 39.10 s

Average speed =

d 23.8 m = = 0.326 s v (263 km/hour × 1000 m/km / 3600 s/hr) d 5280 km 5280 km Use the formula: t = . tair = = 5.87 hr. tshiip = = 151 hr. 900 km/hr v 35 km/hour 100 m 200 m v1 = = 9.49 m/s. v2 = = 9.37 m/s 10.54 s 21.34 s 1 d = v i t = 6 m/s × (s) = 0.06 m 100 Time taken for the arrow to reach the deer is d 50 m 10 t= = = s = 0.77 s. v 65 m/s 13 t =

In this time the deer traveled from 40 m to 50 m, i.e., 10 m. Thus d 10 m v = = = 13 m/s. t (10/13s) †2-19.

(a) Take x = 0 to be the cheetah’s starting position. Then the cheetah’s position is given by xc = vc t. The antelope’s starting position is 50 m from the cheetah’s starting position, so the position of the antelope is given by xa = va t + 50. When the cheetah catches the antelope, their positions are the same, and we get vc t = va t + 50. The speeds are νc = 101 km/h = 28.1 m/s and 50 m 50 m va = 88 km/h = 24.4 m/s. Solving the equation for t gives t = = = 28.1 m/s − 24.4 m/s vc − va

13.8 s, or 14 s to two significant figures. During this time, the cheetah travels (28.1 m/s)(13.8 s) = 380 m. (b) The cheetah must catch the antelope within 20 s. Call the antelope’s initial position x0. We use the same equation that says the cheetah catches the antelope, vc t = va t + x0 , but now we set t = 20 s and calculate what head start x0 the antelope needs. We get x0 = (vc − va )t = (28.1 m/s − 24.4 m/s) (20 s) = 72 m. If the antelope is farther away than 72 m, 2-20. †2-21.

the cheetah will not be able to catch it. d 100 m Average speed = = = 10.1 m/s t 9.86 s d = 26 mi × 1.6 × 103 m/mi + 385 yd × 0.9144 m/yd = 4.195 × 104 m t = 2 hr 24 min 52 s = 2 hr × 3600 s/hr + 24 min × 60 s / min + 52 s = 8692 s d 4.195 × 104 m = = 4.83 m/s 8692 s t 2π × 0.9 cm = = 0.094 cm/s 60 s

average speed = 2-22.

vsecond =

d second tsecond

vminute =

d minute 2π × 0.9 cm = = 1.6 × 10−3 cm/s 60 min × 60 s/min tminute


CHAPTER vhour =

†2-23.

2-24.

2 d hour 2π × 0.5 cm = = 7.3 × 10−5 cm/s 12 hr × 60 min/hr × 60 s/min thour

x = 4.0t − 0.50t 2. To find the maximum value of x, differentiate with respect to t and set the dx = 4.0 − t = 0. The result is t = 4.0 s. (This is the point at which the derivative equal to zero: dt runner turns around and moves back toward the starting line.) The distance traveled at this time is x = 4.0(4.0) − 0.50(4.0) 2 = 8.0 m. At t = 8 seconds, x = 4.0(8.0) − 0.5(8.0) 2 = 0; that is when he comes back to the starting line. The total distance traveled is 16 m. Then average speed = distance 16 m = = 2.0 m/s. time 8s 50 km 50 km + = 1.46 hr average speed Distance = 100 km time = 60 km/hour 80 km/hr distance 100 km = = = 69 km/hr. The average speed is not exactly 70 km/hr because the car time 1.46 hour

moves at 80 km/hr for a shorter period of time than it does at 60 km/hr. †2-25.

Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

Orbit circumference (km) 3.64 × 108 6.79 × 108 9.42 × 108 1.43 × 109 4.89 × 109 8.98 × 109 1.80 × 1010 2.83 × 1010 3.71 × 1010

Period (s) 7.61 × 106 1.94 × 107 3.16 × 107 5.93 × 107 3.76 × 108 9.31 × 108 2.65 × 109 5.21 × 109 7.83 × 109

Speed (km/s) 47.8 35.0 29.8 24.1 13.0 9.65 6.79 5.43 4.74

1 The slope of the line through the nine points is − . 2


log speed 1.68 1.54 1.47 1.38 1.11 0.985 0.832 0.735 0.676

log radius 8.56 8.83 8.97 9.16 9.69 9.95 10.26 10.45 10.57

CHAPTER

2

1 log radius + log C. 2 −1/2 Therefore log speed = log[C(radius) ]. Thus Speed = C(radius) −1/2 where C is some constant. This means that log speed = −

2-26.

2-27.

2-28.

†2-29.

2-30.

†2-31.

distance 8 + 8 = 6.27 m/s = 2.55 time displacement Avg velocity = = 0 m/s time 200 = 20 m/s Avg speed (for t = 0 to t = 10 s) = 10 270 − 200 = 16.3 m/s Avg speed (for t = 10 to t = 14.3 s) = 14.3 − 10 Avg speed =

Distance = (8 floors + 4 floors + 7 floors) × 4 m/floor = 76 m. distance 76 m = = 1.5 m/s. The total change in position is Average speed = time 50 s ∆x = (12 floors − 1 floors) × 4 m/floor = 44 m. The average velocity is ∆x 44 m v = = = 0.88 m/s. ∆t 50 s Distance = (12 blocks + 6 blocks + 3 blocks) × 81 m/block = 1701 m. The elapsed time = 14 min 5 s + 6 min 28 s + 3 min 40 s = 23 min 73 s = 1453 s. Then average distance 1701 m speed = = = 1.17 m/s. The total displacement is ∆x = (12 blocks − 6 blocks + time 1453 s ∆x 729 m = = 0.502 m/s. 3 blocks) × 81 m/block = 729 m. The average velocity is v = ∆t 1453 s x(t = 0) = 0; x(t = 8) = 4 × 8 − 0.5 × 82 = 0; x(t = 10) = 4 × 10 − 0.5 × 102 = −10 m. x(t = 8) − x(t = 0) Average velocity between t = 0 to t = 8.0 s is: = 0; 8−0 x(t = 10) − x(t = 8) −10 m − 0 = = −5.0 m/s. Average velocity between t = 8.0 s to t = 10.0 s is: 10s − 8s 2s Total distance = 3 × 0.25 mile × 1609 m/mile = 1207 m. Displacement = 0 m because the horse returns to the starting point. Total time = 1 min 40 s = 100 s. Then average speed = distance 1207 m ∆x = = 12.1 m/s. The average velocity is v = = 0. time 100 s ∆t

2-32.

Total distance = 35 m + 22 m = 57 m. The total time is = 4.5 s + 3.6 s = 8.1 s. The average speed distance 57 m = = 7.0 m/s. The total displacement is ∆x = 35 m − 22 m = 13 m. Then is time 8.1 s average velocity is v =

∆x 13 m = = 1.6 m/s. ∆t 8.1 s


CHAPTER †2-33.

2-34.

2-35.

2-36. †2-37. 2-38.

From the graph, the total distance traveled by the squirrel is 6 m + 6 m + 2 m + 6 m = 20 m. The distance 20 m = = 0.67 m/s. The total total elapsed time is 30 s. Then average speed = time 30 s ∆x 13 m displacement is ∆x = 16 m, so the average velocity is v = = = 0.53 m/s. ∆t 8.1 s displacement 25 m = = 12.5 m/s. For time 2s 2.0 ≤ t ≤ 4.0 s, displacement = 40 m − 25 m = 15 m. (This requires an estimate for the position displacement 15 m = = 7.5 m/s. at 4.0 s. Your value may be slightly different.) v = time 2s To find the instantaneous velocity at any time, draw a tangent to the position vs time curve at that time and determine the slope of the line. Your numbers may be slightly different from the ones given here. At 1.0 s, we get a tangent passing through points with coordinates (0.3 s, 0 m) and (2.5 s, 25 m). This gives a slope of 11 m/s. At 3.0 s, the position vs time graph is a straight line, so the instantaneous velocity will be the same as the average velocity between 2.0 s and 4.0 s, or 7.5 m/s. Again, your value may be slightly different if you estimate a different position at 4.0 s. 82.6 gee × (9.807 m/s 2 ) a = ∆ν/∆t ⇒ ∆ν = a ∆t = gee × 0.04 s ∆v = 32.4 m/s

For 0 ≤ t ≤ 2 s, displacement = 25 m. v =

a = ∆ν/∆t = [(96 − 0) km/h]/2.2 s × 1 h/3600 s × 1000 m/km = 12 m/s 2 a =

2-40.

v2 − v1 27 m/s − 0 = = 3.4 × 103 m/s 2 . −3 8.0 × 10 s t2 − t1

vi = 80 (km/hr) = 22.22 m/s, v f = 0, t = 2.8 s. a=

2-39.

2

v f − vi t

=

0 − 22.22 m/s = −7.94 m/s 2 2.8 s

(a) t(s) a (m/s 2 )

a (in gees) 0 6.1 0.62 10 1.4 0.14 20 0.83 0.085 30 0.56 0.057 40 0.49 0.050 2 a (gees) (b) t(s) a (m/s ) 0 −0.74 −0.075 10 −0.44 −0.045 20 −0.44 −0.045 30 −0.31 −0.032 40 −0.22 −0.022 dx x = 2.5 t + 3.1 t 2 − 4.5 t 3; v = = dt

Method: i) Draw tangent to curve. ii) Get slope of line by counting squares to find ∆ν and ∆t. iii) Convert from km/h to m/s.

2.5 + 6.2 t − 13.5 t 2 ; a =

dv = 6.2 − 27.0 t. dt

dx |t = 0 = 2.5 + 6.2 × 0 − 13.5 × 02 = 2.5 m/s. Instantaneous dt acceleration = 6.2 m/s2. At t = 2. 0 s, instantaneous velocity =

At t = 0 s, instantaneous velocity =


CHAPTER

2

dx |t = 2 = 2.5 + 6.2 × 2 − 13.5 × 22 = −39.1 m/s (−39 m/s to two significant figures). dt Instantaneous acceleration = 6.2 − 27.0(2) = −48 m/s 2 . x(t = 2 s ) − x(t = 0) 2.5(2) + 3.1(2) 2 − 4.5(2)3 = = −9.3 m/s. 2−0 2 v(t = 2) − v(t = 0) −39.1 m/s − 2.5 m/s = = – 21 m/s2. a = 2−0 2s dx x = 3.6 t 2 − 2.4 t 3 , v = = 7.2 t − 7.2 t 2 . v = 0 if dt 7.2 t − 7.2 t 2 = 0 ⇒ t = 0 s or t = 1.0 s. At t = 0, x = 0. At t = 1.0 s, x = 3.6 − 2.4 = 1.2 m. To make a sketch, 3.6 consider that x = 0 when t = 0 and t = = 1.5 s. 2.4 Also, dx/dt = 0 when t = 0 and t = 1s.

For 0 ≤ t ≤ 2 s, v =

†2-41.

2-42.

v = Bt − Ct 2 . V = 0 if t = 0 or t = dv B 6.0 m/s 2 = B − 2Ct. a = 0 if = = 3.0 s. a = 3 dt C 2.0 m/s B = 1.5 s. To make a sketch: consider the 2C above information, plus when t = 1.5 s, ν = (6.0 m/s 2 ) (1.5 s) − (2.0 m/s3 ) (1.5 s) 2 = 4.5 m/s. v(t = 5 s) − v(t = 0) 5 m/s − 0 ≈ = 1 m/s2. (Your value may be slightly For 0 ≤ t ≤ 5.0 s, a = 5s 5s different depending on how you read the values of ν at 0 and 5 s.) For 5.0 ≤ t ≤ 10.0 s, v(t = 10 s) − v(t = 5 s) 9.5 m/s − 5.0 m/s a = ≈ = 0.9 m/s 2 . (Again, your value may be slightly 5s 5s t =

†2-43.

2-44.

different depending on how you estimate the values of ν at 5 and 10 s.) To find the instantaneous acceleration at 3 s, draw a tangent to the curve at that time. Your estimate may be slightly different from ours. We get a tangent line that passes through the points (1 s, 0 m/s) and (5 s, 5 5 m/s − 0 m/s). The slope of this line is the instantaneous acceleration a = = 1.3 m/s 2. 5 s −1 s v0 − v f − t / 2.5 s dv d . At t = 0, a= = [v f + (v0 − v f )e − t / 2.5 s ] = − e (2.5 s) dt dt v0 − v f m/s dv 200 km/hr − 18 km/hr a= |t = 0 = − =− dt 2.5 s 2.5 s 182 km/hr × 1000 m/km =− = −20 m/s 2 . 2.5 s × 3600 s/hr


CHAPTER

†2-45.

2-46.

2

⎤ 2 Av0t v0 dv d ⎡ v0 d = =− ( At 2 ) = − . At t = 0, a = 0. ⎢ 2 ⎥ 2 2 dt dt ⎣ (1 + At ) ⎦ (1 + At ) dt (1 + At 2 ) 2 2 Av t 2(25 m/s)(2 s −2 )(2 s) at t = 2 s, a = − = − 2.5 m/s 2 . As t → ∞, a → − 2 04 , so a → 0. −2 2 2 At ⎡⎣1 + (2 s )(2 s) ⎤⎦ (a) Estimated average velocity (by mid-point method) is about 30 km/h. Because v = ∆x/∆t, we 5 m/s have ∆x = v ∆t ≈ 30 km/h × 5 s × ≈ 42 m. 18 km/h ⎡ 5 m/s ⎤ v (km/h) ∆x(m) ⎢ by v km / h × 5s × (b) Time interval (s) ⎥ 18 km / h ⎦ ⎣ a=

5−10 70 100 10−15 90 120 15−20 110 150 20−25 130 180 25−30 140 190 30−35 150 210 35−40 160 220 40−45 170 240 (c) Total distance traveled is the sum of the last column plus the 42 m traveled in the first 5 s: d = 1450 m = 1.45 km. †2-47.

(a)

(b)

Time interval (s) Avg speed (m/s) 0−0.3 647.5 0.3−0.6 628.5 0.6−0.9 611.5 0.9−1.2 596.0 1.2−1.5 579.5 1.5−1.8 564.0 1.8−2.1 549.5 2.1−2.4 535.0 2.4−2.7 521.0 2.7−3.0 508.0 Total distance traveled = 1722 m

Distance traveled (m) 194 189 183 179 174 169 165 161 156 152


CHAPTER

2-48.

2

(c) Counting the number of squares under the ν versus t curve gives the same answer within about ± 2 m. 2 ν = 655.9 − 61.14t + 3.26t dv acceleration, a = = −61.14 + 6.52t dt a |t = 0 s = −61.14 m/s 2 a |t = 1.5 s = −61.14 + (6.52) (1.5) = −51.36 m/s 2 a |t = 3.0 s = −61.14 + (6.52) (3.0) = −41.58 m/s 2

†2-49.

(a)

(b) Calculus method: If x = 0, then cos t = 0, so t = π/2 or 3π/2 s. Particle crosses x = 0 at t = 1.6 s and 4.7 s. ν = dx/dt = −2.0 sin t ν (π/2 s) = −2.0 sin(π/2) = −2.0 m/s ν (3π/2 s) = −2.0 sin(3π/s) = 2.0 m/s a = dv/dt = −2.0 cos t a(π/2) = −2.0 cos(π/2) = 0 m / s 2 a(3π/2) = −2.0 cos(3π/2) = 0 m / s 2 (c) Maximum distance achieved when cos t = ± 1, i.e., when t = 0, π, 2π, or t = 0 s, 3.1s and 6.3s. v = −2.0 sin t v(0) = −2.0 sin(0) = 0 m/ s v(π) = −2.0 sin(π) = 0 m/ s v(2π) = −2.0 sin(2π) = 0 m/ s a = dv/dt = −2.0 cos t a(0) = −2.0 cos(0) = −2.0 m / s 2 a(π) = −2.0 cos(π) = 2.0 m/ s 2 a(2π) = −2.0 cos(2π) = −2.0 m / s 2 2-50.

x = uext + uex(1/b − t) ln(1 − bt ) (a) Instantaneous velocity, v = dx/dt [1 − ln(1 − bt) − 1] −b ⎤ dx ⎡ = uex = uex + uex ⎢ − ln(1 − bt ) + (1/ b − t ) ⎥ 1 − bt ⎦ dt ⎣ v = −uex ln(1 − bt ) (b) a =

u b d 2x = −uex[−bt(1 − bt)] = ex 2 dt 1 − bt


CHAPTER

2

dx = −3.0 × 103 ln(1 − 7.5 × 10−3t) m/s dt 3 ν(t = 0) = −3.0 × 10 ln 1 = 0 m/s

(c)

ν(t = 120) = −3.0 × 103 ln(1 − 7.5 × 10−3 × 120) m/s ν(120) = −3.0 × 103 ln(1 − 0.9) m/s = −3.0 × 103 × (−2.80) m/s v(120) = 6.9 × 103 m/s (d) a =

dυ d 2x 3.0 × 103 × 7.5 × 10−3 22.5 m/s2 = 2 = m/s2 = dt dt 1 − 7.5 × 10−3 t 1 − 7.5 × 10−3 t

a(0) = 22.5 m/s2 a(120) =

22.5 22.5 m/s2 = m/s2 −3 1 − 0.9 1 − 7.5 × 10 × 120

a(120) = 225m/s 2 2-51.

1 2 (v − v02 ) 2 5 m/s = 100 m/s; v0 = 0 x − x0 = 2100 m; v = 360 km/h × 18 km/h v 2 − v02 1 1 1002 = × m/s2 a= × x − x0 2 2 2100

Use Equation (25): a(x − x0) =

a = 2.4 m/s 2 v 2 − v02 (657) 2 = = 3.26 × 104 m/s 2 2( x − x0 ) 2 × 6.63 v − v0 Time to travel the length of the barrel, t = a 657 t= = 2.02 × 10−2 s 4 3.26 × 10

2-52.

Acceleration, a =

2-53.

4.2 ly = 4.2 ly × 9.46 × 1015 m/ly = 3.97 × 1016 m 1 1 Use x − x0 = ν0t + (at 2 ) with x0 = ν0 = 0 and x = (4.2ly) 2 2 t 1/2 =

2x = a

3.97 × 1016 m = 6.36 × 107 s 9.807 m / s 2

Because the magnitude of the acceleration is the same for both parts of the trip, the time for the second half is identical to that of the first half. Thus, the total time for the trip, T, is T = 2t 1/2 = 2 × 6.36 × 107 s = 1.3 × 108 s ≈ 4.0 yr 1 2 (v − v02 ) 2 ν0 = 0; x − x0 = 1.99 × 1016 m; a = 9.807 m/s2 Then v22 = 2a(x − x0) = (2 × 9.807 × 1.99 × 1016) m2/s2 ν2 = 3.90 × 1017 m2/s2 v = 6.2 × 108 m / s (This exceeds the speed of light!!) Speed at midpoint: Use a(x − x0) =


CHAPTER 2-54.

2a( x − x0 ) = v 2 − v02 ⇒ v0 =

−2a( x − x0 ) From the information given in the problem, x − x0 =

290 m, and a = −10 m/s2. Then v0 = 2-55.

2

−2(−10 m/s 2 )(290 m) = 76 m/s, which corresponds to

about 270 km/h, or 170 mph. 1 5 m/s Use (v 2 − v02 ) = a(x − x0) with v = 0; ν0 = 80 km/h × = 22.2 m/s; x − x0 = 0.7 m. 2 18 km/h 1 2 1 (v − v02 ) (−22.2) 2 2 2 . Therefore a = = x − x0 0.7m/s 2

a = −350m/s 2 (will probably survive) 2-56.

2a( x − x0 ) = v 2 − v02 ⇒ v0 =

−2a ( x − x0 ) From the information given in the problem, x − x0 =

9.6 km = 9.6 × 103 m, and a = −5 m/s2. Then v0 =

−2(−5 m/s 2 )(9.6 × 103 m) = 3.1 × 102 m/s,

which corresponds to about 700 mph. The time to stop is t = †2-57.

v − v0 0 − 3.1 × 102 m/s = = 62 s. a −5 m/s 2

1 2 (v − v02 ) = a(x − x0) with x − x0 = 50 m; v = 0; 2 v0 = 96 km/h = 26.67 m/s 1 2 1 (v − v02 ) ( −26.67) 2 a= 2 = 2 50 m/s 2 x − x0

Use

a = −7.1m/s 2

2-58.

Use v = v0 + at with v = 0 m/s, v0 = 96 km/h = 26.7 m/s, and a = −7.1 m/s2. Then v t = − 0 = 3.8 s. a v 2 − v02 (260 × 103 / 3600) a= =− 2( x − x0 ) 2 × 1500 a = − 1.74 m/s 2 (The minus sign denotes deceleration.)

2-59.

2-60.

Use v = v0 + at with v = 0 m/s, v0 = 260 km/h = 72.2 m/s, and a = −1.74 m/s2. Then v t = − 0 = 42.5 s. a v = v0 + at = 1.5 m/s 2 × 20 s = 30 m/s 1 1 x = v0t + at 2 = 0 + × 1.5m/s 2 × (20 s) 2 = 300 m. 2 2 1 1 x − x0 = v0t + at 2 ⇒ 1.0 m − 5.0 m = 3.0 m/s i 4.0 s + a i (4.0 s) 2 2 2 2 ⇒ −4 m = 12 m + 8 a ⇒ a = −2.0 m/s . The velocity is v = v0 + at = 3.0 m/s − (2.0 m/s 2 )(4.0 s) = − 5.0 m/s.

2-61.

x = v0 t +

1 2 1 2 at = at (since v0 = 0). t = 2 2

2x = a

2 × 150 m = 16 s. 1.2 m/s 2


CHAPTER

2-62.

2

v0 = 550 km/hr =

550 km/hr × 1000 m/km = 153 m/s. v = v0 + at = 3600 s/hr

153 m/s + (0.60 m/s 2 )(90 s) = 2.1 × 102 m/s. †2-63.

The sketch should be based on the following: For 0 ≤ t ≤ 6 s, a = 3.0 m/s2; v = v0 + at = 0 + 3t ; x =

t

∫

t

v dt =

0

3

∫ 3t dt = 2 t . At t = 6s, 2

0

3 2 i 6 = 54 m. For 6 ≤ t ≤ 10 s, a = −4.5 m/s2, and 2 v = v0 + a(t − 6) = 18 − 4.5 (t − 6) m/s = 45 − 4.5t m/s. 1 1 x = x0 + v0 (t − 6) + a(t − 6) 2 = 54 + 18(t − 6) − i 4.5(t − 6) 2 = − 2.25(t − 6) 2 + 18(t − 6) + 54 2 2 v = 3 × 6 = 18 m/s; x =

At t = 10 s, v = 45 − 4.5 × 10 = 0; x = −2.25 i 42 + 18 i 4 + 54 = 90 m.

2-64.

x = v0 t +

1 2 at ⇒ v0 = 2

x−

at 2 2 2 2 2 = 700 m − 0.5 × 0.05 m/s × 30 s = 22.6 m/s. t 30 s

v = v0 + at = 22.6 m/s + 0.05 m/s 2 i 30 s = 24 m/s. †2-65.

x = v0 t +

1 2 at ⇒ v0 = 2

x−

at 2 2 2 2 2 = 550 m − 0.5 × 0.5 m/s × 15 s = 32.9 m/s. t 15s

v = v0 + at = 32.9 m/s + 0.5 m/s 2 i 15 s = 40.4 m/s. 2-66.

Speed at the end of the 440-yard mark, 250.69 × 1760 × 3 v= = 367.7 ft/s 60 × 60


CHAPTER ∆v 367.7 = = 65.23ft/s 2 ∆t 5.637 (b) For a constant acceleration, the distance traveled would be = avg speed × time = ⎛ 367.7 ⎞ ⎜ ⎟ (5.637) = 1036.36 ft = 345.45 yd. ⎝ 2 ⎠ Therefore the acceleration was not constant. v (c) Assuming constant acceleration, distance = 440 × 3 = t (5.637) 2 2 × 440 × 3 = 463 ft/s = 319 mi/h. vt = 5.637 The distance traveled by the elevator is 1 1 ⎛ ⎞ x = at12 + vt2 + ⎜ vt3 − at32 ⎟ 2 2 ⎝ ⎠ where v = constant speed reached at at1; t1, t2 and t3 are the times spent in the following, respectively: accelerating, traveling at constant speed, and decelerating. 1 1 21 × 2.5 m = a(5)2 + (a5)7 + (a5)5 − a52 2 2 52.5 m = 60 a a = 0.875m/s 2 The maximum speed of the elevator is vmax = at1 vmax = 0.875 m/s2 × 5 s vmax = 4.4 m/s (a) Average acceleration, a =

2-67.

2-68.

400 = 7.3 m/s 55 (b) For minimum values of acceleration and deceleration, the elevator should travel half the distance in half the time. Therefore 2 1 ⎛ 55 ⎞ 200 = a ⎜ ⎟ 2 ⎝ 2 ⎠ a = 0.53 m/s 2 where a is the acceleration and −a is the deceleration.

(a) Average speed =

55 = 14.6 m/s 2 (This is twice the average speed. What is its significance?)

Maximum speed is given by vmax = at = 0.53 ×

2-69.

v0 (km/h)

v0 (m/s)

v0 ∆t (m)

15 30 45 60 75 90

4.17 8.33 12.5 26.7 20.8 25.0

8.3 16.7 25.0 33.3 41.7 50.0

v02 (m) 2a 1.1 4.3 10 18 27 39 −


Total stopping distance (m) 9.4 21.0 35.0 51.3 68.7 89.0

2

CHAPTER 2-70.

2-71.

Table on Page 47: v2 v0 ∆t = 0 ⇒ v0 = 2at = 2 × 8 m/s 2 × 0.75 s = 12 m/s. 2a 5 m/s = 13.9 m/s; 2.0 ft = (2 × 0.3048) m = 0.610 m. In the inertial frame that is 50 km/h = 50 × 18 traveling at constant velocity with the car, all velocities are zero. In this frame, a = 200 m/s2 also. Therefore the speed with which the dashboard hits the passenger is v=

2-72.

2

2a ( x − x0 ) =

2(200 m/s 2 )(0.610 m) = 15.6 m/s.

Consider the position of the car in the reference frame of the truck x0c = −12 − 17 = −29 m The final position of the car is xc = 17 m, therefore, 1 xc = x0c + v0c t + at2 2 1 17 = −29 + 0 + at2 2 1 2 46 = at . (i) 2 The final speed of the car relative to the truck is 24 km/h 24 × 103 = 6.67 m/s. = 60 × 60

†2-73.

at = 6.67 m/s Divide equation (i) by (ii) to get t = 13.8s and 6.67 a= = 0.48 m/s 2 . 13.8 v = −gτ + gτe−t/τ dv (a) acceleration, = ge − t / τ dt Lim −t/τ = 0 , therefore e (b) t→∞ Lim Lim − t/τ = − gt e v = −gτ + t→∞ t→∞ d (c) v = ( − gτ t − gt 2e−t / τ + gτ 2 + x 0 ) dt τ 2 − t/τ = − gτ + gτ e −t / τ e = −gτ + g

τ

(d) for t << τ, the exponential e−t/τ can be expanded as t 1 t2 +.... e−t/τ = 1 − + τ 2τ2 t 1 t2 ) + gτ2 + x0 Therefore, x = −gτ t − gτ2(1 − + τ 2τ2 1 = −gτ t − gτ2 + gt τ − gt2 + gτ2 + x0 = − 1 gt 2 + x0 2 2


(ii)

CHAPTER

2-74.

x − x0 = v0t −

1 2 gt with x − x0 = −380 m; v0 = 0; g = 9.8 m/s2 2

1 −380 m = − (9.8 m/s 2 ) t2 2 2 2 t = 77.6 s t = 8.8s For the impact velocity use v = v0 − gt v = 0 − 9.8 m/s2 × 8.8 s = −86 m/s †2-75.

130 km/h = 36.1 m/s. The acceleration of the freely falling falcon is g, so use v 2 − v02 (36.1 m/s) 2 − 0 v 2 − v02 = 2 g ( x − x0 ) ⇒ x − x0 = = = 66.5 m. 2g 2(9.81 m/s 2 )

2-76.

v 2 = v02 + 2ax. Here v0 = 0, a = g. v =

2-77.

v 2 = v02 + 2ax. Here v = 0, a = −g because the displacement is upward and the acceleration is downward. v0 =

−2 gh =

2 gh =

2(9.81m s 2 )(8.7 m) = 13 m/s.

−2(−9.81m s 2 )(1.9 m) = 6.1 m/s.

1 2 mv , where h is the maximum height reached 2 1 ⎛ v2 ⎞ (366) 2 = 6834 m h= ⎜ ⎟ = 2 ⎝ g ⎠ 2 × 9.8

2-78.

mgh =

2-79.

Neglecting air resistance, we have 1 x − x0 = v0t − gt2 with v0 = 0; g = 9.8 m/s2; t = 3.0 s 2 1 x − x0 = 0 − (9.8 m/s 2 ) (3.0 s) 2 = 44 m 2 2 2 Use v − v0 = −2g(x − x0) with v = 0; g = 1.80 m/s2; x − x0 = 200,000 m

2-80.

v02 = 2 × 1.80 m/s2 × 2 × 105 m = 7.2 × 105 m2/s2 v0 = 849 m/s

2-81.

v = 45 km/h = 12.5 m/s. h =

v2 (12.5 m/s) 2 1 floor = = 8.4 m × = 3 floors. 2 2 g 2(9.81 m/s ) 2.9 m

v = 75 km/h = 20.8 m/s. h =

v2 (20.8 m/s) 2 1 floor = = 22.1 m × = 8 floors. 2 2 g 2(9.81 m/s ) 2.9 m

v = 105 km/h = 29.2 m/s. h =

2-82. †2-83.

t= tup =

2h = g 2h = g

v2 (29.2 m/s) 2 1 floor = = 43.4 m × = 15 floors. 2 g 2(9.81 m/s 2 ) 2.9 m

2(10 m) = 1.43 s. This is 1.43 s/(3 half turns) = 0.48 s per half turn. 9.81 m/s 2 2(9.5 m) = 1.39 s. Total time = 2(1.39 s) = 2.8 s. 9.81 m/s 2

v0 = gtup = (9.81 m/s 2 )(1.4 s) = 14 m/s. This is the initial speed. The initial velocity is 14 m/s

up.


2

CHAPTER 2-84.

†2-85.

2-86.

†2-87.

2

v0 = 1.0 m/s downward, and the ball is initially some distance h above the ground. After falling that distance h, the ball will strike the ground with some speed v and will rebound (reverse its direction of motion) at the same speed if the collision with the ground is elastic. (This concept will be introduced in a later chapter.) Since it starts moving back up with the same speed it had just before it hit the ground, the time required to return to its starting point a distance h above the floor will be the same as the time required for it to reach the floor in the first place, and it will arrive at the distance h above the ground with same speed with which it was initially thrown. Thus the downward and upward travel times are each equal to half the total travel time. The speed t of the ball just before striking the ground is v = v0 + g , where t is the total travel time. Then 2 v 2 − v02 2 2 . Substituting, we get the height h can be found using v = v0 + 2 gh, which gives h = 2g

(4.68 m/s) 2 − (1 m/s) 2 ⎛ 0.75 s ⎞ Then h = = 1.1 m. 4.68 m/s. v = 1 m/s + (9.81 m/s 2 ) ⎜ = ⎟ 2(9.81 m/s 2 ) ⎝ 2 ⎠ Comment: This requires conservation of momentum! The ball will collide with the ground and rebound with a speed equal to its speed just before it hit, a topic that obviously isn’t covered in this chapter. Take the coordinate axis to point up. Then the final displacement is −9.2 m, a = −g, t = 2.5 s. at 2 d− 2 2 at 2 = d − at = −9.2 m − (−9.81 m/s )(2.5 s) = 8.57 m/s, or 8.6 ⇒ v0 = d = v0t + 2 2 2.5 s 2 t t m/s (to two significant figures). The height the penny reaches above its launch point is v2 v02 v2 (8.57 m/s) 2 h=− 0 =− = 0 = = 3.7 m. 2a 2(− g ) 2 g 2(9.81 (m/s 2 ) v + v0 v0 = 0, a = −g, x = −h ⇒ v = v02 + 2ax = 2 gh . v = for constant acceleration, so 2 2 gh v = . 2 Assume that the collision of the ball with the floor simply reverses the direction of the ball’s velocity. Then the time for the ball to reach the floor is the same as the time for the ball to return to its starting point, and its speed upon returning will be the same as the speed with which it 0.90 s = 0.45 s. Since the ball begins by moving down, choose the started. Thus tdown = 2 direction of the axis for the motion to point down. Then the acceleration and initial velocity are 1 represented by positive numbers. h = v0t + gt 2 2 2 2 2h − gt 2(1.5 m) − (9.81 m/s )(0.45 s) 2 ⇒ v0 = = = 1.13 m/s, or 1.1 m/s to two significant 2t 2(0.45 s) figures. The velocity just before hitting the floor is v = v0 + gt = 1.13 m/s + (9.81 m/s 2 )(0.45 s) = 5.5 m/s.


CHAPTER

2-88.

2-89.

2-90.

2-91.

1 2 gt . At t = 0,1,2,3 s the distance fallen is 0, 4.91 m, 19.6 m, 44.1 m, 78.5 m. The distance 2 traveled from 0 − 1 s is 4.91 m; from 1 s to 2 s the distance traveled is 19.6 m − 4.91 m = 14.7 m. Likewise the distance traveled from 2 s to 3 s is 24.5 m. Dividing each of these by 4.91 m gives the ratios 1:3:5 and so on. at 2 2h . v0 = 0 ⇒ a = 2 . h = v0t + 2 t 10 cm 4 fingers × 2∆h ∆a ∆h 4 fingers ∆a = 2 = = × 100% = 0.22%. 46 cm t a h 100 cubits × cubit Time taken to fall through a distance of 45 m, 1 y2 = y1 + v1t + at2 2 1 0 = 45 + 0 − (9.8) t2 2 t = 3.03 s 2 y1 , to measure y1 within 10%, the time must be known to within 5% or within 0.15 s. Since t = g Therefore a stopwatch must be used. An ordinary wristwatch will have an uncertainty of ±1 s. x=

Velocity of ball on impact = − 2 gx1 = − 2(9.8 m/s 2 ) (1.5 m) = −5.42 m/s Velocity of ball after impact = a =

2-92.

2

2gx2 =

2(9.8m/s 2 ) (1.1m) = 4.64 m/s

∆v 4.64 m / s − (−5.42m/s) = = 1.6 × 104 m/s 2 ∆t 6.2 × 10−4 s

(a) Impact speed, v = (b) a =

2 2

2

2gh =

2 × 9.8 × 96 = 43.4 m/s

2

v −v −(43.4) = −254.5m / s 2 = 2 × 3.7 2( x2 − x)

(The minus sign denotes deceleration.) †2-93.

The muzzle speed is given by v0 =

2 gh =

2(9.81 m/s 2 )(180 × 103 m) = 1.88 × 103 m/s. To

find out how long the projectile remains above 100 km, we can use the fact that the time for the projectile to climb from 100 km to 180 km is the same as the time for it to fall from 180 km back to 100 km. So we can just calculate the time to fall a distance of 80 km from rest and double that value. The time to fall can be calculated from y = 2-94.

gt 2 ⇒t = 2

2y = g

2(80 × 103 m) = 128s, 9.81 m/s 2

so the total time above 100 km is 2t = 256 s. Given, (a) g = 978.0318 cm/s2 × (1 + 53.024 × 10−4 sin2 Θ − 5.9 × 10−6 sin2 2 Θ) for Θ = 45° g = 978.0318 cm/s2 × (1 + (53.024 × 10−4)/2 − 5.9 × 10−6) = 980.6190 cm/s 2 At the pole Θ = 90°, g = 978.0318 cm/s2 × (1 + 53.024 × 10−4) = 983.2177 cm/s 2


CHAPTER

2

(b) Let g = A(1 + B sin2 Θ − C sin2 2 Θ) where A = 978.0318 cm/s2, B = 53.024 × 10−4, C = 5.9 × 10−6. The condition dg = A( B sin 2Θ − 4C sin 2Θ cos 2Θ) dΘ = A (B − 4C cos 2 Θ)sin 2 Θ = 0 gives extrema at Θ = 0, ±

π

2 π d2g π To distinguish between extrema at Θ = 0, ± , evaluate at Θ = 0, ± 2 d Θθ 2 2

d 2g Θ = 0 = A(2 B cos 2Θ − 8C cos 4Θ sin 4Θ ) θ = 0 d Θ2 = 2AB > 0 (Thus at Θ = 0, g has a minimum.) Similarly d 2g = −2 AB < 0 d Θ 2 θ =±π / 2

2-95.

shows g has maxima at ± π/2 (poles). At the equator, Θ = 0, so g = 978.0318 cm/s2. 300 km/h = 83.3 m/s. If down is negative, then v0 = −83.3 m/s for bomber. Projectile speed relative to ground is −700 m/s − 83.3 m/s, or −783.3 m/s. 1 Use (v 2 − v02 ) = −g(x − x0) with g = 9.8 m/s2; x − x0 = −1500 m 2 2 v = v02 − 2g(x − x0) = (783.3)2 − 2 × 9.8(−1500 m) m2/s2 v = ±802 m/s To find the time, use x − x0 = v0t −

2-96.

1 2 gt , with x − x0 = −1500 m; v0 = −783.3 m/s 2

1 ⎛ ⎞ −1500 m = ⎜ −783.3t − 9.8t 2 ⎟ m 2 ⎝ ⎠ 2 4.9t + 783.3t − 1500 = 0 t = 1.9s For the elevator, v = 370 m/min = 6.17 m/s. Its distance above the ground is given by ye = vt. The height of the penny above the ground is y p = y0 − and penny meet, ye = yp ⇒ vt = y0 − root is t =

−2v ± (2v) 2 + 8 gy0 2g

=

gt 2 , where y0 = 335 m. When the elevator 2

gt 2 , which can be rearranged as gt 2 + 2vt − 2 y0 = 0. The 2

−2(6.17 m/s) ± 4(6.17 m/s)2 + 8(9.81 m/s 2 )(335 m/s) 2(9.81 m/s 2 )

t = 7.66 s (after dropping the negative root). gt 2 9.81 × 7.662 ye = yp = y0 − = 335 − = 47.2 m. 2 2


CHAPTER †2-97.

tup = 1.0 s, h = 10 m, a = − g = 9.81 m/s2. h = v0t +

2-98.

2

1 2 at ⇒ v0 = 2

h−

at 2 ( −9.81 m/s 2 )(1.0 s) 2 10 m − 2 = 2 = 14.9 m/s. The impact speed is t 1.0 s

v = v0 + at = 14.9 m/s − (9.81 m/s 2 )(1.0 s) = 5.1 m/s. (a) The trajectory of the first stone is 1 x1 = x0 + v0t − gt2 = 15t − 4.9t2. The second stone is thrown 1.00 s later, 2 so x2 = v0t′, where t′ is the time after the second stone is thrown, so t′ = t − 1.00 s. Therefore x2 = v0(t − 1) − 4.9(t − 1)2 We want the stones to collide at a height of 11 m. Therefore x1 = 15t 0 − 4.9t 02 = 11 m = v0(t 0 − 1) − 4.9(t 0 − 1)2 = x 2 Solving for t 0 gives t0 = 1.84 or1.22 s Using these values for t 0, we get for v0: 11 + 4.9 (t0 − 1) 2 1.84 s: v0 = = 17.2 m/s 11 + 4.9 (0.84) 2 t0 − 1 = 0.84 m/s 11 + 4.9(0.22) 2 = 51.1m/s 0.22 The 1.84 s corresponds to hitting the stone on its way down, whereas the 1.22 s corresponds to hitting it on the way up. (b) If the second stone is thrown 1.30 s after the first one, the first stone has already passed 11 m on the way up, so the collision can take place only on the way down. 11 + 4.9(t0 − 1.30) 2 11 + 4.9 (1.84 − 1.30) 2 v0 = = = 23.0 m/s 1.84 − 1.30 m/s t0 − 1.30 m/s 1.22 s:

2-99.

v0 =

(a) Time taken for drops to fall is t =

2h / g . If there are n drops per second, the number of

drops in the air at any one time is nt = n(2h / g )1 / 2 . (b) Since the drops are falling at a constant rate, the median height is the distance fallen by a drop in half the time needed for it to hit the ground. Calling this time τ, we have 1 τ = (2h / g ) 1/2 = (h/2g)1/2. 2 1 1 1 Then x = v0t − gt 2 = − g(h/2g) = − h 2 2 4 1 Thus the position of the median is h below the edge of the spout 4 3 or h above the ground. 4


CHAPTER

2

(c) The density of drops is proportional to 1/v, where v is the velocity of a particular drop. Thus 1 1 . density ∼ = v 2 g ( h − x) The average height calculation must take into account the weighting factor of the density of drops. Thus x ∫ 2 g (h − x) dx h = . dx ∫ 2 g ( h − x) Let ξ = h − x, so that h ( h − ξ )dξ h =

∫

0

∫

h

0

2-100.

a=

ξ dξ ξ

dv = Ct 2 . dt

[2 / 3ξ 3 / 2 ]

=h− [2ξ 1 / 2 ] v

∫

8.0m/s

h 0

h 0

=h−

2 / 3h3 / 2 1 2 = h − h = h. 1/ 2 2h 3 3

t

dv = ∫ Ct 2 dt ⇒ v − 8.0 m/s = 0

4

Ct 3 3

t

⇒ v = 8.0 m/s + 0

Ct 3 3

C = 0.25 m/s ⇒ v = 8.0 m/s + (0.0833 m/s )t , At t = 3.0 s, v = 10.25 m/s. Final result = 10 4

3

m/s. To find x, integrate v: t

x=

∫ 0

†2-101.

t

vdt = ∫ ⎡⎣8.0 m/s + (0.0833 m/s)t 3 ⎤⎦ dt = (8.0 m/s )t + 0

(0.0833 m/s 4 )t 4 4

= (8.0 m/s )t + (0.0208 m/s 4 )t 4 At t = 1.0 s, x = 8.0 m. (This is also the change in position, since x = 0 at t = 0.) dv t2 At t = 1 s, a = = a0 (1 − ) dt 4.0 s 2 v

1s

0

0

v = ∫ adt =

∫ a0 (1 −

t2 t3 1 ) dt = a ( t − ) |10 s = 20 m/s 2 (1 − ) s = 18.3 m/s. 0 4.0 s 2 12.0 s 2 12

After 2 s, the acceleration becomes zero, so the velocity becomes constant at whatever value it had at t = 2 s. So to find v after a long time (t >> 2 s), find its value at 2 s: 2s v t2 t3 8 v = ∫ adt = ∫ a0 (1 − ) dt = a ( t − ) |02 s = 20 m/s 2 (2 − ) s = 26.7 m/s 0 2 2 4.0 s 12.0 s 12 0 0 The distance traveled is 2s

x − x0 =

∫ 0

2s

vdt = ∫ a0 (t − 0

2 t3 t2 t4 (2 s) 4 ⎤ 2s 2 ⎡ (2 s) ) dt a ( ) | (20 m/s ) = − = − 0 0 ⎢ ⎥ 12.0 s 2 2 48 s 2 48 s 2 ⎦ ⎣ 2

= 33.3 m. 2-102.

v

2s

0

0

v = ∫ adt =

2 ∫ ( At + Bt ) dt =

At 2 Bt 3 2 s 1 1 + |0 = i 15 m/s 2 i 22 s 2 + i 25 m/s 4 i 23 s3 2 3 2 3

= 96.7 m/s


CHAPTER 2s

Distance traveled ∆x =

∫

1s

2s

vdt =

∫

1s

(

2

At 2 Bt 3 At 3 Bt 4 2 s + )dt = + |1 s 2 3 6 12

15 m/s3 (23 − 1) s3 25 m/s 4 (24 − 1) s 4 + = 48.75 m 6 12 v dv dv dv a= = g − Av ⇒ =t = dt ⇒ ∫ dt g − Av g − Av v0

= †2-103.

1 g − Av g − Av ln =t⇒ = e − At ⇒ g − Av = ( g − Av0 )e− At A g − Av0 g − Av0 1 g ⇒ v = [ g − ( g − Av0 )e − At ] = (1 − e − At ) + v0 e − At A A g After a long time, t >> 1/ A, e − At → 0, v → . A v dv dv 1 a= = − Bv 2 ⇒ ∫ 2 = − B ⋅ ∆t ⇒ − |vv0 = − B∆t dt v v v0 ⇒−

2-104.

∆t =

⎛ ⎞ 1 1 1 − ⎜ ⎟ × 3600 s/hr −1 6.1 × 10 m i 1000 m/km ⎝ 90 km/hr 120 km/hr ⎠ = 16.4 seconds 1 mile 1 mile Cram’s speed is vC = = . The time difference for the two runners 3 min 46.32 s 226.32 s getting to the finish line is ∆t = 3 min 46.32 sec − 3 min 44.39 sec = 1.93 s. So Cram is behind 1 mile 1609 m by a distance of vC ∆t = = 13.7 m. × 1.93 s = 8.53 × 10−3 mile × 226.32 s mile distance 1500 m 1500 m v= = = = 14.43 m/s time 1min 43.95sec 103.95 s distance (100 + 100) m = = 1.3 m/s Average speed = time (10 + 60 + 80) s =

†2-105.

2-106. 2-107.

2-108.

†2-109.

⎛ ⎞ 1 ⎛1 1 ⎞ 1 1 1 − ⎜ − ⎟= ⎟ −4 −1 ⎜ B ⎝ v v0 ⎠ 6.1 × 10 m ⎝ 90 km/hr 120 km/hr ⎠ −4

Average velocity = 0, because the woman returns to her starting point. 1 × 30 min = 17.5 km; (a) d1 = 35 km/hr × 30 min = 35 km/hr × 60 min/hr 1 d 2 = 85 km/hr × 30 min = 85 km/hr × × 30 min = 42.5 km; 60 min/hr d + d2 (17.5 + 42.5) km = = 1.0 km/min, or 60 km/hr (b) Average speed = 1 time 60 min Define: t = time from when the sailfish spots the mackerel to when it catches the mackerel. Then: distance for the sailfish = 109 km/hr × t , distance for the mackerel = 33 km/hr × t. The separation between the fish is 20 m, so the time for the sailfish to catch the mackerel is given by 109 km/hr × t − 33 km/hr × t = 20 m


CHAPTER

2

20 m 20 m i 3600 s/hr = 0.95 s. During this time, the sailfish travels a = 76 km/hr 76 km/hr i 1000 m/km 109 km/hr × 1000 m/km × 0.95 s distance d = 109 km/hr × 0.95 s = = 28.8 m. 3600 s/hr Distance traveled by the first plane = d1 = 720 km/hr i t , where t is measured beginning at 10:00. Distance traveled by the second plane = d2 = 640 km/hr i (t − 1 hr), since that plane left one hour (1286 + 640) km later. d1 + d 2 = 1286 km ⇒ 720t + 640(t − 1) = 1286 km ⇒ t = = 1.42 hr 1360 km/hr = 1 hr 25 min. The distance traveled by the first plane is d1 = (720 km/hr)(1.42 hr) = 1022 km, ⇒t=

2-110.

†2-111.

so the planes meet 1022 km north of San Francisco. Since the total elapsed time is 1 hr 25 after departure of the first plane, they meet at 11:25 AM. Distance that my car travels = 80 km/hr i t Distance that the other car travels = 50 km/hr i t. To go from 10 m behind the slower car to 10 m ahead of it requires traveling a total relative distance of 10 m + 10 m + 4 m, because of the length of the car. Thus 80 km/hr i t − 50 km/hr i t = (10 + 10 + 24 m 24 m i 3600 s/hr 4) m ⇒ t = = = 2.9 s. 30 km/hr 30 km/hr i 1000 m/km

2-112.

†2-113.

2-114.

v 2 − v02 (1052 − 0) km 2 /hr 2 = = 1.72 × 103 km/hr 2 2x 2(3.2 km) v − v0 105 km/hr − 0 t = = = 0.0610 hr, or 219 seconds. a 1.72 × 103 km/hr 2 a=

(a) x = 2.0 + 6.0t − 3.0t 2 . At t = 0.50 s, x = 2.0 + 6.0 × 0.50 − 3.0 i (0.50)2 = 4.3 m. dx (b) v = = 6.0 − 6.0t. At t = 5.0 s, v = 6.0 − 6.0 i 0.50 = 3.0 m/s dt dv d (c) a = = (6.0 − 6.0t ) = −6.0 m/s2 at all times. dt dt (a) 100 km/hr = 27.8 m/s. The position of the speeder after 8.0 s is xs ,1 = vt1 = 27.8 m/s × 8 s = 222 m from the starting point. (b) The speed of the police cruiser goes from 0 to 120 km/hr (33.3 m/s) in 10 s, so its acceleration is 3.33 m/s2. The position of the police cruiser after reaching its final speed is at 2p ,1 (3.33 m/s 2 )(10 s) 2 x p ,1 = = = 168 m from the starting position. At this time the position of 2 2 the speeder is xs ,2 = vts ,2 = 27.8 m/s × 18 s = 500 m, so the speeder is 332 m ahead of the police cruiser. (c) Let t be the time from when the cruiser reaches its final speed of 120 km/hr until it catches up to the speeder. When the cruiser catches up to the speeder, both vehicles have traveled the same distance from the point where the cruiser reached 120 km/hr. Mathematically this means (33.3 m/s)t = (27.8 m/s)t + 332 m, which gives t = 60 s. So the total time that has elapsed since the cruiser began pursuit is 70 s. During this time the cruiser traveled a total distance of 168 m + (33.3 m/s)(60 s) = 2.18 × 103 m, or 2.18 km.

†2-115.

The initial speed of the car is v0 = 90 km/hr = 25.0 m/s. The distance traveled during the reaction time t1 = 0.75 s is d1 = v0t1 = (25 m/s)(0.75 s) = 18.8 m. The remaining distance to the cow is d2 = 30 m − 18.8 m = 11.2 m. The car’s acceleration as it travels this distance is a = −8.0 m/s2. Its


CHAPTER

2

final speed when it hits the cow is given by v 2 − v02 = 2ad 2 ⇒ v = 2-116.

v02 + 2ad 2 =

(25 m/s) 2 + 2(−8.0 m/s)(11.2 m) = 21.1 m/s, or

76 km/hr. Use v2 − v 02 = −2g(x − x0) with v = 0; g = 9.8 m/s2; v0 = 26 m/s x − x0 =

(26 m/s) 2 = 34m −2 × 9.8 m/s 2

The total time of flight for any particle of water is: 2v 2 × 26 m/s = 5.3 s = 8.84 × 10−2 min t= 0 = g 9.8 m/s 2 The discharge rate is 280 l/min so the total amount of water in the air after 5.3 s is vol = 280 l/min × 8.84 × 10−2 min vol = 25l 2-117.

Speed upon impact, v =

2-118.

v 33.1 = = 2209 m/s 2 t 0.015 The final speed of the part with acceleration g is v = 200 km/hr = 55.6 m/s. v − v0 55.6 m/s The time with acceleration is t1 = = = 5.66 s. The distance she falls during this 9.81 m/s 2 g

2gh =

2 × 9.8 × 56 = 33.1m/s

Average deceleration, a =

time period h1 =

†2-119.

v 2 − v02 55.62 (m/s) 2 = = 158 m. The rest of the height is h2 = 1000 − 158 m 2g 2 × 9.81 m/s 2

= 842 m. She falls the distance h2 with a constant speed of 55.6 m/s. The time for this part of the h 842 m fall is t2 = 2 = = 15.2 s ⇒ total time = (5.66 + 15.2) s = 20.9 s. v 55.6 m/s 1 (a) ∆t = 1 s. The distance the first ball falls in that interval is y1 = gt 2 2 1 = (9.81 m/s 2 )(1 s) 2 = 4.90 m, so the first ball is 13 m − 4.9 m = 8.1 m above the ground when 2 the second ball is released. The time for the first ball to fall the total distance of 13 m is 2h 2 × 13 m = = 1.63 s. When the first ball hits the ground, the second ball has been t1 = 9.81 m/s 2 g falling for t2 = 0.63 s. The distance second ball has fallen during this time is 1 1 y2 = gt22 = (9.81 m/s 2 )(0.63 s) 2 = 1.95 m, so the second ball is 13 m − 1.95 m = 11.1 m 2 2 above the ground when the first ball lands. (b) v (1st) = gt1 = 9.81 m/s 2 i 1.63 s = 16.0 m/s

v(2nd) = gt2 = 9.81 m/s 2 i 0.63 s = 6.18 m/s ⇒ instantaneous velocity of the first ball relative to the second just before the first hits the ground is: 16.0 m/s − 6.18 m/s = 9.8 m/s down. (c) Both balls have same acceleration, (9.81 m/s2 down,) so the relative acceleration is zero.


CHAPTER 3

VECTORS


RGP = RGA + RAP = 10.2 xˆ + 5.9 yˆ

| RGP | =

(10.2) 2 + (5.9) 2

= 11.8 km 5.9 θ = tan−1 = 30°N of E 10.2

3-2.

The two displacements are OA = 240(sin 29° xˆ + cos 29° yˆ ) = 116.4 xˆ + 209.9 yˆ

AB = 560(−cos 29° xˆ + sin 29° yˆ ) = − 489.8 xˆ + 271.5 yˆ The resultant displacement vector is: OB = OA + AB = −373.4 xˆ + 481.4 yˆ | OB | =

(373.4) 2 + (481.4) 2

= 609.2 m

⎛ 481.4 ⎞ θ = tan−1 ⎜ ⎟ ⎝ 373.4 ⎠ = 52.2° N of W †3-3.

A reduced copy of the diagram is shown. In the actual diagram, 1 cm = 1 km. From the diagram, the line representing the resultant R is 11.2 cm long, so the length of R is 11.2 km. The angle θ is measured to be 27.5°. Graphically we find R = 11.2 km @ 27.5° S of E. To do the problem trigonometrically, take x to point east and y to point north. Then


CHAPTER

3

r1 = 18.0 sin 60°i + 18.0 cos 60°j = 15.6i + 9.0 j km r2 = 9.5 cos 60°i − 9.5sin 60°j = 4.75i − 8.23 j km r3 = − 12.0 sin 60°i − 12.0 cos 60°j = − 10.4i − 6.0 j km Adding gives R = 9.95i − 5.23j km. The magnitude is R = 9.952 + 5.232 = 11.2 km, and the −5.23 direction is θ = tan −1 = − 27.7°. In terms of compass directions, this is R = 11.2 km @ 9.95 27.7° S of E.

3-4.

We can combine the two vectors at 45° E of N to make a vector 5.8 km at 45° E of N (see diagram). Then R2 = 5.82 + 4.52 − 2(5.8)(4.5) cos 85° R=

49 m 2 = 7.0 km

By the law of sines: sin θ sin 85° = 4.5 7.0 ⎛ 4.5 ⎞ θ = sin−1 ⎜ sin 85° ⎟ = 40° ⎝ 7.0 ⎠

Therefore, the angle between R and N is (45 − 40)° = 5°. Thus, the resultant is magnitude 7.0 km, 5° E of N. †3-5.


CHAPTER

3-6.

The graph is drawn with a scale of 1 cm = 20 m. Measuring the distance from South Bay to Mosquito Rock, we get: r ≈ −0.6 cm i − 3.0 cm j on the graph = −120 m i − 600 m j. 120 = The distance between them is (−120) 2 + (−600) 2 = 612 m at an angle θ = tan −1 600 11.3° west of south. r1 = 2.5 km j, 1.5 km r2 = (1.5 km) sin 30° i + (1.5 km) cos 30° j = 0.75 km i + 1.30 km j 30° r = r1 + r2 = 0.75 km i + 3.80 km j. |r| =

0.752 + 3.802 = 3.87 km

The angle of this vector with the x axis: θ = tan −1

3.80 = 0.75

2.5 km

78.8° north of east.

†3-7.

A reduced copy of the graphical solution is shown in the diagram. In the actual graph, the scale is 1.0 cm = 100 km, so the line for the first displacement is 4.8 cm long and the line for the second displacement is 3.7 cm long. The line for the total displacement is measured to be 4.4 cm long, corresponding to a magnitude of 440 km. The direction is measured to be 7.5° W of N. The total displacement is D = 440 km @ 7.5° W of N. Check analytically. Let r1 = 480 sin (40°) km i + 480 km cos (40°) j = 308.5 km i + 367.7 km j and r2 = −370 cos (10°) km i + 370 km cos (10°) j = −364.4 km i + 64.2 km j, where E corresponds to x and N corresponds to y. Then D = r1 + r2 = −55.9 km i + 431.9 km j. |D| =

3-8.

(−55.9) 2 + 431.92 km = 436 km. To get the direction relative to the x direction, note that

the resultant vector is in the third quadrant (negative x component, positive y component). The standard method of finding the direction will give an angle measured from E, the x axis. Using a calculator is used to find the inverse tangent will give a negative angle, so the correct angle (in the second quadrant) is found by adding 180º to the calculator result: 431.9 θ = 180° − tan −1 = 97.4°, which measured N of E. This is 7.4° W of N. The analytical 55.9 result is D = 436 km @ 7.4° W of N, which agrees almost exactly with the graphical result. Distance east = 14 cos 60° = 7 km Distance south = 14 sin 60° = 12.1 km


3

CHAPTER

3

†3-9.

The resultant displacement vector, R = (5 m). The given displacement vector, A = (2.2 m) sin 35° i + (2.2 m) cos 35° j = (1.26 m) i + (1.8 m)j. Let the other displacement vector be B = Bxi + Byj. Therefore, R = A + B = (1.26 m + Bx) i + (1.8 m + By) j = 5 mj. Comparing i and j components gives 1.26 m + Bx = 0 or Bx = −1.26 m 1.8 m + By = 5 or By = 3.2 m Thus, B = (−1.26 m)i + (3.2 m) j.

3-10.

Graphs were drawn at actual size. Reduced copies are shown.

A + B + C = 13.7 cm @ 8.0° E of N A + B − C = 7.7 cm @ 14.0° E of N

†3-11.

The easiest method is to write the vectors in component form, with ˆj as N and î as E. Then R = 1.2 ˆj + (sin 38° î cos 38° ˆj )6.1 + (sin 59° î − cos 59° ˆj )2.9 + (sin 89° î + cos 89º ˆj ) 4.0 + (sin 31° î − cos 31° ˆj )6.5 = 1.2 ˆj + 3.76 î − 4.81 ˆj + 2.49 î − 1.49 ˆj + 4.00 î + 0.07 ˆj + 3.35 î + 5.57 ˆj = 13.6 î + 0.54 ˆj This translates into a vector of length13.6 nmi at 88° E of N.

3-12.

(a) (b)

(b) (cont.) The radius of the earth is 1.49 × 1011 m, so using the Pythagorean theorem, we get for the displacement vector d = (1.49 × 1011 ) 2 + (1.49 × 1011 ) 2 = 2.1 × 1011 m


CHAPTER 3-13.

The straight-line distance is d2 = R2 + R2 − 2R2 cos 178° = R2 + R2 + 1.999 R2 = 3.999 R2 d = 2.000 R = 12,750 km Distance around the equator is S=

2π × 178° × R 360°

S = 19,810 km

3-14.

By completely filling in all of the angles in the triangle, we find that it is an isosceles triangle, so that P′ P = AP′.

3-15.

Position vector at 11h10m is 4.2 km at 33° E of N. Displacement vector from position at 10h30m to 11h10m is A2 = 9.52 + 4.22 − 2(9.5)(4.2) cos 27° = 36.8 km2 A = 6.07 km. sin θ/4.2 = sin 27°/6.07 θ = sin−1 (4.2/6.07 sin 27°) The displacement vector then is A = 6.07 mi, 78.3° W of S. In another 20 min, it would have traveled a distance of 60 = 9.1 km. 6.07 × 40 Therefore, its position at 11h30m should be

B 2 = 9.52 + 9.102 − 2(9.5)(9.1) cos 18.3° = 8.90 km2

B = 3.0 km sin ø = 9.10/3.0 sin 18.3° = 0.952 ⇒ ø = 72.3° Thus, the position vector is 3.0 km, 12.3° W of N


3

CHAPTER

3

3-16.

If the bearing of each ship remains constant with respect to the other, the angles θ1 and θ2 are constant in the diagram where the two ships have been denoted by A and B. A′, B ′ shows their positions at a later instant. For θ1, θ2 constant, the separation between the ships will eventually go to zero, otherwise they will miss each other, assuming they are point masses.

3-17.

N: 12.0 cos 40° = 9.19 km W: 12.0 sin 40° = 7.71 km

3-18.

Ax = 5.0 cos 30° = 4.3 m Ay = 5.0 sin 30° = 2.5 m

†3-19.

Place the x axis along the sloping line as shown. Then: x component = − (4.0 m) sin(25°) = − 1.7 m.

y

x 25°

4m 25°

3-20.

By the Pythagorean theorem, the distance along the ground is:

D = 152 − 0.52 km ≈ 15 km x : 15.0 km × cos 56° = 8.4 km. y : 15.0 km × cos 35° = 12.3 km. z : 0.5 km = 500 m


CHAPTER †3-21.

3-22.

†3-23.

(a)

(b)

A = 3i + 2j cm B = −1i + 3j cm A + B = 3i + 2j − 1i + 3j cm = (2i + 5j cm)

Ax = 8.0 cos 52° = 4.9 units Ay = 8.0 sin 52° = 6.3 units Az = 8.0 cos 90° = 0 units x : 6.0 cos 45° = 4.2 units y : 6.0 cos 85° = 0.5 units Since the magnitude of the vector is 6.0 units, we know that Ax2 + Ay2 + Az2 = 6.02. Thus Az = ±

A=

6.02 — 4.22

—

0.52 = ± 4.2 units

Az can be either ± 4.2 units, so it is not uniquely determined. 32 + 22 + 12 = 14 units = 3.7 units

3-24.

Magnitude =

3-25.

(a) A + B = (5i − 3j + k) + (2i + j − 3k) = −3î − 2ˆj − 2kˆ (b) A − B = (−5i − 3j + k) − (2i + j − 3k) = 7i − 4 j + 4k (c) 2A − 3B = 2(−5i − 3j + k) − 3(2i + j − 3k) = –16i – 9j + 11k

3-26.

r=6i−8jm |r| = 62 + 82 = 10 m. The angle between r and +x is: θ = tan −1

−8 = −53.1° 6

3-27.

x = 14 × cos(135°) = –9.9 m y = 14 × sin(135°) = 9.9 m

3-28.

E = (2.5 + 1.0 + 1.5 + 0) i + (3.5 + 4.5 + 2.0 + 3.0) j + (0 + 2.5 + 3.0 + 1.5) k = 5.0 i + 13.0 j + 7.0 k F = (2.5 −1.0 + 1.5 − 0) i + (3.5 − 4.5 + 2.0 − 3.0) j + (0 − 2.5 + 3.0 − 1.5) k = 3.0 i − 2.0 j − 1.0 k

†3-29.

|E| =

52 + 132 + 7 2 = 15.6

|F| =

32 + (−2) 2 + (−1) 2 = 3.74

rˆ =

A 2i + 4 j + 4k 1 2 2 = = i+ j+ k 3 3 3 |A| 22 + 42 + 42


3

CHAPTER 3-30.

3

A = 2.0 i + 3.0 j + 1.0 k, B = −1.0 i + 2.0 j + Bz k, A + B = (2.0 − 1.0) i + (3.0 + 2.0) j + (1.0 + Bz) k |A + B| = 12 + 52 + (1 + Bz ) 2 = 6.0 ⇒ 26 + (1 + Bz ) 2 = 36

⇒ 1 + Bz = ± 10 ⇒ Bz = 10 − 1, or, Bz = − 10 − 1 †3-31.

3-32.

c1A + c2B = c1(2.0 i + 3.0 j) + c2(1.0 i + 5.0 j) = (2.0 c1 + 1.0c2)i + (3.0 c1 + 5.0c2)j c1A + c2B = C = −1.0 i + 3.0 j ⇒ 2.0c1 + 1.0c2 = −1.0 (1) 3. 0c1 + 5.0c2 = 3.0 (2)

9 8 Solve for c1, c2: (1) × 3 − (2) × 2 : ⇒ −7c2 = −9 ⇒ c2 = , c1 = − 7 7 Let the planes be denoted by A and B and î → east and ˆj → west. A = 120 × 103(cos 20° î − sin 20° ˆj ) + 2500 kˆ B = 110 × 103(cos 25° î − sin 25° ˆj ) + 3500 kˆ ˆ ˆ A = 112.8 × 103 i − 41.0 × 103 j + 2500 kˆ ˆ ˆ B = 99.7 × 103 i − 46.5 × 103 j + 3500 kˆ

The position of B relative to A is ˆ ˆ ˆ B − A = −13.1 × 103 i − 5.5 × 103 j + 103 k The displacement vector is: horizontal distance =

(

The bearing is 5.5 = 23° W of S θ = tan−1 13.1 Altitude difference = 1000 m = 1 km 3-33.

)

(13.1) 2 + (5.5) 2 103 m = 14.2 km

Let the directions of i, j and k be as shown at the right, where GM is the Greenwich meridian. New York will have a position vector:


CHAPTER

NY: R(0.728 î + 0.209 ˆj + 0.653 kˆ ) where R = radius of the earth. And for Cape Wrath:

CW: R(0.068i + 0.779j + 0.624k) The displacement vector is = [(0.068 − 0.728) i + (0.779 − 0.209) j + (0.624 − 0.653)k]R = (−0.66 î + 0.570 ˆj − 0.029 kˆ )R The magnitude is = (0.6582 + 0.5702 + 0.0292)1/2R = 0.873R = 5550 km 3-34.

†3-35.

5.02 + 3.02 + 1.02 = 35 = 5.9 units A 5.0 5.0 = 32° x-axis : cos θx = x = ⇒ θx = cos−1 A 5.9 5.9 Ay 3.0 3.0 = −60° y-axis : cos θy = = ⇒ θy = cos−1 A 5.9 5.9 A 1.0 1.0 = 80° z-axis : cos θz = z = ⇒ θz = cos−1 5.9 A 5.9

Magnitude =

32 + 62 + 22 = 49 = 7 units 1 The unit vector in this direction is ( 3i − 6 j + 2k ) 7 Therefore the vector with magnitude 2 is 2 × (unit vector) 6 12 4 = i − j+ k 7 7 7 3i − 6j+ 2k has magnitude


3

CHAPTER 3-36.

3

3A − 2C = 4B. Then 3A − 4B = 2C 3 A − 2B 2 3 C = ( 6i – 2 j ) – 2 ( –4i − 3j + 8k ) = 17i + 3j – 16k 2 ( 5i + 2 j + k ) i ( 2i − k ) = (5 × 2) + (−2 × 0) + (1 × [−1]) = 9

C=

3-37. 3-38. †3-39.

The vectors A and B in Example 1 are perpendicular to each other, so the angle between them is 90º. Because cos 90º = 0, A • B = 0, regardless of their magnitudes.

A=

22 + 12 + 22 = 3

B=

32 + 62 + 22 = 7

Dot product A • B = (−2)(3) + (1)(−6) + (2)(2) = −8

3-40. †3-41. 3-42. †3-43.

A • B = AB cos ø AiB 8 = 112° = cos−1 ø = cos−1 AB 7×3 A i B = ( Ax i + Ay j + Az k ) i ( Bx i + By j + Bz k ) = Ax Bx + Ay By + Az Bz A i i 3 i1 + 4 i 0 + 2 i 0 = = 0.557 ⇒ θ = cos −1 0.557 = 56.1° 2 2 2 |A| 3 +4 +2 AiB = − 4 + 6 + 2 = 4 A × B = −1i − 6 j + 11k

cos θ =

A i B = | A || B | cos θ A × B = | A || B | sin θ If A × B = A • B, then A

B cos θ = |A| |B| sinθ ⇒ cos θ = sin θ

ο

3-44.

which gives: θ = 45 A = 50 sin 30° i + 50 cos 30° j = 25 i + 43.3 j B = −35 sin 70° i + 35 cos 70° j = −32.9 i + 12 j A × B = (25)(12) i × j − (43.3)(32.9) j × i = 1725 k (magnitude 1725 m 2 ; direction along the z -axis) B × A = −1725 k (the same magnitude as A × B; direction along the − z − axis)

†3-45.

Take north to be j and east to be i. Then A × B = (2180i ) × ( − 1790j) km 2 = −3.90 × 106 (i × j) km 2 = −3.90 × 106 k km 2

3-46.

See Eqs. (3.31) through (3.33). (2i − 5 j + 3k ) × (i + 0 j − 2k ) = [(−5)(−2) − (3)(0)]i + [(3)(1) − (2)(−2)]j + [(2)(0) − (−5)(1)]k = 10i − 5 j + 3k

3-47.

A = cos ω t i + sin ω t j dA = −ω sin ω t i + ω cos ω t j dt dA = −ω sin ωt cos ωt + ω sin ωt cos ωt = 0 A• dt dA Since both A and are nonzero, they must be perpendicular to each other. dt


CHAPTER 3-48.

N

B

N y

B

20°

x

20°

20º 20°

20º

E

E

20°

A

A

(a) (b) Define B to be the y axis as shown in (a). The x axis is shown perpendicular to B. The angle between A and B is 130° measured from the y axis, so A = 30sin130°i + 30cos130°j = 23.0i – 19.3j. The component of A along B is −19.3 m.

†3-49.

Next, define A to be the x axis as shown in (b). The y axis is shown perpendicular to A. Now the angle between A and B is 130° measured from the x axis, so B = 40cos130°i + 40sin130°j = –25.7i + 30.6j. The component of B along A is −25.7 m. A × B = ( Ay Bz − Az By )i + ( Az Bx − Ax By ) j + ( Ax By − Ay Bx )k

A × B = (5.0i − 2.0 j + 3.0k ) × ( Bx i + 3.0 j + Bz k ) = (−2 Bz − 9)i + (3Bx − 5Bz ) j + (15 + 2 Bx )k A × B = C = 2.0 j + C z k. (A × B) x = Ay Bz − Az By = C x ⇒ –2 Bz − 9 = 0 ⇒ Bz = −4.5 ( A × B) y = Az Bx − Ax Bz = C y ⇒ 3Bx − 5Bz = 2 ⇒ Bx =

1 (2 + 5Bz ) 3

1 [2 + 5 × (−4.5)] = −6.83. 3 ( A × B) z = Ax By − Ay Bx = C z ⇒ 15 + 2 Bx = C z ⇒ C z = 15 + 2(−6.83) = 1.34 =

3-50.

A = Ax i + Ay j, B = Bx i + By j A i B = | A || B | cos θ = Ax Bx + Ay By

(1)

A × B = | A || B | sin θ = Ax By − Ay Bx

(2)

(2) ÷ (1) ⇒ tan θ = 3-51.

Ax By − Ay Bx Ax Bx + Ay By

A = 2i − j − 4k 22 + 12 + 42 = 21 1 unit vector is ( 2i − j − 4k ) = 0.44i − 0.22 j − 0.87k 21

A=

3-52.

Let A = j + 2k and B = 3i − j + k. Let the unit vector along A be u A = along B be u B =

3i − j + k . 5


i + 2j and the unit vector 5

3

CHAPTER

3

A vector C pointing from 0 to the midpoint of the difference vector u A − u B will bisect the angle between A and B, 1 1 C = u B + ( i1 − i 2 ) = ( u1 + u 2 ) 2 2 =

uBuB 00

1⎛ 3 1 ⎞ 1 ⎞ ⎞ ⎛ 1 ⎛ 2 − + i+⎜ j+⎜ ⎜ ⎟ ⎟k ⎟ 2 ⎝ 11 11 ⎠ 11 ⎠ ⎠ ⎝ 5 ⎝ 5

C C

uuA A−- uB

uA uA

Dividing C by its magnitude (|C| = 1/ 2 ) gives the required unit vector uC = 2C = ( 0.64i + 0.10 j + 0.81k ) 3-53.

3-54.

The displacement vector pointing from A to B is B − A. The vector that points halfway between A and B is (B − A)/2 from the tip of A, which makes its position vector at A + (B − A)/2 = 1 1 1 (A + B)/2. Here, ( A + B ) = ( 4i + 2 j − i + 3j + 2k ) = ( 3i + 5 j + 2k ) . Its magnitude 2 2 2 38 1 2 is 3 + 52 + 2 2 = . The unit vector that points in this direction is 2 2 1 ( 3i + 5 j + 2k ) = 0.49i + 0.81j + 0.32k. 38 Edges of the cube are parallel to the x-y-z axes. The diagonal of the unit cube is then AC = i + j + k whose magnitude is 12 + 12 + 12 =

3.

Let AB be the unit vector in the

x-direction. Then, cos ø = ø= †3-55. 3-56.

î i (î + ˆj + kˆ ) AC i AB 1 = = | AC || AB | 3 1i 3

cos −1 3

= 54.7°

Since |A| ≠ 0, |B| ≠ 0, yet A i B = 0, we can conclude the angle between A and B is 90°. So | A × B | = | A || B | sin 90° = 4 × 6 = 24 (a) Right-hand-rule: fingers wrap from the first vector toward the second vector, the thumb points at the direction of the product. (b) A × B = ( Ax i + Ay j + Az k ) × ( Bx i + By j + Bz k ) = Ax Bx i × i + Ay Bx j × i + Az Bx k × i + Ax By i × j + Ay By j × j + Az By k × j + Ax Bz i × k + Ay Bz j × k + Az Bz k × k = ( Ay Bz – Az By )i + ( Az Bx – Ax Bz ) j + ( Ax By – Ay Bx )k


CHAPTER

(c) If A || B,

A B A B Az B = z, x = x, x = x Ay By Ay By Az Bz

⇒ Ax By − Ay Bx = 0

Ax By − Ay Bx = 0 Az Bx − Ax Bz = 0 ⇒ A× B =0 3-57.

i j k Ay Az Ax Ay Ax Az Ax Ay Az = i– j+ k By Bz Bx Bz Bx By Bx By Bz 3-58. †3-59.

= (AyBz − AzBy) i + (AzBx − AxBz) j + (AxBy − AyBx) k [(i × j) × i ] × i = (k × i ) × i = j × i = − k [(i × j) × j] × j = (k × j) × j = − i × j = − k A = 2i − 3 j + 2k. B = −3i + 4k. A × B = (2i − 3 j + 2k ) × (−3i + 0 j + 4k ) = ( Ay Bz − Az By )i + ( Az Bx − Ax Bz ) j + ( Ax By − Ay Bx )k = [( −3)(4) − (2)(0)]i + [(2)(−3) − (2)(4)]j + [(2)(0) − (−3)(−3)]k = −12i − 14 j − 9k.

3-60.

(a) A • (B + C) = (3î − 2ˆj + 2kˆ ) • (2î − 3ˆj + 4kˆ ) (b) A × (B + C) = (3î − 2ˆj + 2kˆ ) × (2î − 3ˆj + 4kˆ )

i

j

k

= 3 −2 2 = (−8 + 6)i− (12 − 4)j + (−9 + 4) k = 2i − 8 j − 5k 2 −3 4

Ax (c) A • (B × C) = Bx

Ay By

Cx

Cy

i

j

3 −2 2 Az Bz = 0 0 4 = 3(12) + 2(−8) + 2(0) = 20 2 −3 0 Cz

k

(d) B × C = 0 0 4 = 12 î + 8 ˆj 2 −3 0

i

j

k

A × (B × C) = 3 −2 2 = −16i + 24 j + 48k 12 8 0 3-61.

The cross product of the two vectors will give a vector perpendicular to both.

i

( 4i + 3 j )

j

k

× (−i − 3j + 2k ) = 4 3 0 = 6i − 8j − 9k −1 −3 2

The magnitude of this vector is

62 + 82 + 92 = 181 = 13.45


3

CHAPTER

3-62.

3

The unit vector perpendicular to the two vectors is then 1 ( 6i − 8 j − 9k ) = 0.44i − 0.59 j − 0.67k 13.45 Volume of parallelopiped is equal to base area × height = Base area × A cos θ The base area = BC sin ø. Hence the base area = B × C. The volume then is | A × B | A cos θ. But B × C points vertically upward and the angle between it and A is θ. So (B × C) i A cos θ = |B×C| A

V = |B × C | A cos θ = | B × C | ⎡ (B × C) i A ⎤ ⎢ ⎥ = A i (B × C) ⎣| B × C | A ⎦ 3-63.

Pick B along the x-axis and C along the x-y plane. B = Bx i C = Cx i + C y i

A = Ax i + Ay i + Az k Then,

i (B × C) = Bx

j 0

Cx

Cy

k 0 = BxCy k 0

i Therefore A × (B × C) = Ax

j Ay

k Az

0

0

Bx C y

= AyBxCy i − AxBxCy j

But B(A • C) − C(A • B) = Bx i ( Ax C x + Ay C y ) − (Cxi + Cyj)AxBx = AxBxCx i + AyBxCy i − AxBxCx i − AxBxCy j = Ay Bx C y i – Ax Bx C y j = A × (B × C)

3-64.

Let x and y be along magnetic east and north, respectively. (a) Coordinates based on magnetic north E comp: 5.0 sin 56° = 4.15 km N comp: 5.0 cos 56° = 2.80 km (b) Coordinates based on geographic north E comp: 5.0 sin 44°5′ = 3.48 km N comp: 5.0 cos 44°5′ = 3.59 km Note: We can also use the rotation angle −11°55′ E comp: 4.15 cos (−11°55′) + 2.80 sin (−11°55′) = 3.48 km N comp: −4.15 sin (−11°55′) + 2.80 cos (−11°55′) = 3.59 km


CHAPTER 3-65.

If the axis of the original coordinate system is rotated so that it points in the direction of this vector, the only non-zero component will be the x′ component. This vector makes an angle tan−1(−3/6) = tan−1(−1/2) = −26.6° with the x-axis. Thus a rotation of −26.6º will do the job. The vector will have coordinates

3-66.

†3-67.

magnitude =

62 + 32 i = 6.71i

902 + 702 = 114 km

⎛ 70 ⎞ direction: θ = tan −1 ⎜ ⎟ = 37.9° west of north. ⎝ 90 ⎠ By the law of cosines: C 2 = A2 + B2 − 2AB cos 115° = 3502 + 1202 + 2(350)(120)(0.4226) = 172,400 m2 C = 415 m

3-68.

By the law of sines, the angle between A and C is given by sin θ sin 115° = 120 415 ⎛ 120 ⎞ θ = sin−1 ⎜ sin 115° ⎟ = 15.2° ⎝ 415 ⎠ Therefore, the angle between C and north is 45° − 15.2° = 29.8° and the resultant has a magnitude of 415 m, at 29.8° W of N. (5 − 2 + 4)i + (0 + 1 + 2) j + (3 − 3 + 1)k 7 1 = i + j+ k rave = 3 3 3 2

2

59 ⎛7⎞ ⎛1⎞ 2 ⎜ ⎟ +1 + ⎜ ⎟ = 3 ⎝ 3⎠ ⎝ 3⎠ Horizontal component = 2.0 × sin(30°) = 1.0 Vertical component = 2.0 × cos(30°) = 1.7 Magnitude = 2 × cos(30°) × 250 = 433 km. The direction is | rave | =

3-69. 3-70.

due south. 30°

30° S


3

CHAPTER †3-71.

3

A = 6.2 cos 30°i − 6.2 sin 30°j B = −9.6j Therefore, A + B = 6.2 cos 30°i + (− 6.2 sin 30° − 9.6)j = 5.4i − 12.7 j and A − B = 6.2 cos 30°i − (6.2 sin 30° − 9.6)j = 5.4i + 6.5 j

3-72.

The displacement D = 4I + 5j + 3k m. | D | =

(4) 2 + (5) 2 + (3) 2 m = 7.1 m.

By trial and error, one discovers that the shortest path across the room starting at the origin is diagonally across the floor (4 m × 5 m) and up the wall (3 m). The distance is

(

)

4 2 + 52 + 3 m =

9.4 m. If the lizard crawls only along the walls as the problem says, the shortest distance will be diagonally along the wall that is 4 m long and 3 m high then horizontally along the 5 m long wall. Now the distance is 3-73. 3-74.

3-75.

3-76.

(

)

42 + 32 + 5 m = 10 m.

22 + 12 + (−4) 2 = 4.58 (a) A + B = i − 6j (b) A − B = 7 i + 2j (c) 3A − B = 15i − 2j A = 50 sin 30° i + 50 cos 30° j B = −35 sin 70° i + 35 cos 70 j A • B = −(50)(35) {sin 30° sin 70° − cos 30° cos 70°} = −1750 (0.47 − 0.30) = −304 m 2 The angle between A and B is 100°. Therefore, the magnitude of A × B is | A × B | = (6.0)(8.0) sin 100° m = 47.3 m By the right-hand rule, the direction of A × B is downward.

†3-77.

3-78.

If φ is the angle between A and B, then the component of A along B is A cos ø. But cos ø = ⎛ AiB ⎞ AiB . B = 12 + 32 + 22 = 14 = 3.74. (A • B)/(AB). Therefore A cos ø = A ⎜ ⎟= B ⎝ AB ⎠ 15 = 4.0. A • B = (3)(1) + (4)(3) + (0) (−2) = 15. Then A cos ø = 3.74 AiB with A = 32 + 42 = 5. Thus, B cos ø Similarly, the component of B along A is A 15 = = 3.0. 5 (a) A + B = 5i + j − k (b) A − B = − i + 3j − k (c) A i B = 6 − 2 + 0 = 4 (d) A × B = [(2)(0) − ( −1)(−1)]i + [(−1)(3) − (0)(2)]j + [(2)( −1) − (2)(3)]k = −i − 3j − 8k


CHAPTER 4

MOTION IN TWO AND THREE DIMENSIONS


(a) Total displacement is (3.2 + 2.6) km 45° E of N + 4.5 km 50° W of N = 7.0 km , 5° E of N (see Problem 3, Chapter 3) (b) Avg vel = ∆r/∆t = 7.0 km/1.25 h, 5° E of N = 5.6 km/h, 5° E of N (c) Avg speed =

4-2.

4-3.

4-4.

total distance 3.2 + 2.6 + 4.5 km = = 8.24 km/h time 1.25 h

avg speed =

distance π × r π × 1.50 × 1011 m = = = 29.9 km/s time 1/ 2 yr 1/ 2 × (365.25 × 24 × 3600) s

| Avg vel | =

∆r 2×r 2 × 1.50 × 1011 m = = = 19.0 km/s ∆t 1/ 2 yr 1/ 2 × 3.16 × 107 s

171 km/h = 47.5 m/s If the bullet is fired at an angle θ, its position vector is given by rbul = 366 sin θ t î + 366 cos θ t ˆj For the bird, rbird = (47.5)t î + 30 ˆj For the bird to be hit rbul = r bird. Comparing the x-components, we have 366 sin θ = 47.5 θ = sin−1(47.5/366) = 7.45° Then d = 30 tan θ = 30 tan 7.45° = 3.93 m. Thus the hunter must aim 3.93 m ahead of the bird in order to hit the bird. At t = 0, the velocity is v = 30 km/h N, or t = 10 s v = 0i + (30 km/h) j. At t = 10 s,

t = 20 s

v = 30 km/h @ 45° W of N, or v = −(21 km/h)i + (21 km/h) j. At t = 20 s, v = 30 km/h W, or v = −(30 km/h)i + 0 j. At t = 30 s, v = 30 km/h @ 45° S of W, or v = −(21 km/h)i − (21 km/h) j. At t = 40 s, the driver is halfway around and v = 30 km/h S, or v = (30 km/h)i + 0 j. 4-5.

t = 30 s

t = 40 s

r = (4 + 2t)i + (3 + 5t + 4t2)j + (2 − 2t − 3t 2)k (a) v = dr/dt = 2i + (5 + 8t ) j − (2 + 6t )k (b) a = dv/dt = 8 j − 6k magnitude, | a | =

(8) 2 + (6) 2 = 10 m/s 2

⎛6⎞ direction θ = −tan−1 ⎜ ⎟ = −37° (37° below the y -axis in the z -y plane) ⎝8⎠


t=0

CHAPTER 4-6.

4

(a) The components of v are vx = dx/dt and vy = dy/dt. Also, the components of a are ax = dvx/dt and ay = dvy/dt. vx = − Ab sin bt v y = Ab cos bt ax = − Ab 2 cos bt

(b) | v | = |a|= †4-7.

a y = − Ab 2sin bt

(− Ab sin bt ) 2 + ( ab cos bt ) 2 =

A2b 2 (sin 2 bt + cos 2 bt = Ab

2 2 2 (− Ab 2 cos bt ) 2 + (− Ab 2sin bt ) 2 = Ab2 cos bt + sin bt = Ab

At t = 2.0 s, the missile has been falling with acceleration a = 0i − (9.81 m/s 2 ) j and horizontal velocity equal to the velocity of the airplane. This means the missile is still directly below the gt 2 (9.81 m/s 2 )(2.0 s) 2 j = 0i − j= airplane. Its displacement relative to the plane is r = 0i − 2 2 0i − (19.6 m) j, or 19.6 m @ 90º below the direction of travel of the airplane. At t = 3.0 s, which is 1.0 s after igniting the engine, the acceleration is a = (6.0 m/s 2 )i − (9.81 m/s 2 ) j. Its ax t 2 gt 2 i− j 2 2 (6 m/s 2 )t 2 gt 2 (6 m/s 2 )(1.0 s) 2 (9.81 m/s 2 )(1.0 s) 2 i− j= i− j = (3.0 m)i − (4.9 m) j. Now = 2 2 2 2 the missile’s total displacement relative to the plane is r2 = (0 m)i − (19.6 m) j + (3.0 m)i − (4.9 m) j = (3.0 m)i − (24.5 m) j. The magnitude is −24.5 r2 = (3.0 m) 2 + (24.5 m) 2 = 24.7 m. The direction is given by θ = tan −1 = − 83.0°, or 3 24.7 m @ 83º below the direction of travel of the plane. dr r = (5t + 4t 2 )i + (3t 2 + 2t 3 ) j + 0k m. v = = (5 + 8t )i + (6t + 6t 2 ) j + 0k m/s. dt d 2 r dv a= 2 = = (8)i + (6 + 12t ) j + 0k m/s 2 . At t = 2.0 s, v = (21 m/s)i + (36 m/s) j + 0k m/s, dt dt displacement during this 1.0 s interval is r1 =

4-8.

which gives a speed of v = (21 m/s) 2 + (36 m/s) 2 = 42 m/s. †4-9.

∆r , where ∆r is the total displacement and ∆t = 1.5 h is the total elapsed time. To find ∆t the total displacement, find the displacement during each part of the trip: ∆r1 = (300 km/h @ 30° N of E) × 0.50 h = 150 km @ 30° N of E. Taking the y direction to point N and the x direction to point E, this is ∆r1 = (150 km)( cos 30°)i + (150 km)( sin 30°) j = (130 km)i + (75 km) j. For the second part of the trip, ∆r2 = (300 km/h @ 30° W of S) × 1.0 h = 300 km @ 30° W of S. In terms of x and y, this is ∆r2 = − (300 km)( sin 30°)i − (300 km)( cos 30°) j = − (150 km)i − (260 km) j. The total ∆r displacement is ∆r = ∆r1 + ∆r2 = −20 km i − 185 km j. The average velocity is v average = ∆t v average =


CHAPTER =

−20 km i − 185 km j = −13.3 km/h i − 123 km/h j. To give this as a 1.5 h

N

speed and heading, use vaverage = 13.32 + 1232 km/h = 124 km/h. Since

4-10.

θ

E

both velocity components are negative, the vector is located in the third vaverage quadrant (between W and S), so we can give the direction 123 = 83.8° S of W. Thus vaverage = 124 km/h @ 83.8º S of as θ = tan −1 13.3 W. ∆v v 2 − v1 The average acceleration is a average = , where v1 is the velocity during the 0.5, v2 is = ∆t ∆t the velocity during the next 1.0 h, and ∆t is the time interval during which the velocity changes. Since no value is given for ∆t, we can calculate ∆v and give the answer symbolically. Using the definitions above, v1 = (300 km/h) cos 30°i + (300 km/h) sin 30° j = (260 km/h)i + (150 km/h) j. 30° W of S is the same as 240° N of E, so v 2 = (300 km/h) cos 240°i + (300 km/h) sin 240° j = −(150 km/h)i − (240 km/h) j. Then ∆v = − (410 km/h)i − (410 km/h) j, −(410 km/h)i − (410 km/h) j . As a magnitude and heading, this and a average = ∆t 410 2 km/h 580 km/h @ 45° S of W. Even though we is a average = @ 45° S of W, or a average = ∆t ∆t don’t have a numerical value, it is important to note that the average acceleration is not zero, even though the speed of the airplane is constant. The acceleration is caused by the change in direction of the velocity. dr d 2r dv r = A cos bti + Btj. v = = − bA sin bti + Bj. a = 2 = = − b 2 A cos bti + 0 j. The equation dt dt dt for the speed is v =

†4-11.

4

vx2 + v y2 =

r = 90ti + (500 − 15t ) j m. v =

(bA sin bt ) 2 + B 2 . dr = 90i − 15 j m/s. The equation dt

for the speed is v=

2 x

2 y

v +v =

y 90 m/s

2

2

(90 m/s) + (15 m/s) = 91 m/s. The direction

of the velocity is θ = tan −1

vy

⎛ −15 ⎞ = tan −1 ⎜ ⎟ = − 9.5°. vx ⎝ 90 ⎠

This is 9.5° below the x axis.


θ

x 15 m/s

CHAPTER 4-12.

4

Measure time from when the ball is thrown. At that instant, the receiver has traveled (7 m/s)(2.0 s) = 14 m along the x direction relative to the quarterback. The receiver is at an initial displacement r0 = (14 m)i + (20 m) j relative to the quarterback.

y vR

Since he’s running at a constant velocity of vR = 7.0i m/s, his total displacement is rR = (14 + 7t )i + 20 j m. Once it’s thrown, the displacement of the ball relative to the quarterback is rB = vB , x ti + vB , y tj = (vB sin θ )ti + (vB cos θ )tj, where vB = 15 m/s.

θ vB

Mathematically, the ball is caught 1 when rB = rR : sin θ = cos θ = . Since the 2 x components must be equal and the y components must be equal, we get two simultaneous equations: (15sin θ )t = 14 + 7t

x

(15 cos θ )t = 20 20 4 = . Substitute this into the first equation to get 15 cos θ 3cos θ ⎛ 4 ⎞ ⎛ 4 ⎞ 15 sin θ ⎜ ⎟ = 14 + 7 ⎜ ⎟ . Simplifying gives 30 sin θ − 21cos θ = 14. ⎝ 3cos θ ⎠ ⎝ 3cos θ ⎠ This transcendental equation can be solved by trial and error, by graphical means, or by use of tools like Solver in Excel or a Solve Block in MathCAD. The result is θ = 57.6°. The flight time 4 for the ball is t = = 2.5 s. 3cos 57.6° v t dv a= = 3i + 2 j ⇒ dv = (3i + 2 j) dt. Integrate to find v: ∫ dv = ∫ (3i + 2 j)dt , which gives v0 0 dt v − v 0 = 3ti + 2tj. The problem states that both components of velocity are initially zero, so we dr , we can follow the same procedure that was used to get v get v = 3ti + 2tj m/s. Since v = dt r t 3t 2 i + t 2 j. The problem says from a: dr = (3ti + 2tj)dt ⇒ ∫ dr = ∫ (3ti + 2tj)dt ⇒ r − r0 = r0 0 2 3t 2 the particle starts moving at the origin, so both components of r0 are zero. r = i + t 2 j m. 2 162.3 km/h = 45.1 m/s 20 = 0.44 s Time to travel 20m: t = 4501 1 1 The distance fallen vertically: z = gt2 = (9.8)(0.44) 2 = 0.95 m 2 2 Use the second equation to get t =

†4-13.

4-14.


CHAPTER †4-15.

Assume the divers launch themselves horizontally from the edge of the cliff, so voy = 0. The time to fall a distance h = 64 m is t =

4-16.

4

2h = g

2(36 m) = 2.71 s. During this time, the diver must 9.81 m/s 2

travel a horizontal distance of at least x = 6.4 m, so the minimum horizontal velocity required x 6.4 m is v0 x = = = 2.4 m/s. t 2.71 s From Example 6, t = 3.19 s (An extra digit has been kept so the final results of this problem can safely be rounded to the two digits required by the data given in the Example.) (a) vx = 83.3 m/s (83 m/s to two digits) (ax = 0, so the horizontal velocity is constant) v y = − gt = −(9.81 m/s 2 )(3.19 s) = −31.3 m/s. To two digits, vy = −31 m/s. (b) v = vx2 + v y2 = (83.3 m/s) 2 + (31.3 m/s) 2 = 89 m/s. The speed has increased by about 6

†4-17.

4-18.

m/s. The stunt car will fall 2 m while it travels 24 m horizontally. 2h 2(2 m) = = 0.639 s. The The time to fall 2 m is t = g 9.81 m/s 2

v0 2m

24 m car must travel 24 m horizontally in 0.639 s, so its horizontal speed must be 38 m/s. (This is about 135 kph, or 84 mph.) v = v0 + at = (3i + 2j) + (i − 4j)t = (3 + t) i + (2 − 4t) j The maximum y-coordinate is reached when the y-component of the velocity is zero. Thus, 2 − 4t = 0 or t = 0.5 s

1 2 a t = 3t i + 2t j + 0.5t2 i − 2t2 j = (3t + 0.5t 2) i + (2t − 2t2) j. 2 At t = 0.5 s, r = (1.625 i + 0.5 j) m r = ro0 + v0t +

4-19.

4-20.

The maximum distance, xmax = v02 /g (i) v0 =

xmax g =

(ii) v0 =

889 × 9.8 = 93.3 m/s (336 km/h)

xmax = v02 sin 2θ/g with θ = 12°, xmax = 250 m gxmax = sin 2θ vo = 77.7 m/s v02 =

max ht = zmax = 4-21.

441 × 9.8 = 65.7 m/s

⎛ 9.81 × 250 ⎞ 2 2 2 2 ⎜ ⎟ m /s = 6030 m /s ° sin 24 ⎝ ⎠

v02 sin 2 θ 6030 sin 2 12° = m = 13.3 m 2g 2 × 9.81

(a) The horizontal range is given by xtarget =

sin 2θ =

gxtarget 2 0

v

. Thus θ =

2v02 sin θ cos θ . Using sin 2θ = 2 sin θ cos θ gives g

1 −1 ⎛ gxtarget ⎞ 1 −1 ⎡ (9.81 m/s 2 )(12 , 500 m) ⎤ sin ⎜ ⎟ , so θ = sin ⎢ ⎥ = 7.25°. 2 2 2 (700 m/s) 2 ⎣ ⎦ ⎝ v0 ⎠


CHAPTER

4

(b) In this part of the problem, it will be necessary to assume that all the digits given are significant, and that the values of g are specified exactly as given. Suppose the gunner uses g = 9.80 m/s2 in the calculations. Then the elevation angle would be calculated using v 2 sin 2θ * xtarget = 0 , where xtarget is the given distance to the target and θ * is the wrong angle 9.80 m/s 2 (9.80 m/s 2 )xt arget 2 (associated with g = 9.80 m/s ). As in part (a), sin 2θ * = . But the actual range v 02 v02 sin 2θ * ⎛ 9.80 ⎞ . Substituting the expression for sin 2θ * gives xmax = xtarget ⎜ ⎟ 2 9.81 m/s ⎝ 9.81 ⎠ ⎛ 9.80 ⎞ = (12, 500 m) ⎜ ⎟ = 12,487 m. The gunner will miss the target by 13 m. ⎝ 9.81 ⎠

will be xmax =

4-22.

Initial speed, v0 =

xmax g = 1500 × 9.8 = 121.2 m/s

v02 sin 2θ x = max = 375 m 2g 4 2v sin θ 2 × 121.2 = 0 = = 17.5 s g 2 × 9.8

zmax = tflight

4-23.

(a) vty = 0 = viy − gt viy ⎛ 87 × 103 ⎞ sin 45 t= =⎜ = 1.74 s ⎟ g ⎝ 60 × 60 ⎠ 9.8 2 viy2 1 ⎛ 87 sin 45 ⎞ (b) zmax = =⎜ = 14.9 m ⎟ 2g ⎝ 3.6 ⎠ 2 × 9.8 2

4-24.

4-25.

v 2 ⎛ 87 ⎞ 1 = 59.6 m (c) xmax = 0 = ⎜ ⎟ g ⎝ 3.6 ⎠ 9.8 Denote Berlin by x1, g1 and Buenos Aires by x2, g2. v 2 sin 2θ v 2 sin 2θ x1 = 0 x2 = 0 g1 g2 x g xg 9.8128 Then 1 = 2 ≥ x2 = 1 1 = 68.11 m × = 69.22 m 9.7967 x2 g1 g2 ∆x = 0.11 m v0 = zmax = tflight =

xmax s = 106 × 9.8 = 3.13 × 103 m/s vy2

2g

=

(3.13 × 103 )2 sin 2 45 = 2.5 × 105 m 2 × 9.8

2v0 sin 45° 2 × 3.13 × 103 sin 45° = = 452 s 9.8 g


CHAPTER 4-26.

(a) xmax = v02 /g ≥ v0 =

xmax g =

4

50 × 9.8 m/s = 22 m/s

(b) The circumference of the circle is 2π × 1.0 m = 6.28 m. dist 6.28 m = trev = = 0.285 s 22 m/s speed 1 1 = s = 3.5 rev/s Thus, number of rev/s = 0.285 trev 4-27.

v02 sin 2θ (26) 2 sin 70° xmax = = = 64.8 m 9.8 g 2v sin θ 2 (26) sin 35° tflight = 0 = = 3.04 s 9.8 g

280 × 3.04 = 14.2 liters 60 730 × 9.8 m/s = 85 m/s

The amount of water in the air at any given instant is 4-28.

xmax = v02 /g ≥ v0 = tflight =

4-29.

4-30.

†4-31.

xmax g =

2v0 sin θ 2 × 85 × sin 45° = s = 12 s 9.8 g

2v02 sin θ cos θ v 2 sin 2 θ 2v 2 sin θ cos θ v 2 sin 2 θ , ymax = 0 . xmax = ymax ⇒ 0 . This = 0 2g 2g g g sin θ gives 2 cos θ = , or tan θ = 4 ⇒ θ = 76°. 2 2v0 y 2(3.43 m/s) v0 y = 2 gymax = 2(9.81 m/s 2 )(0.60 m) = 3.43 m/s. t flight = = = 0.70 s. (The 9.81 m/s 2 g xmax =

exact angle at which the balls are thrown is not relevant.) dv a= = 2.0i − 4.5 j ⇒ dv = (2.0i − 4.5 j)dt. Integrate to find v following the method used in dt Problem 4-13: v − v 0 =

∫

t

0

(2.0i − 4.5 j)dt = 2.0ti − 4.5tj. The problem states

that v 0 = − 10i + 25 j, so we get v = 2.0ti − 4.5tj + (−10i + 25) j = (2.0t − 10)i + (25 − 4.5t ) j m/s. At t = 3.0 s, the velocity is v = (6.0 − 10)i + (25 − 13.5) j = −4.0i + 11.5 j m/s. The speed is v = vx2 + v y2 = (4.0 m/s) 2 + (11.5 m/s) 2 = 12 m/s.

Since v =

dr , we can follow the same procedure that was used to get v from a: dt

dr = [(2.0t − 10)i + (25 − 4.5t ) j]dt ⇒

∫

r

r0

dr =

∫

t

0

[(2.0t − 10)i + (25 − 4.5t ) j]dt ⇒

r − r0 = (t 2 − 10t )i + (25t − 2.25t 2 ) j m. The problem says the particle starts moving at the origin,

so both components of r0 are zero. r = (t 2 − 10t )i + (25t − 2.25t 2 ) j m. At t = 3.0 s, r = −21i + 55 j m.


CHAPTER

4-32.

t flight

4 gt flight 2v sin θ 2v 2 sin θ cos θ 2 sin θ cos θ = . xmax = 0 = 0 ⇒ v0 = 2 sin θ g g g

2

gt 2flight ⎛ gt flight ⎞ = ⇒ ⎜ ⎟ 2 tan θ ⎝ 2 sin θ ⎠

2 2 ⎛ gt 2flight ⎞ −1 ⎡ (9.81 m/s )(6.0 s) ⎤ θ = tan . ⎟⎟ ⎢ ⎥ = 67°. 2(75 m) ⎣ ⎦ ⎝ 2 xmax ⎠

θ = tan −1 ⎜⎜

2v02 sin θ cos θ ⇒ v0 = g

gxmax = 2 sin θ cos θ

4-33.

xmax =

4-34.

The time to fall a distance h is t =

(9.81 m/s 2 )(15 m) = 21 m/s. 2 sin10° cos10°

2h . The object must also g

v0

travel the same horizontal distance h during this time, h gh so v0 = = . After time t the vertical velocity will 2 t be v y = − gt = − 2 gh , so the final velocity at time of impact is

v= †4-35.

h h

gh i − 2 gh j. 2

The time to reach maximum height is theight =

v0 y 2.25 s = 1.125 s. theight = ⇒ 2 g

v0 y = gtheight = (9.81 m/s 2 )(1.125 s) = 11.0 m/s. If we say the launch point was at y = 0, then the

equation for the position below the launch point is y = v0 y t −

gt 2 with v0 y = 11.0 m/s. This 2

(9.81 m/s 2 )(4.00 s) 2 = − 34.5 m. The cliff is 34.5 m high. Since 2 = v0 y = 11.0 m/s. The total horizontal distance traveled since

gives y = (11.0 m/s)(4.00 s) − the launch angle was 45°, v0 x

launch is x = v0 x ttotal = (11.0 m/s)(2.25 s + 4.00 s) = 68.8 m. 4-36.

Choose the origin of the coordinate system to be at the bottom of gt 2 the cliff. Then y = y0 + v0 y t − . y0 = h = 30 m; v0 = 25 m/s. 2 v0 y = v0 sin 30° = 12.5 m/s. At the bottom, y = 0, so we must solve this quadratic equation: 0 = 30 + 12.5t − 4.91t 2 . The roots are given

4-37.

v0 30° h

−12.5 ± (12.5) 2 + 4(4.91)(30) = − 1.51 s, 4.06 s. The by t = −9.81 final result must be positive, so t = 4.06 s. x 25 m The flight time to the goal is t = = = 0.391 s. The height of the puck at v0 cos θ (65 m/s) cos10° gt 2 (9.81 m/s 2 )(0.391 s) 2 = (65 m/s)(sin10°)(0.391 s) − = 3.66 m. 2 2 The puck will pass 2.2 m above the goal.

this time is y = v0 sin θ t −


CHAPTER

4-38.

(a) xmax =

2v02 sin θ cos θ ⇒ v0 = g

gxmax = 2 sin θ cos θ

4

(9.81 m/s 2 )(240 m) = 70.8 m/s. 2(sin14°)(cos14°)

2(71.4 m/s) 2 sin14° cos14° = 244 m. The ball will overshoot by about 4 m. 9.81 m/s 2 2(70.8 m/s) 2 sin14.5° cos14.5° (c) xmax = = 248 m. The ball will overshoot by about 8 m. 9.81 m/s 2 2v 2 sin θ cos θ v 2 sin 2 θ 1 xmax = 0 , ymax = 0 . For θ = 45°, sin θ = cos θ = . Substituting in the 2g g 2 (b) xmax =

4-39.

equations for xmax and ymax gives xmax =

4-40.

which is the desired result. v 2 sin 2θ zmax = 0 = 8.667 ft 2g xmax =

2v02 v02 v2 y 1 = , ymax = 0 . Taking the ratio gives max = , 2g 4g g xmax 4

(i)

v02 sin 2θ = 37 ft g

(ii)

sin 2 θ = 0.2342 sin 2θ sin 2 θ sin 2 θ 1 = = tan θ = 0.2342 sin 2θ 4 sin θ cos θ 4 θ = tan−1 (4 × 0.2342 ) = 43.1°

Dividing (i) by (ii) gives

From (ii) v02 = 37 × 32.2/sin 86.2° = 1194 ft2/s2 v0 = 34.6 ft/s †4-41.

4-42.

†4-43.

xmax =

2v02 sin θ cos θ v 2 sin 2 θ , ymax = 0 . Setting these equal to each other gives 2g g

2v02 sin θ cos θ v 2 sin 2 θ sin θ , from which we get = 0 = tan θ = 2, or θ = tan −1 2 = 63.4°. 2g g cos θ 2v sin θ 2 × 38 sin 52° tflight = 0 = = 6.1 s g 9.8 In order to catch the ball, the fielder must run with a minimum speed of 45/6.1 = 7.4 m/s. Therefore with 8 m/s the fielder will be able to catch the ball. A rough sketch (not to scale) is shown. The coordinates of y the projectile are gt 2 v0 y = h + v0 (sin θ )t − 2 θ x = v0 (cos θ )t

The coordinates of the target are (xmax,0). The flight time xmax then is t = . Set y = 0 and substitute this time. v0 cos θ


h xmax

x

CHAPTER

4

⎛ x2 ⎞ ⎛ x ⎞ After some rearranging, the equation becomes g ⎜ 2 max2 ⎟ − 2v0 sin θ ⎜ max ⎟ − 2h = 0. We ⎝ v0 cos θ ⎠ ⎝ v0 cos θ ⎠ can convert this to a quadratic equation with tan θ as the variable by using the following sin θ sin 2 θ 1 − cos 2 θ 1 . tan 2 θ = = = − 1. Thus trigonometric steps: tan θ = 2 2 cos θ cos θ cos θ cos 2 θ 2 ⎛ gxmax ⎞ 1 2 2 1 tan , = + θ and we have ⎜ 2 ⎟ (1 + tan θ ) − 2 xmax tan θ − 2h = 0. Multiplying by 2 cos θ v ⎝ 0 ⎠ ⎛ v02 ⎞ ⎜ 2 ⎟ and rearranging finally gives this equation: ⎝ gxmax ⎠ ⎛ v2 ⎞ ⎛ v2 ⎞ tan 2 θ − 2 xmax ⎜ 02 ⎟ tan θ + 1 − 2h ⎜ 02 ⎟ = 0. Using xmax = 12 × 103 m, v0 = 600 m/s, and g = ⎝ gxmax ⎠ ⎝ gxmax ⎠ ⎛ v2 ⎞ 9.81 m/s2 gives ⎜ 02 ⎟ = 2.548 × 10-4 m-1. Substituting h = 50 m and doing the rest of the ⎝ gxmax ⎠ arithmetic finally gives tan 2 θ − 6.116 tan θ + 0.9745 = 0. The solution is

4-44.

6.116 ± 6.1162 − 4(0.9745) tan θ = = 5.953, 0.1635. The two angles are the inverse tangents 2 of these: θ = 80.5°, 9.29°. It turns out that most projectile aiming problems like this have two solutions. The difference between the two angles is that the larger elevation will give a longer flight time and the projectile will strike the target at a larger angle than for the smaller elevation. gt 2 (a) y = − , x = v0 t. The distance d down the slope is related v0 2 to x and y by the slope angle of 45°: x = d cos 45°, y = d sin 45°. Substitute these into the equations for x and y. Solve the x d cos 45° equation for t: t = , Substitute into the y equation: v0 2

d sin 45° = d =

g ⎛ d cos 45° ⎞ ⎜ ⎟ . Solve for d: 2⎝ v0 ⎠

2v02 sin 45° 2 sin 45° . = 2, so d = g cos 2 45° cos 2 45°

45˚ x

2v02 . v0 = 110 km/h = 30.6 m/s g

2(30.6 m/s) 2 = 135 m. 9.81 m/s 2 (b) Skiers are able to land farther down the slope by arching their bodies forward over the skis. This moves their center of gravity forward, reduces aerodynamic drag, and actually provides some aerodynamic lift which carries them farther down the slope than would be predicted by projectile motion analysis. x = 90 = (70 cos θ) t (i) 1 2 y = 0 = (70 sin θ) t − gt (ii) 2 v 2 sin 2θ Substitute for t from (ii) in (i) to get x = 0 g

⇒d =

4-45.

d

y


CHAPTER

4

⎛xg⎞ 1 ⎛ 90 × 9.8 ⎞ 1 sin−1 ⎜ 2 ⎟ = sin−1 ⎜ ⎟ = 5.18° 2 2 ⎝ (70) ⎠ ⎝ v0 ⎠ 2 when the arrow is misaimed by 0.03° in the vertical direction, 2 2 1 ⎛ 90 ⎞ 1 1 1 ⎛9⎞ y = x tan θ − g ⎜ 70 ⎟ 90 tan (5.21 ) 9.8 = ° − ⎜ ⎟ 2 2 2 ⎝ ⎠ cos θ 2 ⎝ 7 ⎠ cos 5.21° −2 = −3.9 × 10 m = −3.9 cm (Arrow hits bull’s-eye) θ=

A misaim of 0.03° in the horizontal direction will give the arrow a component of velocity in the z-direction v0z = v0 cos 5.18° sin 0.03 The deviation in the horizontal direction from the center of the bull’s-eye is vozt = (70 cos 5.18° sin 0.03°) = 4.7 × 10−2 m = 4.7 cm 4-46.

90 70 cos 5.18°

Therefore the arrow will hit the bull’s-eye. v 2 sin 2θ xmax g ⇒ sin 2θ = (a) xmax = 0 g v02 sin 2 θ =

(700 m × 9.8 m/s 2 ) / (630 m/s) 2 = 0.01728

1 sin−1 (0.01728) = 0.50° 2 If d is the distance above the target, then tan θ = d/700 or d = 700 m tan 0.5° d =6m θ=

v02 sin 2θ 6302 sin 2 (0.5) = 1.5 m = 2 × 9.8 2g 2v sin θ 2 × 630 × sin (0.5) (c) tflight = 0 = 1.1 s = 9.8 g

(b) zmax =

4-47.

xmax = v02 sin 2θ/g. Let the target distance be x x − 180 = v02 sin 14.667°/g

(i)

2 0

x + 120 = v sin 15.167°/g

(ii)

Dividing (i) by (ii) gives x − 180 sin 14.667° = ⇒ (x − 180) sin 15.167° = (x + 120) sin 14.667° x + 120 sin 15.167° 77.48 = 9180 m x= 0.00844


CHAPTER

4

From (i) 9180 m = 180 m = v02 sin 14.667°/g v02 = 3.483 × 105 m2/s2 xg 9180 × 9.8 = = 0.23 v02 3.48 × 105 θ = 1/2 sin−1 0.23 = 7.49° = 7°29 ' sin 2θ =

4-48.

xmax =

2v02 sin θ cos θ v 2 sin 2 θ , ymax = 0 . Dividing the second equation by the first gives g 2g

⎛ 4y ⎞ ymax sin 2 θ tan θ ⎡ 4(2.5) ⎤ , from which we get θ = tan −1 ⎜ max ⎟ = tan −1 ⎢ = = ⎥ = 63.4°. 4 sin θ cos θ 4 xmax ⎣ 5 ⎦ ⎝ xmax ⎠ 2(9.81 m/s 2 )(2.5 m) = 7.8 m/s. sin θ sin 63.4° Consider the motion of two of the projectiles. Call their positions (x1, y1) and (x2, y2) and assume that they were launched with elevation angles θ1 and θ2, respectively. The positions are given by gt 2 gt 2 x1 = v0 cos θ1t , y1 = v0 sin θ1t − and x2 = v0 cos θ 2t , y2 = v0 sin θ 2t − . For the projectiles 2 2 to collide, they must have the same positions at the same time: x1 = x2 and y1 = y2. This gives the gt 2 gt 2 = v0 sin θ 2t − . Canceling following pair of equations: v0 cos θ1t = v0 cos θ 2t , v0 sin θ1t − 2 2 terms gives cos θ1 = cos θ 2 , sin θ1 = sin θ 2 . Dividing the second equation by the first gives tan θ1 = tan θ 2 as the requirement for the projectiles to collide. Since we assumed that the v0 =

†4-49.

4-50.

†4-51.

2 gymax

=

projectiles were launched at different angles, we arrive at a contradiction, and we conclude that the individual projectiles will never collide during their flight. Neglecting air resistance, the net force on the projectile is always − mgj (taking the +y direction to point up). If the motion is instantaneously uniform circular motion at the top of the trajectory, then the magnitude of the centripetal force must be mg, and the centripetal acceleration must be g. v2 v2 Thus 0 x = g , which gives r = 0 x for the radius of the osculating circle. r g Take the x and z directions to point as shown. (The y direction is up, out of the page.) The projectile has two horizontal components of velocity: v horizontal = v0 cos θ i + vS k , where v0 is the muzzle speed of

x

the projectile, θ is the elevation angle of the projectile, and vS is the speed of the ship. Neglecting air resistance, the flight time for the 2v sin θ 2(720 m/s)(sin 30°) = = 73.5 s. The projectile is t flight = 0 g 9.81 m/s 2 speed of the ship is 45 km/h = 12.5 m/s, so the net horizontal displacement of the projectile is r = v horizontal t flight

z

= [(720 m/s)(cos 30°)i + (12.5 m/s)k ](73.5 s) = (45.8 × 103 m)i + (918 m)k. The total horizontal range is r =

rx2 + rz2 = 45.8 km. The change in the total range due to the ship’s motion is

negligible to three significant digits; however, failure to compensate for the ship’s forward speed would cause the projectile to miss its target by nearly 1 km.


CHAPTER

4-52.

x = v0t cos θ (i) z = v0t sin θ − t=

x [from (i)] (iii). v0 cos θ

1 g t2 2

(ii)

Substituting (iii) into (ii) gives z = x tan θ −

1 gx 2 2 2 v0 cos 2 θ

1 (9.8)(50) 2 sec2 θ. 2 252 13 = 50 tan θ − 19.6(1 + tan2 θ) 19.6 tan2 θ − 50 tan θ + 32.6 = 0 This equation has imaginary roots, so stone cannot hit the window. z = 50 tan θ − 19.6(1 + tan2 θ) 2 Let y = tan θ, z = −19.6y + 50y = 19.6 dz ⎛ dz ⎞ ⎛ dy ⎞ dz =⎜ ⎟ ⎜ = − 39.2 y + 50 ⎟ and dθ ⎝ dy ⎠ ⎝ dθ ⎠ dy 13 = 50 tan θ −

At maximum, dz/dθ = 0, so that dz/dy = 0 = −39.2y + 50 y = 50/39.2 = 1.28 = tan θ z = 50(1.28) − 19.6(1 + 1.282) = 12.3 m †4-53.

From the diagram v0x = v0 cos θ (i) v0y = v0 sin θ In the xy-frame (ii) x = v0x t 1 2 y = v0y t − gt (iii) 2 Also y = l sin α x = l cos α (iv) l When the projectile hits the incline 1 2 gt (v) y = l sin α = v0yt − 2 from (ii) t = x/v0x = l cos α/v0x (vi) Substitute t from (vi) in equation (v) to get the range as measured along the incline. 1 ⎧ l cos α 1 l 2 cos 2 α ⎫ l= − g 2 ⎨ v0 sin θ i ⎬ sin α ⎩ v0 cos θ 2 v0 cos 2 θ ⎭ Solving for l, gives 2v02 l= cos θ sin θ − tan α cos 2θ } { g cos α l=

2v02 cos 2 θ ( tan θ − tan α ) g cos α

l will be maximum when

dl =0 dα

2v02 dl = ( cos 2θ + (tan α) 2 cosθ sinθ ) = 0 dθ g cos α


4

CHAPTER

4-54.

4

or cos 2θ = − tan α sin 2θ, cot 2θ = − tan α = cot (α ± π/2) 1 Thus θ = (α ± π / 2) 2 This problem is almost identical to 4-12. Follow the procedure given there. Using the slightly different numerical data for this problem leads to these equations for the position of the receiver and the position of the ball relative to the quarterback after the ball is thrown: rR = (16 + 8t )i + 15 j, rB = (20 sin θ )ti + (20 cos θ )tj. The condition rB = rR gives this pair of equations: (20 sin θ )t = 16 + 8t ; (20 cos θ )t = 15. Use the second equation to 15 3 . Substitute this into the first equation to get = get t = 20 cos θ 4 cos θ ⎛ 4 ⎞ ⎛ 4 ⎞ 20 sin θ ⎜ ⎟ = 16 + 8 ⎜ ⎟ . Simplifying gives 15 sin θ − 16 cos θ = 6. Solving the ⎝ 3cos θ ⎠ ⎝ 3cos θ ⎠ transcendental equation gives 62.7°, or θ = 63° to two digits. The flight time for the ball is 3 t = = 1.64 s. 4 cos 62.7° Call the elevation angle α. To find this angle, use the fact that we know both the range and the flight time. In terms of the initial speed of the ball and the angle α, we have 2v 2 sin α cos α 2v sin α r = 0 ; t = 0 . Solve the second equation for v0 and substitute that g g 2

gt 2 cos α gt 2 ⎛ gt ⎞ ⎛ sin α cos α ⎞ . The = = expression into the first equation: r = 2 ⎜ ⎟ ⎟ ⎜ g 2 sin α 2 tan α ⎝ 2 sin α ⎠ ⎝ ⎠ gt 2 range can also be written in terms of vB and t: r = vB t , so we have vB t = , from which we 2 tan α 2 ⎛ gt ⎞ −1 ⎡ (9.81 m/s )(1.64 s) ⎤ tan = get α = tan −1 ⎜ ⎟ ⎢ ⎥ = 21.9°, or α = 22°. 2(20 m/s) ⎣ ⎦ ⎝ 2v B ⎠ †4-55.

The total instantaneous velocity of a point on the rim of the tire is the vector sum of the forward velocity of the tractor and the tangential velocity of the point, v tot = v tan + u. This

u u cosθ θ

total instantaneous velocity must be zero when the point is R sinθ θ on the road (θ = 270°), or the tire will be slipping. Thus the tangential speed of a point on the rim must be equal to the R forward speed u of the tractor. In the reference frame of the tractor, the horizontal component of velocity of a glob is always zero and the vertical component of the velocity is v y = u cos θ , which becomes the y component of the initial velocity v0y of a launched glob. The maximum height above the launch point is v02y u 2 cos 2 θ . To find the maximum height above the ground, we must add the height = ymax = 2g 2g of the launch point above the ground, which is R + R sin θ as shown in the diagram. So the u 2 cos 2 θ height above the ground is h = R(1 + sin θ ) + . 2g


CHAPTER

4

The maximum height depends on θ. To find the maximum possible height reached by a glob, differentiate h with respect to θ and set the derivative equal to zero: dh u 2 cos θ sin θ gR = R cos θ − = 0, from which we get sin θ max = 2 . For R = 0.80 m and u = 30 dθ g u km/h = 8.33 m/s, this gives θmax = 6.49°, from which we get (8.33 m/s) 2 (cos 6.49°) 2 = 4.4 m. h = (0.80 m)(1 + sin 6.49°) + 2(9.81 m/s 2 ) 4-56.

40 km/h = 11.11 m/s (a) The equation for the projectile is 1 ⎡ ⎤ (700 cos θ)t i + ⎢ (700 sin θ ) t − gt 2 ⎥ j = rproj 2 ⎣ ⎦ The equation for the ship is (1500 − 11.11t) i = rship When the projectile hits the ship, the coefficient of j = 0, so 1 (700 sin θ) t − gt 2 = 0 2 Also, the coefficients of i must be equal, so (700 cos θ) t = 15,000 − 11.11t 15, 000 t= 700 cos θ + 11.11 Substituting this into (i) gives 1 1 15, 000 700 sin θ = gt = 2 g 2 (700 cos θ + 11.11) sin θ (700 cos θ + 11.11) = 105 An iterated solution gives θ ≈ 8.6° (b) The time interval is t =

4-57.

15,000 = 21.3s (700 cos 8.6° + 11.11)

If the ship does not move, the time of flight is given by tflight = 2v0 sin θ/g θ is given by xmax = 17,000 m = v02 sin 2θ/g = 7002 sin 2θ/9.81 9.81 × 17, 000 = 0.34 sin 2θ = 7002 1 θ = sin−1 (0.34) = 9.94° 2 ⎛ 2 × 700 m × sin 9.94° ⎞ tflight = ⎜ ⎟ s = 24.67 s 9.81 ⎝ ⎠ In this time the ship would have moved 8.33 m/s × 24.67 s = 205.5 m. Thus the new xmax (by the Pythagorean theorem) is 17, 0002 + 205.42 = 17,001 m 1 ⎛ 17001 × 9.8 ⎞ sin−1 ⎜ ⎟ = 9.94° 2 7002 ⎝ ⎠ so the time of the flight is still the same, so the projectile will arrive there when the ship does. θ = 9.94° = 9°56 '

xmax = 17001 m = v02 sin 2θ/g ≥ θ =


(i)

CHAPTER 4-58.

4

Let ω stand for the angular speed of the disk. Then v = ω R = (32.5 rad/s)(4.0 cm) = 130 cm/s. The rotation rate in rev/min is ω = (32.5 rad/s)(1 rev/2π radians)(60 s/min) = 310 rev/min.

†4-59.

4-60.

v2 ⇒ v = gR = (9.81 m/s 2 )(200 m) = 44.3 m/s. Use ω as defined in the solution to R 4-58 to find the number of revolutions per minute required to give this v 44.3 m/s 60 s 1 rev × × = = 2.1 rev/min. acceleration: ω = R 200 m min 2π radians v 2v , where D = diameter of hole. For D = 3 mm, v =3.0 m/s ⇒ ω = = R D 2(3.0 m/s) 60 s 1 rev × × ω = = 1.9 × 104 rev/min. For D = 25 mm, −3 3 × 10 m min 2π rad a= g =

ω = †4-61.

4-62. †4-63.

4-64.

2(3.0 m/s) 60 s 1 rev × × = 2.3 × 103 rev/min. 25 × 10−3 m min 2π rad

2π R v2 4π 2 R . For an orbit with a ⇒a= = T R T2 4π 2 (6500 × 103 m) radius of 6500 km and T = 87 min, this gives a = = 9.4 m/s 2 . 2 2 (87 min) (60 s / min)

Let T stand for the time for one complete orbit. v =

v = ω R = (45 rad/s)(0.80 m) = 36 m/s. v2 . Let f stand for the number of rotations per second. Then v = 2πrf = 2π × 0.1 m × 1000/s r v2 (200π m/s) 2 = = 3.95 × 106 m/s 2 . This is = (3.95 × 106 m/s2)/(9.81 m/s2) g = 200π m/s. a = (0.10 m) r a=

= 4.0 × 105 g. 1 rev. of blade = π × 0.2 = 0.63 m.

7000 = 73.3 m/s 60 2 /r = (73.3)2/0.1 = 5.4 × 104 m/s 2 The centripetal acceleration is ac = vtip

Turning (7000/60) rev/sec, the speed is vtip = 0.63 ×

4-65.

v = 0.9999995 c = 0.9999995 × 2.998 × 108 m/s = 2.998 × 108 m/s r = 1.0 km = 1000 m a = v2/r = (2.998 × 108 m/s)2/1000 m = 8.99 × 1013 m/s 2 = 9.16 × 1012 gee

4-66.

v = ω R = (33.33 rev/min)(1 min/ 60 s)(2π rad/rev)(15 cm) = 52 cm/s. r = 1.50 × 1011 m 2π r 2π × 1.50 × 1011 m v= = = 2.99 × 104 m/s 365.25 × 24 × 3600 T

4-67.

ac = 4-68.

(2.99 × 104 ) 2 m/s2 = 5.9 × 10−3 m/s 2 1.50 × 1011

v = 95 km/h = 26.4 m/s. This is also the tangential speed of a point on the rim of the wheel, so v2 (26.4 m/s) 2 = = 2.2 × 103 m/s 2 . a= (0.32 m) r


CHAPTER

4-69.

v= aeq

4

2π R v2 4π 2 R . T = 23h 56 min = 86160 s. At the equator, ⇒a= = T R T2 4π 2 R 4π 2 (6.37 × 106 m) = = = 3.39 × 10−2 m/s 2 . At a latitude of 45°, the radius is the T2 (86160 s) 2

equatorial radius multiplied by cos 45°, so a45 = aeq cos 45° = 2.40 × 10−2 m/s 2 .

†4-71.

2 vtop

(97.2 m/s) 2 = 18.9 m/s2. At the bottom, 500 m r v2 (172 m/s) 2 abottom = bottom . 620 km/h = 172 m/s. abottom = = 59.3 m/s2. 500 m r At the top, consider the free body diagram for the pilot, who is upside down so the normal force exerted by the seat points down. Taking the + direction to point down, Ntop N top = acceleration the pilot feels = mg Newton’s Second Law says N top + mg = matop . m atop − g = (18.9 − 9.81) m/s2 = 9.09 m/s2. At the bottom, the pilot is upright so the normal force points up. Taking the + direction to point up, Newton’s Second Law gives N N bottom − mg = matop . Now bottom = g + abottom = (9.81 + 59.3) m/s2 = 69.1 m/s2. m 4π 2 R . In the table shown, the acceleration was calculated using a = T2 1/R2 (m−2) Planet R (m) T (yr) a (m/s2) Mercury 5.79E+10 0.241 0.0395 2.98E-22 Venus 1.08E+11 0.615 0.0113 8.57E-23 Earth 1.50E+11 1.000 0.00595 4.44E-23 One way to see if the centripetal acceleration is inversely proportional to R2 is to plot a graph of a vs 1/R2. The resulting plot is a straight line, proving the proportionality. (In the graph, the values of 1/R2 have been multiplied by 1022 to eliminate large negative exponents for the x axis values.) At the top, atop =

. 350 km/h = 97.2 m/s. atop =

a vs 1/R

a (m/s2)

4-70.

2

0.045 0.04 0.035 0.03 0.025 0.02 0.015 0.01 0.005 0 0.0

0.5

1.0

1.5 2

2.0 22

1/R x 10

2.5

3.0

3.5

-2

(m )

Another approach to the analysis is to calculate the log of a and R and plot a graph of log a vs log R. If a is inversely proportional to 1/R2, then this graph should be a straight line with a slope of −2.


CHAPTER

4

log a −1.4034 −1.94692 −2.22548

log R 10.76268 11.03342 11.17609

Log a vs Log R 0

Log a

-0.5 log a = -1.991(logR) + 20.024

-1 -1.5 -2 -2.5 10.7

10.8

10.9

11

11.1

11.2

Log R

4-72.

The graph was plotted in a spreadsheet, which was used to find the equation for the resulting line. The slope turns out to be almost exactly −2, again proving the desired proportionality. 30 km/h = 8.33 m/s v = v′ + V = −10 j + 8.33 i

v = 102 + 8.332 = 13.0 m/s ⎛ 8.33 ⎞ The angle θ is θ = tan−1 ⎜ ⎟ = 40° ⎝ 10 ⎠

4-73.

v ground = v walk + v passenger . If the velocity of the walkway and the velocity of the passenger are in the same direction, then vground = vwalk + v passenger = 1.5 m/s + 4.0 m/s = 5.5 m/s. If the passenger is walking in the opposite direction from the walkway’s velocity, the speed of the passenger is vground = vwalk − v passenger = 1.5 m/s − 4.0 m/s = 2.5 m/s.

4-74.

vrain = −10 j m/s; vcar = 25 i m/s v′ = vrain − vcar = −25 i − 10 j m/s

v′ = 252 + 102 = 26.9 m/s θ = tan−1 25/10 = 68° 4-75.

4-76.

vship = 13 i m/s vmuzzle = v cos θ i + v sin θ j = 660 cos 20° i + 660 sin 20° j = 620 i + 226 j m/s vshot = vship + vmuzzle = (620 + 13) i + 226 j m/s = 633 i + 226 j m/s (magnitude: vshot = 672 m/s, θ = tan−1 226/633 = 19.6°) v = v′ + V By the Pythagorean theorem:

v=

3302 − 302 = 329 m/s


CHAPTER 4-77.

V = v B + vW ⇒ V =

vB2 + vW2 =

(1.5 m/s) 2 + (12 m/s) 2

vW

v 12 V = 12 m/s. θ = tan −1 W = tan −1 = 83°. vB 1.5 4-78.

vB θ

V

The speeds add going downstream, so vdown = 6.0 km/h. In 40 min, you can go d = 6.0 km/h × 40 min × 1 h / 60 min = 4.0 km. Upstream, the speed of the current must be subtracted from the speed of the boat, so vup = 1.0 km/h. The time to make the return trip of 4.0 d 4 km km is t = = = 4.0 h. vup 1 km/h

†4-79.

V = v train + v flea , where vflea is the velocity of the flea relative to the train. The cat is moving backwards relative to the train and the flea is moving backwards relative to the cat. The velocity of the flea relative to the train is 0.50 m/s − 0.10 m/s = 0.40 m/s backwards relative to the train. So the flea’s speed relative to the ground is V = 5.00 m/s − 0.40 m/s = 4.60 m/s.

4-80.

The position of any point on the shore is given by r = di + yj, where y is the displacement (positive or

y

vR

negative) along the y axis of the landing point on the opposite shore. The velocity of the boat relative to the shore is V = vB cos θ i + (vR − vB sin θ ) j. The time to reach any point on the opposite shore is given by r = Vt , or di + yj = (vB cos θ )ti + (vR − vB sin θ )tj. This is equivalent to two equations: d = (vB cos θ )t

x

θ vB d

y = (vR − vB sin θ )t To reach a point directly opposite the launch point, y = 0, which means sin θ =

vR . This vB

describes a right triangle with a hypotenuse vB and an opposite side vR, so we know the adjacent side is

t =

vB2 − vR2 . That means cos θ =

d = vB cos θ

d 2 B

v − vR2

vB2 − vR2 . Use this in the first equation to get t: vB

.

To find the time to reach any point on the opposite shore, begin with cos θ = we deduce sin θ =

( vB t ) 2 − d 2

vB t

d , from which vB t

. Substitute this into the y equation to get

⎛ (vB t ) 2 − d 2 ⎞ ⎜ ⎟ t = vR t − (vB t ) 2 − d 2 . Rearrange to get (vB t ) 2 − d 2 = vR t − y. y = vR − vB ⎜ ⎟ vB t ⎝ ⎠ Square both sides and rearrange: (vB2 − vR2 )t 2 + 2vR yt − ( d 2 + y 2 ) = 0. The roots of this equation are t =

−2vR y ± (2vR y ) 2 + 4(vB2 − vR2 )(d 2 + y 2 ) 2(vB2 − vR2 )

. The time must be positive, so choose the +


4

CHAPTER

4

sign in front of the square root. Rearrange and simplify to get t =

4-81.

(50 cm/s) 2 + (33.3 cm/s) 2 = 60 cm/s. The

direction of the velocity relative to the floor is θ = tan −1

33.3 = 34°. 50

v0 = initial velocity relative to the car. If the projectile is v0 observed to move straight up and down relative to an observer on the ground, then the horizontal component of the θ projectile’s initial velocity must have the same magnitude as the velocity of the car and point in the opposite direction. Thus v 30 v0 cos θ = vcar ⇒ θ = cos −1 car = cos −1 = 53.1°. The maximum height reached is 50 v0

ymax = 4-83.

v02 y 2g

=

(v0 sin θ ) 2 [(50 m/s) sin 53.1°]2 = = 81 m. 2g 2(9.81 m/s 2 )

v = v′ + V

v=

202 + 152 − 2(20) (15) cos100° = 27 km/h

15/sin θ = 27/sin 100° 15 sin 100° = 0.547 sin θ = 27 θ = 33° 4-84.

(vB2 − vR2 )

. As

a quick check, set y = 0 and we get the same result we had in the first part of the problem. 30 cm 20 cm v esc = i+ j = (20 cm/s)i + (13.3 cm/s) j. The rider is walking up with a velocity 1.5 s 1.5 s 30 cm 20 cm v rider = i+ j = (30 cm/s)i + (20 cm/s) j. The rider’s total velocity relative to 1.0 s 1.0 s the floor is 30 cm 20 cm V = v esc + v rider = i+ j = (20 cm/s + 30 cm/s)i + (13.3 cm/s + 20 cm/s) j = 1.0 s 1.0 s (50 cm/s)i + (33.3 cm/s) j. The speed is v floor =

4-82.

(vB2 − vR2 )d 2 + vB2 y 2 − vR y

v = v′ + V, where v′ = velocity of wind relative to boat V = velocity of boat v = velocity of wind relative to ground

v=

(14) 2 + (32) 2 − 2(14) (32) cos 40° = 23.1 km/h

sin θ sin 40° = 32 23.1 ⎛ 32 ⎞ θ = sin−1 ⎜ sin 40° ⎟ = 117° ⎝ 23.1 ⎠ Wind is coming from 13° W of N


vcar

CHAPTER

4-85.

1500 = 3000 s 0.5 (a) Horizontal distance travelled before reaching the ground, ⎛ 60 × 103 ⎞ 3 x= ⎜ ⎟ (3000) = 50 × 10 m 60 × 60 ⎝ ⎠

Time of descent t =

(b) Relative velocity with respect to still air 60 − 20 = 40 km/h 40 × 103 Therefore x = × 3000 = 33 × 103 m 60 × 60 In the upwind direction, the relative velocity is 60 + 20 = 80 km/h, and 80 × 103 x= × (3000) = 67 × 103 m 60 × 60 4-86.

sin θ/50 = sin 135°/250 ⎛1 ⎞ θ = sin−1 ⎜ sin 135° ⎟ ⎝5 ⎠ −1 θ = sin (0.1414) = 8° α = 180° − (135° + 8°) = 37° Pilot must point the plane (45° − 37°) = 8° W of N v/sin α = 250/sin 135° v = sin 37° × 250/sin 135° km/h = 213 km/h

†4-87.

v=

4.22 + 162 − 2(4.2) (16) cos 70°

= 15.1 km/h x = 4.2 sin 20° = 1.44 km/h y = (16 − 1.44) km/h = 14.56 km/h sin θ = 14.56/v = 14.56/15.1 θ = sin−1 (0.96) = 75° (15° E of N) 4-88.

(a) average speed = ∆rrel/∆t = 1 km/2 min = 1 km/(1/30h) = 30 km/h (b) v = v′ + V with V = 90 km/h; v′ = 30 km/h v = (90 + 30) km/h = 120 km/h

4-89.

Let the origin be fixed on car 1. Then in that frame, car 2 moves at a speed of 2v0.


4

CHAPTER

4

If they are “together” (at t = 0), then after time t, they are separated by r=

h 2 + (2v0t ) 2

Then, dr d 2 = ( h + 4v02t 2 ) 1/2 dt dt 2 1 = ( h 2 + 4v02t 2 ) −1/2 ( 8v0 t ) 2 4v02t vrel = 1/ 2 ( h2 + 4v02t 2 ) vrel =

=

2v0 1/ 2

⎛ h ⎞ ⎜ 2 2 + 1⎟ ⎝ 4v0 t ⎠ [Note: as t → ∞, vrel → 2v0] time (s) vrel (m/s) 2

10 20 30 40 50 60 70 80 †4-90.

9.6 17.1 22.3 25.6 27.7 29.1 30.0 30.8

vg = v + V

(a) vg2 =

v2 − v2

Total distance across and back is 2d. t = dist/vg = 2d / v 2 − V 2 (b) vup = v − V vdown = v + V tup = d/(v −V); tdown = d/(v + V) ttot = d/(v −V) + d/(v + V) = 2dv /(v 2 − V 2 ) The trip up and back takes longer (the denominator is larger)


CHAPTER †4-91.

4

The velocity of the AWACS relative to the ground is vAW = 150 i + 750 j km/h Relative to this, the UFO has velocity vUFO = −950 cos 45° i = 950 sin 45° j km/h = 672 i − 672 j km/h The velocity of the UFO relative to the ground is v g (UFO) = (150 i + 750 j ) + ( −672 i – 672 j ) km/h = 522 i + 78 j km/h

ν g (UFO) = 5222 + 782 = 528 km/h ⎛ 78 ⎞ The bearing is θ = tan−1 ⎜ ⎟ = 8.5° N of W ⎝ 522 ⎠

4-92.

h j, where h is the vertical distance traveled in time t, and t 2340 m j = ( −669 m/h) j. No information is the positive direction is assumed to be up. v y = − 3.5 h given about the horizontal distance traveled, so there is no way to determine the average

The average vertical velocity is v y = −

horizontal component of velocity. Since the speed is 4-93.

vx2 + v y2 , it’s not possible to find the

average speed from the information given. Assume the airplane is gliding at a constant velocity, so its horizontal and vertical components of velocity are constant. (a) Taking directions as shown in the diagram, vx = (240 km/h) cos15° = 232 km/h. The vertical component of

x

15° y

240 km/h

velocity is v y = (240 km/h) sin15° = 62.1 km/h. (b) The time to fall a distance h = 2000 m = 2 km is given by h 2 km t= = = 0.0322 h = 1.9 min. v y 62.1 km/h 4-94.

a=

v f − vi

∆t

=

(−25 m/s)j − (25 m/s)j = 0i − (4.2 m/s 2 ) j. 12 s

vi

vf


CHAPTER 4-95.

4

vE = (4.2 km/h) cos 20° = 3.9 km/h.

N

vN = −(4.2 km/h) sin 20° = − 1.4 km/h. 20°

E

4.2 km/h 4-96.

v = (15 km/h)i + (15 km/h) j ⇒ v =

direction is θ t = tan −1 †4-97.

vy vx

(15 km/h) 2 + (15 km/h) 2 = (15 km/h) 2 = 21 km/h. The

= 45°, or 45° E of N.

(a) r = (6.0 + 2.0t 2 )i + (3.0 − 2.0t + 3.0t 2 ) j. v = v = 8.0i + 10 j m/s, so the speed is v =

(b) a = a =

dr = 4.0ti + (−2.0 + 6.0t ) j. At t = 2.0 s, dt

vx2 + v y2 =

(8.0 m/s) 2 + (10 m/s) 2 = 13 m/s.

dv = 4.0i + 6.0 j m/s 2 at all times. The magnitude of the acceleration is dt

ax2 + a y2 =

θ = tan −1

vy vx

(4.0 m/s 2 ) 2 + (6.0 m/s 2 ) 2 = 7.2 m/s 2 . The direction relative to the x axis is

= tan −1

6.0 = 56.3°. 4.0

2h 2(1.5 m) = = 0.553 s. Final result: t = 0.55 s. g 9.81 m/s 2 (b) x = v0 t = (60 m/s)(0.553 s) = 33 m.

4-98.

(a) t =

4-99.

(a) ymax = ymax =

v02 sin 2 θ . The maximum possible height would occur for θ = 90°, so 2g

(700 m/s) 2 = 2.5 × 104m, or 25 km. 2(9.81 m/s 2 )

(b) xmax =

2v02 sin θ cos θ . The maximum horizontal range would occur for θ = 45°, so g

(700 m/s) 2 = 5.0 × 104 m, or 50 km. 2 9.81 m/s (c) It is not reasonable to ignore air resistance because at such speeds, air friction would be considerable. This is why the rocks do not rise to 25 km or cannot be thrown to distances approaching 50 km. 2v 2 sin θ cos θ gxmax ⇒ v0 = xmax = 0 . For θ = 45°, sin θ = cos θ = 1/ 2, so g 2 sin θ cos θ xmax =

4-100.

v0 =

(9.81 m/s 2 )(180 m) = 42 m/s.


CHAPTER

†4-101.

y = y0 + v0 y t −

4

gt 2 , x = v0 x t. When y = 0, x = xmax. For θ = 45°, v0 x = v0 y = v0 / 2. Then the 2

⎛ v0 ⎞ ⎜ ⎟+ 2⎠ ⎝ time to reach y = 0 is t =

v02 + 2 y0 g v + v02 + 4 y0 g 2 = 0 . (There is another value of t g g 2 that solves the equation for y = 0, but it will be negative because the square root in the quadratic formula will be larger than v0. Choose the positive value.) The horizontal range is 2 2 2 v ⎛ v + v0 + 4 y0 g ⎞ v0 + v0 v0 + 4 y0 g ⎟= xmax = 0 ⎜ 0 . Rearrange: 2 gxmax − v02 = v0 v02 + 4 y0 g . ⎜ ⎟ 2g 2⎝ g 2 ⎠ 2 − 4 gxmax v02 + v04 = v04 + 4 gy0 v02 . Cancel the fourth degree terms and Square both sides: 4 g 2 xmax

the remaining common factor of 4 and solve for v0: v0 =

2 xmax g = xmax + y0

(70.87 m) 2 (9.81 m/s 2 ) = 70.87 m + 2.0 m

26 m/s. 4-102.

For a horizontal initial velocity, the time for the water to fall a distance h is t = horizontal range is xmax = v0t = v0

2h and the g

2h g 9.81 m/s 2 = (4.0 m) = 8.86 . This gives v0 = xmax 2h 2(1.0 m) g

m/s. Assuming that the speed of the water stays the same when the hose is pointed straight up v2 (8.86 m/s) 2 gives ymax = 0 = = 4.0 m. 2 g 2(9.81 m/s 2 ) 4-103. 4-104.

v2 (25 m/s) 2 = = 8.9 m/s 2 . r 70 m Relative to the ground, the concrete has velocity v = 2ax = 2 × 9.8 × 5 m/s = 9.90 m/s 90 km/h = 25 m/s. a =

The velocity of the car is 90 km/h = 25 m/s. If v′ is the velocity of the concrete in the frame of the car, then v′ = v − V = −9.9 j − 25 i m/s v′ = 9.92 + 252 = 27m/s

†4-105.

⎛ 25 ⎞ Angle of impact = θ = tan−1 ⎜ ⎟ = 68° ⎝ 9.9 ⎠ (a) V = vcar − vtruck = 90 km/h − 60 km/h = 30 km/h = 8.33 m/s.

(b) In the reference frame of the truck, the car has to travel a total distance of 90 m: It must travel 40 m to catch up to the truck, 10 m to pass the truck, and then travel another 40 m to get ahead of 90 m = 10.8 s. the truck. The time for the car to travel 90 m at 8.33 m/s is t = 8.33 m/s


CHAPTER 5

NEWTON’S LAWS OF MOTION

Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions Manual, available for purchase. Answers to all solutions below are underscored. 1kg = 442 kg. 2.205 lb

5-1.

m = 975 lb ×

5-2.

matom = m p + me = 1.673 × 10−27 kg + 9.11 × 10−31 kg = 1.67 × 10−27 kg. To three significant digits the mass of the electron is negligible compared to the mass of the proton. 1.0 kg = 6.0 × 1026 atoms. N = 1.67 × 10−27 kg/atom

5-3.

matom = 8(m p + mn + me ) = 8(1.673 × 10−27 kg + 1.675 × 10−27 kg + 9.11 × 10−31 kg) = 2.68 × 10−26 kg. N =

5-4.

1.0 kg = 3.7 × 1025 atoms. 2.68 × 10−26 kg/atom

The forces must have equal magnitudes. Thus mboy aboy = mgirl agirl . mgirl =

†5-5.

5-6.

a =

mboy aboy a girl

=

50 kg × 7 m/s 2 = 43 kg. 8.2 m/s 2

v2 − v1 ( 80 km/h − 0 km/h ) 1000 m 1h × × = = 3.83 m/s 2 . Final answer ∆t 5.8 s km 3600 s

= 3.8 m/s 2 . To find the magnitude of the average force, use the three-digit intermediate result for m a : F = ma = 1620 kg × 3.83 2 = 6205 N. Final answer = 6.2 × 103 N. s a = F/m = 2.7 × 105 N/1.6 × 104 kg = 17 m / s 2

5-7.

∆v = 50 km/h − 0 km/h = 50 km/hr = 13.9 m/s F = m a = m∆v/∆t = 57 kg(13.9 m/s)/0.12 s = 6.6 × 103 N, which is 12 times the weight.

5-8.

m = 16 000 metric tons × 103 kg/metric ton = 1.6 × 107 kg. F 6.7 × 105 N a = = . a = 0.0419 m/s 2 . Final result = 0.042 m/s2. 7 m 1.6 × 10 kg v − v0 50 km/h − 0 m 1h = = 332 s. Final result: t = 5.5 min. × 1000 × v = v0 + at ⇒ t = 2 0.0419 m/s km 3600 s a

†5-9.

Vector note: “Decelerates” is a nontechnical way of stating that the acceleration points in the opposite direction from the velocity, causing the speed to decrease. If we take the positive direction for vectors to be in the direction of motion, then a must be represented by a negative number, and the net force will also be a negative number. A “free-body” diagram is shown to illustrate these concepts. Thus Fnet = ma = 1500 kg × (−8.0 m/s 2 ) = −1.2 × 103 N. Again, the − sign means the force points in the opposite direction from the original motion.


Fnet Direction of motion

CHAPTER

5-10.

5

0 − (13.9 m/s ) v 2 − v02 v = v + 2ax. v0 = 50 km/h = 13.9 m/s , v = 0 ⇒ a = = 2x 2(0.40 m) 2

2

2 0

= − 241 m/s 2 . (Vector note: The − sign means the acceleration is in the opposite direction from the motion.) The magnitude of the net force is F = m a = (1400 kg)(241 m/s 2 ) = 3.4 × 105 N. 5-11.

v0 = 174 km/h = 48.3 m/s. Since v2 − v 02 = 2a(x − x0), −v02 −(48.3 m / s) 2 = −1.8 × 103 m / s 2 a= = 2 × 0.66 m 2( x − x0 ) F = ma = −75 kg × 1.8 × 103 m/s2 = −1.3 × 105 N

5-12. 5-13. 5-14.

†5-15.

We calculated that it had an acceleration = 3.26 × 104 m/s2 F = ma = 45 kg × 3.26 × 104 m/s2 = 1.46 × 106 N. ∆v 30 m/s − 0 m/s = = 500 m/s 2 . F = ma ⇒ F = 0.07 kg × 500 m/s 2 = 35 N ∆t 0.060 s ∆v 22 m/s − 0 m/s a = = = 110 m/s 2 . F = ma = 0.035 kg × 110 m/s 2 = 3.9 N. Vector note: ∆t 0.20 s a =

Becasuse both a and F are positive, they point in the same direction, as required by Newton’s Second Law. ∆v 10.0 m/s − 15.0 m/s m = − 4.17 m/s 2 . Final result = −4.2 2 . Second a = . First interval: a1 = ∆t 1.2 s s 5.0 m/s − 10.0 m/s interval: a2 = = − 2.38 m/s 2 . Final result = −2.4m/s 2 . During the first 2.1 s interval, F1 = ma1 = 240 kg × (−4.17 m/s 2 ) = −1.0 × 103 N. During the second interval,

F2 = ma2 = 240 kg × (−2.38 m/s 2 ) = −5.7 × 102 N. Vector note: The − signs imply that the 5-16.

accelerations and forces point in the opposite direction from the motion. F 55 N a = = = 2037 m/s 2 . v final = vinitial + a ∆t = 0 + 2037 m/s 2 × 0.13 s = 264.8m/s. m 0.057 kg Final answer = 2.6 × 102 m / s.

5-17.

The forces must have equal magnitudes, so Fastronaut = Fsatellite .

mastronaut aastronaut = msatellite asatellite . asatellite =

mastronaut aastronaut msatellite

=

95 kg × 0.50 m/s 2 = 750 kg

2

0.063 m/s . 5-18.

t(s) 0 10 20 30 40

a(ms−2) 0.74 0.44 0.44 0.31 0.22

F = ma(N) 860 510 510 360 260


Velocity (km/hr) 130 105 85 75 50

CHAPTER

†5-19.

For the first 0.30-s interval, a =

5

∆v 638 m / s − 657 m/s = = −63.3 m/s 2 . The mass is ∆t 0.30 s

1 kg = 45.4 kg. F = ma = 45.4 kg × (−63.3 m/s 2 ) = −2.9 × 103 N . 2.205 lb 502 m/s − 514 m/s For the last 0.30 s interval, a = = −40.0 m/s 2 . 0.30 s

m = 100 lb ×

F = ma = 45.4 kg × (−40.0 m/s 2 ) = −1.8 × 103 N . Vector note: The − signs imply that the 5-20.

accelerations and forces point in the opposite direction from the motion. dr d ⎡⎣ (5.0 × 104 t )i + (2.0 × 104 t − 2.0 × 105 t 2 ) j − (4.0 × 105 t 2 )k ⎤⎦ = v= dt dt dv = ( 5.0 × 104 ) i + ( 2.0 × 104 − 4.0 × 105 t ) j − ( 8.0 × 105 t ) k m/s. a = dt d ⎡ ( 5.0 × 104 ) i + ( 2.0 × 104 − 4.0 × 105 t ) j − ( 8.0 × 105 t ) k ⎤ = 0i − ( 4.0 × 105 ) j − ( 8.0 × 105 ) k . = ⎦ dt ⎣ a = − ( 4.0 × 105 ) j − ( 8.0 × 105 ) k m/s 2 . F = ma = (1.7 × 10−27 kg) ⎡⎣ − ( 4.0 × 105 ) j − ( 8.0 × 105 ) k m/s 2 ⎤⎦ = − (6.8 × 10−22 ) j + (1.36 × 10−21 )k N. Final result = − (6.8 × 10−22 ) j + (1.4 × 10−21 )k N. F =

†5-21.

5-22. 5-23.

(6.8 × 10−22 ) 2 + (1.36 × 10−21 ) 2 N = 1.5 × 10−21 N.

dx d = x0 ⎡1 − cos ( bt ) ⎦⎤ = 0 + bx0 sin ( bt ) . v = bx0 sin ( bt ) . dt dt ⎣ dv d a= = bx0 sin ( bt ) = b 2 x0 cos ( bt ). F = ma = mb 2 x0 cos ( bt ). dt dt

v=

x = x0 ⎡⎣1 − cos ( bt ) ⎦⎤ ⇒ x0 cos ( bt ) = − ( x − x 0 ) . F = − mb 2 ( x − x0 ). This type of motion, in which the magnitude of the force is proportional to the distance from an equilibrium point and always points in the opposite direction from the displacement, turns out to be very important in science. F Fnet = ma = 1200 kg × 7.8 m/s 2 = 9.36 × 103 N. Fwheel = net = 2.3 × 103 N. 4 2300 children means 1150 children on each side. Therefore, the force exerted by each side is 1150 × 130 = 150, 000 N tension in rope. No, it was not safe.


CHAPTER 5-24.

5

The component of each pull along the direction of motion is 460 cos 30° N = 398 N . The components perpendicular to the direction of motion cancel. The net force on the barge is F 596 N 2 × 298 N = 596 N. The acceleration of the barge is a = net = = 0.331 m/s 2 . 1800 kg m

†5-25.

Find the east and south components of P that will make the net force zero: Fnet , E = ∑ F E = PE − (240 N) cos 30° = 0 ⇒ PE = 208 N.

Fnet , N = P =

∑F

N

= 270 N − PS − (240 N) sin 30° = 0 ⇒ PS = 150 N .

PE2 + PS2 =

θ = tan −1

(208 N) 2 + (150 N) 2 = 256 N.

PS 208 N = tan −1 = 54.2° (54.2o S of E). 150 N PE

N

N 270 N E 30

o

θ

240 N 5-26.

PE

θ

E PS

P

P

The net north (y) component of the total force is Fnet , N = (25 N) cos 30° − (35 N) cos 30°. = − 8.66 N. (The − sign means the net north-south force component points south.) The net east (x) component of the force is Fnet , E = (25 N) sin 30° + (35 N) sin 30° = 30 N. The magnitude of the net force is Fnet =

Fnet2 , E + Fnet2 , N =

(30 N) 2 + (−8.66 N) 2 = 31 N . The direction is given

by

θ = tan −1

FS −8.66 N = tan −1 = −16°. The direction of the net force is 16o S of E. FE 30 N

N

N

30o

25 N 30 N

E 30o

θ

35 N


8.66 N

E

CHAPTER †5-27.

5

The net force in the east direction is Fnet,E = (2500 N) sin15° + (3200 N) sin 30° = 2247 N . The net force in the north direction is Fnet,N = (2500 N) cos15° + (3200 N) cos 30° = 5186 N. To find the east and north components of acceleration, divide each force component by the mass: F F 2247 N 5186 N aE = net,E = = 1.605 m/s 2 aN = net,N = = 3.704 m/s 2 m m 1400 kg 1400 kg aE2 + aN2 =

a=

θ = tan −1

(1.605 m/s ) + ( 3.704 m/s ) 2 2

2 2

= 4.037 m/s 2 .

aE 1.605 = tan −1 = 23.43°. aN 3.704

This is 23.43o E of N.

N 3200 N

N

30o

θ

15o

aN

E aE

5-28.

Fnet = F1 + F2 = [ (4.0 N)i + (3.0 N) j] + [ (2.0 N)i − (5.0 N) j] = (6.0 N)i − (2.0 N) j a=

Fnet (6.0 N)i − (2.0 N) j = = (1.2 m/s 2 ) i − ( 0.40 m/s 2 ) j m 5 kg

a =

(1.2 m/s ) + ( 0.40 m/s )

a 2x + a 2y =

θ = tan −1

ay ax

= tan −1

2 2

2 2

= 1.3 m/s 2 .

−0.40 = −18°. 1.2

This is 18 below the +x axis. By vector addition Fnet = FSun + FMoon Fnet =

θ

m s2

x

0.40

o

5-29.

1.2

(4.3 × 1020 ) 2 + (2.0 × 1020 ) 2 N

= 4.7 × 1020 N And the angle it makes with line of force to Sun is θ = tan−1 2.0/4.3 = 25° .


m s2

CHAPTER 5-30.

5

The net force is simply given by Fwater + Fwind Then magnitude, by law of cosines, is 22002 + 19902 − 2(2200)(1990)cos 25° N = 930 N . Let angle that net force makes with the wind force vector be θ. Then, by law of cosines, 19902 = 22002 + 9302 − 2(2200)(930) cos θ cos θ = (22002 + 9302 − 19902)/2(2200)(930) = 0.426 θ = 65° The net force vector is pointing in the direction of the longitudinal axis of the boat, with magnitude 930 N.

†5-31.

5-32.

Define x and y axes with the x direction perpendicular to the dock and the y direction along the dock. Resolve the two 360-N forces into x and y components. By symmetry the y components cancel. The resultant force in the x direction is Rx = 2 × 260 N + 2 × (360 N) sin 20° = 766 N. The resultant expressed as a vector is R = 766i + 0 j N.

Take the x direction to point right. The net force on the tractor is FT = (1.60 × 104 N)i − (1.50 × 104 N)i = (1.0 × 103 N)i. (The vertical forces add to zero, so they aren’t included here.) F (1.0 × 103 N)i The acceleration of the tractor is a = T = 2000 kg mT = (0.50 m/s 2 )i. Because the rope connecting the tractor to the Jeep doesn’t stretch or break and has negligible mass, the magnitude of the tension at the Jeep must be the same as at the tractor, and the Jeep’s acceleration must be the same as the tractor’s acceleration. Thus the net force on the Jeep points to the right and has a magnitude given by ∑ F = mJ a = (1.50 × 104 N − FJ )i = (1400 kg)(0.50 m/s 2 )i = (700 N)i , where we have assumed that the force on the tractor points to the left. Solving for FJ gives FJ = 1.43 × 104 N, which is the magnitude of the force. Because we assumed it points to the left, the vector is FJ = − (1.43 × 104 N)i .


y

20

360 N

x

2 × 260 N 20o

360 N

1.50×104N

FJ

1.60×104 N

CHAPTER W 25 N = = 2.6 kg. W = mg = (3.5 kg)(9.81 m/s 2 ) = 34 N. g 9.81 m/s 2

5-33.

m=

5-34.

WPluto = mg Pluto = 60 kg × (0.045 × 9.81 m/s 2 ) = 26 N .

†5-35.

m=

5

WEarth 750 N = = 76.5 kg . Mass is an intrinsic property of the object, so its mass is 9.81 m/s 2 g Earth

76.5 kg on Earth, Mars, and Jupiter. Its weight depends on the value of g. The calculations can be simplified using proportional reasoning: g WMars = mg Mars = WEarth Mars = 0.38(750 N) = 285 N g Earth g Jupiter WJupiter = mg Jupiter = WEarth = 2.53(750 N) = 1.90 × 103 N g Earth 5-36.

5-37.

1 lb = 0.4536 kg, 1 lbf = 4.4482 N. Find the weight in New York and then use proportional reasoning to find weights in Hong Kong and Quito. 0.4536 kg 1 lbf WNY = 1 lb × = 0.9996 lbf . × 9.803 m/s 2 × lb 4.4482 N 9.788 WHK = WNY × = 0.9981 lbf . 9.803 9.780 WQ = WNY × = 0.9973 lbf . 9.803 (a) ∆W = m ( g Paris − g SF ) = (0.50000 kg) ( 9.8094 m/s 2 − 9.7996 m/s 2 ) = 4.900 × 10−3 N.

5-38.

⎛g ⎛ − g SF ⎞ g SF ⎞ ∆W −3 = m ⎜ Paris ⎟ = m ⎜1 − ⎟ = 9.058 × 10 . WParis g g Paris Paris ⎠ ⎝ ⎠ ⎝ (b) The value of the gold depends on its mass, not its weight. Its mass doesn’t change, so its value doesn’t change in the move. The chair exerts 60 × 9.8 = 588 N on the woman. The floor exerts (60 + 20)9.8 = 784 N on the chair.

†5-39.

Draw two “free−body” diagrams: Bottom chandelier:

T2

Top chandelier:

T2

m2g

T2 − m2 g = 0 T2 = m2 g = 29.4 N

T1

m1g

T1 − T2 − m1 g = 0 T1 = T2 + m1 g = m2 g + m1 g = 128 N


CHAPTER 5-40.

5

Call the tension in the 3-m cable T3 and the tension in the 4−m cable T4. The two cables form a 3−4−5 right triangle. The trig functions for the angles θ1 and θ2 are then easy to as figure out. Resolve the forces into components along the x and y as axes as shown and set the net force along each direction equal to zero: ∑ Fy = T4 sin θ1 + T3 sin θ 2 − mg = 0

∑F

x

y

T3

θ2

θ1

T4 x

= T4 cos θ1 − T3 cos θ 2 = 0

Rearrange: T4 sin θ1 = mg − T3 sin θ 2

mg

T4 cos θ1 = T3 cos θ 2 Divide the top equation by the bottom equation and solve for T3: mg − T3 sin θ 2 tan θ1 = T3 cos θ 2

mg tan θ1 cos θ 2 + sin θ 2 Use the trig functions: [tan θ1 = 0.75, sin θ2 = 0.8, cos θ2 = 0.6] to find T3: 200 kg × 9.81 m/s 2 T3 = = 1.57 × 103 N . Use this value and (0.75)(0.6) + 0.8 T3 =

3

θ2

5

θ1 4

cos θ1 = 0.8 to calculate T4: T cos θ 2 (0.6)(1.57 × 103 N) = = 1.28 × 103 N . T4 = 3 cos θ1 0.8 †5-41.

“Free-body” diagrams:

m1 T3

F

T2

m2 m2g m3

T3

m1g

T2

m3g

The system is described as being stationary, so it must be in equilibrium. Thus the net force on each mass must be zero. For m3, T3 − m3 g = 0 ⇒ T3 = m3 g . For m2, T2 − T3 − m2 g = 0 ⇒ T2 = T3 + m2 g = ( m2 + m3 ) g . For m1, F − T2 − m1 g = 0 ⇒ F = T2 + m1 g = ( m1 + m2 + m3 ) g .


CHAPTER 5-42.

5

“Free-body” diagram:

F

N

Ny

50o

Nx

F

mg The net force on the box is zero. The x and y components of the normal force N are shown in the “free-body” diagram. N y − mg = 0 ⇒ N y = mg = (5 kg) ( 9.81 m/s 2 ) = 49.1 N. Because N is perpendicular to the ramp and the ramp is inclined at 50o, the angle between N and Nx is 40o. Thus Ny 49.1 N = = 76.4 N. To find F, set the sum of forces in the x N y = N sin 40° ⇒ N = sin 40° sin 40° direction equal to zero: F − N x = 0 ⇒ F = N x = N cos 40° = (76.4 N) cos 40° = 58.5 N. †5-43.

Suppose another sailor pulls on the free end of the rope with a force P as shown. Then the “freebody” diagram for the sailor in the seat has two forces, and the net force on the sailor is Fnet = P − Mg . The reaction from the rope acts on the puller. For the sailor to move up, P must at least be equal to Mg. If the sailor in the seat reaches over and pulls on the rope, the reaction from the rope pulls up on the sailor in the seat. Now the “free-body” diagram shows two upward forces on the sailor in the seat, and the net force is Fnet = 2 P − Mg . Now the minimum force required for the sailor to lift himself is Mg/2.

PP

P

P

P

M

M

Mg

Mg

The sailor in the seat pulls himself up.

Another sailor pulls on the rope.

5-44.

T2

T2

T1

T1

T

In the “free-body” diagrams, T =12 000 N. T1 is the tension in the coupler between cars 1 and 2, and T2 is the tension in the coupler between cars 2 and 3. All three cars have the same mass M and the same acceleration a. (The vertical forces on the cars have been ignored because they add to zero.) Writing Newton’s Second Law for each car gives T − T1 = Ma, T1 − T2 = Ma, T2 = Ma.


CHAPTER

5

Adding the second and third equations gives T1 = 2 Ma. Substituting this into the first equation T 2 gives T = 3Ma. This means T1 = T = 8000 N. From the third equation we get T2 = 1 = 2 3 4000 N.

5-45.

5-46.

Suppose a friction force f acts on each car. The direction will be to the left in the diagram: T − T1 − f = Ma, T1 − T2 − f = Ma, T2 − f = Ma. Subtracting the third equation from the second gives T1 − 2T2 = 0 , which is the same relation we had before. Subtracting the second 3 equation from the first gives T − 2T1 − T2 = T − T1 = 0, which is the same result we had before. 2 So adding the same friction force to each car does not change the tensions. (Our intuition would tell us that adding a constant friction force should not change the result, but it’s good to demonstrate the result rigorously.) At every point along the cable, the net force must be zero. That means the upward tension must equal the weight of the cable below that point. At the very top the upward force must be equal to π d 2l ρ g the total weight of the cable, so T = . At the midpoint, the tension must be equal to half 4 π d 2l ρ g the weight: T = . 8 By Newton’s Third Law, the forces the car and truck exert on each other must have equal magnitudes and opposite directions. Thus mtruck atruck = − mcar acar . Then acar

2800 kg × ( −500 m/s 2 ) mtruck atruck =− =− = 1.17 × 103 m/s 2 . The final speed of the car is mcar 1200 kg

v = acar t = (1.17 × 103 m/s 2 ) (0.20 s) = 233 m/s. †5-47.

The 60-N force acts on a total mass of 50 kg, so a =

F 60 N = = 1.2 m/s 2 . To find the force mtotal 50 kg

each mass exerts on the other, use these “free-body” diagrams with R as the unknown force, assuming the 60-N force points to the right:

60 N

30

R

The vertical forces aren’t shown because they add to zero. Applying Newton’s Second Law to the 30-kg mass gives R = 30a = 36 N. The 20-kg box exerts a 36-N force to the right on the 30-kg box, and the 30-kg box exerts a 36-N force to the left on the 20 kg box. (Note: To check, note that the net force on the 20 kg box is (60 − 36) N = 24 N. This gives an acceleration of 1.2 m/s2.)


CHAPTER 5-48.

5

The minimum distance corresponds to the maximum frictional force, F = µsmg = ma or a = µs g v0 = 90 km/h = 25 m/s v 2 − v02  = 2ax = 2 µs gx x=

5-49.

2µ s gx (see solution to question 26)

v0 = =

5-50.

v02 (25) 2 = = 39.9m 2 µs g 2 × 0.8 × 9.8 2 × 0.8 × 9.8 × 290 = 67.4 m/s = 243 km/h

Find the x and y components of the resultant force R: Rx = 1.2 × 104 N − (1.2 × 104 N ) cos 40° = 2.81 × 103 N

y

Ry = (1.2 × 104 N ) sin 40° = 7.71 × 103 N

R = (2.81 × 103 N)i + (7.71 × 103 N) j R =

R + R = 8.2 × 10 N θ = tan 2 x

2 y

3

−1

Ry Rx

R = 70°

θ 40o 1.2 × 104

5-51.

5-52.

The net force on the point where the rope is being pushed must be zero. By symmetry the x components of the tension cancel. The sum of the y components of the forces must also add to zero: 900 N 5o ∑ Fy = 900 N − 2T sin 5° = 0 ⇒ T = 2 sin 5° = T 5.16 × 103 N (a) Forces are as shown on diagram. For forces to balance, T sin θ = N but sin θ = s/l, so that N = Ts/l ≥ T = Nl / s. (b) Using formula T = 150 N × 1.5 m/0.02 m = 1.1 × 104 N


x

y

900 N 5o

x T

CHAPTER †5-53.

5

The net force is the vector sum of the force exerted by the wind plus the tension in the string. (The weight of the balloon is assumed to be much less than the other forces acting and can be neglected.) Take the +y direction to point up and the +x direction to point right. Then Fnet , x = 200 N − (130 N) cos 70° = 156 N Fnet , y = 67 N − (130 N) sin 70° = − 55.2 N

Fnet =

2 x

5-54.

5-55.

Ry Rx

200 N 130 N 70o

2 y

R + R = 165 N

θ = tan −1

67 N

= −19.4° (19.4o below horizontal, or +x direction)

(A dynamics note: because the net force is not zero, the balloon is not in equilibrium at the instant described in this problem.) Because there cannot be a net force in direction perpendicular to bank, the components of force in this direction must cancel. Therefore, the component in this direction of force from horse = 300 sin 30° = 150 N. Thus, the transverse force of the rudder is 150 N, , pointing away from the bank. The first diagram shows the forces on a small segment of rope. From the second diagram we see that balance of these forces requires ⎛ dθ ⎞ 2T sin ⎜ ⎟ = dN, ⎝ 2 ⎠

or because dθ is small, 2T d N = Tdθ .

dθ ≅ dN 2

The vertical component of this force is = cos θ Tdθ, where the angle θ is measured from the vertical. Integrate this from θ = –90° to θ = 90°. 90° F=∫ Tcos θ dθ = Y ∫ cos θdθ −90° = Tsin θ

90D −90D

= 2T


CHAPTER 5-56.

5

N = normal force. Then net upward force = N − mg N − mg = ma ⇒ N = ma + mg = m(a + g) N = 80 kg(9.8 + 1.8) m/s2 = 928 N Normal force exceeds weight by N = mg = 80(1.8)N = 144 N

5-57.

5-58.

Deceleration a =

vi2 − v 2f 2s

=

25 = 12.5m/s 2 2 ×1

The force exerted by the ground on the feet is N, where N − mg = ma N = m(g + a) = 80(9.8 + 12.5) = 1784 N. m − m1 From Example 10, a1 = 2 g , where m1 is the mass of the elevator cage and m2 is the mass m1 + m2 of the counterweight. Here m1 = 1280 kg, the mass of the empty cage plus the mass of four people 1100 kg − 1280 kg at 70 kg each. a = × 9.81 m/s 2 = − 0.742 m/s 2 . The − sign means the car 2380 kg is moving down because in the example the positive direction for vectors was chosen to be up. To find the speed, use v 2 = v02 + 2ax with caution. a and x are both vector quantities, and their algebraic signs must agree. In this problem, we already know that the car is accelerating downward. Because it began moving from rest, its displacement x is also downward and would be represented by a negative number. Thus the product ax is positive, giving v=

†5-59.

2ax =

2 × ( −0.742 m/s 2 ) × (−10 m ) = 3.85 m/s.

A diagram and three “free-body” diagrams are shown. Write Newton’s Second Law for each of the three masses: F − m 1 g − T2 = m1a1

m1

T3

T2

F

T2 − m 2 g − T3 = m2 a2 T3 − m 3 g 2 = m3 a3 All three masses must have the same acceleration: a 1 = a2 = a3 , which we’ll just call a. To find a, add all three equations together. Then T2 and T3 cancel, giving F − ( m1 + m2 + m3 ) g = ( m1 + m2 + m3 ) a

⇒a=

F − ( m1 + m2 + m3 ) g m1 + m2 + m3

m2

m3

m2g

T3

m3g

.

Then the tension in the first cable is F, and the tensions in the second and third cables are ⎛ m2 + m3 ⎞ T1 = F − m1 ( g + a) = F ⎜ ⎟ ⎝ m1 + m2 + m3 ⎠ and

⎛ ⎞ m3 T2 = m3 ( g + a ) = F ⎜ ⎟. ⎝ m1 + m2 + m3 ⎠


m1g

T2

CHAPTER

5

5-60.

6.0 N

3 kg

4 kg

4.0 N

R

6.0 N

4.0 N R

A sketch shows the two “free-body” diagrams with the force each box exerts on the other called R. Write Newton’s Second Law for each box, recognizing that both boxes must have the same acceleration and taking the positive direction to point to the right: 6.0 − R = 3a

R − 4.0 = 4a 2m = 0.29 m/s 2 . (The acceleration points to 7 s2 the right.) Now we can solve either of the individual equations to find R: 36 N = 5.1 N. The 3-kg mass exerts a force of 5.1 N to the right on the 4-kg R = 4a + 4.0 N = 7 mass, and the 4-kg mass exerts a force of 5.1 N to the left on the 3-kg mass. The reading on the scale is the normal force exerted by the scale. The net N force on the passenger is N − mg = ma ⇒ N = mg + ma , choosing up Add the two equations to find a: 2.0 = 7 a ⇒ a =

†5-61.

to be the positive direction for vectors. The man’s mass is 220 lbm = 99.8 kg, and his weight is mg = 220 lbf = 979 N. If a = 1.6 m/s 2 , then N = (979 N + 99.8 kg × 1.6 m/s 2 ) = 1139 N = 256 lb. If the elevator has a downward acceleration, then a = −1.6 m/s 2 , and N = 819 N = 184 lb. 5-62.

T - mg = ma ⇒ a = m(a + g) = 9.5 kg (1.9 m/s 2 + 9.81 m/s 2 ) = 111 N.

mg

T

mg


CHAPTER †5-63.

A “free-body” diagram is shown for one of the dice. From the diagram, T cos θ − mg = 0, T sin θ = ma

a mg

⎛ 2.5 m/s 2 ⎞ = 14.3°. The final result = 14o. The tension in 2 ⎟ 9.81 m/s ⎝ ⎠ mg 0.025 kg × 9.81 m/s 2 the string for each die is T = = = cos θ cos14.3° 0.25 N.

θ = tan −1 ⎜

5-64.

T

θ

where the positive horizontal direction is assumed to point to the right, and the positive vertical direction is up. Dividing the bottom a equation by the top equation gives tan θ = , so g

5

x

y N

mg cos 50o

50o

25.0 N

50

o

mg sin 50

mg

o

50o

25.0 N 25 cos 50o o 25sin 50

mg Three diagrams are given. One is a “free-body” diagram showing the three forces acting on the box along with definitions of the x and y directions pointing up the incline and perpendicular to the incline. The other two show resolution of the vertical weight mg and the horizontal 25 N force into components along the x and y directions. Now write Newton’s Second Law for the x direction: 25 cos 50° − mg sin 50° = ma. (The sum of forces in the y direction is zero because there is no motion perpendicular to the incline.) The acceleration is m⎞ ⎛ (25.0 N) cos 50° − (3.50 kg) ⎜ 9.81 2 ⎟ sin 50° m s ⎠ ⎝ = −2.92 2 . The − sign means the mass is a= 3.50 kg s

5-65.

sliding down the incline—the 25-N force is not large enough to cause it to slide up. Force = mg sin θ F 1 a= = 1.9m/s 2 = g sin θ = 9.8 × m 26 Speed after 50 m; use v2 – v02 = 2a(x – x0) v=

2 × 1.9 × 50 m/s = 14 m / s


CHAPTER 5-66.

5

(a) See drawing of skier on the right. (b) mg = 75 × 9.8 N = 735 N Normal force = mg cos θ = 735 cos 35° = 602 N Magnitude of resultant = mg sin θ = 735 sin 35° = 422 N F 75 × 9.8 × sin 35° m/s2 = 75 m = 9.8 sin 35° m/s2 = 5.62 m / s 2 (c) a =

5-67.

a = g sin θ = 9.8 × sin 30° m/s2 = 4.9 m/s2 90 km/hr = 25 m/s v2 – v02 = 2a(x − x0) 252 m = 64 m 2(4.9) v − v0 25 m/s Time taken: t = = = 5.1s a 4.9 m/s 2

x − x0 =

5-68.

N = the force the shoulder exerts on the backpack, which is equal and opposite to the force the backpack exerts on the shoulder. N − mg = ma ⇒ N = m(a + g). m m⎞ ⎛ N = (20 kg) ⎜ 2.0 2 + 9.8 2 ⎟ = 236 N. The force exerted s s ⎠ ⎝ by the backpack on the shoulder is 236 N down.

N

mg 5-69.

∑Fx = 0 = T sin θ − N

(i)

∑Fy = 0 = T cos θ = mg (ii) Dividing (i) by (ii) we get tan θ = N/mg N = mg tan θ but R2 + y2 = (e +R)2 ⇒ y2 = e2 + 2eR = e(e + 2R) R tan θ = e(e + 2 R ) N =

mgR e(e + 2 R )

(Note that N → 0 as e → ∞)


CHAPTER 5-70.

By the equation in example 10, ⎛ m2 m1 ⎞ a= ⎜ ⎟ g (where g is the unknown ⎝ m1 + m2 ⎠ acceleration due to gravity) 402.0 − 400.0 = × g = 2.4938 × 10–3 × g 400.0 + 402.0 1 But x − x0 = v0t + at2 2 2[( x − x0 ) − v0 t ] 2(0.50 m) 2 = s = 0.0244 m/s2 = 2.4938 × 10–3 × g a= 6.42 t2 0.0244 g= m/s2 = 9.79 m / s 2 −3 2.4938 × 10

5-71.

F cos 30 − 588 sin 30° = 0 F = 588 tan 30° = 339.5 N N − F sin 30° − 588 cos 30° = 0 N = F sin 30° + 588 cos 30° = 679 N

5-72.

Since the watch is at rest in the frame of the airplane, it is accelerating at 1.2 m/s with respect to the inertial frame (i.e., earth). This force is transmitted to watch via chain. The components of this force are as shown in the diagram. The angle θ that the chain makes with the vertical is tan–1 (a/g) = tan–1 (1.2/9.8) = 7.0°

5-73.

(a) Inclined forwards. (b) The component of force acting in the horizontal direction is T sin 20° where T is tension in string. Therefore T sin 20° = ma T cos 20° = F where F is the net force on the balloon in the vertical direction and is the difference in the buoyant force and the weight of the helium. The mass of the fabric is assumed negligible. This is equal to V(ρa − ρ=)g, where V = volume of balloon; ρa, ρh are densities of air and helium respectively. T cos 20° = V(ρa − ρh)g (i)


5

CHAPTER

5

T sin 20° = ma = vρha

(ii) V ρh a ρh (ii) + (i) gives tan 20° = a = V ( ρa − ρh ) g ( ρa − ρh ) g ⎛ ρ − ρh ⎞ a = tan 20° ⎜ a ⎟g ⎝ ρh ⎠ ρa = 1.293 kg/m3 ρh = 0.1785 kg/m3 ⎛ 1.2930. − 0.1785 ⎞ 2 2 a = tan 20° ⎜ ⎟ 9.8 m/s = 22.3m / s 0.1785 ⎝ ⎠ 5-74.

Three “free-body” diagrams are shown beside the figure. Note that a “free-body” diagram is included for pulley 2. Write Newton’s Second Law for each diagram: T1 − m1 g = m1a1

T1 T1

1

T1 − 2T2 = 0

m1

T2 − m2 g = m2 a2

2

T2

2

The second equation is tricky. It describes m2g the acceleration of pulley 2, but because m2 m1g we assume the pulley is massless, it says T2 T2 the sum of forces must be zero. To solve the equations, we must figure out the relation between the accelerations a1 and a2. If mass m1 falls a distance h, then pulley 2 will be pulled up the same distance. Because one end of the string over pulley 2 is fixed, m2 will be raised twice as far, 2h. Thus the magnitude of a2 is twice the magnitude of a1. We also see that if m1 moves down, m2 moves up. Thus a2 = −2a1. Making these substitutions in the third equation gives T2 = m2 g − 2m2 a1 . Substituting T1 = 2T2 from the second 2m2 − m1 g . The acceleration of pulley 2 is −a1 , and the equation in the first equation gives a1 = m1 + 4m2 acceleration of m2 is −21 .


CHAPTER 5-75.

Along the incline mg sin θ = ma, a = g sin θ

distance, l moved along the incline in time t l=

1 g sin θ • t2 2

horizontal distance x = l cos θ =

1 g sin θ cos θ t2 2

t=

2x g sin θ cosθ dt =0= dθ = 45°

2 x 1/ 2(cos 2θ − sin 2θ ) if θ = tan–1 (l ) g (sinθ cosθ )3 / 2

Minimum time, tmin =

5-76.

2x x =2 g sin 45°cos 45° g

The equation of motion of each mass is m1a1 = T1 − m1g

(i)

m3(a2 − a1) = T2 − m3g

(ii)

m2 (–a1 − a2) = T2 − m3g

(iii)

2 T2 = T1

(iv)

The extra a1 term in equations (ii) and (iii) occur because a2 is the acceleration of the masses m2 and m3 with respect to the small pulley. But the small pulley is also accelerating with respect to the inertial frame downwards at a rate of a2!

(iv) comes about because the pulley is massless, hence the net force on the pulley must be zero. (ii) − (iii) gives: m3(a2 − a1) + m2(a1 + a2) = (m2 − m3)g Substituting (iv) into (i) and (i) − 2(ii) gives: m1 a1 − 2m3(a2 − a1) = –m1 g + 2m3 g or (m1 + 2m3)a1 − 2m3 a2 = (2m3 − m1)g (v) From above: (m2 − m3)a1 + (m2 + m3)a2 = (m3 − m3)g From this: ⎡ (4m2 m3 − m1m2 − m1m3 ) ⎤ a1 = ⎢ ⎥g ⎣ (m1m2 + 4m2 m3 + m1m3 ) ⎦ ⎡ ⎤ 2m1 (m2 − m3 ) a2 = ⎢ ⎥g ⎣ (m1m2 + 4m2 m3 + m1m3 ) ⎦ a1 is acceleration of mass m1 (upwards) Acceleration of m2 is (relative to fixed frame)


(vi)

5

CHAPTER

5

⎡ (4m2 m3 − m1m2 − 3m1m3 ) ⎤ a1 + a2 = ⎢ ⎥ g (down) ⎣ (m1m2 + 4m2 m3 + m1m3 ) ⎦

Acceleration of m3 is (relative to fixed frame) ⎡ (3m1m2 − m1m3 − 4m2 m3 ) ⎤ a2 − a1 = ⎢ ⎥ g (up) ⎣ (m1m2 + 4m2 m3 + m1m3 ) ⎦ From (i),

⎡ ⎤ m1 (8m2 m3 ) T1 = m1(a1 + g) = ⎢ ⎥g ⎣ (m1m2 + 4m2 m3 + m1m3 ) ⎦

⎡ ⎤ 4m1m2 m3 T2 = ⎢ ⎥g ⎣ (m1m2 + 4m2 m3 + m1m3 ) ⎦

5-77.

From Example 10, a =

m2 − m1 g . If the left end of the chain is a distance x above its starting m1 + m2

m⎛ l m⎛ l ⎞ ⎞ ⎜ − x ⎟ and m2 = ⎜ + x ⎟ . The total mass is m1 + m2 = m. Substituting l ⎝2 l ⎝2 ⎠ ⎠ 2g x. To find the equation for x(t), use the derivative these into the equation for a gives a = l

point, then m1 =

d 2 x 2g definition of a: 2 = x. According to the hint, the solution to this is x(t ) = Ae dt l A, use the fact that at t = 0, x = x0 ⇒ x(t ) = x0 e

5-78.

mtruck atruck acar

.

=

(2000 kg) (1.2 m/s 2 ) 2.5 m/s 2

= 960 kg.

m = 3400t = 3.4 × 106 kg a = F/m = 270 N/3.4 × 106 kg = 7.9 × 10−5 m / s 2

Then x − x0 = v0t + 5-80.

. To find

The forces on the car and truck must have equal magnitudes, so mcar acar = mtruck atruck . Thus mcar =

5-79.

2g t l

2g t l

1 2 1 at = 0 + (7.9 × 105 m / s 2 ) × (60 s)2 = 0.14 m 2 2

dv d = ( 655.9 − 61.14t + 3.260t 2 ) = − 61.14 + 6.520t. dt dt (b) F = ma = (45.36 kg) ( −61.14 + 6.520t ) = − 2773 + 295.7t.

(a) a =

The equation for v predicts that at some time the velocity would change sign, which implies that the projectile would reverse its motion and start speeding up. Of course this is not what would happen in the real world. Eventually the horizontal velocity would reach zero and the projectile would stop moving horizontally—its velocity will not begin to increase in the opposite direction. So we might expect that when the speed becomes very small, the equations in this problem will no longer accurately describe the motion of the projectile.


CHAPTER †5-81.

Ftotal = F1 + F2 = (2 − 4)i + (−5 + 8) j + (3 + 1)k = −4i + 3j + 4k N. a=

Ftotal 2 1 2 −4i + 3 j + 4k N = = − i + j + k m/s 2 = −0.67i + 0.50 j + 0.67k m 6.0 kg 3 2 3 2

a = 5-82.

5

ax2 + a y2 + az2 =

2

m/s 2 .

2

⎛2⎞ ⎛1⎞ ⎛ 2⎞ 2 ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = 1.1m/s . 3 2 3 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(a) Force on diver = ma = 75 kg × 9.81 m/s2 = 736 N. By the third law, the force on the Earth is 736 N (b) Acceleration of diver = 9.8 m / s 2 = 735 N/mass of Earth = 735 N/5.98 × 1024 kg ≈ 1.3 × 10−22 m / s 2 .

†5-83.

(a)Represent the boat as a point mass. The water exerts an upward force with magnitude B on the boat. The tension T is shown resolved into x and y components. The angle θ is found from the fact that the rope is 50 m long and the depth of the ⎛ 10 m ⎞ water is 10 m: θ = sin −1 ⎜ ⎟ = 11.5°. ⎝ 50 m ⎠

y

B

T sin θ

T cos θ

x

θ

7000 N

T mg

(b) The boat is in equilibrium so the sum of all the forces on the boat must be zero. The sum of x components will give the tension T: 7000 N = 7.14 × 103 N. 7000 N − T cos θ = 0 ⇒ T = cos11.5° (c) To find B, use the sum of the y components: B − mg − T sin θ = 0 ⇒ B = mg + T sin θ = (2500 kg) ( 9.81 m/s 2 ) + ( 7.14 × 103 N ) (sin11.5°) = 2.59 × 104 N.

5-84.

Vector note: B = (2.59 × 104 N)j. (a)In the diagram, the x and y components of the 80-N force are shown. (b)In the x direction, there is only one force acting. Fnet , x = ∑ Fx = (80 N ) cos 30° = max , which gives ax =

y

N

(80 N)cos 30° = 2.77m/s 2 . Final answer = 2.8 m/s 2 . 25 kg

(c)There is no motion in the y direction, so

∑F

y

30o

= 0. This will

80 N

give the normal force N: N − mg − (80 N ) sin 30° = 0 ⇒ N = (25 kg) ( 9.81 m/s 2 ) + (80 N)(sin 30°) = 285 N.


mg

x

CHAPTER †5-85.

5

(a) Represent each child by a block, and represent each end of the rope by points between them. Only the horizontal forces are shown. The vertical weight of each child and the vertical normal force exerted by the ground on each child have been omitted from the diagrams.

Boy

Fboy on rope

Frope on boy

Girl Fgirl on rope Frope on girl

T T

Fground on girl

Fground on boy

(b) By Newton’s Third Law, Frope on boy = Fboy on rope . Because Fboy on rope = 250 N, the magnitude

5-86.

of Frope on boy = 250 N. The boy doesn’t move, so the net horizontal force on him must be zero, and that means the magnitude of Fground on boy = 250 N. By similar reasoning, the magnitude of Fground on girl = 250 N. (c) The net horizontal force on the rope must be zero, so T = Fboy on rope = 250 N. (d) Nothing changes except the origin of the force on the girl’s end of the rope. Suppose the rope passed through a hole in a wall so the boy couldn’t see what was at the other end. He would not be able to tell whether there was a person at the other end or if the end was tied to something immovable. Total mass of the 250 cars is (250 × 64 × 1000)kg = 1.6 × 107 kg. The tension on the coupling between locomotive and first car = force that pulls all the cars = mass of 250 cars × accel. = 1.6 × 107 kg × 0.043 m/s2 = 6.9 × 105 N. Similarly, tension on coupling on last car = force accelerating last car = mass of 1 car × accel. = 64000 kg × 0.043 m/s2 = 2.8 × 103 N.

†5-87.

N

P

1

θ

P

11

mg cos θ

θ mg mg sin θ (a) A sketch is shown, and the “free-body” diagram is shown on the right. The weight of the boxcar has been resolved into components along the incline and normal to it. The angle θ can be 1 = 5.19°. found from the slope: θ = tan −1 11 (b) For the boxcar to move at constant speed, its acceleration along the incline must be zero. Thus the sum of forces along the incline must be zero: P − mg sin θ = 0 , from which we get P = mg sin θ = (20 × 103 kg) ( 9.81 m/s 2 ) (sin 5.19°) = 1.8 × 104 N. Vector note: As shown in

the diagram, P = 1.8 × 104 N up the incline.


CHAPTER

5

(c) In the absence of any friction forces, it makes no difference whether the car is moving at constant velocity up the incline or down the incline. The force required is 1.8 × 104 N up the incline in either case. 5-88.

N

T

+

m1 30o

m2

T m1g

m1g cos 30o 30o

m2g m1 g sin 30

o

(a) “Free-body” diagrams are shown for the log (m1) and the counterweight (m2). The weight m1g of the log has been resolved into components along and normal to the incline. The accelerations of the two masses must have equal magnitudes. Following Example 11, if we choose the positive direction for components along the incline to point right in the expected direction of motion, and choose the positive direction for the vectors associated with m2 to point up, then when m1 is accelerating in its positive direction, m2 is accelerating in its negative direction. If the acceleration of m1 is represented by a, the acceleration of m2 must be −a. Write Newton’s Second Law for the forces along the incline that act on the log and for the forces that act on the counterweight: T − m1 g sin 30° = m1a, T − m2 g = −m2 a. Subtract the second equation from the first to get an equation for a: m − m1 sin 30° 300 kg − (500 kg)(sin 30°) a= 2 g = ( 9.81 m/s2 ) = 0.613 m/s2 . m1 + m2 500 kg + 300 kg (b) The minimum value for m2 would be one for which the acceleration would be nearly zero. This would give T = m2 g , which means m2 − m1 sin 30° = 0 , or m2 = (500 kg)(sin 30°) = 250 kg. †5-89.

The scale reading is equal to the normal force exerted on the woman by the scale. Newton’s Second Law says ∑ F = N − mg = ma, where m is the woman’s mass

N

(60 kg). The scale reading is N = m( g + a ). Choose the positive direction to be up. (a)At rest ⇒ a = 0 ⇒ N = mg = (60 kg)(9.81 m/s2) = 589 N (b)Accelerating up ⇒ N = (60 kg) ( 9.81 m/s 2 + 1.8 m/s 2 ) = 697 N. (c)Constant velocity ⇒ a = 0 ⇒ N = mg = (60 kg)(9.81 m/s2) = 589 N. (d)Free fall ⇒ a = −g ⇒ N = 0 N.


mg

CHAPTER 5-90.

5

The “free-body” diagram is shown beside the sketch of the system. The tension T is resolved into x and y components. The crate is in equilibrium, so the sum of forces on it is zero. Fnet , x = ∑ Fx = 1800 N − T sin θ = 0 Fnet , y =

y

θ

∑ Fy = T cos θ − mg = 0

T sin

θ

T θ T cos x

Rearrange: T sin θ = 1800 N

1800

T cos θ = mg

Divide the first equation by the second: ⎡ ⎤ 1800 N 1800 N ⎥ = 5.24°. tan θ = ⇒ θ = tan −1 ⎢ 2 mg ⎢⎣ (2000 kg) ( 9.81 m/s ) ⎥⎦ †5-91.

(a)A sketch and “free-body” diagram are shown with the initial velocity of the car shown in the sketch. Choose the +x direction to point leftward up the incline in the direction of the original motion. The weight of the car has its x component pointing in the −x direction. The sum of forces normal to the incline is zero, and there is only one force along the incline: Fnet , x = ∑ Fx = − mg sin θ = max . The angle θ

mg

N

v

1

θ

1 mg cos

θm

mg sin 1 = 5.71°. 10 The acceleration is ax = − g sin θ = − ( 9.81 m/s 2 ) sin 5.71° = −0.976 m/s 2 . The − sign means the is found from the slope: θ = tan −1

acceleration points down the incline, opposite the direction of the car’s initial velocity. In non– technical language, this is the car’s “deceleration.” km m = 13.9 . Solving for x gives (b) We can use v 2 = v02 + 2ax x , where v = 0 and v0 = 50 h s 2 2 (13.9 m/s ) = 99.0 m. v x=− 0 =− 2a x 2 ( −0.976 m/s 2 ) (c) Even when the car momentarily comes to rest and then begins to roll back down the incline, it m continues to have an acceleration ax = − 0.976 2 . To find its speed when it gets back to its s 2 2 starting point, we can again use v = v0 + 2ax x, but caution must be used in dealing with the vector quantities in the equation. Now v0 = 0, and both ax and x point down the incline and must be entered as negative quantities in the equation. We get v=

2ax x =

2 ( −0.976 m/s 2 ) (−99.0 m) = 13.9 m/s. (This is the car’s speed. Its velocity is

−(13.9 m/s)i because it is now moving down the incline.)


CHAPTER

5-92.

(a) From Example 10, a =

5

m2 − m1 1000 kg − 1200 kg × (9.81 m/s 2 ) = −0.892 m/s 2 . g = m2 + m1 2200 kg

The - sign means the acceleration of the cage points down. v − v0 0 − 1.5 m/s v = v0 + at ⇒ t = = = 1.7 s. a −0.892 m/s 2 (b) From Example 10, T − m1 g = m1a ⇒ T = m1 ( g + a ) = (1200 kg ) ( 9.81 m/s 2 − 0.892 m/s 2 ) = 1.07 × 104 N. (c) When the brakes are locked, the acceleration of the car and the counterweight is zero. The net force on each object must be zero, so the upward force on the counterweight is equal to its own weight. Therefore the tension in the cable is 1000 kg × 9.81 m/s2 = 9.81 × 103 N. †5-93.

N

T T

m2g

m1g

(a) Take the x direction to point right and the y direction to point up. Newton’s Second Law for m1 gives ∑ Fy = N − m1 g = 0, ∑ Fx = T = m1a1 . For m2, ∑ Fy = T − m2 g = m2 a2 . If the string passing over the pulley does not stretch and has negligible mass, then the magnitude of a2 must be the same as the magnitude of a1. If m1 moves in the +x direction, then m2 must move in the −y direction, which means a2 = −a1. Then the equations for ∑ Fx for m1 and ∑ Fy for m2 can be combined to give an equation for a1: a1 = (b) The equation for 5-94.

∑F

x

m2 g . m1 + m2

gives the tension: T = m1a1 =

T2

N2

N1 m1

m1m2 g . m1 + m2

m2 T1

m3

T1

T2 m2g

m1g

m3g

Three “free-body” diagrams are required. Take the x direction to point right and the y direction to point up. The sums of forces in the y direction are zero for m1 and m2 and give no useful information about the motion. ∑ Fx for m1 and m2 and ∑ Fy for m3 will give the acceleration and tensions.


CHAPTER

5

(a) T1 = m1a1 T2 − T1 = m2 a2 T2 − m3 g = m3 a3 If the strings don’t stretch, then all three accelerations must have the same magnitude. The accelerations for m1 and m2 must also have the same direction, and if m1 and m2 are accelerating in the +x direction then m3 must be accelerating in the −y direction. Thus we have a2 = a1 and a3 = –a1. Subtracting the second equation from the first eliminates T1, and subtracting the third m3 g equation from that result eliminates T2, leaving an equation for a1: a1 = . m1 + m2 + m3 (b) The first equation gives T1: T1 = T2 = T1 + m2 a1 =

m1m3 g . The second equation gives T2: m1 + m2 + m3

m1m3 g m2 m3 g m (m + m2 ) g + , or T2 = 3 1 . m1 + m2 + m3 m1 + m2 + m3 m1 + m2 + m3


CHAPTER 6

FURTHER APPLICATIONS OF NEWTON'S LAW

Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions Manual, available for purchase. Answers to all solutions below are underscored. †6-1.

Pull = nPEgyption, where n = number of Egyptians. Assume that the obelisk is being pulled at a constant speed. The total pull must be equal to the kinetic friction force acting on the obelisk: Pull = f = µ k N . For an object resting on a horizontal surface (flat ground) with no vertical

forces other than gravity and the normal force acting, the normal force N is equal to the object’s µ mg (0.30)(7000 kg)(9.81 m/s 2 ) weight mg. Thus nPEgyption = µ k mg ⇒ n = k = = 5.7 × 103 PEgyption 360 N Egyptians. 6-2. 6-3.

f s ,max = 4 µ s N bolt = 4(0.40)(2700 N) = 4.3 × 103 N. Let M be mass of automobile. 65 km/hr = 18.06 m/s 1 2 ( v − v02 ) 12 (18.06)2 2 2 a= = m/s = 8.15m/s 2 x − x0 20

But F = Ma where F = frictional force. a 8.15m/s 2 = 0.83 = F = Ma = µsMg ⇒ a = µs g ⇒ µs = 9.8m/s 2 g 6-4.

†6-5.

The force that decelerates the crate is the frictional force between it and the platform of the truck. The frictional force is F ≤ µsmg (m = mass of crate). When the truck decelerates at a, if the crate is not to slip, there must be a frictional force F so that F/m = a. Therefore, at most, a = F/m ≤ µs g. a ⇒ µs g ⇒ 0.4 × 9.8 m/s2 = 3.9 m/s 2

Assume that that the car has antilock brakes so it stops without sliding. Then the force between the tire and the road is always the force of static friction. The minimum stopping distance will occur for the largest friction force, which is f s ,max = µ s N . On a level surface with no vertical forces acting other than gravity, N = mg ⇒ f s ,max = µ s mg . The friction force is opposite the motion, so madry = − f s ,max = − µ s , dry mg ⇒ adry = − µ s ,dry g . adry = means xicy xdry

=

v 2 − v02 v2 = − 0 , which 2 xdry 2 xdry

v02 v2 = µ s ,dry g . For the icy road, 0 = µ s ,icy g. Combining the two equations gives 2 xdry 2 xicy

µ s ,dry µ s ,dry ⎛ 0.85 ⎞ 2 , from which we get xicy = xdry = (38 m) ⎜ ⎟ = 1.6 × 10 m. µ s ,icy µ s ,icy ⎝ 0.20 ⎠


CHAPTER

6-6.

†6-7.

v02 = µ s g . v0 = 2µ s xg = 2(0.80)(290 m)(9.81 m/s 2 ) = 67.5 m/s. Convert 2x m 1 mi s this to mph: v = 67.5 × × 3600 = 152 mph! s 1601 m h The technique is similar to that outlined in Problem 6-5 except that we use the force of kinetic v02 v02 2 2 friction, which means a = − µ k g . 2ax = v − v0 ⇒ x = − = . v0 = 10 km/h = 25 m/s 2a 2 µ k g

From Problem 6-5,

⇒x= 6-8.

6

(25 m/s) 2 = 53 m. 2(0.60)(9.81 m/s 2 )

Take the x direction to point down the slide and the y direction to be normal to the ramp in the free body diagram. There is no motion in the y direction, so the sum of forces in that direction must be zero. The sum of forces in the x direction must be equal to max: ∑ Fy = N − mg cos30° = 0, ∑ Fx = mg sin 30° − f k = max . Using f k = µ k N , solving the y equation for N, and substituting into the x equation gives max = mg sin 30° − µ k mg cos30°, or

ax = g (sin 30° − µ k cos30°) = (9.81 m/s ) [sin 30° − (0.15)cos30°] = 2

N

fk m1g cos 30o

3.63 m/s . Call h the height of the slide. The total distance traveled h 3.5 m = = 7.0 m. Since the initial along the slide is x = sin 30° sin 30° speed is zero, the speed at the end of the slide will be

v = 2ax x = 2(3.63 m/s 2 )(7.0 m) = 7.2 m/s.

mg

30o

2

mg sin 30o x h 30o

†6-9.

fk µ mg =− = − µ g , where the – sign means the acceleration points in the m m opposite direction from the velocity. Since the player is on level ground with no vertical forces other than gravity and the normal force acting, N = mg . µk = 0.30 ⇒ a = − (0.30)(9.81 m/s 2 ) = − 2.94 m/s 2 . The distance required for him to fk = µk N ⇒ a = −

stop is x =

v 2 − v02 0 − (4.5 m/s) 2 = = 3.4 m. Since he only has to slide 2.8 m to reach home, he 2a 2( − 2.94 m/s 2 )

will easily reach the plate. His speed when he reaches the plate will be

v = v02 + 2ax = (4.5 m/s) 2 + 2(−2.94 m/s 2 )(2.8 m) = 1.9 m/s. 6-10.

3

⎛ 102 cm ⎞ ⎛ 10−3 kg ⎞ g f s ,max = µ s N = µ s mg ⇒ µ s = . m = 2.33 3 × (10 × 10−6 m)3 × ⎜ ⎟ ×⎜ ⎟ mg cm ⎝ m ⎠ ⎝ g ⎠ 0.5 × 10−9 N = 22. = 2.33 × 10–12 kg. µ s = (2.33 × 10−12 kg)(9.81 m/s 2 )

f s ,max


CHAPTER 6-11.

For an object sliding on a horizontal surface with no downward forces acting except gravity, f a ∆v µ N µ mg v − v0 35 m/s = µk g ⇒ µk = . a = k = k = k = = = 4.67 m/s 2 . m m m g ∆t ∆t 7.5 s

µk = 6-12.

†6-13.

6-14.

4.67 m/s 2 = 0.48. 9.81 m/s 2

C ρ Av 2 . f road − f air = ma ⇒ f road = f air + ma. At constant speed, a = 0, and f road = f air . 2 C ρ A (0.35)(1.3 kg/m3 )(3.4 m 2 ) = = 0.7735 kg/m. At 20 m/s, froad = (0.7735 kg/m)(20 m/s)2 = 2 2 3.1 × 102 N. If the speed is doubled to 40 m/s, the force is quadrupled: froad = 4( 3.1 × 102 N) = 1.2 × 103 N. C ρ Av 2 The net horizontal force acting on the car is f road − f air = ma ⇒ f road = ma + f air = ma + . 2 Substituting the numerical data from the problem, (0.30)(1.3 kg/m3 )(2.8 m 2 )(25 m/s)2 f air = (900 kg)(2.0 m/s 2 ) + = 2.1 × 103 N. 2 f air =

At terminal speed vT, the upward drag force equals the downward weight of the falling object.

C ρ AvT2 = mg ⇒ vT = 2 6-15.

For a sphere, vT =

6-16.

vT =

†6-17.

6

2mg = Cρ A

2mg = C ρπ r 2

2(2.5 × 10−3 kg)(9.81 m/s 2 ) = 9.6 m/s. (0.51)(1.3 kg/m 3 )(π )(1.6 × 10−2 m) 2

2mg 2mg 2(0.045 kg)(9.81 m/s 2 ) ⇒ = = = 0.27. C C ρπ r 2 πρ r 2 vT2 (π )(1.3 kg/m3 )(0.02 m) 2 (45 m/s) 2 2(70 kg)(9.81 m/s 2 ) = 50 m/s. (0.42 m 2 )(1.3 kg/m3 )

2mg = (CA) ρ

A free body diagram for the sled is shown. P is the pull exerted by the girl, m is the mass of the sled, N is the normal force exerted by the ground on the sled, and f is the kinetic friction force between the sled and the ground. Take x to point right and y to point up. Then ∑ Fy = N + P sin 30° − mg = 0

N

P 30º

f mg

because the sled does not move in the y direction. ∑ Fx = P cos30° − f = 0 because the sled is being pulled at constant velocity. From the equation for

∑F , x

∑F , y

N = mg − P sin 30°. From the equation for

P cos30° − f = P cos30° − µ k N = P cos30° − µ k N = 0. Thus

P cos30° − µ k (mg − P sin 30°) = 0, which gives P = =

µ k mg cos30° + µ k sin 30°

(0.60)(40 kg)(9.81 m/s 2 ) = 2.0 × 102 N. cos30° + (0.60)sin 30°


CHAPTER 6-18.

6

The maximum frictional force between the driving wheels and the track = µsmg (mg is the weight µ mg (0.25)(1.36 × 105 kg)(9.81 m/s 2 ) = on the driving wheels). µsmg = mtotala. a = s (2.00 × 105 + 100 × 1.80 × 103 ) kg mtotal = 0.88 m/s 2 .

†6-19.

“Deceleration” is a nontechnical way of saying that the speed box v0 of the truck is decreasing. That means the magnitude of the velocity is decreasing. Suppose the truck’s initial velocity was to the right (+ x) as shown. If it is slowing down, then its acceleration must point to the left (–x). If the box doesn’t N slide on the back of the truck, then it must always be at rest relative to the truck and must have the same acceleration as f the truck. If the box is sliding, that means its acceleration is different from the truck’s. The free body diagram for the box shows that there are only three forces acting on it: its weight mg mg, a normal force N, and a friction force f between it and the truck. To determine the direction of the friction force, suppose there was no friction between the box and the truck. Then when the truck began to slow down, the box would continue moving to the right at its original speed. That means the friction force must point to the left as shown in the diagram. This is very important: There is no horizontal force in the direction of motion! The vertical forces add to zero, so N = mg. The box is sliding, so f is a force of kinetic friction. Since f is the only horizontal force acting, − f = − µ k N = max . Since N = mg, this gives

ax = − µ k g = − (0.50)(9.81 m/s 2 ) = − 4.91 m/s 2 . The acceleration of the box relative to the truck is a′ = a x − atruck , which gives a′ = − 4.91 m/s 2 − (−7.0 m/s 2 ) = 2.01 m/s 2 . (The + sign means the box is moving to the right relative to the truck.) If the box slides a distance x along the back of the truck, then its speed relative to the truck is given by (v ') 2 − (v ')02 = 2a′x, where v0’ = 0 because the box was initially at rest relative to the truck (it was moving with the same velocity as the truck). Thus v′ = 2ax = 2(2.01 m/s 2 )(2.0 m) = 2.8 m/s. 6-20.

35 × 1.852 = 18 m/s 3.6 Therefore, the angle of descent 0.46 θ = sin–1 = 1.46° 18 Ff = 7350 sin 1.46° = 188 N

35 knots =

FLIFT = 7350 cos 1.46° = 7348 N 6-21.

Force of gravity (along direction of road) must equal total frictional force. F = mg sin θ = 500 N, 500 where θ is the angle of the road to the horizontal sin θ = = 0.0340 1.5 × 103 × 9.8 θ = 1.9°


CHAPTER 6-22.

Choose the x axis to point down the slope and the y direction to point up and normal to the slope. Then ∑ Fy = N − mg cos 40° = 0, ∑ Fx = mg sin 40° − f k = max . Using f k = µ k N , solving the y equation for N, and substituting into the x equation gives max = mg sin 40° − µ k mg cos 40°, or ax = g (sin 40° − µ k cos 40°) = (9.81 m/s 2 ) [sin 40° − (0.10) cos 40°] 2

2 0

2

v −v (36.1 m/s) − 0 = = 2a 2(5.5 m/s 2 ) v − v0 36.1 m/s − 0 117 m. Final result = 1.2 × 102 m. t = = = 6.5 s. a 5.55 m/s 2 = 5.55 m/s2. 130 kph = 36.1 m/s. x =

†6-23.

A free body diagram is shown. As the plate is tilted, the block remains at rest as the static friction force increases from zero (at zero elevation) to its maximum value of µsN at some angle θ, and then the block begins to slide. At θ, since the block is at rest, the sums of forces along the plate and perpendicular to the plate are N = mg cosθ Dividing the second equation by the first gives sin θ µs = = tan θ . For θ = 38°, µs = 0.78. cosθ

6-24.

vT =

2mg = C ρ air A

N

fk mg cos 40o

mg 40o

⎛ 4πρ steel r 3 ⎞ 2⎜ ⎟g 3 ⎝ ⎠ = 2 C ρ airπ r

40o

mg sin 40o

N

fs mg cos q

µ s N = mg sin θ

mg

θ

θ

mg sin θ

8 ρ steel rg . The terminal speed is proportional to the square 3C ρ air

root of the ball’s radius, so proportional reasoning can be used: vT , 5 cm (5 cm) = ⇒ vT , 5 cm = (88 m/s) 50 = 3.9 × 102 m/s. (0.25 cm) vT , 0.25 cm †6-25.

6-26.

6

At terminal speed, a = 0, which means viscous drag force = weight ⇒ bvT = mg. mg (3.9 × 10−9 kg)(9.81 m/s 2 ) = = 1.4 × 10–3 m/s (1.4 mm/s). vT = −5 2.8 × 10 kg/s b Choose the + direction to point down. Newton’s Second Law dv dv bv gives ∑ F = mg − bv = ma = m ⇒ = g − . Rearrange dt dt m v t dv and integrate: ∫ = dt. Either look up the integral in a bv ⎞ ∫0 0 ⎛g − ⎜ ⎟ m⎠ ⎝ bv to do the integration table or use the substitution u = g − m v

m ⎛ bv ⎞ directly. The result is − ln ⎜ g − ⎟ = t , which b ⎝ m ⎠0


bv

mg

G v

CHAPTER

6

bv ⎞ ⎛ bt ⎜g− m ⎟ − bv bt m becomes ln ⎜ = − . Taking the antilog of both sides gives g − = ge . Rearranging ⎟ m g m ⎜⎜ ⎟⎟ ⎝ ⎠ bt − ⎞ mg ⎛ gives v: v = ⎜ 1 − e m ⎟. Note that this goes to mg/b as t → ∞, in agreement with the result in b ⎝ ⎠

Problem 6-25. The characteristic time is the time required for the speed to reach (1 – e–1)vT, which 3.9 × 10−9 kg = is tcharacteristic = m/b. For the data in Problem 6-25, this is tcharacteristic = 2.8 × 10−5 kg/s †6-27.

1.4 × 10–4 s (0.14 ms). Recall that fk = µkN. On a level road with no vertical forces acting except gravity, N = mg, so alevel = –µkg if friction is the only horizontal force acting. On a slope of angle θ, N = mg cosθ . If the car is initially moving down the slope, then there is a component of the car’s weight pointing down the slope and the friction force points up the slope. Taking the + direction to be downward (in the direction of the initial velocity), then the acceleration along the slope is aslope = g (sin θ − µ k cosθ ). If the car is going to slow down, the acceleration must be negative (point in the opposite direction from the initial velocity). For the slope, this means µ k cosθ > sin θ , or the car won’t be able to slide to a stop. For both cases, use v 2 = v02 + 2ax with v0 = 90 kph = 25 m/s and v = 0. Assume µk is the same on the level road and the slope. Solve for a on the level road and use that to find µk, then find a and x on the slope. Level road: v2 a (25 m/s) 2 8.93 alevel = − 0 = − = − 8.93 m/s 2 . µ k = − level = = 0.910. For a 1:10 slope, 2 xlevel 2(35 m) 9.81 g 1 θ = tan −1 = 5.71°. aslope = (9.81 m/s 2 )[sin 5.71° − (0.910) cos5.71°] = − 7.91 m/s 2 . 10 v2 (25 m/s) 2 = 39.5 m. xslope = − 0 = 2aslope 2(7.91 m/s 2 )

6-28.

Construct a free-body diagram as shown. T + m1 g sin θ – m1 g µ1 cos θ = m1 a (1) –T + m2 g sin θ – m2 g µ2 cos θ = m2 a (2) Adding (1) + (2) gives (m1 + m2)g sin θ – g(µ1m1 + µ2m2) cos θ = (m1 + m2)a g [(m1 + m2)sin θ – (µ1m1 + µ2m2) cos θ] (m1 + m2 ) 9.81 = (4.0 sin 50° − [0.60(2.0) + 0.40(2.0)]cos50°) m/s2 = 4.4 m/s 2 2.0 + 2.0 T = m1(a = g µ1 cos θ – g sin θ) from (1) = 2.0(4.36 + 9.81(0.60) cos 50° – 9.8 sin 50°) N = 1.3N ⇒a=


CHAPTER 6-29.

(b) N = F sin θ µs N + F cos θ = mg from (i) and (ii) mg F= µs sin θ + cos θ

6

(i) (ii)

⎛ µs cos θ − sin θ ⎞ ⎜ 2 ⎟ ⎝ ( µs sin θ + cos θ ) ⎠ = 0 if µs cos θ – sin θ = 0 or θ = tan −1µs

(c)

∂F = –mg ∂θ

(d) µs F cos θ = mg + F sin θ mg F= µs cosθ − sinθ for θ = tan–1 µs F → ∞ and it will become impossible to hold the book.

6-30.

Forces shown on free-body diagram. Normal force N = mg – t sin θ. Frictional force F = µk N = µk (mg – T sin θ ). To keep the box moving at constant velocity the net force = 0. Therefore F = T cos θ ⇒ µk (mg – T sin θ) = T cos θ ⇒ µk mg = T (cos θ + µk sin θ) ⇒ T = µk mg [cos θ + µk sin dT θ]–1 = – µk mg[cos θ + µk sin θ]–2[–sin θ + µk cos θ] = 0 at minimum tension This condition is dθ fulfilled at – sin µ + µk cos θ = 0 ⇒ tan θ = µ k . T=

µ k mg = cosθ + µ k sinθ

µ mg k

1 1 + µk2

= †6-31.

+

µk2 1 + µk2

µ k mg 1 + µ k2

In Example 3, change the angle from 30° to some arbitrary value θ . Then requirement for the minimum magnitude of P is P cosθ − µ k P sin θ − µ k mg = 0, or P (cosθ − µ k sin θ ) − µ k mg = 0. If cosθ − µ k sin θ < 0, then both terms in the equation will be negative and it will not be possible to find a value for P that satisfies the equation. If µ k sin θ − cosθ > 0, the crate will not move no 1 matter how hard it is pushed. This can be rewritten as µ k sin θ > cosθ , or tan θ > , or

µk

⎛ 1 ⎞ ⎟. ⎝ µk ⎠

θ > tan −1 ⎜


CHAPTER

6-32.

6

Let F1, F2 be frictional forces of masses as shown. F1 = µ1 N1 = µ 1 m1 g. F2 = µ 2 N2 = µ 2(m1 + m2)g Acceleration of masses: F − F1 Net Force = m1 : a = m m1 F − µ1m1 g = = F / m1 − µ1 g m1 F − F2 µ m g − µ 2 (m1 + m2 ) g m2 : a = 1 = 1 1 m2 m2 g = m2 ( µ1m1 − µ 2 [m1 + m2 ])

†6-33.

Free body diagrams for the two masses are shown. Let the tension on the string be T. Take the +x direction for m1 to point up the ramp. For m2, take the + direction to point up. The sum of forces perpendicular to the ramp is zero because there’s no motion in that direction. Thus N = m1 g cos35°. The magnitude of the kinetic friction force is µkN¸so the sum of forces along the x direction is T − µ k m1 g cos35° − m1 g sin 35° = m1a1 . For m2, the sum of forces gives T − m2 g = m2 a2 . If m2 moves down, then m1

N

T + T

f

m1g cos 35o m1g

35o

m2g

must move up the ramp. Thus the accelerations a1 and a2 m1 g sin 35o must have opposite signs. If the string connecting them doesn’t stretch and has negligible mass, then the accelerations must have the same magnitude. Thus we conclude that a2 = –a1, which we’ll just call a. Then the two force equations become T − m1 g (sin 35° + µ k cos35°) = m1a and m2 g − T = m2 a. Adding the g[m2 − m1 (sin 35° + µ k cos35°)] two equations eliminates T and gives a = m1 + m2 = 6-34.

9.81 m/s 2 [3.0 kg − (1.5 kg)(sin 35° + 0.40 × cos35°)] = 3.6 m/s 2 . 4.5 kg

Let the pull of the man be T. Then the free-body diagram is as shown. This means for constant velocity, the net force along the incline = 0. Therefore mg sin 25° + µk N = T cos 35°


CHAPTER ⇒ mg sin 25° + µk (mg cos 25° – T sin 35°) = T cos 35° ⇒ T(cos 35° + µk sin 35°) = mg(sin 25° + µk cos 25°) T= 6-35.

mg (sin25° + µk cos25°) 80 × 9.8(sin25° + 0.7cos25°) = = 680 N cos35° + 0.7sin35° cos35° + µk sin35°

Let T be the tension on the string. Let the trailing block have mass m1, the leading block mass m2. Draw the free-body diagram. Using F = ma gives (i) T cos θ + m1 g sin θ – 2µk(m1 g cos θ + T sin φ φ) = m1 a (ii) m2 g sin θ – T cos φ – µk(m2 g cos θ – T sin φ) = m2 a

or (1) T(cos φ – 2µk sin φ + m1 g sin θ – 2µk m1 g cos θ = m1 a (2) T(µk sin φ – cos φ) + m2 g sin θ – µk m2 g cos θ = m2 a To solve for T multiply (1) by m2, (2) by m1 Tm2(cos φ – 2µk sin φ) + m1 m2 g sin θ – 2µk m1 m2 g cos θ = m1 m2 a Tm1(µk sin φ – cos φ) + m1 m2 g sin θ – µk m1 m2 g cos θ = m1 m2 a. Subtracting one from the other gives T[m2(cos φ – 2µk sin φ) – m1(µk sin φ – cos φ) ]– µk m1 m2 g cos θ = 0 T(m2 cos φ – 2m2 µk sin φ – m1 µk sin φ + m1 µ cos φ) = µkm1 m2 g cos θ T = ( µ k m1 m2 g cosθ ) /[(m1 + m2 ) cos φ − (m1 + 2m2 ) µ k sin φ ]. To solve for a, multiply (1) by (cos φ – µk sin φ), (2) by (cos φ – 2µk sin φ) T(cos φ – 2µk sin φ)(cos φ – µk sin φ) + m1 g sin θ (cos φ – µk sin φ) – 2µk m1 g cos θ (cos φ – µk sin φ) = m1(cos φ – µk sin φ)a – T(cos φ – µk sin φ)(cos φ – 2µk sin φ) + m2 g sin θ (cos φ – 2µk sin φ) – µk m2 g cos θ (cos φ – 2µk sin φ) = m2 (cos φ – 2µk sin φ)a. Adding the latter two equations gives ⇒ m1 g sin θ (cos φ – µk sin φ) + m2 g sin θ (cos φ – 2µk sin φ) – 2 µk m1 g cos θ (cos φ – µk sin φ) – µk m2 g cos θ (cos φ – 2µk sin φ) = m1 (cos φ – µk sin φ)a + m2 (cos φ – 2µk sin φ)a a = g[ m1 (sinθ cosφ − µ k sinθ sinφ − µ k cosθ cos φ + 2 µ k2 cosθ sinφ )] + m2 (sinθ cosφ 2 µk sinθ sinφ µ k cosθ cosφ + 2µ k2 cosθ sinφ )] +{m1 (cosφ − µ k sinφ ) + m2 (cosφ − 2 µ k sinφ )}.


6

CHAPTER

6

6-36.

y

y

NB

Nm P 30o P cos 30o

P sin 30o

x

fs,m

fs,B

P cos 30o x 30o P sin 30o P Mg

mg

Box

man

Free-body diagrams are shown for the man and the box. Note that because of Newton’s Third Law, if the man pushes on the box with a force P directed towards the right at 30o below horizontal, then the box must push back on the man with a force P directed towards the left at 30o above the horizontal. The mass of the box is given as M, and the mass of the man is m. Since the box is being pushed toward the right, the static friction force on it points to the left. Because the box is pushing toward the left on the man, the static friction force on him points to the right. The normal forces on the man and the box are labeled Nm and NB, respectively. The diagrams show the resolution of the P force into x and y components. The heaviest box the man can push will result in the largest friction forces acting on the man and the box, so each friction force will be equal to the appropriate coefficient of friction multiplied by the appropriate normal force. (a) Box: ∑ Fy = N B − P sin 30° − Mg = 0; ∑ Fx = P cos30° − f s , B = 0 Man:

∑F

y

= N m + P sin 30° − mg = 0;

∑F

x

= f s , m − P cos30° = 0

Solve for NB and use f s ,B = µ s ,B N B to find P in terms of M: P =

µ s , B Mg . Now solve cos30° − µ s , B sin 30°

for Nm and use f s ,m = µ s ,m N m : µ s , m mg − µ s ,m P sin 30° − P cos30° = 0. Substitute the expression

µ s , B Mg (cos30° + µ s ,m sin 30°) = µ s ,m mg . g cancels cos30° − µ s , B sin 30° µ (cos30° − µ s , B sin 30°) m. Now substitute from the equation, which can be solved for M: M = s ,m µ s , B (cos30° + µ s ,m sin 30°) for P gotten from the equations for the box:

the numerical data: M =

(0.80)[cos30° − (0.20)sin 30°] (75 kg) = 182 kg. (0.20)[cos30° + (0.80)sin 30°]

(b) If the man is pushing upward on the box, the signs of the sin 30o terms are reversed; nothing else changes in the problem. The equation for the maximum mass becomes µ (cos30° + µ s , B sin 30°) M = s ,m m, and substituting the numerical data gives M = 622 kg. µ s , B (cos30° − µ s ,m sin 30°)


CHAPTER †6-37.

The force exerted by the spring is F = –kx, where x is the displacement of the end of the spring. By Newton’s Third Law the force required to change the length of the spring is equal in magnitude and opposite in direction to the force exerted by the spring. Therefore the magnitude of the force required to stretch the spring to twice its length is: ⎜F⎪= k⎜x⎪ = 150 N/m × 0.15 m = 23 N. To compress it to one-half its length, x = (0.15 m)/2 = 0.075 m. The magnitude of this force is ⎜F⎪ = kx = 150 N/m × 0.075 m =11 N.

6-38.

Spring constant k =F/x Force due to gravity = mg Therefore F mg 1.5 × 9.8 k= = = N/m = 73.5 N / m 0.20 x x 1.0 N In stretching from 6.3 to 1.2 cm, k = = 0.26 N/cm = 26 N/m. In stretching from (10.2 − 6.3) cm 2.0 N 6.3 to 16.5 cm, k = = 0.20 N/cm = 20 N/m. Since k is not constant, the spring (16.5 − 6.3) cm

†6-39.

does not obey Hooke’s Law. F

=

mg (70 kg)(9.81 m/s 2 ) = = 4.3 × 103 N/m. ∆x 0.16 m

=

3.5 N = 50 N/m. 0.070 m

6-40.

k=

6-41.

k=

6-42.

F = k ∆x = (4.8 × 10−2 N/m)(2 × 10−11 m) = 9.6 × 10–13 N/m.

6-43.

k=

6-44.

k=

6-45.

x=

∆x F ∆x

F ∆x F ∆x F k

=

mg (0.250 kg)(9.81 m/s 2 ) = = 8.8 × 102 N/m. 2.8 × 10−3 m ∆x

=

mg (75 kg)(9.81 m/s 2 ) = = 2.5 × 102 N/m. ∆x 2.9 m

=

mg (1500 kg)(9.81 m/s 2 ) = = 1.1 × 10–3 m (1.1 mm). 7 k 1.4 × 10 N/m


6

CHAPTER 6-46.

6

k is the spring constant of the original piece. |F | = k ∆x. Now for some |F |, the shorter piece stretches 1/3 ∆x, the longer 2/3 ∆x. Let ks , ke be the spring constants for the shorter and longer pieces, respectively. ⎛1 ⎞ For the shorter piece: |F | = ks ⎜ ∆x ⎟ = k∆x ⇒ k s = 3k ⎝3 ⎠ 3 ⎛2 ⎞ For the longer piece: |F | = ke ⎜ ? x ⎟ = k∆x ⇒ ke = k 2 ⎝3 ⎠ Therefore the shorter piece has ks = 360 N/m. The longer piece has ke = 180 N/m.

6-47.

Since the springs are in parallel, they both must suffer identical extensions. Let this be x. Let the force due to the spring with constant k1 be F1, and that for k2 be F2. Then, by Hooke’s law, F1 = k1 x, F21 = k2 x. But by Newton’s law, F = F1 + F2. Therefore, F = F1 + F2 = k1 x + k2 x = (k1 + k2)x. For the two springs F = kx therefore, k1 + k2 = k .

6-48.

Suppose spring 1 stretches ∆x1 and spring 2 stretches ∆x2. The force pulling on the end of each spring is F1 = k1∆x1 and F2 = k2 ∆x2 , respectively. Assume spring 1 is at the end where the external force F is applied, so F = F1. Then the magnitude of the net force on that spring would be F2 − F . If the springs are in equilibrium, then the net force on each must be zero, which means

⎛1 1⎞ = F ⎜ + ⎟ , since the k1 k2 ⎝ k1 k2 ⎠ forces must be the same. This means the combined springs act like a single spring with an F ⎛1 1⎞ 1 1 1 = F ⎜ + ⎟ , or = + . effective force constant given by ∆x = k k k1 k2 ⎝ k1 k2 ⎠ F2 = F . The total stretch of both springs is ∆x = ∆x1 + ∆x2 =

6-49.

Let the force that the feet exert on the on the floor be N. Then the force that the floor exerts on the feet (and hence the man) = N. The diagram shows the net force on the man at the top and at the bottom of the ferris wheel. 2π (30) m = 3.14 m/s. So that with F = mv2/r, we have v= 60 s F = 80 × 3.142/30 = 26.3 N At top: mg – Ntop = F; Ntop = mg – F = (9.8 × 80) – 26 = 757 N At bottom: Nbot – mg = F; Nbot = mg + F = (9.8 × 80) + 26 = 810 N


F

+

F2

CHAPTER 6-50.

†6-51.

6-52.

Centripetal force = mv2/r ; v = 2πr/T (T = period of rotation) mv 2 m(2π r )2 4π 2 mr 4π 2 (7.3 × 1022 ) × 3.8 × 108 kg m = = = 2.0 × 1020 N F= = rΤ 2 Τ2 (27 × 24 × 60 × 60) 2 s 2 r The motion described in this problem isn’t really uniform circular motion because the woman’s speed varies during the swing. However, she is moving in a circular arc, so the centripetal force mv 2 , where v is her instantaneous acting on her at any point is the swing is still given by F = r speed . Strictly speaking, the forces acting on the woman are her weight and the normal force exerted by the seat. However, if we neglect the mass of the seat, then the normal force will be the same as the total tension in the two ropes that support the seat. So at the bottom of the swing, the net vertical force acting on the woman is 2T – mg, which must be the centripetal force at that mv 2 v 2 + gr ⇒T =m . The radius is the length of the rope (5.0 m), and her point: 2T − mg = r 2r (5.0 m/s) 2 + (9.81 m/s 2 )(5.0 m) speed is 5.0 m/s, so T = (60 kg) = 4.4 × 102 N. 2(5.0 m) The condition for not sliding is that the maximum force of friction, mv 2 or v ≤ µ s rg Ff = µs mg ⇒ F = r At v = µs rg = 0.5 × 90 × 9.8 = 21m/s the coins will start to slide (the frictional force will be inadequate to keep the coins accelerating towards the center with the car).

†6-53.

6

The net force in the vertical direction is zero because the car doesn’t move in that direction. However, the net horizontal force cannot be zero because a centripetal force is needed to make the car go in a circle. That force comes from the horizontal component of the normal force. From the free-body diagram, N cos θ = mg, N sin θ = mv2/r. (Note that even though the diagram looks very much like one for an object on an inclined plane, the motion being described is quite different and requires a different approach to the analysis.) Dividing v2 . The the second equation by the first gives tan θ = rg required speed is 75 kph = 20.8 m/s. Then

⎛ v2 ⎞ ⎞ (20.8 m/s) 2 –1 ⎛ = tan = 6.3° ⎟ ⎜ 2 ⎟ ⎝ rg ⎠ ⎝ (400 m)(9.81 m/s ) ⎠

θ = tan–1 ⎜


N N cos θ

N sin θ

θ mg

CHAPTER 6-54.

6

The normal force N provides the centripetal force keeping the rider moving in a circle. The minimum value of N is the one for which the static friction force is at its maximum value µsN. If the rider doesn’t slide down, then fs,max = mg = µsN. Since N is the centripetal mg mv 2 rg force, N = = ⇒v= is the minimum r µs µs

fs,max

N

linear speed the rider must have not to slide down.

mg (6.0 m)(9.81 m/s 2 ) = 15.3 m/s. Let T stand for the time for one complete revolution. The 0.25 2π r 2π (6.0 m) = = 2.46 s, from rider travels through one circumference in time T, so T = v 15.3 m/s 1 which we get a rotation frequency f = = 0.41 rev/s. T Static friction provides the centripetal force that keeps the ant moving in a circle. The maximum force the surface can exert will occur at the maximum distance the ant reaches N mv 2 . The before sliding, which is f s ,max = µ s N = µ s mg = rmax v=

†6-55.

fs

speed is the circumference of the circle divided by the time T for one revolution, and T is the reciprocal of the 2π r revolution rate f ⇒ v = = 2π rf . Combining all these T µs g (0.30)(9.81 m/s 2 ) = = 0.13 m gives rmax = 2 4π 2 f 2 2 2 ⎛ 1 min ⎞ 4π (45 rev/min) ⎜ ⎟ ⎝ 60 s ⎠ (13 cm).

6-56.

When the skateboarder is at the bottom, the net vertical force must provide the centripetal acceleration. Thus mv 2 N − mg = , where h is the radius of the circle. h According to the hint, v = 2 gh , so N = mg +

m(2 gh) = 3mg . The normal force is 3 × the h

weight.

mg

N h

h

mg


CHAPTER †6-57.

6-58.

†6-59.

The time T for one complete orbit is 24 h = 8.64 × 104 s, and the satellite’s speed is the circumference of its orbit divided by T. The “weight” is the centripetal force, so ⎛ 2π r ⎞ m⎜ ⎟ 2 2 2 7 mv T ⎠ 4π mr 4π (1 kg)(4.23 × 10 m) = ⎝ = = = 0.224 N. w= r r T2 (8.64 × 102 s) 2 This is analogous to a car going around a banked curve. The horizontal component of the normal force provides the centripetal acceleration, and the vertical component of the normal force balances the passenger’s weight: mv 2 N sin θ = r N cosθ = mg ⎛ v2 ⎞ ⎡ ⎤ (140 m/s) 2 Divide the top equation by the bottom: θ = tan −1 ⎜ ⎟ = tan −1 ⎢ ⎥ = 18°. 2 3 ⎝ gr ⎠ ⎣ (9.81 m/s )(6.0 × 10 m) ⎦ The vertical component of the tension in the cable balances the weight of the rider, and the horizontal component provides the centripetal acceleration. Call the mass of the rider m and the cable tension T. Then 7m θ mv 2 T sin θ = ; T cosθ = mg , where v is the tangential r 3m speed of the rider. Dividing the first equation by the second v2 r gives tan θ = . The radius r is (3 + 7sin θ ) m, so gr v2 = tan θ . This gives v = g (3 + 7sin θ )

g tan θ (3 + 7sin θ ).

For θ = 35°, v = 6.9 m/s. 6-60.

6

For the cars to travel at maximuim speed without skidding mvA2 mvB2 = Fs = µs mg = RA RB vA2 R = A 2 vB RB since RA < RB, vA < vB Therefore,

Time taken by the two cars to travel through the curved paths ⎛Rθ⎞ Rθ tA = A = ⎜ A2 ⎟ vA vA ⎝ vA ⎠ ⎛Rθ⎞ RB θ = ⎜ B2 ⎟ vB vB ⎝ vB ⎠ Since the quantities in brackets are equal and VA < VB, tA < tB, therefore car A emerges from the curve first. tB =


CHAPTER 6-61.

6

The net force on the car is mg – N downward where N = normal force of road on car. At the speed where the car just loses contact with the road, N = 0, so that mg = mv2/r ⇒ v2/r = g or v = gr = 9.8 × 50 m/s = 22 m/s = 80 km/hr

6-62.

6-63.

If the water is moving fast enough to stay in the bucket at the highest point in the swing, the bucket will be pushing down on the water with a normal force N. The total force, which is the mv 2 . At the centripetal force, at the top will be N + mg = r minimum speed, N = 0 at the top. (If the speed is less than this, N will go to zero before the top, which means the water is not touching the bottom of the bucket—it’s falling out.) Thus 2 mvmin ⇒ vmin = gr = (9.81 m/s 2 )(0.90 m) = 3.0 m/s. mg = r

mg

N

60 mi/hr = 88 ft/s. From the free-body diagram, with N being the normal force acting on the motorcycle (at angle θ with the vertical), mg = N cos θ (1) 2 mv (2) centripetal force = N sin θ = r mv 2 N= r sin θ Therefore from (1) mv 2 cos θ mg = N cos θ = r sin θ v2 cot θ r v2 tan θ = gr g=

2

⎛ 96 ⎞ 2 ⎜ ⎟ ⎛ ⎞ v 3.6 ⎠ θ = tan–1 ⎜ ⎟ = tan–1 ⎝ = 67° 9.8 × 30 ⎝ gr ⎠ 6-64.

The maximum friction force is µsmg, which gives an acceleration of a = – µsg. (The – sign means the acceleration points in the opposite direction from the velocity.) The distance to stop is v2 v2 v2 = . The radius of the turn required is r = from Example 10. This is twice the x=− 2a 2 µ s g µs g stopping distance, so it’s safer for the driver to stop than to attempt the turn.


CHAPTER †6-65.

A free body diagram is shown. T is the tension in the string and v is the speed of the ball as it moves in the circle. The sum of forces in the vertical direction is zero, and the horizontal component of the tension provides the centripetal force that keeps the ball moving in mv 2 , its circle. Thus T cosθ = mg , T sin θ = r where r = l sin θ is the radius of the circle. Divide the second equation by the first: sin θ v2 = tan θ = . This means the speed of cosθ gr the ball must satisfy v =

6

T cos θ T

θ θ

v

T sin θ

mg

gr tan θ .

Substituting for r gives v =

gl sin θ tan θ .

6-66.

Let N and F be the normal and frictional forces on the car, respectively. Summing forces horizontally and vertically gives N cos θ = mg + F sin θ (1) N cos θ + F cos θ = m(v2/r) = centripetal force (2) These give 0.9945 N – 0.1045 F = 11760 (3) 0.1045 N + 0.9945 F = 1875 (4) Solving these simultaneously gives N = 1.19 × 104 N F = 640 N

6-67.

See Problem 6-58. tan θ =

(88.9 m/s) 2 v2 v2 = ⇒r= . v = 320 kph = 88.9 m/s. r = (9.81 m/s 2 )(tan 30°) gr g tan θ

1.40 × 103 m (1.40 km). 6-68.

Both balls must take the same time to go around their respective circles. If we call this time t, r 2π r1 2π r2 and v2 = then v1 = , which says v2 = v1 2 . For m1, the centripetal force is the tension t t r1 in the string: T = T sin θ =

m1v12 . For m2, the centripetal force is the horizontal component of the tension: r1

m2 v22 . Dividing the second equation by the first to eliminate T and using the relation r2

⎛m ⎞ between the speeds gives r2 = r1 ⎜ 1 ⎟ sin θ . ⎝ m2 ⎠


CHAPTER 6-69.

6

When the balls are first released, their instantaneous velocities are both zero. That means the centripetal acceleration of each ball at that instant is equal to zero, so the tension in each string must balance the component of that ball’s weight that points along the string. This will be mg cos θ for each ball: T − m1 g cosθ1 = 0 T − m2 g cosθ 2 = 0

T

θ mg

Rewrite each of these with the T on one side of the equal sign and divide one by the other. The cosθ1 m2 result is = . Because m1 > m2, we conclude that cosθ2 < cosθ1. That means θ2 > θ1. cosθ 2 m1 When the masses reach the bottom of their swing after being released, v2 > v1 because both masses take the same time to reach the bottom and m2 must move farther than m1. Now for the tension to be the same in both segments of string when the segments are vertical, Newton’s Second Law requires m v2 T − m1 g = 1 1 l m v2 T − m2 g = 2 2 l where l is the length of the string. If we subtract the first equation from the second, we get m v 2 − m2 v22 (m1 − m2 ) g = 1 1 . However, this condition can’t be satisfied because m2 − m1 > 0 but l m1v12 − m2 v22 < 0 since m2 > m1 and v2 > v1. Thus it’s not possible for these masses to swing without having the heavier one fall down. 6-70.

2 ⎛ dθ ⎞ σ R dθ v –N + 2T sin ⎜ ⎟ = R ⎝ 2 ⎠ dθ dθ Using sin ≅ 2 2

N = T dθ – σ dθv2

= (T – σ v2) dθ N = 0 for v =

T

σ


CHAPTER 6-71.

Take an element of string, making an angle dθ with the center. Let T be the tension. As shown, the net effect of the tensions at the ends of the element is to produce a net force inward equal to 2 T sin (dθ/2) = T dθ because sin dθ/2 ≈ dθ/2, when dθ/2 is small. Mass of the element is rdθ dθ s m= m= m 2π r 2π 2π Then (dθ/2π)m v2/r = T dθ gives T = mv 2 / 2π r

6-72.

The speed of a point at a distance r from the dm = ρdr 2π r hub is v = , where t is the time for one t r complete revolution, 1/320 min. Let ρ = mass per unit length of material. Then the centripetal force required to accelerate element dm at distance r from the hub is v 2 dm v 2 ρ dr (2π ) 2 ρ rdr dF = . Therefore the tension = = r r t2 r=L (2π ) 2 ρ L (2π ) 2 ρ L2 (2π ) 2 ML rdr . t = 1/320 min = 0.1875 s, so = = is T = ∫ dF = r =0 2t 2 2t 2 t 2 ∫0 2π 2 (140 kg)(3.6 m) T = = 2.83 × 105 N. 2 (0.1875 s)

6-73.

Let the x- and y-axes be as shown. Look at a plumb-line at a point at latitude θ. The force due to gravity on the bob is –mg cos θ î – mg sin θ ˆj Let the plumb-line be at an angle φ to direction of earth’s radius as shown. Let the tension on string be T. Then the force on the bob due to the tension on the string is T cos (θ + φ) î + T sin (θ + φ) ˆj Thus, the net force on the bob is (T cos (θ + φ) – mg cos θ) î + (T sin (θ + φ) – mg sin θ) ˆj


6

CHAPTER

6

Thus, the net force on the bob is (T cos (θ + φ) – mg cos θ) î + (T sin (θ + φ) – mg sin θ) ˆj The net force in the ˆj direction must be 0, so T sin (θ + φ) =

mg sin θ (1) ˆ The net force in the i direction must equal the centripetal force. Therefore, mv 2 ˆ i –mg cos θ + T cos (θ + φ) î = – r m(2π R cosθ/t )2 ˆ =− i R cosθ (t = period of rotation, R = radius of earth) m 4π 2 R cos θ – mg cos θ + T cos (θ + φ) = – t2 ⎛ 4π 2 R ⎞ (2) gives T cos (θ + φ ) = m cos θ ⎜ g − 2 ⎟ t ⎠ ⎝

(2) (3)

(1) divided by (3) gives g tan θ tan (θ + φ) = 4π 2 R g− 2 t ⎛ ⎞ ⎡ ⎛ ⎞⎤ ⎜ ⎟ ⎢ ⎥ ⎜ ⎟ g tan θ 9.81tan θ −1 ⎜ ⎟ −θ  ⎥ − φ = ⎢ tan −1 ⎜ θ = tan ⎟ 2 ⎜ 9.81 − 4π 2 6,378,000 ⎟ ⎢ ⎜ g − 4π R ⎟ ⎥ ⎜ ⎟ ⎜ ⎢⎣ (24 × 60 × 602 ) ⎟⎠ t 2 ⎠ ⎥⎦ ⎝ ⎝ = tan −1 (1.00345 tan θ ) − θ

At θ = 45°

φ = tan–1 (1.00345 tan 45°) – 45° = tan–1 1.00345 – 45° = 0.099° 6-74.

Draw the free-body diagram, including all forces acting on the car—weight, normal, frictional. The net force on the car is: [N cos 10° – mg – F sin 10°] ˆj + [N sin 10° + F cos 10°] î

But the maximum value for F, which will give the maximum possible speed, is F = µs N.


CHAPTER

6

Therefore, Fnet = [N cos 10° – mg – µs N sin 10°] ˆj + [N sin 10° + µs N cos 10°] î (at maximum speed) The net force in the ˆj direction is zero, so that N (cos 10° – µs sin 10°) = mg mg N= (1) cos 10°µs sin10° The force in the î direction must equal the centripetal force: mv 2 r Substituting (1) into (2) gives mv 2 mg (sin10° + µs cos10°) = cos10° − µs sin10° r

N (sin 10° + µs cos 10°) =

(2)

1/ 2

⎡ rg (sin10° + µs cos10°) ⎤ v= ⎢ ⎥ ⎣ cos10° − µs sin10° ⎦ = 39 m/s = 140 km/hr †6-75.

=

120 × 9.8(sin10° + 0.9cos10°) cos10° − 0.9sin10°

In Problem 6-60, an equation was derived relating the speed of the mass at the end of the pendulum and the angle the pendulum makes with the vertical. Using the variables in this v2 . However, the speed we need is v0, which is the speed at a problem gives tan α = g ( R + l sin α ) distance R from the center. To figure out v0 in terms of v, we must recognize that the time for one complete revolution is the same for all parts of the apparatus. This time is the circumference of 2π R 2π ( R + l sin α ) = , from which we one of the circles divided by the corresponding speed, so v0 v get v = v02 =

6-76.

( R + l sin α ) 2 v02 ( R + l sin α )v02 R + l sin α v0 . Then tan α = = , which gives R gR 2 ( R + l sin α ) gR 2

gR 2 tan α . For the numerical data given, v0 = R + l sin α

(9.81 m/s 2 )(0.20 m) 2 (tan 45°) = 2.2 m/s. 0.20 m + (0.30 m)(sin 45°)

(a) minimum thrust = 2.45 × 9.8 × 106 = 2.4 × 107 N (b) 3.3 × 107 – 2.4 × 107 = 2.45 × 106 a (3.3 − 2.3)107 = 3.67 m/s 2 a= 6 2.45 × 10 (c) Assuming the thrust is constant during the flight, 9 × 106 = 12 m/s 2 a= 0.75 × 106


CHAPTER †6-77.

6

f s , max = µ s N = µ s mg .

∑F

N

= − µ s mg = max ⇒ ax = − µ s g .

x

v 2 − v02 = 2ax x ⇒ x =

v 2 − v02 v2 v2 = − 0 = 0 . v0 = 90 2a x 2ax 2 µ s g

fs

kph = 25 m/s. (25 m/s) 2 x= = 40 m. 2(0.80)(9.81 m/s 2 ) t= 6-78.

v0 mg

v − v0 v 25 m/s = 0 = = 3.2 s. ax µ s g (0.80)(9.81 m/s 2 )

(a) Take the x direction to point up the slope. If there’s no friction, ∑ Fx = −mg sin θ = max ⇒ ax = − g sin θ . The – sign

N

means the acceleration points in the opposite direction from the 1 car’s initial velocity. θ = tan −1 = 9.46°, so 6 ax = − (9.81 m/s 2 )sin 9.46° = 1.61 m/s 2 . Final result = –1.6 m/s2. v 2 − v02 v2 (b) v − v = 2ax x ⇒ x = = − 0 . v0 = 48 kph 2ax 2ax 2

2 0

mg

mg cos θ mg sin θ

= 13.3 m/s. (13.3 m/s) 2 x=− = 55 m. 2(−1.61 m/s 2 ) †6-79.

θ

(a)The free-body diagram is shown at the right. It is assumed that the initial velocity points to the right (+ x), so the kinetic friction force points left (– x). (b) ∑ Fy = N − mg = 0 ⇒ N = mg = (40 kg)(9.81 m/s 2 ) = 392 N. Final result N = 3.9 × 102 j N. (c) f k = µ k N = (0.80)(392 N) = 314 N. Final result f k = − 3.1 × 102 i N.

N

fk

(d) w = –mgj = 3.9 × 102 j N. Fnet = ∑ Fx i + ∑ Fy j = − f k i = −314i N. Final result Fnet = −3.1 × 102 i N. F −314i N (e) a = net = = − 7.85i m/s 2 . Final result a = – 7.9i m/s2. m 40 kg x=

v 2 − v02 0 − (22.2 m/s) 2 . v0 = 80 kph = 22.2 m/s. x = = 31 m. 2ax 2(−7.85 m/s 2 )


mg

CHAPTER 6-80.

(a) The free-body diagram is shown. Take the x direction to point up the slope and the y direction to be pointing rightward and normal to the slope (in the direction of N in the diagram). (b) ∑ Fy = N − mg cos30° = 0 ⇒ N = mg cos30°

N

fs

= (2.0 kg)(9.81 m/s 2 )cos30° = 17.0 N. Final result N = 17 N. (c) ∑ Fx = f s − mg sin 30°. Since we’re told the block is resting on the slope, this sum of forces must be zero ⇒ f s = mg sin 30° = (2.0 kg)(9.81 m/s 2 )sin 30° = 9.81 N. Final result for this part = 9.8 N. As a check, let’s verify that this is less than the maximum possible static friction force: f s ,max = µ s N = (0.90)(17.0 N) = 15.3 N. This is greater than the

30° mg

mg sin 30°

calculated friction force, so our result of 9.8 N is reasonable. (d) Fplane on box = N + f s = 9.81i + 17.0 j N. The direction of this force is tan −1 6-81.

6-82.

17.0 = 60° above the slope or 90° above the horizontal. 9.81

Let the spring be compressed ∆x. Then, to balance forces 4k ∆x = mg (weight of car) mg (1200 × 9.8)N = ∆x = = 0.15 m (4 × 2.0 × 104 )N/m 4k The paper is, for all practical purposes, massless. The net force on paper always must add to zero. The maximum frictional force of paper on surface is µs mg cos θ = 0.5 mg cos θ. Therefore, the maximum frictional force of the paper on the block = 0.5 mg cos θ. Block slips (together with paper) when weight force mg sin θ = 0.5 mg cos θ ⇒ θ = tan −1 (0.5) = 27°

†6-83.

N fk

T T

m1g


mg cos 30°

m2g

6

CHAPTER

6

(a) Take the x direction to point right and the y direction to point up. Newton’s Second Law for m1 gives ∑ Fy = N − m1 g = 0, ∑ Fx = T − f k = m1a1 . For m2, ∑ Fy = T − m2 g = m2 a2 . We see that N = m1 g ⇒ f k = µ k mg ⇒ T − m1 g = m1a1 . If the string passing over the pulley does not stretch, then the magnitude of a2 must be the same as the magnitude of a1. If m1 moves in the +x direction, then m2 must move in the –y direction, which means a2 = –a1. Then the equations for m −µ m ∑ Fx for m1 and ∑ Fy for m2 can be combined to give an equation for a1: a1 = m2 + mk 1 g. 1 2 6-84.

This is exactly the same as 6-36 but with P at an angle of 0° instead of 30°. We can use either of the equations derived for M from 6-36 with 30° changed to 0°: µ (cos 0° − µ s , B sin 0°) µ m (0.80)(75 kg) = 300 kg. M = s ,m m = s ,m = µ s , B (cos 0° + µ s , m sin 0°) µs,B 0.20

†6-85.

The wording in the problem implies the springs are in series with one spring hanging from the end of the other. Suppose spring 1 has force constant k1 = 2.0 × 103 N/m and spring 2 has force constant k2 = 3.0 × 103 N/m. Let us further suppose that spring 1 is at the bottom, with the mass m hanging from it. We assume that the mass and the springs are in equilibrium, so the force stretching spring 1 is mg. Then spring 1 stretches ∆x1 because of that force. Spring 1 exerts a force on spring 2 causing it to stretch by ∆x2. If the springs and the mass are in equilibrium, then the net force on each object must be zero. Since there is a force mg pulling down on spring 1, this same force must be pulling on spring 2. Then for each spring, mg = k1∆x1 and mg = k2 ∆x2 , ⎛1 1⎞ mg mg + = mg ⎜ + ⎟ . k1 k2 ⎝ k1 k2 ⎠ This means the combined springs act like a single spring with an effective force constant given by ⎛1 1⎞ mg 1 1 1 ∆x = = mg ⎜ + ⎟ , or = + . k k k1 k2 ⎝ k1 k2 ⎠

respectively. The total stretch of both springs is ∆x = ∆x1 + ∆x2 =

For the springs in this problem, k =

k1k2 6 × 106 N N = = 1.2 × 103 . The magnitude of the 3 m k1 + k2 5 × 10 m

total force stretching the “spring” is mg = (5 kg)(9.81 m/s 2 ) = 49.1 N. The total stretch of both springs is ∆x = ∆x1 = 6-86.

F k1

=

F k

=

49.1 N = 0.0409 m (4.1 cm). The individual stretches are 1.2 × 103 N/m

F 49.1 N 49.1 N = 2.5 cm, and ∆x2 = = = 1.6 cm, which do indeed 3 2.0 × 10 N/m k2 3.0 × 103 N/m

give a total stretch of 4.1 cm to two significant figures. (a) If the spring is massless, then the force the spring exerts on the block will be the same as the force exerted on the spring. To start the block moving, the pull P must exceed the maximum static friction force µ s N . Since the only vertical forces acting are the weight mg and the normal force N, N = mg = (1.5 kg)(9.81 m/s2) = 14.7 N and f s ,max = µ s mg = (0.60)(14.7 N) = 8.83 N. The stretch of the spring under this force will be P 14.7 N ∆x = = = 7.4 × 10−3 m (7.4 mm). k1 1.2 × 103 N/m


N f

P

mg

CHAPTER

6

(b) When the block begins to move, the force changes from static friction to kinetic friction. Take the x direction to point to the right. Then ∑ Fx = P − f k = P − µ k mg = max , which gives P − µ k mg 8.83 N − (0.40)(14.7 N) = = 2.0 m/s 2 . m 1.5 kg (c) “Constant speed” means ax = 0, so P = µ k mg = (0.40)(14.7 N) = 5.88 N. Now the stretch of ax =

the spring is ∆x = †6-87.

P k1

=

5.88 N = 4.9 × 10–3 m (4.9 mm). 1.2 × 103 N/m

(a)If the spring is massless, then the force the spring exerts on the block will be the same as the force exerted on the spring. To start the block moving, the pull P must exceed the total force pointing down the slope, which is the sum of the maximum static friction force µ s N and the component of the block’s weight parallel to the slope. Adding forces normal to the slope gives ∑ Fy = N − mg cos30° = 0 ⇒ N = mg cos30°

N P f 30°

= (1.5 kg)(9.81 m/s 2 ) cos30°

mg cos 30°

mg

= (14.7 N)cos 30° = 12.7 N. Parallel to the slope,

∑F

x

= P − mg sin 30° − µ s N = 0

⇒ P = (14.7 N)sin 30° + (0.60)(12.7 N) = 15.0 N. The stretch of the spring under this force will

be ∆x =

P k1

=

15.0 N = 1.2 × 10−2 m (1.2 cm). 1.2 × 103 N/m

(b) When the block begins to move, the force changes from static friction to kinetic friction. Now P − mg sin 30° − µ k N P − mg sin 30° − µ k N = max , which gives ax = . m 15.0 N − (14.7 N)sin 30° − (0.40)(12.7 N) = = 1.7 m/s2. The positive result means the 1.5 kg acceleration points up the incline. (c) “Constant speed” means ax = 0, so P = mg sin 30° + µ k N = 7.34 N + (0.40)(12.7 N) = 12.4 N. Now the stretch of the spring is ∆x = 6-88.

P k1

=

12.4N = 1.0 × 10–2 m (1.0 cm). 3 1.2 × 10 N/m

Free-body diagrams are shown at the right. For m1, the vertical forces add to zero and the tension in the string provides the centripetal acceleration. For m2, the net force is m v2 zero, so T = m2g. Thus for m1, T = 1 = m2 g ⇒ r m2 gr . v= m1


T

N

T

m1g

m2g

CHAPTER 6-89.

6-90.

6

The maximum frictional force, Fs = µs mg = 0.8 mg = 7.84 m N. The centripetal force mv 2 m ⎛ 75 ⎞ 2 = ⎜ ⎟ = 9.64 m N 45 ⎝ 3.6 ⎠ R mv 2 Since > Fs, the car will skid. R Centripetal force = mv2/r ; at top of circle, the tension = 0 so the only force on the stone is the weight of the stone = mg. Then mg = mv2/r ⇒ v = gr. Let T be tension on the string at the bottom. Thus T – mg = centripetal force = mv2/r = mg Therefore: T = 2 mg = 2 × 0.9 × 9.8 = 17.6 N


CHAPTER 7

WORK AND ENERGY


7-2.

†7-3.

7-4.

7-5.

The work done by a force F is W = Fs cos θ, where s is the displacement and the angle θ is the angle between the force and the displacement vectors. Since the force applied to the car is in the same direction as the displacement, the angle is 0° and cos 0° = 1. W = Fs cos θ = (300 N)(5.0 m)cos 0° = 1500 J = 1.5 × 103 J During the lift, the force exerted by the weight lifter must be directed upward and is equal in magnitude to the downward force of gravity on the barbell, assuming the lift is done at constant velocity. The magnitude of the weight of the barbell is mg. Therefore, the work done is W = Fy∆y = mg∆y = (254 kg)(9.81 m/s2)(1.98 m) = 4930 J = 4.93 × 103 J. Since the force applied by the weight lifter is in the same direction as the barbell is displaced, the work done on the barbell is positive. With each movement, either forward or backward, the displacement and the force are in the same direction. Therefore, positive work is done on the saw. W = Fx∆x = (35 N)[(30 × 0.12 m) + (30 × 0.12 m)] = 252 J The work done on the bucket to raise it is W = Fy∆y, where the upward force on the bucket is Fy and it is raised a distance ∆y from the bottom of the well to the top. Assuming the bucket is raised at a constant speed, the magnitude of the upward force is equal to the weight mg of the bucket. Using the definition of work, solve for ∆y. W 2200 J ∆y = = = 15 m Fy (15 kg)(9.81 m/s 2 )

W = Fx ∆x = f k ,1∆x1 + f k ,2 ∆x2 = N [ µ k ,1∆x1 + µ k ,2 ∆x2 ] = mg[ µ k ,1∆x1 + µ k ,2 ∆x2 ] = (20 kg)(9.81 m/s2)[(0.25)(10 m) + (0.55)(30 m)] = 3700 J

7-6.

†7-7.

7-8. †7-9.

7-10.

The forward and backward motions are represented in parts (a) and (b) of the figures, respectively. In both cases, the horizontal component of the force is in the direction of the displacement and positive work is done on the vacuum. The total work is then W = F s cos θ = 300 (40 N)(1.0 m) cos 60° = 6000 J. W = F∆y = mg∆y = (75 kg)(9.81 m/s2)(320 m) = 235 000 J The average rate of doing work is W 235 000 J = = 357 J/s . ∆t 659 s W = F s cos θ = (50 N)(1.6 m) cos 60° = 40 J The first tugboat does positive work W1 on the barge, where W1 = F1 s cos θ1 = (2.5 × 105 N)(100 m) cos 30° = 2.2 × 107 J. Similarly, the second boat does positive work W2 on the barge, where W2 = F2 s cos θ2 = (1.0 × 105 N)(100 m) cos 15° = 1.0 × 107 J. The total work done by both tugboats is then W = W1 + W2 = 3.2 × 107 J. Wg = Fg∆y = − mg∆y = −(2.0 kg)(9.81 m/s2)(4.0 m) = −78 J


F

(a)

60°

s

(b)

F 60° s

CHAPTER †7-11.

7-12.

7-13.

7-14.

7

N P is the push applied by the man. From the free-body diagram, N = mg cos θ and P = mg cos θ + f, where P f = µkN = µk mg cos θ is the frictional force. Therefore, h P = mg cos θ + µk mg cos θ. To raise the box to a height mg cos θ f + mg sin θ h, the box is pushed up the incline a distance s = h/sin θ. θ The work done on the box by the man is W = Ps = (mg cos θ + µk mg cos θ)(h/sin θ), which may be simplified to W = mgh(1 + µk cot θ) = (60 kg)(9.81 m/s2)(2.5 m)[1 + (0.45) cot 30°] = 2.6 × 103 J. The free-body diagram shows the three forces acting on the car as N it coasts downhill: the normal force N, the gravitational force mg, Ff and the combined frictional forces Ff. The angle θ is found from the given slope, θ = tan−1(1/20). Since the car coasts at constant mg sin θ mg cos θ θ speed, the magnitude of the frictional force is equal to the θ component of the gravitational force directed down the incline, Ff = mg sin θ = (1200 kg)(9.81 m/s2) sin[tan−1(1/20)] = 590 N. This assumes that the mass of the driver is included in the 1200 kg. The work done by the frictional forces is W = Ff s = (−590 N)(1000 m) = −5.9 × 105 J. W = Fgs = (mg sin θ)s = (1700 kg)(9.81 m/s2)(sin 25°)(10 m) = 7 × 104 J The work done by the frictional force is −7 × 104 J, since the net work on the car is 0 J. Then 0 J = Wg + Wf + WN = 7 × 104 J + Wf + 0 J and Wf = −7 × 104 J The frictional force due to air resistance is f air = 12 C ρ Av 2 , where the aerodynamic constant for

the car is C = 0.30, ρ is the density of air, the cross-sectional area for this car is A = 2.8 m2. Using this information, the work done on the car by air resistance is W20 = f air ∆x = 12 C ρ Av 2 ∆x

= 12 (0.30)(1.3 kg/m3 )(2.8 m3 )(20 m/s) 2 (250 000 m) = 5.5 × 107 J. When the speed is increased to 30 m/s, the work done increases to W30 = 12 (0.30)(1.3 kg/m3 )(2.8 m3 )(30 m/s)2 (250 000 m) = 1.2 × 108 J.

7-15. 7-16.

†7-17.

θ = cos−1(W/Fs) = cos−1[(175 J)/(25 N)(12 m)] = 54° As the man walks north, the component of the wind force F that acts on him is F cos 30°. When he walks east, the component is F sin 30°. In both cases, positive work will be done. The total work is the sum of the work done during each segment. W = F1s1 + F2s2 = (F cos 30°)s1 + (F sin 30°)s2 = (150 N)(cos 30°)(100 m) + (150 N)(sin 30°)(200 m) = 2.8 × 104 J (a) If the cart is to move at constant velocity, the magnitude of the applied force T1 must equal that of the kinetic frictional force fk, which is oppositely directed. The work the man does on the cart is W1 = T1∆x = (250 N)(50 m) = 1.3 × 104 J. T (b) When the force is applied in the direction shown, work is 30° fk done by the horizontal component of the tension T. Because the vertical component of the tension would reduce the normal force and also the frictional force. In this problem, additional mass has been added so that the frictional force is the same in both cases. To determine the tension, we assume that the resulting horizontal acceleration is equal to zero, so that T2 cos 30° − fk = 0. Solving for T gives, T2 = fk/(cos 30°) = (250 N)/(cos 30°) = 290 N.


CHAPTER

7

The work the man must now do is W2 = T(cos 30°)∆x = (290 N)(cos 30°)(50 m) = 1.3 × 104 J, which is the same as W1. 7-18. †7-19.

W =

∫

x2

x1

y2

Fx dx + ∫ Fy dy = y1

∫

2

0

−1

3dx + ∫ 2dy = 3 x 0 + 2 y 0 = 6 J − 2 J = 4 J 2

−1

0

(a)The net force on the elevator is T1 − Mg = Ma, where T1 is the tension in the elevator cable and M is the elevator mass including its load. T1 = M(g + a) = (1200 kg)(9.81 m/s2 + 1.5 m/s2) = 1.4 × 104 N For the counterweight, the tension is T2 = m(g − a) = (1000 kg)(9.81 m/s2 − 1.5 m/s2) = 8.3 × 103 N. (b)To determine the distance traveled in the 1.0 s interval, apply the kinematic relation, y = 12 at 2 = 12 (1.5 m/s 2 )(1.0 s) = 0.75 m.

T1 T1

Mg Assuming the elevator is moving upward during the interval, the work done by the pulley motor is W = T1y + T2y = (1.4 × 104 N)(0.75 m) − (8.3 × 103 N)(0.75 m) = 4300 J. (c) For motion at constant speed, T1 = Mg = (1200 kg)(9.81 m/s2) = 1.2 × 104 N and T2 = mg = (1000 kg)(9.81 m/s2) = 9.8 × 103 N The work done during the upward displacement is W = (T1 − T2 ) y = (1.2 × 104 − 9.8 × 103 )(10.0 m) = 2.2 × 104 J.

7-20.

T2

T2

mg

Since the sled has no vertical acceleration, the vertical component T N of the net force on the sled can be used to find the magnitude of fk 30° the normal force, N = mg − T sin 30°. Since the sled moves with constant speed, the magnitude of the horizontal component of the mg tension equals that of the kinetic friction force, fk = T cos 30°. By applying the relationship between the normal force and friction, the tension in the rope can be determined. µ k N = T cos30°

µ k (mg − T sin 30°) = T cos30° Solving for the tension, gives µ k mg (0.10)(150 kg)(9.81 m/s 2 ) (1000 m) = 160 N. = T = cos30° + µ k sin 30° cos30° + (0.10)sin 30°

†7-21.

The work done by the girl on the sled is equal to the product of the horizontal component of the tension and the distance traveled, W = (T cos 30°)∆x = (160 N)(cos 30°)(1000 m) = 1.4 × 105 J. (a) The angle the rope makes with respect to the horizontal direction is θ = sin−1(10 m/50 m). Since the boat is stationary, the horizontal component of the tension must be equal in magnitude to the wind force. Thus, T cos θ = 7000 N, which can be solved for the tension 7000 N 7000 N T = = = 7100 N. cos θ ⎡ −1 ⎛ 10 m ⎞ ⎤ cos ⎢sin ⎜ ⎟⎥ ⎝ 50 m ⎠ ⎦ ⎣ (b) Since the wind force is 7000 N, the sailors must exert 7000 N in the direction of motion assuming the boat is to be pulled at constant speed. If 30 m of rope is pulled in, the hypotenuse is reduced to 20 m. The horizontal distance is found from the Pythagorean theorem, x2 = (20 m) 2 − (10 m) 2 = 17 m


CHAPTER

7

Similarly, the original distance was x1 = (50 m) 2 − (10 m) 2 = 49 m So, the net displacement of the boat is (49 m − 17 m) = 32 m and the work done is W = (7000 N)(32 m) = 2.2 × 105 J. In this new position, the tension is 7000 N 7000 N T = = = 8100 N. cos θ ⎡ −1 ⎛ 10 m ⎞ ⎤ cos ⎢sin ⎜ ⎟⎥ ⎝ 20 m ⎠ ⎦ ⎣ 7-22.

To compress the spring 0.10 m, the work done on the spring is W = 12 k ( x22 − x12 ) = 12 (3.5 × 104 N/m)[(0.10 m) 2 − (0 m) 2 ] = 180 J. To compress the spring by another 0.10 m, requires additional work equal to W = 12 k ( x22 − x12 ) = 12 (3.5 × 104 N/m)[(0.20 m)2 − (0.10 m)2 ] = 530 J.

†7-23.

7-24.

The work is equal to the area under the Fx versus x curve. The work done in moving between 0 and 6 m is + 8 J. Between 6 and 8 m, the work done is −2 J. Therefore, the total work done is W = +8 J − 2 J = +6 J. FS The spring constant can be determined from the information given 0.25 kg since the sum of the vertical forces is k(∆y) − mg = 0, k = mg/(b − a), mg where a and b are the initial and final positions, respectively. The free body diagram shows the two forces acting on the object hung from the spring. The work done by gravity on the object is Wg =

∫

b

a

Fy ( y )dy =

∫

b

a

b

−mg dy = − mg ∫ dy a

= − mg (b − a) = − (0.25 kg)(9.81 m/s 2 )[(−0.18 m) − (0 m)] = 0.44 J. The work done by the spring is b b mg WS = ∫ Fy ( y )dy = ∫ − ky dy = − 12 k (b 2 − a 2 ) = − (b 2 − a 2 ) a a 2(b − a )

=− †7-25.

(0.25 kg)(9.81 m/s 2 ) [(−0.18 m) 2 − (0 m) 2 ] = 0.22 J. 2( − 0.18 m − 0 m)

The spring force is a restoring force, which is directed so as to return the spring to its equilibrium position. In this problem, the mass attached to the end of the spring is displaced in the positive x direction, but as the mass passes x = 0 m, the spring force changes direction. Since the force is varying, the work done by the spring is found through integration. WS =

∫

b

a

Fx ( x)dx =

∫

0.40 m

−0.20 m

− kx dx = − 12 kx 2

0.40 m −0.20 m

2

= − (440 N/m)[(0.40 m) − (−0.20 m) 2 ] = −26 J 1 2

7-26.

Someone applies a force on the spring to stretch it by 0.027 m, so positive work is done on the spring. The force is equal in magnitude to the restoring force of the spring, kx. Therefore, the work done on the spring is WS =

∫

b

a

Fx ( x)dx =

∫

0.027 m

0

kx dx = 12 kx 2 2

0.027 m 0 2

= (4500 N/m)[(0.027 m) − (0 m) ] = 1.6 J. 1 2


CHAPTER

†7-27.

7

1 To stretch the spring from 0 to d, requires work W0 = − kd 2 . The amount of work to further 2 stretch the spring to 2d is W =

∫

2d

d

Fx ( x)dx =

∫

2d

d

− kx dx = − 12 kx 2

2d d

= − 12 k[(2d ) 2 − (d ) 2 ] = − 32 kd 2 = 3W0 . If we now wish to generalize this for greater distances, the work done would be W =

∫

( N +1) d

Nd

kx dx = 12 kx 2 2

( N +1) d Nd

= − k[( N + 1) d − ( Nd ) 2 ] = − 12 k (2 N + 1)d 2 = (2 N + 1)W0 . 1 2

7-28.

2

The work done by the non-ideal spring is WNI =

∫

b

a

Fx ( x)dx =

∫

b

a

b

(6 x − 2 x 3 ) dx = 3 x 2 − 12 x 4 . a

We will compare this with the work done by an ideal spring, which is WI = For x = 0 to 0.50 m, WNI = 3x 2 − 12 x 0.50 m

and WI = 12 kx 2

0

4 0.50 m 0

∫

b

a

b

kx dx = 12 kx 2 . a

= 3(0.50 m) 2 − 12 (0.50 m) 4 = 0.72 J.

= 12 (6.0 N/m)(0.50 m) 2 = 0.75 J.

For x = 1.0 to 1.50 m, WNI = 3x 2 − 12 x 4 and WI = 12 kx

1.5 m

1.0 m 0.50 m 2 0

= [3(1.5 m) 2 − 12 (1.5 m) 4 ] − [3(1.0 m) 2 − 12 (1.0 m) 4 ] = 1.7 J = 12 (6.0 N/m)[(1.5 m) 2 − (1.0 m) 2 ] = 3.8 J.

For x = 2.0 to 2.50 m, WNI = [3(2.5 m) 2 − 12 (2.5 m) 4 ] − [3(2.0 m) 2 − 12 (2.0 m) 4 ] = 3.2 J and WI = 12 (6.0 N/m)[(2.5 m) 2 − (2.0 m) 2 ] = 6.8 J. †7-29.

In the vertical direction, the magnitude of the external force is equal to twice the vertical component of the restoring force of the spring Fy. Consider one-half of the spring as represented in the triangle below the figure to the right. The extension of one-half of the spring from its equilibrium length of l/2 is equal to ∆l =

(l 2)

2

2Fy F F

l 2

+ y 2 − l 2.

The applied force is P = 2 F sin θ = 2k (∆l )

θ

Fx

Fx

y

(l 2)

2

+y

2

( )

l 2 2

.

θ +y

y

2

where F is the magnitude of the force exerted by the spring. Substituting the equation for ∆l gives ⎡ ⎤ y y (l / 2) ⎥ 2 2 ⎡ ⎤ ⎢ P = 2k ( l 2 ) + y − l 2 = 2k y − 2 2 ⎢⎣ ⎥⎦ ⎢ (l 2) + y2 ( l 2 ) + y 2 ⎥⎦ ⎣

⎡ P = 2ky ⎢1 − ⎢ ⎣

⎤ ⎥. 2 2 ⎥ l 2 + y ( ) ⎦ (l / 2)


CHAPTER

7

The work done on the spring by the force P is ⎡ ⎤ y y (l / 2) ⎥ dy W = ∫ Pdy = 2k ∫ y ⎢1 − 2 ⎢ 2 ⎥ 0 0 (l 2) + y ⎦ ⎣ ⎧ y2 l = 2k ⎨ − ⎡ ⎩ 2 2 ⎢⎣

(l 2)

2

⎫ ⎡ l2 l + y 2 − (l / 2) ⎤ ⎬ = k ⎢ y 2 + + ⎥⎦ ⎭ 2 2 ⎣

(l 2)

⎤ + y 2 ⎥. ⎦

2

As a check, the work done by the external force on the spring in displacing it by a distance y and extending it by ∆l is equal to the change in the spring potential energy, W = 12 k (∆l ) 2 . Then, the total work done on the whole spring is twice this value, 2 2⎞ 2 ⎛1 W = 2 ⎜ k ( ∆l ) ⎟ = k ⎡ ( l 2 ) + y 2 − l 2 ⎤ ⎣⎢ ⎦⎥ ⎝2 ⎠

7-30. †7-31.

which is the same result obtained when the quantity in the square brackets is expanded. 2m 2 x2 2 3 2 2 W = ∫ F ( x) dx = ∫ (2 x + 8 x) dx = x + 4 x = 21 J 0 x1 3 0 Consider the path from the origin to the point (2 m, 2 m, 0). This is a straight line path on which x = y and, therefore, dx = dy. The work done on this path is W = =

∫

x2

∫

x2

x1

0

y2

y1

z1

y2

(4x 2 + 1) dx + ∫ 2 x dy = 0

= 43 x 3 + x + x 2 7-32.

z2

Fx dx + ∫ Fy dy + ∫ Fz dz

2m 0

∫

2m

0

(4x 2 + 1) dx + ∫

2m

0

2 x dx

= 17 J.

Consider the free body diagram for this situation. The horse is pulling upward along the arc with a force F1 for a distance x at constant speed. The magnitude of F1 is equal to the magnitude of F2, which is the sum of frictional force f = µkN and the component of the weight directed downward along the arc, mg sin θ. The magnitude of the normal force is equal to the component of the weight perpendicular to the arc, N = mg cos θ. Thus, F1 = µk mg cos θ + mg sin θ. The work done by the horse is W =

∫

b

F ( x) dx =

∫

x

θ R F1

N F2

mg ( µ k cosθ + sin θ ) dx .

mg

θ Since x = Rθ, where the angle θ goes from 0 to 45° (π/4 rad) and dx = Rdθ. The work is then, a

W = mgR ∫

π /4

0

= mgR ( µ k

0

π /4

( µ k cosθ + sin θ ) dθ = mgR ( µ k sin θ − cosθ ) 0 ⎡ ⎤ 2 2 2 − + 1) = mgR ⎢1 + ( µ k − 1) ⎥. 2 2 2 ⎣ ⎦

Now, consider the same sled pulled up a 22.5°-incline to a height h = R(1 − distance up the incline is x = R(1 − W =

∫

b

a

F ( x) dx =

∫

x

0

2 2

2 2

). In this case, the

)/(sin 22.5°). The work done in this case is then, x

mg ( µ k cosθ + sin θ ) = mg ( µ k cosθ + sin θ ) ∫ dx 0

R (1 − 22 ) = mgR ( µ k cot 22.5° + 1)(1 − 22 ). sin θ Comparing the two expressions for the work done, we find they are nearly identical. = mg ( µ k cosθ + sin θ )


CHAPTER †7-33.

7

(a) When the two atoms are close to each other, the force is repulsive. At longer distances, the force is attractive; and at the equilibrium separation, the force is zero. Setting the force equal to zero and solving for x, gives F = Ax −13 − Bx −7 = 0

x=

A . B

6

(b) The work done is equal to the W =

∫

∞

xeq

Fx dx =

∫

∞

xeq

∞

Ax

−13

⎛ x −12 ⎞ ⎛ x −6 ⎞ B2 B2 B2 − Bx dx = A ⎜ − = − = − . B ⎟ ⎜ ⎟ 12 A ⎝ 12 ⎠ ⎝ 6 ⎠ xeq 12 A 6 A −7

2

7-34. †7-35.

⎡⎛ 160 km ⎞⎛ 1000 m ⎞⎛ 1 h ⎞ ⎤ K = mv = (0.060 kg) ⎢⎜ ⎟⎜ ⎟⎜ ⎟ ⎥ = 59 J ⎣⎝ h ⎠⎝ 1 km ⎠⎝ 3600 s ⎠ ⎦ The Earth-Sun distance is R =1.5 × 1011 m and the orbit is approximately circular, so the circumference of the Earth’s orbit is C = 2πR = 9.4 × 1011 m. The Earth travels this distance in one year = 3.156 × 107 s, so the average speed of the Earth is v = C/t = 3.0 × 104 m/s. The Earth’s mass is 5.98 × 1024 kg. The kinetic energy of the Earth in its orbit is 2

1 2

1 2

K = 12 mv 2 = 12 (5.98 × 1024 kg) ( 3.0 × 104 m/s ) = 2.7 × 1033 J. 2

7-36. 7-37.

K = 12 mv 2 = 12 (9.11 × 10−31 kg) ( 2.2 × 106 m/s ) = 2.2 × 10−18 J 2

KSkier

⎞⎤ 5 ⎟ ⎥ = 1.3 × 10 J ⎠⎦

⎡⎛ 44.88 km ⎞⎛ 1 h ⎞⎛ 1000 m 1 1 = mv 2 = (75 kg) ⎢⎜ ⎟⎜ ⎟⎜ 2 2 1h ⎠⎝ 3600 s ⎠⎝ 1 km ⎣⎝ 1.3 × 105 J = = 22 5.8 × 103 J

⎞⎤ 3 ⎟ ⎥ = 5.8 × 10 J ⎠⎦

K Runner K Skier K Runner 7-38.

2

⎡⎛ 212.52 km ⎞⎛ 1 h ⎞⎛ 1000 m 1 1 = mv 2 = (75 kg) ⎢⎜ ⎟⎜ ⎟⎜ 2 2 1h ⎠⎝ 3600 s ⎠⎝ 1 km ⎣⎝

2

K = 12 mv 2 = 12 (1770 kg)(120 m/s) 2 = 1.3 × 107 J ⎛ 1 kg TNT ⎞ mTNT = 1.3 × 107 J ⎜ ⎟ = 2.8 kg 6 ⎝ 4.6 × 10 J ⎠

†7-39.

7-40.

1 2 (1600 kg) ( 22.2 m/s ) = 4.0 × 105 J 2 1 2 (b) 20 km/hr = 5.56 m/s, so K = 12 mv 2 = (1600 kg) ( 5.56 m/s ) = 2.5 × 104 J 2 1 2 (c) 140 km/hr = 38.9 m/s, so K = 12 mv 2 = (1600 kg) ( 38.9 m/s ) = 1.2 × 106 J 2 ⎛ 95 km/h ⎞ At 95 km/h, the probability P95 = 3% ⎜ ⎟ = 4.2%. ⎝ 80 km/h ⎠ ⎛ 110 km/h ⎞ ⎛ 125 km/h ⎞ Similarly, P110 = 3% ⎜ ⎟ = 5.7% and P125 = 3% ⎜ ⎟ = 7.3% . ⎝ 80 km/h ⎠ ⎝ 80 km/h ⎠ (a) 80 km/hr = 22.2 m/s, so K = 12 mv 2 =


CHAPTER

7-41.

7-42.

7

⎛ 1 kg ⎞ 2 6 K 0 = 12 mv 2 = 12 (100 lb) ⎜ ⎟ (657 m/s) = 9.8 × 10 J 2.205 lb ⎝ ⎠ 1 kg ⎛ ⎞ 2 6 K 3 s = 12 mv 2 = 12 (100 lb) ⎜ ⎟ (502 m/s) = 5.7 × 10 J 2.205 lb ⎝ ⎠ ∆K = −4.1 × 106 J 2 K Bullet = 12 mBullet vBullet = 12 (0.015 kg)(630 m/s) 2 = 3.0 × 103 J 2 K Ball = 12 mBall vBall = 12 (15 kg)(6.3 m/s) 2 = 3.0 × 102 J

†7-43.

The kinetic energy of the bullet is ten times greater than that of the bowling ball. 2 K Ball = 12 mBall vBall = 12 (0.045 kg)(45 m/s) 2 = 46 J 2 K Person = 12 mPerson vPerson = 12 (75 kg)(1.0 m/s) 2 = 38 J

7-44.

The kinetic energies of the golf ball and person are on the same order of magnitude. The work done on the ball by the spring will be equal to the stored potential energy of the compressed spring. W = 12 kx 2 = 12 (30 N/m)(0.080 m) 2 = 0.096 J Assuming all of the spring potential energy is used to launch the ball, the kinetic energy will equal the work done on the ball, 0.096 J. 2K 2(0.096 J) The initial speed of the ball when it is launched will be v = = = 3.1 m/s. m 0.020 kg

†7-45.

7-46.

The kinetic energy of the object as it is launched is K = 12 mv 2 = 12 (0.150 kg) ( 5.0 m/s ) = 1.9 J. 2

The work done on the object is equal to the change in kinetic energy of the object. Since the object started from rest, the work is W = ∆K = 1.9 J. This energy was also the stored energy in the compressed spring. Therefore, 2W 2(1.9 J) x= = = 0.44 m. 20 N/m k Assuming no other forces are acting on the player, 2K 2(250 J) ∆K = W = F ∆x = (500 N)(0.50 m) = 250 J and v = = = 2.9 m/s. m 60 kg

7-47.

K = 12 mv 2 = 12 (1300 kg) ( 3100 m/s ) = 6.2 × 109 J

7-48.

Assume the initial potential energy of the stone is zero at the point it is released. The initial total mechanical energy is equal to the initial kinetic energy, K(0). All of this energy becomes potential energy at the maximum height, h. Therefore, K(0) = mgh. At some intermediate point y, the stone has one-half its initial kinetic energy plus a potential energy that must be one-half its final potential energy. The total mechanical energy is K(0) = K(y) + mgy = mgh h Therefore, mgy + mg h2 = mgh which gives y = 2 Let d be the depth the bullet penetrates when it is fired with a speed v. Since d is assumed to be proportional to the kinetic energy, we may write the relationship as d = kv2, where k is some constant. The two cases may then be compared to find the speed of the second bullet.

†7-49.

2


CHAPTER

7

d1 = kv12 or 0.8 cm = k(160 m/s) d 2 = kv22 or 1.2 cm = k v22

Taking the ratio of these two equations yields, v2 = 160 7-50.

(a) W =

∫

x2

x1

Fx dx =

∫

x2

x1

1.2 = 196 m/s. 0.9

− ax + bx3 dx

= − a ∫ x dx + b ∫ x 3 dx = − 12 a ( x22 − x12 ) + 14 b ( x24 − x14 )

†7-51.

x2

x2

x1

x1

(b) The work found in (a) is equal to the change in kinetic energy, ∆K. The stored potential energy in the spring is equal to the work done in moving it to its initial position, W = 12 kx12 . After the cantilever is released, the cantilever has both kinetic energy and spring potential energy, the sum of which equals the work done. Therefore, 2 2 1 1 2 kx1 = K + 2 kx2 K = 12 k ( x12 − x22 ) = 12 (2.5 × 10−2 N/m)[(3.0 × 10−8 m) 2 − (2.5 × 10−8 m) 2 ] = 3.4 × 10−18 J

7-52.

(a) F = ma = (1500 kg)(−8.0 m/s 2 ) = 1.2 × 103 N (b) Using 90 km/h = 25 m/s and applying the kinematic formula, a( x − x0 ) = 12 (v 2 − v02 ) , x=

v 2 − v02 0 − (25 m/s) 2 = = 39 m 2a 2(−8.0 m/s 2 )

(c) Assuming the braking force is constant, the work is W = Fx cos180° = − (1200 N)(39 m) = −4.7 × 104 J. (d) The change in the kinetic energy is the same as the work done by the braking force, ∆K = W = −4.7 × 104 J. †7-53.

(a) ∆K = K2 − K1 = ½ (40 kg)(2.0 m/s)2 = 80 J (b)Wf = fk • x = −fkx=−µkNx Applying Newton’s second law in the vertical direction, N = mg. The sum of the horizontal forces is P − fk = ma or P − µ k mg 250 N − (0.60)(40 kg)(9.81 m/s 2 ) a= = = 0.36 m/s 2 . (40 kg) m We can find the sliding distance from x =

7-54. †7-55.

v 2 − v02 (2 m/s) 2 − 0 = = 5.5 m. 2a 2(0.36 m/s 2 )

The work done by the friction force is Wf = −µkNx = −µkmgx = −(0.60)(40 kg)(9.81 m/s2)(5.5 m) = −1295 J. (c) Ww = P • x = Px = (250 N)(5.5 m) = 1375 J. Note that the total work done on box is W = Ww + Wf = 80 J = ∆K. W =∆U = mg∆y = (3200 kg)(9.81 m/s2)(6.0 m) = 1.9 ×105 J W = mg∆y = (75 kg)(9.81 m/s2)(10 m) = 7.4 ×103 J The metabolization of one apple is 4.6 ×105 J, so in climbing the stairs (7.4 ×103 J)/(4.6 ×105 J) = 1.6 % of the energy stored in an apple is used.


N fk

P

mg

CHAPTER

7-56.

†7-57.

7-58.

7

⎛ 30 km ⎞⎛ 1 h ⎞⎛ 1000 m ⎞ v=⎜ ⎟⎜ ⎟⎜ ⎟ = 8.3 m/s ⎝ 1 h ⎠⎝ 3600 s ⎠⎝ 1 km ⎠ K = 12 mv 2 = 12 (6.0 kg)(8.3 m/s) 2 = 210 J U = mgh = (6.0 kg)(9.81 m/s2)(90 m) = 5300 J The gravitational potential energy of the water is U = mgh. The mass of the water can be determined from its density ρ = m/V = 1000 kg/m3. To determine the volume, combine these two relations, m U 2.0 × 1013 J = = 8.2 × 106 m 3 V = = ρ ρ gh (1000 kg/m3 )(9.81 m/s 2 )(250 m)

The car has a total energy E = K + U. Since frictional effects are ignored, the energy of the car will be conserved. So, at the bottom of the hill, the car will have the same total energy as it had when it was released. Because the reference height is arbitrary in the determination of gravitational potential energy, it is convenient to choose the bottom of the hill as h = 0 m. The speed of the car when it is at a height of 60 m and heading downward is 13 km/h = 3.6 m/s. 2 2 1 1 2 mv1 + mgh1 = 2 mv2 + mgh2 total energy at release 1 2

total energy at bottom

2 1

mv + mgh1 = 12 mv22

Divide through by the mass m and solve for the speed v2. v2 = v12 + 2 gh1 = (3.6 m/s) 2 + 2(9.81 m/s 2 )(60 m) = 34.5 m/s or 124 km/h

†7-59.

Applying the principle of the conservation of mechanical energy to this situation 1 mv12 + mgh1 = 12 mv22 + mgh2 2 1 2

mv12 = mgh2

h2 =

7-60.

v12 (10 m/s) 2 = = 5.1 m 2 g 2(9.81 m/s 2 )

However, the athlete actually reaches a height of 5.7 m. In the process of performing the vault, the athlete exerts an additional amount of internal energy in pulling himself upward that results in the additional 0.6 m height. 1 mv12 + mgh1 = 12 mv22 + mgh2 2 total energy at top

total energy at bottom

2 2

mgh1 = mv 1 2

v2 = 2 gh1 = 2(9.81 m/s 2 )(45 m) = 29.7 m/s

†7-61.

Use conservation of energy, since friction effects are negligible. 1 mv12 + mgh1 = 12 mv22 + mgh2 2 mgh1 = 12 mv22

which gives

v2 = 2 gh1 = 2(9.81 m/s 2 )(500 m) = 99 m/s

The kinetic energy of the 2.0 × 107 kg of snow is K = 12 mv 2 = 12 (2.0 × 107 kg)(99 m/s) 2 = 9.8 × 1010 J. The equivalent amount of TNT is 9.8 × 1010 J N= = 23 tons of TNT. 4.2 × 109 J/ton


CHAPTER 7-62.

7

The difference in her total energy between her jump and her landing is equal to the amount of work done by frictional forces, W = ∆E. Therefore, W = E2 − E1 = ( K 2 + U 2 ) − ( K1 + U1 ) On the aircraft flying parallel to the ground below, she has a velocity equal to that of the aircraft and a corresponding kinetic energy. However, we are interested in her speed in the vertical direction, so we’ll assume her kinetic energy as she leaves the airplane is K1 = 0 J. Choose the ground as our potential energy reference, U2 = 0 J. W = K 2 − U1 = 12 mv 2 − mgh = 12 (60 kg)(5.0 m/s) 2 − (60 kg)(9.81 m/s 2 )(800 m) = −4.7 × 105 J

†7-63.

Consider the free-body diagram superposed on the drawing. In the direction perpendicular to the inclined plane, N − mg cos θ = 0 or N = mg cos θ. The distance the block slides is s = (1.5 m)/sin(15°) = 5.8 m; and the work done by the frictional force on the block is W = E2 − E1 = ( K 2 + U 2 ) − ( K1 + U1 )

fk 1.5 m

N mg

15°

= K 2 − U1 = − fs

which is then used to find the coefficient of kinetic friction, since fk = µkN. mgh1 − 12 mv22 = µ k Ns

µk = 7-64.

1 2

mgh1 − 12 mv22 mgh1 − 12 mv22 2 gh1 − v22 2(9.81 m/s 2 )(1.5 m) − (3.5 m/s) 2 = = = = 0.16 Ns (mg cosθ ) s 2( g cosθ ) s 2(9.81 m/s 2 )(cos15D )(5.8 m)

mv12 + mgh1 = 12 mv22 + mgh2

mgh1 = 12 mv22 v2 = 2 gh1 = 2(9.81 m/s 2 )(148 m) = 53.9 m/s †7-65.

The difference in the total energy between the release from rest of the ball and its landing is equal to the amount of work done by frictional forces, W = ∆E = (K2 − U2) − (K1 − U1). The fraction lost to air friction is equal to W/U1. W = K 2 − U1 = 12 mv22 − mgh1 W = U1

7-66.

1 2

mv22 − mgh1 v 2 − 2 gh1 (9.0 m/s)2 − 2(9.81 m/s 2 )(20 m) = 2 = = 0.79 or 79%. 2 gh1 2(9.81 m/s 2 )(20 m) mgh1

Since frictional effects are ignored, conservation of energy applies. 1 mv12 + mgh1 = 12 mv22 + mgh2 2 mgh1 = 12 mv22 + mgh2 h2 =

†7-67.

gh1 − 12 v22 (9.81 m/s 2 )(45 m) − 12 (15 m/s) 2 = = 33.5 m (9.81 m/s 2 ) g

Since frictional effects are ignored, conservation of energy applies. 1 mv12 + mgh1 = 12 mv22 + mgh2 2 mgh1 = 12 mv22 + mgh2 v2 = 2 g ( h1 − h2 ) = 2(9.81 m/s 2 )(3.0 m) = 7.7 m/s


CHAPTER 7-68.

7

(a) Consider the free-body diagram superposed on the N T drawing. Ignoring frictional forces, in the direction parallel to the incline, the sum of the forces is T − mg sin 12 m mg θ θ = 0. Since the distances are given, the angle θ = sin−1(12 m/500 m) = 1.38°. Each metric ton is 1000 kg. ⎛ 1000 kg ⎞ 2 4 T = mg sin θ = (70 t + 35 t) ⎜ ⎟ ( 9.81 m/s ) sin (1.38° ) = 2.5 × 10 N ⎝ 1t ⎠ (b) The work done is equal to the change in the gravitational potential energy of the barge and carriage, W = ∆U = mg(h2 − h1) = (1.05 × 105 kg)(9.81 m/s2)(12 m) = 1.2 × 107 J. (c) Using conservation of energy, 12 mv12 + mgh1 = 12 mv22 + mgh2 mgh1 = 12 mv22 v2 = 2 g ( h1 − h2 ) = 2(9.81 m/s 2 )(12 m) = 15 m/s

†7-69.

When the ball is displaced from its equilibrium position, it is raised to a height h relative to its lowest position before it is released from rest. To find the height h, a horizontal line is drawn from the ball’s initial position to the vertical line. A right triangle is then formed, from which we can find the length of the side labeled b. The initial height is then, h = (10 m) − b = (10 m) − (10 m) cos 35° = (10 m)(1 − cos 35°). We can then apply the principle of conservation of mechanical energy to find the ball’s kinetic energy at its lowest point where its potential energy U2 = 0. K 2 + U 2 = K1 + U1

35° b h

K 2 = U1 K 2 = mgh1 = (600 kg)(9.81 m/s 2 )(10 m)(1 − cos35°) = 1.1 × 104 J 7-70.

When the stone is launched, it has a total energy E1 = K1 + U1 = K1 = 12 mv12 . If frictional forces do work W1 on the stone as it flies upward, then the total energy is reduced to a new value E2 = K2 + U2 = mgh2 = E1 + W1 = 12 mv12 + W1= 12 mv12 − fh2 (the maximum height the stone reaches is less than it would have attained without air friction). In the downward portion of the journey, the energy of the stone is further reduced by frictional forces and the stone reaches the ground with a total energy E3 = K3 + U3 = K3 = 12 mv12 − 2fh2. As a result the landing speed v3 < v1. To understand the time difference, we need to determine the net acceleration during each part of the flight. During the upward portion, we have 2f f ⎞ ⎛ mgh2 = 12 mv12 − fh2 or v12 = 2 gh2 − h2 = 2h2 ⎜ g − ⎟ m m ⎝ ⎠ v2 − v2 From kinematics, for constant acceleration, we have the relation, a = 2 1 . 2y So, during the upward portion of the flight, the acceleration is a =

−v12 . The net acceleration is 2h2

negative since the frictional force and gravitational are both directed downward as the stone moves upward. The time for the upward flight can be found from another kinematic relation,


CHAPTER

7

4h22 2h2 . = f ⎞ f ⎞ ⎛ ⎛ 2h2 ⎜ g − ⎟ ⎜g − ⎟ m⎠ m⎠ ⎝ ⎝ The ratio of f/m will always be less than g. Applying a similar analysis to the downward portion of the flight, 4f 2f ⎞ ⎛ v32 = 2 gh2 − h2 = 2h2 ⎜ g − ⎟ m m ⎠ ⎝ y = 12 at 2 or, in this case, t1 =

†7-71.

4h22 = v32

4h22

4h22 = v12

2h2 2f ⎞ 2f ⎞ ⎛ ⎛ 2h2 ⎜ g − ⎟ ⎜g − ⎟ m ⎠ m ⎠ ⎝ ⎝ To see that t2 > t1, note that the denominator under the radical for t1 is larger than that for t2. 1 At position 1, choosing the downward direction as positive, the sum of the forces on the stone is mv 2 T1 + mg = 1 2 r Since the tension is about zero, the speed at the top of the circle is v12 = rg. Therefore, the total energy is and t2 =

2h2 = a

2h2 = a

=

E = K1 + U1 = 12 mv12 + mgh1 = 12 mrg + 2mrg = 52 mrg

4

3

At position 3, the tension is directed upward and the sum of the forces on the stone is mv32 T3 − mg = . To find v3, consider the total energy at position 3. r E = K 3 + U 3 = 12 mv32 + mgh3 = 12 mv32 = 52 mrg

v32 = 5rg

7-72.

Therefore, the tension at position 3 is mv 2 m5rg T3 = mg + 3 = mg + = 6mg = 6(0.90 kg)(9.81 m/s 2 ) = 53 N. r r At its highest point, the velocity of the ball is in the horizontal direction and has the same magnitude as the horizontal component of the initial velocity, v = v0 cosθ . So, the kinetic energy at the highest point is K f = 12 mv 2 = 12 m(v0 cosθ ) 2 = 12 (0.17 kg)(28 m/s cos 30°) 2 = 50 J. The final potential energy can be found using conservation of energy. The initial kinetic energy is 2 K i = 12 mv0 = 12 (0.17 kg)(28 m/s) 2 = 67 J.

Pf = ∆K = 67 J − 50 J = 17 J 7-73.

2 vtop = vbottom − 2 gh = (172 m/s)2bottom − 2(9.81 m/s 2 )(1000 m) = 1.0 × 102 m/s or 360 km/h

7-74.

At any point along its trajectory, the pendulum bob will have the same total energy as it has at its starting position. The height above the lowest position is h = L(1 − cos 30°), where L is the pendulum length. For a given angle θ < 30°, the total energy is


CHAPTER

7

mgh = 12 mv 2 + mg (1 − cosθ ) L. At the lowest point, θ = 0° and the speed is vbottom = 2 g (cosθ − cos30°) = 2(9.81 m/s 2 )(1 − cos30°) = 1.6 m/s. The angle at which the speed is one-half this value is v2 2 g (cosθ − cos30°) = v 2 = bottom . 4 2 ⎛v ⎞ ⎛ (1.6 m/s) 2 ⎞ + cos30° ⎟ = 26° θ = cos −1 ⎜ bottom + cos30° ⎟ = cos −1 ⎜ 2 ⎝ 8(9.81 m/s ) ⎠ ⎝ 8g ⎠ 7-75.

At position 1, choosing the downward direction as positive, the sum of the forces on the stone is mv 2 T1 + mg = 1 r Since the tension is about zero, the speed at the top of the circle is v12 = gr. Therefore, the total energy is

1

2

E = K1 + U1 = 12 mv12 + mgh1 = 12 mgr + 2mgr = 52 mgr

4

3

At position 3, the tension is directed upward and the sum of the forces on the stone is mv32 T3 − mg = . To find v3, consider the total energy at position 3. r E = K 3 + U 3 = 12 mv32 + mgh3 = 12 mv32 = 52 mgr v32 = 5 gr or v3 = 5 gr 7-76.

At the top of the loop, the net force acting on a car is equal to N + mg and both forces are directed downward. This force is the cetripetal force, F = mv2/R, where R is the radius of the loop. For the car to stay on the track, the normal force N ≥ 0 N. The minimum speed that the car must be traveling to stay on the track of the top of the loop occurs when N = 0 N. mv 2 = N − mg = mg or v = Rg = (10 m)(9.81 m/s 2 ) = 9.9 m/s R To find the minimum speed at the bottom, apply conservation of energy. K 2 + U 2 = K1 + U1 1 2

mv22 + mgh2 = 12 mv12 + mgh1

1 2

mv22 + mgh2 = 12 mv12

Then, solve for v1. v22 + 2 gh2 = v12 v1 = v22 + 2 gh2 = (9.9 m/s) 2 + 2(9.81 m/s 2 )(40 m) = 29.7 m/s 7-77.

(a)

1 2

mv12 + mgh1 = 12 mv22 + mgh2

mgh1 = 12 mv22 v2 = 2 gh1 = 2(9.81 m/s 2 )(30 m) = 24 m/s (b) At the bottom of the circular valley, the sum of the vertical forces on the car is mv22 N − mg = = ma R


CHAPTER

7

The passengers feel an acceleration of 8g. Therefore, v22 v2 (24 m/s) 2 = 8g and R = 2 = = 7.3 m R 8 g 8(9.81 m/s 2 ) mv32 R If the passengers feel a normal force of zero N, then, v3 = (c) At the top of the hill, N + mg =

gR

To find the height h′, reapply the conservation of energy. 1 mv22 + mgh2 = 12 mv32 + mgh′ 2 1 2

mv22 = 12 mv32 + mgh′

h′ = 7-78.

v22 − v32 v22 − gR (24 m/s) 2 − (9.81 m/s 2 )(7.3 m) = = = 25.7 m 2g 2g 2(9.81 m/s 2 )

The initial speed at the bottom of the circle at point P1 is v1 = (2gr)1/2. The cart moves along the circle to the point P2 where the normal force goes to zero N. At the position P2, the sum of the forces in the radial direction is N + mg sin θ = mv22 / R. To find

mg

θ R

the angle θ, the height is found using h = R(1 + sin θ). Begin by applying conservation of energy, 1 2

P2

N

P1

mv12 = 12 mv22 + mgh

(

mv22 = mv12 − 2mgh = m 2 gR

)

2

− 2mgR(1 + sin θ ) = 2mgR(1 − sin θ )

This relationship can be substituted into the centripetal force equation. mv22 2mgR (1 − sin θ ) = mg sin θ = R R sin θ = 2(1 − sin θ )

θ = sin −1 ( 23 ) = 42° †7-79.

The particle will slide along the sphere until the normal force N goes to zero. At the angular position shown, the sum of the forces on the particle in the radial direction is mg cos θ − N = mv2/R, which gives v 2 = gR cosθ . The conservation of energy is then

N mg

applied to the situation, mgR = mgR cosθ + 12 mv 2 = mgR cosθ + 12 m( gR cosθ ). . Solving for θ gives, θ = cos −1 ( 23 ) = 48.2°.

7-80. 7-81. 7-82.

W = F • y = mgh = (0.20 kg)(9.81 m/s2)(35 m) = 69 J W = F • s = (150 N)(20 m)cos 45° = 2100 J The sum of the forces on the crate in the vertical direction is N − mg = 0, so N = mg. In the horizontal direction, the sum of the forces is F − fk = 0, so F = fk = µkN = µkmg. The crate is pushed for a distance s = 15 m, so the work done by the man is W = F • s = µkmgs = (0.50)(120 kg)(9.81 m/s2)(15 m) = 8800 J.


N fk

F mg

CHAPTER 7-83.

7

In the first case, the automobile accelerates from 20 to 25 m/s. W = ∆K = 12 mv22 − 12 mv12 = 12 m(v22 − v12 ) = 12 (1500 kg) ⎡⎣(25 m/s) 2 − (20 m/s) 2 ⎤⎦ = 1.69 × 105 J In the second case, the automobile accelerates from 25 to 30 m/s. W = ∆K = 12 mv22 − 12 mv12 = 12 m(v22 − v12 ) = 12 (1500 kg) ⎡⎣ (30 m/s) 2 − (25 m/s) 2 ⎤⎦ = 2.06 × 105 J

7-84.

†7-85.

In raising the books, she applies an upward force F. The books are assumed to move upward at a constant speed, so the sum of the vertical forces equals F − mg = 0 and F = mg. The work done by the woman on the books is Wup = F • s = Fs cos θ = Fs = mgs = (20 kg)(9.81 m/s2)(1.8 m) = +350 J. When the books are lowered at a constant rate, the same analysis applies. However, in this case the force continues to be applied upward and the displacement is downward. Therefore, Wdown = F • s = Fs cos θ = −Fs = −mgs = −(20 kg)(9.81 m/s2)(1.8 m) = −350 J. The total work done is W = Wup + Wdown = + 350 J − 350 J = 0 J. While the woman’s muscles are causing work to be done in the upward and downward motions, there is nothing in the analysis to indicate what is happening internally. It will depend upon the details of how the woman does the work over time. (a) Frictional forces do work on the car over the stopping distance, W = −fs. This results in a change in kinetic energy ∆K. W = − µ mgs = ∆K = 12 mv22 − 12 mv12 2 1 0 − 12 mv12 2 (25 m/s) = = 35.4 m (0.90)(9.81 m/s 2 ) − µ mg To determine the acceleration, assuming the acceleration is constant as the car is slowing, we can use the kinematic relation v22 − v12 = 2as.

s=

a=

7-86.

v22 − v12 0 − (25 m/s) 2 = = −8.8 m/s 2 2s 2(35.4 m)

(b) f=µN=µmg = (0.90)(1200 kg)(9.81 m/s2) = 1.06 × 104 N W = −fs = −(10 600 N)(35.4 m) = 3.75 × 105 J (a) K1 = 12 mv12 = mgh1 = (0.050 kg)(9.81 m/s 2 )(1.5 m) = 0.74 J (b) K 2 = 12 mv22 = mgh2 = (0.050 kg)(9.81 m/s 2 )(1.0 m) = 0.49 J

7-87.

(c) K1 − K2 = 0.74 J − 0.49 J = 0.25 J (a) U1 = mgh = (1200 kg)(9.81 m/s2)(2000 m) = 2.35 × 107 J 2

⎡⎛ km ⎞⎛ 1000 m ⎞⎛ 1 h ⎞ ⎤ 6 K1 = 12 mv12 = 12 (1200 kg) ⎢⎜ 250 ⎟⎜ ⎟⎜ ⎟ ⎥ = 2.89 × 10 J h 1 km 3600 s ⎠⎝ ⎠⎝ ⎠⎦ ⎣⎝ 2 (b) U 2 = mgh2 = (1200 kg)(9.81 m/s )(1500 m) = 1.77 × 107 J

K 2 = K1 + U1 − U 2 = 2.86 × 106 J + 2.35 × 107 J − 1.77 × 107 J = 8.7 × 106 J v2 =

2K = m

2(8.7 × 106 J) = 120 m/s or 430 km/h 1200 kg


CHAPTER 7-88.

7

Assuming no losses due to frictional forces, apply conservation of energy to determine the kinetic energy at the bottom of valley and the corresponding speed. 1 mv12 + mgh1 = 12 mv22 + mgh2 2

mgh1 = 12 mv22 v2 = 2 gh = 2(9.81 m/s 2 )(30 m) = 24 m/s The speed at the top of the second hill can be determined in a similar manner. 1 mv12 + mgh1 = 12 mv32 + mgh3 2

mgh1 = 12 mv32 + mgh3 v3 = 2 g (h1 − h3 ) = 2(9.81 m/s 2 )(10 m) = 14 m/s 7-89.

(a) Wpull = F • s = Fs cos θ = Fs = (300 N)(0.50 m) = 150 J (b) The bow exerts a constant force of 300 N on the arrow over the distance of 0.50 m. Assuming any other forces are negligible, the bow does the work that changes the kinetic energy of the arrow from zero J to the value it has at the release. W = ∆K = 150 J (c) K = 12 mv12

v1 =

2K = m

2(150 J) = 122 m/s (0.020 kg)

(d) Assuming the arrow lands at the same height as it was launched and there are no losses due to air friction, 2 v 2 sin 2θ (122 m/s) sin [ 2(45°) ] = = 1520 m xmax = 1 g (9.81 m/s 2 )

7-90.

(e) Assuming the arrow lands at the same height as it was launched and there are no losses due to air friction, the kinetic energy it has when it lands will equal its kinetic energy at launch. Therefore, the landing speed will be 122 m/s. (a) From the range, the initial speed and, thus, the initial kinetic energy can be calculated, assuming no losses due to air friction. v 2 sin 2θ xmax = 0 g

gxmax (9.81 m/s 2 )(180 m) = = 42 m/s sin 2θ sin 90° K = 12 mv02 = 12 (77 kg)(42 m/s) 2 = 6.8 × 104 J

v0 =

(b) At the highest point, the vertical component of the stone’s velocity is equal to zero m/s. The kinetic energy is then, K top = 12 m(v0 cosθ ) 2 = K (cos 45°)2 = 12 K = 3.4 × 104 J †7-91.

The energy lost to friction is the difference between the final and initial mechanical energy of the luger. The initial mechanical energy is equal to the potential energy because the luger starts from rest.


CHAPTER

7

∆E = mgh1 − ( 12 mv22 + mgh2 ) = mg (h1 − h2 ) − 12 mv22 ⎡ ⎛ 1 h ⎞⎛ 1000 m ⎞ ⎤ = (95 kg)(9.81 m/s )(350 m − 240 m) − (95 kg) ⎢(130 km/h) ⎜ ⎟⎜ ⎟⎥ ⎝ 3600 s ⎠⎝ 1 km ⎠ ⎦ ⎣ = 4.1 × 104 J 2

7-92.

2

1 2

Before the release, the total energy of the pendulum bob is equal to its potential energy, mgl. At the bottom of the swing, conservation of energy will give the kinetic energy, and thus, the speed. 1 mv12 + mgh1 = 12 mv22 + mgh2 2

mgh1 = 12 mv22

θ

v2 = 2 gl At the bottom of the swing, the sum of the vertical forces acting on the bob is mv22 T − mg = r where T is the tension in the string. ⎛ v2 ⎞ mv22 ⎛ 2 gl ⎞ T = + mg = m ⎜ 2 + g ⎟ = m ⎜ + g ⎟ = 3mg r ⎝ l ⎠ ⎝ l ⎠

7-93.

The tension in the string is three times greater when the pendulum is swinging from an initial angle θ = 90°, than when it is simply hanging at rest its equilibrium position. (a) 12 mv12 + mgh1 = 12 mv22 + mgh2

mgh1 = 12 mv22 v2 = 2 gh = 2(9.81 m/s 2 )(34 m) = 25.8 m/s (b) 100 km/h = 27.8 m/s 1 mv12 + mgh1 = 12 mv22 + mgh2 2 1 2

mv12 + mgh1 = 12 mv22

v12 = v22 − 2 gh1 v1 = v22 − 2 gh1 = (27.8 m/s) 2 − 2(9.81 m/s 2 )(34 m) = 10.3 m/s 7-94.

(a)

1 2

mv12 + mgh1 = 12 mv22 + mgh2

mgh1 = 12 mv22 + mgh2 v2 = 2 g (h1 − h2 ) = 2(9.81 m/s 2 )(6.0 m − 1.0 m) = 9.9 m/s (b) The horizontal distance is determined by the time she is in the air before hitting the water. The flight time is the time it takes to fall 1.0 m. y = 12 gt 2 t=

2y = g

2(1.0 m) = 0.45 s (9.81 m/s 2 )

x = v0 t = (9.9 m/s)(0.45 s) = 4.5 m


CHAPTER 8

CONSERVATION OF ENERGY

Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions Manual, available for purchase. Answers to all solutions below are underscored. kx 2 ⇒x= 2

2U = k

2(100 J) = 0.076 m (7.6 cm) . 3.5 × 104 N/m

8-1.

U =

8-2.

(a) Let the particle go from x1 and back to x1 again. Then W =

∫

x2

x1

x1

F ( x)dx + ∫ F ( x)dx =

∫

x2

x2

x2

x1

x1

(2 x3 + 1)dx + ∫ (2 x3 + 1) dx x2

x1

⎡1 ⎤ ⎡1 ⎤ = ⎢ x 4 + x ⎥ + ⎢ x 4 + x ⎥ = 0 for any x1 , x2 2 2 ⎣ ⎦ x1 ⎣ ⎦ x2 (b) 8-3.

∫

x2

x1

x1

F ( x)dx + ∫ F ( x)dx = x2

total work is W =

†8-5.

†8-7.

F ( x)dx =

∫

x2

x1

F ( x)dx = 0

∫

r2 r1

∫

r2 r1

F ( v1 ) i dr. On the return, W2 =

∫

r1 r2

r2

F ( v 2 ) i dr = − ∫ F( v 2 ) i dr. The r1

r2

F ( v1 ) i dr − ∫ F( v 2 ) i dr. Since F( v1 ) ≠ F ( v 02 ), the two terms generally r1

will not add to zero, as stated in the problem. x x K K K K U ( x) = − ∫ F ( x ')dx ' = − ∫ dx ' = 3 − 3 . Choose x0 = ∞. Then U ( x) = 3 . x0 x0 ( x ') 4 3 x 3 x0 3x Ax 4 Ax 4 Ax04 . Choose x0 = 0. Then U ( x) = . The law − x0 x0 4 4 4 of conservation of energy states that if no nonconservative forces act, U 2 + K 2 = U1 + K1 . Let point 1 be x with K1 = 0 (the particle is at rest) and point 2 be x = 0 with a kinetic energy x

x

U ( x) = − ∫ F ( x ')dx ' = − ∫ A( x ')3 dx ' =

K2 = 8-6.

x2

x1

Suppose the particle moves with constant velocity v1 from r1 to r2 and then moves with a different velocity v2 from r2 back to r1. (The velocities must at least have opposite directions on the two parts of the trip, so the assumption of different velocities for each half of the round trip is not unreasonable.) Since the force is velocity dependent, assume that F( v1 ) ≠ F ( v 2 ). Then on the first part of the trip, W1 =

8-4.

∫

mv 2 mv 2 Ax 4 . Then 0 + = +0⇒ v= 2 2 4 x

x

x0

x0

Ax 4 = 2m

(50 N/m3 )(0.50 m) 4 = 5.6 m/s. 2(0.050 kg)

U ( x) = − ∫ F ( x ')dx ' = − ∫ (− F0 )dx ' = F0 ( x − x0 ). Choose x0 = 0. Then U ( x) = F0 x. x4 , where it has been assumed that 0 0 4 x0 = 0. (A note about units: the coefficient 2 of the x term in U has units of N/m, while the coefficient of the x3 term is actually 1 N/m3. In the integral, dx has units of m, so the coefficient of the x2 term is actually 1 N/m, and the coefficient of the x4 term is now 0.25 N/m3. It’s important to remember that apparently abstract operations like integration can involve units.) The table that follows evaluates the equation for the values of x requested in the problem. x

x

U ( x) = − ∫ F ( x ')dx ' = − ∫ [−2 x '− ( x ')3 ]dx ' = x 2 +


CHAPTER

8 x (m) 1.0 2.0 3.0

8-8.

U (J) 1.3 8.0 29.3

x

U ( x) = − ∫ F ( x ')dx '. Take x0 = 0. Suppose –a < x < 0. In this range, F(x) = 0, so U = 0. x0

For x < –a, the integral must be evaluated in two parts: −a

x

U ( x) = − ∫ 0dx ' − F0 ∫ dx ' = − [ F0 x − F0 (− a )] = − F0 ( x + a). For 0 < x < a, U(x) = 0. For −a

0

a

x

0

a

a < x < 0, U ( x) = − ∫ 0dx ' + F0 ∫ dx ' = F0 x − F0 a = F0 ( x − a ). Summarizing all the steps gives

⎧− F0 ( x + a), x ≤ − a ⎪ − a < x < a. U ( x) = ⎨0, ⎪ F ( x − a), x≥a ⎩ 0 8-9. 8-10.

8-11.

8-12.

dU d = − (2 x 2 + x 4 ) = − 4 x − 4 x 3 . dx dx x F F U ( x) = − ∫ F0 sin( ax ') dx ' = 0 [cos( ax) − cos(ax0 )]. Choose x0 = 0, so U ( x) = 0 [cos(ax) − 1]. x0 a a 160 Spring constant of ‘spring’ F / x = = 307.7 N/m 0.52 1 Potential energy stored in the spring = kx 2 2 At equilibrium all of the energy is kinetic energy in the arrow. Therefore, 1 2 1 kx = mv 2 2 2 k 307.7 v= x= (0.52) = 64.5 m/s m 0.02 (a) F = kx, or x = F/k = mg/k (b) Work done against gravity = F ∆z = m2 g2/k F =−

Work against spring =

∫ F dz = − ∫ z

x

0

kz dz

= − 1/ 2 kx 2 = − 1/ 2 k (mg / k ) 2 = −1/ 2 m 2 g 2 / k †8-13.

Since F is a conservative force, the work done by it as a particle moves from one point to another is independent of the path followed. The path suggested by the hint makes evaluation of the integral easy. Part I of the path goes from (0,0) to (x,0) (x varies while y is held constant at 0), and part II goes from (x,0) to (x,y) (y varies while x is held constant). Do the integral for part I of the path: dW = F i dr ⇒ WI =

∫

( x ,0)

(0,0)

(0i + bx ' j) i (dx ' i + dy ' j) = bx ' ∫

y '= 0 y '= 0

dy ' = 0

(since y doesn’t change along part I of the path). Another interpretation is that along I F points in the y direction and the displacement points in the x direction. Since F is perpendicular to dr along I, the work must be zero. Along part II of the path, we get


CHAPTER WII =

∫

( x, y)

( x ,0)

(by ' i + bxj) i (dx ' i + dy ' j) = bx ∫

y '= y y '= 0

8

dy ' = bxy. The total work

is W = WI + WII = bxy. The potential energy is U = − W = − bxy. 8-14.

At maximum deformation, all of the energy is contained in the springs. Since there are four springs, E = 4(1/2 kx2) = 2 kx2 where x is the deformation. By conservation of energy this is equal to the potential energy of the car = mgh. Therefore, 2kx 2 = mgh x = mgh / 2k =

8-15. 8-16.

1200 kg × 9.8 m/s 2 × 0.8 m = 0.26 m 2 × 7.0 × 104 kg/s 2

1 2 1 ⎛ 58000 ⎞ 2 6 kx = ⎜ ⎟ (90) = 2.61 × 10 J 2 2 ⎝ 90 ⎠ When the string breaks, the energy is released as kinetic energy. Each brake pad exerts a normal force of 1.0 × 105 N on the rail. Since µk = 0.15, the magnitude of the friction force exerted by each brake is 1.5 × 104 N, for a total friction force of 3.0 × 104 N acting to slow the elevator. If the elevator is moving down, the friction force will point up. The work done by friction will be W f = f i ∆x = − f ∆x, where ∆x is the distance the elevator moves Elastic energy =

while it stops. The work is negative because the displacement vector and the force vector point in opposite directions. According to the law of conservation of energy, K 2 + U 2 = K1 + U1 + W f , where 1 refers to the point where the brakes first start to act and 2 refers to the point lower down mv 2 where the elevator comes to rest. K and U are both zero at point 2, so 0 = 1 + mg ∆x − f ∆x, 2 2 2 (2000 kg)(10 m/s) mv from which we get ∆x = = = 9.63 m. 2( f − mg ) 2[(3.0 × 104 N) − (2000 kg)(9.81 m/s 2 )] The energy dissipated by friction is the magnitude of the total work done by friction, which is also the total initial kinetic plus potential energy: W f = K1 + U1 = f ∆x = (3.0 × 104 N)(9.63 m) = 2.89 × 105 J. To find the time required to stop, use a =

8-17.

v22 − v12 (10 m/s) 2 =− = − 5.19 m/s 2 . (The – sign 2∆x 2(9.63 m)

means the acceleration points in the opposite direction from the initial velocity.) Then v −v −10 m/s ∆t = 2 1 = = 1.93 s. a −5.19 m/s 2 x x A B U ( x) = − ∫ F ( x ')dx '. Take x0 = ∞. Then U ( x) = − ∫ [ A( x ') −13 − B ( x ') −7 ]dx ' = − 6. 12 x0 ∞ 12 x 6x

∫

( x, y, z )

F( x ', y ', z ') i dr = − k ∫

( x, y,z )

8-18.

W =

8-19.

k k ( x ' dx '+ y ' dy '+ z ' dz ') = − ( x 2 + y 2 + z 2 ). U = − W = ( x 2 + y 2 + z 2 ). 2 2 Call the length of the rope L. The climber starts a distance L above the fixed end of the rope and ends a distance L + ∆x below the fixed end, where ∆x is the stretch of the rope. The total distance fallen is 2L + ∆x. Take the kinetic and potential energies to be zero at the beginning of the motion. At the end of the fall, the climber’s kinetic energy is zero again but his potential energy is

(0,0,0)

= − k∫

(0,0,0)

( x ' i + y ' j + z ' k ) i (dx ' i + dy ' j + dz ' k )

( x, y, z )

(0,0,0)


CHAPTER

8

k ∆x 2 , where k is the force constant for the rope. Since the total initial 2 energy was zero, conservation of energy says the total final energy must also be zero, which gives k ∆x 2 2mg 4mgL − mg (2 L + ∆x) = 0, or ∆x 2 − ∆x − = 0. Use the quadratic formula to solve for 2 k k U f = − mg (2 L + ∆x) +

2

⎛ 2 mg ⎞ ⎛ 2 mg ⎞ ⎛ 4 mgL ⎞ ⎜ ⎟± ⎜ ⎟ + 4⎜ ⎟ ⎝ k ⎠ ⎝ k ⎠ ⎝ k ⎠ . The value of k depends on L, with k = 4.9 × 103 ∆x: ∆x = 2 N/m for the 10 m rope and 9.8 × 103 N/m for the 5 m rope. 2 mg 2(80 kg)(9.81 m/s 2 ) 4mgL (a) For L = 10 m, = = 0.320 m and = 6.40 m 2 . The solutions k 4.9 × 103 N/m k are ∆x = 2.70 m, – 2.38 m. The negative root has no physical meaning, so we keep only the positive root. The force exerted by the rope at this stretch is F = k ∆x = (4900 N/m)(2.70 m) = 1.3 × 104 N. 2 mg 2(80 kg)(9.81 m/s 2 ) 4 mgL (b) For L = 5 m, = = 0.160 m and = 1.60 m 2 . The solutions k 9.8 × 103 N/m k are ∆x = 1.35 m, – 1.19 m. Again keeping only the positive root we find the force exerted by the rope is F = k ∆x = (9800 N/m)(1.35 m) = 1.3 × 104 N . The force exerted by the rope is the

8-20.

8-21.

same, whether a long rope or a short rope is used, as long as both ropes are made from the same material. In the frame of the conveyor belt, the package is moving at velocity v (to the left). The deceleration of the package in this frame is µkg (to the right). So the time that the package slides is vƒ = v0 + at ≥ 0 = –v + µkgt ⇒ t = v/µkg. In the lab frame, this acceleration is also constant. Therefore, the distance 1 1 moved = 1/2 at 2 = ( µ k g )(v / µ k g ) 2 = (v 2 / µ k g ) 2 2

In conveyor belt’s frame, all of the kinetic energy of the package gets converted into friction in 1 the conveyor belt. Therefore, the energy dissipated = mv 2 2 −∂U ˆ ∂U ˆ ∂U ˆ F= i− j− k where U = a ( x 2 + y 2 ) −1/ 2 ∂x ∂y ∂z 2 = − a( − 1/2)(x + y 2 ) −3/ 2 2 x î − a (−1/ 2)( x 2 + y 2 )−3/ 2 2 y ˆj + 0 kˆ = ( x î + y ˆj) a /( x 2 + y 2 )3 / 2

8-22. 8-23.

(The force vector is along the line through the origin and [x,y].) dU 2 K K = 3. U ( x) = 2 . Fx = − x dx x U = ηr. F = – dU/dr = –η, so the magnitude of the force is η = 1.18 × 1024 eV/m = (1.18 × 1024 eV × 1.6 × 10–19 J/eV)/m = 1.89 × 105 N.


CHAPTER (a) The graph was plotted using a spreadsheet. (b) The equilibrium points are the locations where the slope of U is zero. There are three such points: x = 0 and x = ± 0.50. The point at 0 is unstable and the points at ± 0.50 are stable.

U(x) vs x 1 U (J)

8-24.

8

0.8 0.6 0.4 0.2 0 -1

-0.5

0

0.5

x (m) 1

0.5

x (m )1

-0.2

(c) The graph at right shows that the turning points for E = – 0.050 J are at x ≈ ± 0.2 m and x ≈ ± 0.65 m. (d) From the first graph we see that the turning points for E = 1.0 J are at x = ± 1.0 m.

1 0.8

U (J)

U(x) vs x

0.6 0.4 0.2 -1

†8-25.

-0.5

0 -0.2 0

(a) K is a maximum when U is a minimum. From the graph in Example 4, this occurs at x ≈ –5 m. To find the exact location, differentiate U(x) and set the derivative equal to zero: 687 dU m = − 4.58 m. This is in very good agreement = 687 + 150 x = 0, which gives x = − dx 150 with the 1-digit estimate from the graph. To find the speed at this point, use E = U + K ⇒ K = E − U . Substitute x = – 4.58 m in the equation for U to get U = –1573 J. Then K = 6180 J + 1573 J = 7753 J. Then v =

2K m

2(7753 J) = 14.9 m/s. 70 kg) dU (b) The force is Fx = − = − 687 − 150 x. For x within the range of the problem, the maximum dx force occurs at the turning point, which is x = –14.7 m. The magnitude of the force at this value of x is F = 1.52 × 103 N, so the magnitude of the acceleration is F 1.52 × 103 N = 21.7 m/s 2 , which is about 2.2 gees. a= = 70 kg m =


CHAPTER 8-26.

8

U ( x) = (2.36 eV) ⎡⎣e −2( x − 0.037) /.034 − 2e − ( x − 0.037) / .034 ⎤⎦ .

U(x) vs x

A plot to get estimates of the turning points for E = –1.15 eV is shown. From the graph, the turning points are x ≈ 0.02 nm and 0.08 nm. With these starting values, more accurate estimates can be obtained by trial and error (plugging in values of x until the right hand side of the equation is close enough to – 1.15 eV) or by using tools such as Solver in Excel or a Solve Block in MathCAD. To 2 significant digits the results are x = 0.019 nm and 0.080 nm.

8-27.

7 6 5 4 3 2 1 0 -1 0 -2

0.05

0.1

0.15

0.2

x ( nm )

-3

(a) U(x) = b/x2 – 2c/x dU/dx = –2b/x3 + 2c/x2 At equilibrium dU/dx = 0 2c/x2 = 2b/x3 ⇒ x = b/c (b) The turning points are where U(x) = E c2 b/x 2 = 2c/x = − 12 c 2 /b ⇒ 12 x 2 − 2cx + b = 0 b 2 = x − 4(b/c) x + 2(b/c) 2 = 0

x = 4(b/c) ± 16(b/c) 2 − 8(b/c) 2 = 2(b/c) ± (b/c) 8/2

( ) 2 ) (b / c), ( 2 − 2 ) (b / c) = 2 ± 2 (b/c)

(

Turning points at x = 2 + (c) U ( x) = b/x 2 − 2c/x =

1 2 c /b 2

1 2 c / bx 2 + 2cx − b = 0 2 ⇒ x 2 + 4(b/c) − 2(b 2 /c 2 ) = 0

Therefore,

−4(b/c) ± 16(b/c) 2 + 8(b 2 /c 2 ) 2 = (b/c) ⎡⎣ −4 ± 24 ⎤⎦ / 2 = (b/c) −2 ± 6 If the particle is moving along the positive x-axis, then it will always be confined there since U(x) → ∞ as x → 0. Thus, there is only one turning point at x = 6 − 2 (b / c). x=

(

)

(

8-28.

)

(a) Since the curve crosses E1 only once, the only turning point for E1 is at x = 0.2 m; for E2 the curve crosses at x = 0.3 m and 3.1 m; for E3 the curve crosses at x = 1.3 m and 0.5 m. (b) The maximum speed occurs at max K, so absolute maximum speed occurs at Umin ⇒ x = 0.9 m. The absolute minima occur when U = Umax, at turning points, A local maximum and minimum occur at x = 2.2 m and 1.7 m, respectively.


CHAPTER

8-30.

†8-31.

8-32.

(c) For E2 there is only one turning point, so the orbit is unbound. For E2 and E3, there are two turning points so orbits are bound. mv 2 mv 2 U(x) E = U + K = A x + K . At x = 0, K = ⇒E= . The 2 2 energy is constant. At the turning points, K = 0, which gives mv 2 mv 2 = A x +0⇒ x = . The final result is that the turning 2 2A x mv 2 points are at x = ± . 2A A graph of U(x) is shown. For the particle to be bound, the total energy must be between –U0 and U0, so the lowest energy is –U0. At the turning points, K = 0, so for E = 0 the turning points are given by cos(ax) = 0, nπ which gives ax = ± , where n = 1, 3, 5, … The 2 nπ turning points then are x = ± , n = 1,3,5, ... 2a The particle will be unbound if E > U0. 2.0 . The graph was plotted using a spreadsheet. 1 + x2 2.0 . Set At the turning points, K = 0, and E = − 1 + x2 E = –1 J. This gives a quadratic equation that can be solved for the values of x: 2.0 ⇒ − 1 − x 2 = − 2 ⇒ x 2 = 1, which gives x = −1 = − 1 + x2 ± 1.0 m. The particle will be unbound for E > 0.

U0

1

0 -6

-4

-2

0

2

4

6

-1

-U0

U =−

U(x) vs x x (m) 0 -3

-2

-1

-0.2 0 -0.4 -0.6 -0.8

1

2

3

-1 -1.2 -1.4 -1.6 -1.8 -2

At the turning points, K = 0 ⇒ E = 2 x 2 + x 4 , or x 4 + 2 x 2 − E = 0. Use the quadratic formula to −2 ± 8 −2 ± 4 + 4 E . For E = 1.0 J, we get x 2 = = 0.414 m 2 . (There is a 2 2 negative root for x2, but it has no physical significance.) Then x = ± 0.64 m. For E = 2.0 J, we get −2 ± 12 x2 = = 0.732 m 2 , again rejecting the negative root for x2. The final result for 2.0 J is 2 x = ± 0.86 m. The large negative exponents and rapid variation of the potential near x = 0 will make this problem difficult to do graphically. However, it can be solved without too much difficulty by using simple algebra and calculus. (a) The equilibrium point is found by setting the derivative of the potential equal to zero. dU = − 12(1.59 × 10−24 ) x −13 + 6(1.03 × 10−21 ) x −7 = 0. x → ∞ satisfies the equation, but this does dx solve this for x2: x 2 =

†8-33.

U(x) vs x

U(x) (J)

†8-29.

8


CHAPTER

8

not give the minimum energy. Multiply through by x13 and 10–21 and divide both sides by 6 to get 1/ 6

⎛ 0.00318 ⎞ this equation: 1.03x 6 = 0.00318, where x is in nm. The result is x = ⎜ ⎟ ⎝ 1.03 ⎠

= 0.382 nm.

(b) The lowest energy occurs at the equilibrium point, so substitute the result from (a) into the 1.59 × 10−24 1.03 × 10−21 equation for U: U = − = − 1.67 × 10−19 J. (0.382)12 (0.382)6 (c) At the turning points, K = 0, so E = U. We need to solve this equation: −2.0 × 10−21 = (1.59 × 10−24 ) x −12 − (1.03 × 10−21 ) x −6 . The turning points would be difficult to locate graphically because E is only about 1% of the minimum value of U, which makes it very hard to see on the graph. But the equation isn’t as bad as it looks. Multiply through by x12 and 1021 and rearrange to get 2 x12 − 1.03x 6 + 0.00159 = 0. This is a quadratic equation for x6, and its roots can be used to find the values of x for the turning points: 1.03 ± (1.03) 2 − (8)(0.00159) = 0.001548 nm 6 , 0.5135 nm 6 . Take the 6th root of each 4 number to get the values for x: x = 0.34 nm, 0.89 nm. (Note: Substituting 0.34 nm in the equation for U gives a result that is wrong. The rapid variation of the potential near x = 0 means that rounding x to even three significant digits gives the wrong answer. To get the exact value of U desired, x must be calculated to five significant figures: x = 0.33979 nm. However, even this result is suspect because the coefficients in the equation are only given to three significant figures.) 1 eV 1 eV = 2.0 × 108 eV. 2.2 × 10−18 J × = 14 eV 3.2 × 10−11 J × −19 1.60 × 10 J 1.60 × 10−19 J x6 =

8-34. †8-35.

8-36.

Mass of one molecule = 5mH + 6mC + 3mN + 6mO = 5(1 u) + 6(12 u) + 3(14 u) + 6(16 u) = 1.66 × 10−27 kg 215 u. In kg this is m = 215 u × = 3.57 × 10−25 kg. u 1 eV Emolecule = (4.6 × 106 J/kg)(3.57 × 10−25 kg) = 1.64 × 10−18 J × = 10.3 eV/molecule. 1.60 × 10−19 J Energy use per day = (8 × 1019/365)J/day Energy obtained from gasoline = 1.3 × 108 J/gal Therefore, the gallons per day consumed = (8 × 1019/365)/1.3 × 108 gal/day = 1.7 × 109 gal/day

8-37. Vehicle

Energy per mile (J/mi)

1/60 gal/mi × 1.3 × 108 J/gal = 2.2 × 106 J/mi Snowmobile 1/12 × 1.3 × 108 = 1.1 × 107 Automobile 1/12 × 1.3 × 108 = 1.1 × 107 Bus 1/5 × 1.3 × 108 = 2.6 × 107 Jetliner 1/0.1 × 1.3 × 108 = 1.3 × 109 Concorde 1/0.12 × 1.3 × 108 = 1.1 × 109 Most efficient = bus, Least efficient = snowmobile.

Energy per passenger per mi (J/passenger-mi)

Motorcycle


2.2 × 106 1.1 × 107 1.1 × 107/4 = 2.7 × 106 2.6 × 107/45 = 5.8 × 105 1.3 × 109/360 = 3.6 × 106 1.1 × 109/110 = 9.8 × 106

CHAPTER

8-38. †8-39.

8

9000 kcal/kg × 1 kg = 12 h. 750 kcal/h 4.187 × 103 J E = 0.2(150 kcal) = 30 kcal × = 1.26 × 104 J. To climb a vertical distance h, this kcal energy must be converted to potential energy (assuming the person climbs slowly so there’s no E . Each person should use his or her significant conversion to kinetic energy). E = mgh ⇒ h = mg t=

own mass in kg to find their personal value for h. Assuming m = 70 kg, 1.26 × 104 J h= = 183 m. (70 kg)(9.81 m/s 2 ) 8-40. 8-41. 8-42.

3.6 × 106 J = 1.08 × 1010 J. kWh 4.187 kJ 2500 kcal × = 1.05 × 104 kJ. kcal E = 3000 kWh ×

ymax =

2g

v0 y

⇒ v0 y = 2 gymax = 2(9.81 m/s 2 )(3 m) = 7.67 m/s.

mv02 (33 × 103 kg)(8.16 m/s) 2 7.67 m/s 1 kcal = 8.16 m/s. K = = × = 263 kcal. sin θ sin 70° 2 2 4.187 J Walking: t = 0.5 h. Ewalk = (3.3 kcal/kg − h)(0.5 h) = 1.7 kcal/kg. v0 =

†8-43.

v02y

=

2.5 km = 0.313 h. tstand = 0.5 h – 0.313 h = 0.187 h. 8 km/h = (8.2 kcal/kg − h)(0.313 h) + (1.3 kcal/kg − h)(0.187 h) = 2.8 kcal/kg.

Slow running plus standing: trun = Eslow

2.5 km = 0.156 h. tstand = 0.5 h – 0.156 h = 0.344 h. 16 km/h = (15.2 kcal/kg − h)(0.156 h) + (1.3 kcal/kg − h)(0.344 h) = 2.8 kcal/kg. The two running

Fast running plus standing: trun = E fast

8-44. 8-45.

strategies both consume about the same amount of energy, while walking consumes the least energy. E 8.4 × 1013 J m= 2 = = 9.3 × 10−4 kg (0.93 g). c 9 × 1016 m 2 / s 2 E = 2m p c 2 = 2(1.67 × 10−27 kg)(9 × 1016 m 2 / s 2 ) ×

1 eV = 1.88 × 109 eV –19 1.60 × 10 J

(1.88 GeV) 1 eV = 1.02 × 106 eV. (1.02 MeV) 1.60 × 10 –19 J

8-46.

E = 2me c 2 = 2(9.11 × 10−31 kg)(9 × 1016 m 2 / s 2 ) ×

8-47.

mthermal = Ethermal/c2 = [2 × 1041/(3 × 108)2]kg = 2.2 × 1024 kg Percentage of mass that is thermal is (2.2 × 1024 kg/2 × 1030 kg) × 100% = 0.0001%. ∆m = mn − m p − me = 1.674 929 × 10–27 kg – 1.672 623 × 10–27 kg – 9.11 × 10–31 kg = 1.39 × 10–

8-48.

30

kg. E = ( ∆m)c 2 = 1.26 × 10−13 J = 7.84 × 105 eV. The mass energy of the electron is

Ee = me c 2 = (9.11 × 10−31 kg)(3 × 108 m/s) 2 = (8.20 × 10−14 J)(1 eV/1.6 × 10−19 J) = 5.11 × 105 eV. The energy released is comparable to the mass energy of the electron.


CHAPTER

†8-49.

8 1 eV 1 keV = 511 keV. × −19 1.60 × 10 J 1000 eV 1 eV 1 MeV = 939 MeV. kg)(9 × 1016 m 2 / s 2 ) × × 6 −19 1.60 × 10 J 10 eV

E = me c 2 = (9.11 × 10−31 kg)(9 × 1016 m 2 / s 2 ) × E = m p c 2 = (1.67 × 10−27

8-50.

m=

E 1000 kWh × 3.6 × 106 J/kWh = = 4.0 × 10−8 kg. 9 × 1016 m 2 / s 2 c2

8-51.

m=

E 1.3 × 108 J = = 1.4 × 10−9 kg. This is about 1/2 of one billionth, or 0.00000005% of c 2 9 × 1016 m 2 / s 2

8-52.

the mass of the gasoline. E = mc 2 = (9.8 × 10−18 kg)(9 × 1016 m 2 /s 2 ) = 0.88 J.

†8-53.

By conservation of energy, the mass energy of the particles produced must be equal to the mass energy of the two original particles plus their kinetic energy. This means K = 2m p c 2 − 2me c 2 = 2( m p − me )c 2 . The mass of the electron is negligible compared to the mass of the proton, so K = 2m p c 2 = 1.88 × 109 eV (see the solution to 8-45). This is the total kinetic energy of the electron and positron. The kinetic energy of the electron is half of this value, or K e = 1.88 × 109 eV/2 = 9.40 × 108 eV.

8-54.

P = Fv. v = 65 km/h = 18.1 m/s. P = (500 N)(18.1 m/s) = 9.03 × 103 W (9.03 kW, or 12.1 hp).

8-55.

E = P∆t. ∆t = 2 h 49 min = 169 min = 10140 s. E = (0.30 hp)(10140 s)(746 W/hp)(1 kcal / 4187 J) = 542 kcal.

8-56.

150 hp = (150 × 745.7)W = 1.12 × 105 W = power of automobile Slaves needed to match car = 1.12 × 105 W/200 W per slave = 560 slaves Slaves needed per capita = 14000 W/200 W per slave = 70 slaves 1 yr = 8766 h. E = P∆t = (0.002 kW)(8766 h) = 18 kWh. If it’s a mechanical clock, some of the

8-57.

8-58.

energy is converted to kinetic energy of the moving parts of the clock. If it’s a digital clock, some of the energy is converted to the light given off by the display. For both kinds of clock the rest of the energy is dissipated as heat generated by work done against friction or because of heat produced by the electrical components. 746 W 20 hp × = 1.49 × 104 W. In one hour, the energy used is E = P∆t hp = (1.49 × 104 W)(3600 s) = 5.4 × 107 J.

8-59. 8-60.

8-61.

E = P∆t = (104 W)(8 hr)(3600 s/hr) = 2.88 × 108 J. ⎛ 3 Btu ⎞⎛ 3 J ⎞⎛ 1 hr ⎞⎛ 1 hp ⎞ ⎜ 170 × 10 ⎟⎜ 1.055 × 10 ⎟⎜ ⎟⎜ ⎟ = 66.8 hp. This is less than the output of hr ⎠⎝ Btu ⎠⎝ 3600 s ⎠⎝ 746 W ⎠ ⎝ the 150 hp engine. 1 hp P = 1.1 W × = 1.5 × 10−3 hp. W = P∆t = (1.1 W)(24 hr)(3600 s/hr)(1 kcal / 4187 J) 746 W = 23 kcal.


CHAPTER 8-62.

8

E = P∆t = (2.0 × 1015 W)(1.0 × 10−9 s) = 2.0 × 106 J. The output from the laser is 1000 times the output of all the power stations, so we infer that Ppower stations = 2.0 × 1012 W. 1 yr = 3.16 × 107 s, which gives E = Ppower stations ∆t = (2.0 × 1012 W)(3.16 × 107 s) = 6.3 × 1019 J.

8-63.

(a) N =

1.0 × 108 Btu = 769 gal. 1.3 × 105 Btu/gal

(b) 7 months ≈ 210 days = 5.04 × 103 hr. P =

1.0 × 108 Btu . 5.04 × 103 hr

Btu 1.055 × 103 J/Btu × = 5.8 × 103 W (5.8 kW). hr 3600 s/hr P = 100 W/kg × 180 kg × 1 hp/746 W = 24 hp. This is about twice as large as the actual power output that is observed. E = P∆t = (0.06 kW)(24 hr/day)(365 days/yr) = 526 kWh/yr. The cost is ($0.15 / kWh)(526 kWh/yr) = $79/yr. = 1.98 × 104

8-64. 8-65. 8-66.

P = 4 × 5000 ft-lb/min = 2 × 104 ft-lb/min. To find the rate at which something is lifted, use P = mg

8-67.

∆h ∆h P 2 × 104 ft-lb/min . mg = 9 tons = 18 × 103 lb, so = = = 1.11 ft/min. Lifting ∆t ∆t mg 18 × 103 lb

the load 15 ft will take (15 ft)/(1.11 ft/min) = 14 min. Power = F • v. Hence, (i) in the frame of ground power = 50 N × 80 km/hr ×

8-68. 8-69. 8-70.

8-71.

5 m/s = 1100 W 18 km/hr

(ii) in frame of automobile, v = 0; therefore, Power = 0 P (100 hp)(746 W/hp) = 1.68 × 103 N. P = Fv ⇒ F = . v = 160 km/hr = 44.4 m/s. F = v 44.4 m/s Power = F • v = 400 N × (5 × 5/18) m/s × cos 35º = 455 W × 1 hp/745.7 W = 0.61 hp m(v 2f − vi2 ) ∆K = K f − K i = . vi = 0, vf = 80 km/hr = 22.2 m/s. The work done is equal to the 2 (900 kg)(22.2 m/s) 2 change in kinetic energy: ∆K = = 2.22 × 105 J. The power is 2 ∆K 2.22 × 105 J 1 hp = × = 39 hp. P= ∆t 7.6 s 746 W The time for one revolution is (3000 rev/min × 1 min/60 s)–1 = 0.02 s. There are six cylinders, and each cylinder fires once every two revolutions, so the time between firings for one of the cylinders is ∆t = 0.02 s/rev × 2 rev/6 cylinders = 6.67 × 10−3 s/cylinder. If E is the energy E ⇒ E = P∆t delivered by each cylinder, then the average power is P = ∆t 746 W = 150 hp × × 6.67 × 10−3 s/cylinder = 746 J/cylinder. hp


CHAPTER

8-72.

8 ρVAv3

(1.3 kg/m3 )(0.30)(2.8 m 2 ) 3 v = 0.546v3 . For v = 30 km/h = 8.33 m/s, P = 316 2 2 W (0.424 hp). For v = 90 km/h = 25.0 m/s, P = 8.53 kW (11.4 hp). At 30 km/h, a 300 km trip takes 10 h = 3.6 × 104 s so the total energy used is E = P∆t

P = fv =

=

= (316 W)(3.6 × 104 s) = 1.14 × 104 J. At 90 km/h, the 300 km trip takes 1/3 as long, or 1.2 × 104 s, and the total energy used is E = (8.54 × 103 W)(1.2 × 104 s) = 1.02 × 108 J. 8-73. 8-74.

P = F i v = Fv cosθ . cosθ = E=

∫

1s

0

Pdt =

∫

1s

0

P P 90 W ⇒ θ = cos −1 = cos −1 = 50°. Fv Fv (40 N)(3.5 m/s)

( P0 − P1 )(t − 1)3 [ P1 + ( P0 − P1 )(t − 1) ]dt = Pt + 1 0 3 2

1s

1s

0

3

(0.75)(0) (0.75)(−1) − = 1.0 kJ . 3 3 P = F i v = [(6.0 N)i + (8.0 N)j] i [(3.0 m/s)i − (2.5 m/s)j] = (6.0 N)(3.0 m/s) + (8.0 N)( − 2.5 m/s)

= 0.75 − 0 + 8-75. 8-76.

= –2.0 W Twenty five percent efficiency gives 110 hp, so 100% efficiency gives 440 hp = 328000 W (from chemical energy). One gallon of gas gives 1.3 × 108 J of energy. Then the rate of gas consumption = 3.28 × 105 J/sec 3600 sec × = 9.1 gal/hr 1.3 × 108 J/gal 1 hr

8-77.

∆E = 12 mv 2 i v = 100 km/hr = 100 × 5/18 m/s = 27.8 m/s

8-78.

1 (11000) kg (27.8) 2 (m/s) 2 = 4.24 × 106 J 2 Average power = ∆E/t = 4.24 × 106 J/10 s = 4.2 × 105 W ∆W ∆m 2 = P = 3.9 × 1026 W = c ∆t ∆t ∆m 3.9 × 1026 3.9 × 1026 = = = 4.3 × 109 kg/s c2 ∆t 9 × 1016 ∆E =

8-79.

In one year, ∆m = 4.3 × 109 × 365 × 24 × 60 × 60 = 1.36 × 1017 kg 8 × 1019 J/yr Power need for U.S. = (365.25 × 24 × 60 × 60)sec/yr = 2.5 × 1012 W = 2.5 × 109 kW At 1kW/m2, the area needed is 2.5 × 109m2 = 2500 km2

8-80.

F = ma = 1770(4.752 – 0.252 t) = 8410 – 446 t v = 4.752 t – 0.126 t2 Power = (8410 – 446 t)(4.752 t – 0.126 t2) = 56.2t3 – 3180 t2 + 40000 t


CHAPTER

8

t Power (W) t Power (W) 0 0 6 138000 1 36900 7 143000 2 67700 8 145000 3 92900 9 143000 4 113000 10 138000 5 123000 dp/dt = 168.6 t2 – 6360 t + 40000 = 0 at max. This occurs at t= †8-81.

6360 ± 63002 − 4(168.6)(40000) = 8.0 (taking root between 0 and 10 s) 2(168.6)

mv 2 (6.50 × 108 kg)(7.22 m/s) 2 = = 1.70 × 1010 J. 2 2 (b) Half of the power output goes toward increasing K, so that power is P = 22 × 103 hp

(a) v = 26 km/h = 7.22 m/s. K =

× 746 W/hp = 1.64 × 107 W. Then ∆t =

K 1.70 × 1010 J = = 1.03 × 103 s (17 min). 7 P 1.64 × 10 W

(c) The time to stop will be the same as the time to go from rest to the final speed, since the magnitude of the change in kinetic energy is the same in both cases. When the ship is slowing v f − vi 0 − 7.22 m/s down, its acceleration is a = = = − 7.00 × 10−3 m/s 2 . (Vector note: the – sign 1.03 × 103 s ∆t means the acceleration points in the opposite direction from the initial velocity.) The distance to v 2f − vi2 0 − (7.22 m/s) 2 = = 3.73 × 103 m (3.73 km, stop is given by v 2f − vi2 = 2ax ⇒ x = 2 −3 2a 2( −7.00 × 10 m/s ) 8-82.

8-83.

8-84.

†8-85.

or 2.3 mi). (a) Potential energy “lost” per second is mgh = 6200 m3/s × 1000 kg/m3 × 9.8 m/s2 × 49 m = 3.0 × 109J/s = 3.0 × 109 W (b) Energy lost per year = 3.0 × 106 k W × (365 × 24) hr = 2.6 × 1010 k W hr (c) This is worth 2.6 × 1010k W hr × $0.05/kW hr = $1.3 × 109 ($1.3 billion) Rate of energy loss per week = mgh = (5 kg × 9.8 m/s2 × 1.5 m) per week = 73.5 J/week = 73.5 J/week × 1/(60 × 60 × 24 × 7) = 1.2 × 104 W ∆x mg sin 10° = 550 × 746 W ∆t 550 × 745 ∆x = 8.93 m/s =v= 27000 × 9.8 × sin10° ∆t Imagine a cylinder of air with cross sectional area A moving with speed v across the windmill. During a time interval ∆t, the length of the cylinder crossing the windmill is v∆t, and the mass of air crossing the area of the windmill is ∆m = ρAv∆t. The kinetic energy carried by the cylinder of (∆m)v 2 ρ Av 3 ∆t air is ∆K = = , so the power delivered to the windmill by the moving air is 2 2 ∆K ρ Av 3 . The windmill absorbs 70% of this power, so the power output from the Pair = = 2 ∆t


CHAPTER

8

0.70 ρ Av 3 = (0.35)(1.29 kg/m3 ) Av 3 = 0.452 Av 3 . The output of the 2 generator is 90% of the power absorbed by the windmill, and the motor converts 90% of the generator power to actual useful work. So the power output of the motor is Pm = (0.90)(0.90)(0.452 Av 3 ) = 0.366 Av 3 , so if the power output of the motor is specified the Pm 746 W . Pm = 2 × 104 hp × = 1.49 × 107 W. required cross-sectional area is given by A = 3 0.366v hp windmill is Pw = 0.70 Pair =

The speed is v = 40 km/h = 11.1 m/s, so A =

1.49 × 107 W = 2.98 × 104 m 2 . This (0.366)(11.1 m/s)3

corresponds to a circle with a diameter of D = 8-86.

8-87.

8-88.

8-89.

8-90.

W = mgh. P =

4A

π

π

= 195 m.

dW dm dm P (15 hp)(746 W/hp) ⇒ = = = 38 kg/s. The density of water is = gh dt dt dt gh (9.81 m/s 2 )(30 m)

1000 kg/m3, and 1 m3 = 1000 l, so 1 kg occupies a volume of 1 l. The pump can lift 38 l/s. (a) If the helicopter is climbing at a constant speed, then the upward force exerted by the engines is equal to the downward weight mg so the net vertical force is zero. Then P = Fv = mgv = (7400 kg)(9.81 m/s 2 )(5 m/s) = 3.63 × 105 W = 487 hp. (b) If the engines are generating maximum power, then the remainder is being used to move the air and by friction. The power lost is Plost = 3080 hp – 487 hp = 2593 hp. Rate of increase of potential energy in travelling uphill ∆h 64 1 = mg = 1500 × 9.8 × × ∆t 3.6 10 = 26133 W = 35 hp.

Therefore, the power needed to travel uphill at 64 km/h is 20 + 35 = 55 hp. The power needed to travel downhill is 20 – 35 = –15 hp. The rate of energy increase from descent ∆x ⎛ 1 ⎞ 95 1 × = mg ⎜ ⎟ = 1500 × 9.8 × = 38792 W = 52 hp ∆t ⎝ 10 ⎠ 3.6 10 This is the power spent to overcome friction. Therefore, on a level road to maintain a constant speed of 95km/h the engine should deliver 52 hp. P = P0 e − t /τ . The energy used between 0 and 5 s is given by E = = − P0τ e − t /τ

5 0

∫

5

0

5

Pdt = P0 ∫ e − t / τ dt 0

= P0τ (1 − e −5 / τ ) . For P0 = 2.0 W and τ = 5.0 s, we get E = (2.0 W)(1 − e −1 ) = 6.3 J.

The total energy used from t = 0 to ∞ is E = − P0τ e − t / τ 8-91.

4(1.49 × 104 m 2 )

=

∞ 0

= P0τ = (2.0 W)(5.0 s) = 10 J.

850 hp × 745.7 W/hp = 633800 W, so that the total power of both engines = 2 × 633800 = 1.27 × 106W. The potential energy gained per second during climb = mgh per second 260 = 10900 kg × 9.8 m/s 2 × m/s = 4.63 × 105 W . 60 Therefore, the percentage engine power for climbing = (4.63 × 105)/(1.27 × 106) × 100% = 37%


CHAPTER 8-92.

8

Volume of water shot out per unit time ∆υ/∆t = Av where A = area, v = velocity. Mass of water shot out is ρAv (per unit time). All of this water reaches height h, so the energy given to the water per unit time is (∆m/∆t)gh = ρAvgh. But v = 2 gh , so that

Power = ρ A

(

)

2gh gh = 2 ρ A( gh)3/ 2

A = π(0.05 m) 2 = 7.85 × 10−3 m 2 Therefore, the power = 2(1000 kg/m3 )(7.85 × 10−3 m 2 )(9.8 × 10 m 2 /s 2 )3 / 2 †8-93.

= 1.1 × 104 W (a) 203.1 km/h = 56.4m/s Rate gravity does work = Fv = (mg sin θ)v = 75 kg (9.8 m/s2)(sin 51º)(56.4 m/s) = 3.2 × 104 J/s (b) Fice = µN = µmg cos θ 0.03 (75 kg) 9.8 m/s2 cos 51º = 13.9 N The power dissipated by ice friction = Ficev = (13.9N)(56.5 m/s) = 780 J/s Power dissipated by air = Powergravity – Powerice = 3.1 × 104 J/s

π D2

π (1.8 m) 2

= 2.54 m 2 . 4 4 The speed of the air is 40 km/h = 11.1 m/s. So the volume that passes over the propeller every dV second is = Av = (2.54 m 2 )(11.1 m/s) = 28.2 m3 /s. The mass passing the propeller every dt dm dV second is =ρ = (1.29 kg/m3 )(28.2 m3 /s) = 36.4 kg/s. The rate at which kinetic energy is dt dt dK 1 ⎛ dm ⎞ 2 (36.4 kg/s)(11.1 m/s) 2 delivered is = ⎜ = 2.25 × 103 W. The actual power output ⎟v = dt 2 ⎝ dt ⎠ 2 200 is 200 W, so × 100% = 8.9% of the power is absorbed from the wind. 2.25 × 103

8-94.

The area of the cylinder of air flowing over the propeller is A =

8-95.

Mass of air ejected per second = m′ 1 min = 8.5 m 2 /min × × 1.3 kg/m3 = 0.184 kg/s 60 sec Therefore, the kinetic energy given per second = power required = 12 m′v 2 = 12 (0.184) × (5.0)2 J/s = 2.3W


=

CHAPTER 8-96.

8-97.

8

Since the kinetic energy remains constant, the loss in potential energy must go in overcoming friction. ∆h ⎛ 280 ⎞⎛ 0.074 ⎞ Rate of energy loss = mg = A × 103 × 103 × 9.8 ⎜ ⎟⎜ 3 ⎟ ∆t ⎝ A ⎠⎝ 10 ⎠ (Where A is the cross-section of the river and we have used m = A × length × density and length = 1 km.) ∆h mg = 2 × 105 W ∆t Mass flowing per day is 2.2 × 103 km3 × 10003 m3/km3 × 1000 kg/m3 = 2.2 × 1015 kg. 1 Thus, the kinetic energy per day = mv 2 2 15 2 2 2 1 = 2 × 2.2 × 10 kg × (4.6 × 5/18) m s = 1.80 × 1015 J/day Power = 1.80 × 1015 J/day ×

1 day = 2.1 × 1010 W = 2.1 × 107 kW 24 × 3600 sec

8-98.

The power will be equal to the magnitude of the rate at which the falling water loses potential energy. The height h the water falls is equal to the diameter of the wheel, so d dm P = ( mgh) = gh = (9.81 m/s 2 )(10 m)(20 kg/s) = 2.0 × 103 W (2.0 kW). . dt dt

8-99.

The power will be equal to the magnitude of the rate at which the falling water loses kinetic (1 kg)(15 m/s) 2 = 112.5 J of kinetic energy when it first hits the energy. Each kg of water has 2 wheel, and it loses essentially all of that kinetic energy by the time it drips off the other side. Since there are 30 kg of water hitting the wheel every second, the magnitude of the total rate of energy loss is (112.5 J/kg)(30 kg/s) = 3.4 × 103 W (3.4 kW). This is the power delivered to the

8-100.

wheel by the flowing water. (a) Weight = 10780 N Flift = 10780 cos 13º = 10503 N Fƒ = 10780 sin 13º = 2425 N (b)Weight = 10780 N Flift = 10503 N Fƒ = 2425 N Fpush = Fƒ + 2425 = 4.85 × 103 N (c) 1 knot = 0.5144 m/s 1hp = 746 W 4.85 × 103 × 90 × 0.5144 Power, P = 300 hp 746 The propeller changes the path of the incoming wind stream. This requires additional work.


CHAPTER 8-101.

8-102.

†8-103.

(a) Mass of 4H atoms = (1.00813)4 u = 4.03252 u Mass of He atom = 4.00388 u ∆m in reaction = 0.02864 u E = ∆mc2 = 0.02864 u × 1.67 × 10–27 kg/u × (3.00 × 108)2 (m/s)2 = 4.30 × 10–12 J = 2.69 × 107 eV (b) Mass 4 H atoms = 4.03252 u × 1.67 × 10–27 kg/u = 6.73 × 10–27 kg gives 4.30 × 10–12 J 1 = 6.39 × 1014 J Therefore, 1 kg H atoms gives 4.30 × 10−12 J × −27 6.73 × 10 (c) 3.9 × 1026 W = 3.9 × 1026 J/s. From part (b), the energy content of H is 6.39 × 1014 J/kg. So the mass consumed = (3.9 × 1026 J/s)/(6.39 × 1014 J/kg) = 6.1 × 1011 kg/s (d) The hydrogen will last (1.5 × 1030 kg)/(6.1 × 1011 kg/s) = 2.5 × 1018 s = 7.8 × 1010 yr x x x 5 x3 3x 2 (a) U ( x) = − ∫ F i dr ' = − ∫ Fx dx ' = − ∫ [5( x ') 2 + 3 x ']dx ' = − − . (Note: The coefficients 0 0 0 3 2 5 and 3 have units of N/m2 and N/m, respectively.) 5 x3 3x 2 5(2.0 m)3 3(2.0 m) 2 (b) W = − ∆U = + −0 = + = 19 J. 3 2 3 2 WI =

∫

1,1

WI =

∫

1,1

0,0

0,0

F i dr =

∫

1,1

0,0

[(4.0i + 2.0 xj) i (dxi + dyj)]. Along path I, x = y, so this becomes

(4.0i + 2.0 yj) i (dxi + dyj) =

1

= 4.0 x 0 + y (b) WII =

∫

2 1

0,1

1,1

0

∫

1,1

0,0

(4.0dx + 2.0 ydy ) =

1

1

0

0

∫ 4.0dx + ∫ 2.0 ydy

= 5.0 J 0,0

0,1

0,0

F i dr + ∫ F i dr = ∫ [(4.0i + 2.0 xj) i (dxi + dyj)] + ∫ [(4.0i + 2.0 xj) i (dxi + dyj)] 0,1

1,1

0,1

From (1,1) to (0,1) y is constant at 1 (dy = 0) as x varies from 1 to 0. From (0,1) to (0,0) x is constant at 0 (dx = 0) as y varies from 1 to 0. WII =

8-104.

8

∫

0

1

0

4dx + ∫ 0dy = −4.0 J. If the force were 1

conservative, we would have WII = – WI. This is not what we find, so we conclude that this force is not conservative. (a) Using the conservation of energy 1 2 mv0 + W = 0 2 1 2 mv0 = µ k mgd = K max 2 K max = 47 J

(b) Once again we use the conservation of energy 1 kx02 = 47 J 2 x02 =

2 × 47 J = 0.78 m 2 120 N/m

x0 = 0.89 m


CHAPTER 8-105.

8-106.

8-107.

8-108.

†8-109.

8

(a) Let W be the maximum weight that can be pulled, then µkW = 0.3 W = 360 × 6000 W = 7.2 × 106 N (b) Power = Force × speed = Fv Power 6000 × 0.2 × 746 ν = = = 0.12 m/s F 7.2 × 106 1 1 60 km/hr = 16.7 m/s. ∆E = mv 2 = (990) (16.7) 2 J = 140000 J. 2 2 Then the energy dissipated = ∆E = 140000 J Average power = ∆E/∆t = 137500/2.1 J/s = 6.5 × 104 W 6.5 × 104 W/745.7 hp/W = 88 hp ∆E (∆m) gh Power wasted = = ∆t ∆t 3 = 13 × 10 × 103 × 9.8 × 33 = 4.2 × 109 W

12 km/h = 3.33 m/s. The rate of potential energy gain = (mg sin θ)( v ) = (70 × 9.8 × 33.3 × 1/10) J/s = 228 J/s = (228 J/s × 3600 s/h)/4.186 × 103 kcal/J = 196 kcal/h Total energy needed = 750 kcal + 196 kcal = 946 kcal/h (a) Since the projectile is traveling horizontally, its potential energy is constant and its total energy is just equal to its kinetic energy. The instantaneous power is then equal to the dK d ⎛ mv 2 ⎞ dv instantaneous rate of change of K: P = ⎜ ⎟ = mv . (Don’t forget the chain rule!) dt dt ⎝ 2 ⎠ dt P = (45.36 kg)(655.9 − 61.1 t + 3.26 t 2 )(−61.1 + 6.52 t ) = − 1.82 × 106 + 3.63 × 105 t − 2.71 × 104 t 2 + 964 t 4

8-110.

(The magnitude of this is the power “removed from” the projectile.) mv(0) 2 (45.36 kg)(655.9 m/s) 2 = = 9.757 × 106 J. At t = 3.00 s, (b) At t = 0, K (0) = 2 2 mv(3.00) 2 (45.36 kg)[655.9 − (61.1)(3.00) + (3.26)(3.00) 2 ] m/s) 2 K (3.00) = = = 5.71 × 106 J. 2 2 ∆K 5.71 × 106 J − 9.757 × 106 J = = −1.35 × 106 W. (c) The average power is P = ∆t 3.00 s (a) W = 2 F ∆x = 2(100 N)(0.50 m) = 100 J. W 100 J = = 1.0 × 102 W. ∆t 1.0 s The athlete has two legs, each of which can deliver energy at a rate of 200 W. The total energy change required is equal to the change in potential energy to climb the stairs: ∆E ∆E 3.36 × 105 J ∆E = mgh = (75 kg)(9.81 m/s 2 )(457 m) = 3.36 × 105 J. P = ⇒ ∆t = = ∆t P 400 W (b) P =

†8-111.

= 8.4 × 102 s (14 min).


CHAPTER

8-112.

The change in kinetic energy for each kg of water is

8

K v 2 (50 m/s) 2 J = = = 1250 . The pump kg 2 2 kg

takes in 0.80 kg/s, so the power is P = (1250 J/kg)(0.80 kg/s) = 1.0 × 103 W (1.0 kW). †8-113.

(a) The total gravitational potential energy stored in the reservoir is U = mgh = ρVgh. Assuming all the energy can be extracted from the falling water with no kinetic energy left over and no energy lost due to friction, the total energy output will be 1 kWh E = U = (1000 kg/m 3 )(2.2 × 107 m3 )(9.81 m/s 2 )(270 m) × = 1.6 × 107 kWh. 3.6 × 106 J (b) Again assuming all the energy is converted to electrical energy, the power will be dU d dV P= = ( ρVgh) = ρ gh , so the volume flow rate will be dt dt dt dV P 109 W = = = 3.8 × 102 m 3 /s. dt ρ gh (1000 kg/m3 )(9.81 m/s 2 )(270 m)

8-114.

(a)

U 2 + K 2 = U1 + K1 . U1 =

kx 2 , U 2 = mgh. K1 = K 2 = 0. 2

2

2

2

kx 2mgh 2(50 kg)(9.81 m/s )(4 m) ⇒x= = = 1.06 m. 2 k 3500 N/m (b) The maximum acceleration occurs when the net force is the largest, and that will be at point 1 where the spring is fully compressed. The spring exerts an upward force with magnitude kx, and gravity exerts a downward force with magnitude mg. The magnitude of the net upward force is Fnet = kx – mg. The magnitude of the acceleration is F (3500 N/m)(1.06 m) − (50 kg)(9.81 m/s 2 ) a = net = = 64 m/s 2 . This is m 50 kg mgh =

h

3 x 1

about 6.6 gee. (c) The maximum speed occurs at point 3, where the spring returns to its equilibrium length and mv 2 its potential energy is zero. U 2 − U 3 + K 2 − K 3 = mgx − mgx + 0 − max = 0 ⇒ 2 vmax = 2 g (h − x) = 2(9.81 m/s 2 )(4.0 m − 1.06 m) = 7.6 m/s.


CHAPTER 9

GRAVITATION

Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions Manual, available for purchase. Answers to all solutions below are underscored. 9-1. 9-2.

†9-3.

F = GMm/r2 = 6.67 × 10–11 Nm2/kg2 × 7 × 108 kg × 7 × 108 kg/(2000 m)2 = 8.2 N The mass of each proton is mp = 1.67 × 10–27 kg. Hence, 2 Gm p m p (1.67 × 10−27 kg)2 −11 N i m F= 6.67 × 10 × = 4.7 × 10−35 N = r2 kg 2 (2.0 × 10−15 m)

If an object of mass m is placed at this special point a distance r1 from the earth and r2 from the Moon, the net force on it will be zero:

GM E m GM M m − = 0. Note that finding the point doesn’t depend on the mass m because m will r12 r22 cancel from the equation. G will also cancel. Let d stand for the Earth–moon distance. Then r2 = d M MM = 0, or r12 ( M E − M m ) − 2dM E r1 + M E d 2 = 0. – r1, and the equation becomes 2E − 2 (d − r1 ) r1 Solving the equation in this form will involve working with very large numbers, which increases the likelihood of making a mistake. The numbers can be made more manageable by dividing through by ME and d2, and defining a new variable x = r1/d, which gives the location of the equilibrium point as a fraction of d. Doing this gives the following equation: x 2 (1 − M m / M E ) − 2 x + 1 = 0. Using the values of the masses from the table in the book finally 2 ± 22 − 4(0.9877) gives 0.9877 x − 2 x + 1 = 0, which has solutions x = = 1.125, 0.900. The 2(0.9877) 2

point must lie between Earth and the Moon, so the correct value of x must be less than 1. That means we reject the first answer and finally get r1 = 0.900d = 0.900(3.84 × 108 m) = 3.46 × 108 m.

9-4.

†9-5.

a = GM/R2 where M, R are mass, radius of planet, respectively. Venus: a = 6.67 × 10–11 Nm2/kg2 × 4.87 × 1024kg/(6.05 × 106)2 m2 = 8.9 m/s2 Mercury: a = 6.67 × 10–11Nm2/kg2 × 3.30 × 1023kg/(2.44 × 106)2m2 = 3.70 m/s2 Mars: a = 6.67 × 10–11 Nm2/kg2 × 6.42 × 1023kg/(3.397 × 106)2m2 = 3.71 m/s2 GMm F= , where M is the mass of the Sun or Moon, m = 70 kg, and r is the distance from the r2 Sun or Moon to the center of Earth. (There are “tidal” effects caused by the fact that objects on one side of Earth are closer to the Sun or Moon than objects on the other side, but those effects will be ignored in this problem.) (6.67 × 10−11 N i m 2 /kg 2 )(1.99 × 1030 kg)(70 kg) For the Sun, Fsun = = 0.41 N (1.50 × 1011 m) 2


CHAPTER

For the Moon, Fmoon =

9-6.

†9-7.

(6.67 × 10−11 N i m 2 /kg 2 )(7.35 × 1022 kg)(70 kg) = 2.3 × 10−3 N (3.84 × 108 m) 2

The force exerted on the person by Earth is mg = (70 kg)(9.81 m/s2) = 687 N. We see that the forces exerted by the Sun and Moon are very small compared to the force exerted by Earth. However, the effect of the forces would be almost unnoticeable anyway, even if they were larger. The reason we notice the force due to Earth is because we are pulled down against the surface, and we feel the reaction force as “weight.” If we were in free fall relative to Earth, that reaction force would be absent and we would feel “weightless.” The Earth (and everything on it) is in free fall around the Sun and Moon, so there is no sense of “weight” from the attractive forces exerted by those bodies. GMm F= . 1 ly = the distance light travels in 1 yr = (3.00 × 108 m/s)(3.15 × 107 s) r2 = 9.45 × 1015 m. (6.67 × 10−11 N i m 2 /kg 2 )(2.0 × 1011 )(3.0 × 1011 )(1.99 × 1030 kg) 2 F= [(2.2 × 106 ly)(9.45 × 1015 m/ly)]2 = 3.8 × 1028 N GMm F= . The force exerted on the sun by Alpha Centauri is r2 (6.67 × 10−11 N i m 2 /kg 2 )(2.0 × 1030 kg)(1.99 × 1030 kg) FAlpha = = 1.5 × 1017 N. The force exerted [(4.4 ly)(9.45 × 1015 m/ly)]2 on the Sun by Earth is (6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg)(1.99 × 1030 kg) FEarth = = 3.5 × 1022 N. The magnitude (1.50 × 1011 m) 2

9-8.

†9-9.

9

of the force exerted by Earth is about five orders of magnitude greater than the force exerted by Alpha Centauri. The Sun–Moon distance, Rsm = 150 × 1011 m The Earth–Moon distance, Rem = 3.84 × 108 m ms = 1.99 × 1030 kg. me = 5.98 × 1024 kg. mm = 7.35 × 1022 kg. Then Fsm = (Gmsmm)/Rsm2 = 4.336 × 1020 N and Fem = (Gmemm)/Rem2 = 1.99 × 1020 N. With Earth between the Sun and the Moon, the direction of the net force acting on the Moon is directly toward Earth (and the Sun!), and has magnitude 6.32 × 1020 N. With the moon between Earth and the Sun, the direction of the force on the moon is toward the sun, and its magnitude is 2.35 × 1020 N. GM g = 2 . Use a spreadsheet to calculate the values of g. R Planet M (kg) R (m) g (m/s2) Jupiter 1.90E + 27 7.14E + 07 24.9 Saturn 5.67E + 26 6.00E + 07 10.5 Uranus 8.70E + 25 2.54E + 07 8.99


CHAPTER 9-10.

9

(See the solution for Problem 9-3.) At ×, the magnitude of the force exerted by the sun on a mass GM s m GM e m m will be the equal to the magnitude of the force exerted by the earth: = 2 x ( R − x)2 ⎛M ⎞ ⎛ M ⎞ ⇒ ( R − x) 2 = x 2 ⎜ e ⎟ . Rearranging gives x 2 ⎜1 − e ⎟ − 2 Rx + R 2 = 0, which has roots ⎝ Ms ⎠ ⎝ Ms ⎠

x=

2 R ± 4 R 2 − 4(1 − M e / M s ) R 2 2(1 − M e / M s )

. This simplifies to x =

R ± (M e / M s ) (1 − M e / M s )

. Choose the – sign

for the root, because the plus sign will give a value of x larger than R. The ratio of the masses is (5.98 × 1024 /1.99 × 1030 ) = 3.01 × 10−6 , so it can be neglected in the denominator. The result is x = (1 − 3.01 × 106 ) R = 0.998R, so the point is 99.8% of the way from the Sun to Earth. 9-11.

9-12.

The weight of 1 kg is 9.81 N at Earth’s surface. The force exerted on the Sun by another sunlike GM s2 (6.67 × 10−11 N i m 2 /kg 2 )(1.99 × 1030 kg) 2 star is F = = = 1 × 109 N. So even though the 2 20 2 (5 × 10 m) R stars are very far apart, their large masses cause the force of one on the otherθto be much larger θ than the force Earth exerts on a 1 kg mass at its surface. We expect that the angle θ will be small, so the distance between the masses is approximately equal to d. (The size of the angle is d exaggerated in the diagram.) There is a weight mg pulling straight Gm 2 down on each mass, a horizontal force directed from each d2 mass toward the other, and a tension T directed along each string. Resolving the tension into vertical and horizontal components for Gm 2 one of the masses gives T cosθ = mg , T sin θ = 2 . Dividing d the second equation by the first gives Gm 2 (6.67 × 10−11 N i m 2 /kg 2 )(1.5 kg) 2 = tan θ ≈ θ = y gd 2 (9.81 m/s 2 )(0.080 m) 2 = 1.6 × 10–9 rad (9.1 × 10–8°).

†9-13.

The 3.0 kg mass is at the origin (0,0). The 4.0 kg mass is located at (0, 2.0 m) and the 7.0 kg mass is at (3.0 m, 0). The net force F on the 3.0 kg mass is given by G (3.0 kg)(7.0 m) G (3.0 kg)(4.0 m) F= i+ j 2 (3.0 m) (2.0 m) 2 = (1.56 × 10−10 N)i + (2.00 × 10−10 N)j. The magnitude of the force is F = (1.56) 2 + (2.00) 2 × 10−10 N = 2.5 × 10 –10 N. The

2.00 = 52°. Note that in the 1.56 calculation of F the common factor of 10–10 was taken out of the square root for simplification, and likewise the common factor of 10–10 was omitted in the calculation of θ. direction is given by θ = tan −1


F

4.0 kg

θ 3.0 kg

x 7.0 kg

CHAPTER

9-14.

F=

9

Gm 2 Gm 2 (1 + cos 60 ° ) i + sin 60° j. The magnitude is a2 a2 1/ 2

Gm 2 Gm 2 F = 2 (1 + cos 60°) 2 + sin 2 60° = 2 a a F=

Gm 2 ⎡ 9 3 ⎤ + a 2 ⎢⎣ 4 4 ⎥⎦

1/ 2

=

⎡⎛ 1 ⎞ 2 ⎛ 3 ⎞ 2 ⎤ ⎢⎜ 1 + ⎟ + ⎜ ⎟ ⎥ ⎢⎝ 2 ⎠ ⎝⎜ 2 ⎠⎟ ⎥ ⎣ ⎦

Gm 2 3 . The magnitude of the force will be the a2

same for the other two masses. †9-15.

GM earth m F . By Newton’s Second Law, a = , so 2 r m −11 2 2 24 (6.67 × 10 N i m /kg )(5.98 × 10 kg) = = 2.71 × 10−3 m/s 2 . As a fraction of g, (3.84 × 108 m) 2

The magnitude of the force on a mass m is F = a=

GM earth r2

2.71 × 10−3 g = 2.76 × 10−4 g. 9.81 GM earth GM wtower = mgtower = m . At the surface, wsurface = mg surface = m 2 earth . The ratio is 2 Rearth ( Rearth + htower )

a=

9-16.

2 wtower Rearth (6.357 × 106 m) 2 = = = 0.578. wsurface ( Rearth + htower ) 2 (6.357 × 106 m + 2.0 × 106 m) 2

†9-17.

Let r stand for the radius of the earth or Io and R stand for the distance from the center of the planet to the center of the moon. Let M stand for the mass of the body causing the tidal force (either the moon or Jupiter), and m stand for the R mass of the body affected by the tidal force R-r (either the earth or Io). A diagram is shown for R+ the case of the Earth–Moon system. The difference in centripetal accelerations from the near side to far side of the affected body is ⎡ ( R + r )2 − ( R + r )2 ⎤ ⎡ 1 ⎡ ⎤ 1 ⎤ 4rR GM . Divide = GM ⎢ ∆a = GM ⎢ − = ⎢ 2 2 ⎥ 2 2 ⎥ 2 2 ⎥ (R + r) ⎦ ⎣ (R − r) (R + r) ⎦ ⎣ (R − r) ⎣ (R − r) (R + r) ⎦ through by R4 to change the variable to r/R: ⎡ ( R + r ) 2 − ( R + r ) 2 ⎤ GM ⎡ ⎤ 4(r / R ) . Note that one factor of R2 was ∆a = GM ⎢ = 2 ⎢ 2 2 ⎥ 2 2 ⎥ − + − + ( R r ) ( R r ) R (1 r / R ) (1 r / R ) ⎣ ⎦ ⎣ ⎦ brought out under GM while the other factor was put under rR in the numerator. 6.378 × 106 m For the Earth–Moon system, r / R = = 0.0166, and at the orbit of the Moon 3.84 × 108 m GM (6.67 × 10−11 N i m 2 /kg 2 )(7.35 × 1022 kg) = = 3.33 × 10−5 m/s 2 . Thus R2 (3.84 × 108 m) 2 ⎡ ⎤ 4(0.0166) ∆a = (3.33 × 10−5 m/s 2 ) ⎢ = 2.21 × 10−7 m/s 2 . The surface gravity on Earth is 2 2 ⎥ ⎣ (0.9834) (1.0166) ⎦


CHAPTER

9

g = 9.81 m/s2, so and

∆a 1.82 × 106 m = 4.31 × 10−3 = 2.25 × 10−7. For the Jupiter–Io system, r / R = 4.22 × 108 m g

GM (6.67 × 10−11 N i m 2 /kg 2 )(1.90 × 1027 kg) = = 0.712 m/s 2 . 2 8 2 R (4.22 × 10 m)

⎡ ⎤ 4(4.31 × 10−3 ) ∆a = (0.712 m/s 2 ) ⎢ = 0.0123 m/s 2 . The surface gravity on Io is 2 2 ⎥ (0.99569) (1.00431) ⎣ ⎦

g=

∆a Gm (6.67 × 10−11 N i m 2 /kg 2 )(8.9 × 1022 kg) = = 1.79 m/s 2 , which gives = 6.86 × 10−3 . 2 6 2 (1.82 × 10 m) r g

We see that the effects (absolute and relative) caused by Jupiter on Io are much larger than the effects caused by the Moon on Earth. 9-18.

F=

6.67 × 10−11 × 1.99 × 1030 × 7.35 × 1022 ˆ i (1.5 × 1011 ) 2

6.67 × 10−11 × 5.98 × 1024 × 7.35 × 1022 j (3.84 × 108 ) 2 = (43.36 × 1019 î − 19.88 × 1019 j) N −

F = (43.36) 2 + (19.88) 2 × 1019 N = 47.7 × 1019 N

⎛ 19.88 ⎞ ⎟ = 24.6° ⎝ 43.36 ⎠ GM (6.67 × 10−11 N i m 2 /kg 2 )(3.8 × 1019 kg) = 0.0406 m/s 2 . This is the largest centripetal g= 2 = 5 2 (2.50 × 10 m) R

θ = tan −1 ⎜ 9-19.

acceleration that can be provided, so the maximum equatorial rotation speed is vmax = 9-20.

gR = (0.0406 m/s 2 )(2.50 × 105 m) = 101 m/s. (Note that we can only find a speed, not a

velocity.) This corresponds to a rotation rate of about one revolution every 14 minutes. Orbital speed = GM E /r , M E = 5.98 × 1024 kg r = 6378 km + 500 km = 6.88 × 106 m Speed v = 6.67 × 10−11 N/kg 2 m 2 × 5.98 × 1024 kg/6.88 × 106 m

9-21.

= 7.6 × 103 m/s T = 2πr/v = 2π(6.88 × 106 m)/7.6 × 103 m/s = 5700s = 1.6 hr 2π r 2π (4.23 × 107 m) v= = = 3.08 × 103 m/s (3.08 km/s). 24 h × 3600 s/h T


CHAPTER

9-22.

9-23.

⎛ r ⎞⎛ T ⎞ 2π rV 2π rE (rV / rE ) = = vE ⎜ V ⎟ ⎜ E ⎟ . From Example 4, vE = 30 km/s. From Example 6, rV TV TE (TV / TE ) ⎝ rE ⎠⎝ TV ⎠ /rE = 0.725 and TE / TV = 1.63. vV = (30 km/s) ( 0.725 )(1.63) = 35 km/s.

vV =

T = 2π is v =

9-24.

9

(3 × 104 ly × 9.45 × 1015 m/ly)3 r3 = 2π = 5.8 × 1015 s, or 1.8 × 108 y. The speed 2 2 41 −11 (6.67 × 10 n i m /kg )(4 × 10 kg) GM

2π R 2π (2.84 × 1020 m) = = 3.1 × 105 m/s. T 5.8 × 1015 s

(a) T = (2π / GM )r 3/ 2

v = 2π r/T .

2π / GM S = T / r

−8

3/ 2

= 3.22 × 10 s / m

Therefore,

3/ 2

Moon Period Orbital speed Tethys 1.89 days 1.14 × 104 m/s Dione 2.73 days 1.00 × 104 m/s Rhea 4.51 days 8.49 × 103 m/s Titan 15.91 days 5.60 × 103 m/s Iapetus 79.10 days 3.27 × 103 m/s (b) Since 2π / GM S = 3.22 × 10−8 s/m3/2 , GM S =

4π 2 m 3 /s 2 −8 2 (3.22 × 10 )

Ms = 4π2/[(3.22 × 10–8)2 × 6.67 × 10–11] = 5.7 × 1026 kg †9-25.

T2 =

4π 2 r 3 ⇒T = GM J

4π 2 r 3 = GM J

(6.67 × 10−11

4π 2 r 3/ 2 , or N i m 2 /kg 2 )(1.90 × 1027 kg)

T = 1.765 × 10−8 r 3/ 2 , where r is in m and T is in s. Use a spreadsheet to do the calculations. Moon Io Europa Ganymede

r (m) 4.22E + 08 6.71E + 08 1.07E + 09

T (s) 1.53E + 05 3.07E + 05 6.18E + 05

T (days) 1.77 3.55 7.15

1/ 3

9-26.

⎛ GM E T 2 ⎞ 4π 2 r 3 ⇒r =⎜ ⎟ 2 GM E ⎝ 4π ⎠

⎡ (6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg)(12 h) 2 (3600 s/h) 2 ⎤ = ⎢ ⎥ 4π 2 ⎣ ⎦ 4 2π r 2π (2.66 × 10 km) which gives r = 2.66 × 107 m. The orbital speed is v = = = 3.87 km/s. If T (12 h)(3600 s/h) 2

T =

the satellite is moving in the same direction as the earth’s rotation, it will pass over any given point on the equator once each day. (This may seem surprising, but a quick sketch shows that if the satellite is above a certain point on the earth at the beginning of an orbit, that point will be on the opposite side of the orbit when the satellite returns to its initial position 12 hours later.) If the satellite is traveling in the opposite direction from the earth’s rotation, it will pass over any given point on the equator three times each day.


1/ 3

CHAPTER

9-27.

9

T2 =

4π 2 r 3 ⇒T = GM S

4π 2 r 3 . r = 4rS, so GM S

T =

4π 2 (4 × 6.96 × 108 m)3 = 8.01 × 104 s = 0.927 day. (6.67 × 10−11 N i m 2 /kg 2 )(1.99 × 1030 kg) 1/ 3

9-28.

9-29.

9-30.

⎛ GM S T 2 ⎞ ⎡ (6.67 × 10−11 N i m 2 /kg 2 )(1.99 × 1030 kg)(26 days) 2 (24 h/day) 2 (3600 s/h) 2 ⎤ r=⎜ ⎟ =⎢ ⎥ 2 4π 2 ⎣ ⎦ ⎝ 4π ⎠ 10 = 2.57 × 10 m.

T =

4π 2 r 3 = GM

4π 2 (2 × 106 m)3 = 8.02 × 103 s. 1 day = 86,400 s, so (6.67 × 10−11 N i m 2 /kg 2 )(7.35 × 1022 kg)

the command module passes over the excursion module about ten times each earth day. 4π 2 r 3 4π 2 r 3 ⇒M = T2 = . 13 yr = 4.11 × 108 s so GM GT 2 4π 2 (8.2 × 1011 m)3 M = = 1.9 × 1030 kg. The mass of 55 Cancri is very 2 2 8 3 −11 (6.67 × 10 N i m /kg )(4.11 × 10 s) similar to the mass of our Sun.

9-31.

T = 2π r 3 /(GM E ) r = 6.67 × 106 m Therefore, T = 2π (6.67 × 106 )3 m3 /(6.67 × 10−11 m3 /s 2 kg × 5.98 × 1024 kg)

9-32.

9-33. 9-34.

= 5420 s = 0.0627 day In this time Earth would have spun around (360°/day × 0.0627 day) = 22.6°. Hence, the satellite is at the same latitude but 22.6º W, i.e., over ≈ Lincoln, Nebraska. Let the radius of the orbit around the CM be R. Then the force between the stars is F = GMm/(2R)2 = centripetal force = mv2/R ⇒ GM/4R = v2; T = 2πR/v. Therefore, T2 = 4π2R2/v2 = 16π2R3/GM M = 16π2R3/GT2 (a) Using this M = 16π2 (8.67 × 108)3 m3/[(6.67 × 10–11 m3/kgs) × (7.75 × 3600s)2] = 1.98 × 1030 kg (b) Speed = 2πR/T = 2π(8.67 × 108 m)/(7.75 × 3600s) = 1.95 × 105 m/s Since center of mass given by m1x1 + m2x2, we have m1/m2 = x2/x1 where x1, x2 are distances to CM, at equal times. This is proportional to the radii; therefore, m1/m2 = 2.2/1.4 = 1.6. The equation of motion of the first star is m1v12 m1m2 =G r1 (r1 + r2 ) 2 and that of the second is m2 v22 m1m2 =G r2 ( r1 + r2 ) 2 with v1 = 2πr1/T, v2 = 2πr2/T, this becomes


1/ 3

CHAPTER

9

4π 2 r1 Gm2 = T2 (r1 + r2 ) 2 and 4π 2 r2 Gm1 = 2 T ( r1 + r2 ) 2 Adding these equations side to side, we find r +r m + m2 4π 2 1 2 2 = G 1 T ( r1 + r2 ) 2 or T2 = †9-35.

4π 2 (r1 + r2 )3 G (m1 + m2 )

r1 10M S 2 ⇒ r1 = r2 . d is the = r2 25M S 5 d 7 separation between the stars, so d = r1 + r2 = r2 . The 5 r1 + 25MS 10MS gravitational attraction between the stars provides the centripetal force that keeps them in orbit: G (25M S )(10 M S ) r2 F= . For the 10MS black hole orbiting at d2 a distance r2 with a period T, we get G (25M S )(10M S ) (10M S )v 2 (10M S )(4π 2 r2 ) . Substituting d in terms of r2 and solving for r2 = = d2 r2 T2

From Problem 9-34,

1/ 3

⎡ 625GM S T 2 ⎤ 5 gives r2 = ⎢ ⎥ . The period is T = 5.6 days = 4.84 × 10 s. The final result is 2 196 π ⎣ ⎦ 1/ 3

9-36.

⎡ 625(6.67 × 10−11 N i m 2 /kg 2 )(1.99 × 1030 kg)(4.84 × 105 s) 2 ⎤ 10 r2 = ⎢ ⎥ = 2.16 × 10 m. The 2 196 π ⎣ ⎦ 7 separation is d = r2 = 3.0 × 1010 m. 5 From Problem 9-14, we know that the net force on any of the masses GM 2 3 . The relationship between the radius R of is given by F = a2 a the circle and the side a of the equilateral triangle is a R 3 GM 2 R , so a = R 3. In terms of R, F = 2 . This = R cos30° = 30º 2 2 R 3 force points toward the center of the circle and is the centripetal force keeping the masses moving in their orbit. GM 2 Mv 2 4π 2 MR , where T is the period of the orbit. Solving = = R T2 R2 3 1/ 2

⎛ 4π 2 R 3 3 ⎞ for T gives T = ⎜⎜ ⎟⎟ . GM ⎝ ⎠


CHAPTER †9-37.

9

(a) “Low altitude” means that the radius of the orbit is essentially the same as the radius of the earth, 6.38 × 106 m. The period of a satellite in such an orbit is T =

4π 2 r 3 = GM

4π 2 (6.38 × 106 m)3 2π r = 5.07 × 103 s, so its orbital speed is vorbit = = 2 2 24 −11 T (6.67 × 10 N i m /kg )(5.98 × 10 kg) 2π (6.38 × 106 m/s) = 7.91 × 103 m/s. A point on the surface of the earth makes one complete 3 5.07 × 10 s rotation every 24 hours, so the tangential speed of a point at latitude 28° is 2π (6.38 × 106 m)cos 28° = 410 m/s. (Note that the tangential speed is much less than the vtan = (24 h)(3600 s/h) orbital speed! Objects on the surface of the earth are not in orbit.) The earth rotates from west to east, so if a satellite is launched toward the east, it gains 410 m/s from the earth, and the required launch speed is reduced by that amount: vlaunch = 7.91 × 103 m/s − 410 m/s = 7.50 × 103 m/s. . If the satellite is launched toward the west, its launch speed must be increased: vlaunch = 8.32 × 103 m/s. 2 mvlaunch (14.0 kg)(7.50 × 103 m/s) 2 = = 3.94 × 108 J. Launching west, 2 2 2 3 2 mvlaunch (14.0 kg)(8.32 × 10 m/s) K= = = 4.85 × 108 J. 2 2 Since angular momentum is conserved, and at these points v vector is perpendicular to r vector mv1r1 = mv2r2 ⇒ v1r1 = v2r2 vr v1 = 2 2 = (8.75 × 1010 m)(5.46 × 104 m/s)/(5.26 × 1012 ) m = 908 m/s r1

(b) Launching east, K =

9-38.

9-39.

9-40.

6.74 × 106 v = 8.91 × 106 × 6.21 × 103 (see eq. 9–19) where ν is the speed at perigee 8.91 × 6.21 × 103 v= = 8.21 × 103 m/s 6.74 By Kepler’s Third Law T 2 ∝ R 3 . Here 1 R = [(6380 + 175) km + (6380 + 181,200)] km = 97,070 km. 2 2 T = 4π 2 r 3 / GM , with r = semimajor axis. Hence,

T = 2π r 3 / GM = 2π

(97070,000)3 m3 6.67 × 10−11 m3 / kg s 2 5.98 × 1024 kg

= 3.01 × 105 s = 3.5 day †9-41.

Even though it’s not explicitly given in the text, the mathematical statement of Kepler’s Third Law for elliptical orbits is T = 106 m, and T =

4π 2 a 3 , where a is the semimajor axis. For Sputnik I, a = 6.97 × GM S

4π 2 (6.97 × 106 m)3 = 5.79 × 103 s (96.5 min). For −11 2 2 24 (6.67 × 10 N i m /kg )(5.98 × 10 kg)


CHAPTER

Explorer I, a = 7.83 × 106 m, and T =

4π 2 (7.83 × 106 m)3 = (6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg)

6.89 × 103 s (115 min). 9-42.

rmax = aphelion distance, rmin = perihelion distance. From the diagram, rmax = a + f , rmin = a − f . Thus r −r 1530 − 1350 × 106 km = 80.0 × 106 km. The distance f = max min = 2 2 from the sun to the opposite focus is 2f 6 = 160 × 10 km. From Table 9.1, Mercury’s mean distance from the sun is 57.9 × 106 km, which is about one third of the interfocal distance for Saturn’s orbit.

+

×

a

f

1/ 3

†9-43.

⎛ GM S T 2 ⎞ Using the equation given for Problem 9-41, a = ⎜ ⎟ , where 2 ⎝ 4π ⎠ 10 T = 2380 yr = 7.326 × 10 s.

+

×

1/ 3

⎡ (6.67 × 10−11 N i m 2 /kg 2 )(1.99 × 1030 kg)(7.326 × 1010 s) 2 ⎤ a=⎢ ⎥ 4π 2 ⎣ ⎦ 13 = 2.67 × 10 m. From the diagram 9-42, we see that rmin + rmax = 2a ⇒ rmax = 2a − rmin

9-44. 9-45.

From Eq. 9.10, v =

GM earth = r

(6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg) 6.6 × 106 m

(6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 2(6.6 × 106 m) U ( r ) = − GMm / r −

GM earth m = 2r kg)(4.7 × 103 kg) = −1.4 × 1011 J.

(6.67 × 10−11 N 2 / kg 2 1030 kg × 5.98 × 1024 kg) = −5.31 × 1033 J 11 1.496 × 10 m 1 1 K = mv 2 = (5.98 × 1024 kg)(29800 m/s) 2 = 2.66 × 1033 J 2 2 Total E = K + U = –2.65 × 1033J 2GM S GM S v vesc = , vorb = ⇒ esc = 2. RS RS vorb =−

9-47.

a

= 2(2.67 × 1013 m) − 1.37 × 1011 m = 5.33 × 1013 m, or 5.33 × 1010 km. This is about 10 times Pluto’s mean orbital distance from the Sun. v r 1.52 rmin vmax = rmax vmin ⇒ max = max = = 1.03. vmin rmin 1.47

= 7.8 × 103 m/s (7.8 km/s). From Eq. 9-24, E = −

9-46.

f


9

CHAPTER

9-48.

9 2GM J = RJ

(a) vimpact = vesc =

2(6.67 × 10−11 N i m 2 /kg 2 )(1.90 × 1027 kg) 7.14 × 107 m

= 5.96 × 104 m/s (59.6 km/s). (b) K impact =

2 mvimpact

2

=

(1.0 × 1010 kg)(5.96 × 104 m/s) 2 = 1.78 × 1019 J. This corresponds to 2

19

1.78 × 10 J = 4.2 × 109 tons (4200 MT of TNT). 4.2 × 109 J/ton 9-49.

U =− v=

GM earth m (6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg)(1 kg) =− = −1.04 × 106 J. 8 3.84 × 10 m r

GM earth = r

(6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg) = 1.02 × 103 m/s, so 8 3.84 × 10 m

mv 2 (1 kg)(1.02 × 103 m/s) 2 = = 5.20 × 105 J. E = U + K = –5.20 × 105 J = –K. 2 2 The apogee of the new orbit is rmax = rsync = 4.23 × 107 m (from Example 6). The new speed is 1 GM earth = 1.54 × 103 m/s. The one-half of the orbital speed for the synchronous orbit: v = rsync 2 K=

9-50.

total energy of an orbit with semimajor axis a is E = −

GM earth m . The total energy of the satellite 2a

⎛ v 2 GM earth ⎞ at its apogee is E = m ⎜ − ⎟ . These energies are equal, so the mass m of the satellite rmax ⎠ ⎝ 2 ⎛ v 2 GM earth ⎞ GM earth cancels and we have ⎜ − , which can be solved for 2a: ⎟=− rmax ⎠ 2a ⎝ 2 2a =

2GM earth rmax 2(6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg)(4.23 × 107 m) = 2GM earth − rmax v 2 2(6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg) − (4.23 × 107 m)(1.54 × 103 m/s) 2

= 4.84 × 107 m. The perigee is rmin = 2a − rmax = 6.04 × 106 m. This is slightly less than the radius †9-51.

of the earth, so the satellite will probably burn up in the atmosphere. 2GM earth 2GM earth ⇒R= . Substituting numerical values For an escape speed of c, vexc = c = R c2 2(6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg) = 8.86 × 10−3 m, or 8.86 mm. gives R = (3 × 108 m/s) 2 1/ 3

9-52.

⎛ GM S T 2 ⎞ GM S m E=− and a = ⎜ ⎟ . There are two orbits with periods T1 and T2. Since all the 2 2a ⎝ 4π ⎠ parameters are constant except for the periods, the ratio of energies is

⎛ 4206 ⎞ =⎜ ⎟ ⎝ 2380 ⎠

2/3

= 1.462. . E2 = 1.462 E1 , so ∆E = 0.462 E1 , or

about 46%.


E2 ⎛ T1 ⎞ =⎜ ⎟ E1 ⎝ T2 ⎠

2/3

∆E = 0.462. The energy increased by E1

CHAPTER

†9-53.

vesc =

2GM moon = Rmoon

9

2(6.67 × 10−11 N i m 2 /kg 2 )(7.35 × 1022 kg) = 2.37 × 103 m/s. 1.74 × 106 m

600 = 0.253, so the average speed is about 25% of the escape speed. But since some 2.37 × 103

9-54.

molecules are moving faster than 600 m/s, some of them will be moving fast enough to escape into space. Over a long period of time, essentially all of the gas molecules will eventually be lost. GMm (a) The magnitude of the force one galaxy exerts on the other is F = , where M is the r2 mass of the Andromeda galaxy and m is the mass of the Milky Way galaxy. The magnitude is (6.67 × 10−11n i m 2 /kg 2 )(6 × 1041kg)(4 × 1041kg) = 3.63 × 1028 N. The magnitude of the F= (2.1 × 1022 m) 2 F acceleration of the Andromeda galaxy is a A = = 6 × 10−14 m/s 2 . The magnitude of the M F acceleration of the Milky Way is aMW = = 9 × 10−14 m/s 2 . m (b) According to the definition of center of mass x m given in the problem, = A . Looking at the M xMW diagram, we see x A + xMW = r , where r is the separation between the galaxies used in (a). Using the definition of CM, this can be rewritten as ⎛ M⎞ x A ⎜ 1 + ⎟ = r. Differentiating both sides with m⎠ ⎝ dx ⎛ m ⎞ dr respect to time gives A ⎜1 + A ⎟ = , or dt ⎝ mMW ⎠ dt

⎛ M⎞ v A ⎜1 + ⎟ = v, where v is the speed of one galaxy m⎠ ⎝ relative to the other and vA is the speed of Andromeda relative to the CM. The ratio of the masses is 3/2 and v = 266 km/s, so v A = 106 km/s. From the definition of CM, M vMW = v A = 160 km/s. m 2 mvMW Mv A2 GmM = − 7.6 × 1050 J. The (c) K A = = 5.1 × 1051 J. U = − = 3.4 × 1051 J. K MW = 2 2 r 51 total energy is E = K A + K MW + U = + 7.7 × 10 J. E > 0, so the galaxies will fly apart. †9-55.

(a) Conservation of energy says K earth + U earth = K moon + U moon . If we take the kinetic energy at the moon to be essentially zero and neglect the potential energy due to the moon’s gravity, then the GM earth m mv 2 GM earth m , where d is the earth– muzzle speed of the cannon must satisfy − =− 2 Rearth d


CHAPTER

9

⎛ 1 1⎞ moon distance. This gives v = 2GM earth ⎜ − ⎟ ⎝ Rearth d ⎠ ⎛ ⎞ 1 1 4 = 2(6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg) ⎜ − ⎟ = 1.11 × 10 m/s. 6 8 6.38 × 10 m 3.84 × 10 m ⎝ ⎠ 2 4 2 mv (2000 kg)(1.11 × 10 m/s) (b) K = = = 1.23 × 1011 J. The mass of TNT required is 2 2 1.23 × 1011 J m= = 29 tons. 4.2 × 109 J/ton (c) v 2 = 2ax ⇒ a = 9-56.

(a) E1 = −

v 2 (1.11 × 104 m/s) 2 . = 1.23 × 105 m/s 2 . = 2x 2(500 m)

GM E m (6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg)(1300 kg) =− = −4.00 × 1010 J. 2r 2(6.378 × 106 m + 1.00 × 105 m)

When the satellite comes to rest on the ground, its energy is GM E m (6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg)(1300 kg) =− = −8.13 × 1010 J. The E2 = − 6.378 × 106 m RE change in energy is E2 − E1 = − 4.13 × 1010 J. The minus sign means the satellite lost energy,

9-57.

which had to be transferred somewhere. (b) The heat of fusion of aluminum is 3.99 × 105 J/kg, which means that much energy is required to melt 1 kg of aluminum. The heat of vaporization is 1.05 × 107 J/kg, which means that much energy is required to vaporize 1 kg of aluminum. The mass of the satellite is 1300 kg, so the amount of energy required to melt it is (1300 kg)(3.99 × 105 J/kg) = 5.19 × 108 J, and the amount of energy required to vaporize it is (1300 kg)(1.05 × 107 J/kg) = 1.37 × 1010 J. The amount of energy lost by the satellite when it came down from orbit is 4.13 × 1010 J. If all of that is absorbed as heat by the satellite, there is enough energy to melt and vaporize it. From Problem 3, the distance where the gravitational pull cancels = (3.84 × 108 m – 3.41 × 108 m) = 4.30 × 107m from moon. Energy at this point = –GMm/r – GM′m/r′ (prime indicates moon) ⎛ 5.98 × 1024 7.35 × 1022 ⎞ ⎛ M M′⎞ −11 6.67 × 10 m = − Gm ⎜ + = − ⎜ ⎟ 8 7 ⎟ r′ ⎠ ⎝ r ⎝ 3.41 × 10 4.3 × 10 ⎠ = –6.67 × 10–11(1.92 × 1016) × m = (–1.28 × 106 × m)J where m is the mass of the material. This must be the energy at the surface of the moon, 1 = − GMm / R = GM ′m / R′ + mv 2 2 24 ⎡ ⎛ 5.98 × 10 7.35 × 1022 ⎞ 1 2 ⎤ = ⎢ −6.67 × 10−11 ⎜ + ⎟+ v ⎥ 8 1.74 × 106 ⎠ 2 ⎦ ⎝ 3.82 × 10 ⎣ = –1.28 × 106 ⎡ ⎛ 5.98 × 1024 7.35 × 1022 ⎞ ⎤ + v2 = 2 ⎢ −1.28 × 106 + 6.67 × 10−11 ⎜ ⎟⎥ 8 1.74 × 106 ⎠ ⎦ ⎝ 3.82 × 10 ⎣ v = 2270m/s


CHAPTER

9

Since particle has energy – (1.28 × 106m) J where m is its mass, this must be its total energy when it hits earth. Therefore, 1 mv 2 − GM E m / RE = − 1.28 × 106 m 2

9-58.

v2 = 2[GME/RE – 1.28 × 106] v = 1.11 × 104m/s (a) From Problem 9-3, the point at which the net gravitational force caused by the earth and moon is a distance r1 = 3.46 × 108 m from earth, or r2 = 3.8 × 107 m from the moon. If the kinetic energy of the spacecraft is zero at the zero force point and it is launched from a point 300 km above the surface of the earth, or 6678 km from the center of the earth, then GM earth m GM moon m GM earth m GM moon m mv 2 − − =− − 6 8 8 2 6.678 × 10 m 3.84 × 10 m 3.46 × 10 m 3.8 × 107 m where v is the launch speed. The exact distance from earth of the spacecraft is negligible compared to the earth-moon distance. Solving for v gives

⎡ ⎛ ⎞ ⎛ ⎞⎤ 1 1 1 1 v = 2G ⎢ M earth ⎜ − − ⎟ + M moon ⎜ ⎟⎥ 6 8 8 7 ⎝ 6.678 × 10 m 3.46 × 10 m ⎠ ⎝ 3.84 × 10 m 3.8 × 10 m ⎠ ⎦ ⎣ = 1.08 × 104 m/s. (b) At the zero force point, K = 0. Using energy conservation there and the moon’s surface, GM earth m GM moon m GM earth m GM moon m mv 2 − − = − − 8 7 8 3.46 × 10 m 3.8 × 10 m 2 3.84 × 10 m 1.74 × 106 m where now v stands for the speed of the craft at the surface of the moon, which has a radius of 1740 km. Solving for v gives

⎡ ⎛ ⎞ ⎛ ⎞⎤ 1 1 1 1 v = 2G ⎢ M earth ⎜ − − ⎟ + M moon ⎜ ⎟⎥ 8 8 6 7 ⎝ 3.84 × 10 m 3.46 × 10 m ⎠ ⎝ 1.74 × 10 m 3.8 × 10 m ⎠ ⎦ ⎣ = 2.27 × 103 m/s. GM s ⎞ 1 2 GM s m ⎛1 = m ⎜ v2 − mv − 2 r r ⎟⎠ ⎝2 ⎡ 6.67 × 10−11 (1.99 × 1030 ) ⎤ = m ⎢ 12 (473002 m 2 / s 2 ) − ⎥ 1.160 × 1011 ⎣ ⎦

9-59.

Net energy of comet = U + K =

9-60.

= –2.56 × 107m < 0. So the orbit is elliptical. (a) At 2 × 107m, velocity for circular orbit is v = GM E / r =

6.67 × 10−11 × 5.98 × 1024 2 × 107

v = 4.47 × 103 m/s = 4.47 km/s Therefore, II will be circular. III will be elliptical. The total energy of I is


CHAPTER 1 2

9

mv 2 − GMm / r = m( 12 v 2 = GM / r )

= 12 (5.57 × 103 ) 2 −

6.67 × 10−11 × 5.98 × 1024 <0 2 × 107

Therefore, I will also be elliptical. (b)

9-61.

(a) By (12) v = GM m / rm =

(6.67 × 10−11 m3 / kg s 2 )(7.35 × 1022 )kg 1.738 × 106 m

= 1680m/s Period T = 2πr/v = 2π(1.738 × 106 m)/(1680 m/s) = 6500 s = 108 min (b) Net energy = 12 mv 2 − GMm /R = m ( 12 v 2 − GM / R )

⎛ 20002 6.67 × 10−11 × 7.35 × 1022 ⎞ = m⎜ − ⎟ <0 1.738 × 106 ⎝ 2 ⎠ So the orbit is closed (elliptical) (c) Net energy = m( 12 v 2 − GM /R ) ⎛ 30002 ⎞ = m⎜ − 2.82 × 106 ⎟ > 0, ⎝ 2 ⎠ so the orbit is not closed.

9-62.

Neglect frictional losses in the earth’s atmosphere. Then the total energy E = U + K is the same at impact as it is a “large” distance from the earth (a “large distance” means that we can neglect the potential energy of the satellite in the earth’s gravitational field). K = (1/2) mv2, and U = –GMEms/r. Then {(1/ 2) mvE2 − GM E ms / rE } = (1/ 2)ms v 2 , where v = speed of the satellite a large distance from earth and vE = 70,000 km/hr = 19446 m/s. Then we have v 2 = vE2 − 2GM E / rE = (19446)2 – 2(6.67 × 10–11) (5.98 × 1024)/6.357 × 106 v = 15895 m/s = 57,200 km/hr


CHAPTER †9-63.

9-64.

9-65.

(a) 140 km/hr = 38.9 m/s. For a stable orbit, the centripetal acceleration is provided by the gravitational attraction, or Fc = mv2/R = GMm/R2. The speed required for a stable orbit is therefore given by v2 = GM/R, where R = 3.76 × 1019 kg, and R = 195 × 103 m. Then v = 113.4 m/s, so that the ball will not orbit around Mimas. (b) The ball will rise until its initial kinetic energy, (1/2)mv2, is equal to the change in potential energy, Ufinal – Uinitial = –GMm/(R + h) + GMm/R. Then (1/2)mv2 = GMm[1/R – 1/(R + h)] solving for h, v2R2 + v2Rh = 2GMh, from which h = 5910 m (a) Escape velocity is given by Etot = K + U = (1/2)mv2 – GMEm/RE = 0. That is, the kinetic energy at the earth’s surface must be just enough to lift the projectile out of the earth’s potential well. Then v2 = 2GME/RE = 2(6.67 × 10–11)(5.98 × 1024)/6.38 × 106 = 11.2 × 103 m/s or 11.2 km/s. Yes, this projectile will escape permanently from the earth. (b) To escape from the solar system, the escape velocity from the sun for a projectile launched from earth would be v2 = 2GMs/R, where R is the earth-sun distance, 1.50 × 1011 m. This gives v = 41.1 km/s to escape from the solar system. The earth has a speed vrel of 30 km/s around the earth. If the projectile from the rail gun is launched parallel to vrel, then the total speed relative to the sun is 15 + 30 = 45 km/s, so that the projectile can escape from the solar system. E = –GMm/2r where r is the semimajor axis of the ellipse. Therefore, E = –GMm/(r1 + r2) where r1 and r2 are the aphelion and perihelion distances, respectively. SPUTNIK I (6.67 × 10−11 )(5.98 × 1024 )83.5 E=− J = −2.39 × 109 J (225 + 6378 + 959 + 6378) × 103 EXPLORER I E=−

9-66.

9

(6.67 × 10−11 )(5.98 × 1024 )14.1 J = −3.59 × 108 J 3 (368 + 6378 + 2540 + 6378) × 10

(a) Velocities are escape velocities because the net energy (U + K) = 0, which is the condition used in calculating escape velocity. And particles with parabolic paths have zero net energy. Therefore, v = 2GM s / Rs = 2 × 6.67 × 10−11 m3 /kgs 2 × 1.99 × 1030 kg/1.496 × 1011m = 4.21 × 104 m/s = 42.1 km/s (b) Head-on collision — velocities add (vE = 2.98 × 104m/s) vrel = 4.2 × 104 + 2.98 × 104 = 7.19 × 104 m/s Overtaking collision — velocities subtract vrel = 4.21 × 104 – 2.98 × 104 = 1.23 × 104 m/s


CHAPTER †9-67.

9

In Problem 9-43, we found the semimajor axis is related to perihelion and aphelion distances by r +r 5.06 + 61.25 a = min max , so a = × 107 km = 3.32 × 1011 m. In terms of a, the total energy 2 2 2 GM S m mvmax GM S m GM S m = − . At perihelion, E = − , so is E = − rmin 2a 2a 2

⎛ 2 1⎞ − ⎟ vmax = GM S ⎜ ⎝ rmin a ⎠ ⎛ ⎞ 2 1 = (6.67 × 10−11 N i m 2 /kg 2 )(1.99 × 1030 kg) ⎜ − ⎟ 10 11 ⎝ 5.06 × 10 m 3.32 × 10 m ⎠ = 6.96 × 104 m/s. To find the speed vmin at aphelion, we could use the energy equation again, but it’s simpler to use Kepler’s Second Law: rmin vmax = rmax vmin ⇒ vmin =

rmin vmax , which gives rmax

(5.06)(6.96 × 104 m/s) = 5.75 × 103 m/s. Note that it was not necessary to include the 61.25 units or the powers of 10 for the perihelion and aphelion distances, because they cancel in the ratio. 2 mv 2p GM E m E va2 GM E E v p GM E . At apogee, . From Kepler’s At perigee, E = = − − ⇒ = − m 2 ra 2 rp m 2 rp vmin =

9-68.

Second Law, rp v p = ra va , so by conservation of energy

v 2p 2

−

2 2 GM E v p rp GM E = − . Rewriting rp ra 2ra2

⎛ 2GM ⎛ 2GM ⎞ 1⎞ this gives ra2 ⎜ 3 2 E − 2 ⎟ − ra ⎜ 2 2 E ⎟ + 1 = 0. From the data given, vp = 1.039 × 104 m/s, rp = ⎜ rv ⎜ r v ⎟ rp ⎟⎠ ⎝ p p ⎝ p p ⎠ 6378 km + 457 km = 6.835 × 106 m. Substituting these numbers and the values for G and ME gives 1.7371 × 10−15 ra2 − 1.5818 × 10−7 ra + 1 = 0, where extra digits have been retained to avoid

†9-69.

errors caused by roundoff. Using the quadratic formula to solve for ra gives ra = 8.422 × 107 m, 6.835 × 106 m. Subtracting 6378 km from each of these gives ha = 7.78 × 107 m, 4.57 × 105 m. The smaller of these is the perigee height, so the correct answer is ha = 7.78 × 107 m. In the solution to Problem 9-50, it was stated without proof that the energy of a body in an elliptical orbit depends on the semimajor axis. In this problem, we’ll prove that statement. In Problem 9-43 we found that 2a = r1 + r2 just by looking at the geometry of an ellipse, so the expression for the energy in this problem is the same as the statement in solution 9-50. In Problem 9-68 it was shown that energy conservation in an elliptical orbit around the sun means v12 GM S v22 GM S rv . Kepler’s Second Law says r1v1 = r2 v2 ⇒ v2 = 1 1 . Substituting this − = − 2 2 r1 r2 r2 into the energy equation gives

⎛1 1⎞ v12 GM E v12 r12 GM E r12 ⎞ 2⎛ − = − ⇒ v 1 − = 2GM S ⎜ − ⎟ . 1 ⎜ 2 2 ⎟ 2 r1 2r2 r2 ⎝ r2 ⎠ ⎝ r1 r2 ⎠


CHAPTER

9-70.

9

⎛1 1⎞ ⎛r −r ⎞ 2GM S ⎜ − ⎟ 2GM S ⎜ 2 1 ⎟ ⎝ r1 r2 ⎠ = ⎝ r1r2 ⎠ = 2GM S r2 ( r2 − r1 ) = 2Gr2 M S . Solving for v1 gives v12 = 2 ⎛ r1 ⎞ ⎛ r1 ⎞⎛ r1 ⎞ r1 ( r2 − r1 )( r2 + r1 ) r1 ( r2 + r1 ) ⎜1 − 2 ⎟ ⎜ 1 − ⎟⎜1 + ⎟ ⎝ r2 ⎠ ⎝ r2 ⎠⎝ r2 ⎠ ⎡ r − ( r2 + r1 ) ⎤ GM S Gr2 M S GM S E v12 GM S = − = − = GM S ⎢ 2 Then . So the ⎥⇒ E=− r2 + r1 m 2 r1 r1 ( r2 + r1 ) r1 ⎢⎣ r1 ( r2 + r1 ) ⎥⎦ statement made in 9-50 is true. GM S m dr . As it falls toward the sun, its speed is v = , The initial energy of the comet is E = − dt r1 2

so its energy at a distance r from the sun is E =

GM S m m ⎛ dr ⎞ . By conservation of energy, ⎜ ⎟ − r 2 ⎝ dt ⎠

2

GM S m m ⎛ dr ⎞ GM S m − = ⎜ ⎟ − . Solving for dt gives dt = r1 r 2 ⎝ dt ⎠

−dr . We must (GM S / r ) − (GM S / r1 )

choose the minus sign in the square root because r decreases as t increases. Integrate to find the r2 dr total time: t = − ∫ . r1 (GM S / r ) − (GM S / r1 ) †9-71.

The maximum distance from the center of the Moon is labeled r in the sketch of the orbit. Using conservation of energy, RM 2 mvlaunch GM M m mv 2 GM M m − = − , where v is the speed of the RM r 2 2 projectile at r. According to Kepler’s Second Law, RM vlaunch = rv. Because r is in the denominator in the energy equation, it will be r easier to find v and then solve for r rather than vice versa. Cancel the mass m of the satellite from the energy equation and use the second equation to get r in terms of v. Substitute that expression into the energy equation to get 2 vlaunch GM M v 2 GM M v . Multiply through by 2 and rearrange to get a quadratic equation − = − 2 2 RM vlaunch RM

.

⎛ 2GM M ⎞ ⎛ 2GM M ⎞ 2 − vlaunch for v: v 2 − ⎜ ⎟v + ⎜ ⎟ = 0. Substituting numerical data and using vlaunch = ⎝ RM vlaunch ⎠ ⎝ Rm ⎠ 2 6 2000 m/s gives v − 2818v + 1.635 × 10 = 0. The roots are 2828 ± 28182 − 4(1.635 × 106 ) = 2000 m/s, 817 m/s. The first root is the launch speed, so 2 the speed at r is v = 817 m/s. R v From Kepler’s Second Law, r = M launch = 4.26 × 106 m. The problem asks for the height, so RM v must be subtracted from this radius: h = 2.52 × 106 m. v=


CHAPTER 9-72.

9

(a) The perihelion speed is given by Problem 9-69. v1 = 2GM s r2 / r1 (r1 + r2 ) =

2 × 6.67 × 10−11 1.99 × 1030 × 2.28 × 1011 1.50 × 1011 (2.28 + 1.50) × 1011

v1 = 3.27 × 104 m/s. The earth is moving at 2.98 × 104 m/s, so the ship is launched at 3.27 × 104 – 2.98 × 104 m/s = 2.9 × 103 m/s relative to the earth. (b) By conservation of angular momentum v1r1 = v2r2, so v2 = v1r1/v2 ⎛ 1.50 × 108 ⎞ v2 = 3.27 × 104 ⎜ m/s = 2.15 × 104 m/s 8 ⎟ 2.28 × 10 ⎝ ⎠ The speed of Mars = v = GM s / r = 1.99 × 1030 × 6.67 × 10−11 / 2.28 × 1011 = 2.41 × 104 m/s Thus the speed relative to Mars = (2.41 × 104 – 2.15 × 104)m/s = 2.6 × 103 m/s (c) T = 2π r 3 / GM s . But T = period for the round trip. We want the time for the one-way trip, i.e., 1/2T. 1 Time taken = T = π r 3 / GM s 2

= π [(1.89 × 1011 )3 ]/(6.67 × 10−11 × 1.99 × 1030 ) = 2.24 × 107 s = 0.710 yr (d) Orbital period Mars = 1.88 year. In 0.710 yr Mars has gone 0.710/1.88 × 360º = 136º. Earth has gone 0.710 × 360º = 256º.

9-73.

(a) Ship is launched at aphelion. The aphelion speed is given by Problem 9-69. v2 = 2GM s r1 / r2 /(r1 + r2 ) =

2 × 6.67 × 10−11 × 7.99 × 1030 × 1.08 × 1011 1.50 × 1011 × ([1.08 + 1.50] × 1011 )

= 2.72 × 104 m/s Relative speed = 2.98 × 104 – 2.72 × 104 = 2.6 × 103 m/s


CHAPTER

9

(b) v1r1 = v2r2. Therefore, v1 = v2r2/r1 ⎛ 1.50 × 1011 ⎞ = 2.72 × 104 ⎜ = 3.780 × 104 m/s 11 ⎟ 2.28 × 10 ⎝ ⎠ Velocity of Venus = GM / r = 6.67 × 1011 × 1.99 × 1030 /1.08 × 1011 = 3.506 × 104 m/s

Therefore, relative velocity = 3.78 × 104 m/s = 3.51 × 104 m/s = 2.74 × 103 m/s (c) Time taken = 12 T = π r 3 / GM s

r = (1.50 × 1.08 × 1011 ) / 2m r = 1.29 × 1011 m

Time taken = π (1.29 × 1011 )3 /(6.67 × 10−11 × 1.99 × 1020 s)

9-74.

= 1.26 × 107s = 0.400 yr (d) Venus orbital period = 0.615 yr. Therefore, in 0.400 yr, Venus has moved (0.400/0.615) × 360° = 234.4° Earth has moved 0.400 × 360º = 144°. 234° – 144° = 90°. At launch, Venus is 90° behind Earth (seen from the Sun). (a) In the reference frame of the planet, the particle follows a hyperbolic path, with the initial and final velocities equal to conserve angular momentum. In the frame of the planet the magnitude of the incoming velocity of the satellite is v = v 2 + u 2 − 2uv cosθ and the magnitude of the velocity of the satellite on leaving is still v 2 + u 2 − 2uv cosθ

Therefore, in the inertial frame, since the planet is moving at velocity u in the direction of leaving satellite, its final speed v′ = u + v 2 + u 2 − 2uv cosθ (b) at θ = 0, v′ = u + v 2 + u 2 − 2uv = u + (v − u ) 2 = u + v − u = v Take dv / dθ = 12 (v 2 + u 2 − 2uv cosθ ) −1/ 2 ,[2uv sin θ ] dv'/dθ = 0 at θ = 0, π. At θ = π, dv'/dθ changes from + to –, indicating a maximum. (c) u = GM s / r = 13.1 km/s, where R is the orbital radius of Jupiter. Then, v′ = u + v 2 + u 2 − 2uv cosθ

= 13.1 + 32 + 13.12 − 2 × 3 × 13.1cos 20° = 23.4 km/s


CHAPTER 9-75.

9-76.

9

Consider an astronaut standing in the space station. The force on his feet is the centripetal force that keeps him moving in a circle. If this force is to be equal to his normal weight on earth, the v2 centripetal acceleration must be equal to g: g = , where v is the astronaut’s tangential speed R and R is the radius of the space station. If the time for one complete revolution is T, then the 2π R tangential speed is v = . Substituting into the acceleration equation and solving for T gives T R 900 m = 2π = 60 s. The time for revolution is 60 s or 1 min, so the rotation rate T = 2π 9.81 m /s 2 g is 1 rev/min. GM m (6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg)(75 kg) = 713 N. Forbit = 2 E = (6.378 × 106 m + 1.00 × 105 m) 2 rorbit Fsurface =

GM E m (6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg)(75 kg) = = 736 N. (6.378 × 106 m) 2 rE2

Gm1m2 (6.67 × 10−11 N i m 2 /kg 2 )(8.0 kg)(0.030 kg) = = 1.6 × 10−9 N. (0.10 m) 2 r2

9-77.

F=

9-78.

gCeres =

GM (6.67 × 10−11 N i m 2 /kg 2 )(7.0 × 1020 kg) = = 0.154 m/s 2 . r2 (5.5 × 105 m) 2

wCeres = wearth

gCeres ⎛ 0.154 ⎞ = (170 lbf) ⎜ ⎟ = 2.7 lbf. g earth ⎝ 9.81 ⎠

9-79.

⎛R ⎞ T22 R23 = 3 ⇒ T2 = T1 ⎜ 2 ⎟ 2 T1 R1 ⎝ R1 ⎠

9-80.

T =

9-81.

T2 =

9-82.

(a) v =

9-83.

(a) v =

4π 2 r 3 = GM

3/ 2

= (1 yr)(2.9)1.5 = 4.9 yr.

4π 2 (1.00 × 103 m)3 = 7.7 × 108 s (24 yr). (6.67 × 10−11 N i m 2 /kg 2 )(1.00 × 103 kg)

4π 2 r 3 4π 2 r 3 ⇒M = . 3.55 days = 3.07 × 105 s. Then M = GM GT 2 4π 2 (6.71 × 108 m)3 = 1.90 × 1027 kg. (6.67 × 10−11 N i m 2 /kg 2 )(3.07 × 105 s)2

GM . r = 250 ly × (9.46 × 1015 m/ly) = 2.37 × 1018 m. ⇒ r rv 2 (2.37 × 1018 m)(5.30 × 105 m/s) 2 = = 1.0 × 1040 kg (about 5 × 109 solar masses). M = (6.67 × 10−11 N i m 2 /kg 2 ) G

GM E GM E . r = 6378 km + 400 km = 6.778 × 106 m. v = = 7.67 × 103 m/s. If the r r objects are moving in opposite directions, then the speed of one relative to the other is twice this value, so vrel = 1.53 × 104 m/s .

(b) m =

2K 2(4.6 × 105 J) = = 3.91 × 10−3 kg (3.91 g). 2 (1.53 × 104 m/s) 2 vrel


CHAPTER

9-84. †9-85.

rp v p = ra va ⇒ va =

rp v p ra

=

(7.02)(8.22 × 103 m/s) = 5.60 × 103 m/s. 10.3

At the lowest point (r1 = 6378 km + 200 km), E = and E = −

9

mv 2 GM E m − . At the highest point, K = 0 2 r1

GM E m GM E m mv 2 GM E m . By conservation of energy, − = − , which gives 2 r2 r1 r2 −1

⎛1 ⎛ ⎞ v2 ⎞ 1 (8.50 × V 103 m/s)2 r2 = ⎜ − = − ⎟ ⎜ ⎟ 6 2 2 24 −11 ⎝ 6.578 × 10 m 2(6.67 × 10 N i m /kg )(5.98 × 10 kg) ⎠ ⎝ r1 2GM E ⎠ = 1.626 × 107 m

−1

The height above the earth is h2 = r2 − RE = 1.626 × 107 m − 6.378 × 106 m = 9.89 × 106 m. 9-86.

(a)

∆v antiparallel to v ∆v parallel to v In each case, the solid circle represents the original orbit and the dashed ellipse is the new orbit of the ejected panel. (b) If the period of the elliptical orbit is a whole number multiple or fraction of the period of the circular orbit, then the space station and the panel will eventually return simultaneously to the 2

†9-87.

3

⎛T ⎞ R ⎛ R⎞ original separation point. T ∝ r , and T ∝ a . ⎜ ell ⎟ = n 2 = ⎜ ⎟ ⇒ = n 2 / 3 is one a ⎝a⎠ ⎝ Tcirc ⎠ a possibility. The other possibility is the reciprocal of this: = n 2 / 3 . In each case, n must be an R integer. GM E m (6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg)(700 kg) =− (a) For a circular orbit, E = − 2r 2(4.23 × 107 m) 2 circ

3

2 ell

3

= −3.30 × 109 J. (b) For a parabolic orbit, E = 0, so the satellite must be given 3.30 × 109 J of extra energy for it to escape. 9-88.

vesc =

2GM moon = Rmoon

2(6.67 × 10−11 N i m 2 /kg 2 )(7.35 × 1022 kg) = 2.37 × 103 m/s. 6 1.74 × 10 m


CHAPTER 10

SYSTEMS OF PARTICLES


No directions are given in the problem, so we’ll only calculate magnitudes of momentum. pbullet = mbullet vbullet = (0.015 kg)(600 m/s) = 9.0 kg i m/s.

parrow = marrow varrow = (0.040 kg)(80 m/s) = 3.2 kg i m/s. 10-2.

No directions are given in the problem, so we’ll only calculate magnitudes of momentum. 65 km/h = 18.1 m/s ⇒ pauto = mauto vauto = (900 kg)(18.1 m/s) = 1.63 × 104 kg i m/s, or p 1.6 × 104 kg i m/s to two significant figures. For the truck, vtruck = auto mtruck =

†10-3.

10-4.

1.63 × 104 kg i m/s = 2.3 m/s. 7.2 × 103 kg

No directions are given in the problem, so we’ll only calculate magnitudes of momentum. A good way to approach this kind of problem is to use a spreadsheet. Fill in the information from the tables in Chapters 1 and 2 and let the spreadsheet do the calculations. Object m (kg) v (m/s) p (kg-m/s) Earth 6.0E+24 3.0E+04 1.8E+29 Jet 1.6E+05 2.7E+02 4.3E+07 Car 1.5E+03 25 3.8E+04 Man 73 1.3 95 Electron 9.1E-31 2.2E+06 2.0E-24 No directions are given in the problem, so we’ll only calculate magnitudes of momentum. 1 0.4536 kg = 6.480 × 10−5 kg. 1 grain = lb × 7000 lb p1 = (90 grains × 6.480 × 10−5 kg/grain)(975 m/s) = 5.7 kg i m/s. p2 = (200 grains × 6.480 × 10−5 kg/grain)(674 m/s) = 8.73 kg i m/s.

†10-5.

p = mv cos 20°i + mv sin 20° j = (9.1 × 10−31 kg)(2.0 × 105 m/s)(cos 20°i + sin 20° j)

y

= (1.6 × 10−25 i + 7.7 × 10−26 j) kg i m/s. 20º

10-6.

Net force = w = mg = (75 kg)(9.81 m/s2) = 7.4 × 102 N. As a vector, we would write dp = 7.4 × 102 N down. dt


x

CHAPTER †10-7.

(a) pi = mv i = (0.43 kg)(26 m/s) @ 30° upward = 11.2 kg i m/s upward. Defining x and y directions as shown, this is pi = 11.2cos30°i + 11.2sin 30° j = 9.7i + 5.6 j kg i m/s.

10

y

30º

x

(b) If there’s no air resistance, the horizontal component of velocity, and hence the horizontal component of momentum, remains constant. Thus at the highest point in the trajectory, the vertical velocity and therefore the vertical momentum are zero, and p highest = 9.7i + 0 j kg i m/s. (c) When the ball reaches the ground again, its horizontal component of velocity is still unchanged but its vertical component of velocity points down. Therefore the ball’s momentum is p f = 9.7i − 5.6 j kg i m/s. The final momentum is not the same as the initial momentum! dp is equal to the magnitude of the force exerted by the sun on the earth: dt dp GM S M E (6.67 × 10−11 N i m 2 /kg 2 )(1.99 × 1030 kg)(5.98 × 1024 kg) = = = 3.53 × 1022 N. 2 11 2 dt r (1.50 × 10 m)

10-8.

The magnitude of

10-9.

p = mv = mgt down = (1 kg)(9.81 m/s 2 )t down = (9.81t kg i m/s) down. At t = 1 s, p = (9.81 kg i m/s) down. At t = 10 s, p = (98.1 kg i m/s) down.

10-10.

Assume the woman throws the life preserver in the i direction. Then the boat (with her in it) must m 3.0 kg (5.0 m/s)i = −0.20i m/s. recoil in the –i direction with velocity v′2 = − 1 v1 = − (20 kg + 55 kg) m2

10-11.

Assume the man’s initial motion is in the i direction. Then the boat must recoil in the –i direction ⎛ 90 kg ⎞ m with velocity v′2 = − 1 v1 = − ⎜ ⎟ (2.0 m/s)i = − 9.0i m/s. m2 ⎝ 20 kg ⎠

10-12.

The problem only gives the magnitude of the momentum, so we’ll find the recoil speed. p photon 7.25 × 10−27 kg i m/s v2′ = = = 4.34 m/s. 1.67 × 10−27 kg matom

10-13.

∆K = K 2 − K1 =

10-14.

(m1 + m2 )(v ') 2 m1v12 (2 × 1500 kg)(12 m/s) 2 (1500 kg)(24 m/s) 2 − = − 2 2 2 2

= –2.2 × 105 J. ∆P = 0 = Prif. + Pbull. Prif. = (10 – 0.015)Vrif. 10 Vrif. = Pbull. = 15 × 10–3 × 650 Therefore, the recoil speed of the rifle, 15 × 650 × 10−3 = 0.98 m/s Vrif. = 10 (EK)bull. = 12 (15 × 10−3 )(650) 2 = 3169 J (EK)rif. = 12 (10)(0.98) 2 = 4.8 J


CHAPTER

10

10-15.

Initial momentum = 0 = final momentum 0 = 15 × 11 × 490 + 4000 × 103vR The recoil speed of the ship, 15 × 11 × 490 vR = − = −2 × 10−2 m/s 6 4 × 10

10-16.

Take the i direction to be the direction of motion of m1. The initial momentum is pi = m1 v1 + m2 v 2 . After

m1

65 km/h

the cars collide, they are stuck together and their final momentum is pi = (m1 + m2 ) v '. By conservation of momentum, (m1 + m2 ) v ' = m1 v1 + m2 v 2 ⇒ v ' =

v′

65 km/h

m1

m2

m2

m1 v1 + m2 v 2 . The m1 + m2

speeds of the cars are equal, so v2 = –v1. Take m1 = 700 kg and m2 = 1500 kg. 65 km/h = 18.1 m/s, so (m − m2 ) v1 (−800 kg)(18.1i m/s) v' = 1 = = −6.58i m/s. 2200 kg m1 + m2 The change in velocities are ∆v1 = v '− v1 = (−6.58 m/s − 18.1 m/s)i = −25 i m/s ∆v 2 = v '− v 2 = ( −6.58 m/s + 18.1 m/s)i = +12 i m/s 10-17.

1 2

mα vd2 = 7.26 × 10−16 J.

Therefore, vα = 2 E / mα = (2 × 7.26 × 10−16 ) /(6.68 × 10−27 )m/s = 4.66 × 105m/s Initial momentum = 0. Let mn, vn be the mass and speed of nucleus respectively. Therefore, vn = mαva/mn 0 = mnvn + mαvα 6.68 × 10−27 kg × 4.66 × 105 m/s =− = 8260 m/s 3.77 × 10−25 kg K.E. of recoil = 12 mn vn2 = 12 (3.77 × 10−25 ) kg (8200) 2 m 2 / s 2 = 1.29 × 10−17 J 10-18.

†10-19.

Since bullets stick in lion, the final velocities of the bullets and the lion = 0, so the final momentum = 0. Therefore, m1v1 + m2v2 = 0 m1 = mass lion = 120 kg v1 = velocity of lion = 12 m/s m2 = total mass of bullets, v2 = velocity of bullets = –630 m/s m2 = –m1v1/v2 = [–120 × (12)]/–630 = 2.286 kg = 2286 gm Therefore, number bullets = 2286 g/15 g/bullet = 150 bullets The projectile velocity is v1 = v1 cos 20°i + v1 sin 20° j. In the y direction the y total mass that recoils is the mass of the cannon plus the mass of the entire earth, since the cannon pushed down on the earth. Because the mass of the earth is so huge (6 × 1024 kg), the y component of the recoil velocity is unmeasurably small. In the x direction, only the cannon recoils. so the x 20˚ component of the recoil velocity is not zero. The total velocity is


x

CHAPTER v′2 = − 10-20.

10

m1 m1 6 v1 cos 20°i − v1 sin 20° j = − (500 m/s)(cos20°)i + 0j = −(1.3 m/s)i + 0j . 2200 m2 m2 + mE

Force of friction, Fƒ = µK2mg Change in momentum ∆P = Fƒ ∆t = 530 kg i m/s ∆P Ff ∆t = 2m 2m 2 µ mg × 0.02 = 0.18m/s = K 2m The change in speed due to friction is not significant. (a) Momentum transferred per min. = 450 bullets/min × 0.014 kg/bullet × 630 m/s = (3970 kg m/s)/min. = 66 kg m/s per s. Since the average force = ∆p/∆t, the average force = 66N. (b) The kinetic energy per min = 12 mv 2

Change in speed = ∆v =

10-21.

= 12 (450 × 0.014 kg/min) × 6302 m 2 /s 2 = 1.3 × 106 J/min 10-22.

†10-23.

p1 = m1 v1 , p 2 = (m1 + m2 ) v′2 . p1 = p 2 ⇒ m1 v1 = (m1 + m2 ) v′2 . All velocities are in the same m1v1 (250 g)(4.0 m/s) = direction, so we can work with the speeds, so v2′ = m1 + m2 300 g = 3.3 m/s. dp Fx = x ⇒ px ,2 − px ,1 = ∆px = dt ∆px =

10-24.

∫

5.0 s

0

∫

t2 t1

Fx dt. Substitute the given expression for Fx:

(2.0t + 3.0t 2 )dt = ( t 2 + t 3 )

5.0 s 0

= 150 N i s. Note that the coefficients 2.0 and 3.0 have

units of N/s and N/s2, respectively, so the two terms in the result of the integration are actually multiplied by 1 N/s and 1 N/s2, respectively. Also note that the unit N • s is the same as kg • m/s. Assuming the two fragments are parallel to floor, by drawing their momentum vectors, the resultant is m 2v at 45° with both. Therefore, third piece has velocity v 2 in opposite direction to conserve (horizontal) momentum.


CHAPTER †10-25.

10

Assuming the nucleus is initially at rest, the initial momentum is zero. Conservation of momentum requires p N + p e + pν = 0. Label the components of the nucleus’s momentum as pN,x = mNvx, pN,x = mNvx. Using these and resolving pν into components gives (mN vx + pe + pν cos30°)i + (mN v y + pν sin 30°) j = 0. Setting p + pν cos30° each component equal to zero gives vx = − e , mN p sin 30° vx = − ν . mN

y

pν = 1.97×10–22 kg m/s pN

30º

x pe = 2.64×10–22 kg m/s

mN = (63.9 u)(1.66 × 10−27 kg/u) = 1.06 × 10−25 kg. Substituting this and the other numerical values in the problem gives vx = − 4.10 × 103 m/s, v y = − 929 m/s. The minus signs mean the components point in the –x and –y directions, respectively. The recoil velocity in component form is v = − (4.10 × 103 i + 929 j) m/s. As a magnitude and direction, the recoil velocity is v = (4.10 × 103 ) 2 + (929) 2 m/s, θ = tan −1

10-26.

10-27.

10-28.

929 = 12.8°. Because both components of v are 1.04 × 103

negative, θ is in the third quadrant. To get the correct angle measured from the +x direction, we must add 180° to the value from the calculator, which gives θ =193°. Another possibility is to say that θ = 12.8° below the –x axis. Volume of wind sweeping past 1m2 area per sec. is velocity × Area = 4 × 105m/s × 1m2 = 4 × 105m3/s. Thus, the number of ions hitting an area 1m2 per second = 107 ions/m3 × 4 × 105m3/s = 4 × 1012 ions/s. Mass hitting area = 1.7 × 10–27kg/ion × 4 × 1012 ion/s = 6.8 × 10–15 kg/s. Momentum transfer per second = ∆p/∆t = Force = (6.8 × 10–15kg/s) × (4 × 105m/s) = 2.7 × 10–9N The volume of rain that hit an area of 1m2 in 1 min is 3.12 cm/min × 1/100 m/cm × 1m3 = 0.0312 m3/min. Hence, the mass of water that hit one m2 = 0.0312 m3/min × 1000 kg/m3 = 31.2 kg/min = 0.52 kg/s. The impact velocity = 10 m/s. Then, since ∆v of raindrop = 10 m/s, m ∆p / ∆t = (∆v) = Average force = 0.52 kg/s × 10 m/s = 5.2N ∆t (a) In the frame of the automobile the rain is hitting the car at an angle tan–1 10/22.2 with respect to the horizontal. The amount of rain hitting the top of the car in one second is


CHAPTER

10

mtop = ρAv = 7 × 10–4kg/m3 × (4 × 2)m2 × 24.4 m/s × sin θ = (7 × 10–4) (8) (10) kg mtop = 5.6 × 10−2 kg The mass hitting the front of the car in one sec is mfront = ρAv = (7 × 10–4) (3) (24.3 cos θ) kg = (7 × 10–4) (3) (22.2) kg mfront = 4.7 × 10−2 kg (b) The total mass hitting the car in one second is mtotal = mfront + mtop = (4.7 × 10–2 + 5.6 × 10–2) kg mtotal = 0.103 kg ∆vrain = − 22.2 m/siˆ − ( − 10 m/s)jˆ

∆v rain = 22.22 + 102 = 24.4 m/s

10-29.

10-30.

The rate at which momentum is given to the rain is ∆prain ∆v m =m = ∆v = 0.103 kg/s × 24.3 m/s = 2.5 kg − m/s ∆t ∆t ∆t Horizontal drag = 2.5N × cos θ = 2.5 N × 22.2/24.3 = 2.3N Volume that the spaceship sweeps through per sec is Av = 25m2 × 1.0 × 106 m/s = 2.5 × 107 m3/s The mass of dust it collides with = 2.0 × 10–18 kg/m3 × 2.5 × 107m3/s = 5 × 10–11 kg/s The change of momentum per second of the dust = ∆p/∆t = F = m∆v/∆t, so that F = (5 × 10–11 kg/s) (1.0 × 106 m/s) = 5 × 10–5 N Suppose the ball is launched in the –x, +y direction with velocity v B = −15cos 45°i + 15sin 45° j = −10.6i + 10.6 j m/s. With a mass of 0.62 kg, the momentum of the ball is p B = mB v B = − 6.58i + 6.58 j kg ⋅ m/s. The total momentum of the player plus ball is zero at the instant the ball is thrown, so the momentum of the player must be p P = − p B = + 6.58i − 6.58 j kg i m/s, so the 75 kg player’s recoil velocity is p v P = P = +0.0877i − 0.0877 j m/s. The player’s feet are a height h = 0.8 m above the floor and mP the player is in free fall with an initial vertical velocity of –0.0877 m/s, so the time for his feet to reach the floor must satisfy 0.8 = − 0.0877t − 4.91t 2 . This gives −0.0877 ± 0.0877 2 + 4(4.91)(0.8) . The positive root is t = 0.395 s, so the player moves a 2(4.91) horizontal distance x = (0.0877 m/s)(0.395 s) = 0.03 m. The player lands about 3 cm from his t=

10-31.

original position. The cart is initially at rest. From conservation of momentum (M – m)v1 – mu = 0 mu where v1 = in the velocity of the cart after the first firing. M −m


CHAPTER

10

Increment in the velocity of the cart after the second firing is v2 where mu (M – 2m)v2 – mu = 0, or v2 = M − 2m Similarly, the increment in velocity after the third firing is mu v3 = M − 3m The velocity of the cart after nth firing is a sum of all the increments n 1 vn = v1 + v2 + v3 + .... = mu ∑ k =1 M − km 10-32.

†10-33.

10-34.

†10-35.

Let the three penny coins be at x = 0 and the fourth coin at x = 20 cm, then 3m(0) + 1m(20) xcm = = 5 cm 3m + 1m The woman (m1) is at x = 0 and the man (m2) is at x = 3.5 m. The location of the center of mass is given by ∑ mi xi = (39 kg)(0) + (72 kg)(3.5 m) = 1.9 m. xCM = 111 kg ∑ mi

Take the center of Earth as the origin. Then the Moon is at xMoon = 3.84 × 108 m. mEarth = 5.98 × 1024 kg mMoon = 7.35 × 1022 kg. Hence, xCM = (mEarth xEarth + mMoon xMoon)/(mEarth + mMoon) 0 + 3.84 × 108 × 7.35 × 1022 kg = = 4.66 × 106 m from center of Earth 5.98 × 1024 + 7.35 × 1022 kg xCM =

M J RJ (1.90 × 1027 kg)(7.78 × 108 m) = 1.99 × 1030 kg MS + MJ

= 7.42 × 105 m. Note that to three significant figures, the mass of Jupiter is negligible compared to the mass of the 7.42 × 105 Sun. The distance to the CM is = 1.07 × 10−3 , 6.96 × 108 10-36.

or 0.107% of the sun’s radius. Assume all three bricks have the same mass m and the same height h. The center of mass of each brick will be at the center of the brick. Choose the origin of the coordinate system to be at the point where the two bottom bricks meet, as shown in the diagram. By ⎛h⎞ ⎛ 3h ⎞ 2m ⎜ ⎟ + m ⎜ ⎟ ⎝ 2⎠ ⎝ 2 ⎠ = 5h . symmetry, xCM = 0. yCM = 3m 6


MS MJ

0

RJ y

x

CHAPTER †10-37.

Assume the base of the triangle has its ends at x = ± a, and the apex y is at y = h. By symmetry, xCM = 0. The right side of the triangle is h defined by the line y = − x + h. Imagine a strip across the triangle a h a distance y up from the base with a length 2x and a thickness dy as base × height = ah, shown. The total area of the triangle is A = 2 since the base has a length of 2a. If the mass M in the triangle is uniformly distributed, then the mass dm in the strip is M 2Mxdy dm = (2 xdy ) = . Inverting the equation for y(x) and A ah substituting in the equation for dm gives a ⎞ ⎛ 2 M ⎜ a − y ⎟ dy 2M ⎛ y⎞ h ⎠ ⎝ dm = = ⎜ 1 − ⎟ dy. The y coordinate of the center of mass is ah h ⎝ h⎠

ydm

x

h

y⎞ y2 ⎞ 2 h ⎛ 2 h⎛ 2 ⎛ y 2 y3 ⎞ 2 ⎛ h2 h2 ⎞ h y ⎜ 1 − ⎟dy = ∫ ⎜ y − ⎟dy = ⎜ − ⎟ = ⎜ − ⎟ = . ∫ M h 0 ⎝ h⎠ h 0⎝ h ⎠ h ⎝ 2 3h ⎠ 0 h ⎝ 2 3⎠ 3 Assume the pyramid has an equilateral triangle for its base. Draw an (x,y,z) coordinate system with the z axis passing through the apex of the pyramid and pointing down, just like the y axis in Example 7. Then at a distance z down from the top, we have a triangular slab like the one shown in the figure. The z axis will point into the page and passes through the center of the triangle, so the base has a length of 4x. If we define x an apex angle φ as seen from the rear of the triangle (looking at the side parallel to the x axis), then x = z tan φ , as in Example 7. The yCM =

10-38.

∫

y =h

a

10

y =0

=

3a 2 , so the area of this 4 triangular slab at a distance z from the top is 3z 2 tan 2 φ . If the slab area of an equilateral triangle of side a is

y

has a thickness dz, its volume is dV = 3 z 2 tan 2 φ dz , and its mass is dm = 3 ρ z 2 tan 2 φ dz , where ρ is the density. Then z =h

zCM

∫ = ∫

zdm

z =0 z =h z =0

dm

h

=

3 ρ tan 2 φ ∫ z 3 dz 0 h

3 ρ tan φ ∫ z dz 2

2

. The common factors of

3ρ tan 2 φ cancel, and we are left

0

with the same integral that was obtained in Example 7. By symmetry, center of mass must be on the axis passing through the apex, so the x and y coordinates are 0. The final result is that the center of mass is on the central axis located a distance of h/4 from the bottom of the pyramid. (Note—it turns out the center of mass of any pyramid with a regular polygon for a base and a triangular cross section has its center of mass at h/4 from the bottom, including a cone.)


CHAPTER 10-39.

10-40.

10-41.

10

Let the center of the wheel be the origin. Then (mwheel xwheel + mlead xlead)/(mwheel + mlead) = 0 xwheel = – mlead xlead/mwheel (0.040 × 0.20) kg m =− 30 kg = –2.7 × 10–4m Therefore, CM of wheel alone is 2.7 × 10–4m on opposite side of center of wheel from lead weight. The CM must lie on the line of symmerty between the two hydrogen atoms. Take the oxygen atom as origin. Then xCM (distance along line of symmerty from “o”) = (mH xH1 + mH xH2 m0 x0)/(2mH + x0) (1.008 u)(0.958cos52.5°) nm + (1.008 u)(0.958cos52.5°) nm + 0 = (16.00 + 2(1.008)) u = 0.00653 nm along axis of symmerty CM must lie on the axis of symmetry (line joining H–O–N).Take the O (between H and N) as the origin. Then xCM

= (mH xH + m01 x01 + mN xN + m02 x02 + m03 x03 ) /(mH + m01 + mN + m02 + m03 ) [1.008 u)(−1.00 nm) + 0 + (14.00 u)(1.41 nm) + 2(16.00 u)(1.41 + 1.22cos 65°) nm] (1.008 + 16.00 + 14.00 + 3200) u = 1.28 nm (from oxygen atom, between it and the nitrogen atom) =

10-42.

Treat the orbits as circular. The position of a planet is given by (r cosθ , r sin θ ). Measure the angles from Figure 9.13 and use the mean radii given in Table 9.1. To make the table easier to read and to simplify the calculations, change the radii from m to AU by dividing the distances by 1.50 × 108 km/AU. (1 AU = mean radius of the earth’s orbit). The use of a spreadsheet to do the calculations is recommended. Your values may be slightly different from the values given here, 3 ⎛ 3 ⎞ ⎜ ∑ mi xi ∑ mi yi ⎟ ⎟ , and (rCM, θCM) can be depending on how you read the angles. The CM is at ⎜ i =13 , i =13 ⎜ ⎟ ⎜ ∑ mi ∑ mi ⎟ 1 1 = = i i ⎝ ⎠ calculated from that.


CHAPTER Planet Mercury Venus Earth

m (kg) 3.30E + 23 4.87E + 24 5.98E + 24

xCM (AU) yCM (AU)

–0.367 0.578

r (km) 5.79E + 07 1.08E + 08 1.50E + 08

r (AU) 0.386 0.720 1.000

θ(˚) 232.0 170.0 95.5

rCM (AU) θCM (°)

0.685 122.4

x (AU) –0.238 –0.709 –0.096

2 2 The polar coordinates were calculated using rCM = xCM + yCM , θ CM = tan −1

10-43.

10

y (AU) –0.304 0.125 0.995

yCM . Note that since xCM

xCM and yCM are both negative, it is necessary to add 180° to the angle given by the inverse tangent to place the angle in the third quadrant. ∑ mi ri = [2 × 1011 (0,0,0) + 3 × 1011 (1640, 290,1440) + rCM = ∑ mi

+ 2.5 × 1010 (8.5,56.7, −149) + 8 × 109 (1830,766,1170)]/(2 × 1011 + 3 × 1011 + 2.5 × 1010 + 8 × 109 ) = (950, 180, 820) (light year) 10-44.

By symmetry, xCM = 0. yCM =

M M M ⎛ ⎞ dm = Rdθ ⎜ dθ . y = R sin θ , so ⎟= π R ⎝ ⎠ π π ⎛M ⎞ ∫0 R sin θ ⎜⎝ π dθ ⎟⎠ R π R π yCM = = ∫ R sin θ dθ = − cosθ 0 0 π π M R 2R = − (−1 − 1) = .

π

†10-45.

10-46.

∫ ydm . From the diagram,

y

dθ R

θ

x

π

Let the axes be as shown. The center of masses of the three sheets are at (L/2, L/2,0), (L/2,0,L/2) and (0,L/2,L/2). Let the mass of each sheet be M. Then center of mass of entire structure is: m(L/2,L/2,0) + m(L/2,0,L/2) + m(0,L/2,L/2)/3M = (L,L,L)/3 = (L/3,L/3,L/3) Let the edges of the box coincide with the xyz-coordinate system. The given mass distribution can ⎛ L L L⎞ be replaced by an intact box of mass 6 M with the center of mass at ⎜ , , ⎟ and a mass of –M ⎝2 2 2⎠ 6 M ( L / 2) − M ( L / 2) L = = yCM xCM = L L ⎛ ⎞ 5M 2 at ⎜ , , L ⎟ where M is the mass of each face. Then − M L ML L 6 ( / 2) 2 2 2 ⎝ ⎠ = zCM = 5M 5 ⎛ L L 2L ⎞ Thus (x,y,z)CM = ⎜ , , ⎟ ⎝2 2 5 ⎠


CHAPTER 10-47.

10

The system can be treated as a solid cube of mass ρvcube and a cylinder of negative mass (–ρvcyl). Let the axes be as shown. By symmetry, the CM must lie along the xaxis. M c × 0 + (− M cyl ) × L / 2 xCM = M cube + M cyl

M cube = ρ L3 ; M cyl = ρπ( L / 4) 2 × L = πρL3 /16 ⎛ L3 L ⎞ ⎛ L3 ⎞ xCM = ⎜ − πρ × ⎟ / ⎜ ρL3 − πρ ⎟ 16 4 ⎠ ⎝ 16 ⎠ ⎝ ⎡⎛ L3 ⎞ ⎤ ⎡ L3 ⎛ 64 ⎞⎤ = ⎢⎜ − πρ ⎟ L ⎥ / ⎢ πρ ⎜ − 4 ⎟ ⎥ 64 ⎠ ⎦ ⎣ 64 ⎝ π ⎠⎦ ⎣⎝ ⎛ ⎛ 64 ⎞⎞ = − L⎜⎜ − 4⎟⎟ ⎠⎠ ⎝⎝ π 10-48.

The center of mass of semicircular element, radius r, is z = (2/π) r (see Problem 10-44). The surface area of the element is πr dr, therefore, its mass is ρπr dr, [ρ = density]. Therefore, the center of mass (which is along z-axis = axis of symmetry)

2 3 4 R / πR 2 / 2 = R 3 3π As in Example 6, take the y axis to point down with the origin at the apex of the cone. Call the vertex angle φ. Then the circular slab at a distance y from the vertex has a radius r = y tan φ and a thickness dy. The mass of = ∫ ρz dA / ∫ ρ dA =

†10-49.

∫

R

r =0

(πr dr ) 2 r / ∫

R

r =0

(πr dr ) =

φ

the slab is dm = ρ (π r 2 )dy = πρ y 2 tan 2 φ dy. Then yCM is given by yCM

∫ =

y=h y =0

ydm

∫ dm

h

∫ πρ y = ∫ πρ y 0 h

0

3 2

tan 2 φ dy tan 2 φ dy

∫ = ∫

h

0 h

0

y 3 dy y 2 dy

=

3h , where h is the 4

height of the cone. Since y points down, this result means the center of mass is h/4 from the bottom of the cone. (This is the same result for the pyramids in Example 6 and Problem 38!) For Mount Fuji, h = 3800 m, so the center of mass is 950 m from the base of the volcano.


y

CHAPTER 10-50.

†10-51.

10

Look at one line from the vertex to the midpoint of the opposite side. Break the triangle into areas dA as shown. Since the line goes through the midpoint on the opposite side, by similar triangles, the distances d on the left and right hand sides as shown are equal, and by uniformity of density, the CM of this dA lies on the line. This is true of all dA’s. By rCM = ∫ r dA / ∫ dA, since all r lie on this line, rCM must lie on this line. Repeat the argument for the other two sides. Do this by finding the change in height of the CM of the arm—ignore the rest of the body. When the arms are hanging at the side of the body, its CM is located at (0.066 M)(0.717 L) + (0.042 M)(0.553 L) + (0.017 M)(0.431 L) h1 = = 0.623 L. When the arms (0.066 M + 0.042 M + 0.017 M) are held horizontally, all the segments are at shoulder height h2 = 0.812 L. The change in potential energy is ∆U = mg (h2 − h1 ) = mg (0.812 − 0.623)L = 0.189 mgL. The total mass lifted is m = 0.066 M + 0.042 M + 0.017 M = 0.125 M ⇒ ∆U = 0.0236 MgL. Substituting the numerical data gives ∆U = 0.0236(80 kg)(9.81 m/s 2 )(1.70 m) = 31.5 J.

10-52.

If the arms are now lifted to a vertical position, the CMs are lifted to heights above the shoulder that are the same as their initial distances below the shoulder. Thus ∆U in lifting the arms from the horizontal position to the vertical position is the same as for lifting the arms from the lowered position to the horizontal position, or 31.5 J. The change in potential energy to lift the arms from the lowered position to the vertical position will be twice this: ∆U = 63 J. Initial height of the CM is (0.215 M)(0.425 L) + (0.096 M)(0.182 L) + (0.034 M)(0.018 L) h1 = = 0.317 L. When one leg is (0.215 M + 0.096 M + 0.034 M) lifted so the thigh is horizontal, the CMs are at the heights shown. The new height of the CM is (0.215 M)(0.521 L) + (0.096 M)(0.418 L) + (0.034 M)(0.544 L) h1 = = 0.466 L. The change in (0.215 M + 0.096 M + 0.034 M) potential energy is ∆U = mg (h2 − h1 ) = mg (0.466 − 0.317)L = 0.149 mgL. The mass lifted per leg is 0.173 M, so for the data given ∆U = 33.1 J per step. Lifting the legs at a total rate of 180 steps/minute give P = (33.1 J/step)(180 steps/min)(1 min/60 s) = 99 W (roughly 1/8 hp).

10-53.

A mass of water = 73 × 9.2 × 3.7 × 103 = 2.48 × 106kg must be lowered an average distance of 1.85 m, because the CM of the full lock is 1.85m above the low level. To empty the lock requires lowering the same mass an additional 1.85 m. Therefore, the wasted potential energy = 2.48 × 106 × 9.8 × 3.7 = 90 × 106 J


CHAPTER

10-54.

†10-55.

(a) By Eq. (31), V = MgzCM and by Example 7, zCM =

1 147 h= m. Thus, the work by laborers 4 4

147 ⎤ ⎡ = MgzCM = (6.6 × 109 kg) × ⎢ (9.8 m/s) × m ⎥ = 2.4 × 1012 J 4 ⎦ ⎣ 12 5 (b) Man hours = 2.4 × 10 J/(4 × 10 J)/man-hour = 6 × 106 man hours The diagram shows a cross section through the shell. Take the x axis to point left as shown. By symmetry, the CM must lie on the x axis. Imagine that the shell is divided in rings. At a distance x from the origin, the radius of a ring will be r = R cos θ, and x = R sin θ. If we imagine a ring with thickness Rdθ, then the mass in the ring is M M Mr dm = dA = 2π r ( Rdθ ) = dθ = M sin θ dθ , where we have 2π R 2 2π R 2 R used the area of a hemisphere to find the mass per unit area. Now xCM

10-56.

10

∫ =

x=R

x =0

xdm

M

= R∫

π /2

0

R sin 2 θ sin θ cosθ dθ = 2

π /2

= 0

y

θ R x

R , where x was given 2

in terms of θ, so going from x = 0 to x = R is equivalent to going from θ = 0 to θ = π/2. Let z = 0 at roof with z being positive downward. Then, after time t from being released from the roof, the drop is at position z = 1/2 gt2 (1). The water drops at a constant rate, so that at any point z′, water passes this point at the rate the roof drops the water (i.e., at constant rate). Let λ(z′) be the mass of water per unit length at point z′. Then λ(z)z (where v = velocity of water at that point) = rate of water falling = k = constant. Then λ(z) = k/z = k/(dz/dt) (2). By (1) dz/dt = gt, and also by (1) t = 2 gz so that dz / dt = g 2 gz = (2 / g ) z1/ 2 Substituting this in (2) gives λ ( z ) = k / (2 / g ) z1/ 2 = Cz −1/ 2 where C = constant. Now zCM = ∑ mi zi / ∑ mi =

∫

e

0

e

λ ( z ) z dz / ∫ λ ( z )dz 0

e

e

0

0

This gives zCM = C ∫ z ( z −1/ 2 )dz / C ∫ z −1/ 2 dz = 23 z 3/ 2 ]e0 / 12 z1/ 2 ]0e = 13 e That is, the CM is 13 e from top of roof or 23 e above ground. (These falling raindrops are just like

†10-57.

a stroboscopic picture of a projectile going up or down—hence the calculations are entirely the same.) According to Eq. 10.37, the velocity of the CM is related to the particle velocities by ∑i mi vi m v v . For two identical particles with one having zero velocity, v CM = P = . v CM = 2mP 2 ∑ mi i


CHAPTER

The speed of the moving proton is given by v =

10

2K 2(1.6 × 10−13 J) = = 1.38 × 107 m/s. The mP 1.67 × 10−27 kg

speed of the CM will be half of this, and the direction will be in the same direction as the velocity of the moving proton. If we assume the proton is moving in the +x direction, then v CM = 6.9 × 106 i m/s. 10-58.

Since xCM is constant, vCM = 0 in inertial frame. mkvk + mBr vBr = 0 m v Br = − k v k . mBr =−

39 u 5.0 × 103 m/s 80 u

= 2.4 × 103 m/s (in opposite direction) †10-59.

10-60.

10-61.

Since there are no external forces, the position of the CM remains unchanged. Let the CM be at the origin. This is where the boat finally meets the shark. Therefore the original positions of the shark and the boat are xsh = 300 – 45 = 255 m xb = –45 m Thus, msh xsh + m b xb = (msh 255 + 5400( −45)) /(msh + 5400) = xCM = 0 msh + m b which gives 5400 × 45 msh = = 953 kg 255 Let recoil distance be r; then mearth r + mman xman = 0. m 75 kg r = − man xman = − × 15m = 1.9 × 10−22 m m earth 5.98 × 1024 kg Because ΣFext = 0, ∆p = 0. Then ∆Pferry = ∆Ptruck mferry ∆xƒ/∆t = mtruck ∆xtruck/∆t While the truck moves 15 m relative to the ferry, it moves (15 – ∆xferry) relative to the water. Then m ferry ∆xferry = m truck (15 − ∆xferry ) ∆xferry

=

m truck (15 − ∆xferry ) m ferry

∆x f

=

mt m 15 − t ∆x f mf mf

∆xf

=

mt m 15 /(1 + t ) = (0.075)(15) /(1.075) mf mf

∆xf

= 1.05 m


CHAPTER 10-62.

10

MaCM = Fext. Hence, since Fext is gravity, the center of mass of the two fragments must continue on its parabolic path, like that of a particle. The CM reaches the ground at t = 2h / a = 5.0 × 104 / 9.8 s = 71.4 s

10-63.

The horizontal position of the CM = 71.4 s × 5.0 × 103 m/s = d = 3.6 × 105 m (in front of where missile explodes). Since one fragment lands 3.6 × 105m before this, for the CM of the two fragments to be at this point, the other must land 3.6 × 105 m after this point. The other fragment lands 7.1 × 105 m in front of the vertically dropping one. Assuming the block is initially at rest, the velocity of the ∑i mi vi mv CM is v CM = , where m is the mass of = ∑ mi m + M

m v

M

i

the bullet, M is the mass of the block, and v is the velocity of the bullet. The mass of the bullet is negligible compared to the mass of the block, so the speed of the CM mv (0.015 kg)(260 m/s) = = 1.6 m/s. If we take is vCM = M 2.5 kg

10-64.

the x direction to point in the direction of the bullet’s velocity, then vCM = 1.6i m/s. Suppose the woman is walking in the +x direction and the man is walking in the –x direction. ∑ mi vi mW vW + mM v M (60 kg)(vi) − (90 kg)(−vi) v Then v CM = i = = = − i. mW + mM 150 kg 5 ∑ mi i

10-65.

10-66.

The CM continues along the original trajectory. According to the problem, the projectile reaches its maximum height at a horizontal distance D from the launch point. Then the range for the projectile is 2D, and the CM must reach the ground at that distance from the launch point. One fragment lands at the launch point (x = 0), and one fragment lands at x = 2D. Since all three fragments have the same mass, the CM must be at the location of the middle fragment when they land, and they must be evenly spaced. That means the third fragment must land at a distance of 4D from the launch point. At the instant of explosion, the projectile’s velocity is y horizontal, so pi = mv0 cosθ i, where v0 is the initial speed and θ is the elevation angle of the initial pi velocity. Immediately after the explosion, the m momentum is p = [(v1, x + v2, x )i + (v1, y + v2, y ) j]. By 2 momentum conservation, x v1, x + v2, x = v0 cosθ v1, y + v2, y = 0


CHAPTER

10

We are told that one fragment (call it fragment 1) lands a distance D from the explosion point D some time t after the explosion. That means v1, x = . At the moment of the explosion, the y t v02 sin 2 θ . If fragment 1 had no vertical component of coordinate of the projectile was h = 2g v sin θ velocity after the explosion, it would have taken a time of 0 to fall back to the ground (the g same time it took to reach h). But we’re told that t is less than this time, so that means fragment 1 received an initial velocity in the –y direction as a result of the explosion. To find that initial gt 2 gt h ⎛ h gt ⎞ , which gives v1, y = − , or v1, y = − ⎜ − ⎟ , to make it clear velocity, use 0 = h + v1, y t − 2 2 t ⎝t 2⎠ that v1,y must point down. Because the net vertical component of momentum was zero and the two fragments have equal masses, fragment 2 must have had an upward component of initial velocity v2,2 y h gt above its starting point, so it reaches a given by v2, y = − . It will rise to a height 2g t 2 2

2

v 2 sin 2 θ v02 sin 2 θ 1 ⎛ h gt ⎞ 1 ⎛ v02 sin 2 θ gt ⎞ maximum height of h2 = 0 + + − ⎟. ⎜ ⎜ − ⎟ = 2g 2g ⎝ t 2 ⎠ 2g 2 g ⎝ 2 gt 2⎠ To find the total horizontal distance traveled by fragment 2, we must determine how far it goes v 2 sin θ cosθ , which is how far it had traveled before the after the explosion and add that to 0 g explosion. The time for fragment 2 to reach its maximum height from the explosion point is v2, y h t t2 = = − , during which time it travels a horizontal distance of v2, x t2 . From the g gt 2 D conservation of momentum equations, we see v2, x = v0 cosθ − v1, x = v0 cosθ − . The horizontal t D⎞ D ⎞⎛ h t ⎞ ⎛ ⎛ distance traveled during t2 is ⎜ v0 cosθ − ⎟ t2 = ⎜ v0 cosθ − ⎟ ⎜ − ⎟ . (The problem indicates t ⎠ t ⎠ ⎝ gt 2 ⎠ ⎝ ⎝ that h should be written in terms of the original variables, and you can do that if you wish. We’ll leave h in the equation because it has been defined above and makes the equations easier to read.) After fragment 2 reaches its maximum height, it takes some time t3 for the fragment to fall back gt 2 to the ground. To find this time, we must solve 0 = h2 + v2, y t3 − 3 . The result is 2

t3 =

v2, y + v2,2 y + 2 gh2 g

, where v2,y and h2 are defined above. We finally get the horizontal

distance traveled by fragment 2 to be x2 =

v02 sin θ cosθ ⎛ D⎞ + ⎜ v0 cosθ − ⎟ (t2 + t3 ), where t2 and t3 g t ⎠ ⎝

are defined above.


CHAPTER †10-67.

10

To solve this problem, we begin with the spreadsheet created for Problem 42. If we assume that 2π r the orbits are essentially circular, then the orbital speed of a planet is v = , where r is the T mean radius of the orbit and T is the period of the orbit. We also know that the velocity of each planet is perpendicular to its radius vector from the sun. The motion of the planets is counterclockwise, so the direction of the velocity for a planet is equal to its position angle plus 90˚. In component form, the velocity is v = v cos(θ + 90°)i + v sin(θ + 90°) j. The information is tabulated in the spreadsheet. Planet Mercury Venus Earth

m (kg) 3.30E + 23 4.87E + 24 5.98E + 24

r (km) θ(°) 5.79E + 07 232.0 1.08E + 08 170.0 1.50E + 08 95.5

T (yr) 0.241 0.615 1.00

v (km/s) 47.77 34.92 29.83

vx (km/s) 37.643 –6.063 –29.688

vy (km/s) –29.410 –34.387 –2.859

(0.33)(37.64) + (4.87)( − 6.063) + (5.98)( − 26.69) = − 17.4 km/s 0.33 + 4.87 + 5.98 (0.33)( − 29.41) + (4.87)( − 34.39) + (5.98)( − 2.859) vCM , y = = − 17.4 km/s 0.33 + 4.87 + 5.98 Note that it was not necessary to carry along the powers of 10 for the masses, since they cancel. These equations give the x and y components of vCM. We can also express the velocity as a vCM , x =

2 2 magnitude and direction. vCM = vCM , x + vCM , y = 24.6 km/s. To calculate the angle, note that

since both vCM,x and vCM,y are negative, the angle is in the third quadrant. This means we must add vCM , y = 225°. 180° to the result a calculator gives to get the correct angle. θVCM = tan −1 vCM , x

∑ m v (1500 kg)(25 m/s + 15 m/s) = = i = 20i m/s. 3000 kg ∑m ( ∑ m ) v = (3000 kg)(20 m/s) = 6.0 × 10 J. = i

10-68.

v CM

i

i

i

K CM

2 CM

i

2

∑m v i

2

∑m v = ∑m i

†10-69.

v CM

i

i

5

2

2 i i

KTOT =

2

=

i

=

(1500 kg)[(25 m/s) 2 + (15 m/s) 2 ] = 6.38 × 105 J. v1′ = 5i m/s, v′2 = − 5i m/s . 2 25 m/s

(1500 kg)(25 m/s − 15 m/s) i = 5.0i m/s. 3000 kg

The kinetic energy carried by the CM is K CM =

∑m v

2 i i

The total kinetic energy is KTOT =

i

2

=

(∑ m )v i

2 CM

2

=

15 m/s

(3000 kg)(5.0 m/s)2 = 3.8 × 104 J. 2

(1500 kg)[(25 m/s) 2 + (15 m/s) 2 ] = 6.38 × 105 J. 2


CHAPTER

10-70.

(45 kg)(640 m/s) 2 = 9.2 × 106 J. 2 2 (b) Since the pieces are both traveling in the same horizontal direction, we can work with speeds. ∑ mi vi = (32 kg)(450 m/s) + (13 kg)(1050 m/s) = 623 m/s. vCM = 45 kg ∑ mi

(a) v CM = v projectile . K =

K CM

(∑ m )v =

10-72.

i

2 CM

2

∑m v

2 i i

=

=

(45 kg)(623 m/s) 2 = 8.7 × 106 J. 2

(32 kg)(450 m/s) 2 + (13 kg)(1050 m/s) 2 = 1.04 × 107 J. The extra energy came 2 2 from the chemical energy released in the explosion. ∑ mi vi (700 kg)(18.1 m/s) − (1500 kg)(18.1 m/s) = i = − 6.58i m/s. Note that Initially, v CM = i 2200 kg ∑ mi KTOT =

10-71.

mv 2projectile

10

=

this is the same as the final velocity in Problem 16, which is required by conservation of 2 ( ∑ mi ) vCM (2200 kg)(6.58 m/s) 2 = = 4.76 × 104 J. momentum. K CM = 2 2 2 2 m v ∑ i i = (700 kg)(18.1 m/s) + (1500 kg)(18.1 m/s)2 = 3.60 × 105 J. KTOT ,initial = 2 2 mTOT (v ') 2 (2200 kg)(6.58 m/s) 2 KTOT , final = = = 4.76 × 104 J. 2 2 Both masses are stationary at the same instant, so vCM = 0. Thus. pTOT = 0. That means the momenta of the masses always have equal magnitudes and opposite directions. So the maximum mv speed of the second mass is 1 1 . The total maximum kinetic energy equal to the maximum m2 potential energy, so U max = K max =

†10-73.

m1v12 m2 v22 m1v12 ⎛ m1 ⎞ + = ⎜1 + ⎟. 2 2 2 ⎝ m2 ⎠

For helium, mHe = 4.00 u × 1.66 × 10−27 kg/u = 6.64 × 10−27 kg. mHe v 2 (6.02 × 1023 )(6.64 × 10−27 kg)(1.4 × 103 m/s) 2 = = 3.9 × 103 J. 2 2 For oxygen molecules, mO2 = 2 × 16.0 u × 1.66 × 10−27 kg/u = 5.31 × 10−26 kg. KTOT = N A

10-74.

mO2 v 2

(6.02 × 1023 )(5.31 × 10−26 kg)(500 m/s) 2 = 4.0 × 103 J. Note that these 2 2 energies are basically identical. 2 MvCM (a) (See diagram (a).) v CM = 0 ⇒ =0 2 KTOT = N A

=

2 MvCM ( M / 2)( −vi + vj) −vi + vj v Mv 2 = ⇒ vCM = ⇒ . = 2 4 M 2 2 2 MvCM ( M / 2)(2vj) Mv 2 = ⇒ vCM = v ⇒ . = M 2 2

(b) (See diagram (b).) v CM = (c) See diagram (c.) v CM


CHAPTER

10

(a) †10-75.

Assume vCM = 0. Since v CM =

(b)

(c)

∑m v ∑m i

i

, this means M J v J = − M S v S . In terms of speeds,

i

vS = 10-76.

†10-77.

10-78.

K M v2 M M J vJ . The ratio of kinetic energies is S = S S2 = J = 9.54 × 10−4 . K J M J vJ M S MS

If we had one mole of atoms all moving at the same speed v, the total kinetic energy would be Mv 2 K= , where M is the molar mass. If we have a mass m of the material, the total kinetic 2 mv 2 energy would the fraction m/M of the kinetic energy of one mole, or K = , as if the entire 2 mass were moving at speed v. Thus for 1 kg of material with the internal particles moving at 360 m/s, K = 6.48 × 104 J. Assume the hunter is at rest when he fires the first bullet. Then he gets a recoil speed of mvB , where m is the mass of the bullet, vB is the speed of the bullet, and M is the mass of v1 = M the hunter. When the second bullet is fired, it has a speed vB relative to the already moving hunter and produces a recoil speed v1 relative to the hunter, so now the hunter’s speed relative to the ground is 2v1. Each bullet adds an increment v1 to the hunter’s speed relative to the ground, so mv (0.015 kg)(600 m/s) = 1.1 m/s. after firing n bullets his speed is vn = n B . Thus vn = 10 × M 80 kg When 1 kg of grain falls a height h and lands on the car, its speed changes from

2gh to 0, so

the magnitude of its momentum change is (1 kg) 2 gh . Since the grain was initially falling down, the direction of this momentum change is up. That means the force exerted by the car on the grain is up, so the grain exerts a downward force of the same magnitude on the car. If the rate at which grain falls on the car is 500 kg/s, then the net downward force exerted on the car has a magnitude of Fdown = (500 kg/s) (2)(9.81 m/s 2 )(4 m) = 4.4 × 103 N (about 990 lbf). †10-79.

The magnitude of the net horizontal force on each child is 200 N, so the magnitudes of their horizontal accelerations are F 200 N aB = net = = 5.0 m/s 2 mB 40 kg F 200 N aG = net = = 6.7 m/s 2 mG 30 kg


mB

x=0

200 N

200 N

4.0 m

mG

CHAPTER

10

The accelerations are in opposite directions. Taking x to point right, a B = 5.0i m/s 2 , aG = − 6.7i m/s 2 . Since the net external force on the pair of children is zero, their mutual CM

remains at rest. That is the point where they will meet. Taking x = 0 at the boy’s initial position, ∑ mi xi = (30 kg)(4.0 m) = 1.7 m. They will meet 1.7 m to the right of the boy. xCM = 70 kg ∑ mi 10-80.

v CM =

∑m v ∑m i

i

†10-81.

i

=

(1200 kg)(60 km/h) + (1500 kg)(80 km/h) i = 71i km/h. 2700 kg

The initial momentum is pi = mcar v car + mtruck v truck . The two vehicles are at rest after the collision, so p f = 0. By conservation of momentum, pi = p f ⇒ mcar v car + mtruck v truck = 0, which means mcar v car m v . Since speed is the magnitude of velocity, vtruck = car car , and we get mtruck mtruck m (40 km/h) = car = 8.0 km/h. 5mcar

v truck = −

vtruck 10-82.

10-83. 10-84.

†10-85.

dp d (mv) dm = =v = (26 m/s)(800 l/min)(1 min/60 s)(1 kg/l) = 347 N. This corresponds dt dt dt roughly to half the body weight of an average man, so you could probably hold the hose. ∑ mi xi = (39.1 u)(0.282 nm) = 0.0927 nm. xCM = 39.1 u + 79.9 u ∑ mi F=

Due to the absence of external forces, the position of the CM is constant. Taking the ship at origin 400 m (4000 t ) xCM = = 50 m (28000 + 4000) t The tugboat reels in 200 m. By inspection one can see that since this is half the original length of towrope, each ship must move half the distance to the CM. Therefore, the ship moves 25 m; the towboat moves 175 m. To have a concrete starting point for the analysis, assume that the cat is initially at the center of the board so she’s at the CM. (It doesn’t really matter, but this assumption makes the calculations simpler.) Call this position x = 0. Since the net force on the cat and board is zero, the CM of the cat-board system must remain fixed as the cat walks. Suppose the cat has walked to a position xC to the right of the initial position. Since the CM of the system must remain at x = 0, the CM of the board must have moved to a position –xB to the left of x = 0. (xB is the distance to the position of m x − mB xB . We’re told that the cat the board’s CM.) Mathematically, this is means xCM = 0 = C C mC + mB walks 1.0 m along the board, so xC + xB = 1.0 m. (Remember that the board must move under the m x − mB (1 − xC ) cat.) Thus xB = 1.0 m-xC . Substituting into the equation for xCM gives C C =0 mC + mB


CHAPTER

10

⇒ xC (mC + mB ) = (1.0)mB ⇒ xC =

xC =

(1.0)mB . Substituting numbers gives ( mC + mB )

(1.0 m)(5.0 kg) = 0.59 m. The cat moves 59 cm relative to the water. 8.5 kg

10-86.

Find the CM along the ladder, ignoring the mass of the ladder. ∑ mi xi = m(5.0 m) + m(15 m) + m(20 m) = 13 m. xCM = 3m ∑ mi

10-87.

Assume the stack of two books is at the apex of the triangle. Also assume the length of a side is a, so the apex is a 3 at y = a sin 60° = . By symmetry, xCM = 0. Then 2 ⎛a 3⎞ 2m ⎜ ⎟ ∑ mi yi = ⎝ 2 ⎠ = a 3 . a = 1.0 m, so the yCM = 4m 4 ∑ mi

y

2m

60˚

⎛ 3 ⎞ location of the CM is ⎜⎜ 0, m ⎟⎟ = (0, 0.43 m). 4 ⎝ ⎠ The given mass distribution can be replaced by the masses as shown. 1/ 2m + 0m + (−1/ 2m) xCM = =0 3m 1/ 2m + 0m + 1/ 2m yCM = 3m = 1/ 3meter

10-89.

Let the axes be as shown, with the x-axis along lighter sheet, mass M, and y-axis along the heavier sheet mass 2 M. The center of masses of the lighter and heavier sheets are (L/2,0,0) and (0,L/2,0) respectively. rCM = M (L/2,0,0) + 2M(0,L/2,0)/(M + 2M) = (L/6,L/3,0)

10-90.

Refer to Problem 55. Instead of dividing a hemispherical shell into thin rings, imagine dividing the solid hemisphere into circular slabs of radius r=

y

R 2 − x 2 and thickness dx. The mass of a slab is

dm = ρ dV = ρ (π r 2 dx) = πρ ( R 2 − x 2 )dx. For a hemisphere of mass M M 3M and radius R, ρ = = , and 3 (2π R / 3) 2π R 3 xCM

m

m

10-88.

x

∫ =

x=R

x =0

xdm

M

3 = 2R3

R x

x

R

∫

R

0

3 ⎛ R2 x2 x4 ⎞ 3R x( R − x )dx = − ⎟ = . 3 ⎜ 2R ⎝ 2 4 ⎠0 8 2

2


dx

CHAPTER 11

COLLISIONS


F=

∆p ∆t

Hits the water with velocity (positive up) v = – 2gh = – 2 × 9.81m s 2 × 11m = –14.7 m/s

11-2.

p = mv = 77 kg × –14.7 m/s = –1.13 × 103 kg-m/s p′ = 0 Impulse = ∆p = 0 – (–1.13 × 103 kg-m/s) = 1.13 × 103 kg-m/s 1.13 × 103 kg-m s = 2.3 × 104 N (up) F= 0.05s 20 km/hr = 5.56 m/s 7.0 × 105 metric tons = 7.0 × 108 kg p = mv = 7.0 × 108 kg × 5.56 m/s = 3.89 × 109 kg-m/s p′ = 0 Impulse = ∆p = 0 – 3.89 × 109 kg-m/s = –3.89 × 109 kg-m/s ∆p −3.89 × 109 kg-m s = –7.8 × 108 N F= = 5s ∆t F 7.8 × 108 N = 1.1 m/s2 a= = 7.0 × 108 kg m

11-3.

It is difficult to obtain precise measurements from the photographs, but the changes in position are approximately v(m/s) a(m/s2) F(N) change distance (m) a→b 0.24 12 250 4.2 × 105N b→c 0.14 7 200 3.4 × 105 c→d 0.07 3 100 1.7 × 105 d→e 0.02 1 1 1.7 × 105 e→f –0.02 –1 Mark the graph into blocks along the tick marks on the axes. Each block then has an area of 103 N • s. The area under the curve is about 7.5 blocks, so Impulse ≈ 7.5 × 103 N • s.

11-4.

Initial velocity v = – 2 gh = – 2 × 9.81m s 2 × 21m = –20.3 m/s final velocity v′ = 0

0 − ( −20.3 m s ) ∆v = 1015 m/s2 = 0.02s ∆t F = ma = 64 kg × 1015 m/s2 = 64,960 N 28 g = 2.8 × 10–2 kg 70 cm = 0.7 m p = mv = 0 p′ = 2.8 × 10–2 kg × 450 m/s = 12.6 kg-m/s Impulse = ∆p = 12.6 kg-m/s – 0 = 12.6 kg-m/s

average acceleration a =

†11-5.


CHAPTER

11

( 450 m s ) = 1.45 × 105 m/s2 v ′2 − v ′2 v′ = v + 2ax so a = = 2 × 0.7m 2x v 450 m s t= = = 3.10 × 10–3 s = time to accelerate a 1.45 × 105 m s 2 ∆p 12.6 kg-m s F= = = 4.06 × 103 N ∆t 0.0031 s (Compare this with F = ma: F = 2.8 × 10–2 kg × 1.45 × 105 m/s2 = 4.06 × 103 N) (a) 55 km/hr = 15.3 m/s ∆v v′ − v 0 − 15.3 m/s a= = = = –139 m/s2 ∆t 0.11 s ∆t 65 km/hr = 18.1 m/s 0 − 18.1m/s a= = –165 m/s2 0.11s 75 km/hr = 20.8 m/s 0 − 20.8 m/s a= = –189 m/s2 0.11 s (b) The crush distance is the distance the automobile travels during stopping. 2

2

11-6.

2

At 15.3 m/s

0 − (15.3m/s ) v ′2 − v 2 x= = = 0.842 m −2(139 m/s 2 ) 2a

At 18.1 m/s

0 − (18.1m/s ) v ′2 − v 2 x= = = 0.993 m −2(164 m/s 2 ) 2a

At 20.8 m/s

0 − ( 20.8m/s ) v ′2 − v 2 x= = = 1.14 m −2(189 m/s 2 ) 2a

2

2

2

†11-7.

11-8.

At 15.3 m/s F = ma = 75 kg × 139 m/s2 = 1.04 × 104 N At 18.1 m/s F = ma = 75 kg × 165 m/s2 = 1.24 × 104 N At 28.8 m/s F = ma = 75 kg × 189 m/s2 = 1.42 × 104 N ∆p F= 50 km/hr = 13.9 m/s ∆t p = mv = 10 kg × 13.9 m/s = 139 kg-m/s p′ = 0 Impulse = ∆p = 0 – 139 kg-m/s = –139 kg-m/s 139 kg-m s F = = 1390 N, or about 310 lb. It is unlikely that the mother could hold on to the 0.10s child. 143 km/hr = 39.7 m/s 0 − ( 39.7m/s ) v ′2 − v 2 a= = = –143 m/s2 2 × 5.5 m 2x 2

pi = mvi = 75 kg × 39.7 m/s = 2977 kg-m/s pf = 0 Impulse = ∆p = 0 – 2977 kg-m/s = –2977 kg-m/s ∆v −39.7 m s Time to stop = = = t = 0.278 s a −143 m s 2 ∆p −2977 kg i m s F= = = –1.07 × 104 N 0.278 s ∆t


CHAPTER †11-9.

11

Because the collision is elastic the initial velocity of the ball will be 15 m/s and the final velocity will be –15 m/s. The mass of the ball is 60 g = 0.06 kg. The impulse is the change in momentum p = mv = 0.06 kg × 15 m/s = 0.9 kg-m/s p′ = 0.06 kg × –15 m/s = –0.9 kg-m/s Impulse = ∆p = –0.9 kg-m/s – 0.9 kg-m/s = –1.8 kg-m/s 0 − (15 m/s ) v ′2 − v 2 To stop the ball a = = = –2.25 × 104 m/s2 2 × 0.005 m 2x 2

∆v −15m s = = t = 6.67 × 10–6 s This is the same time needed a −2.25 × 104 m s 2

Time to stop the ball is

to accelerate the ball. ∆p −1.8 kg-m s F= = = –1.35 × 103 N -4 ∆t 2 × ( 6.67 × 10 s )

11-10.

†11-11.

11-12.

(Check with F = ma: F = 0.06 kg × –2.25 × 104 m/s2 = –1.35 × 103 N) p = mv = 0.50 kg × 2.0 m/s = 1.0 kg-m/s p′ = 0 From momentum conservation, impulse delivered to board is the negative of the impulse delivered to the hammer Impulse to hammer = ∆p = 0 – 1 kg-m/s = –1.0 kg-m/s ⇒ impulse to board = 1.0 kg-m/s ∆p 1 kg-m s F= = = 50 N ∆t 0.020s p = mv = 0 p′ = 0.45 kg × 18 m/s = 8.1 kg-m/s Impulse to ball = ∆p = p′ – p = 8.1 kg-m/s – 0 kg-m/s = 8.1 kg-m/s ∆p 8.1 kg-m s Impulse is F∆t = ∆p so ∆t = = = 0.045 s 180 N F We need to find the height that will result in a velocity for the egg such that the momentum change when the egg stops in 0.020 s will result in a force greater than 3 N For a falling body (down is negative) v = – 2 gh A force of 3 N applied for 0.020 s gives an impulse of F∆t = (3 N)(0.020 s) = 0.06 kg-m/s Impulse = ∆p = 0 – mv so 0.06 kg-m/s = m 2 gh

⎛ 0.06 kg-m ⎜ ⎝ 0.05kg †11-13. 11-14.

Impulse I =

2

s⎞ 1.44 m 2 s 2 = 0.073 m (7.3 cm) = 2 gh or h = ⎟ 2 × 9.8m s 2 ⎠

∫

∆t

0

Fdt =

∫ ( 3.0t + 0.5t )dt = ∫ 3s

2

0

3s

0

3.0tdt + ∫

3s

0

3s

3t 2 0.5t 3 0.5t dt = + 20 30 2

3s

= 18 kg • m/s

Impulse = ∆p = p′ – p Before: px = mvx = 0.15 kg × 35 m/s = 5.25 kg • m/s py = 0 2 2 v sin (100 ) 110m × 9.81m s = 33.1 m/s After: Range = ⇒v= g sin (100 )

p′y = mv′y = 0.15 kg × (33.1 p′x = –mv′x = –0.15 kg × (33.1 m/s)cos(50) = –3.19 kg • m/s m/s)sin(50) = 3.80 kg • m/s ∆px = –3.19 kg • m/s – 5.25 kg • m/s = –8.44 kg • m/s ∆py = –3.8 kg • m/s – 0 = –3.8 kg • m/s ∆p =

( −8.44 kg i m/s )

2

+ ( 3.8 kg i m/s ) = 9.3 kg • m/s 2

⎛ 3.8 ⎞ tan–1 ⎜⎝ 8.44 ⎟⎠ = 24° above the horizontal


CHAPTER †11-15.

11-16.

†11-17.

11

120 km/hr = 33.3 m/s Time interval of contact 2.5 m 2.5 m = = 7.52 × 10–2 s ∆t = horizontal speed 33.3 m/s × cos(3) Magnitude of average force 2 ⎡ 600 kg × 33.3 m/s × sin ( 3) ⎤⎦ ∆p F= = 2.8 × 104 N = ⎣ ∆t 7.52 × 10−2 s 1 1 1 1 In an elastic collision kinetic energy is conserved so m1v12 + m2v22 = m1v1′2 + m2v2′2 2 2 2 2 1 Canceling the factors of and rearranging m(v12 – v1′2) = m(v2′2 – v22) 2 Factoring m1(v1 – v1′)(v1 + v1′) = m2(v2′ – v2)(v2′ + v2) Momentum is conserved in all collisions so m1v1 + m2v2 = m1v1′ + m2v2′ Rearranging m1(v1 – v1′) = m2(v2′ – v2) Since m1 = m2 all the masses cancel. We can divide the energy equation by the momentum equation and get v1 + v1′ = v2′ + v2 Enter the original numbers into the momentum conservation expression and the derived eq. 10 m/s + 5 m/s = v1′ + v2′ 10 m/s + v1′ = v2′ + 5 m/s Solve simultaneously to get v1′ = 5.0 m/s and v2′ = 10 m/s (a) Using equations 11.13 and 11.14 (pp. 345-346), velocity of the projectile after collision m − m2 −0.06 kg v1 = v1′ = 1 × 0.8 m/s = –0.27 m/s 0.18 m1 + m2 velocity of target after collision 2m1 v1 = 0.53 m/s v2′ = m1 + m2

11-18.

11-19.

(b) Initial kinetic energy of projectile: 1 1 K1 = m1v12 = (0.06 kg)(0.8 m/s)2 = 1.9 × 10–2 J 2 2 Target: K2 = 0 J Final kinetic energy of: m (v ′ ) 2 1 Projectile: K′1 = 1 1 = (0.06 kg)(0.27 m/s)2 = 2.2 × 10–3 J 2 2 2 m (v ′ ) 1 Target: K′2 = 2 2 = (0.12 kg)(0.53 m/s)2 = 1.7 × 10–2 J 2 2 Using equation 11.13, the velocity of the bullet after the collision will be m − m2 0.015 kg- 0.04 kg v1 = v1′ = 1 × 600 m/s = –2.7 × 102 m/s 0.055 kg m1 + m2 m − m2 2m1 v1 ; v2′ = v1 From Equations 11.13 & 11.14, v1′ = 1 m1 + m2 m1 + m2 Where v1 = velocity of club v2′ = velocity of ball after impact


CHAPTER

11

m1 = mass of club m2 = mass of ball 0.15 kg + 0.045 kg v1 = × 60 m/s = 39 m/s 0.15 kg 11-20.

The mass of a neutron is mm = 1 u, and the mass of a carbon nucleus is mC = 12 u. If the initial m − mC 1 − 12 velocity if the neutron is v, its final velocity is given by v′ = n v= v = − 0.846v. In 1 + 12 mn + mC terms of energy, the initial speed is given by

v=

2

K = m

2 × 4.8 × 10−13 J 9.8 × 10−7 = mn mn

The desired final speed is given by

v′ =

2

K′ = m

2 × 1.6 × 10−19 J 5.66 × 10−10 = mn mn

After each collision the speed is reduced by a factor of 0.846. Therefore, after the nth collision, v′ = (0.846)nv We want to know what n is required so that v′ = (0.846)nv. Therefore,

v′ =

†11-21.

11-22.

5.66 × 10−10

mn

=

( 0.846 )

n

× 9.8 × 10−7

mn

or (0.846)n = (5.66 × 10–10/9.8 × 10–7) n log(0.846) = log(5.66 × 10–10/9.8 × 10–7) n = log(5.66 × 10–10/9.8 × 10–7)/log(0.846) = 46 times In an elastic collision kinetic energy and momentum are conserved and the velocity of the nail just after impact is given by eq. 11.14: 2m1 2 × 0.5 kg v1 = v2′ = × 5.0 m/s = 9.77 m/s 0.012 kg + 0.5 kg m1 + m2 1 1 The kinetic energy of the nail KN = m2 v2′2 = (0.012 2 2 kg)(9.77 m/s)2 = 0.57 J

In an elastic collision kinetic energy and momentum are conserved and we can use eqs. 11.13 and 11.14 mQ = 0.0056 kg mD = 0.0023 kg m − m2 2m1 v1 v2′ = v1 v1′ = 1 m1 + m2 m1 + m2 If the quarter is initially stationary, its velocity after the collision is 2 × 0.0023 kg v2′ = × 2.0 m/s = 1.2 m/s 0.0023 kg + 0.0056 kg


CHAPTER

11

The velocity of the dime is 0.0023 kg − 0.0056 kg v1′ = × 2.0 m/s = –0.84 m/s 0.0023 kg + 0.0056 kg If the dime is initially stationary, its velocity after the collision is 2 × 0.0056 kg v2′ = × 2.0 m/s = 2.8 m/s 0.0056 kg + 0.0023 kg The velocity of the quarter is 0.0056 kg − 0.0023 kg v1′ = × 2.0 m/s = 0.84 m/s 0.0023 kg + 0.0056 kg †11-23.

The first ball hits a stationary block, but after that first collision the block is moving and Equations 11.13 and 11.14 are not applicable. We must instead solve the problem of a mass m1 moving with initial velocity v1 striking a mass m2 moving with velocity v2. After the collision the masses have velocities v′1 and v′2 respectively. Then conservation of momentum and kinetic energy require m1v1 + m2 v2 = m1v1′ + m2 v2′

m1v12 m2 v22 m1 (v1′ ) 2 m2 (v2′ ) 2 + = + 2 2 2 2 The solution can be simplified by describing the motion in a frame of reference moving at velocity v2. In this frame, m2 is at rest and Equations 11.13 and 11.14 can be used. Define new velocities: V1 = v1 − v2 V1′ = v1′ − v2

V2′ = v2′ − v2 In this frame V2 = 0 and the final velocities are m − m2 V1′ = 1 V1 m1 + m2

2m1 V2 m1 + m2 To get the final velocities in the original frame, substitute the velocity definitions: m − m2 (m − m2 )v1 + 2m2 v2 v1′ − v2 = 1 (v1 − v2 ) ⇒ v1′ = 1 m1 + m2 m1 + m2 V2′ =

2m1 2m1v1 + (m2 − m1 )v2 (v1 − v2 ) ⇒ v2′ = m1 + m2 m1 + m2 (If v2 = 0, these reduce to Equations 11.13 and 11.14.) In this problem, v1 = v and m2 = 40m1 so the sum of the masses is 41m1 and the magnitude of the difference is 39m1. For the first collision, v2 = 0 and 2m1v 2 ′ = v2,1 = v = 0.0488v 41m1 41 v2′ − v2 =

For the next collision, this becomes the initial velocity of m2 and the final velocity is 2m1v + 39m1 (0.0488v) ′ = v2,2 = 0.0952v 41m1


CHAPTER

11

Continuing the process gives 2m1v + 39m1 (0.0952v) ′ = v2,3 = 0.139v 41m1

11-24.

′ = v2,4

2m1v + 39m1 (0.139v) = 0.181v 41m1

′ = v2,5

2m1v + 39m1 (0.181v) = 0.222v 41m1

Since the collision is elastic and one body is initially at rest we can use m − m2 2m1 v1′ = 1 v1 v2′ = v1 m1 + m2 m1 + m2 v2′ = 1.5 m/s m2 = 0.15 kg v1′ = 0.5 m/s m − 0.15 kg 2m1 v1 and 1.5 m/s = v1 0.5 m/s = 1 m1 + 0.15 kg m1 + 0.15 kg eq. 1 (0.5 m/s)(m1 + 0.15 kg) = (m1 – 0.15 kg)v1 eq. 2 (1.5 m/s)(m1 + 0.15 kg) = 2m1v1 Solving eq. 2 for v1 and substituting in eq. 1 gives m1 = 0.45 kg Substituting this in eq. 2 yields v1 = 1.0 m/s

11-25.

Smaller marble exits in one-third the time it takes larger marble to exit means v1′ = –3v2′. Since the collision is elastic and one body is initially at rest we can use m − m2 2m1 v1′ = 1 v1 v2′ = v1 m1 + m2 m1 + m2 v′ m−M 2m v1 and – 1 = v1 v1′ = 3 m+M m+M m−M M so m = Divide the second equation into the first equation –3 = 2m 7

11-26.

The mass of the alpha particle (m1) = 4 × 1.66 × 10–27 kg = 6.64 × 10–27 kg. With 1 Kα = 1.6 × 10–13 J = m1v12 v1 = 6.9 × 106 m/s 2 For silicon m2 = 4.6 × 10–26 kg m − m2 6.64 × 10−27 kg − 4.6 × 10−26 kg × 6.9 × 106 m/s = –5.2 × 106 m/s v1′ = 1 v1 = 6.64 × 10−27 kg + 4.6 × 10−26 kg m1 + m2 1 1 Kα = m1v12 = (6.64 × 10–27 kg)( –5.2 × 106 m/s)2 = 9 × 10–14 J 2 2 For copper m2 = 1.0 × 10–25 kg m − m2 6.64 × 10−27 kg − 1.0 × 10−25 kg × 6.9 × 106 m/s = –6.0 × 106 m/s v1′ = 1 v1 = 6.64 × 10−27 kg + 1.0 × 10−25 kg m1 + m2 1 1 Kα = m1v12 = (6.64 × 10–27 kg)( –6.0 × 106 m/s)2 = 1.2 × 10–13 J 2 2 The collision is elastic. In a reference frame moving at 55 km/hr the front car is at rest and we can use m − m2 2m1 v1 v2′ = v1 v1′ = 1 m1 + m2 m1 + m2

†11-27.

60 km/hr = 16.7 m/s 55 km/hr = 15.3 m/s This give us


CHAPTER

11

v1 = 1.4 m/s v2 = 0 m1 = 1200 kg m2 = 1000 kg 1200 kg − 1000 kg 2(1200 kg) × 1.4 m/s = 0.13 m/s and v2′ = × 1.4 m/s = 1.5 m/s v1′ = 1200 kg + 1000 kg 1200 kg + 1000 kg Changing back to the original frame of reference (adding 15.3 m/s to each velocity) v ′ = 15 m/s and v ′ = 17 m/s 1

11-28.

2

The kinetic energy is the same (since same amount of energy expended). 1 1 Stationary gun: K = mv2 = (45kg)(656.6 m/s)2 = 9.70 × 106J 2 2 Recoiling gun: v1 = velocity of gun and v2 = velocity of the projectile m Conservation of momentum m1v1 + m2v2 = 0 v1 = − 2 v2 m1 1 1 Conservation of kinetic energy m1v12 + m2v22 = 9.70 × 106 J 2 2 2

1 ⎛ m2 v2 ⎞ 1 m1 ⎜ + m2v22 ⎟ 2 ⎝ m1 ⎠ 2 2 ⎛ m2 ⎞ 1 K = v22 ⎜ m + m ⎟ 2 ⎠ 2 ⎝ 1 K=

v2 =

†11-29.

2K = 2 ⎡⎣ (m2 m1 ) + m2 ⎤⎦

2 × 9.7 × 106 J

⎡( 45 kg )2 6.6 × 103 kg + 45 kg ⎤ ⎣ ⎦ This is an elastic collision and we can use equations 11.13 and 11.14 m − m2 2m1 v1 v2′ = v1 v1′ = 1 m1 + m2 m1 + m2

= 654 m/s

Collision between first and second balls: m − m2 2m1 v1 = 0 and v2′ = v1 = v1 Equal masses gives v1′ = 1 m1 + m2 m1 + m2 Collision between second and third balls: m − m3 2m2 v2 = v1 v2′ = 2 v2 = 0 and v3′ = m2 + m3 m2 + m3 11-30.

Last ball moves with velocity v1 = v, other two are stationary. This is an elastic collision and we can use equations 11.13 and 11.14 m − m2 2m1 v1 v2′ = v1 v1′ = 1 m1 + m2 m1 + m2 Collision between first and second balls: m − 2m v 2m 2v v = – and v2′ = v = v1′ = 3 m + 2m m + 2m 3 Collision between second and third balls:


CHAPTER

v2′ = 11-31.

2m − m 2v 2v 2 × 2m 2v 8v × = and v3′ = × = 2m + m 3 2m + m 3 9 9

From conservation of energy, the velocity of the ball at impact is v = 2 gh This is an elastic collision and we can use equations 11.13 and 11.14 We get v m − 2m 1 2m vm′ = 2 gh v2m′ = vm = m = vm = 3 m + 2m m + 2m 3 2v2m 2 = 2 gh 3 3 Potential energy equals kinetic energy: 1 2 v2 mv = mgh so h = 2g 2 2 gh h Height mass m reaches is hm = = 9 × 2g 9 Height mass 2m reaches is hm =

4 × 2 gh 4h = 9 × 2g 9

(b) By conservation of energy, at the next impact the balls have velocities 1 2 1 2 gh and − 2 gh , respectively. Transfer to the frame moving to the right at 2 gh in 3 3 3 this frame, the ball of mass m is stationary, the ball of mass 2m is moving at v2m = − 2 gh The velocities after the collision in this frame are: 2 × 2m 4 2 gh vm′′ = − 2 gh = − 3 m + 2m 2m − m 1 2 gh − 2 gh = − v2′′m = 2m + m 3 So in the lab frame 4 1 2 gh + 2 gh = − 2 gh or height reached = h vm′′ = − 3 3 1 1 2 gh + 2 gh = 0 or height reached = 0 v2′′m = − 3 3 Treat this as an elastic collision between a moving satellite and a stationary Jupiter. The satellite is traveling at –10 km/s. Converting to Jupiter′s reference frame gives vs = –10 km/s – 13 km/s = –23 km/s. Because the mass of the planet >> mass of the satellite, eq. 11.13 m − mJ m vs′ = s vs′ ≈ − J vs = − vs vs becomes eq. 11.15 ms + mJ mJ

(

)

(

11-32.

)

After the collision with Jupiter the satellite will rebound at minus its incoming speed.


11

CHAPTER

11-33.

11

In the reference frame of Jupiter the satellite is traveling at –23 km/s initially. vs′ = 23 km/s We subtracted the velocity of Jupiter to convert to the reference frame of a stationary Jupiter. To return to the reference frame of the Sun we must add 13 km/s. The final speed of the satellite in the reference frame of the Sun is 23 km/s + 13 km/s = 36 km/s. The change in velocity is vsf – vsi = 36 km/s – (–10 km/s) = 46 km/s Move to the frame of reference of the turbine. Let the turbine be moving at vt, with the water moving at vw. Then in the reference frame of the turbine, the turbine is stationary and the water is moving at vw – vt = uw (i) m − m2 uw By Equation 11.13 uw′ = 1 m1 + m2 The mass of the turbine >> the mass of the water particle, m − m2 m2 so that 1 ≈ = − 1. Therefore, uw′ = −uw in the m1 + m2 m2 frame of the turbine. In the frame of the ground this is vw′ = vt + uw′ = vt − uw . For maximum transfer of kinetic energy: vw′ = 0 relative to ground, so that

11-34.

vt – uw = 0 ⇒ from (i) vt – uw = vt – (vw – vt) = 0 ⇒ 2vt = vw 1 ⇒ vt = vw . Therefore, the velocity of the turbine ideally 2 = (1/2) vw = 27/2 = 13.5m/s (a) As in Example 4, the velocity of the recoiling neutron is m − m2 1.01 u − 2.01 u v1′ = 1 v1 = v1 = −0.331v1 1.01 u + 2.01 u m1 + m2 so the speed has been reduced by a factor or 0.331. (See Appendix 7 for the mass of the neutron in u.) (b) In a single head-on collision with a proton, the velocity of the recoiling neutron is m1 − m2 1.0087 u − 1.0073 u v1 = v1 = 6.9 × 10−4 v1 . In each collision with a deuterium nucleus, m1 + m2 1.00087 u + 1.0073 u the speed is reduced by a factor of 0.331. Hence the number n of such collisions required to attain a factor of 6.9 × 10–4 is given by (0.331)n = 6.9 × 10–4 or log ( 6.9 × 10−4 ) –4 n log(0.331) = log(6.9 × 10 ) ⇒ n = = 6.6 seven collisions are required. log ( 0.331)

†11-35.

The 1:10 slope implies a right triangle with an angle of tan–1(1/10) = 5.71°. With the hypotenuse as the ramp, moving 12 m down the hypotenuse means a drop h = 12 × sin(5.71°) = 1.19 m. In 1 dropping 1.2 m the automobile gains kinetic energy of mgh = mv 2 . At the bottom the first 2 automobile has a speed v1 =

2gh =

2 × 9.81 m/s 2 × 1.19 m = 4.83 m/s. For the second

automobile v2 = 0. After the collision eqs. 11.13 and 11.14 give m − m2 1400 kg − 800 kg × 4.83 m/s = 1.32 m/s and v1 = v1′ = 1 1400 kg + 800 kg m1 + m2


CHAPTER

v2′ =

11-36.

11

2m1 2 × 1400 kg × 4.83 m/s = 6.15 m/s v1 = 1400 kg + 800 kg m1 + m2

Frictional force for automobile 2: f = µN = 0.9 × 800 kg × 9.81 m/s2 × cos(5.71°) = 7828 N Gravitational force on automobile 2: Fg = 800 kg × 9.81 m/s2 × sin(5.71°) = 781 N Net force on automobile 2: F2 = f + Fg = –7828 N + 781 N = –7047 N, where the positive direction is taken to point down the slope. Acceleration of automobile 2: a2 = −7047 N/800 kg = −8.81 m/s2 Acceleration of automobile 1: a1 = g sin(5.71º) = 9.81 m/s2 × sin(5.71°) = 0.976 m/s2 at 2 The cars will collide when they have traveled the same distance x = vt + 2 2 2 at at 0.976t 8.81t ⇒ v1t + 1 = v2t + 2 ⇒ 1.32 + = 6.15 − 2 2 2 2 9.66 t= s = 0.987 s 9.79 (.976)(0.987) 2 = 1.78 m Automobiles travel x = (1.32 m/s)(0.987 s) + 2 (a) From conservation of momentum m1v1 + m2v2 = m1v′1 + m2v′2 Rewrite m1(v1 – v′1) = m2(v′2 – v2) (1) From conservation of energy m1v12 + m2v22 = m1v′12 + m2v′22 Rewrite m1(v12 – v′12) = m2(v′22 – v22) m1(v1 – v′1)(v1 + v′1) = m2(v′2 – v2)(v′2 – v2) (2) Divide Eq (2) by Eq (1) v1 – v2 = v′2 – v′1 for v2 = 0 v′ −v ′ = −v 1

2

1

(b) Kinetic energy is the sum of the center of mass kinetic energy and the internal kinetic energy 1 1 m1m2 2 K = MvCM2 + ( v1 − v2 ) 2 2 m1 + m2 With coefficient of restitution e v′1 − v′2 = −e(v1 − v2)

1 m1m2 2 ( v1 − v2 ) 2 m1 + m2 1 m1m2 1 m1m2 2 2 2 e ( v1 − v2 ) Final internal kinetic energy KI′ = ( v1′ − v2′ ) = 2 m1 + m2 2 m1 + m2

Initial internal kinetic energy KI =

Therefore, K I′ = e2 K I (c) For the case where particle 2 is at rest conservation of momentum m1v′1 + m2v′2 = m1v1 But v1′ − v2′ = −ev1 So v2′ = v1′ + ev1. Substituting this into the conservation of momentum expression m1v′1 + m2(v1′ + ev1) = m1v1 or (m1 + m2) v1′ = m1v1 − m2ev1 m − em2 ∴ v1′ = 1 v1 m1 + m2 v1′ = v2′ + ev1 Substituting this (1 + e ) m1 v v2′ = 1 m1 + m2


CHAPTER †11-37.

11

Suppose the fist and the block begin moving together at the same speed after impact so the collision is completely inelastic. The final m1v1 , where m1 = mass of fist velocity is v′ = m + m2 and m2 = mass of block. The energy available to break the block is the difference between the initial kinetic energy and the final kinetic energy: 2

m v 2 (m + m2 )v′2 m1v12 (m1 + m2 ) ⎛ m1v1 ⎞ m1m2 v12 E = K − K′ = 1 1 − 1 = − = ⎜ ⎟ 2 2 2 2 2(m1 + m2 ) ⎝ m1 + m2 ⎠

11-38.

11-39.

11-40.

†11-41.

11-42.

Since the block is considerably heavier than the fist, i.e., m2 >> m1, this becomes m v 2 0.4 × 122 J = 29 J, which is enough to break the block. E≈ 1 1 = 2 2 Conservation of momentum m1v1 + m2v2 = m1v1′ + m2v2′ In this case m1v1 + m2v2 = (m1 + m2)v′. Let north (toward Munchausen′s city) be the positive direction. Then 90 kg × (–150 m/s) + 20 kg × 300 m/s = (90 kg + 20 kg)v′ −7500 kg i m/s v′ = = –68 m/s. Since he is still traveling south, it is not possible for him to arrive 90 kg + 20 kg back in his own city! Conservation of momentum for a totally inelastic collision m1v1 + m2v2 = (m1 + m2)v′ (a) Two automobiles of equal mass v1 = 15 m/s, v2 = 0 m(15 m/s) + 0 = (m + m)v′ v′ = 7.5 m/s ∆v = –7.5 m/s (b) Two automobiles of equal mass v1 = 15 m/s, v2 = –15 m/s m(15 m/s) + m(–15 m/s) = (m + m)v′ v′ = 0 ∆v = –15 m/s (c) Automobile and stationary wall v1 = 15 m/s, v2 = 0, mcar << mwall mcar mcar(15 m/s) = (mcar + mwall)v′ v′ = × 15 m/s ≈ 0 ∆v ≈ –15 m/s mcar + mwall Conservation of momentum for a totally inelastic collision (25 kg + 10 kg) × 3.0 m/s + 30 kg × 0 = (35 kg + 30 kg)v′ 105 kg i m/s v′ = = 1.6 m/s 35 kg + 30 kg

m1v1 + m2v2 = (m1 + m2)v′

Conservation of momentum for a totally inelastic collision m1v1 + m2v2 = (m1 + m2)v′ For the woman, the first throw just causes her to recoil: 0 = m1v1′ + m2v2′ = 5 kg × 2.5 m/s + 75 kg × vw′ vw′ = –0.167 m/s For the man, first throw 0 + 5 kg × 2.5 m/s = (65 kg + 5 kg) × vm′ vm′ = 0.179 m/s For the man, second throw. The ball′s velocity relative to the man is –3 m/s, so its velocity relative to the ice is (–3 + 0.179) m/s = –2.82 m/s. Then 70 kg × 0.179 m/s = 5 kg × –2.82 m/s + 65 kg × vm′ vm′ = 0.410 m/s For the woman, second throw 75 kg × (–0.167 m/s) + 5 kg × –2.82 m/s = 80 kg × vw′ vw′ = –0.333 m/s The mass of the oxygen atom is m1 = 16 × 1.66 × 10–27 kg = 2.66 × 10–26 kg. Conservation of momentum m1v1 + m2v2 = (m1 + m2)v′


CHAPTER

11-43.

11

2.66 × 10–26 kg × 600 m/s + 0 = (2.66 × 10–26 kg + 2.66 × 10–26 kg) × vO2′ 1.60 × 10−23 kg i m/s vO2′ = = 300 m/s 5.32 × 10−26 kg 1 1 Kinetic energy before is KB = mOvO′2 = (2.66 × 10–26 kg)(600 m/s)2 = 4.8 × 10–21 J 2 2 1 1 Kinetic energy after is KA = mO2vO′2 = (5.32 × 10–26 kg)(300 m/s)2 = 2.4 × 10–21 J 2 2 Kinetic energy changed to internal energy = 4.8 × 10–21 J – 2.4 × 10–21 J = 2.4 × 10–21 J 1 Fraction of initial kinetic energy transferred is . 2 Set the zero level for potential energy at the cannon. From launch to just before collision, conservation of energy yields 1 1 mv12 = mv22 + mgh 2 2

v2 =

v12 − 2 gh =

(12 m/s)2 − 2 × 9.81 m/s 2 × 3.5 m = 8.7 m/s

For collision conservation of momentum yields m1v2 + 0 = (m1 + m2)v3 v3 = v2/2 = 4.3 m/s 2 4.3 m/s ) ( v32 1 2 h= = (2m)v3 = 0 + 2mgh = 0.94 m 2g 2 2 × 9.8 m/s 2

(m1 = m2 = m)

Total height to fall = 3.5 m + 0.94 m = 4.44 m Speed after falling back to height of cannon v = 11-44.

2 gh =

2 × 9.8 m/s 2 × 4.44 m = 9.3 m/s

Conservation of momentum for a totally inelastic collision m1v1 + m2v2 = (m1 + m2)v′ (a) Two automobiles, one at rest: m1 = 800 kg, v1 = 15 m/s, m2 = 1400 kg, v2 = 0 (800 kg)(15 m/s) + 0 = (800 kg + 1400 kg)v′ v′ = 5.5 m/s ∆v1 = − 9.6 m/s ∆v2 = 5.5 m/s (b) Two automobiles, 1400 kg auto moving at –15 m/s: m1 = 800 kg, v1 = 15 m/s, m2 = 1400 kg, v2 = –15 m/s ∆v1 = − 19 m/s (800 kg)(15 m/s) + (1400 kg)(–15 m/s) = (800 kg + 1400 kg)v′ v′ = –4.1 m/s ∆v2 = 11 m/s (c) Automobile and stationary wall v1 = 15 m/s, v2 = 0, mcar << mwall mcar mcar(15 m/s) = (mcar + mwall)v′ v′ = × 15 m/s ≈ 0 ∆v ≈ –15 m/s for any mass car mcar + mwall

†11-45.

(a) 80 km/h = 22.2 m/s. By conservation of momentum, m v − m2 v2 1400 kg − 540 kg v′ = 1 1 × 22.2 m/s = 9.84 m/s = 1400 kg + 540 kg m1 + m2 (b) Before collision: 1 1 1 K = m1v12 + m2v22 = (1400 kg + 540 kg)(22.2 m/s)2 = 4.78 × 105 J 2 2 2 After collision: 1 K′ = (m1 + m2)(9.84 m/s)2 = 9.4 × 104 J 2 (c) To determine the accelerations relative to the ground, it is necessary to know how far each car moves during the collision. This is difficult to determine in the frame relative to the ground but it


CHAPTER

11

is easy to do in a reference frame moving with the center of mass. In the CM frame, the distance each car travels during the collision is the crumple distance of 0.60 m. The center of mass moves with constant velocity, so the accelerations in the center of mass frame will be the same as the accelerations relative to the ground. The final velocity of each car in the CM frame is zero, and the initial velocities are found by subtracting the velocity of the CM relative to the ground from the initial velocities of the cars. The velocity of the CM is the same as the final velocity of the combined mass after the collision which was found in (a). Thus vCM = 9.84 m/s in the direction of motion of the 1400 kg car. Then the initial velocities in the CM frame are v1 = (22.2 – 9.84) m/s = 12.4 m/s for the 1400 kg car and and v2 = (–22.2 – 9.84) m/s = – 32.0 m/s for the 540 kg car. The magnitudes of the accelerations are v 2f − v12 0 − 12.42 a1 = = = 128 m/s 2 2x 2(0.60) a2 = 11-46.

†11-47.

11-48.

v 2f − v22 2x

=

0 − 32.02 = 853 m/s 2 2(0.60)

(a) The frictional force on the car while skidding is f = –µmg. (The minus sign indicates that the force is in the opposite direction from the initial motion.). Therefore, acceleration a = f/m = –µmg/m = –µg and v0 = −2a ( x − x0 ) = 2µ g ( x − x0 ) = 2 × 0.95 × 9.81 × 18 = 18 m/s (b) Using conservation of momentum m1v = (m1 + m2)v0 v = (m1 + m2)v0/m1 = [(2200 + 1400) kg • 18 m/s]/2200 kg = 29 m/s The “energy available for inelastic reactions” is that part of the total kinetic energy that is not 2 MvCM associated with the motion of the center of mass. The kinetic energy of the CM is K CM = , 2 v where M is the total mass of the particles in the system. By equation 11.27, vCM = for two 2 identical particles when one is at rest and the other is moving with speed v. Since one proton is mpv2 . The kinetic energy of the CM is moving, the total kinetic energy is K = 2 (2m p )(v / 2) 2 m p v 2 K K CM = = = . Thus the available energy is 2 4 2 −13 K 8.0 × 10 J = = 4.0 × 10−13 J E= 2 2 The key to the substitution is to transfer the same energy for inelastic processes (energy in excess of the kinetic energy) to the automobile in both situations. For inelastic collisions conservation of momentum gives m1v1 + m2v2 = (m1 + m2)v′ where m1 = barrier mass, m2 = automobile mass. mv +0 1800 kg × 2.2 m/s = 1.2 m/s With the moving barrier the common velocity is v′ = 1 1 = 1800 kg + 1400 kg m1 + m2 1 1 Kbefore = m1v12 = (1800 kg)(2.2 m/s)2 = 4356 J 2 2 1 1 Kafter = (m1 + m2)v′2 = (1800 kg + 1400 kg)(1.2 m/s)2 = 2304 J 2 2 ∆K = 2052 J, which is the excess energy transferred to the automobile. Stationary barrier:


CHAPTER

†11-49.

11-50.

11-51.

11-52.

11

The barrier has no initial kinetic energy and the final kinetic energy is zero, so all the energy comes from the initial kinetic energy of the automobile: 2 × 2052 J 1 1 K = m2v22 = (1400 kg)v22 = 2052 J ⇒ v2 = = 1.7 m/s 1400 kg 2 2 m v + m2 v2 and vCM′ = vCM From eq. 11.27 the velocity of the center of mass is vCM = 1 1 m1 + m2 m1v1 m1v2 = 16 m/s, vCMy = = 8.1 m/s ⇒ vCM = 162 + 8.12 = 17.9 m/s vCMx = m1 + m2 m1 + m2 (a) In the CM frame of reference: Before: 1 (2400 kg)(17.9 m/s) 2 2 K CM = (m1 + m2 )vCM = = 3.9 × 105 J 2 2 Since the CM travels at constant velocity, KCM is still 3.9 × 105 J after the collision. (b) Before the collision: 1 1 (1100 kg)(34 m/s) 2 (1300 kg)(15 m/s) 2 K = m1v12 + m2 v22 = + = 7.8 × 105 J 2 2 2 2 After the collision, the vehicles are moving at vCM, so their total kinetic energy is the same as KCM found in (a): K′ = 3.9 × 105 J. Zero vertical velocity landing means the cat′s initial vertical speed was the same as if falling from same height. The time to reach the height of the chair is 2h 2 × 0.45 m t= = = 0.30 s g 9.81 m/s 2 1.2 m The cat covers 1.2 m horizontally in 0.3 s, so the horizontal velocity is v = = 4.0 m/s. The 0.30 s cat has an inelastic collision with the chair. From conservation of momentum m1v1 + m2v2 = (m1 + m2)v′ (4.5 kg)(4 m/s) = (4.5 kg + 12 kg)v′ v′ = 1.1 m/s Magnitude of work done against friction = K.E. immediately after collision 1 mg µs = W = K = mv2 ⇒ v = 2µ gs = 2(0.60)(9.81 m/s 2 )(0.060 m) = 0.840 m/s 2 From conservation of momentum (m1 + m2)v′ = m1v1 + m2v2 (3 kg + 0.012 kg)(0.84 m/s) = 0.012 kg × v1 v1 = 211 m/s The forces from the bullets and gravity (shown) are in F = I/t equilibrium with the string tension. The resultant of these 24º forces is at an angle of 24° from vertical. Fg = Mg, Fbullets = Impulse/time = mv/(0.5 s) = 2mv /s. where the mass increase caused by the bullets is neglected. ( 2 × 0.015 kg × v ) s −1 F tan(24°) = bullets = mg 4 kg × 9.81 m/s 2 Fg v=

0.445 × 4 kg × 9.81 m/s 2 = 582 m/s (2 × 0.015 kg) s −1


CHAPTER †11-53.

11-54.

11

(a) Treat this as an inelastic collision where the two bodies do not stay together. Conservation of momentum m1v1 + m2v2 = m1v1′ + m2v2′ 9 kg i m/s − 2.4 kg i m/s = 440 m/s (0.015 kg)(600 m/s) = (0.015 kg)v1′ + (0.3 kg)(8 m/s) v1′ = 0.015 kg ∆v1 = 440 m/s – 600 m/s = –160 m/s (b) Bullet: 1 1 1 1 Kbefore = m1v12 = (0.015 kg)(600 m/s)2 = 2700 J Kafter = m1v1′2 = (0.015 kg)(440 m/s)2 2 2 2 2 = 1452 J ∆K = K2 − K1 = 1452 J – 2700 J = –1248 J (c) Wood: 1 1 Kafter = m2v2′2 = (0.3 kg)(8 m/s)2 = 9.6 J. Kbefore = 0 2 2 ∆K = K2 − K1 = 9.6 J – 0 J = 9.6 J (d) The missing energy (about 1240 J) is energy that shows up as heat in the bullet and block, work done in compression/deformation, and as noise. Conservation of momentum yields the speed of the joined automobiles after the collision (1300 kg)(13.9 m/s) + 0 = (1300 kg + 1300 kg)v′ m1v1 + m2v2 = (m1 + m2)v′ 18070 kg i m/s v′ = = 6.95 m/s 2600 kg The force to stop the automobiles is µmg = 0.90 × 1300 kg × 9.8 m/s2 = 1.1 × 104 N

( 6.95 m/s ) = 5.8 m v ′2 1.1 × 104 N = Acceleration a = x= = 4.2 m/s2 v2 = v′2 +2ax 2a 2 × 4.2 m/s 2 2600 kg x 5.8 m = vavg = 3.5 m/s t= = 1.7 s v′ 3.5 m/s Once we know the speed of block and bullet after the collision we can use conservation of momentum to find the original speed of the bullet. The block travels horizontally 1.4 m during the time it takes to fall 1.8 m 2h 2 × 1.8 m Time to fall: t = = = 0.606 s g 9.81 m/s 2 x 1.4 m = x = vt so v = = 2.31 m/s t 0.606 s (m1 + m2)v′ = m1v1 + m2v2 (0.015 kg + 4 kg)(2.31 m/s) = (0.015 kg)v + 0 v = 619 m/s From equation 11.27 and vCM′ = vCM we have (60 kg)(22 m/s)sin(45°) = 8.5 m/s vCMx = 50 kg + 60 kg (50 kg)(20 m/s) + (60 kg)(22 m/s)cos(45°) vCMy = = 17.6 m/s 50 kg + 60 kg v 8.5 m/s (North is reference) tan θ = CMx = = 0.48 θ = 25° 17.6 m/s vCMy 2

†11-55.

11-56.

v=

(8.5 m/s )

2

+ (17.6 m/s ) = 20 m/s 2


CHAPTER †11-57.

Momentum is conserved in the x and y directions. Since the masses and speeds of the hydrogen atoms are both the same mvx = 2mvx′ and mvy = 2mvy′ so vx′ = vx/2 and vy′ = vy/2. Since vx = vy = v the final speed will be 2

2

2 ⎛v⎞ ⎛v⎞ v v′ = ⎜ ⎟ + ⎜ ⎟ = 2 ⎝ 2⎠ ⎝ 2⎠ The kinetic energy before and after the collision is 2

11-58.

⎛ 2 ⎞ 1 2 1 2 1 v ⎟⎟ Since 6.1 × 10–22 J is transferred to internal energy mv + mv = (2m) ⎜⎜ 2 2 2 ⎝ 2 ⎠ 1 1 mv2 – m(v2/2) = 6.1 × 10–22 J or (1.66 × 10–27 kg)v2 = 6.1 × 10–22 J ⇒ v = 860 m/s 2 2 Let the axes be as shown, with cars, mass m, traveling at velocities v1i and v2 j. By conservation of momentum, v v mv1i + mv2 j = 2m(vx′ i + v′y j), which gives v′x = 1 ; v′y = 2 . 2 2 1 2 The speed immediately after the collision: vafter = v1 + v22 2 If both v1, v2 < 14m/s, then 1 vafter < 142 + 142 m/s = 9.90 m/s. 2 Acceleration of wreck = –f/2m = –µ2mg/2m = –µg = –7.84 m/s2. But the length of the skidmarks = 18 m so vafter = −2a( x − x0 ) = 2 × 7.84 m/s 2 × 18 m

†11-59.

= 16.8m/s, indicating that at least one car′s speed > 14m/s (a)Let one player have velocity v1 = 7.0cos 65°i − 7.0sin 65° j and other v 2 = 7.0cos 65°i + 7.0sin 65° j. From the problem, the velocities make an angle 130º with respect to each other. Then, by conservation of momentum m1 v1 + m2 v 2 = (m1 + m2 ) v′ m1 v1 + m2 v 2 m1 + m2 (here m1 = m2 = 80 kg.) Therefore, 1 v′ = m( v1 + v 2 ) / 2m = ( v1 + v 2 ) 2 1 = (14.0cos 65°i + 0 j) = 7.0cos 65°i = 3.0i m/s 2 (b) ∆v = v1 − v′ or v 2 − v′ = ± 7.0sin 65° j m/s v′ =

The magnitude of the average acceleration = ∆v / ∆t ≈

7.0 × sin 65° = 79 m/s2 0.080


11

CHAPTER 11-60.

11

(a) By conservation of momentum m1 v1 + m2 v 2 = (m1 + m2 ) v′ m1 = 40000 tons, v1 = (22 knots)(sin15°i − cos15° j) m2 = 20000 tons v 2 = (19 knots)(sin 48°i − cos 48° j) Therefore,

v′ =

m1 v1 + m2 v 2 m1 + m2

(40000 tons)(22 knots)(sin15°i − cos15° j) + (20000 tons)(19 knots)(sin 48°i − cos 48° j) 60000 tons = 8.50i − 18.4 j knots =

The resultant velocity magnitude = 8.52 + 18.42 = 20.3 knots Direction tan–1 (8.5/18.4) E of S = 25º East of South (b) 22 knots = 11.3 m/s; 19 knots = 9.77 m/s 1 1 Original K.E. = mv12 + m2 v22 2 2 1 1 (20000 × 1000 kg)(9.77 m/s)2 = 3.51 ×109 J = (40000 × 1000 kg)(11.3 m/s)2 + 2 2 1 1 2 Final K.E. = (m1 + m2 )v′ = (60000 × 1000)(10.4 m/s) 2 2 2 9 = 3.24 × 10 J

†11-61.

∆K = 3.24 × 109 J – 3.51 × 109 J = –2.7 × 108 J kg of TNT = 2.7 × 108 J/(4.6 × 106 J/kg) = 59 kg Let the x-axis point East, y-axis North. Let v2 be the velocity of the other car. Then by the conservation of momentum m1 v1 + m2 v 2 = (m1 + m2 ) v′. v 2 = v2 cos 40°i − v2 sin 40° j. Therefore, (m v + m2 v 2 ) v′ = 1 1 m1 + m2 900 ⎛ ⎞ ⎛ 1200 ⎞ =⎜ ⎟14m/s i + ⎜ ⎟ (v2 cos 40°i − v2 sin 40° j) + 900 1200 ⎝ ⎠ ⎝ 900 + 1200 ⎠ 3 ⎛ 4⎞ = (14m/s)i + ⎜ ⎟ (v2 cos 40°i − v2 sin 40° j) 7 ⎝7⎠ 2

Magnitude of v′ is

⎡ ⎛4 ⎞ ⎤ ⎡4 ⎤ ⎢6 + ⎜ 7 cos 40° ⎟ v2 ⎥ + ⎢ 7 sin 40°v2 ⎥ ⎠ ⎦ ⎣ ⎦ ⎣ ⎝

2


CHAPTER

11

1/ 2

2 2 ⎡ ⎧⎪⎛ 4 ⎞ ⎛4 ⎞ ⎫⎪ 2 ⎤ = ⎢36 + 5.25v2 + ⎨⎜ cos 40° ⎟ + ⎜ sin 40° ⎟ ⎬ v2 ⎥ ⎠ ⎝7 ⎠ ⎭⎪ ⎦⎥ ⎩⎪⎝ 7 ⎣⎢

= [36 + 5.25v2 + 0.327v22 ]1/ 2 The acceleration of the cars is a = − µ g . From the skid marks, the magnitude of v′ is

v′ = −2a( x − x0 ) = 2( µ g )( x − x0 ) = 2(0.85 × 9.81 m/s 2 )(17.4 m) = 17.0 m/s By equating the two, we get 36 + 5.25v2 + 0.327v22 = (17.0) 2 = 290

⇒ 0.327v22 + 5.25v2 − 254 = 0

11-62.

This gives v2 = ⎡ −525 ± (5.25) 2 + 4(0.327)(254) ⎤ /(2 × 0.327) = 21 m/s (taking the positive root ⎣ ⎦ only). So the speed of the other car is 21 m/s. Because the incoming ball moves exactly toward the center of the two stationary ones, by symmetry the angle at which the velocity of the two balls make with the x-axis are equal. Let v′ be velocity of incoming ball after the collision, and u′ be the magnitude of the velocities of the other two. Then, mv = mv′ + mu cosθ + mu cosθ = m(v′ + 2u cosθ ). Also by conservation of kinetic energy: 1 2 1 mv = mv′2 + mu 2 2 2 We can deduce what θ is. Since all three balls are identical, on impact the centers of the three alls make an equilateral triangle. The impulse one ball gives to the other is along the line of the enters (because this is the only point where two perfectly hard spheres meet); hence, θ = 30º. Then our two equations are: v = v′ + 2u cos30° = v′ + 3u (i) v 2 = v′2 + 2u 2

(ii)

Substituting u = (v − v′) / 3 into (ii) 2

(v − v′) 2 = v′2 + (v 2 − 2vv′ + v′2 ) 3 3 2 4 2 1 5 4 v 2 = v′2 + v 2 − vv′ + v′2 ⇒ v 2 = v′2 − vv′ 3 3 3 3 3 3 2 2 = 5v′ − 4vv′ − v = 0 v 2 = v ′2 + 2

4v ± 16v 2 + 20v 2 4v ± v 36 4v ± 6v 1 = = = v or − v 10 10 10 5 If v′ = v, then u = 0 which is not a physical situation. v′ =


CHAPTER

11

1 Therefore, v′ = − v 5

11-63.

1 ⎞ 2 ⎛ Therefore, u = (v − v′) / 3 = ⎜ v + v ⎟ / 3 = 3v 5 ⎠ 5 ⎝ Thus, the incoming ball rebounds at – (1/5)v along its line of motion, and the other two move at 2 3 v at + 30° to line of motion of incoming ball. velocities 5 (a)The impact parameter is the perpendicular distance between the centers of the balls. Because Impulse = ∆p, and the impulse is directed along the centers of the balls, ∆p and hence ∆v must be along the centers of the balls. Because the collision is elastic, if the speed of the incoming ball is v, then the speed of the outgoing one must be v. The diagram shows ∆v. ø = angular deflection = π - 2θ. But θ = sin–1 (b/2R), so that ø = π – 2 sin–1 [b/(2R)](b < 2R) (b)The magnitude of the momentum change = m∆v. From the diagram |∆v| = 2v cos θ. But cosθ = 1 − sin 2 θ = 1 − b 2 /(4 R 2 )

Therefore, the magnitude of the momentum change = m∆v = 2mv 1 − b 2 /(4 R 2 )

11-64.

1 2 1 1 mv = mv12 + mv22 ⇒ v 2 = v12 + v22 (1) 2 2 2 Conservation of momentum: mv = mv1 + mv 2 ⇒ v = v1 + v 2 (2) Let v1 point at an angle 60º above the x axis and v2 point at an angle θ below the x axis. Then the angle between v1 and v2 is θ + 60º. From (2), v i v = v 2 = ( v1 + v 2 ) i ( v1 + v 2 ) = v12 + v22 + 2 v1 i v 2

Elastic collision:

From (1), v 2 = v12 + v22 ⇒ v1 i v 2 = v1v2 cos(θ + 60°) = 0 ⇒ θ + 60° = 90° ⇒ θ = 30°. Conservation of momentum in the x-direction: mv + 0 = mv1cos(60°) + mv2cosθ (3) Conservation of momentum in the y-direction: 0 = mv1sin(60°) – mv2sinθ (4) From (4), v1sin(60°) = v2 sin30°) sin ( 30 ) v1 = v2 = 0.577v2 sin ( 60 )


CHAPTER

†11-65.

11

From (3), v = v1cos(60°) + v2cos(30°) = 0.577v2cos(60°) + v2cos(30°) = 1.15v2 v1 = 0.5v v2 = 0.87v For particles of equal mass, conservation of momentum and kinetic energy give mv = mv1 + mv 2 ⇒ v = v1 + v 2

mv 2 mv12 mv22 = + ⇒ v 2 = v12 + v22 2 2 2 where v is the initial velocity (assuming the target is at rest) and v1 and v2 are the final velocities. Assume the angle between the final velocities is θ. Taking the dot product of the momentum equation gives v i v = v 2 = ( v1 + v 2 ) i ( v1 + v 2 ) = v12 + v22 + 2 v1 i v 2 = v12 + v22 + 2v1v2 cosθ Combining this with the energy equation gives v1v2 cosθ = 0 ⇒ θ = 90° 11-66.

11-67.

11-68.

The final velocities are perpendicular. Conservation of kinetic energy gives 2mv 2 2mv12 m(3v1 ) 2 11 = + ⇒ 2v 2 = 2v12 + 9v12 = 11v12 ⇒ v 2 = v12 (1) 2 2 2 2 Using the procedure from 65, v 2mv = 2mv1 + mv 2 ⇒ v = v1 + 2 2 v ⎞ ⎛ v ⎞ v2 v2 13 ⎛ v i v = v 2 = ⎜ v1 + 2 ⎟ i ⎜ v1 + 2 ⎟ = v12 + 2 + v1 i v 2 = v12 + 2 + v1v2 cosθ = v12 + 3v12 cosθ 2 2 4 4 4 ⎝ ⎠ ⎝ ⎠ where θ is the angle between the final velocities. Using (1) gives θ = cos–1(9/12) = 41.4° Impulse = ∆p = 0.045 kg × 70 m/s = 3.2 kg • m/s ∆p 3.2 kg i m/s = = 3.2 × 103 N F= −3 ∆t 1 × 10 s (a) v =

2gh =

2

(b) v = 2ax a = †11-69.

2 × 9.81 m s 2 × 28 m = 23.4 m/s

( 23.4 m/s )

2

2 × 0.15 m

= 1.83 × 103 m/s2

(c) F = ma = 52 kg × 1.83 × 103 m/s2 = 9.52 × 104 N Using equations 11.13 and 11.14, with v1′ = velocity of (initially) moving car. v2′ = velocity of stationary car,

⎛ 1200 − 700 ⎞ v1′ = [(m1 − m2 ) /(m1 + m2 )]v1 = ⎜ ⎟10 km/h = 2.6 km/h ⎝ 1200 + 700 ⎠ v2′ = [2m1 /( m1 + m2 )]v1 = [2(1200) /(1200 + 700)]10 km/h = 13 km/h 11-70.

Conservation of momentum m1v = (m1 + m2)v′ 2 g (366m/s) v′ = m1v /(m1 + m2 ) = = 14m/s (50 + 2) g


CHAPTER †11-71.

11-72.

†11-73.

11-74.

11

Since kinetic energies are given, it will be convenient to work directly with those energies instead of velocities. Since the amount of kinetic energy lost is greatest in a totally inelastic collision, the magnitude of this energy loss is the total “energy available for inelastic processes”. Since the protons are moving in opposite directions, conservation of momentum says v −v mv1 − mv2 = 2mv′ ⇒ v′ = 1 2 2 The final kinetic energy is 2 1 m(v12 − 2v1v2 + v22 ) ⎛ v1 − v2 ⎞ 2 ′ ′ K = (2m)v = m ⎜ ⎟ = 2 4 ⎝ 2 ⎠ 2 K1 2 2 K 2 , v2 = . Substituting gives In terms of the initial kinetic energies, v12 = m m K + K2 K′ = 1 − K1 K 2 2 The change in kinetic energy is ⎛ K + K2 ⎞ ∆K = K ′ − ( K1 + K 2 ) = − ⎜ 1 + K1 K 2 ⎟ 2 ⎝ ⎠ Thus the total energy available for inelastic processes is K + K2 E = ∆K = 1 + K1 K 2 2 (8.0 + 4.0) × 10−13 ⇒E= + 8 × 4 × 10−13 J = 1.2 × 10−13 J 2 (a) Treat this as an elastic collision between a moving ball and a stationary bat. The bat frame of reference is traveling at –30 m/s. vbat = –30 m/s – (–30 m/s) = 0 vball = 40 m/s – (–30 m/s) = 70 m/s (b) Use Equations 11.13 & 11.14 m − mbat 2mball vball vbat′ = vball vball′ = ball mball + mbat mball + mbat 0.15 kg − 0.85 kg 2(0.15 kg) vball′ = × 70 m/s = –49 m/s & vbat′ = × 70 m/s = 21 m/s 0.15 kg + 0.85 kg 0.15 kg + 0.85 kg (c) In the ground frame of reference vbat = 21 m/s + (–30 m/s) = –9 m/s vball = –49 m/s + (–30 m/s) = –79 m/s (a) For an elastic collision between two equal masses where the moving mass has v1 = 20 m/s, m − m2 2m1 v1 v2′ = v1 v1′ = 1 m1 + m2 m1 + m2 m−m 2m v1′ = × 20 m/s = 0 v2′ = × 20 m/s = 20 m/s. The balls just exchange velocity. m+m m+m 2h 2 × 1.5 m = = 0.553 s to fall and (b) Ball 1 lands at the base of the fence. Ball 2 takes t = g 9.81 m/s 2 moves horizontally x = vt = (20 m/s)(0.553 s) = 11 m from the base of the fence. Momentum is conserved in all collisions. For the inelastic collision 1200 kg × 45 km/h = (1200 kg + 400 kg)v′ m1v + 0 = (m1 + m2)v′ v′ = 34 km/h


CHAPTER †11-75.

11-76.

†11-77.

11

40 km/hr = 11.1 m/s. The velocity after the collision = v′ = m1v1/(m1 + m2) = [(3.0 × 104t)(11.1)m/s]/(3.0 × 104 + 8.0 + 105)t = 0.40 m/s Therefore, ∆K/original K = fraction converted to inelastic energy 1 (m1 + m2 )v′2 8.3 × 105 (0.40) 2 ⎡1 2 2⎤ ⎛ 1 2⎞ ′ ( ) / 1 1 m v m m v m v − + = − = − 2 ⎢2 1 1 2 1 ⎥ ⎜2 1 1 ⎟ 3.0 × 104 (11.11) 2 m1v12 ⎣ ⎦ ⎝ ⎠ = 1 − 0.036 = 0.964 Fraction remaining as K = 0.036 (a) For the inelastic collision momentum conservation gives marrowv + 0 = (marrow + mapple)v′ 0.040 kg × 80 m/s = (0.040 kg + 0.200 kg)v′ v′ = 13.3 m/s The apple and arrow will travel horizontally during the time it takes the apple to fall 2h 2 × 1.4 m = = 0.534 s (b) Time to fall t = g 9.81 m/s 2 Horizontal distance x = vt = (13.3 m/s)(0.534 s) = 6.9 m (a) v′ = m1v/(m1 + m2) = (2 × 109kg × 104m/s)/ (2 × 109 + 5.98 × 1024)kg = 3.3 × 10–12 m/s (b) The final recoil velocity is essentially zero, so all the initial kinetic energy is released for inelastic processes. This energy is m v 2 (2 × 109 kg)(104 m/s) 2 1 ton TNT E= 11 = = 1 × 1017 J × = 2.4 × 107 tons TNT (24 MT) 9 2 2 4.2 × 10 J (c) Assume the meteorite came to rest over a distance equal to the depth of the crater, and v assume that it was accelerated uniformly to a stop. Its average acceleration is a = − 1 . It comes t 2 1⎛v ⎞ 2 x 2(180 m) at = v1t − ⎜ 1 ⎟ t 2 ⇒ t = = = 3.6 × 10−2 s. The to rest in a distance x = v1t + 4 2 2⎝ t ⎠ 10 m/s v1 magnitude of the impulse is ∆p = m1v1 = 2 × 109 kg × 104 m/s = 2 × 1013 kg i m/s.

∆p 2 × 1013 kg i m/s = = 5.6 × 1014 N 0.036 s t Both automobiles skid to a stop. The acceleration of each car is a = − µ g = − 0.90 × 9.81 m/s 2 = − 8.83 m/s 2 . The final speed of each car after the collision is 0, F=

11-78.

so the initial speed of each car just after the collision is given by v′ = −2ax , where x is the distance each car slid after the collision. For each car

vB′ = 2 × 8.83 m/s 2 × 1.0 m = 4.20 m/s vW′ = 2 × 8.83 m/s 2 × 2.0 m = 5.94 m/s By conservation of momentum mB vB + 0 = mB vB′ + mW vW′ m v′ + mW vW′ 1400 × 4.20 + 800 × 5.94 m/s = 7.59 m/s just before the collision vB = B B = mB 1400


CHAPTER

11

The pre-collision speed of the black automobile is given by v02 − vB2 = − 2ax, where x is the distance the car skidded before the collision. Thus 11-79.

v0 = vB2 − 2ax = (7.59 m/s) 2 + 2 × 8.83 m/s 2 × 5 m = 12.1 m/s (a)The steel ball is lifted a height h = l (1 − cosθ ). Therefore, its potential energy = mgh = mgl (1 − cosθ ). Just before the collision, its speed is given by energy conservation: mv12 = mgl (1 − cosθ ) 2 v1 = 2 gl (1 − cosθ ) When it collides elastically with the other ball, according to Eqs. 11.13 and 11.14 it stops and the other ball moves off at the same speed. Thus v′ = v1 = 2 gl (1 − cosθ ). Because no energy is lost in an elastic collision, the kinetic energy of the second ball is equal to that of the first ball, and the second ball will rise to the same height that the first ball was dropped from, or h′ = h. (b) In the inelastic collision, momentum is conserved but kinetic energy is not conserved. From (a), the speed of the first ball just before the collision is v1 = 2 gl (1 − cosθ ) Conservation of momentum gives mv1 = (m + m)v′

v1 2 Since energy is conserved after the collision, (2m)v′2 = (2m) gh′ = (2m) gl (1 − cosθ ′) 2 v′2 (v1 / 2) 2 1 ⎛ v12 ⎞ 1 = = ⎜ h′ = ⎟ = [l (1 − cosθ )] 2g 2g 4 ⎝ 2g ⎠ 4 v′ =

From (a), [l (1 − cosθ )] is just the original height h, so h′ = 11-80.

Speed of the peregrine just before collision is v =

2gh =

h 4 2 × 9.81 m/s 2 × 40 m = 28.0 m/s.

Taking the positive direction to be up, the peregrine′s velocity is v1 = − 28.0 j m/s. The pigeon′s initial velocity is v 2 = 15i m/s. Momentum is conserved: m1 v1 + m2 v 2 = (m1 + m2 ) v′ 1.5(−28.0 j) + 0.40(15i ) = 1.95(vx′ i + v′y j) 0.40 × 15 m/s = 3.08 m/s 1.95 1.5 v′y = × − 28 m/s = − 21.5 m/s 1.95 vx′ =

v′ =

( −21.5 m/s )

θ = tan −1

vy vx

2

+ ( 3.08 m/s ) = 21.7 m/s 2

= − 81.8° (81.8° below horizontal)


CHAPTER †11-81.

11

(a) 70 km/h = 19.4 m/s, 100 km/h = 27.8 m/s. Momentum is conserved in both the N-S and E-W directions. Take N to be plus and E to be plus. From equation 11.27 and vCM′ = vCM we have (1500 kg)( − 27.8 m/s)sin(30°) vCME-W = = –4.17 m/s 1500 kg + 3500 kg (3500 kg)(19.4 m/s) + (1500 kg)(27.8 m/s)cos(30°) = 20.8 m/s vCMN-S = 1500 kg + 3500 kg v′ =

( 20.8 m/s )

tan θ =

2

+ ( −4.17 m/s ) = 21.2 m/s

vCME-W −4.17 m/s = 20.8 m/s vCMN-S

2

θ = –11.3º, or 11.3° West of North

m1v12 m2 v22 3500 × 19.42 1500 × 27.82 + = + J = 1.24 × 106 J 2 2 2 2 ( m1 + m2 )v′2 5000 × 21.22 K′ = J = 1.12 × 106 J = 2 2 ∆K = K ′ − K = − 1.2 × 105 J, so the KE lost is 1.2 × 105 J. (a) If both bodies are assumed to be initially at rest at an infinite separation where the Gm1m2 , where r gravitational potential energy is zero, conservation of energy gives K = − U = r is the separation between their centers (100 m + 200 m = 300m). Thus (6.67 × 10−11 kg × m 2 /s 2 )(1.0 × 107 kg)(8.0 × 107 kg) = 178 J is the total KE. K= 300 m The velocity of the CM is zero, so the total momentum is zero: m m1v1 + m2 v2 = 0 ⇒ v1 = − 2 v2 = − 8v2 m1 (b) K =

11-82.

Use the total KE: m1v12 m2 v22 m1 (8v2 ) 2 m2 v22 (72 × 107 kg)v22 + = + = = 178 J ⇒ v2 = 7.03 × 10−4 m/s 2 2 2 2 2 v1 = 8v2 = 5.63 × 10−3 m/s. The velocity of m1 is –5.63 × 10–3 m/s relative to m2. The individual KEs are m v2 K1 = 1 1 = 158 J 2 m v2 K 2 = 2 2 = K − K1 = 20 J 2 (b) The velocity immediately after the collision is equal to the velocity of the CM. Since vCM = 0, the velocity of the joined asteroids is zero.


CHAPTER 12

ROTATION OF A RIGID BODY


ω=

dφ 2π = = 1.7 × 10−3 rad/s. dt 60 × 60 s

Therefore v = Rω = 0.20 m ×

π

12-2.

rad/s = 3.5 × 10−4 m/s. 1800 The angular velocity is the same for each dφ 2π 2π = = = (7.3 × 105) / s = dt 1 day 24 × 60 × 60 s

12-3.

R = radius of earth = 6.38 × 106 m The linear speed of Quito is ωR = 7.3 × 10–5/s × 6.38 × 106 m = 460 m/s. The linear speed of New York = ωR = ωR cos 41° (see diagram) = 7.3 × 10–5/s × 6.38 × 106 m × cos 41° = 350 m/s. v = Rω 88 m/s v = 88 km/h = 3.6 v 88 Therefore ω = = = 81.5 rad/s R 3.6 × 0.3 ω 81.5 v = frequency = = = 13 rev / s 2π 2π v = 2π Rf = (2π )(0.35 m)(37 × 103 rev/min)(1 min/60 s) = 1.4 × 103 m/s.

12-4.

†12-5.

(a) ω = 2π f = (2π )(33.33 rev/min)(1 min/60 s) = 3.49 radians/s. At the rim of the record, v = ω R = (0.15 m)(3.49 radians/s) = 0.52 m/s . At the edge of the label, R = 0.050 m, and v = 0.17 m/s. (b) acentripetal = ω 2 R. At the rim, acentripetal = (3.49 radians/s)2 (0.15 m) = 1.8 m/s2. At the edge of

the label, R = 0.050 m, and acentripetal = 0.61 m/s 2 . 12-6.

f = (5000 rev/min)(1 min/60 s) = 83.33 rev/s. T =

1 1 = = 0.012 s. f 83.33 s −1

ω = 2π f = (2π )(83.33 rev/s) = 524 radians/s. †12-7.

ω = 2π f = (2π )(210 rev/min)(1 min/60 s) = 22.0 radians/s. v = ω R = (0.058 m)(22.0 radians/s) = 1.28 m/s. At R = 2.3 cm, ω = f =

ω 2π

=

55.7 radians/s = 8.83 rev/s. 2π


v 1.28 m/s = = 55.4 radians/s. R 0.023 m

CHAPTER

12-8.

†12-9.

d (π D / 4) π R π (50 m) = = = = 8.73 m/s, or 8.7 m/s to two significant figures. t t 4t 4(4.5 s) v 8.73 m/s ω= = = 0.35 radians/s. R 25 m ω = 2π f = (2π )(90 rev/min)(1 min/60 s) = 9.42 radians/s, or 9.4 radians/s to two significant figures. v = ω R = (0.10 m)(9.42 radians/s) = 0.94 m/s. v=

∆ω 9.42 radians/s = = 1.9 radians/s 2 . ∆t 5.0 s ω = 2π f = (2π )(3450 rev/min)(1 min/60 s) = 361 radians/s. ∆ω 361 radians/s = = 226 radians/s 2 . v = ω R = (0.065 m)(361 radians/s) = 23.5 m/s. α = ∆t 1.6 s ∆ω −361 radians/s α= = = − 10.3 radians/s 2 . ∆t 35 s v For aluminum, vAl = (100 m/min)(1 min/60 s) = 1.67 m/s . v = 2π fR ⇒ f = 2π R 1.67 m/s = = 88 rev/s. For steel, vst = (20 m/min)(1 min/60 s) = 0.333 m/s . (2π )(3.0 × 10−3 m)

α=

12-10.

†12-11.

f = 12-12.

v 2π R

=

0.333 m/s = 2.1 rev/s. (2π )(2.5 × 10−2 m)

∆ω 500 radians/s = = 625 radians/s 2 , or 6.3 × 102 radians/s2 to two significant figures. ∆t 0.80 s atangential = α R = (625 radians/s 2 )(0.030 m) = 18.8 m/s 2 , or 19 m/s2 to two significant figures.

α=

2 2 acentripetal = ω 2 R = (50 radians/s)2 (0.030 m) = 75 m/s 2 . anet = atangential + acentripetal

= (18.8 ) 2 + (75) 2 m/s 2 = 77 m/s 2 . †12-13.

⎛

φ = C ⎜t2 −

⎛ dω t3 ⎞ dφ 3t 2 ⎞ 3t ⎞ ⎛ = C ⎜ 2t − = C ⎜ 2 − ⎟. ⎟. ω = ⎟, α = dt 2⎠ 4⎠ 4 ⎠ dt ⎝ ⎝

⎝ At t = 0, φ = 0; ω = 0; α = 40 radians/s2. ⎡ (1.0 s)3 ⎤ At t = 1.0 s, φ = (20 radians/s 2 ) ⎢(1.0 s) 2 − ⎥ = 15 radians; 4s ⎦ ⎣ ⎡ 3(1.0 s) 2 ⎤ ω = (20 radians/s 2 ) ⎢ 2(1.0 s) − ⎥ = 25 radians/s; 4s ⎦ ⎣ ⎡ ⎣

α = (20 radians/s 2 ) ⎢ 2.0 −

3(1.0 s) ⎤ = 10 radians/s 2 . 2 s ⎥⎦

⎡ (2.0 s)3 ⎤ At t = 2.0 s, φ = (20 radians/s 2 ) ⎢(2.0 s) 2 − ⎥ = 40 radians; 4s ⎦ ⎣ ⎡ 3(2.0 s) 2 ⎤ 2 ω = (20 radians/s2 ) ⎢ 2(2.0 s) − ⎥ = 20 radians/s ; 4 s ⎣ ⎦


12

CHAPTER

12

3(2.0 s) ⎤ ⎡ 2 ⎥ = −20 radians/s . 2 s ⎣ ⎦ Note that the factor of 1/4 in the second term of φ must have units of s-1 for the expression to be dimensionally correct. When the aircraft is directly overhead, the velocity vector is perpendicular to the position vector. Therefore, we can use v = R ω ⇒ ω = v/R.

α = (20 radians/s 2 ) ⎢ 2.0 −

12-14.

v = 900 km/hr = 250 m/s. Therefore, ω = v/R 250 m/s = = 0.025 rad/s. When the aircraft is not 1000 m directly overhead, the velocity component that is perpendicular to the position vector is v cos φ , where φ is an angle as shown. Then in time dt, it moves along this direction a distance ds = v cos φ dt. Therefore, dφ = ds / R = v cos φ dt / R and dφ / dt = v cos φ / R. But R sin φ = d = 1000 m. Therefore, R = d / sin φ . Then dφ v cos φ v =ω = = cos 2 φ . In 3 minutes, the dt d / cos φ d plane has flown 250 m/s × 180 s = 45000 m.

12-15.

Then cos 2 φ = (10000 / 46100)2 = 0.0471. Therefore dφ 250 m/s =ω = × 0.0471 = 1.2 × 10−3 rad/s . dt 10000 m 100 1 ω = 2π v = 2π × = 3.49 rad/s × 3 60 6.35 − 14.6 (i) The radial speed of the needle = = − 5.5 × 10−3 cm/s 25 × 60 (ii) the velocity of the outer edge = Rω = −3.49 × 14.6 cm/s j = 51.0 cm/s j Therefore, the speed of the outer edge = ( −5.5 × 10−3 ) 2 + (51.0) 2 = 51 cm/s. (iii) The velocity of the inner edge = −3.49 × 6.35 cm/s j = −22.2 cm/s j The speed of the inner edge =

12-16.

(−5.5 × 10−3 ) 2 + (−22.2) 2 = 22.2 cm/s

The record makes 100/3 rev/min and plays 25 min. Therefore, the number of grooves on the record is 100/3 × 25 = 833 grooves. Assume equal spacing, ∆R between the grooves. Let R = (14.6 + 6.35)/2 = 10.5 cm be the average of the outer and the inner radius. The length of the grooves L = 2π Σ R i L = 2π[(R + 416∆R) + (R + 415∆R) + · · · (R + ∆R) + (R) +(R – ∆R) + · · · (R – 415∆R) + (R – 416∆R)] We note that all the positive and negative contributions to R pair up and cancel. Thus L = 2π (R + R + · · · R) 833 = 2π 833R = 2π 833 (10.5) = 55 × 103 cm = 550 m.


CHAPTER †12-17.

12-18.

†12-19.

12

The final frequency f = 7000 rev / min = 7000 / 60 Hz = 116.7 Hz ⇒ final angular speed ω = 2π f = 2π rad × 116.7 Hz = 732.9 rad/s ω − ω 0 732.9 rad/s α= = = 611 rad/s2 t 1.2 s The total angular displacement: 1 1 φ = ω 0t + α t 2 = × 611 rad/s 2 × 1.22 s 2 = 440 rad = 70 revolutions. 2 2 ∆ω (1.0 rev/s)(2π radians/rev) = = 0.314 radians/s 2 . (a) α = ∆t 20 s α t 2 (0.314 radians/s 2 )(20 s) 2 1 rev (b) φ = = = 62.8 radians × = 10 rev. 2 2 2π radians (c) s = rφ = (0.60 m)(62.8 radians) = 38 m.

ω = (7200 rev/min)(2π radians/rev)(1 min/60 s) = 754 radians/s. α = 754 radians/s-0 = 1.5 × 102 radians/s 2 . 5.0 s 1 ω 2 − ω 02 , (φ − φ0 ) = 25 rev = 50π radians α= 2 φ − φ0

∆ω ω − ω 0 = ∆t ∆t

=

12-20.

( 10π ) 2 100 2π 10π = −0.039 rad/s 2 × rad/s = rad/s. Therefore, α = 12 9 50π 3 60 9 The carousel makes one revolution every 9 s, so its rotation frequency is 1/9 rev/s. ω 0 = (1/ 9 rev/s)(2π radians/rev) = 0.698 radians/s. “Decelerate” is a nontechnical way of stating

ω0 =

12-21.

12-22.

†12-23.

that the angular speed of the carousel is decreasing, which means the acceleration is negative. ω 2 − ω02 ∆ω ω − ω 0 0 − 0.698 radians/s α= = = = − 0.0436 radians/s 2 . ω 2 − ω02 = 2αφ ⇒ φ = ∆t ∆t 2α 16 s 0 − (0.698 radians/s) 2 1 rev = 5.58 radians × = 0.89 rev. = −2(0.436 radians/s 2 ) 2π radians v 1.0 m/s x 3.0 m = 66.7 radians/s. φ = = = 200 radians. v0 = 1.0 m/s ⇒ ω0 = 0 = r 0.015 m r 0.015 m ω 2 − ω 02 0 − (66.7 radians/s)2 α= = = −11 radians/s 2 . 2φ 2(200 radians)

ω 2 − ω02 . ω 0 = 6.0 rev/s × 2π radians/rev = 12π radians/s. 2φ 0 − (12π radians/s) 2 φ = 5.0 rev × 2π radians/rev = 10π radians. α = = − 22.6 radians/s 2 , or 2(10π radians) ω − ω0 ω − ω0 −12π radians/s ⇒ ∆t = = –23 radians/s2 to two significant figures. α = = 1.7 s. α ∆t −22.6 radians/s 2 ω 2 − ω02 = 2αφ ⇒ α =


CHAPTER

12-24.

12-25.

12-26.

12 ω − ω0

0 − (188 radians/s) 25 s ∆t ω 2 − ω 02 2 2 = −7.54 radians/s , or –7.5 radians/s to two significant figures. φ = 2α 0 − (188 radians/s) 2 1 rev = 2356 radians × = 3.7 × 102 rev. = −2(7.54 radians/s 2 ) 2π radians

ω 0 = 30 rev/s × 2π radians/rev = 188 radians/s. α =

=

⎛1 1⎞ ⎛ 1 1⎞ 2π (−1.01) ω – ω0 = 2π ⎜ − ⎟ = 2π ⎜ − ⎟ = Τ02 ⎝ T T0 ⎠ ⎝ T0 + 1.01 T0 ⎠ ω − ω0 2π (−1.01) α= = 365 × = − 9.6 × 10−22 rad/s 2 t 77T0 × T02 ω0 = 200 rev/min = 21 rad/s (a) ω = 3000 rev/min = 314 rad/s The angular acceleration, α is constant ω − ω0 (314 − 21) α= = = 42 rad / s 2 t 7 (b) Tangential acceleration = rα At t = 0, 7 s; at = rα = 0.18 × 42 = 7.6 m / s 2

Radical acceleration = rω2 At t = 0 s, ar = 0.18 × (21)2 = 79.4 m / s 2 At t = 7 s, ar = 0.18 × (314)2 = 1.77 × 104 m/s 2

⎛a ⎞ (c) tan θ = ⎜ t ⎟ (see diagram) ⎝ ar ⎠ At t = 0, ⎛ 7.6 ⎞ θ = tan–1 ⎜ ⎟ = 5.5° ⎝ 79.4 ⎠ At t = 7 s ⎛ ⎞ 7.6 θ = tan–1 ⎜ = 0.025° 4 ⎟ ⎝ 1.77 × 10 ⎠ †12-27.

α=

dω ⇒ dt

ω

t

0

0

∫ω dω ' = ∫ α dt ' ⇒

ω = 8.0 radians/s +

ω = ω0 + 4

t

∫ C (t ') 0

2

dt ' = ω0 +

Ct 3 . At t = 3.0 s, 3

3

(0.25 radians/s )(3.0 s) = 10.25 radians/s, or ω = 10 radians/s to two 3

⎡ C (t ')3 ⎤ Ct 4 ω ω dt ' t + = + 0 ∫0 ⎢⎣ 0 3 ⎥⎦ 12 Ct 4 (0.25 radians/s 4 )(1.0 s) 4 ∆φ = ω0t + = (8.0 radians/s)(1.0 s) + = 8.0 radians. 12 12

significant figures. ω =

φ t dφ ⇒ ∫ dφ ' = ∆φ = ∫ ω dt ' ⇒ ∆φ = 0 φ0 dt


t

1.0

0

CHAPTER

12-28.

t

⎛

0

⎝

ω (t ) = ω 0 + α 0 ∫ {1 − [(t ') 2 / 4]}dt ' = α 0 ⎜ t −

12

t3 ⎞ ⎟ , 0 ≤ t ≤ 2.0 s. Since α = 0 for t > 2.0 s ω will 12 ⎠

⎛ (1.0 s)3 ⎞ become constant at ω(2.0 s) for t > 2.0 s. At t = 1.0 s, ω = (20 radians/s 2 ) ⎜ 1.0 s − ⎟ 12 s 2 ⎠ ⎝ (2.0 s)3 ⎞ 2 ⎛ = 18 radians/s. At t = 2.0 s, ω = (20 radians/s ) ⎜ 2.0 s − ⎟ = 26.7 radians/s. For t > 2 s, ω 12 s 2 ⎠ ⎝ becomes constant at this value, so after a “long time,” ω = 27 radians/s (to two significant t ⎛ ⎛ t2 t4 ⎞ t3 ⎞ = − figures). ∆φ = ∫ α 0 ⎜ t − ' α dt ⎟ ⎟ , 0 ≤ t ≤ 2.0 s. After 1.0 s, 0⎜ 0 12 ⎠ ⎝ ⎝ 2 48 ⎠

†12-29.

⎛ (1.0 s) 2 (1.0 s) 4 ⎞ 1 rev ∆φ = (20 radians/s 2 ) ⎜ − = 9.58 radians × = 1.5 rev. 2 ⎟ 48 s ⎠ 2π radians ⎝ 2 ω dω ' t dω dω = − Aω ⇒ = − Adt. Integrating both sides gives ∫ = − A∫ dt ', from which α= 0 ω0 ω ' dt ω ω

we get ln ω ' ω = − At ⇒ ln 0

12-30. †12-31.

ω = − At. Inverting the natural log gives ω = ω 0 e − At . ω0

I = 52 MR 2 = (0.4)(0.300 kg)(0.045 m) 2 = 2.4 × 10−4 kg i m 2 . I = MR 2 = (1.9 × 106 kg)(38 m) 2 = 2.74 × 109 kg i m 2 . 2π radians 0.050 rev 1 min = 5.24 × 10−3 radian/s, so × × rev min 60 s I ω 2 (2.74 × 109 kg i m 2 )(5.24 × 10−3 radian/s) 2 = = 3.8 × 104 J. K= 2 2 About its longitudinal axis, I1 = 12 MR 2 = (0.5)(0.50 kg)(0.025 m)2 = 3.9 × 10−5 kg i m 2 . About a

ω = 2π f =

12-32.

1 (0.50 kg)(1.5 m)2 = 0.094 kg i m 2 . Ml 2 = 12 12 I = ∑miR i2 where Ri is the distance from the CM to the transverse axis through the middle, I 2 =

†12-33.

oxygen atoms; R1 = R2 = R; m1 = m2 = m. So I = 2mR2 R=

I = 2m

1.95 × 1046 kg m 2 2 × 2.66 × 10−26 kg

= 6.05 × 1011 m. Thus, distance between atoms = 2R = 1.21 × 1010 m. 12-34.

†12-35.

The moment of inertia is smaller because the core is much denser than the mantle and the crust. If the earth were a uniform sphere, then 5 I = 52 M E R 2 = 0.331M E RE2 ⇒ R = 0.331i i RE = 0.910RE 2 2 I = ∑ mi Ri , where Ri is the perpendicular distance to the axis. The only contribution is from two oxygen atoms off-axis. This gives: I = 2 × 2.66 × 10−26 kg × (1.22 × 10−10 × sin 65°) 2 m 2 = 6.50 × 10−46 kg i m 2


CHAPTER 12-36.

12

About the axis AA', IAA' = 2mHR2 = 2mHd 2 sin2

θ

where mH = mass of 2 hydrogen atom. About BB', IBB' = 2mHR′2 = 2mH d 2 cos2

θ

2

. Therefore, IAA' + IBB'

θ θ⎞ ⎛ = 2mH d 2 ⎜ sin 2 + cos 2 ⎟ = 2mH d 2 2 2⎠ ⎝ Therefore, d = I AA ' + I BB ' / 2mH =

(1.93 × 10−47 + 1.14 × v10−47 ) kg m 2 2(1.67 × 10−27 kg)

= 9.5 × 10−11 m I AA ' = tan2 ( θ2 ) I BB ' Therefore θ = 2 tan–1 †12-37. 12-38.

I AA ' = 2 tan–1 I BB '

1.93 × 10−47 = 105° 1.14 × 10−47

I = MR2 = 4 kg × (0.33) 2 m2 = 0.44 kg m2 1 I = 3 × Ml 2 = Ml 2 = (6.0 kg)(1.2 m)2 = 8.64 kg i m 2 . 3 ω = (2500 rev/min)(2π radians/rev)(1 min/60 s) = 262 radians/s. I ω 2 (8.64 kg i m 2 )(262 radians/s) 2 = = 3.0 × 105 J. 2 2 The moment of inertia of a cylinder about its longitudinal axis is I = 12 MR 2 . (The length doesn’t K=

12-39.

matter.) Thus I1 = (0.5)(70 kg)(0.23 m) 2 = 0.46 • kg • m2. 12-40.

I = I CM + Md 2 . For this case, d = R, the radius of the disk. Thus I = 12 MR 2 + MR 2 = 23 MR 2 .

†12-41.

R

The rotational frequency is 1 rev/day, so the angular speed is ω = (1 rev/day)(2π radians/rev)(1day/86,400 s) = 7.27 × 10−5 radian/s. The moment of inertia is given, so K =

I ω 2 (0.331)(5.98 × 1024 kg)(6.378 × 106 m) 2 (7.27 × 10−5 radians/s) 2 = = 2 2

2.13 × 1029 J. 12-42.

I1 = 12 MR 2 = (0.5)(120 kg)(0.60 m) 2 = 21.6 kg i m 2 , or I = 22 • kg • m2.

ω = (2.0 rev/s)(2π radians/rev) = 12.6 radians/s. K =

I ω 2 (21.6 kg i m 2 )(12.6 radians/s) 2 = = 2 2

1.7 × 103 J.


CHAPTER

12-43.

1 MR 2 = (0.5)(75 kg)(0.16 m) 2 = 0.96 kg i m 2 . 2 ω = (53 × 103 rev/min)(2π radians/rev)(1 min/60 s) = 5.55 × 103 radians/s. I=

I ω 2 (0.92 kg i m 2 )(5.55 × 103 radians/s) 2 = = 1.4 × 107 J. 2 2 The total surface area of the can = 2πRL + 2πR2 where 2πRL is the surface area of the curved part, πR2 is the surface area of each end. Assuming uniform density of the sheet metal, the mass is proportional to the area. Thus, the mass of the curved surface 2π RL L 12 M= = M= 0.05 kg = 0.0392 kg. 2 L+R 15.3 2π ( RL + R ) K=

12-44.

The mass of the ends is ⎡ ⎤ π R2 3.3 R 0.050 kg = 0.0054 M= ⎢ ⎥ M= 2(15.3) 2( L + R ) ⎣ 2π R( L + R ) ⎦ kg I of can is simply the sum of the individual parts. By tables ⎛1⎞ I = Mcurved R2 + 2 ⎜ ⎟ Mend R2 ⎝ 2⎠ = 0.0392 kg × (0.033)2 m2 + 0.0054 kg × (0.033)2 m2 = 4.9 × 10−5 kg m 2 12-45.

∆I = IHolland – IVenezuela; I is given by moil × (RE cos θ2 RE = radius of earth; θ = latitude (see diagram) Therefore ∆I = moilR 2E cos2 53° – moilR 2E cos2 10° = 4.4 × 108 kg × (6.39 × 106m)2(cos2 53° – cos2 10°) = −1.1 × 1022 kg m 2

12-46.

(a) Perpendicular to rod: (by table 12.1) 1 m(l – 2R)2 12 2 ⎛2 2 l2 ⎞ ⎛l⎞ 2 Isphere = MR + M ⎜ ⎟ = M ⎜ R 2 + ⎟ 5 4⎠ ⎝ 2⎠ ⎝5 Irod =

(parallel axis theorem) ⎛2 1 l2 ⎞ Therefore, I dumbell = m(l − 2 R) 2 + 2M ⎜ R 2 + ⎟ 12 4⎠ ⎝5 (b) Along axis of rod: Irod = 0; Isphere =

2 4 MR2; Idumbell = MR 2 5 5


12

CHAPTER †12-47.

12

The solution to this problem requires a result derived in Problem 53 and again in 57. The moment 2M ( R25 − R15 ) of inertia of a thick spherical shell with inner radius R1 and outer radius R2 is I = . 5 ( R23 − R13 )

The moment of inertia of the earth is I earth = I inner core + I outer core . The inner core is a sphere, so 2 (0.22 M E )(0.54 RE ) 2 = 0.0257 M E RE2 . The outer core is a thick shell, so 5 2(0.78M E ) ⎡⎣ RE5 − (0.54 RE )5 ⎤⎦ I outer core = = 0.353M E RE2 . Thus I earth = 0.379 M E RE2 . 3 3 5 ⎡⎣ RE − (0.54 RE ) ⎤⎦ With arms out Using parallel axis theorem: ⎛1 ⎞ Iarms = 2 ( I cm + Mx 2 ) = 2 ⎜ ML2 + Mx 2 ⎟ ⎝ 12 ⎠ ⎡1 ⎤ = 2 ⎢ (2.8) × 0.62 + 2.8 × 0.52 ⎥ kg m2 ⎣12 ⎦ 2 = 1.568 kg m I inner core =

12-48.

With arms in: I = ΣmR2 = 2 ( 2.8 × 0.22 ) kg m2 = 0.224 kg m2 Therefore ∆I = 1.568 – 0.224 kg m2 = 1.34 kg m 2 †12-49.

I = ∫R2 dm. Let the little mass element be ρdx where ρ is the linear density of the material. Then, R = ⎢x sin θ ⎢ (see diagram), hence, 2 2 2 L/2 ∫R dM = ∫ x =− L / 2 x sin θ ρdx 2 /2 = ρ sin2 θ ∫ Lx=− L / 2 x dx L/2

⎛ x3 ⎞ sin 2θ ⎛ L3 L3 ⎞ = ρ ρ sin2 θ ⎜ ⎟ + ⎜ ⎟ 3 ⎝ 8 8 ⎠ ⎝ 3 ⎠− L / 2 L ⎛ L2 ⎞ ML2 sin 2θ = ρ ⎜ ⎟ sin2 θ = 3⎝ 4 ⎠ 12 12-50.

By parallel axis theorem, I = Icm + Md 2. Icm = I axis thru center =

2 MR2 5

Md 2 = MR2 Therefore, I =

2 7 MR2 + MR2 = MR 2 5 5


CHAPTER †12-51.

12-52.

Take the y axis to be the axis of rotation. Imagine a strip of length l and width dx a perpendicular distance x from the axis. The mass of the strip is dm = ρldx, where ρ is the mass per unit area of the plate. The moment of inertia of the strip is dI = x2dm. To find the total moment of inertia, add up the dI’s by integrating from x = 0 to x = l: x =l x =l x =l ρl 4 I = ∫ dI = ∫ x 2 dm = ρ l ∫ x 2 dx = . ρ = M / l 2 , so x =0 x =0 x =0 3 Ml 2 . I= 3

y dx

0

s

θ

where ρ =

h

r dr = sin θ , ds = s sin θ R R dr 2πρ 3 πρ R 4 π R 4 m 2 ⇒ I = ρ ∫ r 2π r = = r dr = ∫ sin θ sin θ 0 2sin θ 2sin θ π R R 2 + h 2 0 π R4 m mR 2 R Since sin θ = , I= = . 2 R 2 + h2 2( R / R 2 + h 2 ) π R R 2 + h 2 Since

†12-53.

x

l

Assume the cone is open at the bottom. I = ∫ r 2 dm = ∫ r 2 ρ dA, mass of the cone m = Area of the cone π R R 2 + h 2 Since ρ is a constant, I = ρ ∫ r 2 dm = ρ ∫ r 2 2π rds

12

ds r R

To solve this problem, we need the moment of inertia of a spherical shell with inner radius R1 and outer radius R2. Assume that the sphere is divided into thin shells of radius r and thickness dr. The volume of one of these shells is dV = 4π r 2 dr , so the mass contained in it is dm = ρ dV = 4πρ r 2 dr , where ρ is the density. The moment of inertia of this thin shell is 2 8πρ r 4 dr ( dm)r 2 = , where we have used the moment of inertia of a thin spherical shell 3 3 from Table 12.3. To find the total moment of inertia, add up the dI’s by integrating from r = R1 to r = R2: r = R2 8πρ R2 4 8πρ 5 I =∫ dI = r dr = ( R2 − R15 ) . ∫ r = R1 R 3 1 15 r = R2 R2 4π 3 The volume of the shell can be found the same way: V = ∫ dV = 4π ∫ r 2 dr = ( R2 − R13 ) , r = R1 R1 3 m 3m , and the moment of inertia of the thick shell is so the density is ρ = = V 4π ( R23 − R13 ) dI =

I=

2m ( R25 − R15 ) 5 ( R23 − R13 )

I peach = I pit

. Now we can proceed to find the moment of inertia of the peach.

2mshell ( R 5 − R 5pit ) 2 2 . Rpit = 0.5R, mpit = 0.05M, mshell = + I shell . Ipit = m pit R pit , Ishell = 5 5 ( R 3 − R 3pit )


CHAPTER

12

⎡ 5 ⎛ R ⎞5 ⎤ 2(0.95M ) ⎢ R − ⎜ ⎟ ⎥ 2 ⎝ 2 ⎠ ⎦⎥ 2 R ⎛ ⎞ ⎣⎢ 0.95M ⇒ I peach = (0.05M ) ⎜ ⎟ + . Doing all the arithmetic finally 5 ⎡ 3 ⎛ R ⎞3 ⎤ ⎝2⎠ 5 ⎢R − ⎜ ⎟ ⎥ ⎝ 2 ⎠ ⎥⎦ ⎢⎣ gives I peach = 0.005MR 2 + 0.421MR 2 = 0.426 MR 2 . (Because the mass of the pit is so small, we 12-54.

need to keep three significant figures to see its effect on the total moment of inertia.) Divide the area of the flywheel into three discs. Then I = I1 + (1/2)I2 + I3, where (1) is a disc of radius R/4, (2) is a disc of inner radius R/4 and outer radius 3R/4, with empty spaces that have 45o opening angles, and (3) is a disc of outer radius R, and inner radius 3R/4. All three discs have density ρ. The density ρ is obtained from ρ = M/V, and if we let the thickness of the disc be unity, then M = ρV = ρAtot (1), where Atot is the total area, and ρ = M/Atot (1). Atot = A1 + A2 + A3. 2 2 2 2 ⎡ 2 ⎛ 3R ⎞ 2 ⎤ π 7Ρ 2 π R2 1 ⎡ ⎛ 3Ρ ⎞ ⎛ R⎞ ⎛Ρ⎞ ⎤ πR ; A = ; A3 = π ⎢ R − ⎜ A1 = π ⎜ ⎟ ⎢π ⎜ ⎟ − π ⎜ ⎟ ⎥ = 2= ⎟ ⎥ = 16 4 16 2 ⎢⎣ ⎝ 4 ⎠ ⎝4⎠ ⎝ 4 ⎠ ⎥⎦ ⎝ 4 ⎠ ⎥⎦ ⎢⎣ 2 2 πR 3π R = Then Atot = πR2 – 4 4 A A1 A 1 4 7 = ; 2 = ; 3 = , and 12 Atot 12 Atot 12 Atot M 4M 7M , M2 = , M3 = therefore M1 = 12 12 12 ⎛1⎞ The moment of inertia of a uniform disc of radius R and mass M is ⎜ ⎟ MR2. The moment of ⎝ 2⎠ inertia of a uniform disc of inner radius R1 and outer radius R2 and of mass M is found from I = R2 2 1 2 2 ∫R1 r ρdv = 2 M ( R1 + R2 )

⎛ R 2 ⎞ ⎛ 1 ⎞⎛ M ⎞ ⎛ R 2 ⎞ MR 2 ⎛1⎞ We then have I1 = ⎜ ⎟ M1 ⎜ ⎟ = ⎜ ⎟⎜ ⎟ ⎜ ⎟ = 384 ⎝ 2⎠ ⎝ 16 ⎠ ⎝ 2 ⎠⎝ 12 ⎠ ⎝ 16 ⎠


CHAPTER

12-55.

12-56.

⎡ 9R2 ⎛ R2 ⎞⎤ 1 4 10 5MR 2 ⎛1⎞ I2 = ⎜ ⎟ M2 ⎢ + ⎜ ⎟⎥ = i M i R2 . = . 2 12 16 48 ⎝ 2⎠ ⎣ 16 ⎝ 16 ⎠ ⎦ ⎛1⎞ 9R2 1 7 25 175 I3 = ⎜ ⎟ M3 [ + R2 ] = i M i R2 = MR 2 2 2 12 16 16 384 ⎝ ⎠ The moment of inertia of the system is: I = I1 + I2 + I3 = 0.56 MR2 2 Let ρ = volume density. Then Mcyl = ρπR2h and Mhemis = πR3ρ. Note that the spacing of the 3 hemispheres is irrelevant if the moment of inertia is to be taken about the axis shown. Then 2 1 2 Ihemis = Isphere = Msphere R2 and Icyl = Mcyl R2 5 2 2 1 Itot = Msp R2+ Mcyl R2 5 2 Now Mcyl = M – Msp, so that 2 1 1 1 Itot = Msp R2 + ( MM sp ) R2 = MR2 – Msp R2 (M = total mass) 5 2 2 10 4 3 πR 1 3 M= But Msp = M 3h 4 1+ π R3 + π R2h 4R 3 ⎛ ⎞ ⎜1 ⎟ 1 2 So I tot = ⎜ − MR 2 (Note that as h → 0, I tot → MR 2 ) ⎟ 5 ⎜⎜ 2 10 + 30 h ⎟⎟ 4 R⎠ ⎝

Treat the hole as a negative mass disk of radius r. Then because moments of inertia are summative : (M = mass of solid disc of radius R) ⎛ r2 ⎞ 1 1 r2 Mr 2 ⎟ I = MR2 – ⎜ 2 Md 2 + 2 2 2R ⎝R ⎠ where the second term is a result of the parallel-axis theorem 1 r2 1 ⎞ ⎛ I = MR 2 − 2 M ⎜ d 2 + r 2 ⎟ R 2 2 ⎠ ⎝

†12-57.

(See also Problem 12-53.) I = ∫ (r sin θ ) 2 dm. dm = ρ i sin θ r 2 drdθ dφ , where Mass of the shell M ⇒ ρ = density = = π 4 Volume of the shell 3 3 ( R2 − R1 ) 3 2π

π

R2

0

0

R1

I = ρ ∫ (r sin θ ) 2 sin θ r 2 drdθ dφ = ρ ∫ dφ ∫ sin 3θ dθ ∫ r 4 dr = 5 2

5 1

4 R −R 3 5 4 R 5 − R15 2 M ( R25 − R15 ) M ⇒ I= i 2π i i 2 = 4π 3 3 5 5 R23 − R13 ( R2 − R13 ) 3

ρ i 2π i i


rsin θ

12

CHAPTER 12-58.

†12-59.

12

Treat the holes as a negative mass disks of radius r. Then because moments of inertia are summative: Let MD be the mass of solid disc of radius R, and MH be the mass of one hole with radius r. Then 1 1 R I = MDR2 – 4 [ M H r 2 + M H ( ) 2 ] 2 2 2 where the second term in the parentheses is a result of the parallel-axis theorem. M R2 Note that D = 2 , (M being the total mass of the disk with four holes.) M R − 4r 2 MH r2 and = 2 M R − 4r 2 1 1 r2 r2 R R2 r2 + 2 ( )2 ] Therefore, I = M R2 – 4M [ 2 2 2 2 2 2 R − 4r 2 R − 4r R − 4r 2 The volume element consisting of a disk of thickness dx has mass dm = ρdx(πr 2). But R r = x tan θ where tan θ = . Then l I=

∫

l

0

x 2 dm = ρπ

R2 l2

∫

l

0

l

x 4 dx = ρπ

R 2 ⎛ x5 ⎞ ⎜ ⎟ l 2 ⎝ 5 ⎠0

R2 l 5 R2 3 l . = ρ • π l2 5 5 1 The volume of a cone is πR2l, so that 3 1 ⎛1 ⎞3 the mass is ρπR2l. Then I = ⎜ ρπ R 2l ⎟ l 2 3 ⎝3 ⎠5 3 2 = Ml 5 From Example 9, the moment of inertia of a spherical shell of inner radius R1 and outer R2 is 2 ⎛ R 5 − R15 ⎞ I = ⎜ 23 ⎟M 5 ⎝ R2 − R13 ⎠ I = ρπ

12-60.

⎡ (6.4 × 106 m)5 − (5.4 × 106 m)5 ⎤ ⎢ 6 3 6 3⎥ ⎣ (6.4 × 10 m) − (5.4 × 10 m) ⎦ = 6.52 × 1012 M e = 0.159 M e Re2

I1 =

2 × 0.28 Me 5

I2 = 0.100 Me Re2 I3 = 0.041 Me Re2 I4 = 0.029 Me Re2 I5 = 0.006 Me Re2 Itot = 0.34 M e Re2 (to 2 significant figures)


CHAPTER 12-61.

12-62.

†12-63.

(a) Since I = ∫R2 dm, and all the mass is located at the radius of the pipe (by assumption that pipe is thin), I = MR2. M = 2000 m × 20 kg/m = 40,000 kg R = 7.5 cm = 0.075 m Therefore, I = MR2 = 40,000 kg (0.075 m)2 900 kg m2 = 225 kg m 2 = 4 1 (b) Kinetic energy = Iω2, ω = 2π rad/s. Therefore, 2 1 K = (225 kg m 2 )(2π ) 2 = 4.4 × 103 J. 2 1 1 1 Kinetic energy, K = Iω2 where I = MR2 = (300 × 103 )(1.8) 2 2 2 2 3000 × 2 π = 100π rad/s = 4.86 × 105 kg m2 and ω = 60 1 4.86 × 105 (100π)2 = 2.4 × 1010 J. Therefore, K = 2 1 kw hr = 103 × 60 × 60 = 3.6 × 106 J 2.4 × 1010 = = 6.7 × 103 kw hr Therefore K = 3.6 × 106 Assume no slipping of the wheels. Translational kinetic energy, 2

1 1 ⎛ 80 ⎞ 5 Mv2 = 1360 × ⎜ ⎟ = 3.36 × 10 J. 2 2 ⎝ 3.6 ⎠ v 80 Angular speed of the wheel ω = = = 58.3 rad/s R 3.6 × 0.381 1 Rotation kinetic energy, ER = 4 ( Iω 2 ) = 2Iω2 2 ⎡1 ⎤ ER = 2 ⎢ (27.2)(0.381) 2 ⎥ (58.3) 2 = 1.34 × 104 J 2 ⎣ ⎦ Total kinetic energy = 3.49 × 105 J. Rotational kinetic energy as percentage of the total energy E 1.34 × 104 = 3.84% = R × 100 = 3.49 × 105 Etot Kt =

12-64.

The total rotational energy in the flywheel is given by K =

1 2 Iω , where the moment of inertia, 2

1 MR2. 2 Then I = 0.5(0.6)2 (1500) = 270 kg/m2 Then K = 0.5(270)[50(2π)]2 = 1.332 × 107 J The bus requires 40 hp = 40 × 746 = 29,840 J/s, and therefore runs for I=


12

CHAPTER

†12-65.

12

1.32 × 107 = 446 s. 29840 Since v = 20 × 0.2778 = 5.56 m/s, the bus moves a distance of 446 × 5.56 = 2480 m 2 1 dE d ⎛1 2⎞ 1 d 2 I of sphere = MR2. E = Iω2, so that = ⎜ Iω ⎟ = I ω 5 2 dt dt ⎝ 2 ⎠ 2 dt ⎡ ⎛ dω ⎞ ⎤ dω 2 dω 2 ⎢ 2ω ⎜ dt ⎟ ⎥ = Iω dt = 5 MR ω dt ⎠⎦ ⎣ ⎝ 30 M = 1.5 × 10 kg, R = 20,000 m, ω = 2π(2.1)/s = 13.2/s dω = 2π(1.0 × 10–15) rev/sec2 = –6.28 × 10–15 rad/sec2 dt ⎢dE/dt ⎢ = rate energy decreasing 2 = (1.5 × 1030 kg)(20,0002 m 2 )(13.2 / s)(6.28 × 1015 / s 2 ) = 2.0 × 1025 W 5 1 1 Pulsar stops when K = 0. K = Iω2 = MR2ω2 2 5 1 = (1.5 × 1030 kg)(20,0002 m 2 )(13.2) 2/s = 2.1 × 1040 J. 5 If dE/dt = constant = ∆E/∆t, then ∆t = time taken for pulsar to stop = E/(∆E/∆t) = (2.09 × 1040 J)/(1.99 × 1025 W) = 1.05 × 1015 s ∆t = 3.3 × 107 yr! dE 1 = I 2 dt

12-66.

†12-67.

1 2 1 mv = 0.0139 kg × 6282m2/s2 = 2740 J. 2 2 The angular velocity of the bullet = ω = 2π(2.47 × 103) rad/sec = 1.55 × 104/s 1 1 I of bullet = MR2 = (0.0139 kg)(3.95 × 103 m) 2 = 1.084 × 10–7 kg m2 2 2 1 Therefore, the rotational K.E. = Iω2 2 1 = (1.08 × 107 kg m 2 )(1.55 × 104 ) 2 / s 2 = 13.1 J 2 13.1 Fraction of total K.E. that is rotational = = 4.7 × 10−3 2740 + 13 Divide the plate into strips as shown. The y axis is the rotation axis, and the perpendicular distance to the strip is x. The length of Translational K.E. =

the strip is 2 R 2 − x 2 , where R is the radius of the plate. The thickness of the strip is dx. The mass of the strip

y

R

dm = ρ (2 R 2 − x 2 )dx, where ρ is the mass per unit area of the

plate. For each strip like the one shown for positive x, there is a corresponding one for negative x. Thus the moment of inertia is given by R

R

I = 2∫ x 2 dm = 4 ρ ∫ x 2 R 2 − x 2 dx. 0

0

The integral can be found in standard tables. The result is


x

dx

CHAPTER

12

R

12-68.

⎡ x R2 ⎛ −1 x ⎞ ⎤ 2 2 2 I = 4 ρ ⎢ − ( R 2 − x 2 )3/ 2 + ⎜ x R − x + R sin ⎟⎥ . R ⎠⎦0 8 ⎝ ⎣ 4 All terms in this are zero except for the inverse sine term evaluated at x = R. The final result is πρ R 4 M I= . The mass per unit area is ρ = , so the final result 4 π R2 MR 2 is I = . 4 Take the y axis to be the rotation axis. The mass element dm is a perpendicular distance x = R sin θ from the y axis. Let ρ stand for the mass per unit length. Then dm = ρ dl = ρ Rdθ . The mass

y x

element is shown in the first θuadrant, but there are a total of four corresponding elements around the ring. Thus I = 4∫

θ =π / 2 θ =0

x dm = 4 ρ R 2

3

∫

π /2 0

θ

dl R x

sin θ dθ 2

The integral is found in standard tables: π /2

⎡θ sin 2θ ⎤ I = 4ρ R ⎢ − = πρ R 3 . ⎥ 4 ⎦0 ⎣2 ρ is the total mass M divided by the circumference 2πR, so MR 2 . I= 2 3

†12-69.

Let z be the rotation axis. The mass element dm is a small strip of length l and width dx located at a perpendicular distance x the rotation axis. The mass of the strip is dm = ρ ldx, where ρ is the mass per unit area of the plate. By the parallel axis theorem, the moment of inertia of the strip for rotation about the z axis is l 2 dm dI = dI CM + (dm) x 2 = + x 2 dm. The strip shown is on the +x 12 side of the axis, and there is another corresponding strip on the negative side. So when we integrate to add up the dI’s, we can integrate from x = 0 to x = l/2 and double the result. 2 x =l / 2 x =l / 2 x =l / 2 ⎛ l ⎛ l2 ⎞ ⎞ I = 2∫ dI = 2∫ dm ⎜ + x 2 ⎟ = 2 ρ l ∫ + x 2 ⎟dx ⎜ x =0 x =0 x =0 ⎝ 12 ⎠ ⎝ 12 ⎠ l/2

⎡ l 2 x x3 ⎤ ⎛ l3 l3 ⎞ = 2ρl ⎢ + ⎥ = 2ρl ⎜ + ⎟ ⎣ 12 3 ⎦ 0 ⎝ 24 24 ⎠ M Ml 2 ρ = 2 , so the final result is I = . l 6


z y

dx

x l

CHAPTER 12-70.

12

Let z be the rotation axis. Imagine that the cube is divided into slabs of thickness dz. The mass of the slab is dm = ρ l 2 dz , where r is the mass per unit volume of the cube. From 12-69, we know that l 2 dm ρ l 4 dz = . If we the moment of inertia of the slab is dI = 6 6 take z = 0 to be at the bottom of the cube, then adding up the dI’s gives z =l ρ l 4 z =l ρl 5 I = ∫ dI = dz = . z =0 6 ∫ z =0 6 M Ml 2 ρ = 3 , so the final result is I = , the same as for the square l 6

†12-71.

12-72.

z

dz

plate. We could set this problem up exactly like 12-67, except that the negative half of the circular plate is missing. However, we can use proportional reasoning to get the result without actually repeating the calculations. Since the negative side of the plate is missing, we don’t double the result of integrating from x = 0 to x = R, and that might lead us to expect that the moment of inertia is only half what was obtained in 12-67. However, the density is twice what it was in 1267 because the area is only half as large, so the final result is that the moment of inertia of the MR 2 semicircular plate is the same as that of the circular plate: I = . (The diagram in the text 4 shows the y axis pointing along the direction labeled as x in 12-67, but this is irrelevant because the actual integration variable does not appear in the final result.) For this problem, assume the thickness t = 1 unit. The object consists of a single disc of radius R and two half-discs, of radius r, as shown. The volume V = π(R2 – r 2), and the density, ρ = M/V = M/π(R2 – r 2). Then the moment of inertia, I = Idisc – 2Ihalf-disc.

For the full disc, mass = πR2ρ = Iy =

π R2Μ R2M = π ( R2 − ρ 2 ) (R2 − r 2 )

1 1 R2 M 2 (mass) R2 = R 4 R2 − r 2 4


CHAPTER

For the half discs,

Iy =

1 mr 2 4

We had Ix from Ix = ∫ y2dm = 2 ∫

π /2

0

Ix = 2 ρr 4 Ix =

∫

π /2

0

(r 2 sin 2 θ )(r cosθ )r cosθ dθ (ρ)

sin 2 θ (1 − sin 2θ ) dθ = 2 ρ r 4

∫

π /2

0

⎞ M ⎛π ⎞ 2 π ⎛ ⎜ ⎜ ⎟ = 2r 2 2 ⎟ 16 ⎝ π ( R − r ) ⎠ ⎝ 16 ⎠

Mr 4 for each half disc θ (R2 − r 2 )

1 1 1⎛ M π r2 ⎞ 2 mr 2 = ( ρ A) r 2 = ⎜ ⎟r 4 4 4 ⎝ π ( R2 − r 2 ) 2 ⎠ 1 Mr 4 = for each half-disc. 8 (R2 − r 2 )

But Iy =

2

⎛h⎞ By parallel axis theorem, IAA = Iy + m ⎜ ⎟ for each half-disc ⎝ 2⎠ 4 2 2 2 Mr h Mr (r 2 + h 2 ) Mr IAA = + = 2 2 2 2 8( R − r ) 8( R − r ) 8( R 2 − r 2 ) For the two half-discs. Mr 2 (r 2 + h 2 ) 2Mr 4 Mr 2 (2r 2 + h 2 ) Iz = IAA + Ix = + = 4( R 2 − r 2 ) 8( R 2 − r 2 ) 4( R 2 − r 2 ) Then from I = Idisc – 2 Ihalf-discs 1 R 2 M 2 Mr 2 (2r 2 + h 2 ) MR 4 − 2Mr 4 − Mr 2 h 2 = = R – 2 2 2 2 4 R −r 4( R − r ) 4( R 2 − r 2 ) 12-73.

Take a little slice of volume element as shown. Its moment of inertia is 1 1 1 1 dM r 2 = dI = ( ρ dV ) r 2 = ρ(πr 2 dz)r 2 = πρr 4dz. 2 2 2 2 R But by similar triangles r = z h 4 1 R therefore dI = πρ 4 z4dz. Then 2 h h1 R4 I of cone = ∫ dI = ∫ πρ 4 z4dz 02 h ⎛ R4 z5 ⎞ h 1 π = πρ ⎜ 4 ρR4h ⎟ = 2 ⎝ h 5 ⎠ 0 10 But the volume of the cone =

1 2 πR h 3

1 πρR2h. Hence, 3 3 ⎛1 3 2 ⎞ 2 2 I= ⎜ πρ R h ⎟ R = MR 10 ⎝ 3 10 ⎠

and its mass =


12

CHAPTER 12-74.

12

Take volume element as shown. dV = Area x dz, with area = πr 2, where r 2 = (R2 – z2). Then dV = π(R2 – z2)dz. The moment of inertia of this representative volume dv, through the z-axis is 1 1 (dm) r 2 = ( ρ dV ) r 2 2 2 1 ρπ(R2 – z2)dz (R2 – z2) 2 1 = ρπ(R2 – z2)2 dz = dI. 2 Therefore moment inertia of sphere R 1 = ∫ dI = ∫ ρπ ( R 2 − z 2 ) 2 dz z =− R 2 R 1 1 ⎛ 2 2 1 5⎞ R 4 2 2 4 = ρπ∫ ( R − 2 R z + z ) dz = ρπ ⎜ R z + z ⎟ R 2 2 ⎝ 5 ⎠R 1 ⎛ 5 2 5 1 5 2 1 ⎞ 8 ρπ ⎜ R − R + R + R 5 − R 5 + R 5 ⎟ = πρR5 2 ⎝ 3 5 5 5 ⎠ 15 4 2⎛ 4 2 ⎞ But M = ρ R3 so that I = ⎜ ρπ R 3 ⎟ R2 = MR 2 3 5⎝ 3 5 ⎠ ω = v/R Therefore velocity at all points at rim of wheel (relative to the center) have magnitude 80 km/hr. But their directions are different. (see diagram). Using axes as shown, the velocities are: top 80 km/hr î bottom −80 km/hr î =

12-75.

front

−80 km/hr ˆj

But these are relative to the center of the wheel, which is moving at (80 km/hr) î. Thus, the velocities relative to the road are: top

(80 + 80) î = 160km / hr î

bottom 80 – 80 = 0 km/hr. front 80 î − 80ˆj km/hr.

(

12-76.

)

(Magnitude 113 km/hr at 45° to horizontal.) The speed of the propeller tip with respect to the center of the propeller is vtip = ω r = 2π fr = (2π )(2500 rev/min)(1 min/60 s)(1.5 m) = 393 m/s. The forward translational speed of the plane is vtrans = 200 km/h = 55.6 m/s. The translational velocity is always perpendicular to the instantaneous tangential velocity of one of the propeller tips, and that tangential velocity is always changing direction as the propeller rotates. This makes it difficult to find an actual direction for the total velocity. However, we can determine a total speed and find a direction relative to the forward motion of the airplane. Since the two velocities are 2 2 perpendicular, the total speed is v = vtip + vtrans = 3932 + 55.62 m/s = 397 m/s. The total


CHAPTER

12

velocity vector sweeps out a cone around the forward velocity of the plane, and the vertex angle vtip 393 of that cone is given by θ = tan −1 = tan −1 = 82°. vtrans 55.6 †12-77.

12-78.

The final tangential speed of a point on the rim of one of the tires is 80 km/h = 22.2 m/s. The final v 22.2 m/s angular speed is ω = = = 74.1 radians/s. The angular acceleration is r 0.30 m ω − ω 0 74.1 radians/s − 0 α= = = 12.3 radians/s 2 , or 12 radians/s2 to two significant figures. The t 6.0 s α t 2 (12.3 radians/s 2 )(6.0 s)2 angle the tires rotated through is φ = = = 221 radians, which is 2 2 1 rev 221 radians × = 35 revolutions. 2π radians ω = (1 rev/h)(2π radians/rev)(1 h/3600 s) = 1.75 × 10−3 radians/s. 1 2 (5.0 × 10−3 kg)(0.15 m) 2 Ml = = 3.75 × 10−5 kg i m 2 . 3 3 I ω 2 (3.75 × 10−5 kg i m 2 )(1.75 × 10−3 radians/s) 2 = = 5.7 × 10-11 J. K= 2 2 1 (0.17 kg)(0.152 m) 2 The moment of inertia of a disk is I = Ml 2 = = 1.96 × 10−3 kg i m 2 . 2 2 ω = (33.333 rev/min)(2π radians/rev)(1 min/60 s) = 3.49 radians/s. I=

12-79.

I ω 2 (1.96 × 10−3 kg i m 2 )(3.49 radians/s) 2 = = 0.0119 J, or 0.012 J to two significant 2 2 figures. To give the record the same translational kinetic energy, it would need a speed given by 2K 2(0.0119 J) v= = = 0.37 m/s m 0.17 kg K=

12-80.

(a) I rim = M rim R 2 = (20 kg)(0.5 m)2 = 5 kg i m 2 . 1 2(0.80 kg)(0.50 m) 2 M spoke R 2 = = 0.133 kg i m 2 . The total moment of inertia is 12 3 + I spokes = 1.38 kg i m 2 , or I = 1.4 kg • m2 to two significant figures.

I spokes = 8 × I = I rim

If the spoke is a rod with length = R and rotating about an axis through its END, Then 1 8(0.80 kg)(0.50 m) 2 I spokes = 8 × M spoke R 2 = = 0.533 kg i m 2 3 3 The moment of inertia of the wheel is the sum of the two, or I = I rim + Ispokes = 5.53 kg • m2 (b) ω = (1 rev/s)(2π radians/rev) = 6.28 radians/s. K=

†12-81.

I ω 2 (5.533 kg i m 2 )(6.28 radians/s) 2 = = 109 J. 2 2

2 4 MR 2 = MR 2 . To find the moment of 5 5 inertia about the transverse axis through the point of contact between the spheres, use the parallel axis theorem to find the moment of inertia for one sphere about an axis tangent to its surface, and ⎛2 ⎞ 14 double that. I 2 = 2( I CM + Md 2 ), where d = R. I 2 = 2 ⎜ MR 2 + MR 2 ⎟ = MR 2 . ⎝5 ⎠ 5

About the longitudinal axis through the center, I1 = 2 ×


CHAPTER

12-82.

12

1 1 M cylinder R 2 + M hemisphere R 2 . Note that only half the moment of 2 5 inertia of a sphere is used. Assuming the bullet has a uniform density, the ratio of the masses of M cylinder π R 2 h 3h the two sections is eθual to the ratio of their volumes: = = , where h is the M hemisphere 2 π R 3 2 R 3 height of the cylinder. Using h = 7.0 mm and R = 2.7 mm gives M cylinder = 3.89 M hemisphere . The

(a) I = I cylinder + I hemisphere =

total mass is 15 g, so M hemisphere + 3.89M hemisphere = 15 g, , which gives M hemisphere = 3.07 g and M cylinder = 11.9 g. Thus I = [(0.5)(0.0119 kg) + (0.2)(0.00307 kg)](0.0027 m) 2 = 4.79 × 10−8 kg i m 2 (b) ω = (1.2 × 103 rev/s)(2π radians/rev) = 7.54 × 103 radians/s. I ω 2 (4.79 × 10−8 kg i m 2 )(7.54 × 103 radians/s) 2 = = 1.36 J. 2 2 The outer part of the wheel is a wide ring with outer radius R and inner radius R/2. The moment 2 1 ⎡ ⎛ R⎞ ⎤ 5 of inertia of a wide ring is derived in Example 11: I ring = M ⎢ R 2 + ⎜ ⎟ ⎥ = MR 2 . To 2 ⎣⎢ ⎝ 2 ⎠ ⎦⎥ 8 determine the moment of inertia of the spokes, assume that the wheel is thin (as indicated in the M M 4M figure), so it has a mass per unit area ρ = = = . Each spoke has a length A ⎡ 2 ⎛ R ⎞ 2 ⎤ 3π R 2 π ⎢R − ⎜ ⎟ ⎥ ⎝ 2 ⎠ ⎥⎦ ⎢⎣ 2 of R/2 and a width of R/12, so the area of a spoke is R /24 and the mass of one spoke is ⎛ R 2 ⎞ 4M ⎛ R 2 ⎞ M M spoke = ρ ⎜ ⎟ = . Each spoke is a rod rotating about one end, and there are ⎟= 2 ⎜ ⎝ 24 ⎠ 3π R ⎝ 24 ⎠ 18π K=

12-83.

eight spokes, so the total moment of inertia of the spokes is I spokes = 8 ×

1 ⎛5 + ⎝ 8 27π

Thus the total moment of inertia is I = I ring + I spokes = MR 2 ⎜ 12-84.

Mv 2 I ω 2 Mv 2 ⎛ I ⎞ + = ⎜1 + ⎟ 2 2 2 ⎝ MR 2 ⎠ I ⎞ I ⎞ (2.0 kg)(1.0 m/s) 2 ⎛ ⎛ = = (1.0 J) ⎜1 + . ⎜1 + 2 ⎟ 2 ⎟ 2 MR ⎠ ⎝ ⎝ MR ⎠ MR 2 1⎞ ⎛ Cylinder: I = ⇒ K = (1.0 J) ⎜ 1 + ⎟ = 1.5 J. 2 2⎠ ⎝ 2 2⎞ ⎛ Sphere: I = MR 2 ⇒ K = (1.0 J) ⎜1 + ⎟ = 1.4 J. 5 5⎠ ⎝ 2 Pipe: I = MR ⇒ K = (1.0 J) (1 + 1) = 2.0 J.

The total kinetic energy for each is K =


1 M R MR 2 × × ( )2 = . 3 18π 2 27π

⎞ ⎟. ⎠

CHAPTER †12-85.

12

When the cylinder is at the top of the ramp, its energy is E = K + U = Mgh, since it’s at rest. At the bottom of the ramp, E = K trans + K rot , since its potential energy is now zero. 2

K trans + K rot

12-86.

Mv 2 Iω 2 Mv 2 1 ⎛ MR 2 ⎞ ⎛ v ⎞ 3Mv 2 . Using conservation of energy, = + = + ⎜ ⎟⎜ ⎟ = 2 2 2 2 ⎝ 2 ⎠⎝ R ⎠ 4

3Mv 2 4 gh 4(9.81 m/s 2 )(1.5 m) = Mgh ⇒ v = . For h = 1.5 m, v = = 4.4 m/s. 3 4 3 (a) vCM = ω RCM = (3000 rev/min)(2π radians/rev)(1 min/60 s)(1.2 m/2) = 188 m/s. Since the blade was horizontal at the instant it broke off, its velocity will be v CM = 188 m/s up. (b) At the instant the blade broke, the speed of the end at the hub is nearly zero, the speed of the CM is 188 m/s, and the speed of the tip is twice the speed of the CM, or 376 m/s. Relative to the CM, the blade is rotating so that its hub end is moving down at 188 m/s and the tip is moving up vtip 188 m/s at 188 m/s. The angular speed of the blade is ω = = = 313 radians/s. 0.60 m R v2 (188 m/s) 2 (c) The height reached by the blade is h = CM = = 1.80 × 103 m. (The size of the 2g 2(9.81 m/s 2 ) propeller blade has been neglected in this calculation because h is much larger than 1.2 m.)


CHAPTER 13

DYNAMICS OF A RIGID BODY


The crane boom of the crane makes an angle φ with respect to the horizontal direction. In the calculation of the torque about R θ F the pivot point P, which is given by τ = FR sin θ, the angle is φ between the force F and the radial line R. The downward force P F is due to the weight mg of the load. Since the load is directed vertically, a right triangle is formed between the boom, the rope supporting the load, and the horizontal direction. Thus, θ = 90° − φ. For φ = 20°, τmax = FR sin θ = mgR sin (90° − 20°) = (500 kg)(9.81 m/s2)R sin 70° = (4610 N • m)R Using this result and assuming the maximum safe load is determined by this maximum safe torque, at φ = 40°, the maximum safe load is τ (4610 N i m)R 4610 N i m FLoad = max = = = 6018 N R sin θ R sin(90° − φ ) sin(90° − 40°)

The mass of this load is F 6018 N m = Load = = 613 kg g 9.81 m/s 2 For φ = 60°, FLoad = and 13-2.

τ max

(4610 N i m)R 4610 N i m = = 9220 N R sin(90° − φ ) sin(90° − 60°) 9220 N = = 940 kg 9.81 m/s 2

=

R sin θ F m = Load g

The torque on the handle must equal the torque of the rope on the drum. τ handle = τ drum

Fhandle L = Frope R 0.04 m R = (2500 N) = 400 N 0.25 m L The pivot point for the wrench is at the bolt itself, so the radial line is along the length L of the handle of the wrench and the force will be applied perpendicular to the end of the wrench. τ = FR sin θ = FL sin 90° = FL τ 62 N i m F= = = 310 N L 0.20 m τ = FR sin θ = mgL sin 90° = mgL = (2.0 kg)(9.81 m/s 2 )(2.0 m) = 39 N i m Fhandle = Frope

†13-3.

13-4.

When the pole makes an angle of 60° with respect to the horizontal direction, the torque will be τ = FR sin θ = mgL sin(90° − 60°) = mgL sin 30° = (2.0 kg)(9.81 m/s 2 )(2.0 m) sin 30° = 20 N i m 13-5.

The torque in the tilted orientation is one-half that in the horizontal orientation. τ = FR sin θ = mgL sin 90° = mgL = (10 kg)(9.81 m/s 2 )(0.60 m) = 59 N i m


CHAPTER 13-6. †13-7.

13-8.

13

τ = FR sin θ = mgL sin 90° = mgL = (48 kg)(9.81 m/s 2 )(0.40 m) = 190 N i m The power delivered by the torque of the engine is P = τω, where ω is the angular speed in rad/s of the engine. When operating at a maximum torque τmax of 203 N • m, the engine delivers a power of rev ⎞⎛ 1 min ⎞⎛ 2π rad ⎞ ⎛ ⎛ 1 hp ⎞ 4 P = τω = (203 N i m) ⎜ 4600 ⎟⎜ ⎟⎜ ⎟ = 9.8 × 10 W ⎜ ⎟ = 130 hp min ⎠⎝ 60 s ⎠⎝ 1 rev ⎠ ⎝ ⎝ 745.7 W ⎠ When running at a maximum power of 142 hp, the torque delivered by the engine is ⎛ 745.7 J/s ⎞ 142 hp ⎜ ⎟ 1 hp ⎠ P ⎝ τ = = = 176 N i m rev ⎛ 1 min ⎞⎛ 2π rad ⎞ ω 5750 ⎜ ⎟⎜ ⎟ min ⎝ 60 s ⎠⎝ 1 rev ⎠ (a) The tangential speeds of the flywheel rims must be equal, if there is no slippage of the belt. Since v = ω r, ω1 R1 = ω2 R2 R ω2 = ω1 1 R2 (b) τ 1 = TR1 − T ′R1 = (T − T ′) R1

and

τ 2 = TR2 − T ′R2 = (T − T ′) R2

(c) P1 = τ 1ω1 = (T − T ′) R1ω1 and P2 = τ 2ω2 = (T − T ′) R2ω2 However, when we apply the result from part (a), we see that R P2 = (T − T ′) R2ω2 = (T − T ′) R2ω1 1 = (T − T ′) R1ω1 = P1 R2

P2 = P1

13-9.

13-10.

⎛ 745.7 J/s ⎞ 850 hp ⎜ ⎟ 1 hp ⎠ P ⎝ τ = = = 2900 N i m rev ⎛ 1 min ⎞⎛ 2π rad ⎞ ω 2100 ⎜ ⎟⎜ ⎟ min ⎝ 60 s ⎠⎝ 1 rev ⎠ rev P = τω = (35 N)(2 × 0.18 m) ( 60 min )( 160mins )( 21πrevrad ) ⎛ 1 J/s ⎞⎛ 1 kcal ⎞⎛ 60 s ⎞ = 79 W ⎜ ⎟⎜ ⎟⎜ ⎟ = 1.1 kcal/min ⎝ 1 W ⎠⎝ 4186 J ⎠⎝ 1min ⎠

†13-11.

In Example 2, the angular speed was found for the meter stick by applying the principle of the conservation of mechanical energy. Every point on the stick, including the top end, has that angular speed as it rotates about the end initially in contact with the floor. The relationship between the tangential speed and the angular speed is v = ω r. Therefore, the tangential speed at the end of the stick is v = rω = (1.0 m)

3g 3(9.81 m/s 2 ) = (1.0 m) = 5.4 m/s 1.0 m l

If the stick is a 2 meter stick, then v = rω = (2.0 m)

3(9.81 m/s 2 ) = 7.7 m/s 2.0 m


CHAPTER 13-12.

†13-13.

13-14.

†13-15.

13-16. †13-17.

13

P = τω ⎛ 745.7 J/s ⎞ 0.050 hp ⎜ ⎟ P ⎝ 1 hp ⎠ τ = = = 2.4 N i m rev ⎛ 1 min ⎞⎛ 2π rad ⎞ ω 150 ⎜ ⎟⎜ ⎟ min ⎝ 60 s ⎠⎝ 1 rev ⎠ The power delivered (in watts) by the motor is the product of the motor torque (in N • m) and the angular speed (in rad/s). Since the angular speed is given in rev/min, a unit conversion is necessary. ⎛ 1 min ⎞⎛ 2π rad ⎞ P = τω = (0.65 N i m)(3450 rev/min) ⎜ ⎟⎜ ⎟ = 230 W ⎝ 60 s ⎠⎝ 1 rev ⎠ Carefully examine Figure 10.17 and its caption. The distances are given as distances from the floor for an average male of height L and mass M. (a) The masses of the parts of the arm that are given must be divided by two since the numbers are given for both left and right arms combined. The data for the shoulder and arm is Distance Distance Mass Body part from floor from shoulder Upper arm 0.717L 0.095L 0.033M Forearm 0.553L 0.259L 0.021M Hand 0.431L 0.381L 0.0085M The shoulder joint is located 0.812L above the floor. The torques acting on the arm with the shoulder as the pivot are τ = (0.033Mg)( 0.095L) + (0.021Mg)( 0.259L) + (0.0085Mg)( 0.381L) = 0.012MgL or 0.12ML (b) The masses of the parts of the leg that are given must be divided by two since the numbers are given for both left and right legs combined. The data for the hip and leg is Distance Distance Body part from floor from hip Mass Upper leg 0.425L 0.096L 0.1075M Lower leg 0.182L 0.339L 0.048M Foot 0.018L 0.503L 0.017M The hip joint is located 0.521L above the floor. τ = (0.1075Mg)( 0.096L) + (0.048Mg)( 0.339L) + (0.017Mg)(0.503L) = 0.035MgL or 0.34ML The work done in rotating the grinding table is equal to the product of the torque and the angle through which the table turns, W = τθ. The angle is measured in radians, so a unit conversion is necessary. ⎛ 2π rad ⎞ 6 W = τθ = (250 N i m)(1200 rev) ⎜ ⎟ = 1.9 × 10 J ⎝ 1 rev ⎠ −24 τ 7.5 × 10 N i m = 3.0 × 10−19 N F= = −6 R 25 × 10 m The work done by the motor is equal to the product of the torque and the angle through which the fan blades turn. Since the fan is starting from rest, the fan is undergoing an angular acceleration, so the angle has the time-dependence given. Therefore, to find the total work done, an integration over the elapsed time must be done.


CHAPTER

13

For t = 0 to 1.0 s, 1.0 s 1.0 s d φ (t ) 1.0 s d 1.0 s (Ct 2 ) dt = 2Cτ ∫ t dt = Cτ t 2 W =τ∫ dt = τ ∫ 0 0 0 dt dt 0 2 2 = (7.5 rad/s )(2.5 N i m)(1.0 s) = 19 J For t = 0 to 2.0 s, W = Cτ t 2 13-18.

†13-19.

2.0 s 0

= (7.5 rad/s 2 )(2.5 N i m)(2.0 s) 2 = 75 J

F = ma = (1500 kg)(8.0 m/s2) = 12 000 N τ = Fl = (12 000 N)(0.6 m) = 7200 N • m The torque will tend to cause the front end to rotate downward, depress it. The power is that required to increase the potential energy of the tractor per unit time. It is given by ∆U ∆h P= = mg = mgv y = mgv sin θ v ∆t ∆t = (4500 kg)(9.81 m/s 2 )(4.0 m/s)sin(tan −1 13 ) = 5.6 × 104 W

1 unit

θ

3 units

The torque exerted by the engine on the rear wheel is P P rP (0.80 m)(5.6 × 104 W) τ = = = = = 1.1 × 104 N i m ω v/r v 4.0 m/s Note that the speed of the tractor up the incline is used rather than the speed in the vertical direction. 2

13-20.

†13-21.

⎡ 1 1 1 ⎛ 1000 m ⎞⎛ 1 h ⎞ ⎤ W = ∆K = m(v 2 − v02 ) = mv 2 = (90 kg) ⎢(12 km/h) ⎜ ⎟⎜ ⎟ ⎥ = 500 J 2 2 2 ⎝ 1 km ⎠⎝ 3600 s ⎠ ⎦ ⎣

The work is also related to the torque, W = τ∆φ. So, the torque is given by 500 J W τ = = = 27 N i m ∆φ ⎛ 2π rad ⎞ 3 rev ⎜ ⎟ ⎝ 1 rev ⎠ As the stick rotates about the pivot from the horizontal position to the vertical position, the center of mass of the meter stick is lowered by 0.1 m. Define the gravitational potential energy at this lowest position to be equal to zero. By applying the principle of the conservation of energy as the potential energy is converted into rotational energy, we find 1 mgh = I ω 2 2 The rotational inertia of the meter stick may be found using the parallel axis theorem. 1 I = I m + md 2 = ml 2 + m(0.10 m) 2 l 2 = 0.093ml 2 12 The angular speed may then be determined using the information given. 2mgh ω2 = I

ω=

2mgh = I

2mgh = 0.093ml 2

2(9.81 m/s 2 )(0.10 m) = 4.6 rad/s 0.093(1.0 m) 2


CHAPTER 13-22.

13

(a) The moment of inertia for the sphere as it rotates about the pivot point where the string is attached to the ceiling is given by 2 I = I m + Md 2 = 52 MR 2 + M ( 23 R ) = 2.65MR 2

3Rcosθ/2

R h

The principle of the conservation of energy may be applied as the potential energy is converted into rotational energy. 1 Mgh = I ω 2 2 U = 0 J at the lowest point that the center of mass of the sphere passes through. As the sphere swings, 3 3 3 h = R − R cos 45° = R (1 − cos 45°) 2 2 2 Then, 2 Mgh ω2 = I

ω=

2Mgh = I

2Mg 32 R(1 − cos 45°) = 2.65MR 2

3(9.81 m/s 2 ) R (1 − cos 45°) 1.80 rad/s = 2.65R 2 R

(b) At the bottom of the arc, the sum of the forces acting on the sphere is Mv 2 T − Mg = ma = R ⎛ ⎛ v2 ⎞ R 2ω 2 ⎞ T = M ⎜g + ⎟ = M ⎜g + ⎟ R⎠ R ⎠ ⎝ ⎝ 2 ⎡ ⎛ 1.80 ⎞ ⎤ 2 rad/s ⎟ ⎥ = M ( g + Rω ) = M ⎢( 9.81 m/s ) + R ⎜ ⎝ R ⎠ ⎥⎦ ⎣⎢ = 13M (in Newtons) 2

†13-23.

(a) The work done by the engine torque on the gear box as the engine makes 2.58 rev is ⎛ 2π rad ⎞ W = τθ = (441 N i m)(2.58 rev) ⎜ ⎟ = 7150 J ⎝ 1 rev ⎠ If all of this work is transferred to the wheels, then a torque τ′ will be applied to the wheels and cause them to rotate one revolution (2π rad). W 7150 J τ′= = = 1140 N i m θ ′ 2π rad (b) The torque on the wheels is equal to the torque due to the force acting at the area of contact between the wheels and the ground. The ground force acts at a distance R, the radius of the wheels. τ ′ 1140 N i m F= = = 3800 N R 0.30 m The acceleration that results from the ground forces on the automobile is F 3800 N a= = = 2.1 m/s 2 M 1770 kg


CHAPTER 13-24.

13

The brake drum of radius r1 shares the same rotational axis as the wheel of radius r2. Consider a wheel on the driver’s side of the automobile as the automobile moves toward the left. The kinetic friction force f of the brake drum applies a clockwise torque as the net ground force F on the wheel applies a counterclockwise torque. For the translational motion of the car, F = ma, where m = 0.25M (M is the mass of the automobile). The torque each brake pad must exert is 1 1 τ = Fr2 = Mar2 = (1200 kg)(7.8 m/s 2 )(0.30 m) = 700 N i m 4 4 The sum of the torques acting on the wheel is Fr2 − fr1 = Iα Since we are to assume the mass of the wheel is zero kg, the rotational inertia is also equal to zero. Then, Fr2 = fr1 Fr2 = µ k Nr1

†13-25.

13-26.

†13-27.

1 Mar2 = µ k Nr1 4 Mar2 (1200 kg)(7.8 m/s 2 )(0.30 m) N= = = 9400 N 4µ k r1 4(0.60)(0.125 m) The downward force F of the piston causes a torque on the crankshaft at a distance l from the pivot point on the crankshaft. The torque is τ = Fl . Therefore, the force may be determined. τ 31 N i m F= = = 820 N l 0.038 m 1 τ = Iα = MR 2α 2 2τ 2(100 Nim) α= = = 110 rad/s 2 2 2 MR (30 kg)(0.25 m) The magnitude of angular acceleration is constant as the bridge opens to 45° and it decelerates at the same rate as it continues to open to 90°. Since the angular acceleration is constant as it opens to 45° in a 30 s interval, the angular displacement as a function of time is given by 1 φ = ω 0t + α t 2 2 and the angular acceleration is 1 φ = αt2 2 ⎛π ⎞ 2 ⎜ rad ⎟ 2φ 4 ⎝ ⎠ = 1.7 × 10−3 rad/s 2 α= 2 = 2 t (30 s) Since the bridge is treated as uniform rod with an axis through its center, I = 121 Ml 2 . The torque

is

τ = Iα =

1 1 ⎛ 1000 kg ⎞ 2 −3 2 4 Ml 2α = (300 t) ⎜ ⎟ (25 m) (1.7 × 10 rad/s ) = 2.7 × 10 N i m 12 12 ⎝ 1t ⎠


CHAPTER

13-28.

13

τ = Iα = MR 2α = MR 2

∆ω ∆t

rev ⎛ 1 min ⎞⎛ 2π rad ⎞ ⎜ ⎟⎜ ⎟ min ⎝ 60 s ⎠⎝ 1 rev ⎠ = (1.9 × 106 kg)(38 m) 2 = 4.8 × 105 N i m 30 s The force required to deliver this torque at the rim of the wheel is τ 4.8 × 105 N i m F= = = 1.3 × 104 N R 38 m Apply equation (13.24), to solve for g, the acceleration due to gravity. m2 − m1 (13.24) a= g m1 + m2 + ( I / R 2 ) 0.050

†13-29.

g=a

13-30.

m1 + m2 + ( I / R 2 ) m2 − m1

The acceleration is found by using the kinematic relation, 1 x = at 2 2 2 x 2(1.6 m) a= 2 = = 5.0 × 10−2 m/s 2 t (8.0 s) 2 1 The rotational inertia of the disk is I = MR 2 , where M is the mass of the disk and R is its 2 radius. ⎛1 ⎞ 1 ⎛ ⎞ m1 + m2 + ⎜ M ⎟ ⎜ (0.4500 kg) + (0.4550 kg) + 2 (0.120 kg) ⎟ 2 ⎠ ⎝ 2 2 −2 g=a = 5.0 × 10 m/s ⎜ ⎟ = 9.7 m/s m2 − m1 0.4550 kg − 0.4500 kg ⎜⎜ ⎟⎟ ⎝ ⎠ −1 The inclined plane makes an angle θ = tan (1/10) = 5.7° with respect to the horizontal direction. The moment of inertia for a hoop for an axis through its center is I = MR2. To find the moment for a rotation about a point on its rim, use the parallel axis theorem to find I 0 = MR 2 + MR 2 = 2 MR 2 The torque on the hoop as it rolls down the incline is τ = I 0α = 2MR 2α = MgR sin θ

a = Rα = R †13-31.

MgR sin θ 1 1 = g sin θ = (9.81 m/s 2 )sin(5.7°) = 0.49 m/s 2 2 2 MR 2 2

1 MR 2 . To find the 2 moment for a rotation about a point on its rim, use the parallel axis theorem to find 1 3 I 0 = MR 2 + MR 2 = MR 2 2 2 The torque on the cylinder as it rolls down the incline is

The moment of inertia for a cylinder for an axis through its center is I =


CHAPTER

3 MR 2α = MgR sin θ 2 MgR sin θ 2 a = Rα = R = g sin θ 3 3 2 MR 2 Since the wheel rolls without slipping down the hill, there is no energy loss due to friction forces and conservation of energy may be applied to this situation. The total kinetic energy of the wheel is 1 1 K = Mv 2 + I ω 2 2 2 1 The moment of inertia for a cylinder for an axis through its center is I = MR 2 . To find the 2 moment for a rotation about a point on its rim, use the parallel axis theorem to find 1 3 I 0 = MR 2 + MR 2 = MR 2 2 2 which when used in conjunction with the relation v = ωR gives 2 1 1⎛ 3 5 ⎞⎛ v ⎞ K = Mv 2 + ⎜ MR 2 ⎟⎜ ⎟ = Mv 2 2 2⎝ 2 4 ⎠⎝ R ⎠ The total energy that the wheel has is equal to the potential energy it had at the time it fell and rolled from the truck. E = K = U = Mgh = (60 kg)(9.81 m/s 2 )(120 m) = 7.1 × 104 J

τ = I 0α =

13-32.

This is also equal to the total kinetic energy at the time of impact at the bottom of the hill. 5 E = Mv 2 4 4E v= = 31 m/s 5M Therefore, the translational portion of the kinetic energy is 1 1 K translational = Mv 2 = (60 kg)(31 m/s) 2 = 2.9 × 104 J 2 2 and the rotational kinetic energy is K rotational = E − K translational = 7.1 × 104 J − 2.9 × 104 J = 4.2 × 104 J †13-33.

13

The sphere is in contact with a plane that is inclined 20° as shown. The torque on the sphere due to the contact force F that is directed up the plane is τ = I 0α = FR = MgR sin θ (i)

N 3.0 m

F

mg 20° The moment of inertia for a sphere for an axis through its center is 2 I = MR 2 . To find the moment for a rotation about a point on the sphere, use the parallel axis 5 theorem to find 2 7 I 0 = MR 2 + MR 2 = MR 2 5 5 Then, using equation (i)


CHAPTER

13

MgR sin θ MgR 2 sin θ 5 = = g sin 20° = 0.244 g (ii) 7 I0 7 2 MR 5 The problem also states that the sphere moves s = 3.0 m in t = 1.6 s. This information can be used to find the acceleration of the sphere down the plane from the kinematic relation 1 x = at 2 2 2x a= 2 (iii) t Subsituting equation (iii) into (ii) and solving for the acceleration due to gravity g gives 2x a = 0.244 g = 2 t 2x 2(3.0 m) = = 9.6 m/s 2 g= 2 2 0.244t 0.244(1.6 s) a = Rα = R

13-34.

The string is oriented vertically as the yo-yo unwinds. For the translational motion, the sum of the net force is F = T − Mg = Ma The rotational inertia of the disk for an axis passing through its center is 1 I = MR 2 2 The equation for the rotational motion is τ = Iα = TR

T

R mg

where the angular acceleration is related to the translational acceleration by a = αR. Therefore, TR α= I TR 2 M ( g − a) R 2 a= = 1 I MR 2 2 a = 2 g − 2a

2 g or 6.54 m/s 2 3 Assume the barrel is not rolling as the two men begin applying the forces. There are two torques being applied to the barrel, which may be treated as a uniform cylinder of mass M and radius R, about the point of contact between the barrel and the road. The rotational inertia for the barrel is I = MR2/2. The net torque will produce an angular acceleration either clockwise (if the net torque is negative) or counterclockwise (if the net torque is positive). τ net = FR − 2 FR = − FR a=

13-35.

The barrel will roll to the right, turning clockwise. If we choose the rotational axis to be through the center of the barrel, the sum of torques yields, fR − FR = Iα The sum of the forces acting on the barrel in the horizontal direction is F − F + f = Ma or f = Ma. Since a is related to the angular acceleration, we have


CHAPTER

13-36.

†13-37.

13

fR − FR = Iα a 1 fR = FR = I = MRa R 2 1 1 f − F = Ma = f 2 2 2 f = F. 3 The rotational work done by the motor of the blender is W = τθ. The angular acceleration is 1 constant, so τ = Iα and θ = α t 2 . Therefore, the work done is 2 ⎛1 ⎞ 1 W = τθ = ( Iα ) ⎜ α t 2 ⎟ = Iα 2t 2 . 2 ⎝ ⎠ 2 The angular acceleration is ω 250 rad/s α= = = 5.0 × 102 rad/s 2 t 0.50 s The instantaneous power delivered to the blades by the motor is found by taking the derivative of the work with respect to time. dW d ⎛1 ⎞ = ⎜ Iα 2t 2 ⎟ = Iα 2t = (2.0 × 10−4 kg i m 2 )(5.0 × 102 rad/s 2 ) 2 (0.50s) = 25 W P= dt dt ⎝ 2 ⎠ The sphere is in contact with a plane that is inclined 15°. The torque on the ball due to the contact force F that is directed up the plane is τ = I 0α = FR = MgR sin θ The moment of inertia for a sphere for an axis through its center is I = 52 MR 2 . To find the moment for a rotation about a point on the sphere, use the parallel axis theorem to find 2 5 I 0 = MR 2 + MR 2 = MR 2 3 3 Then, using the torque equation, MgR sin θ MgR sinθ 3g g α= (sin15°) = 0.155 = = 5 5R I0 R MR 2 3 Use the rotational kinematic relation to find the angle φ through which the ball rolls. 1 1 1⎛ g⎞ φ = ω 0t + α t 2 = α t 2 = ⎜ 0.155 ⎟ t 2 2 2 2⎝ R⎠

13-38.

⎛ ⎞ ⎛ 1rev ⎞ 1 ⎜ (0.155)(9.81m/s 2 ) ⎟ (4.0s) 2 = 106 rad ⎜ = ⎜ ⎟ ⎟ = 17 rev 1 2⎜ 2π rad ⎠ ⎝ ⎟ (0.23m) ⎜ ⎟ 2 ⎝ ⎠ The circumference of the shaft is C = 2πR = 2π(0.0020 m) = 0.0125 m. Since the string is 0.25 m long, it can wrap around the shaft (0.25 m)/(0.0125 m) = 20 times. The string does not slip as it is pulled, so the string will be in contact with the shaft for 20 rev = 126 rad. The torque applied tangentially to the shaft is τ = Iα. The angular acceleration of the shaft is α = FR/I , which can be used to find the elapsed time for the pull.


CHAPTER

13 1 2 2θ

θ = αt2 t=

α

=

2 Iθ = FR

2(5.0 × kg i m)(126 rad) = 1.1s (5.0 N)(0.0020 m)

This can then be used to find the angular velocity at the time when the string is pulled out of the shaft. FR (5.0 N)(0.0020 m)(1.1s) ω = αt = = 220 rad/s t= I 5.0 × 10−5 kg i m 2 †13-39.

13-40.

†13-41.

The turntable is rotationally accelerated from rest to an angular speed rev ⎛ 1 min ⎞⎛ 2π rad ⎞ ω = 33.3 ⎜ ⎟⎜ ⎟ = 3.48 rad/s min ⎝ 60 s ⎠⎝ 1 rev ⎠ The torque is applied at the outer rim of the disk of radius R and mass M. The torque required is 3.48rad/s ⎛ω ⎞ 1 ⎛ω ⎞ 1 = 0.024 N i m τ = Iα = I ⎜ ⎟ = MR 2 ⎜ ⎟ = (1.2 kg)(0.15m) 2 2.0s ⎝t ⎠ 2 ⎝t ⎠ 2 The tangential force applied at the rim is τ 0.024 N i m = 0.16 N F= = R 0.15 m Assume the ball is initially at rest when the subway car is either at rest or moving at constant velocity. Then, when the subway car accelerates, the static friction force applies a torque to the ball, so it rolls in the direction opposite to the direction of the car’s acceleration. In an inertial reference frame at the contact point between the ball and the floor of the car, the equation for the horizontal translational motion is f = Ma′. For the rotational motion of the ball about its center of mass, 2 τ = Rf = Iα = MR 2α (i) 5 The translational acceleration of the center of mass of the ball relative to the car is −αR; and relative to the ground the acceleration of the ball is a′ = a − αR, where a is the acceleration of the subway relative to the ground. Equation (i) then becomes 2 2 a − a′ 2 = MR (a − a′) τ = Rf = MR 2α = MR 2 R 5 5 5 which can be rearranged to solve for a′. 2 RMa′ = RM (a − a′) 5 2 a′ = a 7 The maximum magnitude of the static friction force at the point of contact of the hoop with the plane is f = µsN = µsMg cos θ (i) The moment of inertia for a hoop for an axis through its center is I = MR 2 . To find the moment for a rotation about a point on its rim, use the parallel axis theorem to find I 0 = MR 2 + MR 2 = 2MR 2 There is also a torque on the hoop about the point of contact of the hoop due to the component of the weight parallel to the inclined plane.


CHAPTER

13

τ = I 0α = 2MR 2α = MgR sin θ MgR sin θ 1 a = Rα = R = g sin θ 2

2MR 2 Combining this result with equation (i) gives 1 f = µs Mg cosθ = Ma = Mg sin θ 2 2µs cosθ = sin θ tan θ = 2µs 13-42.

†13-43.

For rotation about an axis through the center of mass of the cylinder, 1 τ = Rf = MR 2α (i) 2 The friction force is f = µkN = µkMg cos θ (ii) Combining equation (ii) with equation (i) gives, 2 µ k Mg cosθ 2µ k g cosθ 2f = = α= MR MR R For the translational motion, the equation is Ma = Mg sin θ − f = Mg sin θ − µkMg cos θ = M(g sin θ − µkg cos θ) Therefore, a = g sin θ − µkg cos θ = g(sin θ − µkcos θ) This is identical to the solution for a block sliding down an incline because the force equation makes no reference to the geometry of the object that is sliding. The car described in problem 12-63 has a mass M = 1360 kg and each of the four wheels has a mass of m = 27.2 kg. The radius of each wheel is R = 0.381 m. Let f represent the friction force at the point of contact between the wheels and the road. As a result, the torque on each wheel is τ = Rf. The torque results in an angular acceleration α of the wheel given by τ = Iα. Therefore, Iα f = 2 (i) R There are five horizontal forces acting on the car, force F pulling the car forward and four friction forces acting at the wheels, in the opposite direction. The sum of these forces acting on the car, applying the result from equation (i) and using the relation a = αR, is ⎛ Ia ⎞ F − 4 f = F − 4 ⎜ 2 ⎟ = Ma ⎝R ⎠ 1 The rotational inertia for each wheel is I = mR 2 . Therefore, 2 ⎛1 2 ⎞ ⎜ 2 mR a ⎟ F − 4⎜ ⎟ = Ma 2 R ⎜⎜ ⎟⎟ ⎝ ⎠ F − 2ma = Ma

F 4000 N = = 2.83 m/s 2 M + 2m 1360 kg + 2(27.2 kg) If the rotational inertia of the wheels is neglected, the acceleration is a=


CHAPTER a=

13-44.

13 F 4000 N = = 2.94 m/s 2 M 1360 kg

The percent difference between these two values of the acceleration is 2.94 − 2.83 × 100 % = 3.89 % 2.83 Let f be the friction force at each wheel. A partial free-body mg sin θ N diagram is shown. In the direction parallel to the inclined plane, the sum of forces is f (4M + m) g sin θ − 4 f = (4M + m)a (i) mg cos θ θ (ii) For each wheel, the torque is τ = fR = Iα 1 where I = mR 2 and the angular acceleration is α = a/R. Solving for the friction force in 2 equation (ii) gives, 1 MR 2 Iα 2 a 1 f = = = Ma (iii) R R R 2 Substituting (iii) into (i), (4M + m) g sin θ − 4( 12 Ma) = (4M + m)a ⎛ 4M + m ⎞ a = g sin θ ⎜ ⎟ ⎝ 6M + m ⎠

13-45.

13-46. †13-47.

13-48.

The angular acceleration of the wheel is fR τ 2 µ N 2(0.80)(2.0 × 105 N) α= = = = = 3330 rad/s 2 1 I MR (160 kg)(0.60 m) MR 2 2 During the time that the wheel is slipping as it is in contact with the ground, the linear speed of the aircraft remains constant. The angular velocity when the wheel is rolling without slipping is km ⎛ 1000 m ⎞⎛ 1 h ⎞ 200 ⎜ ⎟⎜ ⎟ v h ⎝ 1 km ⎠⎝ 3600 s ⎠ ω= = = 92.6 rad/s R 0.60 m The time the wheel takes to accelerate to this angular speed is ω 92.6 rad/s = 0.0278 s t= = α 3330 rad/s 2 During this time, the magnitude of the displacement of the plane is x = vt = (55.6 m/s)(0.0278 s) = 1.6 m 2 rev ⎞⎛ 2π rad ⎞ ⎛2 ⎞ ⎛ −4 2 L = Iω = ⎜ MR 2 ⎟ ω = (0.070 kg)(0.025 m)2 ⎜ 5.0 ⎟⎜ ⎟ = 5.5 × 10 kg i m /s 5 5 s 1 rev ⎝ ⎠ ⎝ ⎠⎝ ⎠ The orbital angular momentum is given by L = mvr. The orbital radius and the orbital period are given for the moon. The average speed of the moon can be determined by dividing the circumference of the moon’s orbit by the period. 2π mr 2 2π (7.35 × 1022 kg)(3.8 × 108 m) 2 ⎛ 2π r ⎞ L = mvr = m ⎜ r = = = 2.8 × 1034 kg i m 2 /s ⎟ t ⎛ 24 h ⎞⎛ 3600 s ⎞ ⎝ t ⎠ 27.3 d ⎜ ⎟⎜ ⎟ ⎝ 1 d ⎠⎝ 1 h ⎠ L = (mv)r = (5.2 × 10−16 kg i m/s)(1.0 × 103 m) = 5.2 × 10−13 kg i m 2 /s


CHAPTER

13-49.

rev ⎞ ⎛ 2π rad ⎞ ⎛ 2 L = mvr = mω r 2 = (0.15 kg)(0.75 m) 2 ⎜ 3.0 ⎟ = 1.6 kg i m /s ⎟⎜ s ⎠ ⎝ 1 rev ⎠ ⎝

13-50.

L = mvr = (100 kg)(490 m/s)(4.22 × 107 m) = 2.0 × 1012 kg i m 2 /s

†13-51.

13

The mass of the electron is m = 9.11 × 10−31 kg. The orbital angular momentum is L = mvr. For the first radius and velocity data given, L1 = mvr = (9.11 × 10−31 kg)(2.18 × 106 m/s)(0.529 × 10−10 m) = 1.05 × 10−34 kg i m 2 /s For the other two “orbits,” L2 = mvr = (9.11 × 10−31 kg)(1.09 × 106 m/s)(2.12 × 10−10 m) = 2.11 × 10−34 kg i m 2 /s L3 = mvr = (9.11 × 10−31 kg)(7.27 × 106 m/s)(4.76 × 10−10 m) = 3.15 × 10−34 kg i m 2 /s

13-52.

In comparing the three results, we note that L2 = 2L1 and L3 = 3L1. One might conclude that the trend would continue and that the orbital angular momentum is quantized, L = nL1. (a) L = mvr = (150 kg)(60 000 m/s)(1.2 × 107 m) = 1.1 × 1014 kg i m 2 /s (b) L = mvr = (5.98 × 1024 kg)(60 000 m/s)(1.2 × 107 m) = 4.3 × 1036 kg i m 2 /s

13-53.

13-54.

⎛ 1000 kg ⎞⎛ km ⎞⎛ 1 h ⎞⎛ 1000 m ⎞ 9 2 L = mvr = (1500 t) ⎜ ⎟⎜ 85 ⎟⎜ ⎟⎜ ⎟ (50 m) = 1.8 × 10 kg i m /s 1 t h 3600 s 1 km ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ Since the point is to the left of the tracks, the direction, using the right hand rule, points upward. When the point is on the track, the radius is zero; and the angular momentum is zero kg • m2/s. Since L = mv1rmin = mv2 rmin = 2 , we can solve for v1 and v2. v1 =

2 2 2(1.05 × 10−34 kg i m 2 /s) = = = 7.6 × 107 m/s mrmin ma (1 − 2 2) (9.11 × 10−31 kg)(5.3 × 10−11 m)(1 − 2 2) 0 3 3

and 2 2 2(1.05 × 10−34 kg i m 2 /s) = = = 2.2 × 106 m/s mrmax ma (1 + 2 2) (9.11 × 10−31 kg)(5.3 × 10−11 m)(1 + 2 2) 0 3 3 2 The angular momentum is L = Iω, where the moment of inertia for a solid sphere is I = mR 2 5 for rotations about an axis that passes through the center of the sphere. The spin angular speed is then L L 5.3 × 10−35 kg i m 2 /s ω= = = = 7.9 × 1022 rad/s 2 2 I 2 −27 −15 2 MR (1.67 × 10 kg)(1.0 × 10 m) 5 5 From the angular speed, the rotational kinetic energy and the linear speed at a point on the equator of the sphere can be found. v = Rω = (1 × 10−15 m)(7.9 × 1022 rad/s) = 7.9 × 107 m/s v2 =

†13-55.

K Rot =

1 2 1⎛2 ⎞ Iω = ⎜ MR 2 ⎟ ω 2 2 2⎝ 5 ⎠ 1 = ⎡ 52 (1.67 × 10−27 kg)(1.0 × 10−15 m)2 (7.9 × 1022 rad/s)2 ⎤ = 2.1 × 10−12 J ⎦ 2⎣


CHAPTER

13

Finally, from Einstein’s theory of relativity (see Chapter 8 or 36), the relationship between an object’s mass and its energy is E0 = mc 2 = (1.67 × 10−27 kg)(3.0 × 108 m/s) 2 = 1.5 × 10−10 J 13-56.

13-57. 13-58.

L = Iω =

1 1 ⎛ rev ⎞⎛ 2π rad ⎞ = 0.3 kg i m 2 /s MR 2ω = (0.2 kg)(0.15 m) 2 ⎜ 20 ⎟⎜ ⎟ 2 2 s ⎠⎝ 1 rev ⎠ ⎝

⎛ rev ⎞⎛ 1 min ⎞ ⎛ ⎛ 2π rad ⎞ ⎜ 78 ⎟⎜ ⎟⎜⎜ ⎟ min ⎠⎝ 60 s ⎠ ⎝ ⎝ 1 rev ⎠ 1 ∆L ∆ω 1 2 ∆ω 2⎝ =I = MR = (1.4 kg)(0.15 m) = 0.051 kg i m 2 /s 2 2 2 2.5 s ∆t ∆t ∆t 1 1⎛1 ⎞ K Rot = Iω 2 = ⎜ MR 2 ⎟ ω 2 2 2⎝ 2 ⎠ 2

=

⎡⎛ 1 rev ⎞⎛ 2π rad ⎞⎛ 1 min ⎞ ⎤ (1200 kg)(0.044 m) 2 ⎢⎜ 200 ⎟⎜ ⎟⎜ ⎟ = 250 J 4 min ⎠⎝ 1 rev ⎠⎝ 60 s ⎠ ⎥⎦ ⎣⎝

1 1 rev ⎞⎛ 1 min ⎞⎛ 2π rad ⎞ ⎛ 2 MR 2ω = (1200 kg)(0.0.44 m) 2 ⎜ 200 ⎟⎜ ⎟ = 24 kg i m /s 2 2 min ⎟⎜ ⎝ ⎠⎝ 60 s ⎠⎝ 1 rev ⎠ The spin angular momentum of the Sun is 1 LS = I ω = MR 2ω 5 1 ⎛ 2π rad ⎞⎛ 1 d ⎞⎛ 1 h ⎞ 41 2 = (1.99 × 1030 kg)(6.96 × 108 m)2 ⎜ ⎟⎜ ⎟⎜ ⎟ = 5.6 × 10 kg i m /s 5 ⎝ 25 d ⎠⎝ 24 h ⎠⎝ 3600 s ⎠ The angular ⎛ 2π rad ⎞ ⎛ 1y L = ∑ MRi 2ωi = (3.30 × 1023 kg)(57.9 × 109 m)2 ⎜ ⎟ ⎜⎜ momentum 7 0.241 y i ⎝ ⎠ ⎝ 3.156 × 10 for the ⎛ 2π rad ⎞ ⎛ 1y planets (4.87 × 1024 kg)(108 × 109 m)2 ⎜ ⎟ ⎜⎜ 7 orbiting the ⎝ 0.615 y ⎠ ⎝ 3.156 × 10 Sun is the ⎛ 2π rad ⎞ ⎛ 1y sum of the (5.98 × 1024 kg)(150 × 109 m)2 ⎜ ⎟ ⎜⎜ 7 ⎝ 1.000 y ⎠ ⎝ 3.156 × 10 individual L = Iω =

†13-59.

angular momenta of the planets. The masses, mean distance from the Sun, and period of the planets is given in Table 9.1.

⎛ 2π rad ⎞ ⎛ 1y (6.42 × 1023 kg)(228 × 109 m)2 ⎜ ⎟ ⎜⎜ 7 ⎝ 1.88 y ⎠ ⎝ 3.156 × 10 ⎛ 2π rad ⎞ ⎛ 1y (1.90 × 1027 kg)(778 × 109 m)2 ⎜ ⎟ ⎜⎜ 7 ⎝ 11.9 y ⎠ ⎝ 3.156 × 10 ⎛ 2π rad ⎞ ⎛ 1y (5.66 × 1026 kg)(1430 × 109 m)2 ⎜ ⎟ ⎜⎜ 7 ⎝ 29.5 y ⎠ ⎝ 3.156 × 10 ⎛ 2π rad ⎞ ⎛ 1y (8.70 × 1025 kg)(2870 × 109 m)2 ⎜ ⎟ ⎜⎜ 7 ⎝ 84.0 y ⎠ ⎝ 3.156 × 10 ⎛ 2π rad ⎞ ⎛ 1y (1.03 × 1026 kg)(4500 × 109 m)2 ⎜ ⎟ ⎜⎜ 7 ⎝ 165 y ⎠ ⎝ 3.156 × 10 ⎛ 2π rad ⎞ ⎛ 1y (1.50 × 1022 kg)(5890 × 109 m)2 ⎜ ⎟ ⎜⎜ 7 ⎝ 248 y ⎠ ⎝ 3.156 × 10


⎞ ⎟+ s ⎟⎠ ⎞ ⎟+ s ⎟⎠ ⎞ ⎟+ s ⎟⎠ ⎞ ⎟+ s ⎟⎠ ⎞ ⎟+ s ⎟⎠ ⎞ ⎟+ s ⎟⎠ ⎞ ⎟+ s ⎟⎠ ⎞ ⎟+ s ⎟⎠

⎞ 43 2 ⎟⎟ = 3.14 × 10 kg i m /s s⎠

CHAPTER

13

The percentage of the total angular mometum of the solar system (including only the Sun and planets) that is in the rotational motion of the Sun is LS 5.61 × 1041 kg i m 2 /s Percentage = 100 × = 100 × = 1.8 % L + LS 3.13 × 1043 kg i m 2 /s + 5.61 × 1041 kg i m 2 /s 13-60.

The figure shows an ellipse with the Sun at one focus, r2 is the aphelion radius, and r1 is the perihelion radius. Applying the conservation of angular momentum, mv2r2 = mv1r1 v r2 = r1 1 (i) v2

ra

rp

The conservation of mechanical energy includes the kinetic energy and the gravitational potential energy of the gravitational force of the Sun acting on the comet. Applying conservation of mechanical energy to this situation gives 1 2 Mm 1 2 Mm mv1 − G = mv2 − G (ii) 2 r1 2 r2 where G is the universal graviational constant and M is the solar mass. Substitute equation (i) into (ii) and divide through by m. 2

1 2 M 1 ⎛r ⎞ M v1 − G = v v12 ⎜ 1 ⎟ − G 2 r1 2 ⎝ r2 ⎠ r2 v12 − 2

GM GM = v22 − 2v2 r1 r1v1

(iii)

To greatly simplify this equation, let V =

GM r1v1

We may then write equation (iii) as v22 − 2Vv2 + v1 (2V − v1 ) = 0 Use the quatratic formula to solve for v2. v2 =

2V ± 4V 2 − 4v1 (2V − v1 )

= V ± V 2 − 2Vv1 + v12

2 The quantity in the square root is just (V − v1)2, so the final result is v2 = 2V − v1 i v1 .

⎛ 2GM S ⎞ ( r1v1 )2 r1v1 = = and r The correct choice is v2 = 2V − v1 = v1 ⎜ − 1 , ⎟ 2 2 ⎜ r v2 ⎟ 2GM S ⎝ 11 ⎠ − v1 2GM S − r1v1 r1v1 †13-61.

1 MR 2ω1 2 Because the child jumps onto the outer edge from along a radial line, she doesn’t apply a torque that would change the angular velocity of the merry-go-round, but the addition of her mass to the merry-go-round does change the angular velocity because the angular momentum must be conserved.

The initial angular momentum is L1 = I1ω1 =


CHAPTER

13

⎛1 ⎞ L1 = L2 = I 2ω 2 = ⎜ MR 2 + mR 2 ⎟ ω 2 2 ⎝ ⎠ 1 ⎛1 ⎞ MR 2ω1 = ⎜ MR 2 + mR 2 ⎟ ω 2 2 ⎝2 ⎠

ω2 =

1 2 1 2

MR 2

MR 2 + mR

ω = 2 1

1 2 1 2

M

M +m

ω1 =

1 (20 2 1 (20 2

kg)

kg) + (25 kg)

(2.0 rad/s) = 0.57 rad/s

The child then applies a torque at the outer edge to increase the angular speed to 2.0 rad/s. In ther process, the angular momentum is increased. As the child walks from the outer edge toward the axis of rotation, the angular velocity must further increase to maintain a constant angular momentum. L3 = L4

13-62.

⎛1 ⎛1 2 2⎞ 2 2⎞ ⎜ MR + mR ⎟ ω1 = ⎜ MR + mr ⎟ ω3 ⎝2 ⎠ ⎝2 ⎠ 1 MR 2 + mR 2 ⎡ 1 (20 kg)(1.5 m) 2 + (25 kg)(1.5 m) 2 ⎤ ω3 = 2 ω = ⎢2 ⎥ (2.0 rad/s) = 5.5 rad/s 1 MR 2 + mr 2 1 1 (20 kg)(1.5 m) 2 + (25 kg)(0.5 m) 2 ⎢ ⎥⎦ 2 ⎣2 To determine the effect of the collision on the rotational period of the Earth, begin by applying conservation of momentum for the inelastic collision in two dimensions. The angular momentum of the Earth before the collision is determined from the current length of the day and its rotational inertia, 2π rad LE = lEω1 = (0.331MR 2 ) 8.64 × 104 s 2π rad = (0.331)(5.98 × 1024 kg)(6.38 × 106 m) 2 = 5.86 × 1033 kg i m 2 /s 4 8.64 × 10 s In the instant before the collision, the angular momentum of the asteroid about the Earth’s rotational axis is LA = m(v cos 235°)R = (5.0 × 1018 kg)(150 000 m/s)(cos 235°)(6.38 × 106 m) = −2.74 × 1030 kg • m2/s The total angular mometum before the collision must equal the total angular momentum after the collision, LE + LA = I Eω 2

ω2 = =

LE + LA LE + LA = I E+A (0.331M + m) R 2 5.86 × 1033 kg i m 2 /s − 2.74 × 1030 kg i m 2 /s ⎡0.331(5.98 × 1024 kg) + 5.0 × 1018 kg ⎤ (6.38 × 106 m)2 ⎣ ⎦

= 7.27 × 10−5 rad/s

The change in the length of the day is then t2 − t1 = †13-63.

2π rad 7.27 × 10

−5

rad/s

− 8.64 × 104 s = 26.2 s

The day would be slightly longer. The initial angular momentum is L1 = ( I 0 + I )ω1 , where I0 is the rotational inertia of the professor and stool and I = 2mr2 is the rotational inertia of the two masses held at a distance r from the


CHAPTER

13

rotational axis. When the masses and arms are pulled in both rotational inertias decrease, so the angular speed must increase to maintain a constant angular momentum. L1 = L2

( I 0 + I )ω1 = ( I 0′ + I ′ )ω 2

( (

) )

6.0kg i m 2 + 2(10 kg)(1.0 m) ⎛ I0 + I rev ⎞ ω2 = ω1 = 0.50 ⎜ ⎟ = 2.2 rev/s (or 14 rad/s) 2 I 0′ + I ′ s ⎠ 4.0kg i m + 2(10 kg)(0.10 m) ⎝ 13-64.

Consider the wheel held such that it has an angular momentum that is directed upward. When the wheel is turned upside down, the angular momentum is now directed downward. Since the angular momentum of the wheel-person-chair system must be conserved, the person and chair will rotate in the direction opposite to the wheel’s rotational direction. L1 = L2 − L1 2 Iω1 = I ′ω 2 For angular momentum to be conserved the angluar momentum of the person and chair must be equal to twice the angular momentum of the wheel. 2 I ω1 2(0.48 kg i m 2 )(18 rad/s) = = 5.8 rad/s ω2 = I′ 3.0 kg i m 2

13-65.

∆LTrain = ∆LEarth m(∆v) R = (0.33MR 2 )∆ω m(∆v) R ∆ω = 0.33MR 2

13-66.

km ⎞⎛ 1000 m ⎞⎛ 24 h ⎞⎛ 1 rev ⎞ ⎛ (7.7 × 106 kg) ⎜ 65 ⎟⎜ ⎟⎜ ⎟⎜ ⎟ m(∆v) h ⎠⎝ 1 km ⎠⎝ 1 d ⎠⎝ 2π rad ⎠ ⎝ = = = 1.5 × 10−19 rev/d 24 6 0.33MR 0.33(5.98 × 10 kg)(6.38 × 10 m) (a) L = m(∆v) R cosθ km ⎞⎛ 1000 m ⎞⎛ 1 h ⎞ ⎛ 6 = (1.1 × 108 )(2000 kg) ⎜ 80 ⎟⎜ ⎟ (6.38 × 10 m)(cos 40°) h ⎟⎜ ⎝ ⎠⎝ 1 km ⎠⎝ 3600 s ⎠ = 2.4 × 1019 kg i m 2 /s (b) I ∆ω = L = 2.4 × 1019 kg i m 2 /s

∆ω = †13-67.

2.4 × 1019 kg i m 2 /s 2.4 × 1019 kg i m 2 /s = = 3.0 × 10−19 rad/s 37 2 I 8.1 × 10 kg i m

The two satellites are in uniform circular motion as they orbit the Earth. They are held in orbit by the Earth’s gravity, which provides the centripetal force. Mm mv 2 G 2 = R R Therefore, at a given radius, the speed will be M v2 = G E R


r1

r2

CHAPTER

13 GM E R

v=

which gives

v1 =

GM E and v2 = r1

GM E r2

For the two satellites, since r1 < r2, v1 > v2 . The satellite closer to the Earth has the greater speed. The angular momentum of each is GM E L1 = mv1r1 = mr1 = m r1GM E r1 L2 = mv2 r2 = mr2

GM E = m r2GM E r2

Since r1 < r2, L2 > L1. The satellite further from Earth has the greater angular momentum. 13-68.

(a) Since the origin for the calculation is the same as the location of the Earth, L = 0. (b) After three months, the Earth will be at ¼ the circumference of its orbit. L3m = M E vR = M E R 2ω 2⎛ 2π rad = (5.98 × 1024 kg) ⎡ 2(1.5 × 1011 m) ⎤ ⎜ ⎣ ⎦ ⎜ 3.156 × 107 ⎝

0

⎞ ⎟ s ⎟⎠

3

9

= 5.4 × 1040 kg i m 2 /s

6

Similarly, for 6 and 9 months, L6m = M E vR = M E R 2ω ⎛ 2π rad = (5.98 × 1024 kg)(4)(1.5 × 1011 m)2 ⎜ ⎜ 3.156 × 107 ⎝

⎞ 41 2 ⎟ = 1.1 × 10 kg i m /s s ⎟⎠

L9m = M E vR = M E R 2ω

13-69.

2⎛ 2π rad ⎞ = (5.98 × 1024 kg) ⎡ 2(1.5 × 1011 m) ⎤ ⎜ = 5.4 × 1040 kg i m 2 /s ⎣ ⎦ ⎜ 3.156 × 107 s ⎟⎟ ⎝ ⎠ Angular momentum is not conserved. 2π ω= T −2π −2π dω = 2 dT = (0.0016 s) = − 1.355 × 10−12 rad/s 4 2 (8.616 × 10 s) T

α=

−1.355 × 10−12 rad/s dω = = −4.3 × 10−22 rad/s 2 dt ⎛ s⎞ (100 y) ⎜ 3.156 × 107 ⎟ y ⎝ ⎠

τ = Iα = 0.331M E RE2 = (0.331)(5.98 × 1024 kg)(6.38 × 106 m)2 ( −4.3 × 10−22 rad/s 2 ) = 3.5 × 1016 N i m P = τω = (3.5 × 1016 N i m)

2π 8.616 × 104 s

= 2.6 × 1012 W


CHAPTER 13-70.

Lm = mvr = I P ∆ω ∆ω =

†13-71.

13

mvr mvr (3 × 1013 kg i m/s)(5.0 × 103 m) = = = 1.1 × 10−6 rad/s 2 MR 2 2 (5.8 × 1015 kg)(7.5 × 103 m) 2 IP 5 5

The total angular momentum of the boat and woman before the woman turns is equal to zero, therefore as she turns, it must remain zero. Let the woman’s angular momentum be Lw = Iw ωw and the angular momentum of the boat be Lb = Ib ωb. Then, 0 = Lw + Lb = I wω w + I bω b If the above equation is multiplied by ∆t, the elapsed time for the woman’s rotation, one obtains the angular displacement of the boat. I wω w ∆t + I bω b ∆t = 0 I wφw + I bφb = 0

φb = − 13-72.

I wφw (0.80 kg i m 2 )(180°) =− = −7° Ib 20 kg i m 2

The rowboat rotates seven degrees in the direction opposite to that of the woman. (a) The initial angular momentum, which is conserved, about a vertical axis at the mid-point of the system is L = m1v1r1 + m2v2 r2 , where the subscripts represent the two vehicles and r1 = r2 = r = 0.5 m, m1 = m2 = m = 1200 kg, and v1 = v2 = v = 30 km/h. The angular velocity of the of the two joined vehicles may then be determined. mvr + mvr = Iω before the collision

after the collision

(

)(

)(

)

ω=

1000 m km 1h 2mvr 2(1200 kg) 30 h 3600 s 1 km (0.50 m) = = 4.0 rad/s I 2500 kg i m 2

(b)

1 2

mv 2 + 12 mv 2 =

before the collision

1 2

Iω 2

after the collision

1 ∆K = Iω 2 − mv 2 2 2

⎡⎛ km ⎞⎛ 1 h ⎞⎛ 1000 m ⎞ ⎤ 1 4 = (2500 kg i m 2 )(4.0 rad/s) 2 − (1200 kg) ⎢⎜ 30 ⎟⎜ ⎟⎜ ⎟ ⎥ = 6.3 × 10 J 2 h ⎠⎝ 3600 s ⎠⎝ 1 km ⎠ ⎦ ⎣⎝ 13-73.

(a) N =

mv 2 = mg r

v = rg = (2.5 m)(9.81 m/s 2 ) = 5.0 m/s (b) The total angular momentum of Skylab before the astronaut begins to run is equal to zero, therefore as he runs, it must remain zero. Let the three astronaut’s angular momentum be La = 3mavar and the angular momentum of Skylab be Ls = Is ωs. Then, 0 = La + Ls = 3ma va r + I sωs

ωs = −

3ma va r 3(70 kg)(5.0 m/s)(2.5 m) =− = −0.009 rad/s Is 300 000 kg i m 2


CHAPTER (c)

13 La ∆t + I sωs ∆t = 0 I wφw + I bφb = 0

φa = − 13-74.

⎛ 300 000 kg i m 2 ⎞ Is I φs = − s 2 φs = − ⎜⎜ 30° = −6860° or 19 rev 2 ⎟ ⎟ Ia 2mr ⎝ 3(70 kg)(2.5) ⎠

(a) As the two flywheels are coupled, the drive belt will slip for a short time while the second wheel accelerates. The first wheel will decelerate during this time. Let T represent the difference in tension between the left and right linear portions of the drive belt. The angular accelerations of the two wheels will be τ 2 = TR2 = I 2α 2 and τ1 = − TR1 = I1α1

α2 =

TR2 = I2

TR2 1 2

M 2 R22

=

2T M 2 R2

α1 =

−TR1 −TR2 −2T = = 2 1 I1 M1R1 M R 2 1 1

The ratio ω1/ω2 must be the same as the ratio α1/α2 since the accelerations are taking place during the same elapsed time. Therefore, −2T ω1 − ω M1R1 M R = =− 2 2 2T M1R1 ω2 M 2 R2

ω1 − ω = −

M 2 R2 ω2 M1R1

When the acceleration of the flywheels is complete and they are rotating smoothly (without slipping), the angular speeds must be related as R1ω1 = R2ω2. Given this additional information, these two equations can be solved for the two unknowns. ω ω R1 and ω1 = ω2 = M M 1+ 2 1 + 2 R2 M1 M1 13-75.

Before the collision, the angular momentum of the ball is mvl and momentum is conserved in the collision. Therefore, the angular speed can be determined

(

)

mvl = ml 2 + 13 Ml 2 ω

ω=

mvl 2

ml + 13 Ml 2

The kinetic energy just after the collision is 2

⎛ ⎞ 1 1 1 mvl m2v 2 ⎟ = K = I ω 2 = (ml 2 + Ml 2 ) ⎜ 2 ⎜ ml + 1 Ml 2 ⎟ 2 2 3 2(m + 13 M ) 3 ⎝ ⎠ Through conservation of energy, the mechanical energy of the ball and rod is entirely graviational potential energy when the ball is at the highest point h. The center of mass of the rod will only increase in height by h/2, however.


CHAPTER 1 1 m2v 2 = mgh + Mgh = gh(m + M ) 1 2(m + 3 M ) 2 2 h= 13-76.

m2v 2 2 g (m + 13 M )(m + 12 M )

The angular momentum about the Earth’s axis is conserved. I 0ω 0 = I ω = ( I 0 + ∆I )ω

2 2π 2π ⎛2 ⎞ MRE2 = ⎜ M ( RE + 10 m)2 ⎟ 5 1 day ⎝ 5 ⎠ 1 day + ∆T 1 1 = ( RE + 10 m)2 RE2 1 day 1 day + ∆T RE2 ( RE + 10 m)

2

=

∆T = ∆T =

1 day 1 day + ∆T ( RE + 10 m) 2 (1 day) RE2

( RE + 10 m) 2 (1 day) RE2

− 1 day =

− 1 day (10 m)2 (1 day) RE2

⎛ 24 h ⎞⎛ 3600 s ⎞ −7 = 2.457 × 10−12 d ⎜ ⎟⎜ 1 h ⎟ = 2.12 × 10 s 1 d ⎝ ⎠⎝ ⎠ 13-77.

13-78.

At the instant of launch the angular momentum is L0 = 0 kg i m 2 /s ⎛ v2 ⎞ mv03 2 = At its highest point, L = mvH = mHv0 cos 45° = m ⎜ 0 ⎟ v0 ⎜ 4g ⎟ 2 g 32 ⎝ ⎠ At the instant the projectile returns to the ground, ⎛ v2 ⎞ mv03 2 = L = mvxmax = mxmax v0 sin 45° = m ⎜ 0 ⎟ v0 ⎜ g ⎟ 2 g 2 ⎝ ⎠ An arbitrary thin plate is rotating around an arbitrary axis with an angular velocity ω as shown. Consider a small mass element represented by the point P in the drawing. At time t = 0, P is at the position shown along vector r from the rotational axis. The tangential velocity of element P is v. As element P rotates around the axis, it sweeps out a circle. We can generalize this to all points that make up the plate. To find the total angular momentum is a matter of summing up the individual contributions to the angular momentum from all of the individual elements. For each of the mass elements, we are only interested in the radial vector component that is perpendicular to the axis of rotation. L = ∑ ri (vi ∆mi ) i

⎛ ⎞ = ∑ r⊥i (vi ∆mi ) = ∑ r⊥i (ω r⊥i ∆mi ) = ω ⎜ ∑ r⊥2i ∆mi ⎟ i i ⎝ i ⎠


v r

P

13

CHAPTER

13-79.

We recognize that ω is the same for all points in the solid plate, so it is a constant that may be taken outside the summation. We also see that the quantity in parentheses is the moment of inertia, which is a scalar quantity. The angular momentum, then must be in the same direction as the angular velocity, which in the case shown is along the +z axis. Another way to show this would be to consider the direction of the angular momentum vector for each mass element, li = m(ri × vi ) The direction of the cross product for all of the mass elements will be the same, in the positive z direction, which is the same direction as the angular velocity vector. mgr mgr ωp = = L Iω =

13-80.

ωp = =

13-81.

13-82.

13

mgr 1 mR 2ω 2

=

2 gr R 2ω

=

2(9.81 m/s 2 )(0.060 m)

(

(0.050 m)2 200

rev s

)( 21πrevrad )

= 0.37 rad/s

mgr mgr = L Iω mgr 1 mR 2ω 2

=

2 gr R 2ω

=

2(9.81 m/s 2 )(0.040 m)

(

(0.030 m) 2 120

rev s

)( 21πrevrad )

= 1.16 rad/s

The angular momentum of the forward-rolling wheel is directed in the same ω, L direction as the angular velocity. If the wheel is tilted an angle θ to the right with respect to the vertical direction, the Earth’s gravity exerts a torque on the θ wheel about the point of contact with the ground. This torque changes the mg direction of the angular momentum that causes a clockwise precession (looking from above the wheel) about the axis that includes the point of contact and the center of the wheel, which turns the bicycle toward the right. To return the bicycle to forward motion, the rider will have to exert a torque to the left with the handle bars to oppose the torque due to gravity and return the wheel to its vertical position, else the wheel will continue to tilt further and turn toward the right. mgr mgr mgr 5 gr ωp = = = = 2 2 mR ω L Iω 2 R 2ω 5

ω= 13-83.

5 gr 2 R 2ω p

(a) L = Iω =

=

(

5 gr T p 5(9.81 m/s 2 )(0.010 m) ⎛ 0.75 s ⎞ = ⎜ 2π ⎟ = 8.1 rad/s 2 R 2 2π 2(0.060 m) 2 ⎝ ⎠ 1 2

MR 2

) Rv =

1 2

MRv = 12 (25 kg)(0.35 m)(25 m/s) = 110 kg i m 2 /s

(b) and (c) dL τ = = ω × L = ω L sin 90° dt v 25 m/s = L= 110 kg i m 2 /s = 34 N i m R 80 m

(

)


τ= L

dL dt R

CHAPTER 13-84.

The propeller from problem 12-38 consists of three radial 1.2-m blades, each with a mass of 6.0 kg. The propeller rotates at 2500 rev/min. L = Iω = Ml 2ω

L

τ=

13 dL dt

R

rev ⎞⎛ 1 min ⎞⎛ 2π rad ⎞ ⎛ 2 = (6 kg)(1.2 m) 2 ⎜ 2500 ⎟⎜ 60 sec ⎟⎜ 1 rev ⎟ = 2260 kg i m /s min ⎝ ⎠⎝ ⎠⎝ ⎠ dL v τ= = ω × L = ω L sin 90° = ω L = L dt R km ⎞⎛ 1 h ⎞⎛ 1000 m ⎞ ⎛ ⎜ 360 h ⎟⎜ 3600 sec ⎟⎜ 1 km ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ 2260 kg i m 2 /s = 450 N i m 500 m L = Iω = Ml 2ω

(

13-85.

)

rev ⎞⎛ 1 min ⎞⎛ 2π rad ⎞ ⎛ = (5 × 105 kg i m 2 ) ⎜ 3000 = 1.6 × 108 kg i m 2 /s ⎟⎜ ⎟⎜ ⎟ min ⎠⎝ 60 sec ⎠⎝ 1 rev ⎠ ⎝ 8 2 L = 1.6 × 10 kg • m /s, east or west The torque due to the Earth’s rotation is 2π dL = ω × L = ωE L = τ = 1.6 × 108 kg i m 2 /s = 1.2 × 104 N i m dt 24 h 3600 s

(

1h

)

(

)

This is equal to the torque that the axle must exert at the bearings. The force F that each axle provides to exert this torque on opposite sides of the flywheel is given by τ = 2 Fl 1.2 × 104 N i m = 1.0 × 104 N 2l 2(0.60 m) l τ middle = F = (200 N)(0.40 m) = 80 N i m 2 τ edge = Fl = (200 N)(0.80 m) = 160 N i m F=

13-86.

τ

=

13-87.

The wind applies a force to the opposite side of the door that results in a torque in the opposite direction as the one you exert. To increase the chance of preventing the door from being blown open, you should push at the outermost edge of the door to maximize your torque. τ = FR = MgR = (900 kg)(9.81 m/s 2 )(0.35 m) = 3100 N i m

13-88.

P = τω

τ =

P

ω

=

700 hp 37 000

(

(

745.7 J/s 1 hp

rev 2π rad min 1 rev

)

)( 160mins )

= 130 N i m

( 21πrevrad ) = 3100 J

13-89.

W = τθ = Flθ = (200 N)(0.25 m)(10 rev)

13-90.

Unlike example 13-2, the end of the meterstick slips as the meter stick falls from its vertical orientation. In this case, the center of mass has a translation in the vertical direction as well. So, the conservation of energy equation is 1 1 mgh = mv 2 + I ω 2 2 2


CHAPTER

13

The partition of energy is expected to be equal between translational and rotational kinetic energies. Therefore, we may write: mgh = I ω 2

ω= †13-91.

mgh = I

mgl / 2 1 ml 2 3

=

3g = 2l

3(9.81 m/s 2 ) = 3.83 rad/s 2(1.00 m)

(a) From the information given, a linear equation for the torque due to the spring as a function of angle in radians can be written. 2.00 × 103 N i m − 0.30 × 103 N i m τ S = 2.00 × 103 N i m − θ π rad 2 mg φ

= 2.00 × 103 N i m − (1.08 × 103 N i m / rad)θ

θ

The angle at which this torque is equal to the torque due to the force of gravity τG on the hatch 1 can be found. Consider the drawing, τ G = lmg cosθ 2 Then, 1 lm gcosθ = 2.00 × 103 N i m − (1.08 × 103 N i m / rad)θ 2 1 (1.2 m)(400 kg)(9.81 m/s 2 ) cos θ = 2.00 × 103 N i m − (1.08 × 103 N i m / rad)θ 2 (1.08 × 103 N i m / rad)θ + (2.35 × 103 N i m)cos θ − 2.00 × 103 N i m = 0 To determine the angle, one must use the graphical method or trial and error. The graph on the left below is a graph of the left hand side of the above equation versus the angle 0 ≤ θ ≤ π/2. From the graph on the right, the value of the function is zero at θ = 1.33 radians = 76.2°. 600

50 40 30 20 10

400

0 -10 -20 -30

200

-40 -50 1.30

1.31

1.32

1.33

θ

0

-200

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

θ


1.6

1.34

1.35

1.36

CHAPTER (b) In the closed position, θ = 0, and the sum of the torques will be; (1.08 × 103 N i m / rad)θ + (2.35 × 103 N i m)cos θ − 2.00 × 103 N i m − F (1.2 m) = 0 ( 2.35 × 103 N i m)cos 0 − 2.00 × 103 N i m = 290 N 1.2 m In the upright position, the force required is (1.08 × 103 N i m / rad)θ + (2.35 × 103 N i m)cos θ − 2.00 × 103 N i m − F (1.2 m) = 0

F=

F=

13-92.

13-93.

(1.08 × 103 N i m / rad) π2 + ( 2.35 × 103 N i m)cos

( π2 ) − 2.00 × 103 N i m = −250 N

1.2 m The minus sign indicates that the direction of the force, and the torque, has changed from the earlier case, as one would expect. τ = FR = Iα FR FR (20 N) α= = = = 15 rad/s 2 2 I (4.0 kg)(0.33 m) MR 1 2 2 + M hoop Rhoop (a) I = M disc Rdisc 2 1 = (0.020 kg)(0.040 m) 2 + (0.015 kg)(0.040 m) 2 = 4.0 × 10−5 kg i m 2 2 rev 2π rad 1 min 0 − 100 min ∆ω 1 rev 60 s (b) α = = = 0.12 rad/s 2 ∆t (90 s)

(

)(

)(

)

(c) τ = Iα = ( 4.0 × 10−5 kg i m 2 )(0.12 rad/s 2 ) = 4.8 × 10−6 N i m

13-94.

1 (d) W = τθ = τ (ω0t − α t 2 ) 2 ⎡⎛ ⎤ rev ⎞⎛ 2π rad ⎞⎛ 1 min ⎞ 1 = ( 4.8 × 10−6 N i m ) ⎢⎜ 100 (90 s ) − (0.12 rad/s 2 )(90 s )2 ⎥ ⎟⎜ ⎟⎜ ⎟ min ⎠⎝ 1 rev ⎠⎝ 60 s ⎠ 2 ⎣⎝ ⎦ = 0.0023 J 1 ∆ω (a) τ = Iα = MR 2 ∆t 2 =

(

0 − 33 13 1 (1.2 kg )( 0.15 m )2 2

rev min

)( 21πrevrad )( 160mins ) = 1.0 × 10−3 N i m

45 s

(b) P = τω

13-95.

⎛ 1 rev ⎞⎛ 2π rad ⎞⎛ 1 min ⎞ = (1.0 × 10−3 kg i m 2 ) ⎜ 33 ⎟⎜ ⎟⎜ ⎟ = 3.7 mW ⎝ 3 min ⎠⎝ 1 rev ⎠⎝ 60 s ⎠ Parallel to the plane, the sum of forces is (i) Mg sin θ − f = Ma The friction force produces a torque on the barrel about the center of mass. τ = Iα = fR ⎛a⎞ MR 2 ⎜ ⎟ Iα ⎝ R ⎠ = 1 Ma, which gives a = 2 f f = = R R M 2 This result is then substituted into equation (i) 1 2


13

CHAPTER

13-96.

13

⎛2f ⎞ f = Mg sin θ − Ma = Mg sin θ − M ⎜ ⎟ ⎝M ⎠ 1 1 = Mg sin θ = (200 kg)(9.81 m/s 2 )sin 40° = 420 N 3 3 a as the object falls with an acceleration The angular acceleration of the disk of radius R is α = R a. Let T represent the tension in the rope. The equations of motion for the disk and falling object are mg − T = ma (i)

τ = Iα =

1 ⎛a⎞ 1 MR 2 ⎜ ⎟ = MaR = TR 2 ⎝ R⎠ 2

1 Ma 2 Substituting equation (ii) into (i) gives 1 mg − Ma = ma 2 ⎛ ⎞ ⎜ ⎟ m a=⎜ ⎟g ⎜⎜ m + 1 M ⎟⎟ 2 ⎠ ⎝ T =

13-97.

13-98.

(ii)

Perpendicular to the plane, the sum of forces is N − Mg cos θ = 0. The normal force is N = Mg cos θ Parallel to the plane, the sum of forces is Mg sin θ − f = Ma The friction force produces a torque on the hoop about its center of mass. τ = Iα = fR ⎛a⎞ MR 2 ⎜ ⎟ Iα ⎝ R ⎠ = Ma, which gives a = f f = = R R M But the friction force is f = µN = µsMg cos θ and f a= = µs g cosθ = 8.50 µs M A static friction force acts at the area of contact between the wheels and road when the wheels roll without slipping. When the wheels are locked and sliding occurs, the friction force is kinetic friction. The normal force acting at each wheel is 0.25 mg. The friction force on the right wheels in this problem is fs = 0.25µsmg; and that acting on the left wheels is fk = 0.25µkmg. The net torque on the car is 1 τ = 2(0.25µs mg )d − 2(0.25µ k mg )d = mgd ( µs − µ k ) 2 1 = (1800 kg)(9.81 m/s 2 )(0.75 m)(0.90 − 0.50) = 2600 N i m 2 The positive value of the torque indicates that the car would spin toward the left (counterclockwise).


CHAPTER

13

The instantaneous angular acceleration is τ 2600 N i m α= = = 1.2 rad/s 2 2 I 2200kg i m 13-99.

MR22

2

⎛ 1.0 × 104 ⎞ = = = 2.0 × 10−10 ⎜⎜ 2 2 8⎟ ⎟ MR1 R1 ⎝ 7.0 × 10 ⎠ (b) The ratio of the angular velocities is given by the conservation of angular momentum, I1ω 1 = I 2ω 2

I (a) 2 = I1

2 5 2 5

R22

ω 2 I1 1 = = = 4.9 × 109 − 10 ω 1 I 2 2.0 × 10 and

ω 2 = ω 1 (4.9 × 109 ) = 1

⎞ ⎟ = 1860 rev/s s ⎟⎠ (a) The center of mass along the rod relative to its center is above its center a distance m 4l + M (0) m xcm = = l 4(m + M ) m+M

or 4.9 × 109

13-100.

rev rev (4.9 × 109 ) = 4.9 × 109 month month

rev ⎛ 12 months ⎞ ⎛ 1y ⎜ ⎟ ⎜⎜ 7 month ⎝ 1y ⎠ ⎝ 3.156 × 10

( )

(b) Apply conservation of linear momentum for the center of mass, mv = (m + M )vcm m v m+M (c) The parallel axis theorem is used to find the moment of inertia about the new center of mass. vcm =

I cm = =

1 ⎛l ⎞ 2 + m ⎜ − xcm ⎟ Ml 2 + Mxcm 12 ⎝4 ⎠

2

1 l 2 (m2 + M 2 ) Ml 2 + 12 16( m + M )

Conservation of angular momentum gives l mv = I cmω 4 mv mv ω= = 4 I cm ⎡1 l 2 (m 2 + M 2 ) ⎤ 4 ⎢ Ml 2 + ⎥ 16(m + M ) ⎦⎥ ⎣⎢12 Then, the angular momentum is 1 m2 v 2 L = Iω = mv = 4 4(m + M )


CHAPTER 13-101.

13-102.

13

Since the satellite is in a stable orbit around the Earth, the angular speed is GM ω = 3/ 2E r Then, the angular momentum is GM L = Iω = mr 2 3 / 2 E = m rGM E r ⎛ N i m2 ⎞ 5.98 × 1024 kg = 1.3 × 1014 kg i m 2 /s = (1000 kg) (4.22 × 107 m) ⎜ 6.673 × 10−11 2 ⎟ ⎜ ⎟ kg ⎠ ⎝ The direction is north. dL ω τ = = ω×L dt ⎛ ⎜ 2π rad = ω L sin β = ⎜ 7 ⎜⎜ 26000 y 3.1561×y10 ⎝

(

s

)

L sin β

⎞ ⎟ 33 2 ⎟ (5.9 × 10 kg i m /s)sin 23° ⎟⎟ ⎠

L

= 1.8 × 1022 kg i m 2 /s β


CHAPTER 14

STATICS AND ELASTICITY


Three forces act in the rope-bucket system shown in the drawing. The forces are the tension T in the rope, the force F the man exerts on the bucket, and the weight W of the bucket. The angle θ is found from the length of the rope, θ 20 m, and the distance from the wall, 2.0 m: θ = sin−1 (2.0/20) θ The sum of the horizontal forces acting on the bucket is T F − T sin θ = 0 F In the vertical direction, the sum is W T cos θ − W = 0 or T = (W/cos θ) Combining these two equations, we can solve for the magnitude of F in terms of the weight of the bucket. ⎛ W ⎞ F −⎜ ⎟ sin θ = 0 ⎝ cosθ ⎠ sin θ F =W = W tan θ = mg tan θ cosθ ⎡ ⎛ 2.0 ⎞ ⎤ = (600 kg)(9.8 m/s 2 ) tan ⎢sin −1 ⎜ ⎟ ⎥ = 590 N ⎝ 20 ⎠ ⎦ ⎣

14-2.

The drawing shows the meter stick with a pivot P chosen 0.1 m to be at the end where the downward force FB of the rear 0.4 m edge of the hand is applied. The force of gravity on the P stick w is applied at the center of mass. Since the system FB FF w is in static equilibrium, the sum of the torques is FF(0.1 m) − w(0.5 m) = 0 where the convention has been applied that torques that would produce counterclockwise motion are positive. The torque equation can be solved for the magnitude of the upward force of the front edge of the hand, FB. 0.5 m ⎛ 0.5 m ⎞ FF = w = ( 0.20 kg ) ( 9.81 m/s 2 ) ⎜ ⎟ = 9.8 N 0.1 m ⎝ 0.1 m ⎠ The sum of the forces in the vertical direction is FF − FB − w = 0 FB = FF − w = 9.8 N − ( 0.20 kg ) (9.81 m/s 2 ) = 7.9 N

14-3.

Let F1 represent the upward force of the left support on the bridge, F2 represent the upward force of the right support on the bridge. The trains have identical mass m = 90 000 kg and are located at 30 m and 80 m from the right support. The mass of the bridge is M = 900 000 kg. The sum of the torques on the bridge, using a pivot chosen at the right support is Mg (45 m) + mg (30 m) + mg (80 m) − F1 (90 m) = 0


CHAPTER

14

Mg (45 m) + mg (110 m) 90 m (900 000 kg)(9.81 m/s 2 )(45 m) + (90 000 kg)(9.81 m/s 2 )(110 m) = = 5.5 × 106 N 90 m The upward force of the right support is found by using the above result in the sum of forces equation. F1 + F2 − Mg − 2mg = 0 F1 =

F2 = ( M + 2m) g − F1 = (900 000 kg + 180 000 kg)(9.81 m/s 2 ) − 5.5 × 106 N = 5.1 × 106 N 14-4.

†14-5.

14-6.

The drawing shows the free-body diagram for the bridge with a slope. The equilibrium condition for rotation about the point P2 is Mg (45 m)sinθ + mg (30 m)sinθ − F1 (90 m)sinθ = 0

F1 90 m

θ

F2 P2

mg Mg Since the sin θ term appears in all three terms on the left side of the equation, we may divide all terms by sin θ. This indicates that the magnitudes of the support forces are not dependent on θ and the same results are found as in Example 1. Note that this solution of the problem assumes implicitly that the piers exert only vertical forces on the bridge, i.e., the joints between the piers and the bridge are designed to avoid any horizontal compression or tension on the bridge. Consider the drawing to the right. Initially the rope is taut between the car and the tree. The driver pushes on the rope with y F the force F shown at the mid-point and the tension in the rope is x T. Because the angle at the top of the triangle is given as 170°, T T 170° the two angles labeled θ are each 5° because the three angles of θ θ a triangle must add to 180°. Since the rope is in stable ca tree equilibrium, the sum of the three forces shown must be equal to zero newtons. The x components of the two tension forces are equal in magnitude, but oppositely directed, so Tx − Tx = (T cos θ) − (T cos θ) = 0 In the y direction, the sum of the forces is F − Ty − Ty = F − (T sin θ) − (T sin θ) = F − 2T sin θ = 0 Solving for T gives, 900 N F = = 5163 N = 5200 N T = 2sin θ 2 sin 5° As long as the rope doesn't break under this tension, this is a highly effective way of pulling the car out of the mud. The sum of the forces in the x direction is 40° 20° cos 40° T1 T2 cos 20° − T1 cos 40° = 0 which leads to T2 = T1 T2 y cos 20° x In the y direction, w T2 sin 20° + T1 sin 40° − w = 0 Now, use the information from the x component to solve for T1.


CHAPTER

cos 40° sin 20° + T1 sin 40° − w = 0 cos 20° ( 90 kg ) ( 9.81 m/s 2 ) w T1 = = = 960 N cos 40° tan 20° + sin 40° cos 40° tan 20° + sin 40° cos 40° cos 40° T2 = T1 = (960 N) = 780 N cos 20° cos 20° A man can lean backward, with his heel as the pivot point, only until his center of mass reaches the limiting point of being directly above his pivot point. Similarly, he can only lean forward until his center of mass passes over his toes. This assumes that he is completely stiff as he leans in either direction. For leaning backward, the horizontal distance from the center of mass to the heel is 26 − 18 cm = 8 cm. For leaning forward, the horizontal distance from the center of mass to the toes is given as 18 cm. F (a) Consider the force diagram shown to the right. The right end of N the log will only be lifted off the ground when the counterclockwise L torque due to the force F exceeds the clockwise torque due to the w weight w. The critical value of F is found by summing these torques. 2 L L FL − w = 0 2 1 F= w 2 Therefore, the right end will be raised when F F > w/2 = 0.5(50 kg)(9.81 m/s2) = 245 N F > 245 N (b) Now, in the torque equation, only the components of the forces in the direction perpendicular to the length of the log N θ w contribute to the torques. L FL cosθ − w cosθ = 0 2 1 F= w 2 So, regardless of the angle, the magnitude of the force is the same, F F = 245 N (c) In this case, the force is applied in the direction perpendicular to the length of the log, so for each case, the sum of the torques is L FL − w cosθ = 0 N θ w 2 1 F = w cosθ 2 For θ = 30°, F = (245 N) cos 30° = 212 N For θ = 60°, F = (245 N) cos 60° = 123 N For θ = 85°, F = (245 N) cos 85° = 21 N T1

14-7.

14-8.

14


CHAPTER †14-9.

14

Assuming the mass of the balance arm is neglected along with the masses of the pans, this problem may be solved by recognizing that the upward force F at the pivot does not contribute when considering the torques on the system. The moment arm for the weight of the small standard mass mg is 49 times longer than the moment arm of the weight of the bag of sugar, so we expect that the mass of the bag of sugar will be 49 times greater than the mass m. The sum of the torques is Mg (0.01 m) − mg (0.49 m) = 0

F mg Mg

M (0.01 m) = m(0.49 m) M = 49m = 49(0.12 kg) = 5.88 kg 14-10.

†14-11.

14-12.

The beam will tip when the man walks a distance x beyond the pivot such that the clockwise torque on the beam due to his weight exceeds the counterclockwise torque due to the weight of the beam. The critical distance is given by the sum of torques equation: Mg (0.50 m) − mgx = 0

M (0.50 m) = mx M (0.50 m) (50 kg)(0.50 m) x= = = 0.3 m m 80 kg The forces acting on the mast are illustrated in the drawing. The weight of the mast is to be neglected. The mast is pivoted at the bottom. Consider the torques acting on the mast. (T2 sin 45°)(10 m) − (T1 sin 30°)(10 m) = 0

N T2 45°

y x

T1 30°

10 m sin 30° sin 30° T2 = T1 = (5.0 × 103 N) = 3500 N sin 45° sin 45° The force that the mass exerts on the boat is equal in magnitude to the normal force the boat exerts on the mast. The normal force can be found by summing the forces in the y direction. N − T1 cos 30° − T2 cos 45° = 0 N = T1 cos 30° + T2 cos 45° = (5.0 × 103 N)cos 30° + (3.5 × 103 N)cos 45° = 6800 N In applying the torque equation for equilibrium, only the F components of the forces in the direction perpendicular to the length L of the sofa contribute to the torques. L FL cosθ − w cosθ = 0 2 θ w N 1 1 2 F = w = (45 kg)(9.81 m/s ) = 220 N 2 2 The result is independent of the angle θ. When there is one person on each end supporting the sofa 1.0 m above the ground. The sum of the forces in the vertical direction is F + F − w = 0 and each person is again supporting one-half of the weight of the sofa, 220 N. The load of each person tall or short is identical. The task is shared equally.


CHAPTER †14-13.

14-14.

In lifting the lid, you can apply the force F vertically as shown or you could imagine applying it at some other angle. To rotate the lid to an angle θ requires a torque to be applied with the lid held at one end at the hinges. The maximum torque occurs when the force F is applied in the direction perpendicular to the moment arm that is the lid. In this case, the sum of the torques acting on the lid is L FL − w cosθ = 0 2 1 F = w cosθ 2 For θ = 30°, F = (0.5)(12 kg)(9.81 m/s2) cos 30° = 51 N For θ = 60°, F = (0.5)(12 kg)(9.81 m/s2) cos 60° = 29 N Choose the pivot point to be the end of the pole where FL the right hand is applying an upward force. The sum of FR the torques acting on the pole is 1.5 FL(1.5 m) − mg(2.25 m) = 0 from which

14 F

θ

N

w

w

2.25

mg (2.25 m) (3.0 kg)(9.81 m/s 2 )(2.25 m) = = 44 N 1.5 m 1.5 m The sum of the forces in the vertical direction is FL − FR − w = 0 FR = FL − w = 44 N − (3.0 kg) (9.81 m/s2) = 14.6 N = 15 N Choose the pivot location to be 1.0 m from the bolted end of the board where the upward support force Fup is applied. Let FB represent the downward force at the bolted end. The sum of the torques acting on the board is FB(1.0 m) − wboard(0.5 m) − wdiver(2.0 m) = 0 This equation may be then solved for FB. w (0.5 m) + wdiver (2.0 m) FB = board 1.0 m (50 kg)(9.81 m/s 2 )(0.5 m) + (60 kg)(9.81 m/s 2 )(2.0 m) = = 1420 N 1.0 m The sum of the forces in the vertical direction is Fup − FB − wboard − wdiver = 0 and Fup = FB + wboard + wdiver = 1420 N + (50 kg)(9.81 m/s 2 ) + (60 kg)(9.81 m/s 2 ) = 2500 N Choose the pivot point P at the right end of the rod where 12 m the support cable exerts a tension force T2. The sum of the torques acting on the rod is T1 T2 wW (10 m) + wR (6 m) − T1 (12 m) = 0 P ww where ww is the weight of the washer and wR is the weight of wR 6m the scaffold rod. The tension in the left cable is found by 10 m solving the above equation. FL =

†14-15.

14-16.


CHAPTER 14 wW (10 m) + wR (6 m) 12 m (90 kg)(9.81 m/s 2 )(10 m) + (110 kg)(9.81 m/s 2 )(6 m) = = 1275 N 12 m Therefore, T1 = 1280 N The tension in the other cable is found by summing the forces in the vertical direction. T1 + T2 − ww − wR = 0 T1 =

14-17.

14-18.

T2 = ww + wR − T1 = (90 kg)(9.81 m/s 2 ) + (110 kg)(9.81 m/s 2 ) − 1275 N = 687 N The pencil is assumed to not slide down the plane as the angle θ is increased. The pencil can rotate about an axis that is parallel θ to its length and located at the lower edge in contact with the N plane. The free-body diagram shows the situation when θ = 30°, 30° the angle at which the weight vector passes through the leading f edge in contact with the inclined plane. This is found from the geometry of a hexagon, which has six equilateral triangles w cos 30° w formed between the sides and the center. For angles smaller than 30°, the forces perpendicular to the plane, N and w cos θ, w sin 30° are equal in magnitude and oppositely directed. In the direction parallel to the plane, the friction force is equal to the weight component w sin θ. When θ > 30°, the weight produces a clockwise torque that causes the pencil to begin rolling. Example 4 provides a basis for understanding this problem. The forces acting on the ladder are the horizontally directed normal force of the wall on the ladder at the top N2, the downward force of the ladder’s weight acting at its center wl, the downward force of the painter’s weight wp acting at some distance x from the end of the ladder in contact with the ground, the upward normal force N1 of the ground, and the friction force f that is directed toward the wall. The ladder will slip when the static friction force reaches its maximum value, f = µN1. Choose a pivot point at the contact point with the ground. The weight of the ladder and the weight of the painter exert counterclockwise torques, while N2 exerts a clockwise torque. The sum of the torques is wl (2.5 m)sin 30° + wp x sin 30° − N 2 (5 m)cos 30° = 0 (i) The sum of the forces in the vertical direction is N1 − wp − wl = 0 which gives N1 = wp + wl The sum of the forces in the horizontal direction is N2 − f = 0 which gives N 2 = f = µ N1 Substituting the results of equations (ii) and (iii) into equation (i) allows one to solve for x. wl (2.5 m)sin 30° + wp x sin 30° − µ N1 (5 m)cos 30° = 0 wl (2.5 m)sin 30° + wp x sin 30° − µ ( wp + wl )(5 m)cos 30° = 0


(ii)

(iii)

CHAPTER 14

x= =

µ (5 m)cos 30°( wp + wl ) − wl (2.5 m)sin 30° wp sin 30°

µ (5 m)cos 30°(mp + ml ) − ml (2.5 m)sin 30° mp sin 30°

(dividing through by g )

(0.35)(5 m)cos 30°(60 kg + 10 kg) − (10 kg)(2.5 m)sin 30° = 3.1 m (60 kg) sin 30° The weight w of the locomotive is equal to the sum of all the downward forces of the wheels on the track. w = 2(74 kN) + 4(117 kN) + 109 kN + 160 kN + 153 kN = 1.038 × 106 N The downward forces of locomotive at the wheels on the track are equal in magnitude, but oppositely directed to the normal forces of the track on the locomotive. The sum of the torques acting on the train must equal zero N • m, if the train is in rotational equilibrium. Let x be the distance of the center of mass from the front wheel of the locomotive. The sum of the torques with the front wheel as the pivot point is equal to the weight of the locomotive times the moment arm x. wx = (74 kN)(17.1 m) + (74 kN)(15.4 m) + (109 kN)(13.0 m) + (160 kN)(10.7 m) + (153 kN)(8.4 m) + (117 kN)(4.8 m) + (117 kN)(3.1 m) + (117 kN)(1.7 m) = 7942 kN • m x = (7.942 × 106 N • m)/(1.038 × 106 N) = 7.65 m Consider the free-body diagram shown. The hinges must provide both FV F B 1.0 m upward forces FV to balance the downward force of the weight and horizontal forces FH to counter to clockwise torque due to the weight. FH These horizontal and vertical forces are the components of forces FA and FB at the lower and upper hinges, respectively. Each hinge bear one-half of the weight. FV = 0.5w = 0.5(18 kg)(9.81 m/s2) = 88 N Using hinge A as the pivot point, the sum of the torques is w FH(2.0 m) − w(0.5 m) = 0 and the magnitude of horizontal components of the forces at the hinges is FV w(0.5 m) (18 kg)(0.5 m) FA FH = = = 44 N (2.0 m) (2.0 m) =

†14-19.

14-20.

2.0 m

The components can then be used to determine the magnitude and direction of the hinges forces. FA = FB =

( 88 N )

2

FH

+ ( 44 N ) = 98 N 2

FV 88 N = tan −1 = 63° 44 N FH FA = 98 N at 63° relative to the + x axis and FB = 98 N at 117° relative to the + x axis

θ = tan −1


CHAPTER 14 †14-21.

14-22.

The forces your fingers apply to the box are the friction force f1 and a normal force N1; and your thumb exerts similar forces as shown. The weight of the box is w. If the box is being held in equilibrium, then the sum of the forces in the vertical direction is f2 − N1 − w = 0 which gives f2 = N1 + w (i) In the horizontal direction, (ii) f1 − N2 = 0 which gives f1 = N2 From the relationship between the friction force and the normal force, we have (iii) f1 = µN1 = N2 Substituting the result from (iii) into (i) gives N1 + w = f2 = µN2 = µ2N1 from which µ can be determined. N1 + w µ= N1

f1 N1

f2 N2

w

Since the weight is always greater than zero N, µ will always be greater than 1. In the limit of the weight going to zero N, µ goes to 1. FH From the diagram, the sum of the torques about the pivot point where the nail is driven is FH (0.25 m) cos 30° – W(0.25 m) sin 30° = 0 FV Solving for the horizontal force gives 0.25 m FH (0.40 kg)(9.81 m/s 2 )(0.25 m) FH = tan 30° = 2.3 N (0.25 m)

w

0.50 m †14-23.

Consider the drawing of the situation. The thin line drawn from the center of the wheel to the pivot point P represents the moment arm, which has a F length equal to the radius of the wheel, R. The horizontal and vertical w R components of the normal force at the pivot point are not shown to avoid a h P cluttered drawing. Also, the wheel as it begins to rotate about the pivot point will lose contact with the ground; and it isn’t necessary to consider the contact force at the point of contact with the ground. For the wheel to begin to roll over the step, the clockwise torque due to F must, however, just exceed the counterclockwise torque due to the weight. Let θ represent the angle between the horizontally applied force F and the moment arm. The critical value of this force may be found by summing these torques about P. wR cos θ − FR sin θ = 0 From the drawing, the sin θ and cos θ terms can be written in terms R 2 − ( R − h) 2 of the radius of the wheel and the height of the step, h. Therefore, θ we can solve for F. R−h

R

R 2 − ( R − h) 2 w Mg F= = = Mg tan θ tan θ ( R − h)


CHAPTER

14

The smallest force that one could apply would be the one applied perpendicular to the moment arm. In that case, the sum of torques is wR cos θ − FR = 0 and w Mg R = = Mg F= 2 cosθ cosθ R − ( R − h) 2

14-24.

Consider the point P located a distance z above the end of the cable. The tension at point P is equal in magnitude to the weight of the material below that point. Therefore, the tension is ⎛ πd2 ⎞ ρz⎟g T = ⎜M + 4 ⎝ ⎠ where the mass of the cable below P is found by multiplying the density and the volume of the cable.

d

T P F

z

M †14-25.

(a) The base of the mast is hinged, so that will be the pivot point. The ropes make an angle θ with respect to the mast. We can find θ from the triangle formed by the rope, the mast, and the base. m θ = tan −1 ( 1.35 = 7.7° 10 m )

The sum of the torques on the mast is F(5.0 m) + T2(sin θ)(10 m) − T1(sin θ)(10 m) = 0 F (5.0 m) (2400 N)(5.0 m) The excess tension is T1 − T2 = = = 9.0 × 103 N (10 m)sin θ (10 m)sin 7.7°

(b) When spreaders are used, the ropes make a new angle with the mast: m θ = tan −1 ( 1.35 = 28° 2.5 m ) and the sum of the torques is now, F(5.0 m) + T2(sin θ)(10 m) − T1(sin θ)(10 m) = 0 The excess tension in this case is F (5.0 m) (2400 N)(5.0 m) = = 2.6 × 103 N T1 − T2 = (10 m)sin θ (10 m)sin 28° 14-26.

Since the excess tension is reduced by using the spreader, this is the preferred arrangement. Two consider the sum for forces acting on the ball, it is convenient to use a coordinate system that is rotated to match the 30° groove. The sum of forces in the x direction is y x NL − w sin 30° = 0 NL This gives the normal force on the left side of the groove. N w R NL = w sin 30° = mg sin 30° = (10 kg)(9.81 m/s2) sin 30° = 49 N Similarly, the normal force on the right side of the groove is NR = w cos 30° = mg cos 30° = (10 kg)(9.81 m/s2) cos 30° = 85 N


CHAPTER †14-27.

14-28.

14

The legs form an equilateral triangle at the ground. Each side of C the triangle has length L, which is also the length of the legs. To determine the tension in each leg, the angle θ between line AB w T and line AC must be determined. For an equilateral triangle, the T distance AB from a vertex to the center of the triangle is given by A θ B T 2 AB = L cos30° 3 So, the angle θ may be found using AB 23 L cos30° 2 cosθ = = = cos30° AC L 3 2 ⎛ ⎞ θ = cos −1 ⎜ cos 30° ⎟ = 54.7° ⎝3 ⎠ For static equilibrium, the sum of the forces in the vertical direction must be equal to zero newtons. 3T(sin 54.7°) − Mg = 0 Therefore, the tension in each leg is Mg T = = 0.408Mg 3sin 54.7° We are given that the length of the rope is a little longer than 3d/4 d/4 the distance between the two ships, 1.2d. Applying the θ1 θ2 x Pythagorean theorem to the two right triangles in the diagram L2 L1 gives: 2 ⎛3 ⎞ 2 2 L1 − x = ⎜ d ⎟ ⎝4 ⎠ 2

⎛1 ⎞ L22 − x 2 = ⎜ d ⎟ ⎝4 ⎠ Subtracting (ii) from (i) gives 1 L12 − L22 = d 2 2 1 ( L1 − L2 )( L1 + L2 ) = d 2 2 2 d d2 d L1 − L2 = = = 2( L1 + L2 ) 2(1.2d ) 2.4

(i)

and L1 + L2 = 1.2d

(ii)

From (i) and (ii), the lengths are L1 = 0.81d and L2 = 0.39d. Knowing these lengths, allows the determination of the angles. ⎛ 3d ⎞ ⎛ 3d ⎞ T1 T1 −1 θ1 = cos −1 ⎜ ⎟⎟ = 22.2° ⎟ = cos ⎜⎜ θ2 θ1 ⎝ 4 L1 ⎠ ⎝ 4 ( 0.81d ) ⎠ ⎛ ⎞ ⎛ d ⎞ d −1 θ 2 = cos ⎜ ⎟⎟ = 50.1° ⎟ = cos ⎜⎜ ⎝ 4 L2 ⎠ ⎝ 4 ( 0.39d ) ⎠ −1


w

T2

CHAPTER The free-body diagram gives, (T1 + T2 ) cosθ 2 − T1 cosθ1 = 0

(iii)

(T1 + T2 )sin θ 2 + T1 sin θ1 − mg = 0

(iv)

14

From equation (iii), T ( cosθ1 − cosθ 2 ) T1 ( cos 22.2° − cos50.1° ) T2 = 1 = = 0.443T1 cosθ 2 cos50.1° Substituting this into equation (iv) and solving for T1 gives (T1 + 0.443T1 )sin θ 2 + T1 sin θ1 = mg T1 (1.443sin 50.1° + sin 22.2°) = mg mg = 0.673 mg 1.443sin 50.1° + sin 22.2° T2 = 0.298mg = 0.3mg T1 = †14-29.

The sum of the forces acting on the disk in the vertical direction is T cos θ − Mg = 0 (i) and in the horizontal direction, it is N − T sin θ = 0 (ii) To find the angle θ, the vertical distance h from the point of contact of the disk and the wall to the point where the string is attached to the wall is found using the Pythagorean theorem. h2 + R 2 = ( L + R)2

L

θ N

h = ( L + R)2 − R 2

T R w

( L + R)2 − R 2 R and sin θ = L+R L+R From equation (i), the tension is found. Then, cosθ =

( L + R)2 − R 2 − Mg = 0 L+R ⎛ ⎞ L+R ⎟ T = Mg ⎜ ⎜ ( L + R) 2 − R 2 ⎟ ⎝ ⎠

T

From equation (ii), the normal force may then be found. ⎛ ⎞ R L+R ⎟ =0 N − Mg ⎜ ⎜ ( L + R) 2 − R 2 ⎟ L + R ⎝ ⎠ MgR N= ( L + R)2 − R 2


CHAPTER 14-30.

14

From the drawing, we can write down three relations. 3.75 w = T1 sin 10° − T2 sin θ (i) 10° T w = 2T2 sin θ (ii) 1 (iii) T2 cos θ = T1 cos 10° T2 θ From equation (ii), w w T2 = 2sin θ which may be substituted into equation (i) to give the tension T1. w T1 sin10° − sin θ = w 2sin θ 3 3 w (20 kg)(9.81 m/s 2 ) T1 = 2 = 2 = 1700 N sin10° sin10° Using equations (ii) and (iii), the angle θ may be determined. w cosθ = T1 cos10° 2sin θ 2 ⎛ ⎞ w −1 ⎛ (20 kg)(9.81 m/s ) ⎞ θ = tan −1 ⎜ tan = ⎟ ⎜ ⎟ = 6.7° ⎝ (1700 N) cos10° ⎠ ⎝ T1 cos10° ⎠

15 m 10°

θ w

w

Now, equation (iii) is used to find the tension T2. mg (20 kg)(9.81 m/s 2 ) T2 = = = 840 N 2sin θ 2(sin 6.7°) The left and right traffic lamps are a distance h1 below the dashed line, where h1 is given by h1 = (3.75 m)(sin 10°) = 0.65 m Similarly, the middle traffic lamp is a distance h2 below the dashed line, where h2 = h1 + (3.75 m)(sin 6.7°) = 1.09 m 14-31.

Consider the forces shown in drawing in figure 14.9. The addition of a vertically downward force F at the top of the ladder introduces an additional counterclockwise torque. The sum of the torques of the ladder becomes 1 mglsin θ + Flsin θ − N 2 l cosθ = 0 (i) 2 The sum of the forces in the vertical direction is N1 − F − mg = 0 which gives N1 = F + mg

(ii)

The sum of the forces in the horizontal direction is N2 − f = 0 which gives N 2 = f = µ N1 Substituting the results of equations (ii) and (iii) into equation (i) allows one to solve for F. 1 mgl sin θ + Fl sin θ − µ N1l cosθ = 0 2 1 2

mgl sin θ + Fl sin θ − µ ( F + mg )l cosθ = 0

1 2

mg tan θ + F tan θ − µ ( F + mg ) = 0

(dividing through by l cosθ )

⎡ ( µ − 12 tan θ ) ⎤ F = mg ⎢ ⎥ ⎣ tan θ − µ ⎦


(iii)

CHAPTER

14-32.

14

Any force greater than this value will lead to instability and the ladder will slip. With θ = 30° and µ = 0.40, the value of the force F is ⎡ ( µ − 12 tan θ ) ⎤ ⎡ ( 0.40 − 12 tan 30° ) ⎤ F = mg ⎢ ⎥ = mg ⎢ ⎥ = 0.62 mg ⎣ tan θ − µ ⎦ ⎣ tan 30° − 0.40 ⎦ The weight of the automobile is 4(3100 N) = 12 400 N and θ = tan−1 (3/10) = 16.7° The normal forces shown represent the forces of the ground at either the left two tires (N1) or the right two tires (N2). The static friction force at each wheel is f. Using the convenient coordinate system shown, the sum of the forces y N1 in the y direction is N2 2N1 + 2N2 − mg cos θ = 0 (i) x f1 Choose the point of contact of the lower wheel as the pivot point. To 3 w f2 θ determine the moment arm for the weight, consider the diagram to the 10 right. The length R of the moment arm is R = (0.65 m)2 + (1.50 m)2 = 1.63 m The component of the weight that produces the torque is mg cos (φ + θ) where the angle φ is given by ⎛ 0.65 m ⎞ φ = tan −1 ⎜ ⎟ = 23.4° ⎝ 1.50 m ⎠ The sum of the torques about the pivot point is (ii) mgR cos (φ + θ) − 2lN1 = 0 which gives the normal force at the upper wheels. mgR cos(φ + θ ) (12400 N)(1.63 m) cos(23.4° + 16.7°) = = 2580 N N1 = 2l 2(3.0 m)

†14-33.

0.65 m

φ R 1.5 m

Substituting this result into equation (i) gives the normal force at the lower wheels. 1 1 N 2 = (mg cosθ − 2 N1 ) = (12400 N) cos 16.7° − 2(2580 N) = 3360 N 2 2 (a) The box tilted upward has a pivot axis that is the remaining bottom edge in contact with the floor. The center of mass moves along the arc of a circle of radius R = (0.30 m) 2 + (0.60 m) 2 = 0.67 m The angle between the bottom of the box and the radial line between a pivot point and the center of mass is θ 0 as the box is tilted at an angle θ with respect to the floor. ⎛ 0.60 m ⎞ θ 0 = tan −1 ⎜ ⎟ = 63.4° ⎝ 0.30 m ⎠ The height of the center of mass of the box is h = R [sin(θ 0 + θ) − sin θ 0] and the potential energy is U = mgh = mgR [sin(θ 0 + θ) − sin θ 0] = (80 kg)(9.81 m/s2)(0.67 m)[ sin(63.4° + θ) − sin(63.4°) ] = 526 J [ sin(63.4° + θ) − sin(63.4°) ] The following figure is a graph of the potential energy as a function of θ.


CHAPTER

14

60 50

U (J)

40 30 20 10 0

0

10

20

30

40

50

θ (degrees)

14-34.

14-35.

(b) The critical angle corresponds to the maximum of U (zero torque). This maximum is determined by sin (θ + 63.4°) = 1 or θ = 26.6° (c) The amount of work done on the box is equal to the potential energy at the maximum, assuming the potential energy was zero J when the box was at rest with the bottom completely in contact with the floor. U = 526 J [ sin(90.0°) − sin(63.4°) ] = 56 J Since the meter stick is suspended by a string, a force F applied to the T bottom of the stick will cause the pivot point P to move to the right as φ shown. The tension in the string is directed along an angle φ with P respect to the vertical direction. From the sum of forces in each 0.20 m direction, θ T cos φ = Mg 1 w T sin φ = Mg 2 F 0.50 m Taking the ratio of these two equations gives φ. ⎛1⎞ φ = tan −1 ⎜ ⎟ = 26.6° ⎝ 2⎠ Now, consider the sum of the torques about the point P. 1 Mg (0.30 m)sin θ − Mg (0.80 m)cosθ = 0 2 Solving for θ gives ⎛ 0.40 m ⎞ θ = tan −1 ⎜ ⎟ = 53.1° ⎝ 0.30 m ⎠ Starting with the top book, the book placed below it must have its right-most edge directly below the center of mass of the top book. The center of mass of the top two books must then line up with the right most edge of the third book, and so on. The optimum arrangement has the right edge of the top book a distance of L/2 from the right edge of the second book, the right edge of


CHAPTER

14-36.

†14-37.

14-38.

14

the second book a distance of L/4 from the right edge of the third book, the right edge of the third book a distance of L/6 from the right edge of the fourth book, and so on. For five books, the maximum protrusion is given by L L L L L⎛ 1 1 1⎞ + + + = ⎜ 1 + + + ⎟ = 1.04 L 2 4 6 8 2⎝ 2 3 4⎠ For an infinite number of books, the sum of series gives N dx L⎛ 1 1 1 1 L L ⎞ L ∞ 1 1 … + + + + + = lim log N = ∞ ⎜ ⎟ = ∑ = Nlim ∫ 2⎝ 2 3 4 5 2 →∞ 1 x 2 N →∞ ⎠ 2 1 N Suppose you push the box at a point a distance x above the floor as L shown such that the box is about to topple instead of slide. If the force F is barely sufficient to initiate the rotational motion, we F have zero torque about the point P, only the lower right edge of the box remains in contact with the floor. This condition of zero torque is x w L LMg Mg − xF = 0 from which x = 2 2F N The equilibrium condition for the horizontal forces is F = f. Since P f the static friction force is f < µsMg, LMg LMg L 0.5 m > = = = 0.31 m x= 2f 2 µs Mg 2µs 2(0.80) and the force required to initiate sliding is F = fmax = µsMg = (0.80)(75 kg)(9.81 m/s2) = 590 N When the car is just about to topple, its normal force is acting at the edge of the outer wheel. The frictional force on the car is the centripetal force F = mv2/R. For the car to balance, the torque about the center of mass must equal zero, that is, 1 Fh = Nl 2 and to balance the vertical forces on the car, N = mg. mv 2 1 1 Fh = h = Nl = mgl R 2 2 Rgl (25 m)(9.81 m/s 2 )(1.5 m) = = 17.5 m/s which gives v = 2h 2(0.60 m) While the car is at rest, the weight is equally distributed at each of the four wheels. The distance between the wheels is L. Let x represent the fraction of the car’s weight, w, on the front wheels as the car accelerates. Choose the center of mass, at height h, as the pivot point in order to calculate the sum of torques. The acceleration is due to the total friction force acting at the wheels, f = ma. 1 1 L fh + mgL = (1 − x)mgL 2 2 1 fh + mgxL = mgL 2 1 1 1 ah h x= ( 2 mgL − mah ) = − 2 gL mgL f f =

1 (6.0 m/s 2 )(0.6 m) − = 0.38 2 (9.81 m/s 2 )(3.0 m)

N1


w

N2

CHAPTER

†14-39.

14-40.

†14-41.

14

Thus, 38% of the weight is distributed on the front wheels and 62% is on the rear wheels. The bicycle is in stable equilibrium as long as the sum of the forces and torques acting on it are zero. However, it is ready to tip when the normal force on the back wheel Nback goes to zero newtons. The friction force is applied tangentially to the rim of the wheel at the radius R of the wheel. This friction force is equal in magnitude to the friction force of the road on the bicycle, which is f = µNfront = ma If the sum of the torques about the center of mass on the bicycle is zero, then Nfront(d sin θ) = f(d cos θ) where d is the distance from the contact point to the center of mass. So, the maximum braking force that can be applied is f = N front tan θ = mg tan θ = ma

a

d

R

w f

Nback

Nfront

⎛ 0.70 m ⎞ 2 Therefore, a = g tan θ = 9.81 m/s 2 ⎜ ⎟ = 7.2 m/s ⎝ 0.95 m ⎠ since the center of mass is located 0.95 m above the road and 0.70 m behind the point of contact of the front wheel, The bicycle is rounding the curve at 20.0 km/h = 5.56 m/s. If the bicycle is to be rotational equilibrium, the sum of torques is zero N • m about the center of mass. The friction force of the wheels on the road provides the centripetal force. mv 2 f = R The sum of vertical forces gives, N = mg. From the diagram, N(h cos α) = f(h sin α) mv 2 cos α f = = mg = mg cot α R sin α ⎡ ⎤ v2 (5.56 m/s) 2 −1 = cot −1 ⎢ α = cot −1 ⎥ = cot (0.5252) = 64.8° 2 gR (9.81 m/s )(6.0 m) ⎣ ⎦ (a) Let N1 and N2 represent the normal forces at the front and rear wheels, respectively. The weight is assumed to be distributed at the four wheels, so N1 + N2 = mg (i) The friction forces are acting at both the front and rear wheels to accelerate the car. Applying Newton’s second law of motion gives, (ii) f = ma = µ N1 + µN2 = µ ( N1 + N2) = µmg 2 2 a = µg = (0.90)(9.81 m/s ) = 8.8 m/s


CHAPTER

14

(b) Consider the sum of torques acting on the car using the center of mass as the pivot point. (1.5 m)N1 + (0.6 m)(0.9)N2 − (1.5 m)N2 = 0 This gives a relationship between N1 and N2. N2 = 1.56N1 = 1.56(mg − N2) where the relationship from equation (i) has been used to eliminate N1 to find N2. N2 = 0.61mg Apply Newton’s second law to find the acceleration in this case. f = ma = µN2 = µ(0.61mg) = (0.90) (0.61)mg = 0.55mg a = 0.55g = (0.55)(9.81 m/s2) = 5.4 m/s2

N1

f

w

N2

(c) When the braking force is only on the rear wheels, the situation is similar to that in part (b). The sum of torques about the center of mass is

14-42.

(0.6 m)(0.9)N1 + (1.5 m)N1 − (1.5 m)N2 = 0 N1 = 0.75N2 = 0.75(mg − N1) = 0.43mg Apply Newton’s second law to find the acceleration in this case. f = ma = µN1 = µ(0.43mg) = (0.90) (0.43)mg = 0.0.39mg a = 0.39g = (0.39)(9.81 m/s2) = 3.8 m/s2 Each of the steel beams has a mass m = M/4. The situation is symmetrical about a line that is parallel to the vertical cable from the crane and extending through the midpoint of beam AB. So, for example, the forces at pin A will be identical in magnitude to those at pin B. Likewise, those at pin C will be identical in magnitude to those at pin D. Therefore, at pin A 1 TAD − mg = 0 2 1 1 TAD = mg = Mg = TBC (tension) 2 8

E

D mg

mg

A

B

2mg Mg = = 0.577Mg (tension) sin 60° 2(sin 60°)

Finally, TDC + TDE cos 60° = 0 TDC = − TDE cos 60° = − (0.577Mg ) cos 60° = − 0.289Mg (compression)


C mg

where the notation is, for example, TAD is the tension in the left rod that is directed upward from pin A toward pin D. To reduce clutter on the drawing, the forces have not been labeled. At pins C and D, TDE sin 60° − 2mg = 0 TDE = TCE =

60°

mg

CHAPTER †14-43.

14

The bottom ball has two normal forces acting on it; and the top ball also has two normal forces acting on it as shown in the drawing. The sum of the horizontal forces is FA − FB = 0 which gives FA = FB For the vertical forces, FC − 2w = FC − 2mg = 0 which gives FC = 2mg Using the center of mass of the lower ball as a pivot point, the sum of the torques is mgR − 3RFA = 0

FA = FB = 14-44.

R R R FA

w

FD FD

3R

FC

mg 3

Consider the drawing of the situation. The sum of the horizontal forces gives, N = T sin θ (i) The sum of the vertical forces gives, mg = T cos θ (ii) Combining these two equations, N tan θ = (iii) mg

θ N

3 2

Given the relationship in equation (iv), 2d d = 2 2 2 9 2 L −d 4 L −d d 7 = L 12 From the drawing, d 7 7 = 49.8° sin α = = gives α = sin −1 L 12 12 The forces acting on the cards are shown in the free-body diagram. The frictional force at the top where the cards are in contact is negligible. Let θ represent the angle that the cards, of length L, make with respect to the vertical direction. The sum of the horizontal and vertical forces acting on one card is:


L

P

α

φ

L

The sum of the torques about the point P is 1 mgL sin α + T sin θ ( L cos α ) − T cosθ ( L sin α ) = 0 2 Substituting the results of equations (i), (ii), and (iii) gives 1 mg sin α + N cos α − mg sin α = 0 2 2N = 2 tan θ (iv) tan α = mg d d and tan θ = From the drawing, tan α = 2 2 2 2 9 L −d 4 L −d

†14-45.

FB

w

T

mg

F1

N

θ w F2

F1 w F2

N

CHAPTER

14

F2 − F1 = 0 or F1 = F2 and N − mg = 0 or N = mg For the maximum static friction force, F2 = µs N = µs mg = F1 , at the maximum angle. The sum or torques about the edge of the card in contact with the floor is 1 mgL sin θ max − F1 L cosθ max = 0 2 1 mgL sin θ max − µs mgL cosθ max = 0 2 1 sin θ max − µs cosθ max = 0 2 tan θ max = 2 µs and θ max = tan −1 ( 2 µs ) 14-46.

Let the mass on one side be m1 and the tension in that side of the rope is T1. Similarly, let the mass on the other side be m2 and the tension in that side of the rope is T2. (i) m1g − T1 = m1a N T2 − m2g = m2a (ii) The tension force is complicated in this problem by the dθ T′ T dθ fact that the rope is in contact with a rough surface. 2 dθ Assume that the tree branch has a circular cross section and that the friction force is at its maximum static value, fs T2 T1 = µsN. The rope makes contact over the semi-circle. Consider the infinitesimal section of rope shown in the force diagram. The magnitude of the vertical component of each of the two tension forces T and T′ is equal to dθ dθ T sin =T (small angle approximation) 2 2 The equilibrium of the radial forces requires that dθ N − 2T = 0 or N = Tdθ 2 The friction force on the section is then given by fs = µsN = µsTd θ The sum of the tangential forces on the section gives dT = µsTd θ or dT = µs dθ T Then, the total tension along the portion of rope in contact with the tree is found through integration. T2 dT θ2 = µs dθ T1 T θ1 ln T2 − ln T1 = µs (θ 2 − θ1 )

∫

or

∫

θ 2 − θ1 =

1

µs

ln

T2 T1

If we take the exponential of both sides of the equation above and rearrange, we get


CHAPTER

14

T2 = e µs (θ 2 −θ1 ) T1 For the problem at hand, θ 2 − θ 1 = π, so T2 = T1e µsπ . Substituting this result into equation (ii) above gives T1e µsπ − m2 g = m2 a m1 ( g − a)e µsπ − m2 g = m2 a a= †14-47.

(the result from equation (i) has been substituted)

µ sπ

m1e − m2 g m1e µsπ + m2

The tension around each semi-circular portion of each flywheel is determined by the total friction force of the belt in contact with the flywheel. The friction force is at its maximum static value, fs = µsN. Consider the infinitesimal section of the belt shown in the force diagram.

T1

T′ dθ N T dθ

dθ T2

2

The magnitude of the horizontal component of each of the two tension forces T and T′ is equal to dθ dθ T sin =T (small angle approximation) 2 2 The equilibrium of the radial forces requires that dθ N − 2T = 0 or N = Tdθ 2 The friction force on the section is then given by fs = µsN = µsTd θ The sum of the tangential forces on the section gives dT = µsTd θ or dT = µs dθ T Then, the total tension along the portion of belt in contact with the flywheel is found through integration. T2 dT θ2 = µs dθ T1 T θ1 ln T2 − ln T1 = µs (θ 2 − θ1 )

∫

or

∫

θ 2 − θ1 =

1

µs

ln

T2 T1

If we take the exponential of both sides of the equation above and rearrange, we get T2 = T1e µs (θ2 −θ1 ) θ 2 − θ 1 = π, so T2 = T1e µsπ . The torque on the flywheel will be τ = R(T2 − T1). Therefore, the tensions may be written in terms of the torque. τ⎛ 1 ⎞ τ⎛ 1 ⎞ T1 = ⎜ µsπ ⎟ and T2 = ⎜ ⎟ R ⎝ e −1⎠ R ⎝ 1 − e − µ sπ ⎠


CHAPTER 14-48.

14

(a) The tension along the belt is determined in similar manner as in problems 46 and 47 above. dT = − µ k dθ (i) T where the minus sign is used because the tension decreases as θ increases in this situation. Integration of equation (i) gives dT ∫ T = µ k ∫ dθ ln T = µ kθ + C where C is a constant or

T = Ce − µkθ

Since T = w for θ = 0, the constant C must equal w. Therefore, T = we − µkθ

(b) The tension in the lower, straight portion of the belt is equal to w. This tension exerts a torque τl = wR on the flywheel. The tension in the upper portion of the belt is T = we − µ (2π ) . It exerts a torque of τ u = wRe −2 µ π . The total torque is τ = τl + τu = wR (1 − e −2 µ π ) k

k

k

(c) The power is the product of the torque and the angular velocity. P = wRω (1 − e −2 µ π ) k

†14-49.

14-50.

The arm is being held in static equilibrium, so the sum of F1 F2 the torques about the elbow (P) is equal to zero. (0.055 m)F2 − (0.355 m)mg = 0 P Solving for the force F2, 30.0 5.5 cm (0.355 m)(25 kg)(9.81 m/s 2 ) F2 = = 1583 N = 1580 N 0.055 m The sum of the vertical forces gives F2 = F1 + w So, F1 = F2 − w = 1583 N − (25 kg)(9.81 m/s2) = 1338 N = 1340 N This problem is a repeat of the previous problem, except F1 F2 45 that the force F2 is now directed at the angle shown in F H the drawing. The sum of torques about the elbow, P, is P (0.055 m)F2 cos 45° − (0.355 m)mg = 0 30.0 5.5 cm Solving for the force F2, (0.355 m)(25 kg)(9.81 m/s 2 ) F2 = = 2239 N = 2240 N (0.055 m)cos 45° The sum of the vertical forces gives F2 sin 45° = F1 + w So, F1 = F2 sin 45° − w = (2239 N)(sin 45°) − (25 kg)(9.81 m/s2) = 1338 N = 1340 N The horizontal force at P is FH = F2 cos 45° = 2239 N cos 45° = 1580 N


w

w

CHAPTER †14-51.

14

The sum of the torques about the center of the cylinder gives F(0.25m) − T(0.040 m) = 0 (0.040 m)(2500 N) F= = 400 N (0.25 m)

F

0.25 T

0.040 14-52.

14-53.

14-54.

14-55.

The contact area the crowbar makes with the ground is the pivot point. The sum of torques about that point gives (75 kg)(9.81 m/s2)(0.60 m) = F(0.040 m) F = 11 000 N The sum of torques about the hinge gives (60 kg)(9.81 m/s2)(0.80 m) = F(2.0 m) F = 240 N (a) The sum of torques about the hinge gives (30 N)(0.040 m) = F(0.12 m) F = 10 N (b) (30 N)(0.040 m) = F(0.010 m) F = 120 N The innermost pan is a horizontal distance x from the fulcrum. The sum of torques about the fulcrum gives m1 gx = m2 g (10 x)

m1 = 10m2 1 m1 10 The maximum value of m1 is 2.0000 milligrams, so the maximum value of m2 is 0.20000

m2 =

14-56.

14-57.

milligrams and we expect the resolution to also increase by a factor of 10 as well from 0.10 micrograms to 0.010 micrograms. See Figure 14.16. If the upward force at the ball of the foot is equal to the man’s weight, then the sum of the torques about the ankle is (73 kg)(9.81 m/s2)(0.14 m) = F(0.05 m) F = 2000 N Consider the sum of the vertical forces in the diagram to the right. T T 4T − w = 0 1 T = (300 kg)(9.81 m/s 2 ) = 736 N 4

T

T

w


CHAPTER 14-58.

†14-59.

14

Similar to the situation shown in problem 14-57, the parbuckle allows the user to apply a force that is a fraction the weight to hold a cylindrical object to hold it in stable equilibrium. The incline that makes an angle θ relative to the horizontal direction also helps the laborer. The sum of the forces parallel to the inclined plane is 2T− w sin θ = 0. Therefore, the laborer need only 1 apply a force equal to T = w sin θ . In other words, the laborer is applying only one-half of the 2 force that the rope is applying, which is w sin θ = 1000 N. The mechanical advantage then is 2:1. (a) Consider the windlass shown in Figure 14-30. Suppose we turn the handle 1 revolution, as indicated, by applying a torque τ. Then the right rope unwinds a distance 2πR1, and the left rope winds up a distance 2πR2. If the original cable length is 2H, then the new cable length will be 2H + 2π(R1 – R2) and ∆L = L2 – L1 = 2π(R1 – R2). The load is moved upward a distance h = π(R1 – R2), so the change in the potential energy of the load is mgπ(R1 – R2), which is equal to the rotational work done. Therefore, since τ(∆φ) = 2πτ, 1 τ = mg ( R1 − R2 ) 2

(b) If the length of the handle is l, then a sum of the torques about the rotational axis gives Fl − τ = 0 So, the tangential force F that must be applied to the handle is (R − R ) 1 F = mg 1 2 l 2 (c) The mechanical advantage is F′ 2l = F R1 − R2 14-60.

†14-61.

The easiest design for the block (pulleys and support structure) and tackle (the rope) is to have n moving pulleys on one axis and n fixed pulleys on a second axis with 2n lines running between the moveable and fixed pulleys. The mechanical advantage of such a system is equal to the number of ropes (2n). Thus, for a mechanical advantage of 4, you could have two pairs of pulleys, one fixed pair above the other, moveable pair with four ropes running between them. To get a mechanical advantage of 5, one way would be to have a set of two fixed pulleys sharing a common axis and a moveable pulley sharing a common axis. Another way would be to use pulleys of unequal diameters. If these two systems are used in tandem, the mechanical advantage would be 9. The total mechanical advantage is a product of the mechanical advantage of each part of the system. The mechanical advantage 0.030 x is }x′ F ′ x′ ⎛ 16.0 cm ⎞⎛ 5.0 cm ⎞ 0.160 = =⎜ MA = ⎟⎜ ⎟ = 8.9 0.050 m 0.030 m F x ⎝ 3.0 cm ⎠⎝ 3.0 cm ⎠


CHAPTER 14-62.

14-63.

14-64.

14

This is a compound lever system. The first part consists of the input crank and the small gear; and the second one consists of the large gear and drum. The total mechanical advantage is a product of the mechanical advantage of each system. F ′ ⎛ 15 cm ⎞⎛ 25 cm ⎞ =⎜ MW = ⎟⎜ ⎟ = 7.5 F ⎝ 5.0 cm ⎠⎝ 10 cm ⎠ Following the discussion related to equation (14.15), the mechanical advantage of the vise is given by the inverse ratio of the distances through which the handle of the vise and screw move with one revolution. F ′ ⎛ 2π (25 cm) ⎞ =⎜ MA = ⎟ = 393 F ⎝ 0.40 cm ⎠ Consider the “scissors” section of the jack. When the screw advances by 2∆x, x changes by ∆x giving rise to a change ∆h in h or 2∆h in the jack height CD. Therefore the mechanical advantage of this section is ∆x/∆h. To evaluate this, note that the side length, L = 25 cm, is fixed, therefore, L = h 2 + x 2 = h ′2 + x ′ 2

and

h 2 + x 2 = h′2 + x ′ 2

which may be rewritten and factored as (h + h′)(h – h′) = (x′ + x)(x′ – x) Since h + h′ 2h; x + x′ 2 x; and x − x′ = ∆x,

†14-65.

θ C L

h 2h∆h = 2x∆x (to the first order in ∆h and ∆x) ∆h h A = . This ratio is found from the triangle with Therefore, x ∆x x hypotenuse L, and sides h and x. Angle ACD is equal to θ /2, where θ is given as 55°. There is an additional contribution to the mechanical advantage due to the screw mechanism as in Problem 14-63. The total mechanical advantage is the product of these two components. The D total mechanical advantage is then 1 ⎛ 2π (18 cm) ⎞ h ⎛ 2π (18 cm) ⎞ ⎛ ⎞ MA = ⎜ ⎟ =⎜ ⎟ tan ⎜ 90° − 55° ⎟ = 434 2 ⎝ 0.050 cm ⎠ x ⎝ 0.050 cm ⎠ ⎝ ⎠ As seen in problem 57, the block and tackle has a mechanical advantage of 4. The total mechanical advantage of the system though is higher than this. As the horizontal bar is pulled down by an amount x, the top of the strings will move down by an 32.5° 72.5° amount x ′ to give a mechanical advantage of x/x′. Suppose we L displace the horizontal bar downward by an amount x while L′ constraining the string joint to remain stationary at the top. This x will increase the string length by L′ – L = x sin 72.5° – x cos 32.5° = 0.110x However, since the string length remains constant, the point will move down by this amount. Therefore, the mechanical advantage is x/(0.11x). Thus, the total mechanical advantage is this system is ⎛ x ⎞ TMA = 4 ⎜ ⎟ = 36 ⎝ 0.110x ⎠


B

17.5°

x

CHAPTER LF (60 m)(1.8 × 104 N) = = 2.26 m YA 0.36 × 1010 mN2 π4 (1.3 × 10−2 m) 2

14-66.

The nylon rope stretches by ∆L =

14-67.

The steel wire stretches by ∆L =

14-68.

The fact given that the cord behaves like a spring is a reference to Hooke’s law, F = k∆x. With this in mind, the Young’s modulus for the cord may be detemined. ∆L 1 F = L Y A L F L F (5.0 m) Y = = = (70 N/m) = 4.5 × 106 N/m 2 ∆L A A ∆L π (0.005 m) 2

†14-69.

14-70.

14-71.

†14-73.

(

)

LF (1.8 m)(70 N) = = 2.0 × 10−3 m YA ( 22 × 1010 N/m 2 ) π (3.0 × 10−4 m) 2

Ideal springs behave according to Hooke’s law, F = k∆x, where a force F applied to one end of a spring with the other end fixed will proportionally stretch the spring by ∆x. The constant of proportionality is the spring constant k. Similarly, in equation 14.18 that ∆L 1 F = L Y A F AY π (3.0 × 10−4 m) 2 (22 × 1010 N/m 2 ) = = = 3.5 × 104 N/m ∆L L 1.8 m This is the same result that is found by substituting F = 70 N and ∆L= 2.0 × 10−3 m (from problem 14-67. F ∆V = −B A V ∆V F/A 6.0 × 109 N/m 2 = − =− = − 3.8% V B 16 × 1010 N/m 2 So, the final volume will be 100 − 3.8% = 96.2% ∆L 1 F (a) = L Y A ⎞ LF ⎛ 10 m 150 N ∆L = =⎜ = 4 × 10−3 m 10 2 ⎟ Y A ⎝ 11.0 × 10 N/m ⎠ π (1.0 × 10−3 m) 2

(b) F = AY 14-72.

14

∆L ⎛ 0.10 m ⎞ 3 = π (1.0 × 10−3 m) 2 (11.0 × 1010 N/m 2 ) ⎜ ⎟ = 3.5 × 10 N L 10 m ⎝ ⎠

∆L 1 F = L Y A ⎞ LF ⎛ 100 m 250 N ∆L = =⎜ = 8.8 m 10 2 ⎟ Y A ⎝ 0.36 × 10 N/m ⎠ π (5.0 × 10−4 m) 2 Assuming the weight of the elevator is equally supported by the upward tension forces in the cables, the tension in each cable is 1 1 T = Mg = (1000 kg)(9.81 m/s 2 ) = 3270 N 3 3 The tension will stretch each steel cable by a distance ⎞ LF ⎛ 300 m 3270 N ∆L = =⎜ = 0.057 m 10 2 ⎟ Y A ⎝ 22 × 10 N/m ⎠ π (5.0 × 10−3 m) 2


CHAPTER 14-74.

14

The compressive force on the femur due to the weight of the second woman is F = mg = (68 kg)(9.81 m/s 2 ) = 668 N The tension will compress the femur by a distance ⎞ 0.38 m 668 N LF ⎛ ∆L = =⎜ = 7.9 × 10−6 m 10 2 ⎟ Y A ⎝ 3.2 × 10 N/m ⎠ 1.0 × 10−3 m 2

†14-75.

4 The volume of the sphere is V = π R 3 . Take the derivative of the volume with respect to R. 3 3 4 dV d ( 3 π R ) = = 4π R 2 dR dR dV = 4π R 2 dR Then, the percent change in the volume is dV 4π R 2 dR dR = 4 =3 3 V R 3π R

14-76.

†14-77.

This shows that the percent change in volume is equal to three times the percent change in the radius, for small changes in the radius. So, if the volume decreases by 0.10%, the radius changes by 0.033%. The change in pressure as a function of depth is given by ∆P = ∆ρgh. The density will change as the pressure increases as the volume is reduced, ∆ρ = m/∆V, where ∆V can be determined from ∆P ∆V P = − = P V B ∆ρ gh ∆ρ P 1.24 × 108 N/m 2 = = = = 0.056 = 5.6 % B 0.22 × 1010 N/m 2 ρ gh ρ The slab is held by two iron bolts with a cross-sectional radius of 0.075 m. A downward shear force on each bolt due to one-half the weight of the slab deflects slightly downward by ∆x. The shear force acts on each bolt at a distance, h = 0.010 m. The shear modulus of wrought iron is 7.7 × 1010 N/m2. ∆x 1 F = h S A h F (0.010 m) 12 (1200 kg)(9.81 m/s 2 ) ∆x = = = 4.3 × 10−6 m S A (7.7 × 1010 N/m 2 )π (0.0075 m) 2

14-78. 9 BS YTable 3B + S Material (× 1010 N/m2) (× 1010 N/m2) Steel 22 21 Cast iron 15 15 Brass 9.0 8.8 Aluminum 7.0 6.8 The breaking strength is equal to the product of the ultimate tensile strength and the crosssectional area of the rope. Thus, if the breaking strength of the two spliced sections is equal, the diameter of the steel rope can be determined from 2 2 π DNylon π DSteel TU, Steel = TU, Nylon 4 4 Y =

†14-79.


CHAPTER

DSteel = DNylonl 14-80.

†14-81.

TU, Nylon

= (1.3 cm)

TU, Steel

3.2 × 108 N/m 2 = 0.52 cm 2.0 × 109 N/m 2

For steel of diameter D and aluminum of diameter D′, an increased pressure P will reduce the diameters by −D − D′ P and ∆D′ = P ∆D = 2B 2 B′ The rod will fit inside the ring when ∆D − ∆D′ = 2 × 10−8 m. Therefore, to find the pressure at which this occurs, use − P ⎛ D D′ ⎞ ∆D − ∆D′ = ⎜ − ⎟ 2 ⎝ B B′ ⎠ −2(∆D − ∆D′) −2(2 × 10−8 m) P= = = 6 × 105 Pa ⎛ D D′ ⎞ ⎛ 0.010 000 00 m ⎞ ⎛ 0.010 000 02 m ⎞ −⎜ ⎜ − ′⎟ ⎜ 10 2 ⎟ 10 2 ⎟ ⎝B B ⎠ ⎝ 16 × 10 N/m ⎠ ⎝ 7.8 × 10 N/m ⎠

Consider the diagram shown. The sum of the vertical forces acting on the beam is 2Tsin θ − w = 0, from which w T = 2sin θ mg (8000 kg)(9.81 m/s 2 ) = = = 4.2 × 104 N m ° 2sin 70.5 ⎤ 2sin ⎡⎣cos −1 ( 1.0 3.0 m ) ⎦ As a result of the tension, the two ropes stretch. The difference in extension of the two ropes is

nylon 3.0 m steel T

T

θ 1.0 m

θ w

⎛

1 1 ⎞ − ⎟ ⎝ YN AN YS AS ⎠

δ L = ∆LN − ∆LS = LT ⎜

14-82.

14

⎛ ⎞ 1 1 = (3.0 m)(4.2 × 104 N) ⎜ − 10 2 2 10 2 2 ⎟ (22 × 10 N/m )π (0.0032 m) ⎠ ⎝ (0.36 × 10 N/m )π (0.0125 m) −2 = 5.3 × 10 m = 5.3 cm The angle the beam makes with the horizontal direction is δL ⎛ 0.053 m ⎞ θ = tan −1 = tan −1 ⎜ ⎟ = 1.5° 2.0 m ⎝ 2.0 m ⎠ The application of the force F to the cast iron rod causes it to stretch by ∆x′ and also causes the shearing of the top of the copper plate by a distance ∆x, such that δx = ∆x′ + ∆x = 3.0 mm. For the copper plate, ∆x 1 F = (i) h S ACu For the cast iron rod, ∆x′YAFe ∆x′ 1 F = and F = L Y AFe L

(ii)

Using equation (ii), equation (i) may then be written,


CHAPTER

∆x =

14 h ∆x′YAFe h YAFe (δ x − ∆x) = hC (δ x − ∆x) = S LACu S LACu

where C =

(iii)

1 YAFe (15 × 1010 N/m 2 )π (0.020 m) 2 = = 3.57 m −1 10 2 S LACu (4.4 × 10 N/m )(2.0 m)(0.060 m)(0.010 m)

Solving equation (iii) for ∆x/h gives, ∆x Cδ x (3.57 / m)(3.0 × 10-3 m) = = = 8.8 × 10−3 h (1 + hC ) 1+ (0.060 m)(3.57 / m) 14-83.

14-84.

†14-85.

14-86.

For the thermal expansion, ∆L = α∆T = (12 × 10−6 / C°)(150 °C) = 0.0018 L F ∆L =Y = (22 × 1010 N/m 2 )(0.0018) = 3.96 × 108 N/m 2 The compressive stress is A L ∆L 1 F = = α∆T L Y A 1 F 2.4 × 108 N/m 2 ∆T = = = 128°C α Y A (1.7 × 10−5 / C°)(11 × 1010 N/m 2 ) Consider the drawing. An element dx is located a distance x from the center of mass through which the rotation axis passes. The centripetal force required to accelerate element dx is dF = dmxω2. If A is the cross-sectional area of the meter stick, then dm = ρA dx. Therefore, the tension on the stick will be the integral of the centripetal force for each element dx. 1/ 2 1 T = ∫ dF = ω 2 ∫ xdm = ρ Aω 2 ∫ xdx = ρ Aω 2 0 8 The maximum angular velocity is then 8 ⎛T ⎞ 8 ω max = = ( 3.8 × 108 N/m2 ) = 624 rad/s ρ ⎜⎝ A ⎟⎠ max 7800 kg/m3

x

ω

T dx

Assume that the pipe has a circular cross-section. Consider a small segment of the pipe that has an arc length s and a thickness h. The pressure exerts an outward force on the interior of the pipe that must be balanced by the inward force due to the tension in the pipe. The vertical component of the tension is dθ ⎛ dθ ⎞ 2T sin ⎜ = Tdθ (for small angles dθ ) ⎟ ≈ 2T dR 2 ⎝ 2 ⎠ Tdθ = PRhdθ R where P is the water pressure. The pipe will burst when the ultimate tensile strength is exceeded. The force per unit area is T PR = T s T δ Rh δ R (0.30 cm)(3.8 × 108 N/m 2 ) Tdθ P= = 3.8 × 106 N/m 2 dθ 30 cm


CHAPTER 14-87.

Consider a small section of the hoop. The centripetal force is equal to the vertical component of the tension is given by v2 ⎛θ ⎞ 2Td sin ⎜ ⎟ ≈ Tdθ = dm R ⎝ 2⎠ where dm = ρSA = ρRAd θ and A is the cross-sectional area of the hoop. Td θ = ρR Av2dθ Then, T = ρ R 2ω 2 where v = Rω A The ultimate tensile strength determines the maximum angular velocity. ⎛T ⎞ 2 2 ⎜ ⎟ = ρ R ω max ⎝ A ⎠ max

ω max = 14-88.

14-89.

14-90.

(T / A) max = ρ R2

7.8 × 107 N/m 2 = 425 rad/s (2700 kg/m3 )(0.40 m) 2

Using the approach taken in problem 14-86, T PR = δ Rh δ R R T PR 2 (2.0 × 10 4 N/m 2 )(0.25 m) 2 ∆R = = = = 1.42 × 10 −6 m 10 2 Y δ Rh Y δ R (22 × 10 N/m )(0.0040 m) The change in the diameter is ∆d = 2∆R = 2.8 × 10−6 m The sum of the vertical forces acting on the stop light gives 2T sin 20° = w w (25 kg)(9.81 m/s 2 ) T = = = 360 N 2sin 20° 2sin 20° Let’s assume the sign of weight w hanging from the boom exerts a torque at the mid-point of the boom. The sum of the torques about the hinged end of the boom is (T sin 45°)L − wL/2 = 0 where L is the length of the boom. Dividing the equation by L gives the tension in the wire.

45° T FH FV w

w (50 kg)(9.81 m/s 2 ) = = 347 N 2sin 45° 2sin 45° The sum of the horizontal forces acting on the boom gives FH = T cos 45° = (347 N) cos 45° = 245 N The sum of the torques about the hinged end of the boom is TL − wL sin 30° = 0 where L is the length of the boom. Dividing the equation by L gives the tension in the wire. T = w sin 30° = (2500 kg)(9.81 m/s2) sin 30° = 1.23 × 104 N The sum of the horizontal forces acting on the boom gives FH = T sin 60° = (1.23 × 104 N) sin 60° = 1.07 × 104 N The sum of the vertical forces acting on the boom gives FV = mg− T cos 60° = (2500 kg)(9.81 m/s2) − (1.23 × 104 N) cos 60° = 1.84 × 104 N The compressional force on the boom is then T =

14-91.

14


CHAPTER

14

F = FH2 + FV2 = (1.07 × 104 N) 2 + (1.84 × 104 N) 2 = 2.13 × 104 N 14-92.

The sum of the torques about the hinged end of the boom is TL − mgL sin 30° − (ML/2) sin 30° = 0 where L is the length of the boom. Dividing the equation by L gives the tension in the wire. T = (0.5M+m)g sin 30° = (400 kg + 2500 kg)(9.81 m/s2) sin 30° = 1.42 × 104 N The sum of the horizontal forces acting on the boom gives FH = T sin 60° = (1.23 × 104 N) sin 60° = 1.07 × 104 N The sum of the vertical forces acting on the boom gives FV = (0.5M + m)g − T cos 60° = (2900 kg)(9.81 m/s2) − (1.23 × 104 N) cos 60° = 2.22 × 104 N The compressional force on the boom is F = FH2 + FV2 = (1.07 × 104 N) 2 + (2.22 × 104 N) 2 = 2.46 × 104 N

14-93.

The engine turns the axle which applies a torque to the wheel. The radius of the axle is much less than the radius of the wheel, so the force applied by the axle on the wheel is much greater than the force that is at the rim of the tire at the contact point between the wheel and the ground. The torque is the same, however, so the torque supplied by the axle is τ = (4000 N)(0.60 m) = 2400 N • m

F

f

14-94.

The sum of the torques about the pivot point where the string is tied to the meter stick gives, w(0.30 m) − F(0.50) = 0 w(0.30 m) (0.24 kg)(9.81 m/s 2 )(0.30 m) F= = = 1.4 N (0.50 m) (0.50 m)

14-95.

The sum of the vertical forces gives 2T sin 60° − w = 0 w Mg T = = = 0.577 Mg 2(sin 60°) 2(sin 60°) The horizontal component of the tension is the compression force on the beam. FH = T cos 60° = (0.577Mg) cos 60° = 0.289Mg

T

θ

θ

T

w 14-96.

The diagram at the right shows only the forces acting on the left leg. Consider the sum of the vertical forces acting on the left leg. Assuming the legs are massless and that each leg must support one-half the weight, T sin 60° − w/2 = 0 F 2 (400 kg)(9.81 m/s ) w T = = = 2260 N 2(sin 60°) 2(sin 60°) Each leg exerts vertical and horizontal forces on the ground equal to the components of the compression forces on the leg. FH = T cos 60° = (2260 N) cos 60° = 650 N FV = T sin 60° = (2260 N) sin 60° = 1960 N


w

T Ty

60° Tx

CHAPTER 14-97.

14-98. 14-99.

14-100.

14

The force parallel to the plane to support the barrel must equal the component of the barrel’s weight down the plane. F = w sin θ = mg sin 30° = (100 kg)(9.81 m/s2) sin 30° = 490 N The mechanical advantage of the pliers is 4, so the force at the jaws will be 800 N. Using the point of contact of the wheel with the ground as a pivot point, the force applied to the top of the wheel at a distance D results in the forward motion of the wagon with a force at the axle that is twice the force the teamster applies. The lever has a mechanical advantage of 2, so the force is 1200 N. The wire makes an angle θ with respect to the pole, where θ = tan−11(0.5L/L) = 26.6° The sum of the torques about the hinge gives TL sin θ − w(0.5L) = 0 θ T FV

FH

T =

w

w Mg = = 1.12Mg 2(sin 26.6°) 2(sin 26.6°)

The sum of vertical forces acting on the pole is T sin θ + FV − w = 0 The vertical component of the force at the hinge support is then FV = w − T sin θ = Mg − (1.12Mg) sin 26.6° = 0.5Mg The horizontal component of this force is FH = T cos θ = (1.12Mg) cos 26.6° = Mg and F = FH2 + FV2 = ( Mg ) 2 + (0.5Mg ) 2 = 1.12 Mg 14-101.

As expected, the magnitude of the force at the hinge must equal the tension in the wire. (a) Hooke’s law may be written, F = k∆L. The spring constant is k = F/∆L. For a wire that is stretched a distance ∆L from its initial length L, we have ∆L 1 F = L Y A F YA Y π R 2 = = k= ∆L L L k=

(b)

Y π R12 Y π R22 = L1 L2

R12 R22 = L1 L2 R2 = R12

L2 4.0 m 2 = (0.5 mm) 2 = mm = 0.71 mm L1 2.0 m 2


CHAPTER 14-102.

14

Since we want the ratio F/∆L to be constant for a given force. If this is the case, then we can write ∆L 1 F = L Y A F YA Y π D 2 = = ∆L 4L L Therefore, for nylon and steel ropes of equal length, the requirement would be satisfied if YN DN2 = YS DS2 DN2 YS = DS2 YN DN YS = = DS YN

†14-103.

22 × 1010 N/m 2 = 7.8 0.36 × 1010 N/m 2

The mass m of the rod is related to its volume V by m = ρV = ρ AL , where ρ is the density, A is the cross-sectional area of the rod, and L is its length. The tensile stress is F w mg ρ ALg = = = = ρ Lg A A A A The length at which the maximum tensile stress will be reached is ( F / A)max 3.8 × 108 N/m 2 L= = = 5.0 × 103 m 3 2 ρg (7800 kg/m )(9.81 m/s )

14-104.

⎛ 1 ∆L ∆LN ∆Ls 1 F 1 F 1 ⎞ = + = + = mg ⎜ + 2 2 ⎟ L LN LS YN AN YS AS ⎝ YNπ RN YSπ RS ⎠ ⎛ 1 1 ⎞ ∆L = mgL ⎜ + 2 2 ⎟ ⎝ YNπ RN YSπ RS ⎠ ⎛ 1 = (4000 kg)(9.81 m/s 2 )(12 m) ⎜ 10 2 2 ⎝ (0.36 × 10 N/m )π (0.0095 m) ⎞ 1 + 10 2 2 ⎟ (19 × 10 N/m )π (0.00475 m) ⎠

= 0.50 m 14-105.

4 The volume of the sphere is V = π R 3 . 3 ∆V 3∆R 1 F = =− V R B A R F 0.10 m ∆R = − = − (5.7 × 107 N/m 2 ) = 2.4 × 10−5 m 10 2 3B A 3(7.8 × 10 N/m ) The change in the diameter is then ∆D = 2∆R = 4.8 × 10−5 m.


CHAPTER 15

OSCILLATIONS


(a) A = 3.0 m. angular frequency ω = 2.0 rad/sec, ω 2 1 = = Hz = 0.318 Hz ⇒ frequency f = 2π 2π π period T = 1/f = π = 3.14 s.

(b) x = 0 when 2.0t = π/2, 3π/2, 5π/2, … The first time this occurs is t =

π

s = 0.785 s. 4 The turning points occur when x = ±A. This happens when 2.0t = 0, π, 2π, 3π, … The first time this occurs after t = 0 is t =

15-2.

π 2

s = 1.57 s.

(a) t = 0, x = 0.6 m. t = 0.50 s, x = 0.6 m × cos

π i 0.5

t = 1.00 s, x = 0.6 m × cos

π

2 2

= 0.6 m × cos

π 4

= 0.42 m.

= 0 m.

dx π πt ⎛ πt ⎞ = − × 0.6 × sin = −(0.3π ) sin ⎜ ⎟ . dt 2 2 ⎝ 2 ⎠ at t = 0 sec, v = 0. π i 0.50 π = −0.3π × sin = −0.67 m/s. at t = 0.50 sec, v = −0.3π × sin 2 4

(b) v =

at t = 1.00 sec, v = −0.3π × sin

π

π

2 2 π πt πt dv =− × 0.6 × cos = −1.48 cos (c) a = dt 4 2 2 at t = 0 sec, a = –1.48 m/s2

at t = 0.50 sec, a = −1.48 cos

†15-3.

π 4

2

= −0.3π × sin

= −0.94 m/s.

= −1.05 m/s 2

at t = 1.00 sec, a = 0. 1 1 Hz = 0.83 Hz (a) f = = T 1.2 angular frequency ω = 2πf =

π 0.6

rad/s = 5.2 rad/s

(b) A = 0.20 m. ⎛ 2π (c) The equation of motion is x = A cos ⎜ ⎝ T ⇒

π 0.6

t=

π 2

π ⎞ ⎛ π ⎞ t) = 0 t ⎟ = 0.20 cos ⎜ t ⎟ . If x = 0, cos( 0.6 0.6 ⎠ ⎝ ⎠

⇒ t = 0.30 s. Because the period is 1.2 s, the particle will also pass through x = 0 at


CHAPTER

15

t = 0.90 s, 1.50 s, 2.10 s, etc. (In SHM the particle passes through the equilibrium position twice during each cycle.)

For x = –0.10 m, −0.10 = 0.2 cos(

π 0.6

t)

π π 2π 4π 8π 10π t = −0.5 ⇒ t= , , , , 0.6 0.6 3 3 3 3 ⇒ t = 0.40 s, 0.80 s, 1.60 s, 2.00 s, .... ⇒

cos

dx π πt π πt = × 0.2 × sin = sin dt 0.6 0.6 3 0.6 From (c) we know that the first time x = 0 corresponds to t = 0.30 s. At this time, π π (0.3) π v = sin = m/s = 1.05 m/s. 3 0.6 3 π π (0.4) The first time x = –0.1 m corresponds to t = 0.40 s. At this time, v = sin = 0.91 m/s. 3 0.6 A = 4.0 cm. 1 60 sec/ min = = 0.10 s period T = 600rev / min f

(d) The speed of the motion is: v =

15-4.

†15-5.

15-6.

frequency f = 10 Hz Angular frequency ω = 2πf = 20π rad/s The equation of motion of the particle is x = cos(100π t ) (a) The speed of the satellite is: vs = rω = 0.8 × 100π = 80π m/s = 251 m/s (b) The speed of the particle at this point is: dx π = 0.8 × 100π × sin(100π t ) t = 0.005 s = 80π × sin = 80π m/s = 251 m/s, v= dt 2 which is the same as the speed of the satellite. x = 1.2 mm cos ωt; ω = 2πf = 800π/s x = (1.2 × 10−3 ) m cos (880 πt) dx = –1.2 × 10–3m 880 π/s (sin 880 πt) dt dx = (1.2 × 10−3 880 π ) m/s = 3.3 m/s dt max d 2x = –1.2 × 10–3m (880 π)2/s cos (880 πt) dt 2 d 2x = 1.2 × 10–3 (880 π)2m/s2 = 9.2 × 103 m/s 2 2 dtmax

†15-7.

frequency f = 80/min = 80 min −1 ×

1 min = 1.33 Hz. ω = 2π f = 8.38 rad/sec. The magnitude of 60 s

the force is Fmax = mamax = mω 2 A = (6.0 kg)(8.38 s −1 ) 2 (0.50 m) = 211 N. There is insufficient information given to find a velocity. The maximum speed is vmax = ω A = 8.38 s −1 × 0.50 m = 4.2 m/s


CHAPTER

15-8.

†15-9.

dx = − Aω sin(ω t + δ ) dt At t = 0, x = 0, which means Acos δ = 0. This means δ = ± π/2, ± 3π/2, etc. v(0) = v0 > 0, which implies sin δ < 0. Thus the smallest possible values are δ = – π/2 or +3π/2. x = A cos(ω t + δ ) x When t = 0, x = x0 = A cos δ ⇒ cos δ = 0 A dx v= = − Aω sin(ω t + δ ) dt v When t = 0, v = v0 = − Aω sin δ ⇒ sin δ = − 0 Aω 2 2 v02 ⎛ v0 ⎞ ⎛ x0 ⎞ 2 2 2 2 Because sin δ + cos δ = 1, ⎜ − ⎟ + ⎜ ⎟ = 1 ⇒ A = x0 + ω 2 ⎝ Aω ⎠ ⎝ A ⎠ x = A cos(ω t + δ ); v =

⇒ A=

x02 +

v02

ω2

,

The phase angle can be expressed three different ways: ⎛ ⎞ ⎜ ⎟ x x0 ⎟ (from the equation for x(0)) δ = cos −1 0 = cos −1 ⎜ 2 ⎟ ⎜ A v 2 0 ⎜⎜ x0 + 2 ⎟⎟ ω ⎠ ⎝ ⎛ ⎜ v0 −1 v0 −1 ⎜ δ = − sin = −sin ⎜ ωA v2 ⎜⎜ ω x02 + 02 ω ⎝

⎞ ⎟ ⎛ v0 ⎟ = − sin −1 ⎜ ⎟ ⎜ (ω x ) 2 + v 2 0 0 ⎝ ⎟⎟ ⎠

⎞ ⎟ (from the equation for v(0)) ⎟ ⎠

⎛ v0 ⎞ ⎟ (by dividing v(0) by x(0)) ⎝ ω x0 ⎠

δ = − tan −1 ⎜ 15-10.

ω = 2π f = 2π × 261.7 = 1.644 × 103 rad/s period T =

1 1 = = 3.821 × 10−3 s 261.7 Hz f

vmax = Aω = 3 × 10−3 m × 1.644 × 103 rad/s = 4.93 m/s amax = Aω 2 = 3 × 10−3 m × (1.644 × 103 rad/s) 2 = 8.11 × 103 m/s 2 †15-11.

f = 250 Hz. ω = 2π f = 2π × 250 rad/s = 500π rad/s amax = Aω 2 If amax = g = 9.81 m/s2, the beads start to lift off. ⇒

15-12.

Aω 2 = 9.81m/s 2 ⇒ A =

9.81 m/s 2

ω2

=

9.81 m/s 2 = 3.98 × 10−6 m. 2 (500π rad/s)

⎛π ⎞ x = 10 cos ⎜ t ⎟ cm ⎝4 ⎠ dx π ⎛π ⎞ ⎛π ⎞ vx = = −10 × × sin ⎜ t ⎟ = −7.85sin ⎜ t ⎟ cm/s dt 4 ⎝4 ⎠ ⎝4 ⎠


15

CHAPTER

ax =

15 dv d 2 x = 2 = −ω 2 x = −0.617 x cm/s 2 dt dt

At t = 1.0 s, x = 10 cos

π

= 7.07 cm 4 ⎛π ⎞ vx = −7.85sin ⎜ ⎟ cm/s = −5.55 cm/s ⎝4⎠ ax = −0.617 x cm/s 2 = −(0.617)(7.07)cm/s 2 = −4.36 cm/s 2

At t = 2.0 s, x = 10 cos

†15-13.

15-14.

15-15.

π

=0 2 ⎛π ⎞ vx = −7.85sin ⎜ ⎟ cm/s = −7.85 cm/s ⎝2⎠ ax = −0.617 x cm/s 2 = 0 dx x = A cos(ω t + δ ), v = = − Aω sin(ω t + δ ) dt at t = 0, x = 0 ⇒ 0 = cos δ ⇒ δ can be π/2, 3π/2. 3π at t = 0, v > 0 ⇒ − Aω sin δ > 0 ⇒ δ = 2 d 2x For simple harmonic motion, x = A cos (ωt + δ) and 2 = –ω2 A cos (ωt + δ). dt 2 2 Hence, the peak acceleration is ω A, which yields ω A = 0.4 g or 0.4 × 9.8 m/s 2 A = 0.4g/ω2 = = 1.1 m 2 ( 2π × 0.3/s ) Thus, the amplitude of the motion is 1.1 m, i.e., 2.2 m from one extreme of the motion to the other. f = 3.0 Hz, so ω = 2πf = 6.0π rad/s. dx = –A sin (ωt + δ). Let x = A cos (ωt + δ) and dt t = 0, x = A cos δ = 0.20 m (i) dx = –Aω sin δ = 4.0 m/s (ii) dt (ii)/(i) gives – ω tan δ = 20 rad/s 20 = – 1.06 tan δ = – 6π δ = –0.815 radian From (i) A = 0.20 m/cos δ = 0.20 m/cos (–0.815) = 0.292 m. ω = 6.0π = 18.8 rad/s Therefore, x = 0.292 cos(18.8t − 0.815) (b) At the turning point dx = –18.8 (0.292) sin (18.8t – 0.815) = –5.48 sin (18.8t – 0.815) = 0 dt Therefore, 0.815 rad = 0.0434 s. 18.8t – 0.815 = 0 ⇒ t = 18.8 rad / s


CHAPTER d 2x = –(18.8 rad/s)2(0.292 m) cos (18.8t – 0.815) dt 2 = –103 cos [18.8 (0.0432) – 0.815] = –103 cos 0 = –103 m/s2. 2π 2π T = 0.70 s ⇒ ω = 2π f = = = 8.97 rad/s T 0.70 s a=

15-16.

k = mω 2 = 70 kg × 8.972 (rad/s)2 = 5.63 × 103 N/m †15-17.

Because F = – kx, F 60 tons × 1000 kg/ton × 9.81 m/s 2 = = 2.80 × 106 N/m k= x 0.21 m angular frequency ω = frequency f =

15-18.

15-20.

19.9 rad/s = 3.16 Hz 2π rad

f =

ω = 2 f one spring = 2 × 8 Hz = 11.3 Hz 2π 2π ⎛ 2π ⎞ 2 2

k 2 m or k = ω m, where ω = . Then k = ⎜ ⎟ m = 4π 2 T T m ⎝ T ⎠ 2 2 Thus, k for all the four springs together = 4π 1100 kg/(0.75 s) = 77,200 kg/s2 Because the springs are in parallel, the spring constant for each k = = 1.9 × 104 kg/s 2 4 k k ω1 = , ω2 = (the k are equal because the "springs" are equal) m2 m ω=

ω1 = ω2

15-21.

=

2.8 × 106 N/m = 19.9 rad/s 7.10 × 103 kg

When two springs are connected in parallel, ktotal = k1 + k2. Since the two springs are identical, so: ktotal = 2k ktotal 2k k the angular frequency ω = = = 2 = 2ω one spring m m m

⇒ 15-19.

ω 2π

k = m

m2 , ω1 = m1

m2 ω2 ⇒ 2πω1 m1

m2 (2πω 2 ) ⇒ f1 = m1

m2 f2 . m1

Let f1, m1 be the frequency, mass of D2, and f2, m1 be that of H. Therefore, 1 f1 = (1.31 × 1014 /s) = 9.27 × 1013 /s 1.998 Because D is two times heavier than H, the ratio of lengths from H to the CM and D to the CM is 2:1. Let ω be the angular frequency. kH kD ⎛3 ⎞ Then ω = = . By their lengths, kH = ⎜ K ⎟ , kD mH mD ⎝2 ⎠ = 3K where k is the spring constant of the entire spring.


15

CHAPTER

ω=

15 3 k = 2 mH

3k = 2mH

3 k 2 mH

But as shown in the example, each hydrogen atom feels a spring constant of 2k. Therefore, 3 k 3 2k 3 ω= = = ωH 2 mH 2 mH 2

15-22.

where ωH is the angular frequency of hydrogen. So, 3 3 f= fH = 1.31 × 1014/s = 1.13 × 1014 /s. 2 2 dx = − Aω sin(ω t + δ ) x = A cos(ω t + δ ), v = dt A Suppose at time t = t1, x = , v = 25 cm/s. 2 A 1 ⇒ = A cos(ω t + δ ) ⇒ cos(ω t1 + δ ) = 2 2 25 cm/s = − A sin(ω t1 + δ ) ⇒ sin(ω t1 + δ ) = −

25 cm/s Aω

Because sin 2 (ω t1 + δ ) + cos 2 (ω t1 + δ ) = 1 25 cm/s 2 1 25 cm/s 2 1 3 ) + ( )2 = 1 ⇒ (− ) =1− = therefore, (− Aω 2 Aω 4 4 4 1 4 1 ⇒ ω = = 1.92 rad/s × 625 cm 2 /s 2 × 2 = × 625 cm 2 /s 2 × 2 3 A 3 15 cm 2 ⇒T =

2π

ω

= 3.27 s

k k 20 N/m , so m = 2 = = 5.43 kg. m ω 1.922 rad 2 /s 2 dx dv x = A cos(ω t + δ ), v = = − Aω sin(ω t + δ ), a = = − Aω 2 cos(ω t + δ ) dt dt Because the mass is released from rest at a displacement of 20 cm from equilibrium, that will be the amplitude: A = 0.20 m. k 8.0 N/m ω= = = 7.30 rad/s m 0.15 kg Because ω =

†15-23.

At t = 0, v = 0, so 0 = − Aω ⋅ sin δ ⇒ δ = 0 dx Therefore, x = A cos ω t , v = = − Aω sin ω t , dt The maximum speed is vmax = Aω = 0.20 m × 7.30 rad/s = 1.46 m/s

a=

dv = − Aω 2 cos ω t dt

The magnitude of the maximum acceleration is amax = Aω 2 = 0.20 m × (7.30 rad/s) 2 = 10.7 m/s2


CHAPTER

15-24.

k=

F 0.2 × 9.81 N = = 13.08 N/m 0.15 m x

⇒ ω = ⇒ T = †15-25.

k 13.08 N/m = = 8.09 rad/s 0.2 kg m 2π

ω

=

2π rad = 0.78 s. 8.09 rad/s

Angular frequency ω = frequency f =

k = m

6.0 × 105 N/m = 3.64 × 104 rad/s −3 0.5 × 10 kg

ω 3.64 × 104 rad/s = = 5.51 × 103 Hz. 2π 2π rad ω 1 = 2π 2π

k , suppose frequency changes a small amount df, then df can be m calculated by taking the differential of f: k df = − dm 4π m3 / 2 k − dm 3/ 2 df dm The ratio = 4π m =− 2m f 1 k 2π m dm df = −2 . Therefore, m f df dm If = −0.01%, then = −2 × ( −0.01%) = 0.02% m f Because f =

⇒ The change of mass: ∆m = 0.02m = 0.02% × 0.5 g = 1.0 × 10−4 g Assume the thickness of the deposited film is t, then ∆m 1 × 10−4 g t × Area × density = ∆m ⇒ t = = = 6.7 × 10−6 cm Area × density 2.0 cm 2 × 7.5 g/cm3 15-26.

15-27.

k=

Ay 9 × (10−3 ) 2 m 2 × 22 × 1010 N/m 2 = = 9.9 ×105 N/m L 2.0 m

f =

ω 1 = 2π 2π

k 1 = m 2π

Equilibrium at x =

9.9 × 105 N/m = 40.9 Hz 15 kg

F mg 2.5 × 9.8 N = 0.27 m = = k k 90 N/m

k 90 = 6/s = m 2.5 Let x be positive upward. Then the ball is let go at x = 0.27 m at v = 0. These initial conditions are satisfied by x = A cos (ωt + δ) = 0.27 cos 6t ω of spring =


15

CHAPTER

15

k ω = (1.1 × 104 N/m ) /14 kg = 28/s. Thus v = = 4.5 Hz. 2¹ m (b) The wheel has to go round at a frequency 4.5 Hz or ω = 28/s. Its speed is then ωr = (28/s) × (0.61/2)m = 8.5 m/s. Oscillations get bigger and bigger.

15-28.

(a) ω =

15-29.

Let x be the direction as shown, with zero at the unstretched point on the spring. Then, Fspring = –kx. Fgravity = mg sin θ. At equilibrium, kx = mg sin θ, or x = (mg sin θ ) / k

⎛ d2x ⎞ (1) Fext = –kx + mg sin θ = m ⎜ 2 ⎟ ⎝ dt ⎠ Let x′ = x – mg sin θ/k. Therefore, x = x′ + mg sin θ/k. Substituting into (1) gives d2 ⎛ mg sin θ ⎞ d2x ' m 2 ⎜ x '+ ⎟ = m 2 = – kx' dt dt ⎝ k ⎠ Therefore, the motion is simple harmonic with frequency

15-30.

1 2π

k m

Let x1, x2 be the coordinates as shown. The force on 1 when displaced x1 is –kx1 + k ( x2 – x1 ) . The force on 2 is –kx2 + k ( x1 – x2 ) . The equations of motion are then d 2 x1 m = –kx1 + k ( x2 − x1 ) = –2kx1 + kx2 (1) dt 2 d 2 x2 m = –kx2 + k ( x1 − x2 ) = kx1 – 2kx2 (2) dt 2 (a) If v1 = –v2, by symmetry, x1 = –x2. Substituting gives: d 2 x1 3k 3k / m , or f = m = –3kx1 . Therefore, ω = 2 dt m 2π The positions are given by xn = A cos(ω t + δ n ), assuming both masses vibrate with the same amplitude. n = 1,2 identifies which mass is which. Assume that when m1 is at its equilibrium position, it’s moving to the right so v1 > 0. Then δ 1 = −

π

, δ2 = +

π

with the angular frequency 2 2 given. The two masses are moving 180º out of phase with each other. (b) If v1 = v2, then both will move together at the same rate to the right ⇒ x1 = x2. By (1) and (2), d 2 x1 k/m k m = – kx, ⇒ ω = , f= 2 dt 2π m The positions are given by xn = A cos(ω t + δ n ), again assuming both masses vibrate with the same amplitude. Assume that when m1 is at its equilibrium position, it’s moving to the right so


CHAPTER v1 > 0. Then δ1 = δ 2 = −

π 2

15

with the angular frequency given. The masses are moving in phase

with each other.

15-31.

Force of spring = – kx, where x = 0 at equilibrium. I of each wheel 1 = MR2. Net force on cart = Fspring + Ffriction. Let the static 2 frictional force on each wheel = F. Then τ = FR and Iα = τ = FR. And for the cart to roll d 2x without slipping, a = 2 = Rα. dt Iα I d2x 1 d 2x = 2 2 = M 2 Thus, F = 2 R R dt dt and the frictional force d2x on all four wheels = 2M 2 . dt The net force on the cart d 2x = – kx – 2M 2 dt (minus because friction opposes motion) d 2x d 2x d 2x Then, Fnet = – kx – 2M 2 = (m + 4M) 2 ⇒ (m + 6M) 2 = –kx dt dt dt Therefore, k /(m + 6M ) k ω= f = (m + 6M ) 2π

15-32.

(a) E =

1 2 4π 2 m 4π 2 0.24 kg kA where k = mω2 = 4π2f 2m = = = 6.58 N/m 2 1.22 s 2 T2 1 Therefore, E = ( 6.58 N/m ) (0.20 m) 2 = 0.132 J 2 (b) Since the mass is at its maximum displacement when t = 0, its kinetic energy is zero whenever x = ± A, or t = 0, T/2, T, 3T/2, … . The potential energy is zero whenever x = 0, or t = T/4, 3T/4, 5T/4, … . (c) When K = U, each type of energy is equal to half the total energy:


CHAPTER

15

E kx 2 kA2 A ⇒ = ⇒x=± 2 2 4 2 A T T ⎛ π ⎞ T ⎛ 3π ⎞ T ⎛ 5π ⎞ cos −1 ±1/ 2 = ± = A cos ω t ⇒ t = ⎜ ⎟, ⎜ ⎟, ⎜ ⎟ , … or 2π 2π ⎝ 4 ⎠ 2π ⎝ 4 ⎠ 2π ⎝ 4 ⎠ 2 t = T/8, 3T/8, 5T/8, 7T/8, … . 1 E = KA2; k = mω2; 4π2f 2m. 2 2 2π 2 (0.60) Then E = 2π2f 2mA2 = (8.0)kg (0.25 m)2 = 3.6 J 2 2 1 2 1 2 Energy E = K + U = mv + kx . 2 2 From energy conservation, this quantity is constant. So we can use the values of K and U at x = 0 (equilibrium position) to find the total energy. 1 1 1 At x = 0, K = mv 2 = × 2 kg × (3 m) 2 = 9 J; U = kx 2 = 0 ⇒ E = 9.0 J. 2 2 2 1 2E 2 × 9.0 J When x = A, K = 0 and E = U, so E = kA2 ⇒ A = = = 0.15 m. 2 8.0 × 103 M/m k U =

(

15-33.

15-34.

†15-35.

)

If mass is added at the turning point, amplitude won’t change. ω 1 k Because f = = , 2π 2π m 1 k 1 k 1 1 × 3.0 Hz = 2.12 Hz. = = if mnew = 2 mold, f new = f old = 2π mnew 2π 2 mold 2 2 1 2 kA won't change. 2 = A(2π f new ) = 0.15 m × 2π × 2.12 Hz = 2.0 m/s.

Because the amplitude does not change, E = max Because vmax = Aω , vnew = Aω new

max 2 Because amax = Aω 2 , anew = Aω new = A(2π f new ) 2 = 0.15 m × (2π × 2.12 Hz) 2 = 27 m/s2

15-36.

E=

1 2 kA for each, so that Etotal = kA2 where k is spring constant for each atom. 2

E = (1.13 × 103 ) N/m × ( 0.5 × 10−10 ) m2 = 2.8 × 10−18 J. No. 2

†15-37.

ω=

2π 2π = = 4.19 rad/sec 1.5 sec T

k = 4.19 rad/sec ⇒ k = mω2 = 0.5 kg × (4.19)2 rad2/sec2 = 8.78 N/m m 1 1 N 2E 2 × 0.5 J ⇒ E = kA2 = × 8.78 i A2 ⇒ A = = 0.34 m = 2 2 m 8.78 N/m k

⇒

15-38.

1 U , x = x0 = 0.15 m, v = v0 = 25 cm/s = 0.25 m/s 2 v2 1 2 1 1 2 1 2 0.252 m 2 /s 2 1 k mv0 = × kx0 = kx0 ⇒ = 2 02 = 2 × = 5.55 2 2 2 m x0 2 2 2 4 0.15 m s

When K =


CHAPTER

15

k 1 = 5.55 2 = 2.36 rad/s s m 2π 2π rad T= = = 2.67 s 2.36 rad/s ω

ω=

15-39.

15-40.

15-41.

After the mass has undergone a displacement d in the direction of the force, the work done by the force is WF = Fd . If there were no friction and the spring were not present, all of this work would be converted to kinetic energy. But some of the work is stored as potential energy in the 1 1 spring: U = kd 2 . Thus the kinetic energy is K = Fd − kd 2 2 2 1 Kinetic energy of mass when it hits the spring K = mv2 = 6.0 J. 2 Using conservation of energy, 1 1 K = Uspring = kx2 = (300) x2 = 6 2 2 x = 0.173 m. The motion of the mass is simple harmonic with period T = 2π m / k = 0.618 s. The time for the mass to stop is T/4 = 0.157 s The lengths of spring from the center of mass to m1 and to m2 are in the ratio m2:m1. Since the center of mass remains fixed, the motion of each mass can be treated as though that mass were attached to one end of a shortened spring, whose other end is fixed (see figure). The segment of spring belonging to mass m1 is a fraction m2 ( m1 + m2 ) of the total spring. The spring constant of this segment is therefore m + m2 k× 1 m2 (see Example 6.10). The frequency of oscillation of this mass is then

ω =

15-42.

( spring constant ) m1

=

⎣⎡ k ( m1 + m2 ) ⎦⎤ m1m2

(a) Take a small element dx of spring at a distance x from the support. That element has mass (dx/e')m', m' = mass of the spring. Then the kinetic energy of that mass element is 2 1 ⎛ dx ⎞ ⎛ x ⎞ m v ⎜ ⎟ ⎜ ⎟ 2⎝ l' ⎠ ⎝l' ⎠ x Since the velocity of that point is v l' 2 2 1 m'x v d(KE) = dx and 2 l '3


CHAPTER

15

1 m′x 2 v 2 1 m′v 2 l ′3 1 dx = = m′v 2 ∫0 ∫x =0 2 l′3 2 l ′3 3 6 1 1 1⎛ 1 ⎞ Then the total kinetic energy = mv2 + m'v2 = ⎜ m + m′ ⎟ v2 2 6 2⎝ 3 ⎠ (b) Let angular frequency = ω. The motion is A cos ωt for some amplitude A. v2 = A2ω2 sin2 ωt, so that 1 ⎞ 1 ⎛ KE = A2 ⎜⎝ m + 3 m′ ⎟⎠ ω2 sin2 ωt. 2

KE of entire spring =

l′

d (KE) =

x =l ′

1 2 2 ⎛⎜ m + 1 m′ ⎞⎟ Aω ⎝ 3 ⎠ 2 This must equal the maximum potential energy 1 2 1 = kmax = kA2. 2 2 Therefore, 1 2 1 2 2 ⎛⎜ m + 1 m′ ⎞⎟ ⇒ 2 k kA = A ω ⎝ ω = ⇒ω= 3 ⎠ 1 ⎞ 2 2 ⎛ ⎜ m + m′ ⎟ 3 ⎠ ⎝ KEmax =

ω (c) new = ω

1 m′) 3 = k/m

k /(m +

m = 1 ⎞ ⎛ ⎜ m + m′ ⎟ 3 ⎠ ⎝

400 5 400 + 3

k 1 ⎞ ⎛ ⎜ m + m′ ⎟ 3 ⎠ ⎝

= 0.998.

Thus, it is (1 – 0.998) × 100% = 0.2% smaller. †15-43. 15-44.

T = 2π T =

l 27 m = 2π = 10.4 s. g 9.81 m/s 2

1 min 60 s = = 7.5 s 8 rev 8 rev

Because T = 2π

l T 2g 7.52 s 2 i 9.81 m/s 2 , therefore, l = = = 14.0 m g (2π ) 2 (2π ) 2

15-45.

T = 2π

l 300 m = 2π = 34.8 s. g 9.81 m/s 2

15-46.

T = 2π

l g

Because the length is the same,

TJupiter TEarth

=

g Earth = g Jupiter

9.81 = 0.629 24.8

⇒ TJupiter = 0.629 T Earth = 0.629 × 2.00 = 1.26 s. 15-47.

T = 2π

l 1 1 , so that f = = g T 2π

g 1 = 2π l

⎛ 9.81 m/s 2 ⎞ ⎜ ⎟ = 0.188 Hz. ⎝ 7m ⎠


CHAPTER

15-48.

Using T = 2π

15

l T2 ⇒e= 2 g g 4¹

Paris

l=

2.0002 s 2 (9.809) m/s2 = 0.9939 m 2 4π

⎛ 9.797 ⎞ l = ⎜ 2 ⎟ m = 0.9926 m ⎝ π ⎠ ⎛ 9.801 ⎞ Washington‚ D.C. l = ⎜ 2 ⎟ m = 0.9930 m ⎝ π ⎠ l l T2 = 2π T1 = 2π g1 g2 Buenos Aires

15-49.

Therefore,

T2 = T1

g1 g2

T2 = period in Austin

T1 = period in N.Y. T2 9.803 = 1.00051 = 9.793 T1 so that “1 min” in Austin is really 1.00051 min. Since 1 day = 24 × 60 min = 1440 min, clock will go x min in 1440 min, where 1.00051x = 1440. 1440 = 1439.27 min or the clock will fall behind 0.73 min/day x= 1.00051 l = 2.000 s. T1 = 2π g1 g 2T12 9.793(2.000 s) 2 = = 0.9922 m. 4π 2 4π 2 m So that ∆e = 0.9932 – 0.9922 m = 6.4 × 10–3 m = 1.0 mm l2 =

15-50.

If the clock runs 1 min late a day, then T1/T2 = 1440/1439 where T1 = period of clock, T2 = correct l period. But T = 2π so, g T1 = T2

l1 ⇒ l2

l2 =

⎛T ⎞ l1 ⎜ 2 ⎟ ⎝ T1 ⎠

2

⎛T ⎞ l2 = l1 ⎜ 2 ⎟ , where l2 = correct length, l1 = actual length ⎝ T1 ⎠ 2 ⎛ 1439 ⎞ l2 = 0.994 m ⎜ ⎟ = 0.9926 m. This must be shortened by about 1.4 mm. ⎝ 1440 ⎠ †15-51.

T = 2π

l T2 102 s 2 ⇒l= = g × 9.81 m/s 2 = 24.8 m g 4π 2 4π 2


CHAPTER

15-52.

15

T = 2π

l g

2 2 TEarth TAsteroid T2 g g g Asteroid = = Earth 4π 2 4π 2 4π 2 T2 102 ⇒ g asteroid = 2Earth × g Earth = 2 × 9.81 = 0.124 m/s2 TAsteroid 89

⇒ l=

†15-53.

T = 2π

I , where d is the distance between the point of suspension and the center of mass. In mgd

this case, d = 0.9 m. I = moment of inertia of the painting =

⇒ T = 2π 15-54. †15-55.

T = 2π

1 mR 2 2

1 mR 2 1 1 2 = 2π R = 2π i 2 m i = 3.0 s mgd 2 gd 2 × 9.81 m/s 2 × 0.9 m

I mR 2 R 1 m = 2π = 2π = 2π = 2.0 s mgd mgR g 9.81 m/s 2

(a) τ = Iα ⇒ −κθ = I

d 2θ , dt 2

where I = moment of inertia of a disc =

1 MR 2 2

1 d 2θ d 2θ 2κ + MR 2 2 + κθ = 0 ⇒ θ =0 2 2 dt dt MR 2 The solution to this differential equation is θ = A cos(ω t + δ ),

⇒

where ω =

2κ MR 2 .

dθ = − Aω sin(ω t + δ ). The maximum angular speed is dt 2κ MR 2

(b) The angular velocity is θ = .

θ max = Aω = A

2κ = θ0 MR 2

Note that θ is not the same as ω! 2κ 2κ = θ0 ⇒ θ 0 = 1 radian (c) For ω = θ max , 2 MR MR 2 15-56.

Because T =

2π

ω

= 0.5 s, and ω =

κ

, then T = 2π 2

MR 2

MR κ −3 2 2 4π MR 4π × 8 × 10 kg × 0.01 m = = 1.26 × 10–4 kg m2/s2 Therefore, κ = 2 2 0.5 s 0.52 s 2 2

2

2


CHAPTER

15-57.

d 2θ ⎛g⎞ = – ⎜ ⎟ θ = –ω2θ dt 2 ⎝L⎠ (for small values of θ). Then 0.5 ⎛g⎞ θ = θ0 cos ⎜ ⎟ t. But the motion ⎝L⎠ here is not truly simple harmonic, as not all values of θ are possible. From the diagram we see that at t = 0, θ0 = 10°, and that the minimum

For a pendulum,

value of θ is 5°, or

θ0 2

0.5

. If we solve

θ0 2

0.5

⎛g⎞ ⎛g⎞ = θ0 cos ⎜ ⎟ t' for t', then ⎜ ⎟ t' = 60°, ⎝L⎠ ⎝L⎠ 0.5

⎛L⎞ or t' = 1.047 ⎜ ⎟ . The ball will bounce ⎝g⎠ back in the same time interval. 0.5

⎛ L⎞ The period of the motion is therefore 2t ′ = 2.09 ⎜ ⎟ s. ⎝g⎠ 15-58.

m m 0.99 θ = A cos ωt = θmax cos ωt g ω= l g t and Then θ = A cos e dθ g g sin t. =–A dt l l g ⎛ dθ ⎞ ⎜ ⎟ max = A l so that ⎝ dt ⎠ 10 ⎛ dθ ⎞ vmax = l ⎜ 9.8 × 0.99 m/s = 0.31m/s ⎟ = A gl = 99 ⎝ dt ⎠

Angle of arc = θ = 0.1

d 2θ ⎛g⎞ = –A ⎜ ⎟ cos 2 dt ⎝l ⎠

⎛ d 2θ ⎞ g ⎛g⎞ = A⎜ ⎟ t so that ⎜ 2 ⎟ l ⎝l ⎠ ⎝ dt ⎠ max

⎛ d 2θ ⎞ d 2x ⎛ 10 ⎞ 2 2 = l ⎜ 2 ⎟ = Ag = θmax g = ⎜ ⎟ 9.8 m/s = 0.99 m/s 2 dt ⎝ 99 ⎠ ⎝ dt ⎠


15

CHAPTER

15

(b) Force exerted = centripetal force exerted on bob + gravity mv 2 v 2 max (0.31) 2 (m 2 /s 2 ) = + mg = m = (0.40 kg) + 9.8 = 4.0 N r l+g 0.99 At the endpoint, the force is simply the component of the force in radial direction = mg cos θ 10 = 0.40(9.8) cos 99 = 3.9 N 15-59.

(a) E =

1 mglA2 2 4 π π= 180 45 2 1 1 ⎛π ⎞ E = mg l ⎜ ⎟ = mglπ2 2 4050 ⎝ 45 ⎠ 3 1 = (0.120 kg) ( 9.81 m/s 2 ) (0.44 m) π2 = 1.26 × 10 J 4050 A = θmax = 40 =

(b)

1 2 mvmax = E, vmax = speed at lowest point 2

vmax = 15-60.

ω=

2E 2 × 1.26 × 103 = = 0.145 m/s m 0.120 I mgl 2π . Therefore, T = = 2π ω mgl I

TNBS = 2π

I I ; TUSCG = 2π mlg NBS mg USGC

T Therefore NBS = TUSCG

g USCG ⇒ gUSCG = gNBS g NBS

⎛ TNBS ⎞ ⎜ ⎟ ⎝ TUSCG ⎠

2

2

†15-61.

⎛ 2.10356 s ⎞ 2 gUSCG = 9.80095 m/s2 ⎜ ⎟ = 9.80104 m/s 2.10354 s ⎝ ⎠ 9.80104 − 9.80095 Percent change = × 100% = 0.0009% 9.80095 I T = 2π mgd 1 2 ml 1 2 2π l . = Because the stick swings around one of its ends, I = m l . Therefore, T = 2π 3 3 mgd 3gd The distance between the point of suspension and the center of mass is d = 0.5 m. The length of the stick is l = 1 m.


CHAPTER Then T = 15-62.

2π l

=

3 gd

2π × 1 m 3 × 9.81 m/s 2 × 0.5 m

= 1.64 s

I of cylinder about CM through axis perpendicular to longitudinal is ⎛ a2 l 2 ⎞ m⎜ + ⎟ where a = radius, l = length. ⎝ 4 12 ⎠ Therefore, I about pivot = Irod + Icylinder ⎛ a2 l 2 ⎞ l ⎞2 1 ⎛1 ⎞ ⎛ = ⎜ md 2 + mb 2 ⎟ + M ⎜ + ⎟ + M⎜d + ⎟ 2⎠ 4 ⎝3 ⎠ ⎝ ⎝ 4 I2 ⎠ where m = mass of rod, M = mass of cylinder. Center of mass of pendulum is m(d / 2) + M (d + l / 2) x= m+M 2 m = πb dρ, M = πa2eρ where ρ = density of brass. 2 2 2 2 ⎡ d l ⎞⎤ ⎛ π b 2 d ) + π a 2l ⎜ d + ⎟ ⎥ ⎡⎢ b d + a l + a 2ld ⎤⎥ ( ⎢ ρ 2 2⎠ ⎝ ⎥ = ⎢ 2 2 2 2 Therefore, x = ⎢ ⎥ 2 2 ρ⎢ b d+a l π (b d + a l ) ⎥ ⎢ ⎥ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎛ a2 l 2 ⎞ 1 ⎞ ⎛1 I = πb2dρ ⎜ d 2 + b 2 ⎟ + πa2lρ ⎜ + + d 2 + dl ⎟ 3 4 ⎠ ⎝3 ⎝ 4 ⎠

ω (m + M )

gx = I

2 ⎛ b2 d 2 ⎞ 2 l a + + a 2 ld ⎟ g ⎜ 2 ⎝ 2 ⎠ 2 ⎛ ⎞ 1 ⎞ a l2 ⎛1 b2 d ⎜ d 2 + b2 ⎟ + a 2l ⎜ + + d 2 dl ⎟ 4 ⎠ 3 ⎝3 ⎝ 4 ⎠

d = 90.00 cm, e = 20.00 cm, a = 3.00 cm, b = 0.50 cm and g = 980.7 cm/s2

(1.865 × 10 ) 7

ω= T= 15-63.

( 6.075 × 10 2π

ω

4

+ 1.806 × 106 ) rad/s

= 3.161 rad/s

= 1.988 s

Tbrass = 2π

l gbrass

Tiron = 2π

l giron

(12 × 60 ) = 267.5 cycles. 1.800 = 2.691 s. Then the number of cycles after 12 min = 2691 9.81 1/4 of a one-way swing is T/8 = 0.13 cycle. Since the pendulums are 1/4 swing out of step, the ratio of one period to the other cannot be different by more than (267.5/267.63). T = 2π


15

CHAPTER

15

T1 267.5 = = 267.63 T2

g2 . Therefore g1

⎛ 267.5 ⎞ 2 g2 = g1 ⎜ ⎟ ⇒ g2 = 0.999 g1 ⎝ 267.63 ⎠ Largest difference is ≈ 0.001 g = 9.8 × 10−3 m/s 2 15-64.

Moment of inertia 2 1 1 d⎞ ⎛ 2 2 I = M1l + M2d + M2 ⎜ l + ⎟ 3 12 2⎠ ⎝ 1 1 (4.1)(0.46) 2 = ( 6.8 × (0.43) 2 ) + 3 12 + 4.1 (0.66)2 = 2.3 kg m2 Center of mass is at ⎡ 1 d ⎞⎤ ⎛ ⎢ M1 2 + M 2 ⎜ l + 2 ⎟ ⎥ ⎝ ⎠⎦ ⎣ = 0.38 m M1 + M 2

ω= =

( M 1 + M 2 ) (0.38)(9.8) I

(10.9 ) (0.38)(9.8) 2.3

Time period T =

2π

ω

= 4.2 rad/s.

= 1.5s which is approximately the

“experimental” value. 15-65.

mgl . Approximate the ruler by a thin rod. Then I through I 1 md 2 + me2 (e,d are as shown) axis (by parallel axis theorem) = 12 mgl ω= ⎛ 1 ⎞ m ⎜ d 2 + l2 ⎟ ⎝ 12 ⎠ ω=

=

ge

⎛ 1 2 2⎞ ⎜ d +l ⎟ ⎝ 12 ⎠

=

Therefore T = period =

9.8 m/s 2 0.2 m = 4.0/s ⎡1 2 2⎤ (1.0) 0.2 + ⎢⎣ 12 ⎥⎦ 2π

ω

= 1.6 s


CHAPTER

15-66.

15

mgl I I = (by parallel axis theorem) 1 = mR2 + ml2 2 mgl Then ω = ⎛1 2 2 ⎞ ⎜ mR + ml ⎟ ⎝2 ⎠

(a) ω =

=

gl 1 ⎛ 2 2 ⎞ ⎜ R +l ⎟ 2 ⎝ ⎠

⎡ ⎤ ⎢ ⎥ dω 1 gl ⎥ (b) = ⎢ de ⎢ 2 ⎛ 1 R2 + l 2 ⎞ ⎥ ⎟⎥ ⎢⎣ ⎜⎝ 2 ⎠⎦

−1 / 2

⎛1 ⎞ g ⎜ R 2 + l 2 ⎟ − 2 ge 2 ⎝2 ⎠ 2 ⎛1 2 2 ⎞ R l + ⎜ ⎟ ⎝2 ⎠

⎛1 2 2 ⎞ 1 2 2 ⎜ R + l ⎟ gR − gl 2 ⎝ ⎠ 2 = g2 2 gl ⎛1 2 2 ⎞ R l + ⎜ ⎟ ⎝2 ⎠ 1 2 2 R This equals zero when R = l or l = 2 2 15-67.

Mgh 2π = where M = m1 + m2, and h as shown in the diagram, is the T I distance from the center of the rod (the axis of rotation), to the center of mass of the system. I is the moment of inertia of the system about the axis of rotation. I = m1L2 + m2L2 But ( m1 + m2 ) h = m1L – m2L

The frequency, ω =

h=

( m1 − m2 ) L . ( m1 + m2 )

Therefore T = 2π T = 2π

I Mgh

(m1 + m2 ) L ( m1 − m2 ) g


CHAPTER 15-68.

15-69.

15

The figure shows a free-body diagram for the rod. The torque of the weight about the suspension point is 1/2 mgL sin θ, and the moment of inertia is 1/2 ml2. Hence the equation of rotational motion is 1 2 1 mL α = mgL sin θ 3 2 3g sin θ α= 2L

The acceleration of the center of mass is a = αL/2, in the transverse direction (see figure: since ω = 0, there is no centripetal acceleration at the initial instant). The equation for translational mα L = mg sin θ – F sin φ. The equation for motion in this transverse direction is then 2 translational motion in the direction along the rod is 0 = mg cos θ – F cos φ. From these equations, we find 3 1 F sin φ = mg sin θ – mg sin θ = – mg sin θ (1) 4 4 and (2) F cos φ = mg cos θ Squaring and adding these, we obtain 1 15 ⎛ ⎞ F2 = (mg sin θ ) 2 + (mg cos θ)2 = (mg)2 ⎜ 1 − sin 2θ ⎟ 16 16 ⎝ ⎠ and 15 15 F = mg 1 − sin 2θ = mg 1 − sin 2 20° = 0.94 mg 16 16 From the ratio of Equations (1) and (2), 1 1 we obtain tan φ = – tan θ = – tan 20° 4 4 φ = – 5.2° Thus the force F has a magnitude of 0.96 mg and a direction of 25.2° with the vertical. When the door is open, τ = 54 = κθ = 30 θ and θ = 1.8 radian. Therefore, for the shut door, θ = 1.8 –

π 2

= 0.23 rad. The change in the potential energy between

the open and shut door 1 1 1 ∆U = κ (0.23)2 – κ (1.8)2 = (30) ⎡⎣ (0.23) 2 − (1.8) 2 ⎤⎦ = 47.8 J. 2 2 2 From conservation of energy, 1 K = Iω2 = – ∆U = 47.8 J 2 (47.8)(2) ω= I


CHAPTER

15

1 2 1 Ml = (27)(0.91) 2 = 7.45 kg m2 3 3 (47.8)(2) = 3.6 rad/s Therefore ω = 7.45 The linear speed of the edge is v = ωl = 3.6 × 0.91 = 3.3 m/s For the door I =

15-70.

15-71.

According to the text just before Checkup 15.4, the increase in the period of a 30° swing 1.7%. T = 2π l / g = 2π 1.5 / 9.81 = 2.457 s. But that with 30° swing has T = 2.499 s. So after 10 min, the “5°” has swung 600 s/2.457 s = 244.2 times, while that with a 30° amplitude has swung 600 s/2.499 s = 240.0 times. Thus the two are out of synch by about 4 cycles out of 240, which would be easy to detect with careful counting. However, it is unlikely that Galileo could keep the amplitude fixed at 30º. Damping would reduce the amplitude, causing the period to decrease to its small amplitude value. Galileo would correctly interpret his result by concluding that the period is independent of the initial amplitude. The restoring force exerted by the rod, F = –kθ. The restoring torque is therefore –kθd, where d is the length of the rod. With a mass m attached to the upper end, the torque exerted by the mass is mgd sin θ, and this torque opposes the restoring torque. The net torque is therefore τ = mgd sin θ – kθ d. For small θ, sin θ ≅ θ so that τ ≅ –θd(k – mg) ⎛ d 2θ ⎞ But from Newton’s Second Law, τ = Iα = I ⎜ 2 ⎟ , where I = md 2 for the system, since we are ⎝ dt ⎠ treating the rod as massless. The equation of motion for the rod is therefore (for small displacements) (md 2 ) α = –θ d(k – mg)

⎡ k ⎛ g ⎞⎤ α+ ⎢ − ⎜ ⎟ ⎥ θ = 0, ⎣ md ⎝ d ⎠ ⎦ which is the equation for simple harmonic motion with ω2 = – k ⎤ ⎡ g ω = ⎢− + ⎥ ⎣ d md ⎦

g k + or d md

0.5

k⎤ ⎡ For stable solutions, ⎢ − g + ⎥ must be greater than 0. Therefore the rod becomes unstable when m⎦ ⎣ k k – g = 0; that is, when m = . m g


CHAPTER 15-72.

15

(a) If the train is at a distance x from the center of the tunnel (see Figure), then the acceleration of gravity is ⎛ GM ⎞ ⎜ 3 ⎟ r and the component of this acceleration in the ⎝ R ⎠ horizontal direction is GM ⎛ GM ⎞ ⎛ GM ⎞ x gx = – ⎜ 3 r ⎟ cos θ = – ⎜ 3 r ⎟ = – 3 x R ⎝ R ⎠ ⎝ R ⎠r (b) The horizontal force on the train is GMm x mgx = – R3 This is a linear restoring force, as in the case of a mass on a spring. The effective “spring GMm constant” is k = . Hence the period of the motion is R3 m m R3 = 2π = 2π 3 GMm / R GM k (c) The time for a trip from San Francisco to Washington is one-half period,

T = 2π

∆t =

T R3 (6.4 × 106 m)3 =π =π −11 2 GM 6.67 × 10 N i m 2 /kg 2 × 5.98 × 1024 kg

= 2.5 × 103 s = 42 min (d) For a mass on a spring, the potential energy is

1 2 kx , and hence the kinetic energy acquired 2

when moving from x = x0 to x = 0 is 1 2 1 mvmax = kx02 2 2 from which vmax =

kx02 = m

x0 m/k

=

2π x0 T

m is the period of the motion. k From Example 1.4, we know that the length of the tunnel is 3.8 × 103 km, i.e., x0 = (3.8 × 103 km) . From part (b) we know that T = 2 × 42 min. Hence, 2 2π × (3.8 × 106 m) / 2 = 2.4 × 103 m/s = 8.5 × 103 km/h vmax = 2 × 42 min

where T = 2π

15-73.

According to Problem 12.59, the moments of inertia of the cone about an axis through its apex is 3/5 ML2. To find the period, we need the distance between the apex and the center of mass. If the half-angle at the apex is θ, the radius of the cone at a distance z below the apex (see Figure) is R ≈ –zθ and the mass in a disk-like segment of radius R and thickness dz is ρπR2dz. Hence the z-coordinate of the center of mass is


CHAPTER − L4 ∫ zρπ R dz = ∫− L z dz = 4 = 3 L zcm = − L0 0 3 2 2 ∫− L ρπ R dz ∫− L z dz L3 4 Equation (48) for frequency of a physical pendulum tells us that 0

ω=

0

2

Mg⏐zcm⏐ = I

Mg (3 / 4 L) = (3 / 5ML2 )

The period is then T =

15-74.

3

2π

ω

= 2π

5g 4L 4L 5g

(a) The force at the midpoint is given by: ⎡ GMm ⎤ GMm + F(x) = – ⎢ 2 ⎥ 2 ⎣ (r + x) ⎦ (r − x) ⎡ 1 1 ⎤ dF d − = GMm ⎢ 2 (r + x) 2 ⎥⎦ dx dx ⎣ (r − x) ⎡ 2 2 ⎤ = GMm ⎢ + 3 (r + x)3 ⎥⎦ ⎣ (r − x)

By Taylor’s expansion, 4 GMm ⎛ dF ⎞ F(x) ≈ F(0) + ⎜ ⏐x = 0 ⎟ x = 0 + x. r3 ⎝ dx ⎠ Thus, the force is 4 GMm|x|/r 3 in the direction of motion of the particle. (b) By symmetry, the force along the line joining the centers of the two spherical bodies is zero. In a direction perpendicular to this line, it is: −GMm −GMm F= 2 sin θ + 2 sin θ 2 (r + r ) (r + r 2 ) =

⎡ ⎤ x −2GMm x = –2GMm ⎢ 2 2 2 2 2 1/ 2 2 3/ 2 ⎥ (r + r ) (r + r ) ⎣ (r + r ) ⎦

3 2 ⎡ 2 ⎤ 2 3/ 2 2 1/ 2 ⎢ (r + x ) − x 2 (r + x ) 2 x ⎥ dF = –2GMm ⎢ ⎥, ( r 2 + x 2 )3 dx ⎢ ⎥ ⎢⎣ ⎥⎦ 3 dF r −2GMm | x = 0 = –2GMm 6 = dx r r3


15

CHAPTER 15-75.

15

(a) The tension in the string obeys the equation. T – mg cos θ = mω 2l = mθ 2l T = mg cos θ + mθ 2l ⎛ g ⎞ g θ = –A sin ⎜⎜ t ⎟⎟ l ⎝ l ⎠ Then ⎡ ⎛ g ⎞ ⎛ g ⎞⎤ t ⎟⎟ T = mg cos ⎢ A cos ⎜⎜ t ⎟⎟ ⎥ + mg A2 sin2 ⎜⎜ ⎢⎣ ⎝ l ⎠ ⎝ l ⎠ ⎥⎦ ⎛ g ⎞ θ2 1 Using cos θ = 1 – + . . . . = 1 – A2 cos2 ⎜⎜ t ⎟⎟ = . . . . 2 2 l ⎝ ⎠ we have ⎡ ⎛ g ⎞ ⎛ g ⎞⎤ A2 T = mg ⎢1 − cos 2 ⎜⎜ t ⎟⎟ + A2sin 2 ⎜⎜ t ⎟⎟ ⎥ 2 ⎢⎣ ⎝ l ⎠ ⎝ l ⎠ ⎥⎦ Using cos2 x = 1 – sin2 x, we have ⎛ g ⎞⎤ ⎛ g ⎞ ⎫⎪ A2 ⎡ ⎪⎧ 2 T = mg ⎨1 − t ⎟⎟ ⎥ + A2 sin 2 ⎜⎜ t ⎟⎟ ⎬ ⎢1sin ⎜⎜ 2 ⎣⎢ ⎝ l ⎠ ⎦⎥ ⎝ l ⎠ ⎭⎪ ⎩⎪

T = mg –

mgA2 3 + mgA2 sin2 2 2

⎡ A2 3 2 2 ⎛ T = mg ⎢1 − + A sin ⎜⎜ 2 2 ⎝ ⎣⎢

⎛ g ⎞ t ⎟⎟ ⎜⎜ ⎝ l ⎠ g ⎞⎤ t ⎟⎥ l ⎟⎠ ⎦⎥

(b)The sin2 term has a maximum when t=

15-76. †15-77.

π 2

g π t = , so that 2 e

l . g

The tension is then ⎛ A2 3 3 ⎞ + A ⎟ = mg (1 + A2 ) Tmax = mg ⎜ 1 − 2 2 ⎠ ⎝ Normal marching pace is around 90 steps per minute, so f ≈ 1.5 Hz. 2π E , where ∆E is the energy lost during As in Example 10, we’ll use Q = ∆E the first cycle and E is the pendulum’s initial energy. To find this quantity, use the hint that ∆E/E is the same for every cycle. If the energy decays by ∆E to a value E1 during the first cycle, then it will decay by ∆E1 to E2 during the second cycle with ∆E1/E1 = ∆E/E. So after one cycle the energy would be E1 = E (1 − ∆E / E ) and after two cycles the energy would be E2 = E1 (1 − ∆E1 / E1 ) = E (1 − ∆E / E ) 2 . After n cycles the energy would be 1/ n

En = E (1 − ∆E / E ) n , from which we get

∆E ⎛E ⎞ =1− ⎜ n ⎟ E ⎝ E ⎠


.

L θ

L h

CHAPTER

15

When a pendulum of length L is raised through some angle θ, the mass at the end is lifted through a height L – L cos θ. The potential energy of the mass is mgh = mgL(1 – cos θ). If the pendulum is released from this height, the equation gives the total energy of the pendulum as it swings. According to the Math Help box following Eq. (15.46), the cosine function can be approximated by 1 −

θ2

for small angles given in radians. Using this in the equation for the energy gives 2 mgLθ 2 E= . For a 1.5-m-long pendulum with an amplitude of 10° (0.175 radian), the initial 2 m(9.81 m/s 2 )(1.5 m)(0.175) 2 = 0.224m J. (No mass is given in the problem, but it energy is E = 2 will cancel in the calculation of Q.) When the amplitude is 4° (0.0698 radian), the final energy is L E final = 0.036m J. The period of the pendulum is T = 2π = 2.46 s, so 12 min represents g 12 × 60 ≈ 293 cycles. The energy loss per cycle is given by 2.46 1 / 293 ∆E 2π ⎛ 0.036m ⎞ =1− ⎜ = 6.22 × 10−3. The final result is Q = = 1.0 × 103. ⎟ −3 E 6.22 × 10 ⎝ 0.224m ⎠ 15-78.

2π E . Since E ∝ A2 , decreasing A by a factor of 2 decreases E by a factor of 4 to E/4. ∆E The period on Earth for a pendulum with a length of 0.994 m turns out to be 2.00 s (see Example 8), so 13 minutes represents 390 cycles. The energy loss per cycle

(a) Q =

1 / 390

∆E 2π ⎛1⎞ =1− ⎜ ⎟ = 0.00355, and Q = = 1.77 × 103 . 0.00355 E ⎝4⎠ (b) For an amplitude of 8° = 0.140 radian, 2 mgLθ max (1.2 kg)(9.81 m/s 2 )(0.994 m)(0.140)2 E= = = 0.114 J. 2 2 ∆E 2π E 2π E The energy loss per cycle is ∆E = . Thus ⇒P= = Q T QT 2π (0.114 J) P= = 2.02 × 10−4 W (1.77 × 103 (2.00 s) is

†15-79.

(a) Follow the method outlined in Problem 77. Here a mass of 25 kg is given. The initial and final amplitudes are 12° (0.209 radian) and 10° (0.175 radian), respectively. 2 mgLθ init (25 kg)(9.81 m/s 2 )(3 m)(0.209) 2 E= = = 16.1 J . 2 2 2 mgLθ final (25 kg)(9.81 m/s 2 )(3 m)(0.175) 2 E final = = = 11.3 J after five swings. 2 2 1/ 5 ∆E ⎛ 11.3 ⎞ =1− ⎜ ⎟ = 0.0684. E ⎝ 16.1 ⎠ 2π E 2π Q= = = 91.9 0.0684 ∆E


CHAPTER

15

(b) The period is T = 2π ∆E = ∆E T 15-80.

=

L 3m = 2π = 3.47 s. The energy loss per cycle is g 9.81 m/s 2

2π E from which we get the rate of energy loss: Q 2π E 2π (16.1 J) = = 0.317 W. QT (91.9)(3.47 s)

2 kA2final kAinitial kA2 E= = . E final = . 2 2 2 1 / 10

1 / 10

15-81. 15-82. 15-83.

⎛ A2final ⎞ ⎛ E final ⎞ ∆E 1/ 5 =1− ⎜ = − = 1 − ( 0.95 ) = 0.0102. 1 ⎜⎜ 2 ⎟⎟ ⎟ E ⎝ E ⎠ ⎝ A ⎠ 2π E 2π Q= = = 616 0.0102 ∆E F Ak 0.4 m × 15 N/m A= 0Q ⇒ Q= = = 30 k F0 0.2 N A=

F0 1 × 10−18 N Q= i 5 × 106 = 1 × 10−9 m −3 k 5 × 10 N/m

k ω 1 k ⇒ f = = ⇒ k = (2π f ) 2 m 2π 2π m m f = 100 GHz = 10 × 109 Hz, m = 1.0 × 10–18 g = 1.0 × 10–21 kg so k = (2π × 100 × 109 Hz) 2 i 1.0 × 10−21 kg = 395 N/m

ω=

⇒ Q= 15-84.

(a)

Ak 1.0 × 10−10 m × 395 N/m = = 395 F0 1.0 × 10−10 N

A1 = 0.5, so one half of the amplitude is lost in its first oscillation. A0

(b) Since E =

15-85.

A2 1 2 E1 kA , = 12 = 0.52 = 0.25 2 E0 A0

Therefore, 75% of the energy is lost in the first oscillation. E 1 (c) Q = 2π = 2π = 8.37 0.75 ∆E E ∝ θ 2 (see problems 77 and 79). Cutting the angular amplitude in half means the energy decreases by a factor of 4, so the final energy is E/4. This loss of energy occurs in about four 1/ 4

∆E E 2π ⎛1⎞ = 1 − ⎜ ⎟ = 0.293. Then Q = 2π = = 21. ∆E 0.293 E ⎝4⎠ (a) x = A cos(ω t + δ ) Since at t = 0, x = xmax , so we should let δ = 0. 2π 2π rad/s = Since T = 0.8 s, so ω = T 0.8 2π ⇒ x = 0.2 cos( t) 0.8

cycles, so

15-86.


CHAPTER (b) At t = 0.1 s, x = 0.2 cos(

15

2π π × 0.1) = 0.2 cos( ) = 0.14 m 0.8 4

2π π × 0.2) = 0.2 cos( ) = 0 m 0.8 2 2π 3π × 0.3) = 0.2 cos( ) = – 0.14 m At t = 0.3 s, x = 0.2 cos( 0.8 4 2π × 0.4) = 0.2 cos(π ) = – 0.20 m At t = 0.4 s, x = 0.2 cos( 0.8 A = 1.5 cm 4000 rev/ min f = 4000 rev/min = = 66.7 Hz. 60 s/min At t = 0.2 s, x = 0.2 cos(

15-87.

15-88. †15-89.

vmax = Aω = 1.0 × 10–5 m × 2π rad/rev × 4000 rev/s = 0.25 m/s

π

(a) Since the first particle moves according to the equation x = 0.30 cos( t ), it reaches its 4 π π π 3π 5π , , ⇒ t = 2 s, 6 s, 10s, ... . At the midpoint when x = 0 ⇒ cos( t ) = 0 ⇒ t = , 4 4 2 2 2 π π dx turning points, = 0 ⇒ sin( t ) = 0 ⇒ t = 0, π , 2π , ⇒ t = 0, 4s, 8s, .... dt 4 4 Midpoint

Turning Point

Particle 1

Satellite

Turning Point

Satellite

Turning Point

Midpoint

Particle 1

Satellite

Turning Point

Particle 1

π

(b) The second particle moves according to the equation x ' = 0.30 sin( t ). At its midpoint, 4

π

π

x ' = 0 ⇒ sin( t ) = 0 ⇒ t = 0, π , 2π , ⇒ t = 0, 4 s, 8s, .... At its turning point, 4 4 dx ' π π π 3π 5π , , ⇒ t = 2 s, 6 s, 10s, ... . = 0 ⇒ cos( t ) = 0 ⇒ t = , dt 4 4 2 2 2 Turning Point

Midpoint

Particle 1

Satellite

(c) Find the time when x = x′:

π

π

Turning Point

Particle 1

π

π

Turning Point

Satellite

π

Midpoint

Particle 1

Turning Point

Satellite

0.3cos( t ) = 0.3sin( t ′) = 0.3cos( t ′ − ) = 0.3cos[ (t ′ − 2)] 4 4 4 2 4 We can see that when t′ = t+2, we will have x = x′, i.e., if the first particle passes a certain point at time t, the second particle passes the same point at time t+2, or 2 s later.


CHAPTER

15-90.

15

(a) f =

ω 3.0 = = 0.48 Hz 2π 2π

k ⇒ k = mω 2 = 6 × 32 = 54 N/m m vmax = Aω = 0.2 × 3 = 0.60 m/s k 54 N/m ω 5.20 (b) ω = = = 5.20 rad/s. f = = = 0.83 Hz 2π 2π m 2.0 kg

ω=

15-91.

vmax = Aω = 0.2 m × 5.2 rad/s = 1.04 m/s 6000 rev/min = 100 rev/s, so that v = 100 Hz. Then ω = wπv = 200 π/s 8.50 cos (200 πt) cm (set δ = 0 since it is arbitrary). Therefore, x = 2 dx = –4.25 cm (200 π)/s sin (200 πt). dt dx Max at |sin 200 πt| = 1 ⇒ = 2670 cm/s = 26.7 m/s dtmax d 2x = ⎡⎣ −4.25 cm(200π ) 2 /s 2 ⎤⎦ cos (200 πt). dt 2 d 2x Max at |cos 200 πt | = 1, 2 = 1.68 × 106 cm/s = 1.68 × 104 m/s dtmax 4 Force = mass × acceleration = 1.2 kg × 1.68 × 104 m/s2 = 2.0 × 10 N max

max

4π mtotal 4π 2 (1.40 kg) . For the 1.00 kg mass, k = = 47.4 N/m. When the unknown mass (1.08 s) 2 k 2

15-92.

T2 =

kT 2 (47.4 N/m)(1.78 s) 2 = = 3.80 kg. Subtracting the mass of the tray and 4π 2 4π 2 straps gives m = 3.40 kg. knew k knew = 2k ⇒ ωnew = = 2 = 2ω old m m ⇒ f new = 2 f old = 2 × 1.5 Hz = 2.12 Hz is used, mtotal =

†15-93.

15-94.

k ⇒ k = mω 2 m Since f = 2.4 Hz, So ω = 2π f = 4.8π rad/s

(a) ω =

⇒ k = 55 kg (4.8π)2 rad2/s2 = 1.25×104 N/m (b) ω = ⇒ f = †15-95.

f =

1 2π

k 1.25 × 104 N/m = = 12.9 rad/s mnew 75 kg

ω 12.9 rad/s = = 2.05 Hz 2π 2π k 1 = m 2π

4.9 × 103 N/m = 1.25 Hz 80 kg

Note that gravity does not affect the frequency of the motion. It only affects the equilibrium point.


CHAPTER

15-96.

(a) Total energy =

1 2 kA 2

If potential energy U equals kinetic energy K, then PE = ⇒

1 1 (total energy) = kA2 2 4

A 1 2 1 2 1 kx = kA ⇒ x2 = A2 ⇒ x = ± 2 4 2 2

i.e., when x = ±

A 2

, kinetic energy equals potential energy.

1 1 1 A2 kA2 A, then U = kx 2 = k = 2 2 2 4 8 1 1 Since E = kA2 , U = E , i.e. 1/4 of total energy is potential energy, 3/4 of total energy is 2 4 kinetic energy. 1 1 N E = kA2 = × 6.0 × 102 × 0.252 m 2 = 18.75 J. 2 2 m 1 2 = 18.75 J At the equilibrium position, U = 0, so Kmax = E = 18.75 J ⇒ mvmax 2 2E 2 × 18.75 J = = 3.54 m/s ⇒ vmax = m 3.0 kg (b) if x =

†15-97.

15-98.

†15-99.

A = 0.24 m, A will double. k ω= ⇒ ω won’t change. m Frequency won’t change. 1 2 Enew = kAnew = 4 Eold ⇒ Energy will quadruple. 2 vmax = Aω ⇒ vmax will double. amax = Aω2 ⇒ amax will double. l 1.5 m T = 2π = 2π = 2.46s. To change to half the period, g 9.8 m/s 2 T l' = 2π ⇒ 2 g

15-100.

l' 1 = g 2

l l ⇒ l ' = = 0.375 m. g 4

Before it hits the peg, T1 = 2 × 2π T2 = 2 × 2π

l1 l = 2 × 2π = 4.01 g 9.8 m/s 2

l2 (1/ 4) l m = 2 × 2π = 2.0 l g 9.8 m/s 2

⇒ period = T1 + T2 = 6.01 l


l

15

CHAPTER 15-101.

15

When the angle is small, τ = − Mg ( R + L) sin θ ≈ − Mg ( R + L)θ , 2 The moment of inertia is I = MR 2 + M ( R + L) 2 5 2 d 2θ Therefore, from Iα = τ , we get: ( MR 2 + ML2 ) 2 = − Mg ( R + L)θ 5 dt 2 dθ g ( R + L) ⇒ + θ =0 2 2 dt 2 2 R + ( R + L) 5 The solution for this differential equation is: θ = A cos(ω t + δ ),

where ω =

g ( R + L) . 2 2 2 R + ( R + L) 5

Therefore, T = 15-102.

2π

ω

= 2π

2 2 R + ( R + L)2 5 . g ( R + L)

τ = −mgx sin θ If the angle θ is small, τ ≈ −mgxθ 1 I = mL2 + mx 2 12 1 From Iα = τ we get: ( mL2 + mx 2 )θ + gxθ = 0 12 d 2θ gx ⇒ θ =0 + 1 dt 2 2 2 mL + mx 12 gx ⇒ω = 1 mL2 + mx 2 12 1 mL2 + mx 2 12 . T = = 2π ω gx 2π

To get the minimum period, we find the x that makes

d ⇒ dx 15-103.

T = 2π

dT =0 dx

1 2 L + x2 L 12 =0 ⇒ x= x 12

l 1 1 ⇒ f = = g T 2π

g 1 = 2π l

9.8 m/s 2 = 0.35 Hz 2m


CHAPTER 16

WAVES


f =

v

λ

⇒ f violet =

3.0 × 108 m/s = 7.5 × 1014 Hz 4.0 × 10−7 m

3.0 × 108 m/s = 4.3 × 1014 Hz −7 7.0 × 10 m 1 1 f = = = 0.114 Hz T 8.77 s ω = 2π f = 0.716 radian/s 2π 2π k= = = 0.0524 m −1 120 m λ λ 120 m v = fλ = = = 13.7 m/s T 8.77 s From the graph, there is a negative peak at about 9 h GMT. There is a positive peak at about 15 h GMT. There are 12.5 cycles during this 6 h interval, so the frequency is about v 740 km/h 12.5 cycles f = = 2.08 h −1 . λ = = = 356 km. 6.0 h f 2.08 h −1 f red =

16-2.

†16-3.

16-4.

16-5.

For λ = 1.0 m, v =

gλ = 2π

(9.81 m/s 2 )(1.0 m) = 1.3 m/s. 2π

For λ = 300 m, v =

gλ = 2π

(9.81 m/s 2 )(300 m) = 22 m/s. 2π

(a) v =

gD = (9.81 m/s 2 )(2.0 m) = 4.4 m/s

(b) v =

gD = (9.81 m/s 2 )(4.3 × 103 m) = 205 m/s (731 km/h).

16-6.

Moving with the crests means vboat = v, where v = wave speed. Thus v = 16 ms. One up and down bob is half of a complete cycle, so bobbing up and down six times per minute means f = 0.050 v 16 m/s Hz. The wavelength is λ = = = 320 m. f 0.050 Hz

16-7.

(a) time = distance/speed = 8000 km/ (740 km/hr) = 10.8 h (b) λ = 300 km. Therefore, f = v/λ = (740 km/hr) / 300 km = 2.5/hr 2π v (2π )(206 m/s) v = 740 km/h = 206 m/s. ω = 2π f = = = 0.0162 s −1 . λ 80 × 103 m

16-8.

v y ,max = ω A = (0.0162 s −1 )(0.30 m) = 4.85 × 10−3 m/s a y ,max = ω 2 A = (0.0162 s −1 ) 2 (0.30 m) = 7.85 × 10−5 m/s

2

These motions are so tiny they probably won’t be noticed.


CHAPTER

†16-9.

16 2π v

(2π )(8.0 m/s) = 22.8 s −1 . λ 2.2 m = ω A = (22.8 s −1 )(0.012 m) = 0.27 m/s. The maximum transverse speed occurs when

ω = 2π f = (a) v y ,max

=

the particle goes through its equilibrium position (between the crests). 2 (b) a y ,max = ω 2 A = (22.8 s −1 ) 2 (0.012 m) = 6.2 m/s . The transverse acceleration has its maximum magnitude at the crests. 16-10.

v y ,max = ω A and a y , max = ω 2 A ⇒ ω =

v y ,max

=

4.0 m/s 2 ω = 20 s −1 . f = = 3.2 Hz. 2π 0.20 m/s

0.20 m/s = 1.0 × 10−2 m = 1.0 cm. ω 20 s −1 Using Eq. (16.11) and following the method presented in Example 3, y = A cos(kx0 − ω t ). Taking x0 = 0, y = A cos ω t. In this problem, y = (0.020 m) cos(9.0t ), with t measured in s. A=

†16-11.

v y ,max

a y , max

=

(a) Comparing the given equation with the standard, A = 0.020 m. (b) ω = 9.0 s −1 ⇒ f =

ω 2π

= 1.43 Hz.

14 m/s v = = 9.8 m/s. f 1.43 Hz (a) y = A cos(kx + ω t + δ ). Thus A = 6.0 × 10–3 m, wave number = k = 20/m, wavelength = λ = (c) λ =

16-12.

†16-13.

16-14. †16-15.

16-16. †16-17.

2π/k = (2π/20) m = 0.31 m, angular frequency = ω = 4.0 radians/s, frequency f = ω/(2π) = 0.64 Hz. The direction of propagation is in the –x direction (because of the + sign). v = ω/k = (4.0 s–1) / (20 m–1) = 0.2 m/s (b) At x = 0, y = 6.0 cos (4.0 t + π/3). Therefore, we have a maximum when 4.0 t + π/3 = nπ, where n is any integer, or t = (πn – π/3) / 4.0. Then ymax occurs at t = –0.26 s, 0.52 s, 1.3 s etc. (a) T = 1/f = λ/v = (1.2 m)/(6.0 m/s) = 0.2 s. f = 1/T = 5.0 Hz. ω = 2πf = 10π s–1 = 31.4 s–1. k = 2π/λ = (2π)/(1.2 m) = 5.2/m (b) y = A cos (kx – ωt + δ), where the minus sign is chosen because the wave is propagating in the +x direction. Since there is a crest at x = 0 at t = 0, δ = 0. A = 2.0 cm = 0.020 m; k = 5.2 m–1; ω = 31.4 s–1 ⇒ y = 0.020 cos (5.2 x – 31.4 t) where length is in meters, time in seconds. f = 12 waves/min = (12/60) Hz = 0.20 Hz. v = f λ = (0.20 Hz)(39 m) = 7.8 m/s. v ∆v λ = . Taking differentials gives ∆λ = because f is constant. As the temperature changes f f from 30ºC to 15ºC, ∆v = v(15°C) − v(30°C) = 1440 m/s − 1530 m/s = −90 m/s. Thus −90 m/s ∆λ = = −0.20 m. The wavelength decreases by 20 cm. 440 Hz λ = v/f. λair = (3.0 × 108 m/s) / (5.5 × 1014 Hz) = 5.5 × 10–7 m λwater = (2.3 × 108 m/s) / (5.5 × 1014 Hz) = 4.2 × 10–7 m A = radius of the Ferris wheel = 6.1 m. f = 6 rev/min = 0.10 Hz. The magnitude of the maximum acceleration is amax = 4π 2 f 2 A = (4π )(0.10 Hz) 2 (6.1 m) = 2.41 m/s 2 . From Problem 4,

λ=

v = f

g λ /(2π ) 9.81 m/s 2 gλ g ⇒ λ2 = ⇒ = = = 156 m. λ f 2π f 2 2π f 2 (2π )(0.10 Hz) 2


CHAPTER

16-18.

†16-19.

16-20.

†16-21.

16

20π s −1 ω 2π 2π = 1.33 m/s. f = = 10 Hz. λ = = = 0.133 m 2π k 15π m −1 k 15π m −1 v y ,max = ω A = (20π s −1 )(0.13 m) = 8.2 m/s. y ( x = 0, t = 0) = (0.13 m)cos(0 + π /4) = 0.092 m. v=

ω

=

(9 / 2)waves = 0.30 Hz. The crest15 s to-trough distance is λ/2, so λ = 2(0.75 m) = 1.5 m. v = fλ = (0.30 Hz)(1.5 m) = 0.45 m/s. M 6.0 × 10−3 kg 2.0 m µ= = = 6.0 × 10−3 kg/m. v = = 60.6 m/s. L 2.0 m 0.033 s M F = µ v2 = = (6.0 × 10−3 kg/m) (60.6 m/s) 2 = 11.0 N. F = mg, where m = 7.0 kg. L F 11.0 N g= = = 1.6 m/s 2 . m 7.0 kg

Nine crests and troughs represent 9/2 complete waves, so f =

M ρ (π R 2 L) = = πρ R 2 = L L F 75 N kg/m. v = = = 104 m/s. µ 6.99 × 10−3 kg/m

Density of copper = ρ = 8.9 g/cm3 = 8.9 × 103 kg/m3. µ =

π (8.9 × 103 kg/m3 )(5 × 10−4 m 2 ) = 6.99 × 10−3

L 100 m = = 0.97 s. v 104 m/s gT 2 (9.81 m/s 2 )(1.0 s) 2 M 5.0 × 10−3 kg = = 0.248 m. µ = = = 0.0201 kg/m. L= 2 2 L 4π 4π 0.248 m

The ∆t = 16-22.

v=

16-23. 16-24.

16-25.

F

µ

=

mg

µ

=

(2.0 kg)(9.81 m/s 2 ) = 31 m/s. (Note that it is important to distinguish 0.0201 kg/m

between the mass M of the string and the mass m hanging at the end. Also note that the mass of the string is negligible compared to the mass hanging at the end of the string, so the string’s mass has no significant effect on the tension.) F F F 150 N L 30 m = ⇒µ = 2 = = 2.0 kg/m. v= = = 8.57 m/s. v = ( M / L) v (8.57 m/s) 2 µ t 3.5 s Y =

( F / A) FL YA∆L (1.5 × 1011 N/m 2 )(3.0 × 10−6 m 2 )(0.02 m) = . F = = = 6.0 × 103 N. L 1.50 m ( ∆L / L) A∆L

µ=

M 0.035 kg = = 0.0233 kg/m. v = L 1.50 m

F

µ

=

6.0 × 103 N = 507 m/s. 0.0233 kg/m

Speed of transverse waves, v = F/µ 12 × 10−3 = 4 × 10−3 kg/m 3 F = 250 N 250 Therefore, v = = 250 m/s 4 × 10−3

µ=

16-26.

v=

F / µ = (50 kg m/s 2 ) /(6.0 × 10−2 kg/m) = 28.9 m/s

time taken = dist/speed = 2 × 10 m/(28.9 m/s) = 0.69 s


CHAPTER

16-27.

16-28.

†16-29.

16-30.

16

d µ t d = 2 × 20 m = 40 m (2 × 20 because the pulse has to travel up and down) F = (d2/t2)µ. t = 1 s µ = 0.8 kg/m. Thus, F = tension = [(40 m)2/(1 s)2] (0.8 kg/m) = 1280 N F / µ and t = d / v = d µ / F . Therefore,

v=

F =

1.3 × 104 × 24 = 340 m/s 2.7 24 time taken = = 0.071 s 340 d d − , where d is the distance to the storm and v20, v100 are the The time difference is ∆t = v20 v100 ∆t = speeds of the 20 m and 100 m waves respectively. Thus d = (1/ v20 ) − (1/ v100 ) (10 h)(3600 s/h) = 1.84 × 105 m. The storm is 184 km away. −1 −1 (2.8 m/s) − (6.2 m/s) F/µ =

v=

Assume the boat is traveling slower than the waves. Then the frequency when the boat is moving v − vboat . When the boat is moving in the opposite in the same direction as the waves is f1 = waves vwaves + vboat

λ

2vboat 2(12 m/s) = λ f 2 − f1 (30 − 6.5) min −1 60 s/min −1 = (61.3 m)(6.5 min ) /(60 s/min) + 12 m/s = 18.6 m/s.

direction from the waves, f 2 =

†16-31.

= 61.3 m. vwave = λ f1 + vboat v 18.6 m/s = 0.304 Hz (about 18 waves/min). f wave = wave = λ 61.3 m F v= . The tension must be the same in both sections of the string,

µ

so v = 16-32.

. Then λ =

F

µ

and v ' =

F

µ'

. Thus

v' = v

µ µ . , or v ' = v µ' µ'

The tension is F = SA, where S is the stress. The mass/length is µ = is v =

F

µ

=

S

ρ

=

ρ LA L

= ρ A. The wave speed

5 × 108 N/m 2 = 253 m/s. The result does not depend on the length or 7800 kg/m3

diameter of the rod.


CHAPTER †16-33.

16-34.

16-35.

F1 = mg, where m = 30 kg. At the point where the strings are connected together, F1 mg 2 F2 cos 22.5° = F1 , or F2 = = . Substituting numbers 2 cos 22.5° 2 cos 22.5° gives F1 = 294 N, F2 = 159 N. The speed of a wave along each string is F1 294 N 159 N = = 271 m/s; v2 = = 200 m/s. v1 = −3 4.0 × 10 kg/m 4.0 × 10−3 kg/m µ

The length of each string is L, so the total travel time for a pulse is L L 1 1 ⎛ ⎞ ∆t = + = (2 m) ⎜ + ⎟ = 0.017 s. v1 v2 ⎝ 271 m/s 200 m/s ⎠ For a wire, the speed of propagation of a transverse pulse is given by µv2 = F, where F is the wire tension, and µ is the mass per unit length of the wire. The mass per unit length of the 0.3 mm wire, µ1 = m1/s1 = 2.2 × 10–3 kg/m For the 0.1 mm wire, µ2 = m2/s2 = 2.45 × 10–4 kg/m. Then v1 = (F/µ1)0.5 = 261 m/s, and v2 = (150/µ2)0.5 = 782.5 m/s. Then the propagation time along both wires, T = T1 + T2 = L1/v1 + L2/v2 = (5/261) + (5/782.5) = 0.0255 s or 25.5 ms. At a height z above the bottom of the rope, the tension is µgz = F, since the weight of a length z of rope is mg. Then F = µv2 = µgz, so that v = (gz)0.5 for the velocity of propagation at that point. Let T = the time for a pulse to travel from the bottom to the top of the rope, where the rope has a length s. Then dt = the time to travel through a length dz of the rope. Then

dt = dz /( gz )0.5, and T = ∫ dt =

∫

l

0

dz /( gz )0.5

l

= 1/( g )0.5 ∫ dz /( z )0.5 = 2 l / g

.

0

16-36.

16

The depth as a function of distance from the left end is D(x) = 4.0 – 0.08x. The speed as a function of x is v =

gdt =

Then ∆t

g ∫ dt = 0

∫

dx = dt

dx , from which we get 4.0 − 0.08 x

50

0

dx 2 4 − 0.08 x = (−0.08) 4.0 − 0.08 x

50

= 50 m1/2 . Thus 0

1/2

∆t =

gD ( x).

50 m

9.81 m/s 2

= 16 s.


CHAPTER 16-37.

16

Take an element of string, making an angle dθ with the center. Let F be the tension. As shown, the net effect of the tensions at the ends of the element is to produce a net force inward equal to 2 F sin (dθ/2) = F dθ because sin (dθ/2) ≈ dθ2 when dθ/2 is small. The mass of the element is mV 2 Rdθ dq dm = m. Then (dθ/2π)m V2/R = F dθ gives F = m= . The 2π R 2π 2π R speed of waves along the taut rope is v =

F

µ

=

F . [m /(2π R )]

Substituting the expression for F gives v = V. 16-38.

The length of the rope is s. Let Fs = the force exerted on the end of the rope by the wall. Then Fs = Fs, where Fe is the tension in the end of the rope. From the diagram, 2Fscos α = µsg, so that Fs = µsg/(2 cos α), where Fs is the tension in the ends of the rope. Because µv2 = F 0.5 Then vend = {F/µ} = {sg/(2 cos α)}0.5 From the second diagram, for the middle of the rope, F∆θ = µ∆sg = F∆s/R and ∆θ = ∆s/R Therefore, F/R = µg and F = µgR Then vmiddle = {F/µ}0.5 = {gR}0.5

16-39.

y1 + y2 = y3 at x = 0 ⇒ Ain cos (–ωt) + Aref cos ωt = Atrans cos (–ωt) ⇒ Ain + Aref = Atrans (1) dy1/dx + dy2/dx = dy3/dx at x = 0 ⇒ – kAin sin (kx – ωt) – kAref sin (kx + ωt)x = 0 = –k'Atrans sin (k'x – ωt)x = 0 ⇒ –kAin (–sin ωt) – kAref (sin ωt) = –k′Atrans (–sin ωt) ⇒ k ( Ain − Aref ) = k ′Atrans Note that v = T/µ

v ′ = T/µ ′

(2)

(v, v′ are velocities in the first, second ropes).

Because k = 2π v / v ⇒ k = 2π v µT = (2π v / T ) and k ′ = (2π v/ T ) µ ′ (i.e., k α

µ , k′ α

µ

µ ′)

′

Now [k' × (1)] + (2) gives 2kAin = k + k Atrans 2k Ain = (2 µ / µ ) + µ ′ Ain (by the fact that k α = Atrans = k + k′ And [k' × (1)] – (2) gives (k' + k)Ain + (k' – k)Aref = 0 µ − µ′ k − k′ Ain = Ain = Aref = k + k′ µ + µ′


µ , k′ α

µ′)

CHAPTER 16-40.

†16-41.

16

(a) The waves are in phase. (b) Since the waves are in phase, their amplitudes simply add. The total amplitude is 2π 0.050 m. The wave number k is 4.0 m–1 for both waves, so λ = = 1.6 m. k (a) y2 = 0.030 sin(4.0 x) = 0.030 cos(4.0 x − π / 2). The sine function is 90°, or π/2 radians, behind the cosine function. (b) Follow the method described for finding the beat frequency in Eqs. (16.16 – 16.20). y = y1 + y2 = 0.030[cos(4.0 x) + cos(4.0 x − π / 2)] ⎧ ⎡ 4.0 x − (4.0 x − π / 2) ⎤ ⎡ 4.0 x + (4.0 x − π / 2) ⎤ ⎫ cos ⎢ = 0.060 ⎨ cos ⎢ ⎥ ⎥⎬ 2 2 ⎣ ⎦ ⎣ ⎦⎭ ⎩

π⎞ ⎛ = 0.060 ⎜ cos ⎟ cos(4.0 x − π / 4) 4⎠ ⎝ The final result is y = 0.042 cos ( (4.0 x − π / 4) ). The amplitude is 0.042 m. The first maximum (crest) in a cosine function occurs when the argument is zero, so the first crest occurs at x=

16-42.

†16-43.

π

m = 0.20 m. 16 The largest amplitude will be the sum of the individual amplitudes, or 1.60 m. The smallest amplitude that can occur is zero. This will happen when the 0.80 and 0.50 m waves are 180° out of phase, and that resultant wave is 180° out of phase with the 0.30 m wave. These waves are exactly 180° out of phase, so the total amplitude is the magnitude of the difference between the individual amplitudes: ATOT = A1 − A2 = A − 2 A = A. Thus ATOT = 6.0 m. The wave number k = 4.0 m–1, so λ =

2π m = 1.57 m. The angular frequency ω = 4.0

5.0 −1 s = 0.796 Hz. At x = 1.0 m, t = 1.0 s, 2π y = (6.0 m)[cos(4.0 − 5.0) + cos(4.0 − 5.0 − π ) = −3.2 m.

5.0 s–1, so f =

16-44. †16-45.

16-46.

fbeat = f 2 − f1 =

ω 2 − ω1 = 1.11 beats/s. 2π

Following the method in Eq. (16.19), ⎧ ⎡ (5.0 x − 6.0t ) − (6.0 x − 7.0t ) ⎤ ⎡ (5.0 x − 6.0t ) + (6.0 x − 7.0t ) ⎤ ⎫ cos ⎢ y1 + y2 = 2 A ⎨ cos ⎢ ⎥ ⎥⎬ 2 2 ⎣ ⎦ ⎣ ⎦⎭ ⎩ x−t⎞ ⎛ ⎛ 11x − 13t ⎞ = 2 A ⎜ cos ⎟ cos ⎜ ⎟. 2 ⎠ 2 ⎝ ⎝ ⎠ The first cosine term represents a long wavelength oscillation with wave number 0.5 m–1 and angular frequency 0.5 s–1, and the second cosine term is a short wavelength oscillation with wave 2π number 5.5 m–1 and angular frequency 6.5 s–1. The short wavelength is λshort = m = 1.14 m. 5.5 2π The long wavelength is λlong = m = 12.6 m. The distance between zeros is λlong /2 = 6.28 m. 0.5 According to Fourier’s Theorem, a periodic disturbance of wavelength λ can be formed by the superposition of harmonic waves with wavelengths λ, λ/2, λ/3,…. Here λ = 2.0 m, so the wavelengths required for the superposition are (2.0 m)/n, where n = 1, 2, 3, ….


CHAPTER †16-47.

16

(a) Using Eq. (16.16), ⎧ ⎡ (16 x − 18t + 1.5) − (16 x − 18t − 2.3) ⎤ ⎡ (16 x − 18t + 1.5) + (16 x − 18t − 2.3) ⎤ ⎫ y1 + y2 = 2(0.030 m) ⎨ cos ⎢ cos ⎢ ⎥ ⎥⎬ 2 2 ⎣ ⎦ ⎣ ⎦⎭ ⎩ = (0.060 m) ( cos1.9 ) cos ( 6 x − 18t − 0.4 ) = −(0.019 m) cos ( 6 x − 18t − 0.4 )

16-48.

A = 0.0194 m. (b) At x = 0, t = 0, y = –(0.0194 m) cos(–0.4 radian) = –0.0179 m. y1 = A1 cos θ , with A1 = 3.0, θ = 5.0x – 8.0t y2 = A2 sin θ , with A2 = 4.0 y = y1 + y2 = A cos(θ + δ ) = A(cos θ cos δ − sin θ sin δ ) y = y1 + y2 = A1 cos θ + A2 sin θ Compare the two expressions for y. The coefficients of the cos θ terms must be equal, and the coefficients of the cos θ terms must be equal. Thus we have this pair of equations that must be satisfied: A cos δ = A1 − A sin δ = A2

16-49.

Dividing the second equation by the first gives A 4 δ = − tan −1 2 = − tan −1 = −53.1° (−0.927 radian). A1 3 A1 3.0 = = 5.0. Then the first equation gives A = cos δ cos 53.1° (a) y = 0.006 sin (π 0.5) cos 400πt – 0.004 sin (3π 0.5) cos 1200πt = 0.006 cos 400πt + 0.004 cos 1200πt (b)

16-50.

fbeat = f 2 − f1 = 0.007 Hz.


CHAPTER †16-51.

16

fbeat = f1 – f2 = 4 Hz. Therefore, the difference in the two is 4 Hz, so the frequency change must be 4 Hz. f = T /ω (λ const.) ⇒ f λ = T / µ f =

λ T µ

Therefore, f1 / f2 = T1 / T2 .

16-52.

16-53.

16-54. 16-55.

But f1/f2 = 298/294 (or 294/290) = 1.014 (both) Thus, (T1/T2) = (1.014)2 = 1.028 Therefore, tension needs to be increased (or decreased) 2.8%, but we don’t know which (increase or decrease) is required. Period of beats is ~14 days, so frequency of beat is ~0.071/day. Period of wave is 8 days for 7.7 periods (between days 12 and 20), so frequency of wave is 0.963/day. Thus, the average of the wave frequencies is 0.96/day and the difference is 0.071/day. (f1 + f2) / 2 = 0.96/day f1 – f2 = 0.071/day This gives f1 = 1.0/day and f2 = 0.92/day, and periods 1.0 day and 1.09 days. Obviously, 1.0 day must correspond to Sun. Hence, 1.09 day must correspond to Moon. [Note: the actual period for Moon is 24h 50m, or 1.03 days. The discrepancy indicates that the naive interpretation of Figure 16.25 as a wave due to the Sun and a wave due to the Moon is not quite correct. Also note that the tides at Pokhoi occur only once per day; this is unusual, but it does happen at a few localities.] nf fn = = nf1 where f n = frequency of (n − 1)th harmonic. 2L First harmonic: 2 × 196 Hz = 392 Hz Second harmonic: 3 × 196 Hz = 588 Hz Third harmonic: 4 × 196 Hz = 784 Hz Fourth harmonic: 5 × 196 Hz = 980 Hz Third overtone. Wavelength = 2L/4 = L/2 = 0.17 m. v 1 F , where µ = M/L. Substituting the The frequency of the fundamental mode is f1 = = 2L 2L µ numerical information, f1 =

16-56.

3.16 Hz. f = F/µ and f = v/λ. f =

F/µ / 2 L =

(

1 500 N = 1.58 Hz. The first overtone is f2 = 2f1 = 2(50 m) 0.020 kg/m λ = 2L (for each string, L and F are equal)

)

F/2 L 1/ µ . Therefore, f αµ −1 / 2

Ratios of frequencies are 196: 294: 440: 659 Ratios of densities are 1.00:0.444:0. 198:0.088 16-57.

f1 =

1 2L

F

µ

=

1 2.2 × 103 N = 28 Hz. 2(3.0 m) 0.080 kg/m


CHAPTER 16-58.

†16-59.

16

For a string, the resonant frequencies are fn = nf1, so the difference between two successive resonances is (n + 1)f1 – nf1 = f1. Here 660 Hz – 440 Hz = 220 Hz, so f1 = 220 Hz. The two frequencies given correspond to n = 2 and 3, so 440 Hz and 660 Hz are the first and second overtones, respectively. y = 2 A cos(ω t ) cos(kx) = 10 cos (3.0πt ) cos (π x). At x = 0.25 m, y = 10 cos (3.0πt ) cos (0.25π ) = 7.07 cos (3.0πt ). The amplitude at this location is 7.07 m.

v y ,max = ω A = (3π s −1 )(7.07 m) = 66.6 m/s a y , max = ω v y ,max = (3π s −1 )(66.6 m/s) = 628 m/s 2 16-60.

†16-61.

16-62. 16-63.

16-64.

16-65.

If the pole end is fixed, the pulses are inverted upon reflection. Then the forward and reflected pulses cancel when they pass each other and the total amplitude is zero. If the pole end is free to slide along the pole, the reflected pulse is oriented the same way as the forward pulse. The amplitudes add as they pass each other giving a total amplitude of 6.0 cm. If the plucked point is exactly at the center then the distance to each end is L/2 and the total distance traveled when the reflections return to the center is L. The time is L 0.65 m ∆t = = = 9.3 × 10−3 s, or 9.3 ms. Since the ends of the string are fixed, the pulses are v 70 m/s inverted upon reflection. If the string was pulled up when it was plucked, it will be displaced down when the reflections return to the center. The wavelength is 2(0.65 m) = 1.30 m, so the v 70 m/s vibration frequency is f = = = 54 Hz. λ 1.30 m 20 × 103 Hz nmax = = 727. Humans can hear about 726 overtones from 27.5 Hz. 27.5 Hz M 0.015 kg n fn (Hz) µ= = = 1.5 × 10−3 kg/m L 10 m 1 8.16 2 16.3 1 F 1 40 N f1 = = = 8.16 Hz 3 24.5 2L µ 20 m 1.5 × 10−3 kg/m 4 32.7 The overtones are nf1. The fundamental and next four overtones are 5 40.8 tabulated. 1 ⎛ 1h ⎞ 6 −5 f = ⎜ ⎟ = 2.32 × 10 Hz. λ = 4(250 km) = 1000 km = 1.0 × 10 m. 12 h ⎝ 3600 s ⎠ v2 v = fλ = 23.2 m/s. D = = 55 m. g

v=

F/µ , λ = 2 L for fundamental

f = v/λ =

F/µ /2 L so that (4 L2 f 2µ ) = F

F = tension = 4(0.34)2m2 (196 Hz)2 (4.0 × 10–3 kg/m) = 71 N


CHAPTER 16-66.

16

The area of the spoke = πR2 = π(0.20 cm)2 = 0.126 cm2. The density per unit length = µ = ρA = (7.8g/cm3) (0.126 cm2) = 0.98 g/cm = 0.098 kg/m v = F/µ , λ = 2 L for fundamental

F/µ /2 L = ⎡⎣ 220 N/ (0.098 kg/m) ⎤⎦ /(2 × 0.09 m) = 830 Hz

f = v/λ = †16-67.

Let mirror be at x = 0. Incoming wave is A sin (kx – ωt). Outgoing is A sin (kx – ωt), so that y = A sin (kx – ωt) + A sin (kx + ωt) = 2A sin kx cos ωt. The nearest antinode is when sin kx = 1 or kx = π/2 ⇒ x = π/(2k) = (π/2)(λ/2π) = λ/4 Nearest node at sin kx = 0, or kx = π ⇒ x = π/k = λ/2. λ = 5.0 × 10–7 m, so that the nearest antinode is at 1.25 × 10–7 m, and the nearest node is at 2.5 × 10–7 m.

16-68.

Let the wall be at x = 0. Then the equation for the incoming wave is A cos (kx + ωt), and the outgoing wave A cos (kx – ωt). Hence, the net wave = 2A cos kx cos ωt. A node appears when cos kx = 0, or kx = π/2 + nπ (n any integer). 1⎞ ⎛ ⎛n 1⎞ x = ⎜ n + ⎟ π /(2π / λ ) = ⎜ + ⎟ λ 2⎠ ⎝ ⎝ 2 4⎠ The nodes are at ⎛n 1⎞ ⎜ + ⎟ 3.0 m = [0.75 + 1.5 n] m where n = 0, 1, 2, . . . . ⎝ 2 4⎠

16-69.

× (cos ω t ). L At x = midpoint = L/2, so that y = A cos ωt |dy/dt|max = ωA |d2y/dt2|max = ω2A. A = 2.0 mm = 2.0 × 10–3 m. ω = 2 πf = 2π(294)/s = 1847/s. vmax = 2.0 × 10–3 × 1847/s = 3.7 m/s amax = (1847)2/s2 2.0 × 10–3 m = 6.8 × 103 m/s2 1 F f = for the fundamental. Let f1, F1 be the correct frequency and tension and f2, F2 be 2L µ

16-70.

y = A sin

π

the present tension and frequency.

f1 = f2

F1 . Then F1 = F2 (f1/f2)2 = (900 N) (261.6/246.6)2 = F2

1013 N. The tension must increase by 113 N.


CHAPTER †16-71.

16

(a) Let θ be the angle of deflection from the normal position. Let T be the tension. Then 2T sin θ = 150 N. Half the length of the rope is 4.5 m, and it deflected 0.070 m at its midpoint. Therefore 150 N sin θ = (0.07/4.5). T = = 4.82 × 103 N. 2(0.070/4.5) (b) For the fundamental, f1 =

1 T 1 T 1 4.82 × 103 N . Therefore, f1 = = 2L µ 2L µ 2(9.0 m) 0.22 kg/m

= 8.2 Hz.

16-72.

16-73.

16-74.

Resonant wavelengths at 2 × 2.5 = 5 m, 2.5 m, 2 × 1 m = 2 m, [2(2.5)/3]m = (5/3) m. f = v/λ, v = speed of sound = 330 m/s. Therefore, 330 m/s f = = (i) 66 Hz λ m (ii) 130 Hz (iii) 165 Hz (iv) 200 Hz f = 440 Hz A = 0.4 mm ⎛ 3π x ⎞ y = (0.4 mm) sin ⎜ ⎟ cos (ω t); ω = 2π f ⎝ 1.5 ⎠ = 0.4 mm (sin 2π x) cos (880 π t ) ⎛ A− B⎞ ⎛ A− B⎞ But sin A + sin B = 2 sin ⎜ ⎟ cos ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ Let A = 2πx – 800 πt B = 2πx + 800 πt Then 0.2 mm sin (2πx – 880πt) + 0.2 mm sin (2πx + 880πt) = 0.4 mm sin 2πx cos (880πt) Amplitudes are 0.2 mm, speed = v = ω/k = 880π/2π = 440 m/s The length of each string is L. The point where the strings are connected must be a node. Assume the other ends of the strings are fixed, so those ends must also be nodes. Then each string must be vibrating in a mode for which its own ends are nodes. The allowed resonant frequencies for the combination are all the common resonances shared by the two strings. The resonant frequencies n T . The tension T is the same in both strings, and the heavier string has are given by f = 2L µ twice the mass of the lighter string. Thus the allowed resonant frequencies of the heavier string are exactly half the frequencies of the lighter string, and the allowed resonant frequencies of the


CHAPTER

†16-75.

combination are all the resonances shared by the strings. This will be all of the resonances for the light string, because they correspond to the even resonances (n = 2, 4, 6,…) of the heavy string. F F (a) v( x) = = µ ( x) A + Bx (b) λ ( x) =

v( x) 1 = f f

F A + Bx

(c) For each eigenfrequency, the wavelength must satisfy λ ( L) =

16-76.

16

2L , n = 1, 2, 3, … . Thus n

2L 1 F n F . ⇒ fn = = 2 L A + BL n f n A + BL (a) r = A cos( kx − ω t ) j + A′ cos(kx − ω t )k. The distance from the origin is

r = [ A cos(kx − ω t )]2 + [ A′ cos(kx − ω t )]2 =

A2 + ( A′) 2 cos(kx − ω t ). The angle of the string

particle’s position measured from the z axis is θ = tan −1

A cos(kx − ω t ) A = tan −1 . Thus the A′ cos(kx − ω t ) A′

string particle oscillates back and forth along a line oriented at an angle θ from the z axis. (b) r = A cos(kx − ω t ) j + A sin(kx − ω t )k. The distance from the origin is

r = [ A cos(kx − ω t )]2 + [ A sin( kx − ω t )]2 = A cos 2 (kx − ω t ) + sin 2 ( kx − ω t ) = A. This means the string particle moves in a circle of radius A around the x axis.

ω . The angle of the string particle’s position measured 2π A sin(kx − ω t ) from the y axis is θ = tan −1 = tan −1 [ tan( kx − ω t ) ] , which means the angular A cos( kx − ω t ) position is θ ( x, t ) = kx − ω t relative to the z axis. The angular velocity of the string particle is ∂θ ( x, t ) = −ω , so the speed of the motion is v = ω A. The centripetal acceleration is

The frequency of the vibration is f =

†16-77.

∂t v2 = ω 2 A. a= A y = y1 + y2 = A1 cos( kx − ω t ) + A2 cos( kx + ω t ) = A1 (cos kx cos ω t + sin kx sin ω t ) + A2 (cos kx cos ω t − sin kx sin ω t )

= ( A1 + A2 ) cos kx cos ω t + ( A1 − A2 ) sin kx sin ω t Because sin θ = cos(θ − π / 2), this represents two standing waves that are π/2 or 90° out of phase with each other. The amplitude of the larger wave is A1 + A2 and these maxima occur where sin kx = 0. The sine function is zero when the argument is nπ, so the locations of the nπ antinodes are xn = . k The smaller standing wave has an amplitude of A1 − A2 , and these occur where cos kx = 0. The cosine function is zero when the argument is π/2, 3π/2, 5π/2,…. The “nodes” (locations of (2n + 1)π minimum amplitude) are xn = , n = 0, 1, 2,…. k


CHAPTER

16-78.

16

A( x) = A sin

πx L

. Each mass element is a simple harmonic

2 (dm)vmax (dm)[2π fA( x)]2 = . 2 2 Substituting for dm and integrating to find the total energy gives L πx E = 2π 2 µ f 2 A2 ∫ sin 2 dx. The integral can be evaluated by 0 L looking it up in a table. Its value is L/2, so we get E = π 2 µ Lf 2 A2 .

dm = µdx

oscillator with total energy dE =

x=0

x=L

Substituting, E = π 2 (2.2 × 10−3 kg/m)(0.18 m)(494 Hz) 2 (3.0 × 10−3 m) 2 = 8.6 × 10−3 16-79.

16-80.

†16-81.

16-82.

y = A cos kx = (0.030 m) cos(1.2 x). (a) A = 0.030 m 2π 2π (b) λ = = = 5.2 m. k 1.2 m −1 (c) Crests and troughs are located where kx = nπ, n = 0, 1, 2, 3,… . The crests and troughs are at x = 0, λ/2, λ, 3λ/2, 2λ,… . (The crests are at 0, λ, etc., and the troughs are at λ/2, 3λ/2, etc.) There are crests at 0, 5.2 m, 10.4 m,…, and there are troughs at 2.6 m, 7.9 m, 13.5 m,… . t = d/v. Let the speed of the p,s waves be vp, vs, respectively. ⎛1 1 ⎞ ts − tF = ∆t = d / vs − d / v p = d ⎜ − ⎟ ⎜ vs v p ⎟ ⎝ ⎠

⎛1 1 ⎞ ⎛1 1⎞ d = ∆t / ⎜ − ⎟ = (9 × 60) s/ ⎜ − ⎟ s/km ≈ 4000 km ⎜v ⎟ ⎝3 5⎠ ⎝ s vp ⎠ Assuming the wave can be described by y = A sin (kx – ωt) dy = ωa cos (kx − ωt ) dt d2y = −ω2 Α sin (kx − ωt ) dt 2 vmax = Aω and amax = Aω2 where A = 23.5/2 m 2π 2π = = 0.419 s ω= T 15 Therefore, 23.5 vmax = × 0.419 = 4.9 m/s 2 23.5 amax = × (0.419) 2 = 2.1 m/s 2 2 1 (a) ( v 2 − v02 ) = a( x − x0 ); here a = g and (x − x0 ) = 2 m. 2 Therefore, v = 2a( x − x0 ) = 2 × 9.8 m/s × 2 m = 6.3 m/s


CHAPTER

16-83.

16

(b) Max speed for S H M (y = A cos ωt) is |dy/dx|max = Aω. Aω = v ⇒ A = v/ω. v = 6.3 m/s, ω = 2πf. ω = 2π/s. A = (6.3/2π) m = 1 m ω = 2π f = 2π (100 Hz) = 628 s −1. v y ,max = ω A = (628 s −1 )(0.020 m) = 13 m/s. a y , max = ω 2 A = (628 s −1 ) 2 (0.020 m) = 7.9 × 103 m/s 2

16-84.

Let L be the length of the ship. Then the wavelength of the wave is λ = L/3. The speed of wave is ⎛ gλ ⎞ gλ g v= . Thus, the frequency of the wave is v / λ = ⎜⎜ . The frequency of the ⎟⎟ / λ = π πλ 2π 2 2 ⎝ ⎠ 1 g Hz = f = λ= 6 2π 3 × 9.81 m/s 2 = = 169 m 2π (1/6) 2 /s 2

waves is 10/min = 1/6 Hz. Therefore, f2 = †16-85.

3g 3g ⇒ L= 2π L 2π f 2

The mass of the rope is M = πρ R 2 L, where ρ is the density, R is the radius, and L is the length. M The mass/length is µ = = πρ R 2 = π (1.1 × 103 kg/m 3 )(4.6 × 10−3 m) 2 = 0.0731 kg/m. The L 5.0 × 103 N = 261 m/s. 0.0731 kg/m µ 0.0060 kg 0.0090 kg µ1 = = 2.0 × 10−3 kg/m. µ 2 = = 3.0 × 10−3 kg/m. The wave speeds are 3.0 m 3.0 m F F 200 N 200 N v1 = = = 316 m/s and v2 = = = 256 m/s. The total 0.0002 kg/m 0.0003 kg/m µ1 µ2

wave speed is v = 16-86.

F

=

travel time for the pulse is t = †16-87.

3g , or πL

(a)

∑F

x

∑F

y

L L 3.0 m 3.0 m + = + = 0.021 s. v1 v2 316 m/s 256 m/s

= T1 cos 45° + T2 cos 30° − T = 0

= T1 sin 45° − T2 sin 30° = 0

sin 45° 1/ 2 = T1 = 2T1. sin 30° 1/ 2 From the equation for the x components, T1 1+ 3 2T ⎛ 2 ⎞ . + 2T1 ⎜ − T = 0 ⇒ T1 = ⎟ − T = 0 ⇒ T1 2 2 1+ 3 ⎝ 3⎠ 2T . Then T2 = 1+ 3

⇒ T2 = T1

(b) The wave speed is vn =

Tn

µ

. All the strings are identical, so

v1 = v

T1 2 = = 0.719 ⇒ v1 = 7.2 m/s T 1+ 3

v2 = v

T2 2 = = 0.856 ⇒ v2 = 8.6 m/s T 1+ 3


CHAPTER

16

16-88.

y = y1 + y2 = 0.012 cos(3.0 x) + 0.030 cos(5.0 x).

16-89.

(a) The resulting wave is periodic because it repeats itself, but it is not harmonic because it can’t be described by a simple sine or cosine function. (b) The largest displacement from equilibrium will be the sum of the individual amplitudes, which is 0.042 m. The resultant wave is a complex waveform that does not have a single wavelength. fbeat = f1 − f 2 = 12 Hz

16-90.

y = 2 A cos(kx) cos(ω t ) = 0.020 cos(15 x) cos(3.0t ) (a) 2A = 0.020 m; f =

ω 3.0 s −1 = = 0.48 Hz 2π 2π

0.020 m 2π 2π = 0.010 m; f = 0.48 Hz; λ = = = 0.42 m 2 k 15 m −1 π 3π 5π π π π , , (c) Nodes are located where 15 x = , ⇒x= m, m, m, … 2 2 2 30 10 6 π 2π m, m,… Antinodes are located where 15x = nπ ⇒ x = 0, 15 15 (d) v y ,max = ω (2 A) = 0.060 m/s (b) A =

a y , max = ω 2 (2 A) = 0.18 m/s 2 †16-91.

(a) The tension at any point along the rope is equal to the weight below that point. Thus at a distance x from the top of the rope, the tension will be F ( x) = µ g ( L − x), where µ is the mass per unit length of the rope. The speed is v( x) =

F ( x)

µ

=

g ( L − x).

For L = 20 m, the speed at the top is v(0) = (9.81 m/s 2 )(20 m) = 14 m/s. At the midpoint, v(10 m) = (9.81 m/s 2 )(20 m − 10 m) = 9.9 m/s. At the bottom, v(20 m) = 0. dx dx (b) v( x) = ⇒ dt = = dt v( x)

dx g ( L − x)

. The time ∆t for a pulse to travel

from the top of the rope to the bottom is given by

∫

∆t 0

dt =

∆t = 2

∫

L 0

L−x ⇒ ∆t = −2 g g ( L − x) dx

L

=2 0

L . The final result is g

20 m = 2.9 s. 9.81 m/s 2

x=L


CHAPTER

16-92.

(a) F = mg ⇒ v = (b) f1 =

mg

µ

=

16

(2000 kg)(9.81 m/s 2 ) = 181 m/s. 0.60 kg/m

v 181 m/s = = 1.51 Hz. 2 L 2(60 m) f 2 = 2 f1 = 3.01 Hz f3 = 3 f1 = 4.52 Hz

†16-93.

16-94.

Since there is no friction on the loop and the loop has no mass, it moves up and down along the rod as a free end. This vibrating string must always have a node at its left end and an antinode at the right end where the loop slides on the rod. Since the distance from a node to the nearest antinode is one quarter of a wavelength, the lowest resonant frequency must have a corresponding wavelength λ1 = 4 L. The next highest resonance must also have a node at the left end and an antinode at the loop end, giving a node-antinode pattern of N-A-N-A. This corresponds to 3/4 of a 3 4L wavelength, which means L = λ2 , or λ2 = . The next possible node-antinode pattern is N4 3 4L . The pattern of A-N-A-N-A, and L now corresponds to 5/4 of a wavelength. Thus λ3 = 5 4L 4L 4L , ,… . allowed wavelengths is given by λ = 4 L, , 3 5 7 The tension in the top ropes is given by 2 F cos 30° = mg , which gives mg (15 kg)(9.81 m/s 2 ) = = 85.0 N. The speed of waves 2 cos 30° 2 cos 30° F 85.0 N along one of those strings is v = = = 84.1 m/s. The µ 0.012 kg/m F =

total length of the string is L = 12.0 m, and the allowed modes must have a node at the center. These correspond all the even-numbered harmonics for the string with a length of 12 m, which will have 2L L wavelengths given by = . The eigenfrequencies are 2n n nv n(84.1 m/s) f = = = 7.01n Hz. Thus the frequencies are 7.01 Hz, L 12 m 14.0 Hz, 21.0 Hz,….


CHAPTER 17

SOUND AND OTHER WAVE PHENOMENA


17-2.

17-3.

17-4.

17-5. 17-6.

17-7.

17-8.

λ = v/f v ≈ 331 m/s f = 20 Hz to 20000 Hz. Therefore, λmax = (331 m/s)/20 Hz = 17 m λmin = (331 m/s)/20000 Hz = 0.017 m (1.7 cm) Going down 4 octaves means multiplying frequency by 1/24 = 1/16, so 440.0 Hz flowest A = = 27.5 Hz. 16 Going up 4 octaves means multiplying frequency by 24 = 16, so f highest C = 261.7 Hz × 16 = 4187 Hz.

Use the speed of sound in sea water from Table 17.3. v 1531 m/s λwhale = = = 765 m 2.0 Hz f v 331 m/s λelephant = = = 166 m 2.0 Hz f v 331 m/s λdog = = = 8.3 × 10−3 m (8.3 mm) 4.0 × 104 Hz f v 331 m/s λbat = = = 4.4 × 10−3 m (4.4 mm) 7.5 × 104 Hz f Size of typical telephone handset is about 20 cm. Comparing this with wavefronts shown in photo, we find λ ≅ 9 cm. 1 F 329.7 Hz Two octaves below means multiply by (1/2)2. f = = 82.4 Hz. f = 4 2L µ ⇒ F = 4 f 2 L2 µ = 4(82.4 Hz) 2 (0.63 m) 2 (5.4 × 10−3 kg/m) = 56 N v λ= f 1500 m/s λ (0.8 MHz) = = 1.9 × 10−3 m (1.9 mm) 6 0.80 × 10 Hz 1500 m/s λ (15 MHz) = = 1.0 × 10−4 m (0.10 mm) 6 15 × 10 Hz 331 m/s 331 m/s In air, λ = = 0.414 m; λ (3500 Hz) = 0.0946 m (9.46 cm). . λ (800 Hz) = 800 Hz f The wavelength ranges are “Low” (< 800 Hz) λ > 0.414 m “Middle” (800 Hz to 3500 Hz) 9.46 cm < λ < 0.414 m “High” (> 3500 Hz) λ < 9.49 cm


CHAPTER

†17-9.

17-10.

†17-11.

vmax =

2I = ρ0 v

17

2(2 × 104 W/m 2 ) = 6.8 m/s. (1.29 kg/m3 )(331 m/s)

This is an extremely loud sound, corresponding to 160 dB. Sounds of this intensity will cause structural damage to people and buildings. (a) Wavelength of fundamental mode = 2L = 2 × 0.326 m = 0.652 m The wavelengths of sound λ = v/f = (331 m/s)/f. From the lowest to the highest frequencies, they are (i) (331 m/s)/196 Hz = 1.69 m (ii) (331 m/s)/294 Hz = 1.13 m (iii) (331 m/s)/440 Hz = 0.752 m (iv) (331 m/s)/659 Hz = 0.502 m (b) v = λf, v = speed must be constant. For the first harmonic λ = L = 0.326 m. Therefore, f = 2 (fundamental frequencies) = 392 Hz, 588 Hz, 880 Hz, 1318 Hz The wavelengths are: λ = v/f = (331/f) m. From the lowest to the highest frequencies they are: (i) 0.844 m (ii) 0.563 m (iii) 0.376 m (iv) 0.251 m 392 (c) 196 Hz = Hz. Therefore, 196 Hz = G. 294 Hz = D 440 Hz = A. 2 659 Hz = (329.5 × 2) = E, which are just the octaves of the four standard violin open-string notes. Since the tension is constant and v = F/µ , the speed of the wave on the string is constant. Since

v = fλ, λ = v/f. Because the string is vibrating in its fundamental mode, λ = 2L, where L = length of string. Then L = v/2f = (v/2)(1/f) or L1/L0 = f0/f1 where the subscript 0 indicates the frequency for the open string, 1 for the fretted string: ⎛ f ⎞ L = ⎜ 0 ⎟ L0 Spacing between Frequency ⎝ f1 ⎠ Tone D D# E F F# G G# A A# B C C# D

(Hz) 293.7 311.2 329.7 349.2 370.0 392.0 415.3 440.0 466.2 493.9 523.4 554.4 587.4

(cm) 34.0 32.1 30.3 28.6 27.0 25.5 24.0 22.7 21.4 20.2 19.1 18.0 17.0

frets (cm) — 1.9 1.8 1.7 1.6 1.5 1.5 1.3 1.3 1.2 1.1 1.1 1.0


CHAPTER

17-12.

17

[intensity level](dB) = 10 log

I . Find x the value of I for which [intensity level](dB) also = x, I0

which means solve the transcendental equation x x = 10 log = 10 log( x × 1012 ) or x = 10 log x + 120. I0

†17-13.

This can be done by trial and error on a calculator or by using a tool like Solver in Excel, Maple, etc. It’s obvious that x must be larger than 120. Here’s the result of trial and error with a calculator: 10 log x + 120 x 120 140.8 130 141.1 140 141.46 141 141.49 142 141.52 To three significant figures, I = 141 W/m2 produces an intensity level of 141 dB. ⎛ ⎞ I 50 dB = 10 log ⎜ −12 ⎟ ⎝ 1.0 × 10 ⎠ ⎛ ⎞ I =5 log ⎜ −12 ⎟ ⎝ 1.0 × 10 ⎠ I = 1.0 × 10−12 × 105 = 1.0 × 10−7 w/m 2 2I = 120 dB + (10 log 2) dB = 123 dB. I0

17-14.

Intensity level of the two rock concerts = 10 log

17-15.

Use a spreadsheet to divide 20 × 103 Hz by all the frequencies in Table 17.1. It turns out that 20 × 103 Hz is very nearly 64 × the frequency of D-sharp in the table. So 20 kHz is 6 octaves above Dsharp near the middle of the piano keyboard. P 1.0 W I = = = 8.0 × 10−4 W/m 2 . To find I at r = 20 m, either use the equation again or 2 2 4π r 4π (10 m)

17-16.

†17-17.

use the inverse square law. The inverse square law gives I(20 m) = I(10 m)(10/20)2 = 2.0 × 10–4 W/m2. The same answer comes from using the basic definition. Assuming each person produces the same intensity, I(25 people) = I(50 people)/2. To find the I (25 people) I (50 people) change in [intensity], use ∆[intensity level](dB) = 10 log − 10 log I0 I0 I (50 people) I (50 people) 1 ⇒ ∆[intensity level](dB) = 10 log − 10 log = 10 log 2I0 I0 2 ⇒ ∆[intensity level](dB) = −3.0 dB.

17-18.

Let I1 be the intensity of one violin and I12 be the intensity of twelve violins. Assuming each violin produces the same intensity, I12 = 12I1. Then I I 12 I1 [intensity level](dB) = 10 log 12 = 10 log = 10 log12 + 10 log 1 I0 I0 I0

⇒ [intensity level](dB) = 10 log12 + 60 dB = 71 dB


CHAPTER †17-19.

17-20.

Let I1 be the intensity of one machine and I2 be the intensity of two machines. Assuming each machine produces the same intensity, I2 = 2I1. Then I I 2I [intensity level](dB) = 10 log 2 = 10 log 1 = 10 log 2 + 10 log 1 I0 I0 I0

⇒ [intensity level](dB) = 10 log 2 + 80 dB = 83 dB I(1 m) = I(50 m)(50/1)2 = 2500I(1 m). I (1 m) I (50 m) 2500 I (50 m) [intensity level](dB) = 10 log = 10 log = 10 log 2500 + 10 log I0 I0 I0 ⇒ [intensity level](dB) = 10 log 2500 + 80 dB = 114 dB

†17-21.

⎛ ⎞ I 70 dB = 10 log ⎜ −12 ⎟ ⎝ 1.0 × 10 ⎠ ⎛ ⎞ I log ⎜ =7 −12 ⎟ ⎝ 1.0 × 10 ⎠ I = 1.0 × 10−12 × 107 = 1.0 × 10−5 w/m 2 P 4π r 2 ⇒ P = 4π r 2 I = 4π (30 m)2 (1.0 × 10−5 W/m 2 ) = 0.11 W I =

17-22.

The area of the hemisphere at 10 m = 2πr2 = 2π(10 m)2 = 628 m2 The power of the sound = 8W × 3/100 = 0.24 W. Therefore, Power 0.24 W = Intensity = Area 628 m 2 = 3.82 × 10−4 W/m 2

This is [intensity level](dB) = 10 log †17-23.

17-24. 17-25. 17-26.

3.82 × 10−4 W/m 2 = 86 dB 1 × 10−12 W/m 2

Intensity α 1/Area, Area α (diameter)2. Then I1 / I 0 = A0 / A1 = D02 / D12 = 82 / 0.7 2 = 130 times more intense ⎛ ⎞ ⎛ ⎞ ⎛ 130 I 0 ⎞ I0 I1 = 10 log ⎜ dB0 = 10 log ⎜ ; dB1 = 10 log ⎜ −12 ⎟ −12 ⎟ −12 ⎟ ⎝ 1.0 × 10 ⎠ ⎝ 1.0 × 10 ⎠ ⎝ 1.0 × 10 ⎠ Then, the increase in intensity level is dB1 – dB0 = 10 log (I1/I0) = 10 log 130 = 21 dB v = 331 m/s, so that the distance = v × time = 331 m/s × 0.5 s = 166 m distance 3 × 103 m Time, t = = = 9.1 s speed 331 m/s

It would be better to use signals transmitted by light. Let d be the distance. Then t1 = d/v1, t2 = d/v2. Therefore, t1 – t2 = d (1/v1 – 1/v2); d = (t1 – t2)/(1/v1 – 1/v2) 1 ⎞ ⎛ 1 d = 5s/ ⎜ − ⎟ s/m = 2100 m (2.1 km) 330 1460 ⎝ ⎠


17

CHAPTER †17-27.

17-28.

17-29.

17-30.

17-31.

17-32.

†17-33.

17

Let d be the distance to the cliff. The time for the sound to travel from you to the cliff is d/v, and it takes the same time for the echo to return. The total time is 2t, so vt (331 m/s)(1.5 s) d = = = 249 m. 2 2 The wavelength is determined by the size of the cavity and does not depend on what kind of gas f v v fills the cavity. He = He ⇒ f He = f air He ≈ 3 f air . vair f air vair v for a tube that is closed at one end. Thus 4L 30 m/s f = = 3.0 × 10−5 Hz 4(250 × 103 m) 1 T = = 3.3 × 104 s, or about 9 hours f (b) The period of the tidal fluctuation due to the moon is 12 hours, which is not the same as the fundamental period for the bay. However, the period of the first overtone is 1/3 of the fundamental period, which is close to 1/4 of the tidal period. So it’s possible that resonant effects can enhance the tidal fluctuation periodically. d = vt. For air, v = 331 m/s, or about 1/3 km/s. Thus in metric units d ≈ t/3 for t in seconds and d in km. In British units, 331 m/s = 1086 ft/s, which is 0.21 mi/s or about 1/5 mile/s. This justifies the rule that d ≈ t/5 for t in seconds and d in miles. (a) Distance to and fro is 1.00 m. The error in that distance is ± 0.04 m. Therefore, ∆d = v∆t so that ∆t = ∆d/v = 0.04 m/(331 m/s) = 1.2 × 10–4 s (b) The camera will focus on the glass. vt (331 m/s)t The camera determines distance using d = = , assuming the distances are 2 2 measured in air. The speed of sound in water is 1500 m/s, which is 4.5 times the speed of sound in air. If the camera is immersed in water, it will arrive at focusing distances that are too close by a factor of 4.5. If it is pointed at something 5 m away, it will focus on a point about 1.1 m away. From Table 17.3, v = 344 m/s at 20ºC. The length of the flute doesn’t change, so v v 344 m/s f 20 = 20 = f 0 20 = (261.7) = 272.0 Hz. 2L v0 331 m/s (a) f1 =

To bring the flute back in tune, it must be lengthened. Taking differentials of the equation for f0 gives Ldv − vdL ⎛ v ⎞ ⎛ dv dL ⎞ df 0 = =⎜ − ⎟⎜ ⎟⇒ 2 L2 L ⎠ ⎝ 2L ⎠ ⎝ v df 0 dv dL = − f0 v L df dv dL . Taking v = 331 m/s with dv = 13 = To bring the flute back in tune, we want 0 = 0 ⇒ f0 v L dL dv 13 = = = 0.039. The flute must be lengthened by 3.9%. m/s, we get L v 331


CHAPTER 17-34.

17

Suppose ∆v is the change in the sound velocity v0 due to a change ∆p in the pressure. Then 1.4 v0 + ∆v = ( p + ∆p) P0 1/ 2

=

∆p ⎞ 1.4 ⎛ p ⎜1 + ⎟ P0 p ⎠ ⎝

1.4 ⎛ 1 ∆p ⎞ p ⎜1 + ⎟ 2 p ⎠ P0 ⎝ assuming ∆p << p and using the binomial expansion. ∆v 1 ∆p Therefore, = v0 2 p ∆v 331.45 − 331.29 × 100 = × 100 = 4.8 × 10−2 %, Since v0 331.45 ∆p × 100 = 9.6 × 10−2 % p =

17-35.

17-36.

17-37. 17-38.

†17-39.

Size of typical telephone handset is about 20 cm. Comparing this with wavefronts shown in photo, we find λ ≅ 9 cm. 331 m/s ≅ 4000 Hz Then f = v/d = 9 cm The distance traveled by the sound is 2d, so vt (331 m/s)(0.20 s) 2d = vt ⇒ d = = = 33 m 2 2 d = vt = (331 m/s)(6.0 s) = 2.0 × 103 m (2.0 km) The frequency of the wave is the same in air and iron, so viron v v 331 = air ⇒ λair = λiron air = (25.5 m) = 1.65 m viron 5130 λiron λair Assume the scuba divers are in sea water. Let t1 = travel time through the pipe, t2 = travel time through the water. Then d t1 = v pipe t2 =

d vwater

∆t = t2 − t1 = d =

17-40.

∆t

d vwater

− =

d v pipe 1.5 s = 3.3 × 103 m (3.3 km) −1 (1539 m/s) − (5100 m/s) −1

⎛ 1 1 ⎞ − ⎜⎜ ⎟⎟ ⎝ vwater v pipe ⎠ The fundamental frequency of a pipe that is open at both ends is twice the fundamental frequency of a pipe of the same length that is closed at one end. The frequency will double to 2f.


CHAPTER 17-41.

(See Problem 39.) d d ∆t = − vair viron d =

17-42.

†17-43.

17

∆t

=

8.0 s = 2.8 × 103 m (2.8 km) −1 (331 m/s) − (5130 m/s) −1

⎛ 1 1 ⎞ − ⎜ ⎟ ⎝ vair viron ⎠ The wavelength of the sound will stay constant. Then vair v v 965 = He ⇒ f He = f air He = (440 Hz) = 1.28 × 103 Hz f air f He vair 331

Let d be the depth of the well. The time for the rock to fall is t1 =

2d . The time for the sound to g

d , where v is the speed of sound through air. The total time is v 2d d + . Move d/g to the other side, square both sides, and rearrange to get g g

travel back is t2 = t = t1 + t2 =

⎛ v2 ⎞ d 2 + 2 ⎜ vt + ⎟ d + (vt ) 2 = 0. Substitute numerical values (t = 4.62 s, g = 9.81 m/s2, v = 331 g⎠ ⎝ m/s) to get d 2 − 25395d + 2338514 = 0. The solutions are 25395 ± 253952 − 4(2338514) = 2.53 × 104 m, 92.4 m 2 The first answer is impossible because the sound could not travel that distance in less than 4.62 s. So the correct answer is d = 92.4 m. Tmax = 3600 N. The speed of a transverse wave, vm = (Tm/µ)0.5, where µ = 0.01 kg/m. Then vm = 600 m/s, which is much less than the speed of sound, 5000 m/s. d =

17-44.

17-45.

Suppose the wind direction is from A (Montmartre) to B (Montlhery). Then vAB = v + vw, where vw is the wind speed. Then tI = d/vAB = d/ (v + vw), (time for the sound from A to B). vBA = v – vw for the speed of sound from B to A, and t2 = d/vBA = d/(v – vw), for the time for the sound from B to A. We need to solve for v. v + vw = d/t1, and v – vw = d/t2 Adding, 2v = d[1/t1 + 1/t2] Therefore v = (d/2)[1/t1 + 1/t2]. t1 = 87.4 s and t2 = 84.8 s, and d = 29 × 103 m. Then v = 336.9 m/s


CHAPTER 17-46.

We can approximate source as infinitely far away and the waves coming in as plane waves. Once one of the waves hits the nearer ear, we can see from the diagram that it still has (0.15 m sin 5º) further to go to reach the other ear. The 0.15 m sin 5° time difference = 331 m/s = 4 × 10–5s.

†17-47.

17-48. †17-49.

Let λn be the wavelength in the tube. Then since the ends of the tube must be pressure nodes, (displacement antinodes), we have 2l ⎛ λn ⎞ ⎜ 2 ⎟ n = l ⇒ λn = n , ⎝ ⎠ which gives v v = n , n = 1, 2, 3 fn = λn 2l

λ1 = 2 L ⇒ f1 =

v 331 m/s = = 34 Hz 2 L 2(4.8 m)

(a) In the fundamental mode f = v/(2l), so that l = v/(2f ) = (331 m/s)/[2(261.7 Hz)] = 0.632 m. (b) Calculate by using l = v/(2f), where f is desired frequency v (m) Spacing (m) Tone Frequency (Hz) Length = 2f C C# D D# E F F# G G# A A# B C

261.7 277.2 293.7 311.2 329.7 349.2 370.0 392.0 415.3 440.0 466.2 493.9 523.4

0.632 0.597 0.564 0.532 0.502 0.474 0.447 0.422 0.399 0.376 0.355 0.335 0.316

— 0.035 = 3.5 cm 0.034 = 3.4 cm 0.032 = 3.2 cm 0.030 = 3.0 cm 0.028 = 2.8 cm 0.027 = 2.7 cm 0.025 = 2.5 cm 0.024 = 2.4 cm 0.022 = 2.2 cm 0.021 = 2.1 cm 0.020 = 2.0 cm 0.019 = 1.9 cm


17

CHAPTER 17-50.

†17-51.

17-52.

17-53.

17-54.

17-55.

17

From Equation (10) fn = nv/(4L), where fn is an eigenfrequency. Here v = 331 m/s, L = 2.7 cm = 0.027 m. 331 m/s so that Then f n = n 4 (0.027) m fn = (3.1 × 103) n Hz (where n = 1, 3, 5,.…) 3000 Hz is about the frequency of fundamental mode. Resonance probably plays a role. Since the tube is closed at the ends, there will be a displacement node at both ends. If λ is the wavelength then, since there must be an integral number of half-wavelengths between ends, (λn/2)n = L ⇒ λn = 2L/n f n = v/λn = nv /(2 L) From the diagram, for small values of θ, OP ≅ AP, and OP1 ≅ AP1 = AP + d , where d is the picket spacing. The distance traveled by sound waves of wavelength λ reflected by P is AP + AP = 2AP. The distance traveled by waves of the same wavelength reflected by P1 is 2AP1 = 2AP + 2d. For constructive interference, the path difference 2d = λ

VR 80 1 × = 580 × = 39 Hz 3.6 331 v Therefore, the frequency heard = 580 + 39 = 619 Hz ⎛ ⎞ 1 By Equation (17), f ′ = f ⎜ ⎟ ⎝ 1 ± VE /v ⎠ ∆f = f

f = 329.7 Hz, VE = 60 km/hr = 16.7 m/s ⎛ ⎞ ⎜ ⎟ 1 f ′ = 329.7 Hz ⎜ On approach, ⎟ = 347 Hz ( ≈ F) ⎜⎜ 1 − 16.7 ⎟⎟ 331 ⎠ ⎝ ⎛ ⎞ ⎜ ⎟ 1 f ′ = 329.7 Hz ⎜ On receding, ⎟ = 314 Hz (≈ D#) ⎜⎜ 1 + 16.7 ⎟⎟ 331 ⎠ ⎝ λ 100 Speed of the waves v = = = 12.5 m/s T 8 Speed with which the waves hit the boat = 9 + 12.5 = 21.5 m/s 21.5 m/s = 0.215 Hz Frequency = 100 m


CHAPTER 17-56.

Speed of 85 km/hr = 23.6 m/s ⎛ ⎞ ⎜ ⎟ 1 ⎜ ⎟ = 560 Hz ⎜⎜ 1 − 23.6 ⎟⎟ 331 ⎠ ⎝ ⎛ ⎞ ⎜ ⎟ ⎛ ⎞ 1 1 Receding f ′ = f ⎜ ⎟ = 485 Hz ⎟ = 520 Hz ⎜ ⎝ 1 + VE /v ⎠ ⎜⎜ 1 + 23.6 ⎟⎟ 331 ⎠ ⎝ The plane is moving with speed VE at an altitude h. At some time it passes over the listener at L. Some time t later the shock wave, also traveling at speed VE, passes L, and the plane has traveled a distance VEt. h From the geometry shown in the diagram, it is seen that the angle the shock wave θ makes with the ground is the same as the L angle of the Mach cone. We have two equations giving the angle θ: h tan θ = VE t ⎛ ⎞ 1 Approaching f ′ = f ⎜ ⎟ = 520 Hz ⎝ 1 − VE /v ⎠

†17-57.

sin θ =

17-58.

θ VE

VEt

v VE

where v is the speed of sound in air. Dividing the bottom equation by the top equation gives vt vt (331 m/s)(18 s) cos θ = ⇒ θ = cos −1 = cos −1 = 60.2° h h 12000 m Then v 331 m/s VE = = = 381 m/s sin θ sin 60.2° ⎛ ⎞ 1 Approaching f1 = f ⎜ ⎟ ⎝ 1 − VE /v ⎠ ⎛ ⎞ 1 Receding f 2 = f ⎜ ⎟ ⎝ 1 + VE /v ⎠ Dividing the top equation by the bottom gives f1 1 + x f − f2 = ⇒x= 1 f2 1 − x f1 + f 2

where x = VE/v. Then 497 − 395 x= = 0.114 ⇒ VE = 0.114(331 m/s) = 37.8 m/s 497 + 395 f = f1 (1 − x) = (497 Hz)(1 − 0.114) = 440 Hz


17

CHAPTER †17-59.

17-60. 17-61. 17-62. †17-63.

17

The pulse arriving at the wall is Doppler shifted because the source (the bat) is approaching. The ⎛ 1 ⎞ frequency of that pulse is f1 = f ⎜ ⎟ , where x = vbat/v. This is the frequency of the pulse ⎝1− x ⎠ reflected back toward the bat. The frequency detected by the bat is Doppler-shifted a second time because the bat is now a receiver approaching the source. The detected frequency is f − f 1+ x 800 Hz . Solving for x gives x = 2 f 2 = f1 (1 + x) = f = = 7.94 × 10−3. 1− x f2 + f 50800 Hz + 50000 Hz Thus vbat = 2.63 m/s. f1 − f 2 f /f − 1 2 −1 =v 1 2 = (331 m/s) = 110 m/s. f1 + f 2 f1 /f 2 + 1 2 +1 v 1 = ⇒ θ = 30° “Mach 2” means VE = 2v ⇒ sin θ = VE 2 v 331 sin θ = = ⇒ θ = 1.3° VE 15 × 103 See Problem 58. VE = v

Let t1 = the time the diver has been in free fall and t2 = time for sound to travel back to the gt 2 listener. At t1 the diver is a distance d = 1 from the top of the cliff and is moving with speed 2 vD = gt1 . The relation between d and t2 is d = vt2, where v is the speed of sound. We want to know what pitch is heard 3.0 s after the diver jumps, so t1 + t1 = 3 ⇒ t2 = 3 − t1 ⇒ d = v(3 − t1 ). Setting the two expressions for d equal gives an equation for t1: gt 2 3v − vt1 = 1 2 or gt12 + 2vt1 − 6v = 0 The solution is t1 =

−2v + 4v 2 + 24 gv −662 + 6622 + 24(9.81)(331) = = 2.88 s. (There is 2g 2(9.81)

also a negative solution that has no physical meaning.) The speed of the diver at this instant is vD = (9.81 m/s 2 )(2.88 s) = 28.2 m/s. The diver is receding from the listeners at the top of the f 440 Hz = = 405 Hz. cliff, so the frequency they hear is f ′ = 1 + vD / v 1 + 28.2 / 331 17-64.

The listener hears two Doppler-shifted tones. Let f be the source frequency. One tone is the f , reflection from the wall. The source is approaching the wall, so this frequency is f1 = 1− x f . These where x = VE/v. The other tone is from the receding car, giving a frequency of f 2 = 1+ x two tones interfere to give an average frequency and a beat frequency: f1 + f 2 = 250 Hz 2 f1 − f 2 = 12 Hz


CHAPTER

17

Using the equations for f1 and f2 gives a pair of equations containing f and x: f = 250 1 − x2 xf =6 1 − x2 Solving gives x = 3/125, from which we get VE = 7.94 m/s. †17-65.

In this problem, each driver hears two Doppler shifts: one caused by the motion of the source (the horn on the other car) and one caused by the motion of the receiver (the motion of each driver toward the other car). Car 1 has speed v1 = 90.0 km/hr = 25.0 m/s, which is the emitter speed for the horn in car 1 and the receiver speed for the sound from the horn in car 2. Car 2 has speed v2 = 60.0 km/hr = 16.7 m/s, which is the emitter speed for car 2 and the receiver speed for the sound from the horn in car 1. fhorn = 524 Hz for both cars. (i) To a stationary observer, horn from car 2 has frequency ⎛ 1 ⎞ f 2′ = f horn ⎜ ⎟ ⎝ 1 − v2 /v ⎠ This frequency is the source frequency for the driver in car 1, who hears 1 + v1 /v 1 + (25.0/331) f 2′′ = f 2′(1 + v1 /v) = f horn = (524 Hz) = 594 Hz 1 − v2 /v 1 − (16.7/331) (ii) Similarly, the frequency heard by car 2 is Doppler-shifted twice because of the motion of the source toward the receiver and the motion of the receiver toward the source. Going through the same analysis, the frequency heard by the driver in car 2 is 1 + v2 / v 1 + (16.7 / 331) f1′′ = f horn = (524 Hz) = 595 Hz 1 − v1 / v 1 − (25.0 / 331) Note that here the source speed is v1 and the receiver speed is v2, the opposite of case (i).

17-66.

In the frame of the wind (i.e., in the frame in which the air is still) car 1 (the faster car) is traveling at (90 – 40) = 50 km/hr, or v1 = 13.9 m/s in the wind frame. Car 2 (the slower car) is traveling at (60 + 40) = 100 km/hr, or v2 = 27.8 m/s in the wind frame. The remaining analysis is the same as in Problem 65, but we use the wind frame values for the speeds. (i) Frequency of horn 2 as heard in car 1: 1 + v1 /v 1 + (13.9/331) f 2′′ = f horn = (524 Hz) = 596 Hz 1 − v2 /v 1 − (27.8/331)


CHAPTER

17

(ii) Frequency of horn 1 horn as heard in car 2: 1 + v2 /v 1 + (27.8/331) f1′′ = f horn = (524 Hz) = 593 Hz 1 − v1 /v 1 − (13.9/331)

17-67.

75 km/hr = 20.8 m/s The frequency of the whistle relative to the mountain is ⎛ 1 ⎞ f′ = f ⎜ ⎟ ⎝ 1 − v E /v ⎠ The frequency of “emission” of the whistle from mountain = f ′ . Then the frequency heard by the train = ⎛ 1 + v R /v ⎞ f ′′ = f ′ (1 + vR /v) = f ⎜ ⎟ ⎝ 1 − v E /v ⎠

17-68.

⎛ 1 + vt /v ⎞ ⎛ v + vt ⎞ Here vE = vR = vt (speed of train) f ′′ = f ⎜ ⎟= f⎜ ⎟ ⎝ 1 − vt /v ⎠ ⎝ v − vt ⎠ ⎛ 331 m/s + 20.8 m/s ⎞ f ′′ = 420 Hz ⎜ ⎟ = 476 Hz ⎝ 331 m/s − 20.8 m/s ⎠ Let vt be the speed of the train, f be the frequency of the source f′ = frequency relative to the ground ⎛ 1 ⎞ f′ = f ⎜ ⎟ ⎝ 1 − vt /v ⎠ v" = frequency heard by receiver = v' (1 – vt/v) ⎛ 1 − vt / v ⎞ f ′′ = f ′(1 − vt /v) = f ⎜ ⎟= f ⎝ 1 − vt / v ⎠

17-69.

17-70. 17-71. 17-72.

If the position of the source/receiver is reversed, the minus signs become plus signs, and we still find f" = f. According to the observer, the speed of sound in air in the direction the sound is coming from is v = 331 m/s + 20 m/s = 351 m/s. Therefore, the frequency he hears is ⎛ ⎞ ⎜ 1 ⎟ ⎛ 1 ⎞ f′ = f ⎜ ⎟ = 481 Hz ⎟ = (440 Hz) ⎜ ⎝ 1 − v E /v ⎠ ⎜⎜ 1 − 30 ⎟⎟ 351 ⎠ ⎝ v 331 m/s The angle is measured to be θ ≈ 54º. VE = = = 409 m/s. sin θ sin 54° v 331 θ = sin −1 = sin −1 = 29.4° 674 VE v 331 θ = sin −1 = sin −1 = 24.4° 800 VE


CHAPTER 17-73.

(a) You hear the sound emitted time t ago, where vt is the distance between the point the sound is emitted and you (see diagram). In this time, the aircraft has moved a distance VEt. Then sin θ = VE t / vt = VE /v (b) VE = v sin 30° =

17-74. †17-75.

17-76. †17-77.

17-78.

†17-79.

λ=

1 (331 m/s) = 165 m/s (600 km/hr) 2

v 331 m/s = = 8.3 × 10−3 m (8.3 mm) 3 f 40 × 10 Hz

The size depends on whether the object is located in air or tissue. Since this is a medical imager, assume that the object is in tissue where v ≈ 1500 m/s (the speed of sound in water). Then the maximum size is close to half the wavelength: λ v 1500 m/s d max = = = = 1.5 × 10−4 m (0.15 mm) 2 2f 2(5.0 × 106 Hz) λ v 5104 m/s d max = = = = 6.4 × 10−7 m (0.64 µm) 4 4f 4(2.0 × 106 Hz) v 1500 m/s λwater = water = = 1.5 × 10−3 m (1.5 mm) 1.0 × 106 Hz f v 331 m/s λair = air = = 3.3 × 10−4 m (0.33 mm) 1.0 × 106 Hz f v v v Or use air = water ⇒ λwater = λair water to get the same answer. λair λwater vair 261.7 Hz = 65.4 Hz (a) f = 4 v 331 m/s λair = air = = 5.06 m 65.43 Hz f (b) For the 0.63 m cello string, λ = 2L = 1.36 m. (The speed of the waves on the string is different from the speed of sound in air.) From the study of simple harmonic motion, vmax = ω A = 2π fA amax = ω 2 A = 4π 2 f 2 A = ω vmax vmax = 2π (10 × 106 Hz)(8.0 × 10−8 m) = 5.0 m/s amax = 2π (10 × 106 Hz)(5.0 m/s) = 3.2 × 108 m/s 2 ≈ 3.2 × 107 g

17-80.

17

[Intensity Level](dB) = 10 log

I 1.0 × 10−3 = 10 log = 90 dB I0 1.0 × 10−12


CHAPTER 17-81.

17-82.

†17-83.

17-84.

17

Intensity, I (W/m2) of sound generated by one woman I = 1.0 × 10–12 × 10115/10 = 0.32 W/m2 Therefore three such women can bring one to threshold of pain (~1 W/m2). I = (1.0 × 10−12 W/m 2 )(10160 / 10 ) = 1.0 × 104 W/m 2 . Then P = IA = (1.0 × 104 W/m 2 )(π )(3.5 × 10−3 m) 2 = 0.38 W. I I I ∆[Intensity Level](dB) = 10 log 2 − 10 log 1 = 10 log 2 . I2 = 2I1, so I0 I0 I1 ∆[Intensity Level](dB) = 10 log 2 = 3.0 dB. I1 = (1.0 × 10−12 W/m 2 )(1020 / 10 ) = 1.0 × 10−10 W/m 2 at r1 = 0.50 m. At r2, I2 = 1 × 10–12 W/m2, so 2

17-85.

17-86.

2

⎛r ⎞ I1 ⎛ r2 ⎞ I = ⎜ ⎟ ⇒ ⎜ 2 ⎟ = 1 = 100 ⇒ r2 = 10r1 = 5.0 m. I 2 ⎝ r1 ⎠ I2 ⎝ r1 ⎠ (a) d = vt, so ∆t = ∆d/v. The time taken for the pulse to return, d 2 × 10 m t= = = 0.061 s. v 330 m/s 2 × (0.5 m) ∆t = ∆d/v = 331 m/s (factor of 2 because we are counting the distance to and fro) Then ∆t = 3 × 10–3 s (b) dD = real distance, v0 = velocity in methane; v1 = velocity in air, d1 = perceived distance In methane t = d0/v0, but bat thinks t = d1/v1. 331 m/s Then d1 /v1 = d 0 /v0 ⇒ d1 /d 0 = (v1/v0 = = 0.77 432 m/s (Bat will think distances are 0.77 real distances.) Let d be the distance (depth). Then the time taken for the stone to reach the bottom t0 = 2d/g (g = acceleration due to gravity). The time taken for the sound signal to reach the physicist is t1 = d/v (where v = velocity of sound). Therefore, t = t0 + t1 = 2d/g + d/v. Then (1/v) d + 2/g d − t = 0 Substituting known parameters gives (3.02 × 10−3 ) d + (0.45) d − 6.0 = 0

)

(

d = [−0.45 ± (0.45)2 + 4(6.0) (3.02 × 10−3 ] /{2 × 3.02 × 10−3 } m1/2 or d = 150 m 17-87.

2160 km/hr = 600 m/s (a) sin θ = v/vE = (331 m/s) (600 m/s) = 0.552 Therefore, θ = sin–1 (0.550) = 33.5º


CHAPTER

17

(b) As shown in the diagram, shock wave hits you when the plane has traveled a distance d as shown. Therefore, (12 × 103 m)/d = tan θ, or d = (12 × 103 m)/tan θ d 12 × 103 m t= = = 30.2 s vConcorde (600 m/s) tan 33.5°

17-88.

†17-89.

VE = 510 km/h = 142 m/s. VE/v = 142/331 = 0.429. For a moving source, f 600 Hz f approach = = = 1.05 × 103 Hz 1 − VE /v 1 − 0.429 f 600 Hz f approach = = = 420 Hz 1 + VE /v 1 + 0.429 (a) The man is at rest relative to the sound source, so he hears f = 660 Hz directly from it. (b) The woman hears a tone Doppler shifted to a higher frequency because the source is approaching her. Let x = vboat/v = 15/331 = 0.0453. Then f 660 Hz f woman = = = 691 Hz. 1 − x 1 − 0.0453 (c) The echo heard by the man has been Doppler-shifted twice. The sound reaching the cliff has been shifted because the boat is approaching the cliff, and the reflected sound has this increased frequency. Then there is a second shift because the man, who is now the receiver, is approaching the cliff. The speed of the source and the speed of the receiver are the same (the speed of the boat), so the reflected frequency finally heard by the man is f 1+ x 1 + 0.0453 f man = × (1 + x) = f = (660 Hz) = 723 Hz. 1− x 1− x 1 − 0.0453


CHAPTER 18

FLUID MECHANICS


1 1 ∆V × rate of flow = A A ∆t –5 3 1 liter/min = 1.67 × 10 m /s

Speed of flow v =

2

⎛ 3.81 ⎞ (i) At hose: A = π ⎜ × 10−2 ⎟ = 1.14 × 10−3 m 2 ⎝ 2 ⎠ −5 95 × 1.67 × 10 2 Therefore, v = = 1.39 m/s At nozzle: A = π R2 −3 1.14 × 10 v=

95 × 1.67 × 10−5 95 × 1.67 × 10−5 = = 22.3 m/s 2 A ⎛ 0.95 −2 ⎞ π⎜ × 10 ⎟ ⎝ 2 ⎠

(ii) At hose v = At nozzle v =

190 × 1.67 × 10−5 = 2.8 m/s 1.14 × 10−3

190 × 1.67 × 105

= 25.1 m/s 2 ⎛ 1.27 −2 ⎞ π⎜ × 10 ⎟ ⎝ 2 ⎠ 284 × 1.67 × 10−5 = 4.2 m/s (iii) At hose v = 1.14 × 10−3

At nozzle v =

18-2. †18-3.

284 × 1.67 × 10−5

= 23.9 m/s 2 ⎛ 1.59 −2 ⎞ π⎜ × 10 ⎟ ⎝ 2 ⎠ dV dV / dt 92 cm 2 /s = Av ⇒ v = = = 20 cm/s π (1.2 cm) 2 dt A

Work to lift a mass m to a height h is W = mgh. The power is P =

dW dm = gh . One liter of dt dt

water has a mass of 1 kg. Then P = (9.81 m/s 2 )(170 m)(26 × 103 liters/min)(1 min/60 s)(1 kg/liter) = 7.23 × 105 W

18-4.

dV 250 liters 1 min = × × 10−3 m3 /liter = 1.04 × 10−3 m 3 /s. dt 4 min 60 s dV / dt 1.04 × 10−3 m3 /s = = 2.1 m/s v= π (1.25 × 10−2 m) 2 A 2

18-5.

2

⎛D ⎞ A ⎛ 10 cm ⎞ v1 A1 = v2 A2 ⇒ v2 = v1 1 = v1 ⎜ 1 ⎟ = (6.0 m/s) ⎜ ⎟ = 12 m/s A2 ⎝ 7.0 cm ⎠ ⎝ D2 ⎠


CHAPTER

18-6.

From conservation of energy,

v12 v2 + gh = 2 ⇒ v2 = v12 + 2 gh , where v1 is the speed of the 2 2

water when it exits the hole. Thus v2 = (0.50 m/s) 2 + 2(9.81 m/s 2 )(4.0 m) = 8.87 m/s. v1 0.50 = (5.0 cm) = 1.2 cm v2 8.87

v1 A1 = v2 A2 ⇒ v2 D12 = v1 D22 ⇒ D2 = D1 †18-7.

dV 0.50 liter 1 min = × × 10−3 m3 /liter = 2.78 × 10−7 m3 /s. In the tube, dt 30 min 60 s dV/dt 2.78 × 10−7 m3 /s = = 8.84 × 10−2 m/s, or 8.84 cm/s. In the needle, v1 = π (1.0 × 10−3 m) 2 A1 2

2

18-8.

⎛D ⎞ ⎛ 2.0 mm ⎞ v2 = v1 ⎜ 1 ⎟ = (8.84 cm/s) ⎜ ⎟ = 884 cm/s, or 8.84 m/s. ⎝ 0.20 mm ⎠ ⎝ D2 ⎠ dm dV =ρ = ρ vA = πρ vR 2 . The mass flow rates are equal for water and propane, giving dt dt 2 ρ propane ⎛ R propane ⎞ v 2.01 2 2 ⇒ water = ρ water vwater Rwater = ρ propane v propane R propane (10) 2 = 0.201. ⎜ ⎟ = 1000 v propane ρ water ⎝ Rwater ⎠

18-9.

(a) Using

( x − x0 )

1 2 ( v − v02 ) = g ( x − x0 ) ; 2 (15 m / s) 2 = v 2 /2g = = 11.5 m 2 × 9.81 m/s 2

(b) Speed at 5 m is: v 2 = v02 − 2 gx, v = v02 − 2 gx v = 152 − 2 × 9.81 × 5 = 11.3 m/s. Therefore, since A1v1 = A2v and A1 ∝ d12 (where d is diameter). v1d12 = v2 d 22

Therefore, d 2 =

(

)

v1 /v2 d1 =

(15 /11.3 ) (7.0 cm) = 8.1 cm

Speed at 10 m is v = 152 − 2 × 9.8 × 10 = 5.39 m/s so that d 2 = 15 / 5.39 (7.0 cm) = 11.7 cm 18-10. 18-11.

F = pA = (0.80 atm)(1.01 × 105 Pa / atm)(π )(0.032m) 2 = 260 N p −p p − p0 = − ρ gy ⇒ y = 0 . The change in pressure is 8.0 mm Hg, so ρg

y= 18-12.

18-13.

18-14.

(8.0 mm Hg)(760 mm Hg/atm)(1.01 × 105 Pa/atm) = 84 m (1.29 kg/m3 )(9.81 m/s 2 )

Fnet = Pnet∆A = [(1010 – 200 millibar] × (12 × 3) m2 = 810 × 100 N/m2 × 36 m2 = 2.9 × 10.6 N Yes, because pressure tends to equalize inside the house, causing less net force. 8 1/2 in × 11 in = 93.5 in2. Atm. pressure = 14.7 lb/in2 Thus, the force down = (93.5 × 14.7) lb = 1370 lb. The paper is not forced against the table top because the layer of air under the paper provides the same force upwards, “cancelling” the downward force. Force = p∆A = 3 × 104 N/m2 × 0.7 m2 = 2.1 × 104 N


18

CHAPTER †18-15.

18-16.

18-17.

18-18.

18

Each tire supports (1300/4 kg)(9.81m/s2) = 3188 N. The normal force acting on each tire is therefore 3188 N, which flattens the tire. If A is the area of contact, then pA = 3188 N 3188 N ⇒ A= = 1.32 × 10−2 m 2 , or 132 cm2. (2.4 atm)(1.01 × 105 Pa/atm) Weight of plane = mg = 10900 kg × 9.81 m/s2 = 1.07 × 105 N. The net force on the wing = 1.07 × 1.07 × 105 N 105 N; therefore p = F/A = = 1.16 × 103 N/m 2 2 92 m 0.95 atm = (0.95)(1.01 × 105 Pa) = 9.60 × 104 Pa. The total pressure in the tire is (9.60 × 104 Pa + 2.4 × 105 Pa) = 3.36 × 105 Pa. Assuming that the total pressure in the tire stays constant as the external pressure rises to 1.01 atm = 1.02 × 105 Pa, the overpressure in the tire will be (3.36 – 1.02) × 105 Pa = 2.34 × 105 Pa. Force = ∆pA. ∆p = (1.00 – 0.28 atm) × 1.01 × 105 Pa = 7.27 × 104 Pa. Net outward force = 7.27 × 104 Pa × 2 m 2 = 1.45 × 105 N. (This is just under 33,000 lb.)

18-19.

Ftotal = pA = (1.01 × 105 Pa)(4π )(0.20 m) 2 = 5.08 × 104 N

18-20.

The pressure due to the weight is p =

†18-21.

F (123 kg)(9.81 m/s 2 ) = = 4.8 × 105 Pa. The end of the A π (0.05 m) 2

bar is probably rough, so there is air under it and there is no contribution from atmospheric pressure. If the end is perfectly smooth and all the air has been forced out, atmospheric pressure must be added and the total pressure is 5.8 × 105 Pa. The pressure on the outside of the suction cup is 1 atm, and the pressure on the inside is nearly zero. The total force caused by the pressure difference must equal the weight of the boy, so mg 50 kg × 9.81 m/s 2 pA = mg ⇒ A = = = 4.86 × 10−3 m 2 (or 48.6 cm2). p 1.01 × 105 Pa

18-22.

50,000 liters of water has a mass of 50,000 kg. F mg 5.0 × 104 kg × 9.81 m/s 2 p= = = = 3.9 × 104 Pa 2 A A π (2.0 m)

18-23.

mg = (60 kg)(9.81 m/s2) = 589 N. When she stands on her shoes, mg 589 N p1 = = = 2.0 × 104 Pa. When she balances on one heel, A1 300 × 10−4 m 2 mg 589 N p2 = = = 7.5 × 106 Pa. A2 π (5.0 × 10−3 m)2

18-24.

The net pressure force = A(∆p) = (1.01 × 105 Pa/atm) × (1 – 0.64) atm × 1 m2 = 3.6 × 104 N (about 4 tons) (a) Force = A(∆p) = (0.06 m)2 × 1.01 × 105 Pa = 360 N. (b) Since the sharpener is resting on a horizontal surface, the magnitude of the normal force is equal to the magnitude of the vertical force caused by the pressure difference. Thus maximum transverse force = µSN = 0.9 × 360 N = 330 N. p = patm + ρgh = 1.01 × 105 Pa + 103 kg/m3 × 9.81 m/s2 × 520 m = 5.2 × 106 Pa. Pressure of water = p = patm + ρgh = 1.01 × 105 Pa + (103 kg/m3 × 9.81 m/s2 × 26 m) = 3.56 × 105 N/m2 Pressure of oil = patm + ρgh = 1.01 × 105 N/m2 + (880 kg/m3 × 9.81 m/s2 × 30 m)

†18-25.

18-26. 18-27.


CHAPTER

18-28.

18-29.

18-30.

18-31.

18-32.

18-33. 18-34.

= 3.60 × 105 N/m2 Net pressure on bottom = 4 × 103 Pa, so the pressure force on 1 m2 = 4 × 103 N. The water pushes up and the oil pushes down, so the net force points downward. (a) At the top of the atmosphere, pressure = 0. At sea level, the pressure = 1 atm, so p0 – p1 = ρgz. The mass of the air in area A is then ρzA = (p0 – p1)A/g = 1.01 × 105 N/m2 × 1 m2/9.81 m/s2 = 1.03 × 104 kg. (b) Surface area of earth = 4πR2 = 4π(6378000 m)2 = 5.11 × 1014 m2 Thus, the mass of the atmosphere ≈ 5.11 × 1014 m2 × 1.03 × 104 kg/m2 = 5.26 × 1018 kg. ∆p = (64 mm Hg)(1 atm/760 mm Hg)(1.01 × 105 Pa/atm) = 8.50 × 103 Pa. Take the reference pressure pref at the surface of the water in the high pressure region. At the same level in the low pressure region, the pressure must also be pref. Let the rise be ∆z so that pref ∆p 8.50 × 103 Pa = = 0.85 m. = pref – ∆p + ρg (∆z) ⇒ ∆z = ρ g (1025 kg/m3 )(9.81 m/s 2 ) (a) p + p0 + ρgz, with ρ = 1055 kg/m3 and g = 9.81 m/s2 Feet: p = 110 mm Hg × 1.33 × 102 N/m2/mm Hg + 1055 kg/m3 × 9.81 m/s2 × 1.40 m = 2.91 × 104 Pa Brain: p = 1.466 × 104 Pa – (1055 kg/m3)(9.81 m/s2)(0.4 m) = 1.05 × 104 Pa (b) Pressure in brain = p0 + ρgz = 2.53 × 104 N/m2 – (1055 kg/m3)(61 m/s2)(0.4 m) = –442 Pa The heart therefore cannot maintain positive pressure at the brain, but since this is only 1.7% below the zero pressure, the value of g is just slightly too large to maintain positive pressure. Pressure at the surface of the water outside the pump = patm = 1.01 × 105 Pa. Assuming that the space in the cavity of the suction pump is almost a vacuum (see diagram), then the pressure at the point level with the surface of the water outside = patm. But the pressure must be due to the column of water of height ∆z. Therefore, ρg(∆z) = 1 atm = 1.01 × 105 Pa ∆z = (1.01 × 105 kg/ms2)/(1000 kg/m3)(9.81 m/s2) = 10.3 m 790 mm Hg = 1.04 atm 760 mm Hg/atm 730 mm Hg = 0.961 atm 760 mm Hg/atm p = ρgh = (1.7 g/cm3)(100 cm/m)3(9.81 m/s2)(2.0 m) = 3.3 × 104 Pa. The pressure is the same on both pistons. The pressure required at the large piston must produce a mg mg = force equal to the weight of the car: p = 2 Alarge π Rlarge =

18-35.

18

(1500 kg)(9.81 m/s 2 ) 1 atm × = 3.2 atm 2 1.01 × 105 Pa π (0.12 m)

p = ρgh = (1000 kg/m3)(9.81 m/s2)(30 m) = 2.94 × 105 Pa.


CHAPTER 18-36.

18

Let h1 be the height of the water on the right side, h2 be the height of the gasoline, and h3 the height of the water on the left side. At the bottom of the tube, the total pressure due to the fluids on each side must be equal:

ρ water gh3 = ρ water gh1 + ρ gasoline g ( h2 − h1 ) ⇒ h3 − h1 =

ρ gasoline (h2 − h1 ). Substituting the given ρ water

value for h2 – h1 and using the densities for water and gasoline from Table 18.1 gives two equations: h2 − h1 = 20 cm h3 − h1 = 14.8 cm Subtracting the second from the first gives h2 − h3 = 5.2 cm. †18-37.

18-38.

∆V 1 = p, where we have used the definition of pressure given in this V B chapter. The effect of atmospheric pressure is negligible at the depth in this problem. Thus ∆V 4.54 × 107 Pa = = 0.021, or 2.1%. V 0.22 × 1010 Pa

From Eq. 14.20,

The pressure difference between the water’s surface and at 4 m deep is 0 assume that the inner ∆p = ρ gh = (1000 kg/m3 )(9.81 m/s 2 )(4.0 m) = 3.92 × 104 Pa.yIf= we surface of the ear drum remains at 1 atm pressure, then the pressure dydifference on the ear drum is 4 also 3.92 × 10 Pa. The pressure force on the ear drum is F = (∆p) A = (3.92 × 104 Pa)(π )(0.25 × 10−2 m) 2 = 0.77 N.

†18-39.

The net force on the dam is caused by the overpressure from the water. The effect of atmospheric pressure is the same on both sides of the dam and causes no net force. For a strip of width dy located a distance y below the surface of the water, the pressure force caused by the water is dF = ρgydA = 70ρgydy. To find the total force, integrate over y from 0 to 30 m: F =

∫

y = 30 m y =0

dF = 70 ρ g ∫

30 m

0

3

ydy = 35 ρ gy 2

70 m

30 m 0

= (35)(1000 kg/m )(9.81 m/s )(30 m) = 3.1 × 108 N 18-40.

2

y = 30 m

2

Centripetal acceleration = ω2r ω = 2πf = 2π(0.70)/s = 4.40/s R = 1.0 m, so that a = ω2r = (4.40/s)2(1.0)m = 19.3 m/s2 We can think of gravity as due to our frame of reference (Earth) accelerating upward at g. (a) Therefore, the point at the lowest point on the circle thinks anet = g + acentripetal = (9.81 + 19.3) m/s2 = 29.1 m/s2. Using p – p0 = ρgeffz, with geff = 29.1 m/s2, we have differences in pressures p – p0 = (1000 kg/m3)(29.1 m/s2)(0.01 m) = 291 Pa. (b) At the highest point, the net acceleration = 9.81 – 19.3 = –9.49 m/s2 ∆p = p – p0 = (1000 kg/m2)(–9.49 m/s2)(–0.01 m) = 95 Pa.


CHAPTER

18-41.

18

(a) The centripetal acceleration felt at a distance r from center is ω2r. By Equation (11), dp = aρdr, where a replaces –g. Here a = ω2r, so that dp = ω2ρr dr Then

∫

r r0

dp =

∫

r r0

ω 2 ρ r dr ⇒ p(r ) − p(r0 )

= p=

1 2 ω ρ ( r 2 − r02 ) 2

1 (3.8 × 104) 2 / s 2 (1000 kg/m3 )(0.132 − 0.102 ) = 5.0 × 109 Pa 2 Imagine the brain to be a cube of 20 cm in each dimension, and to have the same density as that of water, 1000 kg/m3. Then ac = v2/R = 25 × 9.81 m/s2. F = mac and since p = F/A, pA = mac, where p is the pressure difference between the front and the back of the brain. Now m = ρV = 0.2ρA = 0.2 × 103A Therefore, pA = 0.2 × 103 A(25 × 9.81), or p = 4.9 × 104 Pa. (a) For case (a), shown in the diagram, Fbottom = Fb = PAb = ρ gh ( π R22 ) (b) p =

18-42.

18-43.

To find Fsides = Fs, we must integrate dFs = PdA = P(h1)(2πrds) = P(h1)(2πrdh1/cos θ). h cos θ = h 2 + ( R2 − R1 ) 2 We must also recognize that as we integrate around the surface, the net horizontal component of the total force is zero. The net vertical component is dF⊥ = dFs sin θ R2 − R1 sin θ = 2 h + ( R2 − R1 ) 2 ∴ dF⊥ = ρ gh1 (dA) sin θ ⎡ 2π r (dh ) h 2 + ( R − R ) 2 1 2 1 = ρ gh1 ⎢ h ⎢⎣ 2πρ gr ( R2 − R1 )h1dh1 = h

⎤⎡ R2 − R1 ⎥⎢ ⎥⎦ ⎢⎣ h 2 + ( R2 − R1 ) 2

⎤ ⎥ ⎥⎦


CHAPTER

18

( R2 − R1 ) h 2πρ g ( R2 − R1 ) ⎡ ( R2 − R1 )h12 dh ⎤ ∴ dF⊥ = + R h dh ⎢ 1 1 1 ⎥ h h 1⎦ ⎣ But r = R1 + h1 tan θ = R1 + h1

2πρ g ( R2 − R1 ) h ⎡ ( R2 − R1 ) 2 ⎤ 2πρ g ( R2 − R1 ) ⎡ R1h 2 ( R2 − R1 )h3 ⎤ R h dh h1 dh1 ⎥ = + + 1 1 1 ⎢ ⎥ ∫0 ⎢⎣ h h h 3h ⎦ ⎣ 2 ⎦ πρ gh( R2 − R1 )( R1 + 2 R2 ) = 3

and F⊥ =

Taking the positive direction to point up, the sum of the net force on the side and the force on the bottom is πρ gh( R2 − R1 )( R1 + 2 R2 ) πρ gh( R12 + R1 R2 + R22 ) Ftotal = − ρ gh ( π R22 ) = − 3 3 π h( R12 + R1 R2 + R22 ) , The volume of a right frustum of a cone of height h and radii R1 and R2 is 3 so this total downward force is just equal to the weight of the fluid in the container.

(b) The analysis is the same as in (a) with two differences. Here the magnitude of the force on the bottom is Fb = PAb = ρ gh ( π R12 ) The element of force on the side is now dF⊥ = − ρ gh1 (dA) sin θ

=−

2πρ gr ( R2 − R1 )h1dh1 h

( R2 − R1 ) , and h 2πρ g ( R2 − R1 ) ⎡ ( R − R1 ) 2 ⎤ dF⊥ = − R2 h1dh1 − 2 h1 dh1 ⎥ ⎢ h h ⎣ ⎦ Carrying out the integration as in (a) gives πρ gh( R2 − R1 )(2 R1 + R2 ) F⊥ = − 3 The total force is

From the figure, r = R2 − h1 tan θ = R2 − h1


CHAPTER

Ftotal = −

πρ gh( R2 − R1 )(2 R1 + R2 )

− ρ gh ( π R12 ) = −

18

πρ gh( R12 + R1 R2 + R22 )

3 3 The result is the same as in (a): The net downward force is equal to the weight of the liquid in the container.

18-44.

18-45.

18-46.

†18-47.

18-48.

In air F = ρair gV = (1.29 kg/m3) (9.81 m/s2)(7.4 × 10–2) m3 = 0.94 N In water F = ρwater gV = (1000 kg/m3)(9.81 m/s2)(7.4 × 10–2) m3 = 726 N As in Example 9, the volume of ice for every 1 m3 submerged is 1025 kg/(920 kg/m3) = 1.114 m3. Then the fraction above the water = 0.114/1.114 = 0.1024. (a) The total volume of the iceberg = (400 × 400 × 30) m3/0.102 = 4.7 × 107 m3 (b) Total mass = ρV = 920 kg/m3 × 4.7 × 107 m3 = 4.3 × 1010 kg. Assume a mass of 80 kg. To walk on water, buoyant force must equal the weight = 80 × 9.81 N = 785 N. Bouyant force = ρwater gV where V is volume of shoes. V = 2(0.3 m)2 L F = (1000 kg/m3) (9.8 m/s2) (2[0.3 m]2 L) = 1.821 L = 785 N L = (785/1821) m = 0.43 m

Mass of barrel = 20 kg + (0.12m3)(730 kg/m3) = 108 kg Volume of barrel ≈ 0.12 m3 Therefore, the mass of water it displaces = 0.12 m3 × 1000 kg/m3 = 120 kg Since the mass of the water displaced by the volume is greater than the mass of the barrel, the barrel floats. Vballoon = 570,000 m3. ρa = 4.3 × 10–3 kg/m3. The buoyant force is equal to the weight of the displaced air = ρgV = 4.3 × 10–3(570,000)(9.81) = 2.40 × 104 N


CHAPTER †18-49.

18

The buoyant force is equal to the weight of the air displaced. The total downward force is the weight of the helium in the balloon plus the weight of the balloon and the load hanging from the balloon. For the balloon to lift the load, the buoyant force must equal that total weight: m + mL FB = ρ airVg = ρ H Vg + mB g + mL g ⇒ V = B ρ air − ρ H where V is the volume of the balloon. Then (4 + 150) × 10−3 kg V = = 0.139 m3 3 (1.29 − 0.18) kg/m Assuming the balloon is spherical, its radius is 1/ 3

18-50.

18-51.

18-52.

†18-53.

18-54.

1/ 3

⎡ 3(0.139 m3 ) ⎤ =⎢ ⎥ = 0.32 m 4π ⎣ ⎦ 3 4π R ρ g FB = ρVg = , where ρ = 1000 kg/m3 and R is the radius of the object. 3 4π (0.020 m)3 (1000 kg/m3 )(9.81 m/s 2 ) = 0.33 N Ping-Pong ball: FB = 3 Baseball: FB = 61 N Basketball: FB = 420 N The net force on the balloon is FB – mHg if the positive direction is assumed to point up. The mass of the balloon has been neglected. Then the acceleration is F − mH g ρ airVg − ρ H Vg ρ air − ρ H a= B = = g ρHV ρH mH 1.29 − 0.18 (9.81 m/s 2 ) = 61 m/s 2 ⇒a= 0.18 The egg first floats when it barely displaces its own volume of water: ρ waterVg = ρeggVg ⇒ ρ water = ρegg . ⎛ 3V ⎞ R=⎜ ⎟ ⎝ 4π ⎠

83 g of salt was added to 1 liter (1 kg) of water. 1 liter = 10–3 m3, so the density is 1.083 kg ρ= = 1083 kg/m3 −3 3 10 m For him to submerge the ball, his weight must be at least equal to the weight of water displaced by the volume of the ball. 4πρ water R 3 g 4πρ water R 3 4π (1000 kg/m3 )(0.30 m)3 ⇒m= = = 113 kg (The mg = ρ waterVg = 3 3 3 friend must be very large!) (a) Mass of water displaced = mass of tanker. Therefore, Vρwater = mass = 2.2 × 108 kg V = (380 × 60 × d) = (22800 × d) m3 (d = draft in m). Therefore, (22800 d)(1020) kg = 2.2 × 108 kg d = 2.2 × 108/(1020)(22800) = 9.5 m (b) Similarly, d = (6.6) × 108/1020 (22800) = 28.4 m


CHAPTER 18-55.

18-56.

18-57.

Mass of water displaced must be the same, so Vsea ρsea = vfresh ρfresh where v = volume of water displaced = Ad, where A is the area of the bottom of the boat. Thus, ⎛ρ ⎞ dsea ρsea = d fresh ρ fresh ; d fresh = dsea ⎜ sea ⎟ ⎝ rfresh ⎠ ⎛ 1.02 × 103 ⎞ = 30 m ⎜ = 30.6 m 3 ⎟ ⎝ 1.00 × 10 ⎠ V = 40,000 m3; m = 1400 kg; ρa = 1.29 kg/m3. For lift-off, the buoyant force required = total weight = 1400(9.81) + 4000(ρi)(9.81) = weight of displaced air = 4000(1.29)(9.81), where ρi = density of air inside the balloon. Then 4000(9.81)ρi = 4000(1.29)(9.81) – 1400(9.81); ρi = 0.94 kg/m3 At the bottom of the balloon, Pinside = Poutside. At the top of the 27-m-high balloon, because the inside density is less than the outside density, there is less decrease in pressure with increasing height inside the balloon than outside. Thus, there is an inside overpressure at the top of the balloon. The pressure difference ∆P = ρ0g(27) – ρig(27) = 1.29(9.81)27 – 0.94(9.81)27 = 92.7 N/m2 By integral calculus, the area of the shaded portion on the diagram = π R2 + 2 R 2 ∫θ0 cos 2 θ dθ 2 where sin θ = z/R ⎛ θ sin 2θ ⎞ + 2R2 ⎜ + ⎟ 2 4 ⎠ ⎝2 π R2 1 ⎛ ⎞ = + R 2 ⎜ θ + sin 2θ ⎟ 2 2 ⎝ ⎠ Since wood has a density of 600 kg/m3, and since vwood/vwater = ρwater/ρwood, vwood/vwater = 1000/6000 = 5/3, where vwood = volume of the log and vwater = volume of water it displaces. 3 Then vwater = vwood . Thus, the area submerged 5 is 3/5 that of area of circle or ⎡ π R2 1 ⎛ ⎞⎤ 3 + R 2 ⎜ θ + sin 2θ ⎟ ⎥ = π R 2 ⎢ 2 ⎝ ⎠⎦ 5 ⎣ 2 =

π R2

1 1 ⎛ ⎞ + ⎜ θ + sin 2θ ⎟ /π = 3 / 5 2 2 ⎝ ⎠ 1 ⎛ ⎞ ⇒ ⎜ θ + sin 2θ ⎟ = 0.1π = 0.314 2 ⎝ ⎠ By numerical methods we find θ = 0.16 rad. The height of the water above the center of the log is R sin θ = 15 cm (sin 0.16) = 2.4 cm. Therefore, the upper surface of the log protrudes 15 cm – 2.4 cm = 12.6 cm above the water. ⇒


18

CHAPTER

18

18-58.

Volume of the balloon 4 V = π (12 × 10−2 )3 = 7.2 × 10−3 m3 3 The net upward force on the balloon F = V(Pair – PHe)g The balloon will reach equilibrium height, h, when F equals the gravitational pull on the combined (balloon + string) system. Therefore, 7.2 × 10–3(1.29 – 0.33)9.8 = (2.5 × 10–3 + h × 2 × 10–3)9.8 or 7.2 × 10−3 (0.96) − 2.5 × 10−3 h= = 2.2 m 2 × 10−3

18-59.

(a) The net force on the sphere = v (ρ – ρ’)g = effective mass of the sphere × acceleration = v (p + 1/2 ρ’)a Thus, ρ − ρ′ a= g 1 ρ + ρ′ 2 (b) For an empty bubble (ρ ~ 0) a = –2g (c) a = ∞

18-60.

Net force of the mass = weight – buoyant force. Let V be the volume of brass weights, V' = volume of the material to be weighed. At balance

Vρg – Vρag = V'ρ'g – V'ρag ⇒ V(ρ – ρa) = V'(ρ′ – ρa) where ρ, ρ′, ρa are the densities of the brass, the material weighed and air, respectively. ( ρ − ρa ) and the mass of the material weighed Then V ′ = v ( ρ ′ − ρa ) ⎛ ρ − ρa ⎞ = V ′ρ ′ = V ⎜ ⎟ ρ′ ⎝ ρ ′ − ρa ⎠ The apparent mass of the brass weights = Vρ = apparent mass of material weighed. Then the ⎛ ρ − ρa ⎞ V⎜ ⎟ ρ′ ρ ′ − ρa ⎠ ⎛ ρ − ρa ⎞ ⎛ ρ ′ ⎞ ⎝ true weight/apparent weight = =⎜ ⎟ ⎜ ⎟ Vρ ⎝ ρ ′ − ρa ⎠ ⎝ ρ ⎠


CHAPTER

18

(i) ρ′ = 1.0 × 103. Then ⎛ 8.7 × 103 − 1.3 ⎞ ⎛ 1.0 × 103 ⎞ true =⎜ ⎟⎜ ⎟ = 1.00115, so the increase = 0.12% apparent ⎝ 1.0 × 103 − 1.3 ⎠ ⎝ 8.7 × 103 ⎠ (ii) ρ′ = 5.0 × 103 ⎛ 8.7 × 103 − 1.3 ⎞ ⎛ 5.0 × 103 ⎞ true =⎜ ⎟⎜ ⎟ = 1.000111 and the increase = 0.01% apparent ⎝ 6.0 × 103 − 1.3 ⎠ ⎝ 8.7 × 103 ⎠ (iii) ρ′ = 10.0 × 103 ⎛ 8.7 × 103 − 1.3 ⎞ ⎛ 10.0 × 103 ⎞ true =⎜ ⎟⎜ ⎟ = 0.99998 and the decrease = 0.002% apparent ⎝ 10.0 × 103 − 1.3 ⎠ ⎝ 8.7 × 103 ⎠ 18-61.

18-62.

†18-63.

Mass of (water + oil) displaced = mass of block. Let the depth of the block in the water be e; then the height of the block in the oil = (0.1 – e) m. Therefore, the mass displaced = ρwater Vwater + ρoil Voil = (1.0 × 103) kg/m3 (0.06 l)m3 + (8.5 + 102) kg/m3 ([0.1 – l]0.06) m3 = 5.5 kg (60 l – 51 l) + 5.1 = 5.5 or 9 l = 0.4. Therefore e = 0.044 m, that is, the block is submerged 4.4 cm in the water. When the device floats in equilibrium in the liquid, the net force on it = weight + buoyancy = 0. Therefore, the weight of the liquid displaced = weight of hydrometer or Vliquid ρliquid = M, where Vliquid is the volume of the liquid displaced by the hydrometer. This is equal to: ⎡4 ⎤ 3 2 ⎢ 3 π R + π R′ (l − h) ⎥ ρliquid = M ⎣ ⎦ Therefore, ⎡4 ⎤ ρliquid = M / ⎢ π R 3 + π R′2 (l − h) ⎥ ⎣3 ⎦ (a) Net force = buoyant force – weight if the positive direction is assumed to point up. Let x be the displacement from the surface of the water to the bottom of the boat. Then d2x ∑ F = − Aρ gx − Mg = Ma = M dt 2 d2x Aρ g ⎛ M ⎞ ⎛ Aρ g ⎞ ⇒ 2 = −⎜ x + g⎟ = − ⎜ x+ ⎟ dt M ⎝ Aρ ⎠ ⎝ M ⎠ Note the minus sign in front of the term with buoyant force term. This is required because x is M measured down from the surface and the buoyant force points up. Let x′ = x + . Then the Aρ equation becomes


CHAPTER

18

d2x ' ⎛ Aρ g ⎞ = −⎜ ⎟ x′, and as shown in Chapter 15, the motion of a particle with this equation is dt 2 ⎝ M ⎠ an oscillation with angular frequency Aρ g ω= . M (b) Frequency f = ω/2π =

18-64.

1 2π

(380 × 60 ) m 2 × (1025 kg/m3 ) × 9.81 m/s 2 = 0.094 Hz 6.6 × 108 kg

When the tip of the pencil is submerged to a depth x, the volume of water it displaces is πR2x, where R = 0.40 cm is the radius, Hence, the buoyant force on the pencil is –ρgπR2x, where ρ is the density of water. This buoyant force can be expressed as the negative of the derivative of an effective potential energy 1 ρ gπ R 2 x 2 . 2 The net potential energy is the sum of this potential energy of the buoyant force and the gravitational potential energy: 1 U = −mgx + ρ gπ R 2 x 2 2 When the pencil comes to rest at its maximum depth, the kinetic energy will be zero, and the energy at this instant is purely potential. By energy conservation, this potential energy must match the initial potential energy mgh, where h = 4.0 cm is the initial height of the tip. Hence, 1 −mgx + ρ gπ R 2 x 2 = mgh 2 or ρπ R 2 2 x − 2 x − 2h = 0 m Inserting numbers, we find 1.0 × 103 kg/m3 × π × (0.40 × 10−2 m) 2 2 x − 2 x − 0.080 = 0 4.0 × 10−3 kg or 12.57x2 – 2x – 0.080 = 0 The solutions of this quadratic equation are 2 ± 4 + 4 × 12.57 × 0.080 x= = 0.19 m, − 0.331 m 2 × 12.57 Only the positive solution makes sense, i.e., the tip penetrates to a depth of 0.19 m, which means the full length of the pencil is just barely immersed.


CHAPTER 18-65.

18-66.

18-67.

18-68.

†18-69.

18

By conservation of energy, if the water leaves the hole with a vertical speed v and reaches a height h, v = 2 gh = 2 × 9.81 × 1.2 m/s = 4.85 m/s. Bernoulli’s equation says 1 2 1 ρ v pipe + ρ gz pipe + p pipe = ρ v 2 + ρ gz pipe + phole , where we assume the points just inside the 2 2 hole and just outside the hole are at the same height. Assume the water inside the pipe is moving very slowly so vpipe ≈ 0. Then 1 (1000 kg/m3 )(4.85 m/s) 2 p pipe = ρ v 2 + phole = + 1.01 × 105 Pa = 1.13 × 105 Pa 2 2 1 From ρ v 2 + p = constant, we have 2 1 pmouth = ρ v 2 + patm 2 1 1 Overpressure pmouth − patm = ρ v 2 = (1.29)7 2 = 32 Pa 2 2 1 2 ρ v + ρ gz + p = constant, so that at region 1, 2 v = 0, and at region 2, ρgz + p = 0 (no pressure). Therefore, 1 2 ρ v = ρ gh + patm 2 vmax = 2 gh + 2 patm / ρ 1 2 1 2 ρ vinside + ρ gyinside + pinside = ρ voutside + ρ gyoutside + poutside 2 2 Choose the two points to be very close to the roof so yinside ≈ youtside . The speed inside the house is zero, so the equation becomes 1 2 1 2 pinside = ρ voutside + poutside ⇒ pinside − poutside = ρ voutside 2 2 Then 1 pinside − poutside = (1.29 kg/m3 )(20 m/s) 2 = 258 Pa 2 The pressure inside is higher than the pressure outside, so there is an outward force of 258 N on each square meter of the roof. Bernoulli’s equation says 1 2 1 2 ρ vabove + pabove = ρ vbelow + pbelow 2 2 where we assume the two points are at essentially the same level. Then 1 1.29 kg/m3 2 2 ∆p = pbelow − pabove = ρ ( vabove − vbelow = [(130 m/s) 2 − (115 m/s) 2 ] = 2.37 × 103 Pa ) 2 2 The net force on the wing is F = (∆p) A = (2.37 × 103 Pa)(8.0 m 2 ) = 1.9 × 104 N

Bernoulli’s equation says


CHAPTER 18-70.

18

During time dt a cylinder of air with radius R = 20 m and length vdt = (5.0 m/s)dt crosses the windmill. The mass in this cylinder is dm = ρ (π R 2 vdt ). The kinetic energy of the cylinder is 1 πρ R 2 v3 dt (dm)v 2 = . The power in the flow is 2 2 dK πρ R 2 v 3 π (1.29 kg/m3 )(20 m) 2 (5.0 m/s)3 = = = 1.01 × 105 W. The windmill absorbs P= dt 2 2 5% of this power, so the output is Pout = 0.05 × 1.01 × 105 W = 5.1 × 103 W dK =

18-71.

Take two points to be at the top of the fluid in the tank and at the nozzle. Assume that the fluid in the tank moves slowly enough that the speed of the fluid there can be neglected relative to the speed of the fluid leaving the nozzle. The diagram shows the nozzle to be at the same height as 1 the top of the fluid. Then ptank + p0 = ρ v 2 + p0 , and 2 2 ptank v= . The distance y to the bottom of the tank is irrelevant.

ρ

18-72.

18-73. 18-74.

3.0 atm = 3.03 ×105 Pa. For the 1.0 cm garden hose, dV π (0.010 m) 4 (3.03 × 105 Pa) = = 0.0119 m3 /s. dt 8(1.0 × 10−3 kg/m i s)(100 m) For the 20-cm water main, we can multiply the hose result by (20/1)4 to get dV = 1.90 × 103 m3 /s. dt Nη (2500)(1.0 × 10−3 kg/m i s) = = 0.013 m/s v= 2ρ R 2(998.2 kg/m3 )(0.10 m) (a) Assume the water leaves the hose vertically with a speed v0 and reaches a height h where its speed is zero. By conservation of energy, v0 = 2 gh =

2(9.81 m/s 2 )(12 m) = 15.3 m/s

(b) 950 liters/min = 1.58 × 10–2 m3/s. For the hose, 2 ⎛ 6.35 ⎞ 2 −2 A = πR = π ⎜ × 10 m ⎟ = 3.17 × 10−3 m 2 . In the hose, ⎝ 2 ⎠ 2 1 dV 1.58 × 10 vhose = = = 4.98 m/s A dt 3.17 × 10−3

18-75.

Write Bernoulli’s equation for two points, one just inside the hose and one just outside the nozzle. Assume the two points are at the same level. The pressure in the hose is p and the pressure outside the nozzle is atmospheric pressure p0. Then 1 2 1 1 2 ρ vhose + p = ρ v02 + p0 ⇒ p = p0 + ρ ( v02 − vhose ) = 1.01 × 105 Pa + (500)(15.32 − 4.982 ) Pa 2 2 2 p = 2.06 × 105 Pa. (a) Since the collision is inelastic, the velocity of the water after the collision is zero. Thus, the magnitude of the force is F = ∆p/∆t = ∆(mv)/∆t = ρ(∆V/∆t)v (v = speed at collision) ∆V/∆t = Av (A = area of water stream) F = ρAv2 = (1000 kg/m3)( π)(0.01252 m)2(26 m/s)2 = 332 N


CHAPTER

18-76.

18

(b) Power = ∆E / ∆t = ∆(mv2)/(2∆t) = (v2/2)(∆m/∆t) = (v2/2)( ρAv) = ρAv3/2 Power = (1000 kg/m3)( π)(0.01252 m)2(26 m/s)3/2 = 4.3 kW (a) Pressure at the surface of both water levels must equal atmospheric pressure. By Bernoulli’s 1 equation, ρ v 2 + ρ gy + p = constant so that patm + ρgy1 2 1 2 = ρ v2 + ρ gy2 + Patm 2 ⇒ v22 = 2 g ( y1 − y2 ) = 2 g (h2 − h1 )

or v =

2 g (h2 − h1 )

The final result is independent of the density ρ. (b) The water velocity along the entire tube is therefore v = 2 g (h2 − h1 ) because the fluid is incompressible. Therefore, by Bernoulli’s equation 1 patm = p + ρv 2 + ρ gy. 2 1 2 p = patm − ρ v + ρ gy 2 p = patm − [ ρ g (h2 − h2 )] − ρ gh1 = patm − ρ gh2 (c) If p = 0, then h2,max = patm/ρg

†18-77.

One way to solve this problem is to look at the physical significance of the terms in Bernoulli’s 1 equation. ρ v 2 is the kinetic energy density (energy per unit volume) of the fluid; ρgh is the 2 potential energy density; and p is the additional energy density produced by an external force doing work on the fluid. Bernoulli’s equation expresses conservation of energy: in the absence of friction, energy density is constant throughout the fluid. In this problem, water is initially at rest under atmospheric pressure in the pond. The pump increases the energy density by lifting the water through a height h, increasing the pressure from 1 atm to 5.4 atm, and increasing the speed from essentially zero to the value that gives a volume flow of 1100 liters/min = 0.0183 m3/s through the 6.35 cm hose. Multiplying the increase in energy density by the volume flow will give the power required. Begin with the speed of water in the hose: 1 dV 0.0183 m3 /s = = 5.78 m/s. v= 2 A dt ⎛ 6.35 ⎞ −2 × 10 m ⎟ π⎜ ⎝ 2 ⎠


CHAPTER

18

Use Bernoulli’s equation to find the increase in energy density compared to the surface of the pond; 1 ∆( E / V ) = ρ v 2 + ρ gh + ∆p 2 = (500 kg/m3 )(5.78 m/s) 2 + (1000 kg/m3 )(9.81 m/s 2 )(4.5 m) + (4.4 atm)(1.01 × 105 Pa/atm) = 5.05 × 105 J/m3

18-78.

The power required is dV P = ∆( E / V ) = (5.05 × 105 J/m3 )(0.0183 m3 /s)(1 hp / 746 W) = 12.4 hp dt In Bernoulli’s equation, the work done by an external agent is represented by the pressure term. If we assume that the upstream and downstream ends of the flow are at the same level, then the decrease in kinetic energy (indicated by the water speed) is due to friction. Then the net work done to overcome friction is ∆W = ( p1 − p2 )∆V ∆W ∆V (2.7 × 105 Pa)(380 × 10−3 m 3 / min) = ( p1 − p2 ) = = 1.71 × 103 W = 2.3 hp ∆t ∆t 60 s/min As in Example 14, 2( p1 − p2 ) v1 = ρ[( A1 /A2 ) 2 − 1] Power =

18-79.

where p1, p2 are the pressures at the positions shown. But p1 will support water of height h1 = p1/ρg, whereas p2 will support that of height h2 = p2/ρg. Then (p1 – p2) = (h1 – h2)ρg 2(h1 − h2 ) ρ g 2( h1 − h2 ) g v1 = = 2 ( A1/A2 ) 2 − 1 ρ[( A1 /A2 ) − 1] =

2 × 3 × 9.81 2

⎛ 30 ⎞ ⎜ ⎟ −1 ⎝ 10 ⎠

= 2.71 m/s

∆V = Av = π R 2 v = π (0.30) 2 (2.71) = 0.192 m3 /s = 192 liters/s ∆t


CHAPTER 18-80.

†18-81.

18-82.

18-83.

18-84.

18

Static pressure of the fluid flowing past = p1. This equals the pressure in the first tube. The pressure in the bent tube p2 can be found by Bernoulli’s relation: 1 p1 + ρ v 2 = p2 2 Therefore, v = 2( p2 − p1 ) / ρ If the speed of the plane relative to the air is v and the cross-sectional area of the window is A, then during time dt a tube of air with a volume of Avdt strikes the window and essentially comes to rest. The force exerted on the window is equal to the rate of change in momentum of the air mass contained in this tube. 950 km/h = 250 m/s, so dp ( ρ air Avdt )v F = = = ρ air Av 2 = (1.29 kg/m3 )(0.30 m) 2 (250 m/s) 2 = 7.3 × 103 N dt dt We can apply Bernoulli’s equation to a small volume of water (i) immediately before, (ii) and immediately after passing through the small hole. Both (i) and (ii) are at the same height z. For (i), V ≅ 0, and P = Patm + pg(h – z). For (ii), P = Pa. From Bernoulli’s equation, pg(h – z) = (1/2)pv2. From this, v = {2g(h – z)}0.5 for the speed of the water emerging from the hole. Since the tank wall is vertical, v is horizontal. The fall time through a height z is given by z = (1/2)gt2, from which t = [2z/g]0.5, and the horizontal distance d = vt = {2g(h – z)}0.5[2z/g]0.5 = 2[z(h – z)]0.5 To find the choice of z that gives the largest horizontal distance dmax, set the partial derivative (dd/dz) = 0 Then 2(1/2)(z–0.5)[h – z]0.5 + 2(z0.5)(1/2)(h – z)–0.5 = 0 (h – z)/z = z/(h – z), or 2z = h; z = h/2 Therefore d max = 2[(h/2)(h − h/2)]0.5 = 2[h/2] = h From Bernoulli’s equation, P + (1/2)pv2 + pgz = constant. Here z is constant; p = 104 kg/m3; v1 = 15 × 103 m/s; v2 = 0; P1 = Patm = 1.1 × 105 N/m2; P2 = ? Then Pa + (1/2)(104)(15 × 103)2 = P2 Therefore, P2 = 1.12 × 1012 N/m 2 . The speed at which the water emerges from the hole =

2gz where z is the height of the surface

of water above the hole. The volume flow rate is dV = − A′v = − A′ 2 gz. dt The minus sign is needed because the volume of water in the tank is decreasing. The height is z = V/A, where V is the instantaneous volume in the tank. Then dV 2 gV = − A′ dt A To find V(t), integrate: V dV 2g t ∫V0 V = − A′ A ∫0 dt 2g 2 ⎡⎣ V (t ) − V0 ⎤⎦ = − A′ t A


CHAPTER

18

where V0 is the initial volume of water in the tank = lA, so g V (t ) − lA = − A′ t 2A When the tank is empty, V(t) = 0. So the time to empty the tank is g t⇒ 0 − lA = − A′ 2A t= 18-85.

lA A'

A 2l 2A = g A' g

The velocity of the water emerging from a thin strip, thickness dz, distance z from surface of water, is given by (as in Example 9) v = 2 gz ∆V/∆t = Av, so that the volume of water emerging from this strip is Area × velocity = (edz ) 2 gz = dV / ∆t

18-86.

Therefore total volume emerging per unit time z = h2 1 1 z = h2 = ∆v = dV e 2 g ∫ z1 / 2 dz ∫ z = h z = h1 1 ∆t ∆t 2 ∆v = e 2 g (2/3) z 3 / 2 ]zz == hh12 = e 2 g ( h23 / 2 − h13 / 2 ) 3 ∆t (a) pwrist = pheart + ρ water gh

= 120 mm-Hg + (1060 kg/m3 )(9.81 m/s 2 )(0.23 m)(760 mm-Hg /1.01 × 105 Pa) = 138 mm-Hg (b) pwrist = pheart − ρ water gh = 120 mm-Hg − (1060 kg/m3 )(9.81 m/s 2 )(0.78 m)(760 mm-Hg /1.01 × 105 Pa) = 59 mm-Hg 18-87.

Since the hydrostatic pressure in the column of saline equals the hydrodynamic pressure in the needle, p = ρ gh = (1060 kg/m 3 )(9.81 m/s 2 )(0.103 m) = 1.07 × 103 Pa = 8.06 mm-Hg

18-88.

p = patm + g ( ρ water hwater + ρoil hoil ) = 1.01 × 105 Pa + 9.81 m/s 2 [(1000 kg/m3 )(0.050 m) + (918 kg/m3 )(0.060 m)] = 1.02 × 105 Pa

†18-89.

When he is 2.0 m deep, the water pressure on his chest is pwater = patm + ρ water gh = 1.01 × 105 Pa + 9.81 m/s 2 (1000 kg/m3 )(2.0 m) = 1.21 × 105 Pa The air pressure inside his chest is 1 atm (1.01 × 105 Pa), so the net pressure force on his chest is F = (∆p) A = (0.20 × 105 Pa)(0.10 m 2 ) = 2.0 × 103 N (about 470 lb).

18-90.

(a) W = mgh = 20 liters × 1 kg/liter × 9.81 m/s 2 × 35 m = 6.86 × 103 J. The power is W 6.86 × 103 J = = 114 W t 60 s (b) (See Problem 77.) The power to move against the overpressure is dV = 3.0 × 105 Pa(20 liters/min)(10−3 m3 / liter)(1 min/60 s) = 1.0 × 102 W Pp = p dt Pg =


CHAPTER

†18-91.

18-92.

18

(c) The total power required is 214 W, which corresponds to 0.29 hp. Yes, a 1/2 hp pump will be adequate. (See Example 9.) The oil sinks until it displaces its own weight of water. Thus ρoilVoil = ρ waterVwater . The volume of the oil above the surface of the water is ρ − ρoil ∆V = water . The oil and water both ∆V = Voil − Vwater ⇒ ρ oilVoil = ρ water (Voil − ∆V ) ⇒ ρ water Voil cover the same surface area A, so the total volume depends on the thickness of each volume. If the thickness of the oil is hoil and the thickness of the displaced water is hwater, ∆h ρ water − ρoil = then ∆V = (hoil – hwater)A = (∆h)A and hoil ρ water 1025 − 950 (10 cm) = 0.73 cm. ⇒ ∆h = 1025 According to Problem 18-91, the fraction of the block’s volume above the water is ∆Vsumberged ρ ρ − ρblock ∆V ∆V . Then the fraction submerged is = water =1− = block = 0.60. ρ water Vblock Vblock Vblock ρ water

18-93.

The barge will barely be floating when its average density is equal to the density of water. The volume of the barge is V = (3 × 5 × 20) m3 = 300 m3. So the maximum load it can carry is given mmax + mbarge by 1000 kg/m3 = ⇒ mmax = 1000V − mbarge . The mass of the barge is 5.0 × 104 kg, V so mmax = 2.5 × 105 kg.

18-94.

(a) The log floats, so it displaces its own weight, or mass, of water. This is 200 kg, which occupies a volume of ∆V = 0.20 m3. The trough is 1.0 m wide and 2.0 m long, so this volume has ∆V 0.20 m3 a height of h = = = 0.10 m, or 10 cm. 2.0 m 2 2.0 m 2 (b) The rock sinks, so it displaces its own volume of water. This is 200 kg ∆V 0.10 m3 3 ∆V = = 0.10 m . The height of this volume is h = = = 0.050 m, 2.0 m 2 2.0 m 2 2000 kg/m3 or 5.0 cm.

18-95.

The volume of tissue in an 80 kg man is V1 =

80 kg = 0.0747 m3 . Let V2 be the volume of 3 1071 kg/m

air in his lungs. If the man is to float, the buoyant force must equal his weight. If he barely floats, the volume of water displaced will be equal to his total volume and m 80 kg − V1 = − 0.0747 m3 = 5.3 × 10−3 m3 . (The mg = ρ water g (V1 + V2 ) ⇒ V2 = 3 ρ water 1000 kg/m 18-96.

additional mass of the air has been neglected.) When an object is immersed in water, its apparent mass will be reduced by the mass of water it displaces. If the mass of the bracelet in air is 0.900 kg and its apparent mass when submerged in water is 0.820 kg, it has displaced 0.080 kg of water and must have a volume of 0.080 kg 0.900 kg V = = 8.0 × 10−5 m3 . Its average density is ρ = = 11.3 × 103 kg/m3 . −5 3 3 1000 kg/m 8.0 × 10 m Assume a fraction x of the bracelet is gold. Then 19.3x + 8.96(1 − x) = 11.3, so x = 0.23. The bracelet contains 23% gold.


CHAPTER 18-97.

18-98.

†18-99.

18-100.

18

Let the person’s volume of tissue be V. The volume of air in the lungs is 1.2 liters = 1.2 × 10–3 m3. As in Problem 96, the person’s apparent mass is reduced by the mass of water displaced if the person sinks. Thus ρ water (V + 1.2 × 10−3 ) = (83.20 − 4.30) kg, which gives V = 0.0777 m3. So the 83.20 kg = 1.07 × 103 kg/m3 . person’s density is ρ = 0.0777 m3 1 2 1 2 + pabove = ρ vbelow + pbelow , where we assume the points above and From Problem 69, ρ vabove 2 2 below the wing are at essentially the same level. Then 1 1.29 kg/m3 2 2 [(95 m/s) 2 − (85 m/s) 2 ] = 1.16 × 103 Pa. ∆p = pbelow − pabove = ρ ( vabove − vbelow = ) 2 2 The net lift force on the wing is F = (∆p) A = (1.16 × 103 Pa)(92 m 2 ) = 1.1 × 105 N. 1 2 1 ρ v1 + ρ gy1 + p1 = ρ v22 + ρ gy2 + p2 . Take point 1 to be at the top of the water in the tank, 2 2 and point 2 is just below the hole in the bottom of the tank where the water runs out. The pressure at both points will be atmospheric pressure, so those terms cancel in the equation. Assume that the area A1 of the water at the top is much larger than the area A2 of the hole at the bottom, so Av v1 = 2 2 ≈ 0. Then v2 = 2 g ( y1 − y2 ) = 2(9.81 m/s 2 )(2 m) = 6.26 m/s. A1 dV2 The flow rate from the hole is = v2 A2 = (6.26 m/s)(π )(0.01 m) 2 = 2.0 × 10−3 m3 / s. dt 1 2 1 ρ v1 + ρ gz1 + p1 = ρ v22 + ρ gz2 + p2 . Take two points on a streamline that are at the same 2 2 height, so z1 = z2. Assume that point 1 is far from the center where the air is nearly calm, so v1 ≈ 2( p1 − p2 ) , 0. Assuming that the density of air is essentially the same at both points, we get v22 =

ρ

18-101.

which agrees with the observations. 1 2 1 ρ v1 + ρ gz1 + p1 = ρ v22 + ρ gz2 + p2 . Take the two points to be at the same height. Point (a) 2 2 1 is outside where the pressure is 1 atm and the speed is v1 = 18 m/s. Point 2 is inside where the speed is essentially zero. Then ρ v2 (1.29 kg/m3 )(18 m/s) 2 = 1.01 × 105 Pa. The pressure inside p2 = p1 + 1 = 1.01 × 105 Pa + 2 2 actually increases by 209 Pa, which is too small to be noticeable to three significant digits. (b) If the air is blowing past the window, it will cause the pressure there to be lower than the pressure inside the house where the air is relatively still. This will cause a movement of air from the inside of the house to the outside, slightly lowering the pressure inside.


CHAPTER 19

THE IDEAL GAS AND KINETIC THEORY


19-2.

†19-3.

⎡9 ⎤ 0°C = ⎢ (0) + 32 ⎥ °F = 32°F 5 ⎣ ⎦ ⎡9 ⎤° −196°C = ⎢ (−196) + 32 ⎥ F = −320.8°F ⎣5 ⎦ ⎡9 ⎤° −253°C = ⎢ (−253) + 32 ⎥ F = −423.4°F ⎣5 ⎦ ⎡9 ⎤° −269°C = ⎢ (−269) + 32 ⎥ F = −452.2°F ⎣5 ⎦ −270°C = −454°F − 273.15°C = −459.67°F 9 5 TF = TC + 32. Therefore, TC = (TF − 32) and T = TC + 273.15 5 9 5 136.4°F = (136.4 − 32)°C = 58.0°C = 331.2°K 9 For iron, M Fe = 56 g/mole.

5 (−126.9 − 32)°C = −88.3°C = 184.9°K 9 m 0.50 g The number of moles n = = = 8.9 × 10−3 moles. So, M Fe 56 g/mole −126.9°F =

N = nN A = ( 8.9 × 10−3 moles )( 6.02 × 10 23 molecules/mole ) = 5.3 × 1021 atoms

19-4.

For NaCl, M = ( 23.0 + 35.5 ) g/mole = 58.5 g/mole. n =

m 10 g = = 0.171 mole. M 58.5 g/mole

The number of atoms of Na is equal to the number of atoms of Cl, N = nN A = ( 0.171 mole ) ( 6.02 × 1023 molecules/mole ) = 1.03 × 1023 molecules †19-5.

From the given information,

0.24mN2 = 0.76mO2 or

mN2 mO2

mN2 mN2 + mO2 =

= 0.76. With some algebraic manipulation this implies

0.76 = 3.2 (1). In general, mN2 = M N2 nN2 = ( 28 g/mole ) nN2 and 0.24

mO2 = M O2 nO2 = ( 32 g/mole ) nO2 . Substituting into (1) and solving, 1− x ⎛ 32 ⎞ = 3.6. Solving for nN2 = 3.2 ⎜ ⎟ nO2 = ( 3.6 ) nO2 (2). Letting x = % O 2 by #, then by (2) x ⎝ 28 ⎠ 1 x= = 0.22 = 22% O 2 and 1 − x = 1 − 0.22 = 78% N 2 4.6


CHAPTER

19-6.

†19-7.

19

⎛n⎞ Assuming that the air behaves like an ideal gas, p = ⎜ ⎟ RT . Substituting n = m / M , produces ⎝V ⎠ ⎛ ( 8.31 J/mole i K )( 500 K ) ⎞ ⎛ m ⎞ RT −4 = (1.5 × 10−9 kg/m3 ) ⎜ p=⎜ ⎟ ⎟ = 2.1 × 10 Pa 0.029 kg/mole ⎝V ⎠ M ⎝ ⎠ p According to the Ideal-Gas Law, = RT . Thus, the ratio of pressure to density is (n /V ) proportional to the temperature. According to Eq. (17.5), v ∝

p0

ρ0

∝ T . If the temperature

263 = 0.947. Since the 293 frequency of a flute is directly proportional to the speed of sound, the frequency drops by a factor of 0.947, from 261.7 Hz to 0.947 × 261.7 Hz = 247.9 Hz. This means that the frequency decreases by 14 Hz. Let T = 85 K + 273 K = 358 K and T ′ = 30 K + 273 K = 303 K. From the Ideal-Gas Law, nR nR p′ T ′ p= T and p′ = T ′, which combines to = or V V p T drops from 293 K to 263 K, then the speed decreases by a factor

19-8.

T′ ⎛ 303 K ⎞ p=⎜ ⎟ (1 atm ) = 0.85 atm. The difference in pressures or underpressure is 1.00 T ⎝ 358 K ⎠ atm – 0.85 atm = 0.15 atm. Pressure in the tire at T2 = 273 K T 273 p2 = p1 2 = (3.2) × = 2.9 atm 273 + 30 T1 p′ =

19-9.

19-10.

The overpressure in the tire = 1.9 atm 5 −6 pV (1.01 × 10 Pa )(1.0 × 10 ) = = 2.7 × 1019 molecules of gas. Using the result from N = −23 kT 1.38 × 10 J/K 273 K ) ( )( problem 5 that air by molecule is 78% nitrogen and 22% oxygen results in there being 2.1 × 1019 nitrogen molecules and 6.0 × 1018 oxygen molecules or a total of 4 ( 2.7 × 1019 ) = 1.1 × 1020 total atoms.

†19-11.

(a) Assume that the nitrogen and oxygen each behave like an ideal gas: pV = nRT , where mRT n = m / M . Therefore, p = . From the periodic table, M N2 = 28 g/mole and MV M O2 = 32 g/mole.

(1.0 g )( 8.31 J/mole i K )( 290 K ) = 7.5 × 104 Pa ( 32 g/mole ) (1.0 × 10−3 m3 ) (1.0 g )( 8.31 J/mole i K )( 290 K ) = 8.6 × 104 Pa = ( 28 g/mole ) (1.0 × 10−3 m3 )

For the oxygen tank, pO2 =

For the nitrogen tank, pN2

(b) When the two gases are combined into one tank at the same temperature, the total pressure p = pN2 + pO2 = ( 7.5 × 104 Pa ) + ( 8.6 × 104 Pa ) = 1.6 × 105 Pa


CHAPTER 19-12.

19

Assume that air behaves like an ideal gas at STP. The Ideal-Gas Law (in terms of number of molecules), pV = NkT , tells us that the amount of volume for a given molecule of gas at STP is

−23 V kT (1.38 × 10 J/K ) ( 273 K ) = = = 3.73 × 10−26 m 3 /molecule. If the molecules are modeled 5 2 N p (1.01 × 10 N/m )

as roughly spherical, then the radius of each sphere of that volume is 1

⎛ 3 ( 3.73 × 10−26 m3 ) ⎞ 3 ⎛ 3V ⎞ ⎟ = 2.07 × 10−9 m. The average distance between molecules r=⎜ ⎟ = ⎜⎜ ⎟ 4 ( 3.14 ) ⎝ 4π ⎠ ⎝ ⎠ would be twice the radius 2r = 2 ( 2.07 × 10−9 m ) = 4.1 × 10−9 m 1 3

19-13.

19-14.

p1V1 pV = 2 2 by the Ideal-Gas Law, so that p2 = p1(T2/T1)(V1/V2). T1 T2 (T2/T1) = (330/280) and (V1/V2) = (1.00/1.05) ⎛ 330 ⎞ ⎛ 1.00 ⎞ Therefore, p2 = 3.0 atm ⎜ ⎟⎜ ⎟ = 3.4 atm ⎝ 280 ⎠ ⎝ 1.05 ⎠ pV 5 × 105 × 0.3 Number of moles n = = = 62 moles RT 8.31 × 293

19-15.

The pressure is the same indoors and outdoors, but Tout = 308 K, and Tin = 294 K. From the IdealGas Law, PV = nRT, Pi/Ti = nR/Vi, and P0/T0 = nR/V0, so that, from dividing the two equations and canceling the two equal pressures, T0/Ti = {nM/Vi}/{nM/v0} = ρi / ρ0 = 308 / 294 = 1.05, that is if ρi = 1.29 kg/m3, then ρ0 = 1.23 kg/m3.

19-16.

From the Ideal-Gas Law pV N = kT ⎡ (1.0 × 10−16 atm ) × (1.01 × 105 Pa/atm ) ⎤ ⎡ (1.0 cm3 ) × (1.0 × 10−6 m3 /cm3 ) ⎤ ⎦⎣ ⎦ = ⎣ −23 (1.38 × 10 J/K ) ( 293 K ) = 2.5 × 103 molecules in 1.0 cm3

19-17. 19-18.

p = NkT/V = (N/V)kT = (1.0 × 1010 m–3)(1.38 × 10–23 J/K)(1.0 × 104 K) = 1.4 × 10–9 Pa p 1 pV = nRT or T = pV/nR = R (n / V ) n/V = number of moles per unit volume 1 kg of air contains 1/0.029 = 34.5 moles/kg Altitude n/V = density (103 m)

× 34.5 moles/kg

20 40 60 80

3.17/m3 0.148/m3 0.013/m3 8.6 × 10–4/m3

⎛ 1 ⎞ T = p / R⎜ ⎟ ⎝ n /V ⎠ 5.6 × 103/8.31 × 3.17 = 212.4 K = –60.7°C 3.2 × 102/8.31 × 0.148 = 259.7 K = –13.4°C 2.8 × 101/8.31 × 0.013 = 257.1 K = –16.0°C 1.3 × 100/8.31 × 8.6 × 10–-4 = 181.5 K = –91.7°C


CHAPTER 19-19.

19-20. †19-21.

19-22. †19-23.

19

pV = nRT. Therefore, (n/V) = p/RT; p = 1.01 × 105 N/m2; T = 4.2ºK Therefore, (n/V) = (1.01 × 105N/m2)/((8.314 J/K)(4.2K)) = 2892 moles/m3 1 mole of helium has a mass of 4 g = 0.004 kg. Then Density = (2892 × 0.004) kg/m3 = 12 kg/m3 pV (3atm × 1.01 × 105 Pa/atm) × ( 2.5 × 10−2 m3 ) Number of moles n = = = 3.1 moles RT 8.31 × 290 K Mass of air in the tire = 3.1 moles × 29 g/mole = 90 g Assume that the air in the tire obeys the Ideal-Gas Law. The volume and number of molecules pf p presumably stays constant throughout the day, so i = = constant. Specifically, Ti Tf 1.0 atm + 4.0 atm 1.0 atm + p f ,over = . Or, p f ,over = 4.3 atm 295 K 311 K (180 atm ) (1.01 × 105 Pa/atm ) ( 0.035 m3 ) pV Using the Ideal-Gas Law, n = = = 257 moles RT ( 8.31 J/mole i K )( 298 K )

(

)

m , and assuming that the temperature is 273 K, M ( 400 atm ) (1.01 × 105 Pa/atm ) ( 0.028 kg/mole ) m pM = = = 500 kg/m3 V RT ( 8.31 J/mole i K )( 273 K )

Using the Ideal-Gas Law, n =

(

19-24.

†19-25.

Assume that the argon in the tube obeys the Ideal-Gas Law. The volume and number of pf p molecules stays constant through the heating process, so i = = constant. Specifically, Ti Tf pf 1 atm . Or, p f = 4.1 atm. = 273 K 1123 K m , Using the Ideal-Gas Law and n = M ( 56 atm ) (1.01 × 105 Pa/atm ) ( 0.044 kg/mole ) m pM = = = 100 kg/m3 i V RT 8.31 J/mole K 298 K ( )( )

(

19-26.

†19-27.

)

)

10 g of water is 10/18 mole of water. Hence, the Ideal-Gas Law tells us nRT 10 /18 × 8.314 J/K × 773 K = = 3.6 × 106 N/m 2 = 35 atm p= V 10−3 m3 The pressure 15 m below the lake is patm + ρgz = 1.01 × 105 N/m2 + (1000 kg/m3)(9.8 m/s2)(15 m) = 2.48 × 105 N/m2 At the surface the pressure = patm = 1.01 × 105 N/m2 ⎛ p ⎞ Since pV = const, p1V1 = p2V2 , or V2 = V1 ⎜ 1 ⎟ . ⎝ p2 ⎠ But V α d 3 where d is the diameter of the bubble. Thus,


CHAPTER

19

1/ 3

⎛ p ⎞ ⎛ p ⎞ d = d ⎜ 1 ⎟ ⇒ d 2 = d1 ⎜ 1 ⎟ ⎝ p2 ⎠ ⎝ p2 ⎠ 3 2

3 1

1/ 3

⎛ 2.48 × 105 ⎞ d 2 = 1.0 cm ⎜ 5 ⎟ ⎝ 1.01 × 10 ⎠

19-28.

= 1.3 cm

Volume of one mole of gas at STP, V =

RT 8.31 × 273 = = 0.02246 m3 1.01 × 105 p

The volume occupied by the He atoms in one mole of gas = 6.02 × 1023 × 3.0 × 10–30 = 1.81 × 10–6 m3 Therefore, fraction of volume occupied by the He atoms = †19-29.

19-30.

19-31.

19-32.

1.81 × 10−6 = 8 × 10−5 = 0.008% 0.0227

Mass of CO2 in the cylinder = 8.2 – 5.9 = 2.3 kg 2.3kg = 52.3 mol Number of moles of CO 2 = (12 + 2 × 16)10−3 kg/mol nRT 52.3 × 8.31 × 293 Therefore, p = = = 4.5 × 107 N/m 2 −3 2.8 × 10 V (a) pV = nRT. Since p, V, R are constants in this problem, nT = constant. Therefore, n1T1 = n2T2 or (n2/n1) = (T1/T2), where n = number of moles. But since the number of moles of air is proportional to the mass, this means (m2/m1) = (T1/T2) = (283/303) = 0.934 Therefore, 0.066 = 6.6% mass of the air escapes (b) Here, V, n, R are constants, so that p α T, p2/p1 = T2/T1, p2 = (T2/T1)p1. In other words, p2 = 1.0 atm (303/283) = 1.07 atm Net pressure on walls and windows is 0.07 atm = 7070 N/m2. Force exerted on window is then 7070 N ≈ 1600 lb. The window probably cannot stand the force. It is equivalent to the weight of about ten 150-lb people. The molecular mass of air is about 29 grams. pV = nRT, n = number of moles = m/µ where m = mass of gas, µ = mass of gas per mole. pV = (m/µ)RT or µ = (m/V)(RT/p) = ρRT/p. Density of gas > density of air, ρ2 > ρ1. Then µ1 = ρ1RT1/p; µ2 = ρ2RT2/p, ρµ2/RT2 = ρ2 > ρ1 = pµ1/RT1. µ2/T2 > µ1/T1 ⇒ µ2 > µ1 (T2/T1) ⎛ 700 + 273 ⎞ µ 2 > 29.0 g ⎜ ⎟ = 96.3 g ⎝ 20 + 273 ⎠ Let ρ′ be the density of the hot air in the balloon, ρ the density of the external air. Buoyant force = (ρ – ρ′)Vg where V = volume of the balloon. To lift off, (ρ – ρ′)Vg ≥ mg or (ρ – ρ′)V > m, which means (ρ – ρ′) > (m/V) (1) As in problem 20 we see that µ = ρRT/p. Here µ (molecular mass) is constant and so is p. Then ρT = ρ′T′; ρ′ = ρ(T/T′). Substituting this into (1) gives


CHAPTER

19

T ⎤ ⎡ ⎢ ρ − T ′ ρ ⎥ > (m / V ) or ⎣ ⎦

T ⎞ ⎛ ⎜ 1 − ′ ⎟ > (m / V ρ ). Then T ⎠ ⎝

⎛ m ⎞ T ⇒ ⎜1 − ⎟ > T / T ′ and T ′ > Vρ ⎠ ⎡ m ⎤ ⎝ ⎢1 − V ρ ⎥ ⎣ ⎦ 293 K = 405 K = 132°C T′ > ⎡ ⎤ 730 kg ⎢1 − 2200 m3 × 1.20kg/m3 ⎥ ⎣ ⎦ †19-33.

19-34.

Number of moles of He contained in the balloon 3.2 × 102 × 8.5 × 105 n = pV / RT = = 1.26 × 105 mol. 8.31 × 260 Net ‘lift’ force on the balloon = Buoyancy – Gravity = n(0.029 – 0.004)g = payload = mg Thus, m = 1.26 × 105 × (0.025) = 3150 kg p ⎛ T1 ⎞ 3.2 × 102 ⎛ 273 ⎞ 5 3 3 V = Volume at STP, V1 = ⎜ ⎟ ⎜ ⎟ 8.5 × 10 = 2.8 × 10 m p1 ⎝ T ⎠ 105 ⎝ 260 ⎠ Buoyant force F = (M – m)g where M = mass of water displaced, m = mass of air pumped into the ship. Let V be the volume of the cavity, ρ = density of water and µ = molecular mass of air. Then, M = ρV (1) mRT (2) pV = nRT = (m/µ)RT which gives V = µp (1) and (2) give M/ρ = mRT/µp or M = m(ρRT/µp). p = pressure = patm + ρgz = 1.01 × 105 N/m2 + 1000 kg/m3 × 9.81 m/s2 × 60 m = 6.90 × 105 N/m2 ⎛ ρ RT ⎞ ⎛ ρ RT ⎞ − m⎟ g = m ⎜ − 1 ⎟ g = M ′g Therefore, F = ( M − m)g = ⎜ m ⎝ µp ⎠ ⎝ µp ⎠ where M′ = mass of the ship. So ⎛ ρ RT ⎞ m = M′/⎜ − 1⎟ ⎝ µp ⎠ −1

†19-35.

⎡ ⎛ (1000 kg/m 2 ) ( 8.314 J/K )( 288 K ) ⎞ ⎤ ⎥ = 4.2 × 105 kg ⎟ − m = 5.0 × 107 kg ⎢ ⎜ 1 ⎟ 0.029 kg × 6.9 × 105 N/m 2 ⎢ ⎜⎝ ⎥ ⎠ ⎣ ⎦ (a) Assume the temperature is constant. Then p1V1 = p2V2. But V (volume of air enclosed) ∝h = height of air in the bell. Therefore, p1h1 = p2h2 or ⎛ p ⎞ h2 = h1 ⎜ 1 ⎟ , where p1 = patm = 1.01 × 105 N/m 2 ⎝ p2 ⎠ p2 = patm + ρgz = 1.01 × 105 N/m2 + (1000 kg/m3)(9.8 m/s2)(15 m) = 2.48 × 105 N/m2 ⎛ 1.01 × 105 ⎞ Therefore, h2 = 2m ⎜ = 0.81 m. This means that the water rises 1.2 m. 5 ⎟ ⎝ 2.48 × 10 ⎠


CHAPTER

19

(b) Pressure of air must equal the pressure of water at the lowest water level, i.e., at the bottom of the bell. The pressure there p3 = patm + ρgZ, where Z = 15m + 1.2m of risen water (so we get to the bottom of the bell) = 1.01 × 105 N/m2 + (1000 kg/m3)(9.8 m/s2)(16.2 m) = 2.6 × 105 N/m2 Original amount of air n = p1V/RT moles. Final amount of air = (p3V/RT) moles. Thus, the number of moles of air added = (p3 – p1)V/RT and the mass of air ( p − p1 )V (2.6 × 105 − 1.01 × 105 ) 3.53 m3 added = 3 M = × 0.029 kg = 6.8 kg RT (8.314) 288°K 19-36.

19-37.

19-38.

pV = NkT. Assuming the temperature is constant, then p α (N/v) = n (number of molecules per ⎛p ⎞ unit volume). Then, n2 = nI ⎜ 2 ⎟ ⇒ n2 = 0.24 n1 ⎝ p1 ⎠ Only 24% of the molecules reach lungs, but there should be 23% oxygen molecules, so 0.23 × 100 = 96% percentage of oxygen in mixture = 0.24 1 mole air contains 0.756 mole N2, 0.231 mole O2, 0.013 mole Ar Its mass is 0.755(28 g) + 0.231(32 g) + 0.013(40) = 29.05 g. Therefore, molecular weight of air is 29 g. (a) 1 mole of the gas contains 0.38 H, 0.62 He, so the mean molecular mass is: 0.38(1.00) + 0.62(4.00) = 2.86 g. m ⎛m⎞ 1 pV = nRT = RT or p = ⎜ ⎟ RT µ ⎝V ⎠ µ 1 p = (1.48 × 105 kg/m3 ) ( 8.314 J/K ) (15.0 × 106 K ) = 6.45 × 1015 N/m 2 −3 2.86 × 10 kg (b) 1 mole of the gas contains 0.71 H, 0.29 He. Therefore, mean molecular mass is 0.71 (1.00) + 0.29 (4.00) = 1.87 g. 1 ⎛m⎞ 1 p=⎜ ⎟ RT = (3.6 × 104 kg/m3 ) ( 8.314 J/K ) ( 9.0 × 106 K ) −3 (1.87 × 10 kg) ⎝v⎠ µ

= 1.4 × 1015 N/m 2 †19-39.

19-40.

m , M m pM pMg dp Mg ρ= = . Substituting ρ into the differential equation: dp = − dy or, =− dy V RT RT p RT

Differentiating p − p0 = − ρ gy yields dp = − ρ gdy. Using the Ideal-Gas Law and n =

From Chapter 18 we know that p = constant –ρgy. Therefore, dp = –ρg dy (1) Also pV = NkT gives p/kT = (N/V) N/V = number of molecules per unit volume = (ρ/m) So, p/kT = (ρ/m) ⇒ ρ = p(m/kT) Substituting this into (1) gives dp = − p ( mg/kT )dy or dp/p = − Integrating both sides gives ln p − ln p0 = ln ( p/p0 ) = − Finally, eln ( p/p0 ) = p/p0 = e

−

mg y kT

⇒ p = p0 e

−

mg y kT

mg y kT


mg dy kT

CHAPTER

†19-41.

Using m = vrms =

19-42.

19 M and vrms = NA

3kN AT = M

Using m =

3kT , m

3 (1.38 × 10−23 J/K )( 6.02 × 1023 molecules/mole ) ( 273 K ) 0.018 kg/mole

M and vrms = NA

3kT , vrms = m

= 615 m/s

3kN AT . M

At T = 4.5 × 103 K vrms =

3 (1.38 × 10−23 J/K )( 6.02 × 1023 molecules/mole ) ( 4500 K ) 0.001 kg/mole

= 1.1 × 104 m/s

7

At T = 1.5 × 10 K

vrms = †19-43.

3 (1.38 × 10−23 J/K )( 6.02 × 1023 molecules/mole )(1.5 × 107 K ) 0.001 kg/mole

Since vrms =

v′ 3kT , then rms = vrms m

= 6.1 × 105 m/s

T′ ′ = 2vrms , then . That is, if vrms T

T ′ = 4T = 4 × ( 300 K ) = 1200 K 19-44.

†19-45.

3kT T′ T ′ = vrms implies that or = , then vrms m′ m m ⎛ m′ ⎞ ⎛ 32 ⎞ T ′ = ⎜ ⎟ T = ⎜ ⎟ ( 293 K ) = 335 K ⎝m⎠ ⎝ 28 ⎠ M 3kT Using m = and vrms = , NA m From vrms =

vrms =

3kN AT = M

3 (1.38 × 10−23 J/K )( 6.02 × 1023 molecules/mole ) ( 673 K ) 1.0 × 10−3 kg/mole

The average kinetic energy is 19-46.

= 4100 m/s

3 3 kT = (1.38 × 10−23 J/K ) ( 673 K ) = 1.4 × 10−20 J 2 2

For electrons: Using vrms =

3kT = m

3 (1.38 × 10−23 J/K ) ( 7000 K )

( 9.1 × 10

−31

kg )

= 5.6 × 105 m/s, The average kinetic

3 3 kT = (1.38 × 10−23 J/K ) ( 7000 K ) = 1.4 × 10−19 J 2 2 For nitrogen ions: M 3kT , and vrms = Using m = m NA

energy is

vrms =

3kN AT = M

3 (1.38 × 10−23 J/K )( 6.02 × 1023 molecules/mole ) ( 7000 K )

The average kinetic energy is

0.014 kg/mole

= 3530 m/s

3 3 kT = (1.38 × 10−23 J/K ) ( 7000 K ) = 1.4 × 10−19 J 2 2


CHAPTER

†19-47.

Using m = vrms =

19-48.

M and vrms = NA

3kN AT = M

3kT , m

3 (1.38 × 10−23 J/K )( 6.02 × 1023 molecules/mole )( 5 × 10−5 K ) 0.0855 kg/mole

= 0.12 m/s

(a) pV = NkT or p = (N/V)kT. But, N/V = ρ/m so that p = (ρ/m)kT and p/ρ = kT/m And, v = 1.4 p/ρ = 1.4 kT/m (b) By Equation (22) kT / m = vrms / 3 so that v v = 1.4 rms = 1.4 / 3 vrms = 0.68 vrms 3 (c) k = 1.38 × 10−23 J/K, m = 0.029 kg/6.02 × 1023 = 4.81 × 10−26 kg

v = 1.4(1.38 × 10−23 ) /(4.81 × 10−26 ) T m/s = 20.0 T m/s T = 0°C = 273 K, v = 20.0 273 m/s = 331 m/s T = 10°C = 283 K, v = 20.0 283 m/s = 337 m/s T = 20°C = 293 K, v = 20.0 293 m/s = 343 m/s

†19-49.

19-50.

T = 30°C = 303 K, v = 20.0 303 m/s = 349 m/s 1 3 Average kinetic energy = m v 2 . This can be expressed as K = kT (translational only). 2 2 3 Therefore, for both oxygen and nitrogen, K = (1.38 × 10−23 J/ °K)(273°K) = 5.65 × 10−21 J 2 23 vrms = 3kT/m ; m = 0.004 kg/(6.02 × 10 ) = 6.64 × 10−27 kg At 0°C, vrms = 3 (1.38 × 10−23 J/K ) ( 273 K ) / ( 6.64 × 10−27 kg ) = 1304 m/s At − 269°C, vrms = 3 (1.38 × 10−23 J/K ) ( 4 K ) / ( 6.64 × 10−27 kg ) = 158 m/s

19-51.

19-52.

3kT , where k = 1.38 × 10–23 J/K. For 02, m m = M/NA = 2 × 0.016/(6.02 × 1023) = 5.32 × 10–26kg For 03, m = 3 × 0.016/(6.02 × 1023) = 7.97 × 10–26kg For 02, vrms = 428 m/s For 03, vrms = 349 m/s For 02, (translational) K = (1/2)mvrms2 = 4.87 × 10–21J For 03, (translational) K = 4.85 × 10–21J vrms = 3kT / m ; for constant T , vrms m = constant. Therefore, T = –38ºC = 235 K. vrms =

v1 m1 = v1 m2 ⇒ v2 = v1 m1 /m2 . Since m is proportional to the atomic mass, v2 = 493 14.007 /1.008 = 1840 m/s 19-53.

As in Problem 52, v1 / v2 =

m2 / m1 , m1 = M1/NA, m2 = M2 /NA, so v1/v2 = 352 / 349 = 1.0043

Therefore, the percentage difference is 0.0043 = 0.43%


19

CHAPTER

19-54.

19 M into vrms = NA

Substituting m =

0.032 kg/mole

For hydrogen, vrms, H2 =

= 482 m/s


The ratio of the velocities, Solving vrms =

3kN AT . M


For oxygen, vrms, O2 =

†19-55.

3kT results in vrms = m

0.002 kg/mole vrms, O2 vrms, H2

=

M H2 M O2

= 1927 m/s

2 g/mole = 0.25 32 g/mole

=

3kT for temperature results in m

2 (1.0 × 10−12 kg )( 2.0 × 10−2 m/s ) = 9.7 × 106 K mvrms T = = 3k 3 (1.38 × 10−23 J/K ) 2

19-56.

Using m = vrms =

†19-57.

19-58.

3kN AT = M

Using m = vrms =

M and vrms = NA

3 (1.38 × 10−23 J/K )( 6.02 × 1023 molecules/mole ) ( 0.90 K ) 0.004 kg/mole

M and vrms = NA

3kN AT = M

Substituting vrms = results in t =

2 pa 2

3kT , m = 75 m/s

3kT , m

3 (1.38 × 10−23 J/K )( 6.02 × 1023 molecules/mole )( 2.0 × 10−4 K ) 0.023 kg/mole

= 0.47 m/s

3kT M N p 2 3 , and the Ideal-Gas Law , m= = into t = V kT NA m ( N / V ) vrms a 2 kMT . NA 1

⎛ (1.38 × 10−23 J/K ) ( 0.002 kg/mole )( 298 K ) ⎞ 2 ⎟ 2⎜ ⎜ ⎟ ( 6.02 × 1023 molecules/mole ) ⎝ ⎠ = 8.1 × 10−10 s For 1.0 atm, t = 5 2 −20 (1.01 × 10 Pa )( 9.0 × 10 m ) 1

⎛ (1.38 × 10−23 J/K ) ( 0.002 kg/mole )( 298 K ) ⎞ 2 ⎟ 2⎜ 23 ⎜ ⎟ 6.02 × 10 molecules/mole ( ) ⎠ = 0.62 s For an “ordinary” vacuum, t = ⎝ 2 −6 −20 (1.0 × 10 torr ) (133 Pa/torr ) ( 9.0 × 10 m )

(

)

1

⎛ (1.38 × 10−23 J/K ) ( 0.002 kg/mole )( 298 K ) ⎞ 2 ⎟ 2⎜ 23 ⎜ ⎟ 6.02 × 10 molecules/mole ( ) ⎠ = 6.2 × 104 s = 17 hours For a UHV, t = ⎝ −11 −20 (1.0 × 10 torr ) (133 Pa/torr ) ( 9.0 × 10 m2 )

(

)


CHAPTER †19-59.

19

Using the hint, the volume swept out per molecule with an effective radius R0 going a distance l is cylindrically shaped with volume V / N = π ( 2 R0 ) l . Solving for l yields the desired result, 2

l= 19-60.

1 4π R

2 0

( N/V )

(a) Using the Ideal-Gas Law

N p 1 results in = in l = 2 V kT 4π R0 ( N / V )

1.38 × 10−23 J/K ) ( 273 K ) ( kT = = 1.8 × 10−7 m l= 4π R02 p 4π (1.3 × 10−10 )2 (1.01 × 105 Pa ) vrms 1 3kT M , = or, using m = l l m NA

(b) The number of collisions per second is vrms 1 3kN AT = l l M

=

3 (1.38 × 10−23 J/K )( 6.02 × 1023 molecules/mole ) ( 273 K ) 1 1.8 × 10−7 m 0.004 kg/mole

= 7.2 × 109 collisions/second (c) From the Ideal-Gas Law, in a 1.0 cm 3 sample, there are N =

†19-61.

5 −6 3 pV (1.01 × 10 Pa )(1.0 × 10 m ) = = 2.7 × 1019 molecules. This means that the number −23 kT (1.38 × 10 J/K ) ( 273 K )

of collisions per second in the sample is 19 9 N ⎛ vrms ⎞ ( 2.7 × 10 )( 7.2 × 10 collisions/second ) = 1.0 × 1029 collisions/second, where the ⎜ ⎟= 2⎝ l ⎠ 2 2 is in the denominator because each collision would otherwise be counted twice. − NVm According to the given equation, the percent correction for the volume would be . V nRT and N = nN A yields Substituting into this expression the Ideal-Gas Law V = p − NVm − nN AVm − pN AVm . = = nRT V RT p 5 23 −29 3 − pN AVm − (1.01 × 10 Pa )( 6.02 × 10 molecules/mole )( 3.7 × 10 m ) = (a) For 1.0 atm, RT ( 8.31 J/mole i K )( 298 K )

= −9.1 × 10−4 = −0.091% (b) For 1000 atm,

(

)

5 23 −29 3 − pN AVm − (1000 ) (1.01 × 10 Pa ) ( 6.02 × 10 molecules/mole )( 3.7 × 10 m ) = RT ( 8.31 J/mole i K )( 298 K )

= −0.91 = −91%


CHAPTER

19-62.

19

v + v0 , where v0 is the vertical speed at the top and v is the 2 vertical speed at the bottom. At the top, the average rate of momentum transfer is mv0 ( v + v0 ) 2mv0 2mv0 N v0 ( v + v0 ) . Similarly, at the = = and the pressure p0 = 2L t 2V 2L ( v + v0 ) / 2

The average vertical speed is

bottom, the average rate of momentum transfer is

2mv = t

mv ( v + v0 ) 2mv = and the 2L 2L ( v + v0 ) / 2

N v ( v + v0 ) . Using the hint v 2 − v02 = 2 gL, then 2V N N p − p0 = ( v + v0 )( v − v0 ) = ( v 2 − v02 ) = ( N / V ) mgL 2V 2V 1 1.0 kg of O 2 contains = 31.25 moles. For a diatomic ideal gas like oxygen the 0.032 kg/mole

pressure p =

†19-63.

thermal kinetic energy 5 U = NkT 2 ⎛5⎞ = ⎜ ⎟ ⎡⎣ ( 31.25 moles ) ( 6.02 × 1023 molecules/mole ) ⎤⎦ (1.38 × 10−23 J/K ) ( 293 K ) ⎝ 2⎠ = 1.9 × 105 J 3 NkT and K rot = NkT , so 2 3/2 Fraction translational = = 0.6 5/2 Fraction rotational = 0.4 5 U = NkT . The average molecular mass of air is 29.0 g/mole. 2 1 × 6.02 × 1023 molecules = 2.08 × 1025 molecules 1.0 kg has 0.029 5 5 ∆U = Nk (∆T ), or ∆U = (2.08 × 1025 ) (1.38 × 10−23 )(1.0°C) = 716 J 2 2 3 For each molecule, K = NkT . For one mole at 300 K, 2 ⎛ 3⎞ K = ⎜ ⎟ ( 6.02 × 1023 molecules )(1.38 × 10−23 J/K ) ( 300 K ) ⎝ 2⎠ = 3.7 × 103 J In this case, K trans =

19-64.

†19-65.

⎛ T′ ⎞ Increasing the temperature to 320 K, K ′ = ⎜ ⎟ K = 1.07 K , a 7% increase in kinetic energy. ⎝T ⎠ Increasing the pressure by 3 atm at constant temperature, K ′ = K , thus there is no change in the kinetic energy.


CHAPTER

19-66.

For a monatomic ideal gas, the change in internal energy ∆E =

19

3 nR ( T − T0 ) . The number of 2

moles of gas can be calculated from the Ideal-Gas Law 5 −3 3 pV (1.01 × 10 Pa )(1.0 × 10 m ) = = 0.045 moles. n= RT ( 8.31 J/mole i K )( 273 K )

†19-67.

Solving the internal energy equation for T and substituting, 3 ∆E + nRT0 2 ( 50 J ) 2∆E 2 = + T0 = + ( 273 K ) = 363 K T = 3 3nR 3 ( 0.045 moles )( 8.31 J/mole i K ) nR 2 To get the new pressure use the Ideal-Gas Law nRT ( 0.045 moles )( 8.31 J/mole i K )( 363 K ) p= = = 1.3 × 105 Pa = 1.3 atm 3 V 0.001 m The number of moles of nitrogen can be calculated from the Ideal-Gas Law (140 atm ) (1.01 × 105 Pa/atm ) ( 3.0 × 10−2 m3 ) pV = = 187 moles. For a diatomic gas, the n= RT ( 8.31 J/mole i K )( 273 K )

(

)

5 5 nR∆T = (187 moles )( 8.31 J/mole i K )( 27 K ) = 1.0 × 105 J 2 2 Since water vapor consists of nonlinear polyatomic molecules, the thermal kinetic energy is E = 3nRT = 3 (1.0 mole )( 8.31 J/mole i K )( 373 K ) = 9.3 × 103 J change in internal energy ∆E =

19-68. †19-69.

19-70.

For seven total degrees of freedom, the total thermal energy is 7 7 ∆E = nR∆T = (1.0 mole )( 8.31 J/mole i K )(10 K ) = 291 J 2 2 Ebefore = Eafter 3 5 Ebefore = N He kTHe + N 02 kT02 2 2 3 5 Eafter = N He kT ′ + N 02 kT ′ 2 2 3⎛ 5 ⎞ = ⎜ N He kT ′ + N 02 ⎟ kT ′ 2⎝ 2 ⎠ 3 5 ⎛5 ⎞ ⎛3 ⎞ T ′ = ⎜ N 02 T02 + N HeTHe ⎟ / ⎜ N He + N 02 ⎟ 2 2 ⎝2 ⎠ ⎝2 ⎠ ⎛ THe pV = NkT, p and V are equal for both He and O2. Then N O2 TO2 = N He2 THe2 or N O2 = ⎜ 2 ⎜ TO ⎝ 2 ⎡ ⎤ ⎢ ⎥ ⎡3 ⎛ THe ⎞ ⎤ ⎢ 4 THe 5 3 5 ⎛ ⎞ ⎥ T ′ = ⎜ THe N He + N HeTHe ⎟ / ⎢ N He + N He ⎜ ⎟⎥ = ⎢ ⎥ 2 2 T ⎛ ⎞ T 3 5 ⎝2 ⎠ ⎢⎣ 2 ⎥ ⎝ 0 ⎠⎦ ⎢ + ⎜ He ⎟ ⎥ ⎢⎣ 2 2 ⎝ T0 ⎠ ⎥⎦ ⎡ ⎛ 3 5 250 ⎞ ⎤ T ′ = [4 × (250)] / ⎢ ⎜ + ⎟ ⎥ = 284 K ⎣ ⎝ 2 2 310 ⎠ ⎦


⎞ ⎟⎟ N He2 ⎠

CHAPTER †19-71.

19

One mole of CH3OH has a mass of (12 + 3 + 16 + 1) g = 32 g m 1.0 × 103 g n= = = 31.25 mole, or 32 g/mole M N = nN A = ( 31.25 mole ) ( 6.02 × 1023 molecules/mole ) = 1.88 × 1025 molecules

19-72.

Slow compression implies that the temperature remains constant. From the Ideal-Gas Law, ⎛V ⎞ ⎛ V ⎞ 5 p′V ′ = pV or p′ = ⎜ ⎟ p = ⎜ ⎟ p = 2 p = 2 atm = 2.02 × 10 Pa. Or, ⎝V′⎠ ⎝V /2⎠ ∆p = p′ − p = 1.01 × 105 Pa. The corresponding force F = ∆pA = ∆pπ r 2 = (1.01 × 105 Pa ) ( π ) (1.25 × 10−2 m ) = 50 N 2

†19-73.

At each temperature, a pressure difference is responsible for the lift force. From the Ideal-Gas nRT nRT ′ and at the higher temperature ∆p′ = Law at the lower temperature ∆p = . From the V V definition of pressure, at the lower temperature the lift force F = 1.2 × 103 N = ∆pA. At the higher temperature, ⎛ nRA ⎞ ⎛ T′ ⎞ ⎛ T′ ⎞ ⎛ 308 K ⎞ 3 3 F ′ = ∆p′A = ⎜ ⎟ T ′ = ( ∆pA ) ⎜ ⎟ = F ⎜ ⎟ = (1.2 × 10 N ) ⎜ ⎟ = 1.3 × 10 N ⎝ V ⎠ ⎝T ⎠ ⎝T ⎠ ⎝ 278 K ⎠

19-74.

The volume of the cylinder V = π r 2 h = π ( 0.1 m ) (1.1 m ) = 3.5 × 10−2 m3 . From the Ideal-Gas 2

pV Law, n = = RT m = nM

†19-75.

5

)

Pa/atm ) ( 3.5 × 10−2 m 3 )

( 8.31 J/mole i K )( 293 K ) = ( 200 moles )( 0.032 kg/mole ) = 6.4 kg

p1V1 = p2V2 V2 =

( (140 atm ) (1.01 × 10

= 2.0 × 102 moles. Using

or

p1V1 (140 atm × 1.01 × 105 Pa/atm) × (3.5 × 10−2 m3 ) = = 4.9 m3 5 p2 (1 atm × 1.01 × 10 Pa/atm)

For this problem use the average density of air ρ = 1.29 kg/m3 .

ρ N = 0.76 ρ = ( 0.76 ) (1.29 kg/m3 ) = 0.98 kg/m3 2

ρO = 0.24 ρ = ( 0.24 ) (1.29 kg/m3 ) = 0.31 kg/m3 2

mN2 = ρ N2 V = ( 0.98 kg/m3 )( 5.0 × 10−3 m3 ) = 4.9 × 10−3 kg mO2 = ρO2 V = ( 0.31 kg/m3 )( 5.0 × 10−3 m3 ) = 1.6 × 10−3 kg For nitrogen n =

m 4.9 × 10−3 kg = = 0.175 mole or M 28 × 10−3 kg/mole

N = nN A = ( 0.175 mole ) ( 6.02 × 1023 molecules/mole ) = 1.1 × 1023 molecules For oxygen n =

m 1.6 × 10−3 kg = = 0.048 mole or M 32 × 10−3 kg/mole

N = nN A = ( 0.048 mole ) ( 6.02 × 1023 molecules/mole ) = 2.9 × 1022 molecules

The total number of molecules is (1.1 + 0.29 ) × 1023 molecules = 1.4 × 1023 molecules


CHAPTER 19-76.

19

The buoyant force is equal to the weight of the water displaced by the life jacket. This is related to the volume of water displaced by w0 = ρ gV0 and w1 = ρ gV1 , where w0 and V0 are the weight and volume respectively of the water displaced near the surface and w1 and V1 are the weight and volume respectively of the water displaced at the depth y. At a depth y the pressure in the water is p1 = p0 + ρ gy. For the air inside the life jacket, p0V0 = p1V1 . Substituting, p0V0 = ( p0 + ρ gy ) V1 . Further substitution for V0 and solving for

w0 ρg V1 = . This in turn can be substituted above to get the buoyant force of the water at p0 + ρ gy p0

1.01 × 105 Pa ) ( 50 N ) ( p0 w0 depth y , w1 = ρ gV1 = = = 42 N p0 + ρ gy (1.01 × 105 Pa ) + (1000 kg/m3 )( 9.8 m/s 2 ) ( 2 m ) †19-77.

(a) For the equal mixture of hydrogen ions and free electrons, the number of particles N in a 2m given mass m is N = , where mH+ and me− refer to the masses of individual mH+ + me− hydrogen ions and free electrons respectively and the factor of two is because there are two particles for each pair of constituent particles. From this we can conclude that N ⎛m⎞ 2 =⎜ ⎟ V ⎝ V ⎠ mH+ + me−

= (1.5 × 105 kg/m3 )

(1.67 × 10

−27

2 = 1.8 × 1032 particles/m3 kg ) + ( 9.11 × 10−31 kg )

Appling this to the Ideal-Gas Law, ⎛N⎞ p = ⎜ ⎟ kT ⎝V ⎠

= (1.8 × 1032 particles/m3 )(1.38 × 10−23 J/K )(1.5 × 107 K ) = 3.7 × 1016 Pa

(b) If the material in the center of the Sun were only hydrogen atoms, the number of particles N m in a given mass m would be N = . From this we can conclude that mH N ⎛m⎞ 1 1 =⎜ ⎟ = (1.5 × 105 kg/m3 ) = 9.0 × 1031 particles/m3 −27 V ⎝ V ⎠ mH (1.67 × 10 kg ) Appling this to the Ideal-Gas Law, ⎛N⎞ p = ⎜ ⎟ kT ⎝V ⎠

= ( 9.0 × 1031 particles/m3 )(1.38 × 10−23 J/K )(1.5 × 107 K ) = 1.9 × 1016 Pa

(c) If the material in the center of the Sun were only diatomic hydrogen molecules, the number of m particles N in a given mass m would be N = . From this we can conclude that mH2 N ⎛m⎞ 1 1 =⎜ ⎟ = (1.5 × 105 kg/m3 ) = 4.5 × 1031 particles/m3 V ⎝ V ⎠ mH2 2 (1.67 × 10−27 kg )


CHAPTER

19

Applying this to the Ideal-Gas Law, ⎛N⎞ p = ⎜ ⎟ kT ⎝V ⎠

= ( 4.5 × 1031 particles/m3 )(1.38 × 10−23 J/K )(1.5 × 107 K ) = 9.3 × 1015 Pa

19-78.

(a) For the air at the top of column, p1V1 = p2V2 and V = Al , where A is the cross-sectional area of the air column and l is its height. Substitution into the first equation yields p1 Al1 = p2 Al2 . This simplifies to p1l1 = p2l2 . From Bernoulli’s Principle, pext = ρ gx + p, where pext is the pressure surrounding the barometer, x is the height of the mercury column, and p is the pressure of the air at the top of the mercury. Knowing that at pext,1 = 760 mm-Hg the barometer reads 750 mm-Hg (or x1 = 750 mm ) implies that p1 = 10 mm-Hg and l1 = 150 mm (because x + l = 900 mm ). When pext,2 = 780 mm-Hg, the mercury in the barometer will be at height x2 and the pressure in the air above the mercury will be p2 = ( 780 − x2 ) mm-Hg. At the same time the height of the air column above the mercury column will be l2 = 900 − x2 . Since p1l1 = p2l2 , 10 (150 ) = ( 780 − x2 )( 900 − x2 ) . The only practical solution to this (from the

quadratic formula) occurs when x2 = 770. So, when pext,2 = 780 mm-Hg the barometer will read 769 mm-Hg 10 (150 ) ( 780 − x2 )( 900 − x2 ) p1V1 pV pl pl = . Or, = 2 2 or 1 1 = 2 2 . So, 300 270 T1 T2 T1 T2 x2 = 770. In other words, the barometer will read 770 mm-Hg, slightly higher than the case with

(b) In this case

†19-79.

constant temperature. In general from the Ideal-Gas Law at constant temperature pV = p′V ′. So, ⎛ p′ ⎞ ∆V V − V ′ V′ p p′ − p ∆p = =1− =1− = = . This can be rearranged to V = ∆V ⎜ ⎟, V V V p′ p′ p′ ⎝ ∆p ⎠ where ∆V is the decrease in volume and ∆p is the corresponding increase in pressure. Furthermore,

⎛ ⎛ p ⎞ p ⎞ p′ p + ∆p p = =1+ and V = ∆V ⎜ 1 + ⎟ ≈ ∆V ⎜ ⎟ for ∆p ∆p ⎠ ∆p ∆p ∆p ⎝ ⎝ ∆p ⎠

p. Using the

⎛ p ⎞ specifics of the problem, Vc = Vs + V or Vs = Vc − V = Vc − ∆V ⎜ ⎟ ⎝ ∆p ⎠ 19-80.

T = 293 K, P = 1.0 atm = 1.01 × 105 N/m2. The molecular mass, m = 2 × 14 × 1.67 × 10–27 kg = 4.676 × 10–26 kg. The mean molecular speed, vrms = 3kT / m , k = 1.38 × 10−23 J/K. The mean x-

⎛ kT ⎞ component of the speed is given by vx = 3 ⎜ ⎟ = 509 m/s. If we imagine a cylindrical tube ⎝ m⎠ with base 80 cm 2 = 8.0 × 10−3 m 2 and height vx (with length units), all the molecules in that cube will impact on the palm of the hand during 1 second. The length of the cube is 509 m = vx . The volume of the cube is therefore ( 8.0 × 10−3 m 2 ) ( 509 m ) = 4.072 m3 . The number of


CHAPTER

19

molecules in this cube is given from the Ideal-Gas Law, PV = nRT = NkT. So, 5 3 pV (1.01 × 10 Pa )( 4.072 m ) = = 1.02 × 1026 impacts per second N = kT (1.38 × 10−23 J/K ) ( 293 K ) †19-81.

19-82.

Using m =

M and vrms = NA

3kT , m

3 (1.38 × 10−23 J/K )( 6.02 × 1023 molecules/mole ) ( 273 K ) 3kN AT vrms = = = 615 m/s M 0.018 kg/mole (a) For the equal mixture of H atoms and He atoms, the number of particles N in a given mass 4m m is N = , where mH and mHe refer to the masses of individual H and He atoms, 3mH + mHe respectively and the factor of four is because there are four particles for each collection of constituent particles in the denominator. From this we can conclude that N ⎛m⎞ 4 =⎜ ⎟ V ⎝ V ⎠ 3mH + mHe = (1.0 × 10−16 kg/m3 )

3 (1.67 × 10

−27

4 = 3.4 × 1010 particles/m3 −27 kg ) + ( 6.64 × 10 kg )

(b) Appling this to the Ideal-Gas Law, ⎛N⎞ p = ⎜ ⎟ kT ⎝V ⎠

= ( 3.4 × 1010 particles/m3 )(1.38 × 10−23 J/K )(1.0 × 104 K ) = 4.7 × 10−9 Pa

(c) For H, vrms = For He, vrms = †19-83.

Since vrms =

3kT = m

3kT = m

3 (1.38 × 10−23 J/K ) (10000 K ) 1.67 × 10

′ vrms = vrms

2T = T

(b)

′ vrms = vrms

T =1 T

(c)

′ vrms = vrms

T =1 T

(d)

′ vrms = vrms

T /4 = 1/ 2 T

kg

3 (1.38 × 10−23 J/K ) (10000 K ) 6.64 × 10

v′ 3kT , then rms = vrms m

(a)

−27

−27

kg

= 1.6 × 104 m/s

= 7.9 × 103 m/s

T′ . T

2


CHAPTER

19-84.

19

+

For H , vrms =

3kT = m

3 (1.38 × 10−23 J/K )(1.5 × 107 K )

The average kinetic energy is For He ions, vrms =

3kT = m

1.67 × 10−27 kg

= 6.1 × 105 m/s

3 3 kT = (1.38 × 10−23 J/K )(1.5 × 107 K ) = 3.1 × 10−16 J 2 2 3 (1.38 × 10−23 J/K )(1.5 × 107 K ) 6.64 × 10−27 kg

= 3.1 × 105 m/s

3 3 kT = (1.38 × 10−23 J/K )(1.5 × 107 K ) = 3.1 × 10−16 J 2 2 In order for K i = K f , then K H2 + K O2 = K H2 O . For the hydrogen and the oxygen (both diatomic), The average kinetic energy is

†19-85.

5 nRT . For the water (nonlinear polyatomic), K = 3nRT . Substituting into the kinetic 2 energy conservation equation: 5 ( 3 moles )( 8.31 J/mole i K )( 300 K ) = 3 ( 2 moles )( 8.31 J/mole i K ) ( T f ) , where ‘3 moles’ was 2 used on the left side of the equation because there were two moles of hydrogen and one mole of oxygen. Solving for T f = 375 K K =


CHAPTER 20

HEAT


Since no heat is lost to the surroundings, the temperature of the water will increase as heat Q is added. The amount of temperature increase ∆T is determined by the mass of the water m and its specific heat c. The heater converts electricity at a rate P = 620 W to increase the temperature of the water. This power P is equal to a rate of 620 J/s. To determine the time required to increase the temperature of the water, we recognize that Q mc∆T P= = t t

20-2.

†20-3.

20-4.

(

)

1.0 kg mc∆T (1.0 liter ) 1.0 liter ( 4187 J/(kg i °C) )( 80 °C ) = = 540 s P 620 J/s Assume each child converts 2000 kcal of food to heat throughout the day. The power associated with 1000 children in the school for seven hours is kcal 4187 J 2000 child ( 1 kcal ) = 3.3 × 105 W or 330 kW Q P= = 1000 children × 3600 s t 7 h( 1h )

t=

Assume that all of the gravitational potential energy (mgh) of the water is converted into heat Q. Then, the temperature change of the water can be determined by rearranging the equation Q = mc∆T. Q mgh gh (9.81 m/s 2 )(120 m) ∆T = = = = = 0.28°C mc mc c 4187 J/(kg i °C)

Note that the mass of the water drops out of the equation. Thus, this would be the temperature change regardless of the actual amount of water that fell. Q m P= = c∆T t t P 1200 × 106 J/s ∆T = = = 4.95°C day ( m/t )c (5.0 × 106 m3 /day) 1000 3kg 861 400 s ( 4187 J/(kg i °C) )

(

20-5.

20-6. †20-7.

1m

)(

)

Therefore, the final temperature is T = T0 + ∆T = 20°C + 4.95°C = 25°C To burn 100 kcal at a rate of 750 kcal/h requires 100 kcal t= = 0.133 h 750 kcal/h If one is running at a rate of 12 km/h, then the distance run would be x = vt = (12 km/h)(0.133 h) = 1.6 km 2000 kcal ⎛ 4187 J ⎞ ⎛ 1 day ⎞ P= ⎜ ⎟⎜ ⎟ = 97 W 1 day ⎝ 1 kcal ⎠ ⎝ 86 400 s ⎠ The power P you exert rotates the paddles, which, in turn, do mechanical work on the water. The process converts chemical energy in your body into mechanical work and then into a temperature increase of the water.


CHAPTER P=

20 Q mc∆T = t t

(

)

1.0 kg mc∆T 4.0 liters 1.0 liter ( 4186 J/(kg i °C) ) (5.0°C) t= = = 750 s W P 0.15 hp 745.7 ( 11 J/sW ) 1 hp

20-8.

(

)

The temperature of the cream must increase after it is added to the coffee. As a result, the mug and coffee must lose heat. Since heat is neither being added nor removed from the system, mcream ccream (T − Tcream ) + mmug cmug (T − T0 ) + mcoffee ccoffee (T − T0 ) = 0 T (mcream ccream + mmug cmug + mcoffee ccoffee ) − mcream ccreamTcream − (mmug cmug + mcoffee ccoffee )T0 = 0 T = =

†20-9.

mcream ccreamTcream + ( mmug cmug + mcoffee ccoffee )T0 mcream ccream + mmug cmug + mcoffee ccoffee (0.015 kg) ( 2900 J/(kg i °C) ) (5 °C) + ⎡⎣ (0.125 kg) ( 840J/(kg i °C) ) + (0.180 kg) ( 4187 J/(kg i °C) ) ⎦⎤ (70°C) (0.015 kg) ( 2900 J/(kg i °C) ) + (0.125 kg) ( 840J/(kg i °C) ) +(0.180 kg) ( 4187 J/(kg i °C) )

= 67°C Assume that all of the energy contained in the snack is converted by the body into gravitational potential energy at a height H above the starting location. Q = mgH = mgnh J 350 kcal ( 4187 Q 1 kcal ) = = 8500 steps mgh (70 kg)(9.81 m/s 2 )(0.25 m) In reality, the body is not completely efficient; and it also uses energy for other purposes, so one would have completely used up the energy in the snack in reaching a much lower altitude than indicated. The temperature of the lead plate must decrease after it is dropped into the water. As a result, the lead must lose heat as the water gains it. Since heat is neither being added nor removed from the system, mw cw (T − Tw ) + mlead clead (T − TL ) = 0

n=

20-10.

T (mw cw + mlead clead ) = mw cwTw + mlead cleadTL m c T + mlead cleadTL T = w w w mw cw + mlead clead =

) ( 4187 J/(kg i °C) ) (25°C) + (20 kg) ( (130 J/(kg i °C) ) (90°C) (200 liter) ( ) ( 4187 J/(kg i °C) ) + (20 kg) ( (130 J/(kg i °C) )

(200 liter)

(

1 kg 1 liter

1 kg 1 liter

= 25.2°C 20-11.

20-12.

∆T =

Q 10(3.6 × 106 J/h)(12 h) = = 0.17°C mc ( 600 m3 ) 1000 3kg ( 4187 J/(kg i °C) ) 1m

(

)

Q mc∆T = t t mc∆T 5.0 kg ( 445 J/(kg i °C) ) (400°C − 25°C) = = 1390 s or 23.2 minutes t= P 600 W ( 11 J/s W) P=


CHAPTER †20-13.

Friction processes generally cause a portion of the mechanical energy of a system to be “lost” in heating the components that are rubbing. Without a means of cooling, this heating could damage components. Therefore, it is necessary to cool the components, in this case, with cooling water. The water passes over the parts and extracts heat at a rate P from them. As a result, the temperature of the water changes by Q ⎛m⎞ = ⎜ ⎟ c∆T P= t ⎝ t ⎠ P (300 W) ∆T = = = 1.7°C 1.0 kg liters ( m / t )c ( 2.5 min ) 1 liter ( 160mins ) ( 4187 J/(kg i °C) )

(

20-14. †20-15.

W fs = = fv = µ Nv = (0.60)(60 N)(0.50 m/s) = 18 J/s t t Since the kinetic energy of the water remains constant, the increase in kinetic energy that would normally accompany a decrease in potential energy is being dissipated by friction. The rate of energy loss per kilometer is mg ∆h ρVg ∆h ⎛V ⎞ P= = = ρ g ∆h ⎜ ⎟ t t ⎝t ⎠ The mass of the water is the product of its density ρ and its volume V. This is also equal to the rate at which heat is added to the water per kilometer. ⎛ V ⎞ mc∆T ⎛ m ⎞ ρ g ∆h ⎜ ⎟ = = ⎜ ⎟ c∆T t ⎝t ⎠ ⎝ t ⎠

ρ g ∆h(V/t ) ( m/t )c

=

Percent difference = =

†20-17.

)

P=

∆T =

20-16.

20

2 g ∆h ( 9.81 m/s ) (0.074 m) = = 1.7 × 10−4°C/km c 4187 J/(kg i °C)

4187 J − mgh

× 100% 4187 J 4187 J − (1.00 kg)(9.81 m/s 2 )(365 m)

× 100% = 14.5% 4187 J The power used is 2.24 × 1010 W. The total incident solar power is 11 m PS = (1000 W/m 2 ) (850 km 2 ) ( 1000 W 1 km ) = 8.5 × 10 2

2.24 × 1010 W = 2.6% 8.5 × 1011 W

20-18.

Therefore, the power used is 2.6% of the incident solar power. This is enough to slightly increase the local temperature. Q mc∆T P= = t t

(

)

1.0 kg mc∆T 10 liters 1 liter ( 4187 J/(kg i °C) ) (50°C − 20°C) t= = = 1.26 × 104 s or 3.49 hours 3 2 2 1 J/s P 1.0 × 10 W/m 0.10 m ( ) ( ) 1W ( )

20-19.

P=

Q mc∆T = t t

1h 1.8 × 108 J/h ( 3600 m P s) = = = (1.09 kg/s) t c∆T ( 4187 J/(kg i °C) ) (88°C − 77°C)


(

1 m3 1000 kg

) = 1.1 × 10

−3

m3 /s

CHAPTER

20-20.

20

Q m = c∆T t t

3 2 2 m Q / t 1.0 × 10 W/m ( 4.0 m ) = = = (0.024 kg/s) t c∆T ( 4187 J/(kg i °C) ) (40°C)

†20-21.

(

1 m3 1000 kg

) = 2.4 × 10

−5

(a) Consider a small amount of water dm approaching the bottom of the water wheel with a speed v1. The angular momentum of dm with respect to the axis of rotation of the wheel is (dm)v1R. As it leaves the wheel, its angular momentum has changed to (dm)v2R. By the principle of the conservation of angular momentum, the angular momentum lost by the water must be gained by the water wheel, assuming frictional losses are negligible. The rate at which the wheel gains angular momentum from the water is dL dm τ = = (v1 − v2 ) R dt dt = (300 kg/s)(5.0 m/s − 2.5 m/s)(2.2 m) = 1650 N i m or 1.7 × 103 N i m

m3 /s

R

(b) The power delivered to the wheel is the product of the torque and the angular speed. P = τω = (1650 N i m)(1.4 rad/s) = 2.3 × 103 W

(c) Each second, 300 kg of water pass through the wheel and lose kinetic energy in the process. Some of this kinetic energy becomes the rotational kinetic energy of the wheel and some goes into heating the water. The rate of energy lost by 300 kg of water each second is 1 m(v2 − v 1 ) 2 ∆K = 2 ∆t ∆t 2 2 1 ⎡ ⎤ 2 (300 kg) ⎣ (2.5 m/s) − (5.0 m/s) ⎦ = = 2813 J/s 1s The difference in the power delivered to the wheel and the power lost by the water is the power that goes into heating the water. Q ⎛m⎞ ∆P = = ⎜ ⎟ c∆T t ⎝ t ⎠ ∆P (2813 W − 2310 W) ∆T = = = 4.0 × 10−4°C ( m/t )c ( 300 kg/s )( 4187 J/(kg i °C) ) 20-22. †20-23.

∆L = α L∆T = (12 × 10−6 / °C)(443 m) [ 35°C − (−29°C) ] = 0.34 m

The Eiffel Tower was built of 7300 tons of iron, which has a coefficient of linear thermal expansion of 12 × 10−6/°C. As the temperature increases, the dimensions of the tower increase. The temperature increase required to increase the height by 0.10 m is found using ∆L = α L∆T ∆L 0.10 m ∆T = = = 27°C α L (12 × 10−6 / °C)(312 m)


CHAPTER 20-24.

20

We want ∆L = α L∆T < 10−6 in. Therefore, ∆T <

10−6 in 10−6 in = = 0.083°C αL (12 × 10−6 / °C)(1.0 in)

∆T < 0.083°C †20-25.

When the copper pipe is heated, its radius will increase linearly with temperature, in similar manner as the length of a heated rod will lengthen. In this case, the diameter of the sleeve must increase by 0.002 cm to just barely fit over the steel rod. The temperature increase required for this thermal expansion can be found from ∆L = α L∆T ∆T =

∆L 0.002 cm = = 118°C α L (17 × 10−6 / °C)(0.998 cm)

20-26.

Therefore, the sleeve must be heated an additional 118°C from 18°C to 136°C. (a) ∆L = α L∆T = (12 × 10−6 / °C)(18 m)(50°C) = 0.011 m

20-27.

(b) ∆L = α L∆T = (12 × 10−6 / °C)(790 m)(50°C) = 0.47 m ∆L = α L∆T ∆T =

20-28. †20-29.

⎛ 1 °C ⎞ ∆L = α L∆T = (12 × 10−6 / °C)(10 m)(120°F) ⎜ 9 ⎟ = 0.008 m ⎝ 5 °F ⎠ When the plastic frame is immersed in the bath, it will linearly expand in each dimension. Thus, the openings for the lenses will also expand. In this case, we want the opening to be 0.75% larger than its original size. This is the same as indicating that ∆L/L is equal to 0.0075. ∆L = α L∆T ∆T =

20-30. †20-31.

1.0 × 10−7 m ∆L = = 0.67°C α L (0.50 × 10−6 / °C)(300 × 10−3 m)

1 ∆L 1 = (0.0075) = 38°C α L 2.0 × 10−4 / °C

∆L = α L∆T = (12 × 10−6 / °C)(1.5 m)(200°C) = 0.0036 m The volume of the alcohol will expand in all three dimensions as the system is heated. At the same time, the volume of the glass container also increases, but not as much since the thermal expansion coefficient is smaller for glass. To determine how much alcohol spills, we must determine the difference in the volume increase of alcohol to that of the container. ∆Valcohol − ∆Vglass = ( β alcohol − 3α glass )V ∆T

= ⎡⎣ (1.0 × 10−3 / °C) − 3(9.0 × 10−6 / °C) ⎤⎦ (1.0 liter)(75°C − (−110°C)) = 0.18 liter 20-32.

∆V = ∆h( π4 d 2 ) = β V ∆T ∆h =

†20-33.

β V ∆T

( π4 ) d 2

(0.182 × 10−3 / °C)(0.20 cm3 ) ( 1001 mcm ) (10°C) 3

=

π

4

(7.0 × 10−5 m) 2

= 0.095 m or 9.5 cm

The work done in expansion is W = ρ∆V, where the volume change is ∆V = βV∆T. W = P∆V = P β V ∆T = P3αV ∆T

= (1.0 atm)

(

1.013 × 105 N/m 2 1.0 atm

) (3)(12 × 10

−6

/°C)(1.0 kg)

(

1 m3 7900 kg


) (60°C) = 0.028 J

CHAPTER

20

The heat added to the cube is Q = mc∆T = (1.0 kg)(445 J/(kg i °C))(60°C) = 2.7 × 104 J The ratio of the heat added to the cube and the work done is Q 0.028 J = ≈ 106 2.7 × 104 J W

20-34.

The amount of work done is about one million times smaller than the amount of heat added to the cube. Consider a square with sides of length L that is at an initial temperature T. The area of the square is A = L2. Assume that the material from which the square is made is uniform, such that its properties are the same in all directions. At temperature T + ∆T, the area changes as

∆L ⎞ ⎛ A′ = ( L + ∆L) 2 = L2 ⎜ 1 + ⎟ L ⎠ ⎝

2

If ∆L/L is small (<< 1), the quantity in parentheses is approximately equal to 1 + 2

∆L . L

Therefore, ∆L ⎞ ∆L ⎞ ⎛ ⎛ A′ = L2 ⎜ 1 + 2 ⎟ = A ⎜1 + 2 ⎟ = A (1 + 2α∆T ) L ⎠ L ⎠ ⎝ ⎝ The change in the area is then, ∆A = A′ − A = A (1 + 2α∆T ) − A = 2α A∆T = α ′ A∆T It is thus demonstrated that α ′ = 2α 20-35.

∆L = α L∆T = (12 × 10−6 / °C)(0.316 m)(150°C − 20°C) = 4.93 × 10−4 m = 4.9 × 10−4 m F = k ∆L = ( 3.5 × 104 N/m )( 4.93 × 10−4 m ) = 17 N

20-36.

†20-37.

Depending on the shape of the wheel, the moment of inertia will be some constant b times the mass and the square of the radius, I1 = bMR2. When the temperature increases, the radius will increase to R + ∆R, where ∆R = αR∆T. So, the moment of inertia will have a new value, I2 = bMR(R + ∆R)2 I 2 = bMR 2 (1 + α∆T ) 2 Since α∆T << 1, we can rewrite this as I 2 = bMR 2 (1 + 2α∆T ) and ∆I = I 2 − I1 = bMR 2 − bMR 2 (1 + 2α∆T ) = 2bMR 2α∆T = 2 Iα∆T (a) The lengths of the brass rod and bob will both increase with temperature as ∆L = αL∆T, so ∆L = α∆T = (19 × 10−6 / o C ) 20°C = 3.8 × 10−4 the fractional change in length is L L The period of the pendulum is related to its length as P = 2π , so the ratio of the periods after g heating to the initial value is P2 L2 L1 + ∆L ∆L 1 ⎛ ∆L ⎞ = = = 1+ =1+ ⎜ ⎟ 2 ⎝ L1 ⎠ P1 L1 L1 L1


CHAPTER

20-38.

20

The fractional increase in period is ∆P = 0.5(3.8 × 10−4) = 1.9 × 10−4. (b) Over the course of 1 day (86 400 s), the pendulum clock loses, ∆t = (1.9 × 10−4)(86 400 s) = 16 s (a) The density ρ = m/V. As the temperature increases, the mass will not change, but the volume will through thermal expansion of the gasoline. Therefore, the new density will be reduced to m m m ρ2 = = = V2 V + β V ∆T V (1 + β∆T ) ⎛ ⎞ 1 1 3 = ρ⎜ = 710 kg/m3 ⎟ = ( 730 kg/m ) −3 + ∆ + T 1 1 (0.95 × 10 /°C)(30°C) β ⎝ ⎠ (b) The price per kilogram at 0 °C is Price Price V Price $0.60 / liter × = = = = $0.82/kg 3 m3 1 kg ρV V m ( 730 kg/m3 ) 101−liter

(

)

At 30°C, the price will be higher since the density is reduced, Price Price $0.60 / liter = = = $0.85/kg 3 m3 1 kg ρV ( 710 kg/m3 ) 101−liter

(

†20-39.

)

It pays to buy gasoline on a cold day! The center of mass of the mercury is located a distance l − h/2 from the pivot point of the pendulum. When the temperature of the system changes, the center of mass changes to a new position l′ − h′/2. The stated requirement is that the distance of the center of mass with respect to the pivot should remain unchanged, therefore l − h/2 = l′ − h′/2 (i) If the temperature changes by ∆T, the length of the brass rod changes to l′ = l(1 + α∆T) (ii) and the volume of the mercury changes to V′ = V(1 + β∆T) (iii) The initial volume of the mercury is V = hA where h is the height of the mercury and A is the diameter of the cylinder. Since the expansion of the glass cylinder is negligible, the height increases with temperature as h′ = h(1 + β∆T) (iv) Substituting equations (ii) and (iv) into equation (i) gives l − 12 h = l (1 + α∆T ) − 12 h(1 + β∆T ) lα∆T = 12 hβ∆T h=

20-40.

2α

β

l

∆Q ∆T = kA , the ratio of the thicknesses of styrofoam to ice (compacted snow) is ∆t ∆x ∆xS kS = ∆xI kI

Since

∆xS =

kS 0.008 J/(s i m i °C) (30 cm) = 1.2 cm ∆xI = 0.2 J/(s i m i °C) kI


CHAPTER

20-41.

20

∆Q ∆T = −kA ∆t ∆x

∆Q ∆x ∆t kA ⎛ ∆Q ∆x ⎞ = 100 °C − ⎜ − ⎟ ⎝ ∆t kA ⎠

∆T = Twater − Thot plate = −

Thot plate

20-42.

†20-43.

20-44.

⎛ ⎞ 0.0010 m ⎟ = 100.28°C = 100 °C − ⎜ − ( 2000 J/s ) 2 2 1m ⎜ ⎟ i i ° 237 J/(s m C) 300 cm ( ) ( ) 100 cm ⎝ ⎠ 2 The cross-sectional area of the inner steal cylinder is A1 = πr1 while the area of the copper sleeve is A2 = π(r22 − r12). The total rate of heat r2 conduction through the steel and copper is ∆Q ∆T ∆T ∆T r1 = k1 A1 + k2 A2 = ( k1 A1 + k2 A2 ) ∆t ∆x ∆x ∆x 2 ⎤ ⎛ 50°C ⎞ ∆Q ⎡ ( 46 J/(s i m i °C) ) π (0.0035 m) + ⎥⎜ =⎢ ⎟ ∆t ⎢⎣ ( 398 J/(s i m i °C) ) π ( (0.0050) 2 − (0.0035 m) 2 ) ⎥⎦ ⎝ 0.01 m ⎠ = 89 W The fraction of heat conducted through the steel is ( 46 J/(s i m i °C) ) π (0.0035 m)2 k1 A1 = k1 A1 + k2 A2 ( 46 J/(s i m i °C) ) π (0.0035 m) 2 + ( 398 J/(s i m i °C) ) π ( (0.0050) 2 − (0.0070 m) 2 )

= 0.099 or 9.9% The remaining fraction, 100% − 9.9% = 90.1%, is conducted through the copper. The rate of heat conducted through the window is ∆Q ∆T ⎛ 39°C ⎞ = kA = (1.0 J/(s i m i °C))(1.0 m)(1.5 m) ⎜ ⎟ = 23 400 W = 23 000 W ∆t ∆x ⎝ 0.0025 m ⎠ The rate of conduction through the wall in example 6 is 1.8 × 103 W. Therefore, the ratio of the heat conducted through the window to that through the walls is 23400 W = 13 1800 W The rate through the window is thirteen times greater than through the walls. As in example 7, 300 cm 2 ) ( 1001 mcm ) (1.2°C) ( ∆Q A∆T = = = 3.1 × 103 W ∆x1 ∆x2 0.050 cm ( 1001 mcm ) 0.030 cm ( 1001 mcm ) ∆t + + k1 k2 46 J/(s i m i °C) 398 J/(s i m i °C) 2 ∆x 0.10 m = = 2.4 m i Js i °C (a) R = k 0.042 J/(s i m i °C) 2

20-45.

(b) R = 2.4 20-46.

m 2 i s i °C J

1 ft 9°F 1055 J 1h ( 0.3048 m ) ( 3600 s )( 5°C )( 1 BTU ) = 13.6 2

ft 2 i h i °F BTU

∆Q ∆T ⎛ (80 − 22)°C ⎞ 2 =k = (0.080 J/(s i m i °C)) ( 1001 mcm ) ⎜ ⎟ = 0.046 W/cm A∆t ∆x ⎝ 1.0 cm ⎠


CHAPTER †20-47.

20

The tank has six glass sides: 2(80 cm × 30 cm), 2(50 cm × 30 cm) and 2(80 cm × 50 cm). Since the temperature inside the tank is higher than that outside the tank, there would be a net flow of heat out of the tank into the surrounding environment without the additional heat added to the tank by the internal heater. To maintain the temperature in the tank, the heater input must match the heat conducted through the walls of the tank. ∆Q ∆T ∆T = kA = k ( A1 + A2 + A3 ) ∆t ∆x ∆x (0.80 m × 0.30 m) + (0.50 m × 0.30 m) ⎤ (26 − 18)°C ⎡ = (1.0 J/(s i m i °C) ) 2 ⎢ ⎥ 0.003 m ⎣ + (0.80 m × 0.50 m) ⎦ = 4.2 × 103 W

20-48. †20-49.

( ∆Q/∆t )glass ( ∆Q/∆t )wood

=

kglass /∆xglass k wood /∆xwood

=

(1.0 J/(s i m i °C) ) /(0.003 m) ( 0.13 J/(s i m i °C) ) /(0.025 m)

= 64

Since heat is conducted through each slab at a constant rate, ∆Q ⎛ dT ⎞ = ki A ⎜ ⎟ ∆t ⎝ dx ⎠i for each slab designated by i. Therefore, the temperature gradient across the ith slab is ∆Q ⎛ dT ⎞ ⎜ ⎟ = dx k ⎝ ⎠i i A∆t Let the temperature difference across the ith slab be ∆Ti. Then, the total temperature difference across n slabs will be n n n ⎛ dT ⎞ ∆Q ∆Q n ∆xi ∆T = ∑ ∆Ti = ∑ ⎜ ∆xi = ⎟ ∆xi = ∑ ∑ A∆t i =1 ki i =1 i =1 ⎝ dxi ⎠ i =1 ki A∆t which can be rewritten ∆Q A∆T = n ∆xi ∆t ∑ i =1 ki

20-50.

∆Q ∆T = kA ∆t ∆x ∆x = kA

20-51.

⎛ 34°C − (−25°C) ⎞ ∆T = ( 0.02 J/(s i m i °C) ) (1.8 m 2 ) ⎜ ⎜ 4.0 × 105 J ( 1 h ) ⎟⎟ ( ∆Q/∆t )max h 3600 s ⎠ ⎝

= 0.019 m or 1.9 cm The rate of heat conduction, ∆Q/∆t, through both rods has the same value. Let T represent the junction temperature. Then, the conduction equation may be written for each rod,

0.060 m 0.060 m

0.0°C


Ag

100.0°C Cu

CHAPTER

20

∆TAg ∆TCu = kAg AAg ∆x ∆x kCu ACu (100°C − T ) = kAg AAg (T − 0°C) kCu ACu

100°C − T kAg AAg = T kCu ACu kAg AAg 100°C =1+ T kCu ACu

20-52.

⇒

T =

100°C 100°C = = 79°C kAg AAg (427 J/(s i m i °C))(0.0025 m) 2 1+ 1+ (398 J/(s i m i °C))(0.0050 m) 2 kCu ACu

The rate of conduction may then be determined. From the information of the silver rod, ∆TAg ∆Q (T − 0°C) = kAg AAg = kAg AAg ∆t ∆x ∆x ⎛ 79°C ⎞ = ( 427 J/(s i m i °C) ) π (0.0025 m) 2 ⎜ ⎟ = 11 W ⎝ 0.060 m ⎠ From problem 49, ∆Q A∆T A∆T = n = ∆ x ∆xi ∆x ∆t 2 glass + Ar ∑ k kglass kAr i =1 i

( 2.0 m ) ( (38 − 25)°C ) 2

†20-53.

(

)

s = 210 sJ 86400 = 1.8 × 107 J/day 1 day ⎛ 0.0032 m ⎞ ⎛ ⎞ 0.0020 m 2⎜ ⎟+⎜ ⎟ ⎝ 1.0 J/(s i m i °C) ⎠ ⎝ 0.017 J/(s i m i °C) ⎠ The rate of heat conduction, ∆Q/∆t, must be constant through any imaginary cylindrical, coaxial surface of radius r between r1 and r2. For the cable length l, the surface area of these r2 T 2 imaginary cylinders is A(r) = 2πrl. For an infinitesimally l r1 small radial distance dr, the conduction equation is T1 ∆Q dT dT = kA = k (2π rl ) ∆t dr dr r or

=

⎛ ∆Q ⎞ dr = 2π kl dT ⎜ ⎟ ⎝ ∆t ⎠ r Integrate both sides of the above equation to obtain, T2 ∆Q r2 dr = 2π kl ∫ dT ∫ r T 1 ∆t 1 r ∆Q (ln r2 − ln r1 ) = 2π kl (T2 − T1 ) ∆t ∆Q 2π kl (T2 − T1 ) = ln(r2 /r1 ) ∆t


CHAPTER 20-54.

20

To maintain the temperature at 0°C, the maximum amount of heat transferred into the box would be just enough to completely melt the ice, ∆Q = mLF, where LF is the latent heat of fusion for water. This would correspond to a minimum thickness of styrofoam to only allow that amount of heat transferred over 4 days. ∆Q ∆T = kA ∆t ∆x ∆T ∆T ∆T ∆t ∆x = kA = kA = kA mLF /∆t mLF ∆Q/∆t s ⎞ ⎛ 30°C (4 d) ( 241 dh )( 3600 1h ) = ( 0.008 J/(s i m i °C) ) (6) ( 0.60 m × 0.60 m ) ⎜ ⎟ ⎜ (20 kg)(3.34 × 105 kgJ ) ⎟ ⎝ ⎠ = 0.027 m or 2.7 cm

†20-55.

Let the rate of growth be δx/δt. For an area A, increasing the thickness by δx means that the volume added per time δt, is δV = Aδx. To bring about this phase change requires the removal of heat ∆Q = ρVLF = ρ(Aδx) LF, where ρ is the density and LF is the latent heat of fusion for water. The magnitude of the rate of heat transfer is ∆Q δx = ρALF δt ∆t This heat transfer is equal to the heat conducted through the ice to the air above the ice from the water below. ∆Q ∆T δx = kA = ρALF ∆t ∆x δt

(1.3 J/(s i m i °C) ) (20°C) δx k ∆T = = δt ρLF ∆x (917 kg/m3 )(3.34 × 105 J/kg)(0.060 m) = 1.3 × 10−6 20-56.

m s

s ( 1001 mcm )( 3600 1 h ) = 0.51 cm/h

The compacted snow on top of the ice only changes the details of the conduction of heat, now through both ice and snow, which is now δx ∆Q ∆T = A = ρALF (∆xice /kice ) + (∆xsnow /ksnow ) δt ∆t δx ∆T = δt ρLF [ (∆xice /kice ) + (∆xsnow /ksnow ) ]

s (20°C) ( 1001 mcm )( 3600 1h )

=

20-57.

(917 kg/m3 )(3.34 × 105 J/kg) ⎡⎣ (0.060 m)/ (1.3 J/(s i m i °C) ) + (0.030 m)/ ( 0.2 J/(s i m i °C) ) ⎤⎦ = 0.12 cm/h (a) Q = mLV = ρVLV 6 14 m = (1.74 × 10−2 kg/m3 )(10.0 km 3 ) ( 1000 J 1 km ) ( 2.45 × 10 J/kg ) = 4.26 × 10 3

(b) n =

4.26 × 1014 J = 5.3 bombs 8.0 × 1013 J/bomb


CHAPTER 20-58.

20

Q = mLF + mc∆T + mc∆T + mLV = m( LF + c∆T + LV )

= (1.0 kg) ⎡⎣ (3.34 × 105 J/kg) + ( 4187 J/(kg i °C) ) (100°C) + (2.26 × 106 J/kg) ⎤⎦ = 3.0 × 106 J 20-59.

For the tea to be cooled, its heat must be transferred to the ice, first to melt the ice and then to raise the temperature of the resulting water until the equilibrium temperature is reached. mice ( LF + c∆Twater ) = mtea c∆Ttea heat gained by the ice/water

(

=

mN LV heat gained by the nitrogen

mN =

)

kg 0.500 liter 11liter ( 4187 J/(kg i °C) )( (50 − 5)°C ) mtea c∆Ttea = = 0.27 kg = 270 g 5 LF + c∆Twater ( 3.34 × 10 J/kg ) + ( 4187 J/(kg i °C) )( (5 − 0)°C )

mice =

20-60.

heat lost by the tea

mCu c∆TCu heat lost by the copper

mCu c∆TCu 20 kg ( 390 J/(kg i °C) )( 20 − (−196)°C ) = = 8.424 kg 2.00 × 105 J/kg LV

8.424 kg 1000 liters = 10.4 liters 3 ρ 807 kg/m3 1 m For liquid helium, we have m c∆TCu 20 kg ( 390 J/(kg i °C) )( 20 − (−196)°C ) mHe = Cu = = 81.786 kg LV 2.06 × 104 J/kg V =

V = 20-61.

m

m

ρ

(

=

=

81.786 kg 125 kg/m3

∆Q = mLF ∆Q mLF = ∆t ∆t

20-62.

1000 liters 1 m3

) = 654 liters

) ( 3.34 × 10 J/kg ) = 1.1 × 10 J 3.5 ton ( ) ( 3.34 × 10 J/kg ) = 1.2 × 10 W or 1.1 × 10 = 1 day ( ) 907.2 kg 1 ton

5

907.2 kg 1 ton

9

5

4

86400 s 1 day

9

J/day

Liquid nitrogen freezes at −210°C. Vaporizing a portion, mliquid of the liquid releases an amount of heat Q = mliquidLV. The removal of this heat reduces the temperature of the liquid and solidifies an amount of the liquid, msolid. msolidLV = mliquidc∆T mliquid c∆T msolid = LV =−

†20-63.

( = 3.5 ton (

)

(1.0 kg) ( 2.0 × 103 J/(kg i °C) ) ( −210°C − ( − 196°C) )

= 1.1 kg 2.6 × 104 J/kg The heat that is removed from the cream to freeze it and further cool it goes into vaporizing the liquid nitrogen. mN LV = mcC ∆TC + mLF + mcI ∆TI heat gained by the nitrogen

heat lost by the cream

heat lost to freeze

heat lost by the ice cream


CHAPTER

mN = =

20

m ( cC ∆TC + LF + cI ∆T )I LV 2.0 kg ⎡⎣ ( 2900 J/(kg i °C) )(10°C ) + 3.34 × 105 J/kg + ( 2200 J/(kg i °C) )(10°C ) ⎤⎦ 2.00 × 105 J/kg

= 3.9 kg 20-64.

The work done against the atmosphere is W = p∆V. The density of water is 1000 kg/m3, so the volume occupied by one kilogram of water is 0.001 m3. This is the initial volume. The final volume of the water vapor may be determined from the ideal gas law, pV = nRT. Since the mass of water in one mole is 18 g, n = (1000 g/18 g) = 55.6 moles. Therefore, the work done is ⎛ nRT ⎞ − V0 ⎟ = nRT − pV0 W = p∆V = p ⎜ ⎝ p ⎠ = (55.6 moles) ( 8.31 J/(K i mole) ) (373 K) − (1.01 × 105 N/m 2 )(0.001 m3 ) = 1.7 × 105 J For a pressure of 0.10 atm = 1.01 × 104 Pa, ⎛ nRT ⎞ − V0 ⎟ = nRT − pV0 W = p∆V = p ⎜ ⎝ p ⎠ = (55.6 moles) ( 8.31 J/(K i mole) ) (373 K) − (1.01 × 104 N/m 2 )(0.001 m3 ) = 1.7 × 105 J As the pressure goes to zero, the work done against the atmosphere goes to zero.

20-65.

11 m (a) m = ρV = (1000 kg/m3 ) ( 2.6 × 103 km 2 ) ( 1000 kg 1 km ) ( 0.076 m ) = 2.0 × 10 2

(b) Q = mLV = (2.0 × 1011 kg)(2.26 × 106 J/kg)(1 cal/4.187 J) = 1.1 × 1017 cal (c) U = mgy = (2.0 × 1011 kg)(9.81 m/s2)(1500 m) = 2.9 × 1015 J = 7.0 × 1014 cal (d) K = 12 mv 2 = 12 (2.0 × 1011 kg)(10 m/s) 2 = 1.0 × 1013 J = 2.4 × 1012 cal

20-66.

The kinetic energy of the drops is less than the potential energy due to frictional loses with the air. mW cW ∆TW = mPb LF + mPb cPb ∆TPb heat gained by the water

heat lost by the lead

mW cW (T − TW ) = mPb LF + mPb cPb (TPb − T ) mW cW T − mW cW TW = mPb LF + mPb cPbTPb − mPb cPbT T = =

20-67.

mW cW TW + mPb ( LF + cPbTPb ) mW cW + mPb cPb 2.5 liter

(

1 kg 1 liter

4

1 kg 1 liter

= 23°C mN LV heat gained by the nitrogen

mN =

) ( 4187 J/(kg i °C) ) (20°C) + (0.50 kg) ⎡⎣ (2.9 × 10 J/kg) + (140 J/(kg i °C) ) (328°C) ⎤⎦ 2.5 liter ( ) ( 4187 J/(kg i °C) ) + (0.50 kg) (140 J/(kg i °C) )

=

mTi c∆TTi heat lost by the titanium

mTi c∆TTi 0.25 kg ( 340 J/(kg i °C) )( 20 − (−196)°C ) = = 0.092 kg LV 2.00 × 105 J/kg


CHAPTER 20-68.

†20-69.

20

Evaporation requires 0.5(750 kcal/h) = 375 kcal/h. The evaporation of 1.0 kg of water requires 580 kcal. Therefore, 375 kcal/h m= (1.0 h) = 0.65 kg 580 kcal/kg Since density is defined as ρ = m/V. The volume of water evaporated per hour is V (m/t ) = ρ t The mass evaporated in time t is determined by the amount of heat absorbed by the water in that time: Q mLV = t t Therefore, the volume of the Mediterranean Sea evaporated per hour is V (m/t ) (Q/t ) A( P/A) = = = ρ ρ LV ρ LV t 2

=

3

6

W m2

)

(1000 kg/m )(2.43 × 10 J/kg)

= 1.2 × 106 20-70.

(

m 2.9 × 106 km 2 ( 1000 1.0 × 103 1 km )

m3 s

s 1 km ( 3600 1 h )( 1000 m )

3

= 4.3 km3 /h

When the tube implodes, air surrounding it is pushed inward, filling the former vacuum, by atmospheric pressure. In this process, the atmosphere does work on the air, W = p∆V = (1.01 × 105 N/m 2 )(0.025 m 2 ) = 2525 J If all of this work becomes the kinetic energy, K = 12 mv 2 , of the glass pieces, then the average speed of the pieces will be 2K 2 vave = m vave =

†20-71.

2K = m

2(2525 J) = 50 m/s 2.0 kg

The air conditioner takes in and sends out air at constant pressure, p = 1.01 × 105 N/m2. Let’s assume that since air is mostly nitrogen that its specific heat is approximately equal to that of nitrogen, cp = 29.1 J/(mol • K). The heat absorbed by the air is Q = ncp ∆T , where n is the number of moles, which can be determined using the Ideal-Gas Law, pV = nRT pV RT where T = 30°C is the initial temperature of the air. Therefore, combining this result with the equation for Q/t, pV Q= cp ∆T RT 7 1h QRT (Q/t ) RT (1.2 × 10 J/h ) ( 60 min ) ( 8.31 J/(K i mol) ) (303 K) ∆T = = = = 11.4 K m3 pVcp p (V/t )cp ( 29.1 J/(K i mol) ) (1.01 × 105 N/m2 ) 15 min n=

(

)

The temperature of the air exiting the fan is Tf = T + ∆T = 30°C + 11.4°C = 41°C.


CHAPTER 20-72.

Cp = 0.75 ( 29.1 J/(mole i K) ) + 0.24 ( 29.1 J/(mole i K) ) + 0.01 ( 20.8 J/(mole i K) ) = 29.0 J/(mole i K) Cv = 0.75 ( 20.8 J/(mole i K) ) + 0.24 ( 20.8 J/(mole i K) ) + 0.01 (12.5 J/(mole i K) ) = 20.7 J/(mole i K)

†20-73.

One kg of helium is equal to 1000 g n= = 249.838 moles 4.00260 g/mole The specific heat of one kilogram of helium is 249.838 moles CV(kg) = nCV(mole) = (12.5 J/(mole i K) ) = 3.12 × 103 J/(kg i K) 1 kg The above calculation may then be repeated for the other gases in Table 20.5 to create the following table of results.

20-74.

Gas

Molecular weight (g)

Moles/kg

CV ( J/(kg i K) )

He Ar N2 O2 CO NH3 CH4

4.00260 39.948 28.0134 31.999 28.0104 17.0305 16.04276

249.838 25.033 35.697 31.251 35.701 58.718 62.333

3.12 × 103 3.13 × 102 7.42 × 102 6.50 × 102 7.39 × 102 1.60 × 103 1.69 × 103

The gas with the highest specific heat per kilogram is helium, while the lowest is for argon. ⎛ ⎞ 1 mole 5 W = p∆V = nR∆T = 1.0 kg ⎜ ⎟ ( 8.314 J/(mole i K) )( 50.0 K ) = 2.06 × 10 J 0.0020158 kg ⎝ ⎠ Under constant pressure conditions, the heat added to the system is ∆Q = nCp ∆T = nCV ∆T + nR∆T Under constant volume conditions, the heat added to the system is ∆Q′ = nCV ∆T = ∆Q − nR∆T = 7.08 × 105 J − 2.06 × 105 J = 5.02 × 105 J

20-75.

For helium, the density is given in Table 18.1 as 0.178 kg/m3. v=

20-76. 20-77.

γp = ρ

(1.66)(1.01 × 105 N/m 2 ) = 971 m/s 0.178 kg/m3

⎛ ⎞ 1 mole 7 ∆Q = nCp ∆T = ( 600 kg ) ⎜ ⎟ ( 20.8 J/(mole i K) ) (20 K) = 6.2 × 10 J ⎝ 0.0040026 kg ⎠ W = p∆V = nR∆T ∆V =

nR∆T (44.6 moles) ( 8.314 J/(mole i K) )(10 K ) = = 0.0367 m 3 = 3.7 × 10−2 m3 p 1.01 × 105 N/m 2

W = p∆V = (1.01 × 105 N/m 2 )( 0.0367 m 3 ) = 3.7 × 103 J


20

CHAPTER 20-78.

20

∆QV = nCV ∆T = (1 mole )(12.5 J/(mole i K) ) (280 K) = 3.5 × 103 J ∆Qp = nCp ∆T = (1 mole )( 20.8 J/(mole i K) ) (280 K) = 5.8 × 103 J

20-79.

CV

20-80.

( 20.8 J/(mole i K) ) + 23 ( 29.1 J/(mole i K) ) = 26.3 J/(mole i K) = 13 (12.5 J/(mole i K) ) + 23 ( 20.8 J/(mole i K) ) = 18.0 J/(mole i K)

Cp =

1 3

(a) The number of moles is given by n =

pV . The heat transferred is given by RT

∆Q = nCp ∆T t (∆Q / ∆t ) = t=

pV Cp ∆T RT pVCp ∆T RT (∆Q / ∆t )

(1.01 × 10 =

5

N/m 2 ) (5.0 m)(5.0 m)(2.5 m) ( 29.1 J/(mole i K) )( 5 K )

( 8.314 J/(mole i K) ) (303 K) ( 8.0 × 106

= 0.0456 h ( 601min = 2.7 min h )

J/h )

(b) In this case, the number of moles is not constant. pV dQ = nCp dT = Cp dT RT pVCp dT pVCp ⎛ T1 ⎞ ⎛T ⎞ ∆Q = = ln ⎜ ⎟ = n1T1Cp ln ⎜ 1 ⎟ ∫ R T R ⎝ T2 ⎠ ⎝ T2 ⎠ The cooling time can be expressed, ⎛T ⎞ n1T1Cp ln ⎜ 1 ⎟ ∆Q ⎝ T2 ⎠ t= = (∆Q/∆t ) (∆Q/∆t ) =

†20-81.

K (2510 mole) ( 303 K )( 29.1 J/(mole i K) ) ln ( 303 298 K )

8.0 × 106 J/h

= 0.046 h ( 601min = 2.8 min h )

There is not an appreciable difference. Each hour that you breathe in the cold dry air, you add to it ∆mvapor ⎛ 0.041 kg vapor ⎞ ⎛ 0.45 kg air ⎞ =⎜ ⎟⎜ ⎟ = 0.1845 kg vapor/h ∆t 1 kg air 1h ⎠ ⎝ ⎠⎝ The heat required to warm and vaporize the air is ∆Q ∆m ∆m = Cp ∆T + LV ∆t ∆t ∆t ⎛ 0.45 kg ⎞ ⎛ 1 h ⎞ =⎜ ⎟⎜ ⎟ (1000 J/(kg i °C) )( 37°C − (−30°C) ) ⎝ 1 h ⎠ ⎝ 3600 s ⎠ + ( 0.1845 kg vapor/h ) ( 2.42 × 106 J/kg )

1 kcal = ( 4.46 × 105 J/h ) ( 4187 = 110 kcal/h J)


CHAPTER 20-82.

(a) For a small adiabatic expansion, the volume of the gas increases by a small amount dV. The work done during this process is dW = pdV. Since the heat absorbed is zero, the change in the energy of the gas is dE = −dW = −pdV. This change in energy can also be written, dE = nCVdT. Therefore, nCVdT = −pdV (i) Differentiating the Ideal-Gas Law gives, Vdp + pdV = nRdT Vdp + pdV (ii) nR Substituting equation (ii) into equation (i), gives ⎛ Vdp + pdV ⎞ nCV ⎜ ⎟ = − pdV nR ⎝ ⎠ − Rp dV = CV pdV + CVVdp dT =

⎛ R + CV ⎞ dV ⎛ Cp ⎞ dV dp dV = −⎜ = −⎜ = −γ ⎟ ⎟ p V ⎝ CV ⎠ V ⎝ CV ⎠ V (b) A monatomic ideal gas, such as helium or argon, has a value of γ of approximately 1.66. Using this value gives, dp dV = −γ = −(1.66)(0.03) = 0.05 or 5.0% p V 20-83.

γ γ According to equation 20.47, pfootVfoot = ptopVtop . This can be rearranged to give 1/ γ

⎛p ⎞ = ⎜ foot ⎟ Vfoot ⎜⎝ ptop ⎟⎠ From the Ideal-Gas Law, ptopVtop p V = foot foot Ttop Tfoot Vtop

Ttop Tfoot

=

ptopVtop pfootVfoot

ptop ⎛ pfoot = ⎜ pfoot ⎜⎝ ptop

Ttop = ( 0.9965 ) Tfoot

1/ γ

1 /(29.1 / 20.8) ⎞ 0.988 ⎛ 1.0 ⎞ = 0.9965 ⎟⎟ = ⎜ ⎟ 1.0 ⎝ 0.988 ⎠ ⎠ = ( 0.9965 ) 293 K = 292 K

The temperature difference between the foot and top of the mountain is only 1 K. 20-84.

20

pV = nRT or p = nRT/V Since pV γ = constant, we have Vγ nRT = constant V TV γ −1 = constant Applying this to this situation gives, γ −1 TV = T2V2γ −1 1 1 γ −1

⎛V ⎞ T2 = T1 ⎜ 1 ⎟ ⎝ V2 ⎠

⎛ 1.0 ⎞ = 298 K ⎜ ⎟ ⎝ 0.1 ⎠

(20.8/12.5) −1

= 1370 K


CHAPTER †20-85.

20

(a) From the ideal gas law, 1 mole nRT (1.0 kg) 0.028 kg ( 8.31 J/(mole i K) ) (293 K) V = = = 0.072 m3 p 1.2 × 106 N/m 2

(

)

(b) Since pV γ = constant, we have p1V1γ = p2V2γ

⎛ p ⎞ V2γ = V1γ ⎜ 1 ⎟ ⎝ p2 ⎠ 1/ γ

1 / 1.40

⎛ p ⎞ ⎛ 1.2 × 106 N/m 2 ⎞ = 0.42 m3 V2 = V1 ⎜ 1 ⎟ = 0.072 m3 ⎜ 5 2 ⎟ p 1.01 × 10 N/m ⎝ ⎠ ⎝ 2⎠ The Ideal-Gas Law is pV = nRT or p = nRT/V; and since pV γ = constant, we have Vγ nRT = constant V TV γ −1 = constant Applying this to this situation gives, γ −1 = T2V2γ −1 TV 1 1 γ −1

⎛V ⎞ T2 = T1 ⎜ 1 ⎟ ⎝ V2 ⎠ 20-86.

⎛p ⎞ T2 = T1 ⎜ 2 ⎟ ⎝ p1 ⎠

1.40 −1

⎛ 0.072 m3 ⎞ = 293 K ⎜ 3 ⎟ ⎝ 0.42 m ⎠

(γ −1) / γ

= 145 K or − 128°C

⎛ 1.0 atm ⎞ = 244 K ⎜ ⎟ ⎝ 300 atm ⎠

(1.40 −1) / 1.40

= 48 K or − 225°C

(γ −1) / γ

20-87. 20-88.

(1.40 −1) / 1.40

⎛p ⎞ 1.0 atm ⎛ ⎞ = 293 K ⎜ = 214 K or − 59°C T2 = T1 ⎜ 2 ⎟ ⎟ ⎝ 1.0 atm + 2.0 atm ⎠ ⎝ p1 ⎠ The total work done in this process is the sum of the work done by you and the work done by the air against the atmosphere (a volume change at constant pressure p1 = 1 atm. V1

W = Wyou + Wair = Wyou + p1∆V = − ∫ pdV V1

Since, p1V1γ = p2V2γ

⎛p ⎞ V1γ = V2γ ⎜ 2 ⎟ ⎝ p1 ⎠ 1/ γ

1 / 1.40

⎛p ⎞ ⎛ 3.4 atm ⎞ V1 = V2 ⎜ 2 ⎟ = ( 0.10 m3 ) ⎜ ⎟ ⎝ 1.0 atm ⎠ ⎝ p1 ⎠ The total work done is then

= 0.224 m3

⎡ ⎛V ⎞ V1 V1 −1 W = − ∫ p1 dV = − ∫ p2V2γ V1−γ dV = ( γ − 1) p2V2 ⎢1 − ⎜ 2 ⎟ V1 V1 ⎢⎣ ⎝ V1 ⎠

γ −1

= (1.40 − 1) 3.4 atm −1

(

1.01 × 105 N/m 2 1 atm

)

⎤ ⎥ ⎥⎦

⎡ ⎛ 0.10 m3 ⎞1.40 −1 ⎤ ( 0.10 m ) ⎢1 − ⎜ 0.224 m3 ⎟ ⎥ = 2.4 × 104 J ⎢⎣ ⎝ ⎥⎦ ⎠ 3


CHAPTER

20

The work done by you is then, Wyou = W − Wair = W − p1∆V = 2.4 × 104 J − (1.01 × 105 N/m 2 )( 0.224 m3 − 0.10 m3 ) = 1.1 × 104 J

20-89.

The amount of heat delivered to the water per minute is ⎛ ∆Q ⎞ 3 5 Q=⎜ ⎟ t = ( 3.0 × 10 J/s ) 60 s = 1.8 × 10 J ⎝ ∆t ⎠ This heating of the water produces an increase in temperature Q 1.8 × 105 J ∆T = = = 3.6 °C kg mc 12 liters 11 liter ( 4187 J/(kg i °C) )

(

20-90.

)

The mass of water in the pond is the product of the water density ρwater and the volume of the pond, which is Vpond = Ad, where A is the area of the pond and d is the depth. The mass of the snow that falls on the pond is msnow = ρsnowAd, since the area and depth are the same as those for the pond. Heat from the pond first melts the snow and then increases the temperature of the resulting water. mwater cwater (Twater − T ) = msnow LF + msnow cwater (T − Tsnow ) = msnow [ LF + cwater (T − Tsnow ) ] heat lost from the water

heat gained by snow

ρ water Adcwater (Twater − T ) = ρsnow Ad [ LF + cwater (T − Tsnow ) ] ρ water cwater (Twater − T ) = ρsnow [ LF + cwater (T − Tsnow ) ]

T = =

20-91.

20-92.

ρ water cwaterTwater + ρsnow ( cwaterTsnow + LF ) cwater ( ρsnow + ρ water )

(

1000 kg 1 m3

) ( 4187 J/(kg i K) )( 279 K ) + (

) ⎡⎣( 2230 J/(kg i °C) ) (273 K) + 3.34 × 10 ( 4187 J/(kg i K) ) ( ) 100 kg

5

1 m3

1100 kg 1 m3

= 274 K or 1.0°C ∆Q ∆m c∆T = ∆t ∆t ⎛ 1 liter ⎞ ∆m ∆Q/∆t 7.5 × 106 J/h = = = 160 liter/h or 0.45 liters/s ∆t c∆T ( 4187 J/(kg i °C) ) (11°C) ⎜⎝ 1 kg ⎟⎠ ∆Q = ( 3.2 × 10−9 J/electron )( 3.0 × 1014 electrons/s ) = 9.6 × 105 J/s = 9.6 × 105 W ∆t ∆T ∆Q/∆t 9.6 × 105 J/s = = = 0.019°C/s (b) ∆t mc 12 m3 1000 3kg ( 4187 J/(kg i °C) )

(a)

(

1m

)

∆Q ∆m ∆m ∆m ∆m = c∆T = c ( T2 − T1 ) = cT2 − cT1 ∆t ∆t ∆t ∆t ∆t ( ∆Q / ∆t ) + (∆m / ∆t )cT1 (c) T2 = (∆m / ∆t )c

( 9.6 × 10 =

5

J/s ) +

(

( ) ( ) ( ) ( 4187 J/(kg i °C) ) (20°C) = 27°C ) ( ) ( ) ( 4187 J/(kg i °C) ) 2.0 m3 1 min

3

2.0 m 1 min

1 min 60 s

1000 kg

1 min 60 s

1 m3

1000 kg 1 m3


J/kg ⎤⎦

CHAPTER †20-93.

20-94. 20-95.

20-96.

20

Assuming the ship is composed of steel, the change in length of the ship will be ∆L = αL∆T = (12 × 10−6/°C)(458 m)[40°C − (−20°C)] = 0.33 m Similarly, the width of the ship will increase by ∆w = αw∆T = (12 × 10−6/°C)(69 m)[40°C − (−20°C)] = 0.050 m As a result of this thermal expansion, the increase in area is ∆A = (L + ∆L) (w + ∆w) − Lw = (458.33 m)(69.050 m) − (458 m)(69 m) = 46 m2 ∆V = 3αV∆T = 3(12 × 10−6/°C)(1.8 × 106 m3)[40°C − (−20°C)] = 3.9 × 103 m3 The rate of thermal conduction into the box is ∆Q ∆T = kA ∆t ∆x where the combined area of the six sides of the box is A = 2(0.30 m)(0.30 m) + 4(0.30 m)(0.40 m) = 0.66 m2 The rate of mass loss due to vaporization of the dry ice is ∆T kA 2 ∆m ∆Q/∆t ∆x = (0.008 J/(s i m i °C)(0.66 m )(99°C/0.040 m) = 2.3 × 10−5 kg/s = = ∆t 5.8 × 105 J/kg LV LV A∆T ∆Q = ∆xwood ∆xbrick ∆xfg ∆t + + k wood kbrick kfg ∆xwood ∆xbrick ∆xfg A∆T + + = k wood kbrick kfg ∆Q / ∆t

⎡ A∆T ∆x ∆x ⎤ ∆xfg = kfg ⎢ − wood + brick ⎥ k wood kbrick ⎦ ⎣ ∆Q / ∆t ⎡ (1.0 m 2 )30°C ⎤ 0.012 m 0.10 m = 0.042 J/(s i m i °C ⎢ − + ⎥ 4 1h ⎣⎢ 4.0 × 10 J/h ( 3600 s ) 0.13 J/(s i m i °C) 0.63 J/(s i m i °C) ⎦⎥

†20-97.

= 0.12 m or 12 cm The ice-water mixture is at 0°C before the dry ice (frozen CO2 at −79°C). For water of mass m to become ice an amount of heat Q = mLF must be removed from the ice-water mixture. This heat comes from the evaporation of the 0.30 kg of dry ice as it is added to the ice-water. The heat required to vaporize the dry ice is Q = mdry ice LF = ( 0.30 kg ) ( 5.80 × 105 J/kg ) = 1.74 × 105 J

This amount of heat removed from the water causes the following mass of water to freeze to ice at 0°C. Q 1.74 × 105 J m= = = 0.52 kg LF 3.34 × 105 J/kg 20-98.

Assume all of the heat to cool the horseshoe to 30°C goes into heating the water to 100°C and then evaporating a portion of the water, mvapor. Since the water is heated to 100°C, the horseshoe will be cooled to 100°C. Then, the amount of water vaporized will be


CHAPTER ∆Q = mshoe cshoe ∆Tshoe = mwater cwater ∆Twater + mvapor LV heat lost by the horse shoe

mshoe cshoe ∆Tshoe − mwater cwater ∆Twater LV

mvapor = =

kg ( 0.70 kg )( 445 J/(kg i °C) )(1100°C ) − ( 0.50 liter ) ( 11liter ) ( 4187 J/(kg i °C) )( 70°C )

2.26 × 106 J/kg

= 0.087 kg 20-99.

heat gained by the water

(

1 liter 1 kg

) = 0.087 liter

∆QV = nCV ∆T = (1.0 mole )(12.5 J/(mole i °C) )( 70°C ) = 8.8 × 102 J ∆Qp = nCp ∆T = (1.0 mole )( 20.8 J/(mole i °C) )( 70°C ) = 1.5 × 103 J

20-100.

(a) We can tell if the gas is monatomic or diatomic by calculating its constant volume specific heat. PV = nRT

(

)

5 2 1m PV (1.01 × 10 N/m ) ( 2.0 liter ) 1000 liters n= = = 0.089 mole RT ( 8.31 J/(mole i K) )( 273 K ) 3

QV = nCV ∆T CV =

20-101.

180 J Q = = 20.2 J/(mole i °C) n∆T ( 0.089 mole )(100°C )

The gas is diatomic. 2.5 g (b) atomic mass = = 28 g/mole 0.089 moles The gas is nitrogen. (a) ∆Qp = nCp ∆T = (1.0 mole )( 29.1 J/(mole i °C) )( 60°C ) = 1.7 × 103 J (b) ∆QV = nCV ∆T = (1.0 mole )( 20.8 J/(mole i °C) )( 60°C ) = 1.2 × 103 J (c) V2 =

nRT2 (1.0 mole )( 8.31 J/(mole i K) ) (353 K) = = 0.029 m3 5 2 p 1.01 × 10 N/m

p3 =

(1.0 mole )( 8.31 J/(mole i K) ) (60 K) = 1.7 × 104 N/m 2 nR∆T == V2 0.029 m3

(d) W1 = p∆V = nR∆T = (1.0 mole )( 8.31 J/(mole i K) ) (60 K) = 5.0 × 102 J W2 = V ∆p = (0.029 m3 )(1.7 × 104 N/m 2 − 1.01 × 105 N/m 2 ) = −2.4 × 103 J


20

CHAPTER 21

THERMODYNAMICS


The change in internal energy for this ideal monatomic gas is the difference between the heat absorbed Q by the gas and the work that it does on the piston. The heat input into the system is Q = nCp ∆T = (1.0 mole) ( 20.8 J/(mole i K) )( 90 K ) = 1.9 × 103 J

where cp is given in Table 20.5. The change in the internal energy is then, ∆E = Q − W = nCp ∆T − W = (1.0 mole) ( 20.8 J/(mole i K) )( 90 K ) − 800 J = 1.1 × 103 J 21-2.

Because there is no temperature change and no pressure change, the process results in no change in the internal energy of the gas. Therefore, the amount of heat transferred to the system must be equal to the work done by the gas as it expands aginst the piston. Q −W = 0

Q = W = p∆V = (1.01 × 105 N/m 2 )(1.50 × 10−5 m3 ) = 1.52 J

†21-3.

(a) In step A, the gas isovolumetrically (∆V = 0) expands, so the work done by the gas W = p∆V = 0 J The heat absorbed by the ideal gas is Q = ncV∆T. From the ideal gas law, we have nRT1 nRT2 p1V1 = nRT1 or V1 = = p1 p2 since the volume is constant. Solving for ∆T = T2 − T1, gives ⎛p ⎞ ∆T = T1 ⎜ 2 − 1 ⎟ ⎝ p1 ⎠

B

P2 P1

A

V1

V2

The system is initially at standard temperature and pressure (STP), T1 = 273 K and P1 = 1.0 atm. One mole of an ideal gas at STP has a volume of 22.4 liters. The heat absorbed in step A is ⎛p ⎞ ⎛ 2p ⎞ 1 mole Q = nCVT1 ⎜ 2 − 1 ⎟ = 4.0 liters ( 22.4 12.5 J/(mole i K) ) (273 K) ⎜ 1 − 1⎟ = 610 J liters ) ( ⎝ p1 ⎠ ⎝ p1 ⎠ The internal energy in this case is equal to the heat absorbed by the system, ∆E = Q − W = Q = 610 J (b) In step B, the gas isobarically (∆P = 0) expands by ∆V = 4.0 liters (10−3 m3/liter) = 4.0 × 10−3 m3, so the work done by the gas is W = p∆V = (2.02 × 105 N/m2)(4.0 × 10−3 m3) = 808 J = 810 J The amount of heat absorbed by the system in this process is again Q = nCV∆T. From the ideal gas law, we have


CHAPTER p2V1 = nRT2

and

p2V1 nR

and

21

p2V2 = nRT3

p2V2 nR pV pV p ∆T = T3 − T2 = 2 2 − 2 1 = 2 ∆V nR nR nR 5 2.02 × 10 N/m 2 = 4.0 × 10−3 m3 ) = 544.5 K ( 1 mole 4.0 liters ( 22.4 liters ) ( 8.31 J/(mole i K) ) T2 =

Then,

21-4.

T3 =

1 mole Q = nCp ∆T = 4.0 liters ( 22.4 20.8 J/(mole i K) ) (544.5 K) = 2022 J = 2.0 × 103 J liters ) (

The internal energy in this case is ∆E = Q − W = Q = 2.0 × 103 J − 810 J = 1.2 × 103 J The free expansion of the air is considered an adiabatic process (Q = 0), so the change in the internal energy is equal to the work done by the gas against the atmosphere, ∆E = Q − W = −1800 J. The change in internal energy is also ∆E = nCV∆T (see section 20.6). Therefore, nCV ∆T = −1800 J/mole

∆T =

1800 J/mole −1800 J/mole = = −86.5 K nCV (1.0 mole) ( 20.8 J/(mole i K) )

T2 = T1 + ∆T = 293 K − 86.5 K = 206.5 K or − 66.5°C †21-5.

21-6.

In an adiabatic process, heat is neither absorbed nor emitted from a system; Q = 0. Since work is being done on the gas instead of by the gas, the internal energy of the system increases ∆E = Q − W = −W Since work is done on the gas, the work done has a negative value and the internal energy is increasing in this situation. According to section 20.6, the change in internal energy is also equal to nCV∆T. Therefore, ∆E = nCV ∆T = ( 0.0300 mole )( 20.8 J/(mole i K) )( (790 − 40) K ) = 470 J The work done is W = −470 J. Since all of the heat generated in the collision with the floor stays in the lead ball, Q = 0 J. Work is done on the ball by the floor causing a change in its kinetic energy and causing an increase in the internal energy of the ball. In the instant before impact, the ball has a kinetic energy K1 equal to its initial energy, which was its gravitational potential energy. Since its final speed is zero, its final kinetic energy is zero. W = ∆K = K 2 − K1 = 0 − mgh = − ( 0.25 kg ) ( 9.81 m/s 2 ) ( 0.80 m ) = −1.96 J The increase in the internal energy of the ball is ∆E = −W = 1.96 J This increase in the internal energy will increase the temperature of the ball by ∆E = mC ∆T ∆E 1.96 J ∆T = = = 0.060°C mC (0.25 kg) (130 J/(kg i °C) )

†21-7.

(a) Assuming the gas obeys the ideal gas law, pV = nRT, 5 2 3 pV (1.01 × 10 N/m )( 0.100 m ) n= = = 4.29 moles RT ( 8.31 J/(mole i K) ) (283 K)


CHAPTER

21

(b) The work done by the gas as it expands is W = p∆V = (1.01 × 105 N/m 2 )( (0.110 − 0.100)m3 ) = 1010 J (c) The change in internal energy is ∆E = Q − W = 3500 J − 1010 J = 2490 J (d) Consider section 20.5 and equations 20.30 to 20.33, in particular. The internal energy of the gas is ∆E = fnR∆T , where f is a fraction related to the nature of the gas (monatomic or diatomic). For the situation given, ∆E 2490 J 5 f = = = nR∆T (4.29 moles) ( 8.31 J/(mole i K) ) (28 K) 2

21-8.

Therefore, CV = 5R/2 and the gas is diatomic. (a) The initial volume of the water is 1.00 liter = 1.00 × 10−3 m3. Assuming the water vapor obeys the ideal gas law, the volume of the steam vapor at 1 atm is given by ⎛ 1 mole ⎞ 1.00 kg ⎜ ⎟ ( 8.31 J/(mole i K) )( 373 K ) 0.018 kg ⎠ nRT ⎝ V = = = 1.705 m3 p 1.01 × 105 N/m 2 In the expansion of the water vapor, work is done by the gas against the atmosphere equal to W = p∆V = (1.01 × 105 N/m 2 ) ( (1.705 − 0.00100)m3 ) = 1.72 × 105 J The change in the internal energy during the expansion is ∆E = Q − W = mLV − W = (1.00 kg)(2.26 × 106 J/kg) − (1.72 × 105 J) = 2.09 × 106 J During the process the internal energy of the system increases due to the heat added to the system, but that energy is reduced by the work done by the vapor. About 8% of the heat input to the system goes into doing work. (b) Because liquid water is incompressible, there is no volume change as the pressure is increased from 1 atm to 2 atm. No work is done, therefore, on the liquid water. Because there is also no temperature change, there is no change in the internal energy of the water. As for water vapor at 1 atm, the heat of vaporization is LV, 1atm = 2.26 × 106 J. Because the temperature is constant at 373 K, ∆E = 0. To compress water vapor from 1 atm to 2 atm, the work done, applying the Ideal-Gas Law, p = nRT/V and p1V1 = p2V2, is V2 V2 dV V W = ∫ pdV = ∫ nRT = nRT ln 2 V1 V1 V V1

21-9.

⎛ ⎞ 1.00 kg 5 =⎜ ⎟ ( 8.31 J/(mole i K) )( 373 K ) ln ( 0.50 ) = −1.19 × 10 J 0.018 kg/mole ⎝ ⎠ The heat of vaporization at 2 atm is LV, 2 atm = LV, 1atm + W = 2.26 × 106 J − 1.19 × 105 J = 2.14 × 106 J (a) W = p∆V = (1.01 × 105 N/m 2 ) (1.000 × 10−3 m3 − 1.091 × 10−3 m3 ) = −9.19 J ∆E = Q − W = mLF − W = (1.00 kg)(3.34 × 105 J/kg) − ( −9.19 J) = 3.34 × 105 J The work done by the atmosphere on the ice is negligible, so the change in internal energy is that of the heat added to the ice to melt it. (b) Since the densities, and hence the volumes, of the water and ice are assumed to remain the same at 2.00 atm as at 1.00 atm, the work done by the atmosphere will be W = p∆V = (2.02 × 105 N/m 2 ) (1.000 × 10−3 m3 − 1.091 × 10−3 m3 ) = −18.4 J


CHAPTER

21-10.

Given the assumptions that the volumes of water and ice do not change between 1.00 and 2.00 atm and the negligible amount of work done by the atmosphere, one concludes that the heat of vaporization remains unchanged. W V′ dV (a) ∫ dW = ∫ nRT V 0 V V′ V ′ dV V′ = nRT ln V = nRT (ln V ′ − ln V ) = nRT ln W = nRT ∫ V V V V (b) In an isothermal expansion, the change in internal energy ∆E = Q − W = 0 J. Therefore, W = Q. Using the result from part (a), V′ W = nRT ln V

eW / nRT = e eW / nRT =

⎛V′⎞ ln ⎜ ⎟ ⎝V ⎠

V′ V

V ′ = VeW / nRT = ( 0.0020 m3 ) e †21-11.

21-12.

21

500 J/ ( (0.10 mole)(8.31 J/mole i K)(700 K) )

= 0.0047 m3

The work done during this process is W = p∆V = 0.25 atm (1.01 × 105 N/m2/1.0 atm)(0.5 × 10−3 m3 − 2.0 × 10−3 m3) = −37.9 J During the process 75 J flows out of the system, therefore Q = −75 J. The change in internal energy is ∆E = Q − W = −75 J − (−37.9 J) = −37.1 J The change in the volume of the cylinder in which the gas is contained decreases by

∆V = Ah = ( 90 cm 2 ) ( 1001 mcm ) ( 0.12 m ) = 1.08 × 10−3 m3 2

where A is the area of the cylinder and h is the displacement of the cylinder. The work done during this process is W = p∆V = 15 atm (1.01 × 105 N/m2/1.0 atm)(−1.08 × 10−3 m3) = −1636 J Because work is done on the gas to compress it and the internal energy decreases, heat must be removed from the system. The quantity of this heat is found by applying the First Law of Thermodynamics, ∆E = Q − W Q = ∆E + W = −15 J + 15 atm †21-13.

(

) ( − 1.08 × 10

1.01 × 105 N/m 2 1.0 atm

−3

m3 ) = −1651 J

In the temperature range of 5°C to 30°C, ethanol is a liquid. The amount of heat added to the ethanol is Q = mc∆T According to Table 20.1, the specific heat of ethanol at 20°C is 2430 J/(kg • °C). To determine the mass of the liquid, we need to know its density, which is given for 20°C. Therefore, we need to know the volume of the ethanol at 20°C. At 5.0°C, the volume is 1.00 liter. As the alcohol is heated, it undergoes volumetric thermal expansion ∆V = β V ∆T = (1.01 × 10−3 / °C)(1.00 liter)

(

1.00 × 10−3 m3 1.00 liter

) (20°C − 5°C) = 1.51 × 10

−5

m3

where β is the volumetric thermal expansion coefficient listed in Table 20.2. The volume at 20°C is V + ∆V = (1.00 × 10−3 m3) + (1.51 × 10−5 m3) = (1.015 × 10−3 m3)


CHAPTER

21

The mass of the ethanol is therefore, m = ρV = (789 kg/m3 )(1.015 × 10−3 m3 ) = 0.801 kg

Then, Q = mc∆T = (0.801 kg) ( 2430 J/(kg i °C) ) (30°C − 5°C) = 4.87 × 104 J The volume expansion between 5°C and 30°C is ∆V = β V ∆T = (1.01 × 10−3 / °C)(1.00 liter)

21-14.

(

1.00 × 10−3 m3 1.00 liter

) (30°C − 5°C) = 2.53 × 10

−5

m3

The work done by the alcohol in its expansion on the surrounding air is W = p∆V = 1.00 atm (1.01 × 105 N/m2/1.00 atm)(2.53 × 10−5 m3) = 2.56 J In an adiabatic compression, Q = 0 J and ∆E = W. Assuming air behaves as an ideal gas, 3 ∆E = W = p∆V = (1.01 × 105 N/m 2 )( 20 cm3 − 80 cm3 ) ( 1001 mcm ) = −6.06 J To determine the new pressure when the gas is compressed, we use the fact that pV γ is constant for an ideal gas, where γ = 1.40 for a diatomic gas. p1V11.40 = p2V21.40 1.40

1.4

⎛V ⎞ ⎛ 80 cm3 ⎞ p2 = p1 ⎜ 1 ⎟ = (1.01 × 105 N/m 2 ) ⎜ = 7.03 × 105 N/m 2 3 ⎟ 20 cm V ⎝ ⎠ ⎝ 2⎠ Using the Ideal-Gas Law, the number of moles of air present and the temperature of the compressed gas may be determined. pV = nRT 5 2 3 1m p1V1 (1.01 × 10 N/m )( 80 cm ) ( 100 cm ) n= = = 0.0032 moles RT1 ( 8.31 J/(mole i K) )( 300 K ) 3

(1.01 × 105 N/m 2 )( 20 cm3 ) ( 1001 mcm ) = 75 K pV T2 = 2 2 = nR ( 8.31 J/(mole i K) )( 0.0032 moles ) 3

The gas then loses heat to its surroundings until the temperature reaches 300 K. This occurs at constant volume, or isovolumetrically. The work done is equal to 0 J, since there is no volume change. The mass of 1 mole of air is 29.0 g. Therefore, the change in internal energy is kg ∆E = Q = mc∆T = 0.0032 moles 0.0029 ( 20.8 J/(mole i K) ) (300 K − 75 K) = 0.043 J 1 mole

(

)

The initial internal energy of the air is found by integrating equation (20.21) dE = nCV dT

∫ dE =nC ∫ dT V

E1 = nCVT1 = (0.0032 moles) ( 20.8 J/(mole i K) ) (300 K) = 19.97 J

As a result of the two steps taken to compress and cool the air, the final internal energy is E2 = E1 + ∆E1 + ∆E2 = 19.97 J − 6.06 J + 0.043 J = 13.95 J †21-15.

This is an example of an isothermal process, ∆T = 0 K. Since the temperature does not change, there is no change in the internal energy of the ideal gas; and W = Q. The work done by the gas in its expansion, as determined in problem 21-10(a), is V ⎛ 10 liters ⎞ 4 W = nRT ln 2 = (10 moles) ( 8.31 J/(mole i K) ) (323 K) ln ⎜ ⎟ = 4.3 × 10 J V1 ⎝ 2 liters ⎠


CHAPTER 21-16.

†21-17.

21

This is an example of an isothermal process, ∆T = 0 K. Since the temperature does not change, there is no change in the internal energy of the ideal gas; and W = Q. V W = Q = nRT ln 2 = (2.0 moles) ( 8.31 J/(mole i K) ) (373 K) ln(10) = 1.4 × 104 J V1 The temperature may be determined from the Ideal-Gas Law, pV = nRT T =

(1.01 × 105 N/m2 )( 0.10 m3 ) = 243 K p2V2 = nR ( 8.31 J/(mole i K) )( 5.00 moles )

Because this is an isothermal process, ∆T = 0 K; and there is no change in the internal energy of the ideal gas; and the work done as the gas expands is V W = Q = nRT ln 2 V1 from which the initial volume may be determined since the quantity of work done by the gas is given. W V = ln 2 nRT V1 W

e nRT =

V2 V1

V1 = V2 e

−

W nRT

= ( 0.10 m3 ) e

− (2.0 × 104 J)/ ⎡⎣ (5.00 moles) ( 8.31 J/(mole i K) ) (243 K) ⎤⎦

= 0.014 m3

The initial pressure was nRT ( 5.00 moles )( 8.31 J/(mole i K) ) (243 K) p1 = = = 7.2 × 105 N/m 2 V1 0.014 m3

21-18.

†21-19.

At 10 m below the surface of the water, the hydrostatic pressure is (see example 6, chapter 18), p1 = p0 − ρ gy = 1.01 × 105 N/m 2 − (1000 kg/m3 )(9.81 m/s 2 )(−10 m) = 1.99 × 105 N/m 2 The bubble forms from the diver’s airline from an initial volume of zero m3. So, the work done in forming the bubble is done against the surrounding water is 3 W = p∆V = (1.99 × 105 N/m 2 ) ⎡ ( 4.0 cm3 ) ( 1001 mcm ) − 0 m3 ⎤ = 0.80 J ⎣ ⎦ As this is an isothermal process, ∆T = 0 K; and there is no change in the internal energy of the ideal gas; and the heat absorbed by the bubble as the gas expands is equal to the work, W = Q = 0.80 J The diagram shows the indirect process Water at Water vapor described in the problem. The net work done Boil 100°C at 100°C against the atmosphere is the same in the indirect process as in the direct process since Cool the changes in the volume are the same. The Heat net changes in the internal energies is also the same. Because the work done and the internal Water at Water vapor energy are the same, then the heat must also be 20°C at 20°C the same. For 1.0 kg of water, the latent heat per kilogram is


CHAPTER

21

LV(20°C) = c∆T + LV(100°C) + cp ∆T = ( 4187 J/(kg i °C) ) (80°C) + 2.26 × 106 J/kg − ( 2010 J/(kg i °C) ) (80°C) = 2.43 × 106 J/kg 21-20.

LV(140° C) = c∆T + LV(100° C) + cp ∆T = ( 4187 J/(kg i °C) ) (−40°C) + 2.26 × 106 J/kg + ( 2010 J/(kg i °C) ) (40°C) = 2.17 × 106 J/kg

21-21.

e=

21-22.

e=

†21-23.

120 hp

(

746 W 1 hp 5

) = 0.203 or 20.3%

4.40 × 10 W

The work done by the runner is W = ∆U = mgh. So, the rate that work is being done is ∆W ∆h = mg = mgv ∆t ∆t The efficiency is determined by the ratio of this work to the total energy generated by the body that includes the work done plus the waste heat generated. (∆W/∆t ) mgv e= = (∆W/∆t ) + (∆Q/∆t ) mgv + (∆Q/∆t ) =

21-24.

W (W/t ) 300 MW = = = 0.35 or 35% Q (Q/t ) 850 MW

e=

( 70 kg ) ( 9.81 m/s 2 ) ( 0.30 m/s ) ( 70 kg ) ( 9.81 m/s 2 ) ( 0.30 m/s ) + (1300 W)

= 0.14 or 14%

W 2.0 × 104 J = = 0.67 or 67% Q1 3.0 × 104 J

The waste heat produced is Q2 = Q1 − W = 1.0 × 104 J †21-25.

According to equation 21.5, e =

Q W = 1 − 2 , the efficiency may be determined in more than Q1 Q1

one way. In this case, we are given Q1 and Q2 and asked to find the efficiency and the work done by the engine. Q 8.0 × 106 J e =1− 2 =1− = 0.60 or 60% 2.0 × 107 J Q1 The work done is W = Q1 − Q2 = 1.2 × 107 J 21-26. 21-27. 21-28.

Q W = e 1 = 0.95(3.0 × 103 W) = 2.85 × 103 W or 2.85 kW t t (Q2/t) = (Q1/t) − (W/t) = 0.15 kW The overall efficiency is the product of the individual efficiencies. e = (0.90)(0.50)(0.99) = 0.445 or 44.5% Summing the individual expenditures of mechanical power gives 2.77 hp. The efficiency of the runner is 2.77 hp e= = 0.21 or 21% 13 hp


CHAPTER 21-29.

21

The maximum efficiency of the body as a heat engine would be T 293 K e =1− 2 =1− = 0.055 or 5.5% T1 310 K The problem asks the student to calculate the work that he or she does to climb to a height of 3.0 m, the work done by the body would be W = ∆U = mgh. The student would insert his/her mass in this calculation. Therefore, the general result for the work done is W = myou ( 9.81 m/s 2 ) ( 3.0 m )

21-30.

With the assumption that the body is an inefficient heat engine, the amount of kcal one would have to consume to climb 3.0 m is 2 W myou ( 9.81 m/s ) ( 3.0 m ) ⎛ 1 kcal ⎞ Q= = ⎜ ⎟ e 0.055 ⎝ 4187 J ⎠ Q 1W 1 The rate at which chemical energy is consumed is 1 = (50 W) = 135 W = 0.37 t e t Q Because, e = 1 − 2 , the rate at which waste heat is produced is Q1 Q2

t

=

Q1

t

(1 − e ) = (135 W)(1 − 0.37) =

85 W

The rate at which glucose is consumed is Q1 135 J/s kg = = 8.7 × 10−6 = 8.7 mg/s 3 kcal 4187 J glucose yield 3.7 × 10 kg ( 1 kcal ) s †21-31.

The engines each do 1100 hp of work per second and the efficiency is 20%, so the heat input to the system to deliver this work is somewhat greater. To find that rate of heat input, remember that efficiency is defined as e = W/Q1. Dividing both factors on the right side of the equation by t gives the ratio of the rate of work to the rate of heat input. W/t e= Q1/t

(

)

745.7 W Q1 W/t 2200 hp 1 hp = = = 8.2 × 106 W = 8.2 × 106 J/s t e 0.20 The rate of fuel input needed to supply this heat input is m Q/t 8.2 × 106 J/s = = = 0.19 kg/s t E/m 4.4 × 107 J/kg

21-32.

(a)

W2

W1 Q 1 , T1 Steam

Q2,T2 High pressure stage

Steam

Low pressure stage

Q3, T3 Water

(b) Because some heat is put back into the system in the low-pressure stage, there is a gain in efficiency because additional work is the result. Therefore, the overall efficiency e is calculated as follows. Given that, T T e1 = 1 − 2 e2 = 1 − 3 T1 T2 and e1T1 = T1 − T2

e2T2 = T2 − T3

The overall efficiency is


CHAPTER

21

e =1−

T3 T1

eT1 = T1 − T3 = ( T1 − T2 ) + ( T2 − T3 ) = e1T1 + e2T2

†21-33.

21-34.

⎛T ⎞ e = e1 + e2 ⎜ 2 ⎟ = e1 + e2 − e1e2 = 0.4 + 0.2 − (0.4)(0.2) = 0.52 or 52% ⎝ T1 ⎠ The efficiency is determined by the temperatures of the high temperature (T1) and low temperature (T2) reservoirs. T 300 K e =1− 2 =1− = 1 − (3 × 109 ) = 0.999 999 997 11 T1 1 × 10 K This is very close to 100% efficiency. However, working with such high temperatures can be a difficult engineering problem in practice. If the efficiency is 33% and the amount of useful power produced is 1 × 109 W, then the rate of waste heat produced is ∆Q/∆t = 2 × 109 W. This the heat that raises the temperature of the water, ∆Q ∆m = c∆T ∆t ∆t ∆m ( ∆Q / ∆t ) 2 × 109 J/s 3 1 m3 = = = 6 × 104 kgs 1000 kg = 60 m /s ∆t c∆T ( 4187 J/(kg i K) )( 8 K )

(

†21-35.

)

The efficiency is determined by the temperatures of the high temperature (T1) and low temperature (T2) reservoirs. The efficiency e is the ratio of the work done W by the engine to the amount of heat absorbed Q1. Therefore, W T e= =1− 2 Q1 T1 5.0 × 104 W = = 1.9 × 105 J T2 273 K ⎞ ⎛ 1− 1− ⎜ ⎟ T1 ⎝ 373 K ⎠ The amount of waste heat produced is the difference between the heat absorbed and the work done, Q2 = Q1 − W = (1.9 × 105 J) − (5.0 × 104 J) = 1.4 × 105 J T 300 K W e= =1− 2 =1− = 0.25 or 25% 400 K Q1 T1 Q1 =

21-36.

Increasing the efficiency by 10%, gives T 1 − 2 = 0.35 T1

†21-37.

T2 = 0.65T1 Therefore, if T2 = 300 K, then T1 = 462 K. Alternatively, if T1 = 400 K, then T2 = 260 K. Q W According to equation 21.5, e = = 1 − 2 , the efficiency may be determined in more than Q1 Q1 one way. In this case, we are given Q1 and Q2 and asked to find the efficiency and the mechanical power output (the work done per second) the engine. Therefore, Q Q /t 1400 J/s e =1− 2 =1− 2 =1− = 0.44 or 44% Q1 Q1/t 2500 J/s


CHAPTER

21

The power output is W W/t e= = Q Q1 /t W = e(Q1/t ) = (0.44)(2500 J/s) = 1100 J/s = 1100 W t The efficiency may be calculated in several ways. Of relevance here is Q W e= =1− 2 Q1 Q1 P=

21-38.

Q2 =1− e Q1 5.0 × 103 J Q2 = = 5880 J = 5.9 × 103 J 1− e 1 − 0.15 Given the heat absorbed for one cycle, we can determine the amount of work done during the cycle. W = eQ1 = (0.15)(5880 J) = 882 J = 8.8 × 102 J Q1 =

Since the rate of work done is P = 12 kW = 1.2 × 103 J/s, the time for one cycle is W 882 J = = 0.074 s P 1.2 × 104 J/s †21-39.

The flowchart for the operation of a heat pump is shown in Figure 21.16. Heat is taken from the inside of the house at T2 and sent to the high temperature reservoir at T1 outside through the work done by heat pump unit. The efficiency of the heat pump is Q T W e= =1− 2 =1− 2 Q1 Q1 T1 To apply the thermodynamic relationships, temperatures must be converted into degrees Kelvin. Each °F = (5/9)°C. Since the freezing point of water is at 32°F, the temperatures given are: ⎛ 5°C ⎞ T2 = ( 75°F − 32°F ) ⎜ ⎟ + 273°C = 297 K ⎝ 9°F ⎠ and ⎛ 5°C ⎞ T1 = (104°F − 32°F ) ⎜ ⎟ + 273°C = 313 K ⎝ 9°F ⎠ Therefore, the heat transferred to the hot reservoir by the heat pump operating in cooling mode each hour will be Q2 T2 = Q1 T1 ⎛T ⎞ ⎛ 313 K ⎞ 6 Q1 = Q2 ⎜ 1 ⎟ = ( 5.0 × 106 J ) ⎜ ⎟ = 5.27 × 10 J ⎝ 297 K ⎠ ⎝ T2 ⎠ Therefore, the work input to the heat pump is W = Q1 − Q2 = 5.27 × 106 J − 5.0 × 106 J = 2.7 × 105 J The rate of work done is W 2.7 × 105 J ⎛ 1 h ⎞ P= = ⎜ ⎟ = 75 W t 1h ⎝ 3600 s ⎠


CHAPTER 21-40.

21

From the efficiency relationship, we may find the temperature of the cold reservoir. T e =1− 2 T1 T2 = T1 (1 − e) = (350 +273 K)(1 − 0.15) = 530 K Then, to increase the efficiency while maintaining the temperature of the cold reservoir, the temperature of the hot reservoir should be increased to T2 530 K T1 = = = 707 K or 434°C (1 − e) 1 − 0.25

†21-41.

When a heat pump is used to provide heat to a home, the cofficient of performance (CP) is defined as energy transferred to the high temperature reservoir Q1 CP = = work done by the heat pump W The efficiency of a carnot engine is Q T T − T2 W =1− 2 =1− 2 = 1 e= Q1 Q1 T1 T1 Therefore, Q T1 293 K CP = 1 = = = 19.5 W T1 − T2 293 − 278 K

21-42.

This value represents the maximum CP for a system operating between these two temperatures. In practice, the CP values for real heat pumps is somewhat smaller than this value. T e =1− 2 T1 T1 =

21-43.

21-44.

T2 273 K = = 403 K (1 − e) 1 − 0.322

W = Q1 − Q2 Q1 = W + Q2 = 2.5 × 103 J + 6.0 × 103 J = 8.5 × 103 J Q 8.5 × 103 J CP = 1 = = 3.4 W 2.5 × 103 J T T − T2 373 K − 273 K W e= =1− 2 = 1 = = 0.268 Q1 T1 T1 373 K W = eQ1 = e(W − Q2 )

†21-45.

⎛ e ⎞ ⎛ 0.268 ⎞ W = Q2 ⎜ ⎟ = Q2 ⎜ ⎟ = 0.366Q2 ⎝1− e ⎠ ⎝ 1 − 0.268 ⎠ For every joule of heat removed from the low temperature reservoir, 0.366 J of work must be done. (a) The work done during each step for the ideal monatomic gas is W = p∆V. Therefore, the work done during step 1 is zero J, since there is no change in volume. During step 2, the work done is W2 = p2 (V2 − V1 ) = (1 × 105 N/m 2 )(0.03 m3 − 0.01 m3 ) = 2 × 103 J

In step 3, both the pressure and the volume change, such processes may be either isothermal or adiabatic. To determine which type of process occurs in this system, consider the beginning and ending points of step 1, applying the Ideal-Gas Law


CHAPTER

21

PV 1 1 = nRT1 T1 =

PV (3 × 105 N/m 2 )(0.01 m3 ) 1 1 = = 361 K nR (1.00 mole) ( 8.31 J/(mole i K) )

P1 P2 = T1 T2 P2 T1 = 13 T1 = 13 (361 K) = 120 K P1 Then, in step 2, there is an isobaric expansion, the temperature at the beginning of step 3 is V2 V3 = T2 T3 T2 =

V3 T2 = 3T2 = 3 ( 13 T1 ) = T1 V2 The temperature in step 3 does not change, so the process is isothermal. The work done in an isothermal process is ⎛ 0.01 m3 ⎞ V = −3.3 × 103 J W3 = p2V2 ln 3 = (1 × 105 N/m 2 )(0.03 m3 ) ln ⎜ 3 ⎟ 0.03 m V2 ⎝ ⎠ T3 =

The total work done during one cycle is W = W1 + W2 + W3 = 0 + 2.0 × 103 J − 3.3 × 103 J = −1.3 × 103 J The negative value of the net work indicates that work is done on the gas during the cycle. (b) During the isobaric step 2, the heat flow Q = mcp∆T. Since the change in temperature is positive, there is a net flow of heat into the system. Heat is absorbed during step 2. During isothermal step 3, the heat flow is equal to the amount of work, since the work is negative, the heat flow is again negative and there is a net heat flow out of the system. Heat is rejected in step 3. (c) During isovolumetric step 1, heat is rejected by the system, since Q = mcp∆T and T1 is greater than T2. (d) Just as in single step processes that occur between high and low temperature reservoirs, the efficiency is still defined in a cyclical processes as the ratio of the work done by the system to the quantity of heat rejected to the high temperature reservoir. Heat is rejected to the high temperature reservoir in step 3. W W W −1.3 × 103 J e= = = = = 0.39 Q3 W3 W3 −3.3 × 103 J 21-46.

(a) In steps 2 and 4, no work is done since these are isovolumetric steps. During steps 1 and 3, the work done depends on whether these are isothermal or adiabatic processes. At the beginning of step 1, the temperature is given by the Ideal-Gas Law, PV 1 1 = nRT1 T1 =

PV (2 × 105 N/m 2 )(0.01 m3 ) 1 1 = = 240 K nR (1.00 mole) ( 8.31 J/(mole i K) )

At the end of step 1, the temperature is PV (1 × 105 N/m 2 )(0.03 m3 ) T2 = 2 2 = = 360 K nR (1.00 mole) ( 8.31 J/(mole i K) ) Step 1 must be an adiabatic process, therefore the work done during step 1 is


CHAPTER

21

W1 =

p1V1 − p2V2

γ

=

(2 × 105 N/m 2 )(0.01 m3 ) − (1 × 105 N/m 2 )(0.03 m3 ) = −600 J 1.66

At the beginning of step 3, the temperature is PV (3 × 105 N/m 2 )(0.03 m3 ) T3 = 3 3 = = 1100 K nR (1.00 mole) ( 8.31 J/(mole i K) ) At the end of step 3, the temperature is PV (4 × 105 N/m 2 )(0.01 m3 ) T4 = 4 4 = = 480 K nR (1.00 mole) ( 8.31 J/(mole i K) ) The work done during step 3 is p V − p4V4 (3 × 105 N/m 2 )(0.03 m3 ) − (4 × 105 N/m 2 )(0.01 m3 ) W3 = 3 3 = = 3000 J γ 1.66 Therefore, the net work done during one cycle is W = −600 J + 0 + 3000 J + 0 = 2400 J (b) Steps 2 and 4 are isovolumetric, so the heat transferred during these steps is Q = nCV∆T. Q2 = nCv ∆T = (1.00 mole) (12.5 J/(mole i K) ) (1100 − 360 K) = 9250 J Q4 = nCv ∆T = (1.00 mole) (12.5 J/(mole i K) ) (240 − 480 K) = −3000 J

21-47.

(c) Since steps 1 and 3 are adiabatic, Q1 = Q3 = 0 J and ∆E = −W. 2.4 × 103 J W (d) e = = = 0.26 Q2 9.25 × 103 J T T − T2 298 K − 278 K W =1− 2 = 1 = = 0.067 (a) e = Q1 T1 T1 298 K (b) W = eQ1 W e Q1 1 ⎛ W ⎞ 1 = ⎜ ⎟= (1.00 × 106 W) = 1.49 × 107 W t e ⎝ t ⎠ 0.067 Q W = 1.49 × 107 W − 1.00 × 106 W = 1.39 × 107 W Pwaste = 1 − t t Q1 m (c) = c∆T t t 1.49 × 107 W m Q1/t = = = 180 kg/s t c∆T ( 4187 J/(kg i °C) )( 20°C ) Q1 =

21-48.

(a)

e=

Q T T − T2 W =1− 2 =1− 2 = 1 Q1 Q1 T1 T1

and

Q2 = Q1 − W

W ⎛1 ⎞ − W = W ⎜ − 1⎟ e ⎝e ⎠ ⎞ Q2 W ⎛ T1 ⎛ 520 + 273 K ⎞ = ⎜ − 1 ⎟ = 5 × 108 W ⎜ − 1 ⎟ = 3 × 108 W t t ⎝ T1 − T2 ⎝ 520 − 30 K ⎠ ⎠ Q2 =

⎛1 ⎞ ⎛ 1 ⎞ (b) Q2 = W ⎜ − 1 ⎟ = 5 × 108 W ⎜ − 1 ⎟ = 1 × 109 W ⎝e ⎠ ⎝ 0.33 ⎠


CHAPTER †21-49.

21

(a) For the carnot heat pump operating in cooling mode, the heat rejected each hour to the high temperature reservoir is found from Q2 T2 = Q1 T1 ⎛T ⎞ ⎛ 300 K ⎞ 6 Q1 = Q2 ⎜ 1 ⎟ = ( 8.4 × 106 J ) ⎜ ⎟ = 8.57 × 10 J ⎝ 294 K ⎠ ⎝ T2 ⎠ Therefore, the work input to the heat pump each hour is W = Q1 − Q2 = 8.57 × 106 J − 8.4 × 106 J = 1.71 × 105 J The rate of work done is W 1.71 × 105 J ⎛ 1 h ⎞ P= = ⎜ ⎟ = 48 W 1h t ⎝ 3600 s ⎠ (b) The air conditioner unit requires 950 W of work input, which is somewhat more than the heat pump from part (a). P 950 W factor = ac = = 20 Php 48 W

21-50.

The heat removed from ice of mass m in cooling it from 30°C to −5°C is Q2 = mcwater ∆Twater + mLF + mcice ∆Tice

Q2 m = ( cwater ∆Twater + LF + cice ∆Tice ) t t 1 day ⎡ 4187 kg Ji ° C ( 30°C ) + 3.34 × 105 = 1 × 104 1 kg day 86 400 s ⎣ = 5.45 × 104 W The maximum efficiency of this ice making process is Q T T − T2 W e= =1− 2 =1− 2 = 1 Q1 Q1 T1 T1

(

e =1−

†21-51.

)(

)

(

J kg

) + ( 2230

J kg i ° C

) ( 5°C ) ⎤⎦

T2 268 K =1− = 0.116 303 K T1

Combining the efficiency relations and the fact that W = Q1 − Q2, then dividing by the time, we find the power required. W Q2 / t 5.45 × 104 W = = = 7.2 × 103 W t ⎛1 ⎞ ⎛ 1 ⎞ − 1⎟ ⎜ − 1⎟ ⎜ ⎝e ⎠ ⎝ 0.116 ⎠ T2 260 + 273 K =1− = 0.34 (a) eturbine = 1 − T1 540 + 273 K eengine = 1 −

T3 38 + 273 K =1− = 0.42 T2 260 + 273 K

(b) The maximum net efficiency for the two components would occur if all of the heat that is rejected from the turbine goes into the stream engine. In practice, heat would be lost to the surroundings. Here, we’ll calculate the maximum efficiency,


CHAPTER

e=

21 Wnet Wturbine + Wengine eturbine Q 1 +eengine Q 2 = = Q1 Q1 Q1

⎛Q ⎞ ⎛T ⎞ ⎛ 260 + 273 K ⎞ = eturbine + eengine ⎜ 2 ⎟ = eturbine + eengine ⎜ 2 ⎟ = 0.34 + 0.42 ⎜ ⎟ = 0.62 ⎝ 540 + 273 K ⎠ ⎝ Q1 ⎠ ⎝T1 ⎠ If a single engine were utilized between the high and low temperature reservoirs, T 38 + 273 K e =1− 3 =1− = 0.62 T1 540 + 273 K 21-52.

The two efficiencies are the same. From the carnot efficiency relationships, the minimum amount of work needed to cool the water from 30°C to 5°C can be determined. Q T T − T2 W and e= W = Q1 − Q2 =1− 2 =1− 2 = 1 Q1 Q1 T1 T1 ⎛T ⎞ ⎛T ⎞ ⎛T ⎞ ∆W = Q2 ⎜ 1 ⎟ − Q2 = Q2 ⎜ 1 − 1 ⎟ = −mc∆T ⎜ 1 − 1 ⎟ ⎝ T2 ⎠ ⎝ T2 ⎠ ⎝ T2 ⎠ This assumes that the heat rejected from the low temperature reservoir has been used to cool the water. Assuming that a sequence of Carnot engines is utilized, the above equation may be written in infinitesimal form. The mass of one liter of water is one kilogram. ⎛T ⎞ dW = −mc ⎜ 1 − 1 ⎟ dT ⎝T ⎠ T1 ⎛ T 303 K ⎞ ∫ dW = − mc ∫T2 ⎜⎝ T1 − 1⎟⎠ dT = mc ( T1 ln T − T ) 278 K ⎡ ⎤ ⎛ 303 K ⎞ = (1 kg) ( 4187 J/(kg i °C) ) ⎢ (303 K) ln ⎜ ⎟ − 303 K + 278 K ⎥ ⎝ 278 K ⎠ ⎣ ⎦

†21-53.

= 4570 J The change in entropy as the water vaporizes is 6 dQ ∆Q mLV (1.0 kg ) ( 2.26 × 10 J/kg ) ∆S1 = ∫ = = = = 6.06 × 103 J/K T T T 373 K In addition, the water vapor expands as the vaporization occurs and does work against the surrounding air. For the expanding gas, dQ = dW = pdV, so the change in entropy due to the expansion is V dQ pdV dV ∆S 2 = ∫ =∫ = ∫ nR = nR ln 2 T T V V1 The initial volume of the water, assuming thermal expansion is negligible, is V1 = m/ρ = 1.0 kg/(1000 kg/m3) = 0.0010 m3 The volume of 1.0 kg of water vapor at 1.0 atm is 1 mole nRT (1.0 kg) 0.018 kg ( 8.31 J/(mole i K) ) (373 K) V2 = = = 1.705 m3 p 1.01 × 105 N/m 2

(

)

Therefore, ∆S 2 = nR ln

V2 = (1.0 kg) V1

(

3

1 mole 0.018 kg

m ) ( 8.31 J/(mole i K) ) ln 1.705 0.001 m

3

= 3.44 × 103 J/K

The net change in entropy is ∆S = ∆S1 + ∆S2 = 6.06 × 103 J/K + 3.44 × 103 J/K = 9.5 × 103 J/K


CHAPTER 21-54.

The net change in entropy is ⎛1 ∆Q1 ∆Q2 1⎞ ∆S = + = ∆Q1 ⎜ + ⎟ T1 T2 ⎝ T1 T2 ⎠ The rate of change of entropy is then, ∆S ∆Q1 ⎛ 1 1⎞ = ⎜− + ⎟ ∆t ∆t ⎝ T1 T2 ⎠ = ( 2.5 × 104

†21-55.

kcal h

)(

4187J J 1 kcal

⎛ ⎞ 1h + )( 3600 ⎟ = 9.6 W/K s )⎜ − 21 + 273 K −5 + 273 K 1

1

⎝ ⎠ From problem 49, the rate of heat removed from the inside is ∆Q2 J⎛ 1h ⎞ 3 = 8.4 × 106 ⎜ ⎟ = 2.33 × 10 W h ⎝ 3600 s ⎠ ∆t The total entropy change is the sum of the entropy change on the inside and outside. The change in heat on the inside is ∆Q2 where the temperature is T2; and ∆Q1 on the outside where the temperature is T1. ∆Q1 ∆Q2 ∆S = − T1 T2

∆S ∆Q1/∆t ∆Q2 /∆t 2.33 × 103 W + 950 W 2.33 × 103 W = − = − = 3.0 W/K ∆t T1 T2 27 + 273 K 21 + 273 K 21-56.

Typical body temperature is 37°C = 310K; and typical room temperature is around 20°C = 293 K. ⎛ 1 ∆Q1 ∆Q2 1⎞ ∆S = + = ∆Q1 ⎜ − + ⎟ T1 T2 ⎝ T1 T2 ⎠ ∆S ∆Q1 ⎛ 1 1⎞ = ⎜ − ⎟ ∆t ∆t ⎝ T2 T1 ⎠ 1 1 ⎞ − ⎟ = 0.44 W/K ⎝ 293 K 310 K ⎠ Using the Carnot efficiency relationships Q T T − T2 W e= =1− 2 =1− 2 = 1 and W = Q1 − Q2 Q1 Q1 T1 T1 = ( 2.0 × 103

21-57.

21

kcal h

)(

4187J J 1 kcal

⎛ 1h )( 3600 s )⎜

we can find the rate of change of entropy for this system. The absolute temperatures are T1 = 480 + 273 K = 753 K and T2 = 27 + 273 K = 300 K ∆Q1 ∆Q2 ∆S = + T1 T2 The rate of change of the entropy for this system is ∆S ∆Q1/∆t ∆Q2 /∆t = + ∆t T1 T2 =

( ∆W/∆t ) /e + (∆Q1 − ∆W )/∆t T1

⎛ 1 ⎞ ⎜ ⎟ (2000 hp) 0.40 ⎠ =⎝ 753 K

T2

(

745.7 W 1 hp

)

=

( ∆W/∆t ) /e + ( (∆W/e) − ∆W ) /∆t T1

(2000 hp) +

(

T2

745.7 W 1 hp

1 ⎞ − 1⎟ ) ⎛⎜⎝ 0.40 ⎠

300 K


= 12400 W/K

CHAPTER

21-58. 21-59. 21-60.

21

5 ∆Q mLF (1.0 kg ) ( 3.34 × 10 J/kg ) = = = 1.22 × 103 J/K 273 K T T 5 ∆Q mLV (1.0 kg ) ( 5.8 × 10 J/kg ) ∆S = = = = 3.0 × 103 J/K T T −79 + 273 K 5 ∆Q mLF (1.00 kg ) ( 2.05 × 10 J/kg ) ∆S melt = = = = 150 J/K Tmelt Tmelt 1083 + 273 K

∆S =

∆S vapor †21-61.

The melting temperatures and heats of fusion are given in Table 20.4. 5 ∆Q mLF (1.00 kg ) ( 3.99 × 10 J/kg ) ∆S Al = = = = 430 J/K Tmelt Tmelt 660 + 273 K

∆S Fe ∆S Ag

5 mLF (1.00 kg ) ( 2.7 × 10 J/kg ) = = = 150 J/K Tmelt 1535 + 273 K 4 mLF (1.00 kg ) ( 9.9 × 10 J/kg ) = = = 80 J/K Tmelt 962 + 273 K

∆S Hg =

21-62.

†21-63.

21-64.

(1.00 kg ) ( 5.2 × 106 J/kg ) mLV ∆Q = = = = 1.8 × 103 J/K Tvapor Tvapor 2567 + 273 K

4 mLF (1.00 kg ) (1.1 × 10 J/kg ) = 47 J/K = Tmelt −39 + 273 K

The change in entropy seems to decrease with increasing atomic number. The largest value occurs for aluminum (Z = 13) and the smallest value occurs for mercury (Z = 80). As the parachute descends, it does work against air friction and absorbs heat. W = fs = ∆Q Because the parachutist is failing at a constant rate, the upward force due to air friction f is equal in magnitude to the downward force due to gravity, mg. The rate of work done is thus, W ∆y ∆Q = mg = mgv = t ∆t ∆t The rate of change of entropy is then 2 ∆S ∆Q / ∆t mgv ( 80 kg ) ( 9.81 m/s ) (5.0 m/s) = = = = 13 W/K ∆t T T 20 + 273 K As the automobile moves, it does work against the various sources of friction and absorbs heat. W = fs = ∆Q The rate of work done is thus, W ∆x ∆Q = f = t ∆t ∆t The rate of change of entropy is then ∆S ∆Q / ∆t 12 × 103 W = = = 41 W/K ∆t T 20 + 273 K (a) The work done by the brakes is equal to the change in kinetic energy. This is also equal to the heat absorbed by the brakes.


CHAPTER W = ∆K = 12 mv1 = ∆Q dQ = T

∫ dS = ∫ = =

1 2

∫

mcdT T ∆Q T2 ln =mc ln 2 = T T1 ∆T T1

mv12 T2 ln ∆T T1

1 2

(2100 kg) ⎣⎡ ( 80

(b) ∆S = ∆Sair

km h

1000 m 1h ⎤ ) ( 3600 s )( 1 km ) ⎦

2

40 K ∆Q ∆Q T2 + ∆S brake = + ln T1 ∆T T1

⎛ 60 + 273 K ⎞ 3 ln ⎜ ⎟ = 1.66 × 10 J/K ⎝ 20 + 273 K ⎠

So, the additional entropy is

21-66.

21-67.

21-68.

†21-69.

(2100 kg) ⎡⎣ ( 80

km h

1000 m 1h ⎤ ) ( 3600 s )( 1 km ) ⎦

2

= 1.8 × 103 J/K 20 + 273 K The potential energy of the water is dissipated as heat at a rate ∆m ∆Q gy = ∆t ∆t The rate of change of entropy is then ⎛ ∆m ⎞ gy ( 5700 m3 /s ) 10001 mkg3 ( 9.81 m/s2 ) (50 m) = 9.5 × 106 W/K ∆S ∆Q / ∆t ⎜⎝ ∆t ⎟⎠ = = = ∆t T T 20 + 273 K Entropy of the water increases as it does work on the diver to change his kinetic energy. That energy is then dissipated as heat in the water. ∆Q mgy (70 kg)(9.81 m/s 2 )(36 m) ∆S = = = = 84 J/K 295 K T T −∆Q1 ∆Q2 ∆S = + T1 T2 ∆Sair =

†21-65.

1 2

∆S −∆Q / ∆t ∆Q / ∆t ∆Q ⎛ 1 1⎞ = + = ⎜− + ⎟ ∆t T1 T2 ∆t ⎝ T1 T2 ⎠ 1 1 ⎛ ⎞ = 1.8 × 103 W ⎜ − + ⎟ = 0.94 W/K ⎝ 21 + 273 K −18 + 273 K ⎠ The melting represents an increase in entropy. The change in entropy is ∆Q mLF (20 kg)2.9 × 104 J/kg ∆S = = = = 965 J/K 328 + 273 K T T In the free expansion of the gas, we remember from the First Law of Thermodynamics that ∆E = ∆Q − W = ∆Q − p∆V Since the process is an isothermal one, ∆E = 0 J and ∆Q = p∆V If we consider the system by adding and removing infinitesimally small amounts of heat, the change in entropy is dQ pdV dV dS = = = nR T T V V′ dV V′ 2.0 liters ∆S = ∫ dS = ∫ nR = nR ln = (1.00 mole )( 8.31 J/(mole i K) ) ln = 5.8 J/K V V V 1.0 liter


21

CHAPTER 21-70.

21

At its initial temperature, the silver is solid since it is below its melting temperature. Heat will be transferred from the silver to the water until the equilibrium temperature is reached. mW cW ∆TW = mAg cAg ∆TAg heat gained by the water

heat lost by the silver

mW cW (T − TW ) = mAg cAg (TAg − T ) mW cW T − mW cW TW = mAg cAgTAg − mAg cAgT T = =

21-71.

mW cW TW + mAg cAgTAg mW cW + mAg cAg 5.0 kg ( 4187 J/(kg i °C) ) (20°C) + (0.50 kg) ( 240 J/(kg i °C) ) (950°C) 5.0 kg ( 4187 J/(kg i °C) ) + (0.50 kg) ( 240 J/(kg i °C) )

= 25.3°C = 298 K The heat lost by the silver has been added to the water, so there is a change in entropy of the water. ∆Q mW cW ∆TW 5.0 kg ( 4187 J/(kg i °C) ) (25.3 − 20°C) ∆S = = = = 370 J/K 298 K T T (a) In the limit of adding or removing infinitesimally small quantities of heat from a system, we have dQ mcdT T ∫ dS = ∫ T = ∫ T = mc ln T12

S 2 − S1 = ∆S = mc ln (b) ∆S = mc ln 21-72.

†21-73.

T2 T1

T2 ⎛ 80 + 273 K ⎞ = (1.0 kg) ( 4187 J/(kg i °C) ) ln ⎜ ⎟ = 780 J/K T1 ⎝ 20 + 273 K ⎠

The total entropy change is ∆S = ∆S1 + ∆S2 where ∆S1 is the change in entropy of the boiling water and ∆S2 is the change in entropy of the ice cubes. T ⎛ 40 + 273 K ⎞ ∆S1 = mc ln 2 = (1.0 kg) ( 4187 J/(kg i °C) ) ln ⎜ ⎟ = −734 J/K T1 ⎝ 100 + 273 K ⎠ mLF T (0.5 kg)(3.34 × 105 J/kg) ⎛ 313 K ⎞ ∆S 2 = + mc ln 2 = + (0.5 kg) ( 4187 J/(kg i °C) ) ln ⎜ ⎟ 273 K T0 T0 ⎝ 273 K ⎠ = 900 J/K ∆S = ∆S1 + ∆S2 = −734 J/K + 900 J/K = 164 J/K Because we are mixing equal amounts of warm and cool water, the final temperature will be 50°C, the mid-point between the two temperatures. The total entropy change is ∆S = ∆S1 + ∆S2 where ∆S1 is the change in entropy of the initially cooler water and ∆S2 is the change in entropy of the initially warmer water. T 50 + 273 K ⎞ kg ∆S1 = mc ln 2 = (1.0 liter) 11liter ( 4187 J/(kg i °C) ) ln ⎛⎜ ⎟ = 408 J/K T1 ⎝ 20 + 273 K ⎠ T 50 + 273 K ⎞ kg ∆S 2 = mc ln 2 = (1.0 liter) 11liter ( 4187 J/(kg i °C) ) ln ⎛⎜ ⎟ = −371 J/K T1 ⎝ 80 + 273 K ⎠

(

)

(

)

∆S = ∆S1 + ∆S2 = 408 J/K − 371 J/K = 37 J/K


CHAPTER 21-74.

(a) In an isobaric process, the heat transferred into or out of a system is given by ∆Q = nCp∆T. If the heat is transferred in infinitesimally small amounts, the the change in entropy is nCp dT T dQ ∆S = ∫ dS = ∫ =∫ = nCp ln 2 T T T1 (b) In an isovolumetric process, the heat transferred into or out of a system is given by ∆Q = nCV∆T. If the heat is transferred in infinitesimally small amounts, the the change in entropy is nC dT T dQ ∆S = ∫ dS = ∫ =∫ V = nCV ln 2 T T T1

21-75.

21-76.

†21-77.

In the free expansion of the gas, we remember from the First Law of Thermodynamics that ∆E = ∆Q − W = ∆Q − p∆V Because the process is an isothermal one, ∆E = 0 J and ∆Q = p∆V If we consider the system by adding and removing infinitesimally small amounts of heat, the change in entropy is dQ pdV dV dS = = = nR T T V ′ V dV V′ ∆S = ∫ dS = ∫ nR = nR ln V V V ∆E = ∆Q − W = 3000 J − 2000 J = 1000 J The work done is W = p∆V = nR∆T, therefore p∆V W 2000 J ∆T = = = = 240 K nR nR (1 mole) ( 8.31 J/(mole i K) ) For n moles of a diatomic gas at temperature T, the internal energy is E =

5 nRT 2

The change in internal energy is ∆E = ∆Q − W = −W Because work is done on the system, the sign for the work is negative, W = −2500 J Combining the above equations, gives 5 ∆E = nR∆T = −W 2 −2 ( −2500 J ) −2W ∆T = = = 120 K 5nR 5 (1 mole )( 8.31 J/(mole i K) ) 21-78.

(a) W = 12 ( p1 − p2 )(V2 − V1 ) + p2 (V2 − V1 ) = 12 ( p1 + p2 )(V2 − V1 ) CV ( p2V2 − p1V1 ) = 32 ( p2V2 − p1V1 ) R (c) Q = ∆E + W = 32 ( p2V2 − p1V1 ) + 12 ( p1 + p2 )(V2 − V1 ) (b) ∆E = CV (T2 − T1 ) =

= 2( p2V2 − p1V1 ) + 12 ( p1V2 − p2V1 ) 21-79.

21

(a) p1 =

nRT (0.20 mole) ( 8.31 J/(mole i K) )( 300 K ) = = 4.16 × 105 N/m 2 3 V1 0.0012 m

p2 =

nRT (0.20 mole) ( 8.31 J/(mole i K) )( 300 K ) = = 2.27 × 105 N/m 2 V2 (0.0012 + 0.0010) m3


CHAPTER

21

(b) In an isothermal compression, the work done on the gas is V W = p2V2 ln 3 V2

21-80.

⎛ 0.0012 m3 ⎞ = ( 2.27 × 105 N/m 2 )( 0.0022 m3 ) ln ⎜ = −3.0 × 102 J 3 ⎟ 0.0022 m ⎝ ⎠ Also, since ∆E is equal to 0 J in such a process, ∆Q = W = − 3.0 × 102 J (a) At 0°C, one mole of an ideal gas occupies 22.4 liters. p2V2 = nRT2 1 mole nRT 4.0 liters ( 22.4 liters ) ( 8.31 J/(mole i K) )( 60 + 273 K ) = = 0.0049 m3 = 4.9 liters V2 = p 1.01 × 105 N/m 2

(b) W = p∆V = (1.01 × 105 N/m2)(0.9 × 10−3 m3) = 91 J ∆E = Q − W = nCp ∆T − W 1 mole = (4.0 liters) ( 22.4 29.1 J/(mole i K) ) (60 K) − 91 J = 220 J liters ) (

1 mole Q = (4.0 liters) ( 22.4 29.1 J/(mole i K) ) (60 K) = 310 J liters ) (

21-81.

(

)

745.7 W W W/∆t 20 hp 1 hp = = = 0.24 = 24% Q Q/∆t 6.3 × 104 W Q e =1− 2 Q1

e=

Q2 =1− e Q1

21-82.

21-83.

Q2 Q = (1 − e ) 1 = (1 − 0.24)(6.3 × 104 W) = 4.8 × 104 W ∆t ∆t 869 W W W / ∆t e= = = = 0.23 = 23 % Q Q / ∆t ⎛ 10.8 liters ⎞ ⎛ 1min ⎞ 4 ( 2.1 × 10 J/liter ) ⎜⎝ min ⎟⎠ ⎜⎝ 60 s ⎟⎠ Q2 T2 = Q1 T1

∆Q2 /∆t T2 = ∆Q1 /∆t T1 ∆Q1 ∆Q2 ⎛ T1 ⎞ 7 = ⎜ ⎟ = ( 3.8 × 10 ∆t ∆t ⎝ T2 ⎠ W = Q1 − Q2

21-84.

J h

)(

1h 3600 s

32 + 273 K

) ⎛⎜ 21 + 273 K ⎞⎟ = 1.095 × 104 W ⎝

⎠

∆W ∆Q1 − ∆Q2 1h = = 1.0950 × 104 W − ( 3.8 × 107 hJ ) ( 3600 = 4.0 × 102 W s) ∆t ∆t Q T 30 + 273 K W (a) e = =1− 2 =1− 2 =1− = 0.61 500 + 273 K Q1 Q1 T1 W (b) e = Q1 Q1 =

W 1.5 × 103 J = = 2460 J e 0.61


CHAPTER †21-85.

(a) Beginning with the point at the upper left, the gas undergoes an isobaric expansion in step 1 as the volume is increased, followed by a isovolumetric reduction of pressure in step 2 as the temperature is reduced. The gas is then compressed isobarically in step 3 by reducing the volume, before an isovolumetric increase in pressure in step 4 by increasing the temperature. (b) Step 1: W1 = p∆V = (1.5 × 105 N/m2)(0.021 − 0.007 m3) = 2100 J Step 2: W2 = 0 J (no work is done in an isovolumetric process) Step 3: W3 = p∆V = (0.5 × 105 N/m2)(0.007 − 0.021 m3) = −700 J Step 4: W4 = 0 J (no work is done in an isovolumetric process) (c) Step 1: At the beginning of step 1, the temperature is PV 1 1 = nRT1 T1 =

PV (1.5 × 105 N/m 2 )(0.007 m3 ) 1 1 = = 126 K nR (1.00 mole) ( 8.31 J/(mole i K) )

At the end of step 1, the temperature is PV (1.5 × 105 N/m 2 )(0.021 m3 ) T2 = 2 2 = = 379 K nR (1.00 mole) ( 8.31 J/(mole i K) ) Q1 = nCp∆T1 = (1 mole)(20.8 J/(mole • K))(379 K − 126 K) = 5260 J Step 2: At the end of step 2, the temperature is PV (0.5 × 105 N/m 2 )(0.021 m3 ) T3 = 3 3 = = 126 K nR (1.00 mole) ( 8.31 J/(mole i K) ) Q2 = nCV∆T2 = (1 mole)(12.5 J/(mole • K))(126 K − 379 K) = −3160 J Step 3: At the end of step 3, the temperature is PV (0.5 × 105 N/m 2 )(0.007 m3 ) T4 = 4 4 = = 42 K nR (1.00 mole) ( 8.31 J/(mole i K) )

21-86.

Q3 = nCp∆T3 = (1 mole)(20.8 J/(mole • K))(42 K − 126 K) = −1750 J Step 4: Q4 = nCV∆T4 = (1 mole)(12.5 J/(mole • K))(126 K − 42 K) = 1050 J In part (b), the net work done is 2100 − 700 J = 1400 J. From part (c), the sum of the four quantities of heat also equals 1400 J. (d) The efficiency of this four-step cycle is W 1400 J e= = = 0.44 or 44% Q2 3160 J Q2 T2 = Q1 T1 Q2 = Q1

21-87.

21

T2 200 K = (1200 J) = 400 J T1 600 K

W = Q1 − Q2 = 1200 J − 400 J = 800 J Q T W 25 + 273 K e= =1− 2 =1− 2 =1− = 0.48 or 48% Q1 Q1 T1 300 + 273 K Q1 W / t 1 × 107 W = = = 2 × 107 W 0.48 t e The rate of waste heat is then, Q2 Q1 W = − = 2 × 107 W − 1 × 107 W = 1 × 107 W t t t


CHAPTER

21-88.

21

Q2 T2 = Q1 T1 Q2 = Q1

( 4 + 273 ) K = 4727 J T2 = (5.0 × 103 J) T1 ( 20 + 273 ) K

W = Q1 − Q2 = 5.0 × 103 J − 4727 J = 270 J †21-89.

The boiling point temperatures and heats of vaporization are given in Table 20.4. 5 ∆Q mLV (1.00 kg ) ( 2.00 × 10 J/kg ) ∆S N = = = = 2600 J/K Tboil Tboil −196 + 273 K ∆SO =

5 ∆Q mLV (1.00 kg ) ( 2.1 × 10 J/kg ) = = = 2300 J/K Tboil Tboil −183 + 273 K

5 ∆Q mLV (1.00 kg ) ( 4.5 × 10 J/kg ) ∆S H = = = = 2.25 × 104 J/K Tboil Tboil −253 + 273 K

21-90.

Hydrogen is largest and oxygen the smallest. (a) Because both the lead and the ice are at their respective melting temperatures, some of the ice will melt into water and the lead will solidify. Because the amount of heat removed from the lead in solidifying is much less than the heat required to melt all of the ice, the equilibrium temperature will be 273 K. As a check, let’s calculate how much of the ice melts. mlead clead (Tlead − T ) + mlead LF, lead = mice LF heat lost from the lead

mice =

(b) ∆S (c) 21-91.

(a) (b)

heat gained by ice

mlead clead (Tlead − T ) + mlead LF, lead LF

(1.0 kg) ⎡⎣ (130 J/(kg i °C) )( 328 °C ) + 2.32 × 104 J/kg ⎤⎦ = = 0.20 kg 3.34 × 105 J/kg The entropy of the lead decreases by T 273 K ⎞ ⎛ = mc ln 2 = (1.0 kg) (130 J/(kg i K) ) ln ⎜ ⎟ = −100 J/K T1 ⎝ 328 + 273 K ⎠ ∆Q mLF (0.20 kg)(3.34 × 105 J/kg) ∆S water = = = = 245 J/K 273 K T T Q 800 J S = 1 = = 1.1 J/K T1 700 K Q2 T2 = Q1 T1 Q2 = Q1

300 K T2 = (800 J) = 340 J 700 K T1

Q2 340 J = = 1.1 J/K 300 K T2 (c) The entropy remains constant during adiabatic processes. ∆S = 0 J/K S =


CHAPTER 22

ELECTRIC FORCE AND ELECTRIC CHARGE


According to Coulomb's law, the magnitude of the electrostatic force one charged object exerts qq on another is given by F = k 1 2 2 , where q1 and q2 are the magnitudes of the two charges, r is the r distance between them, and k is the Coulomb constant. In this particular problem, we have

F = k

⎛ N i m 2 ⎞ ( 40 C )( −40C ) q1q2 = ⎜ 8.99 × 109 = 5.8 × 105 N ⎟ 2 2 2 C ⎠ ( 5000 m ) r ⎝

22-2.

2 −1.60 × 10−19 C )( +1.60 × 10−19 C ) ⎛ q1q2 9 Nim ⎞ ( F = k 2 = ⎜ 8.99 × 10 = 2.90 × 10−9 N ⎟ 2 −10 2 r C ⎠ (2.82 × 10 m) ⎝

22-3.

2 −1.60 × 10−19 C )( +1.60 × 10−19 C ) ⎛ q1q2 9 Nim ⎞ ( F = k 2 = ⎜ 8.99 × 10 = 57.7 N = 58 N ⎟ r C2 ⎠ (2.0 × 10−15 m) 2 ⎝

a= 22-4.

F 57.7 N = = 3.5 × 1028 m/s 2 m 1.67 × 10−27 kg

2 2 1.60 × 10−19 C ) 92 (1.60 × 10−19 C ) ⎛ q1q2 9 Nim ⎞ ( F = k 2 = ⎜ 8.99 × 10 = 17 N ⎟ r C2 ⎠ (5.0 × 10−14 m) 2 ⎝

a=

F 17 N = = 2.5 × 1027 m/s 2 m 4(1.67 × 10−27 kg)

†22-5.

The magnitude of the electric force between these two quarks is given by Coulomb's law. Since the two particles have the same sign, they will repel each other with a force −19 −19 2 1 ⎛ qq N i m 2 ⎞ 3 (1.60 × 10 C ) 3 (1.60 × 10 C ) F = k 1 2 2 = ⎜ 8.99 × 109 = 51 N ⎟ r C2 ⎠ (1.0 × 10−15 m) 2 ⎝

22-6.

Each electron carries a charge of 1.602 × 10−19 C. To determine how many electrons, divide the total charge by the charge on an individual electron. −1.0 × 10−6 C N = = 6.25 × 1012 electrons −19 −1.60 × 10 C

†22-7.

Each electron carries a charge of 1.60 × 10−19 C. To determine how many electrons, divide the total charge by the charge on an individual electron. −25 C N = = 1.6 × 1020 electrons −19 −1.60 × 10 C e −1.602 177 × 10−19 C = = 9.109 386 × 10−31 kg 11 e / me −1.758 820 × 10 C/kg

22-8.

me =

22-9.

2 +1.60 × 10−19 C ) (17) ( +1.60 × 10−19 C ) q1q2 ⎛ 9 Nim ⎞ ( = 2.39 × 10−7 N F = k 2 = ⎜ 8.99 × 10 ⎟ 2 −10 2 r C (1.28 × 10 m) ⎝ ⎠


CHAPTER

22-10.

22 2 1.67 × 10−27 kg )(1.67 × 10−27 kg ) m1m2 ⎛ −11 N i m ⎞ ( 6.67 × 10 = ⎜ ⎟ r2 kg 2 ⎠ (1.0 × 10−12 m) 2 ⎝ = 1.9 × 10−40 N

(a) Fg = G

2 +1.60 × 10−19 C )( +1.60 × 10−19 C ) q1q2 ⎛ 9 Nim ⎞ ( (b) Fe = k 2 = ⎜ 8.99 × 10 ⎟ C2 ⎠ (1.0 × 10−12 m) 2 r ⎝ = 2.3 × 10−4 N

Fe 2.3 × 10−4 N = = 1.2 × 1036 −40 Fg 1.9 × 10 N (c) r =

k

−19 −19 ⎛ N i m 2 ⎞ ( +1.60 × 10 C )( +1.60 × 10 C ) q1q2 = ⎜ 8.99 × 109 ⎟ C2 ⎠ 1.9 × 10−40 N Fg ⎝

= 1.1 × 106 m †22-11.

Although the gravitational and electric forces have similar formulas, the magnitudes of these forces are very different. The electric force is a much stronger force that that of gravity. −11 −11 ⎛ mm N i m 2 ⎞ (1.0 × 10 kg )(1.0 × 10 kg ) Fg = G 1 2 2 = ⎜ 6.67 × 10−11 ⎟ r kg 2 ⎠ (1.0 × 10−10 m) 2 ⎝ = 6.7 × 10−13 N 2 +1.60 × 10−19 C )( +1.60 × 10−19 C ) q1q2 ⎛ 9 Nim ⎞ ( Fe = k 2 = ⎜ 8.99 × 10 ⎟ r C2 ⎠ (1.0 × 10−10 m) 2 ⎝ = 2.3 × 10−8 N

22-12.

2 −1.60 × 10−19 C ) 82 ( +1.60 × 10−19 C ) ⎛ q1q2 9 Nim ⎞ ( F = k 2 = ⎜ 8.99 × 10 = 0.044 N ⎟ r C2 ⎠ (6.5 × 10−13 m) 2 ⎝

a= †22-13.

22-14.

†22-15.

22-16.

F 0.044 N = = 4.8 × 1028 m/s 2 m 9.11 × 10−31 kg

The electric charge on a proton is +1.60 × 10−19 C. One mole contains Avogadro's number of particles, NA = 6.02 × 1023 particles. Therefore, the total charge of mole mole of protons is Faraday constant = ( +1.60 × 10−19 C/proton)( 6.02 × 1023 protons) = 9.63 × 104 C Each second, there are 1.3 C of electrons moving in the lightbulb filament. Each electron carries a charge of 1.602 × 10−19 C. Therefore, 1.3 C corresponds to −1.3 C N = = 8.1 × 1018 electrons −19 −1.60 × 10 C If there are 7.5 × 10−6 C of excess electrons on the sphere; and each electron carries a charge of 1.602 × 10−19 C, then −7.5 × 10−6 C N = = 4.7 × 1013 electrons −19 −1.60 × 10 C The number of moles of iron is 0.30 g n= = 5.38 × 10−3 moles 55.8 g/mol


CHAPTER

†22-17.

22

Each iron atom has 26 electrons, so the total number of electrons is N = 26nNA = 26(5.38 × 10−3 mol)(6.022 × 1023 atoms/mol) = 8.4 × 1022 electrons The number of moles of copper is 2.7 g = 4.25 × 10−2 moles n= 63.5 g/mol Each copper atom has 29 electrons (see the periodic table), so the total number of electrons is N = 29nNA = 29(4.25 × 10−2 mol)(6.022 × 1023 atoms/mol) = 7.42 × 1023 electrons The total charge for these electrons is q1 = (7.42 × 1023 electrons)(−1.60 × 10−19 C) = −1.19 × 105 C An equal quantity of positive charge exists due to the protons in the remaining nucleii, q2 = +1.19 × 105 C Therefore, the magnitude of the electric force on the electrons due to the nucleii is 5 5 ⎛ qq N i m 2 ⎞ ( −1.19 × 10 C )( +1.19 × 10 C ) F = k 1 2 2 = ⎜ 8.99 × 109 = 3.2 × 1019 N ⎟ 2 2 r C (2.0 m) ⎝ ⎠

22-18. Element

Percentage of body (%)

Mass (kg)

O

70

51.1

C

20

14.6

nC =

14600 g = 1220 moles 12.0 g/mol

H

10

7.3

nH =

7300 g = 7230 moles 1.01 g/mol

Number of moles 51100 g nO = = 3190 moles 16.0 g/mol

Each mole contains Avogadro's number of atoms with each atom containing it’s atomic number of electrons and protons. Therefore, the total number of electrons or protons in the body is approximately, N = 6.022 × 1023 atoms/mol [ (3190 mol O)(8) + (1220 mol C)(6) + (7230 mol H)(1) ] = 2.4 × 1028 electrons or protons

†22-19.

One mole of sodium chloride, NaCl, has a mass which is the sum of the molar masses of Na and Cl, (22.99 g Na/mol) + (35.45 g Cl/mol) = 58.44 g NaCl/mol. The number of moles to be disolved in 100 g of water is 36 g nNaCl = = 0.62 moles 58.44 g/mol Each sodium atom has 11 electrons and 11 protons and each chlorine atom has 17 electrons and 17 protons. The total number of electrons or protons added to the water will be N NaCl = 6.022 × 1023 atoms/mol [ (0.62 mol Na)(11) + (0.62 mol Cl)(17) ] = 1.0 × 1025 electrons or protons The number of electrons or protons in the 100 g of water is found in similar manner. One mole of water, H2O, has a mass which is the sum of the molar masses of two hydrogen atoms (2 × 1.01 g H/mol) and one oxygen atom (16.0 g O/mol) = 18.0 g H2O/mol. The number of moles in 100 g of water is


CHAPTER

22

nH2O =

100 g = 5.6 moles 18.0 g/mol

Each hydrogen atom has 1 electron and 1 proton and each oxygen atom has 8 electrons and 8 protons. The total number of electrons or protons in the water will be N H2O = 6.022 × 1023 atoms/mol [ 2(5.6 mol H)(1) + (5.6 mol O)(8) ] = 3.3 × 1025 electrons or protons Therefore, the factor by which the number of electrons or protons increases is N H2O + N NaCl 3.3 × 1025 + 1.0 × 1025 = = 1.3 N H2O 3.3 × 1025

22-20.

22-21.

22-22.

2 −2.0 × 10−6 C )( −2.0 × 10−6 C ) q1q2 ⎛ 9 Nim ⎞ ( Fe = k 2 = ⎜ 8.99 × 10 ⎟ r C2 ⎠ (1.0 m) 2 ⎝ = 3.6 × 10−2 N (repulsive)

The ratio of the exact value to the simpler value is N i m2 8.987 551 787 × 109 C2 = 0.9986 = 99.9% 2 Nim 9.0 × 109 C2 The total charge on each disk is the product of the area and the charge per unit area. q = π (0.01 m) 2 (2.5 × 10−8 C/m 2 ) = 7.85 × 10−12 C The electric force that each plate exerts on the other is 2 7.85 × 10−12 C )( 7.85 × 10−12 C ) q1q2 ⎛ 9 Nim ⎞ ( Fe = k 2 = ⎜ 8.99 × 10 ⎟ C2 ⎠ (2.0 m) 2 r ⎝ = 1.4 × 10−13 N (repulsive)

†22-23.

22-24.

Through the ionization, an atom at one end has one excess electron, giving it a −e net negative charge; and an atom at the other end has lost an electron, giving it a +e net positive charge. The electric force then compresses the chain molecule by ∆L = (−0.012)(1.9 × 10−6 m). The molecule is to be modeled like a spring. The electric force compresses the spring until the electric force and the spring force are equal in magnitude, but oppositely directed. Applying Hooke's law, we find F F k= S = e ∆L ∆L 2 −1.60 × 10−19 C )( +1.60 × 10−19 C ) ( ⎛ 9 Nim ⎞ ⎜ 8.99 × 10 ⎟ C2 ⎠ ⎡ (1.9 × 10−6 m) + (−0.012)(1.9 × 10−6 m) ⎤ 2 ⎝ ⎣ ⎦ = 2.9 × 10−9 N/m = (−0.012)(1.9 × 10−6 m) 2 9.11 × 10−31 kg )( 2.0 × 1015 kg ) m1m2 ⎛ −11 N i m ⎞ ( 6.67 × 10 = ⎜ ⎟ kg 2 ⎠ (1.0 × 105 m) 2 r2 ⎝ = 1.2 × 10−35 N

Fg = G

For the electric force to be equal in magnitude, but oppositely directed to that of the gravitational force,


CHAPTER

22

Fe = Fg k

q1q2 mm = G 12 2 2 r r kQe = GMm

2 ⎛ −11 N i m ⎞ −31 15 ⎜ 6.67 × 10 ⎟ ( 2.0 × 10 kg )( 9.11 × 10 kg ) 2 kg ⎠ GMm ⎝ = = −8.4 × 10−17 C Q= 2 ke ⎛ 9 Nim ⎞ −19 ⎜ 8.99 × 10 ⎟ ( −1.60 × 10 C ) 2 C ⎠ ⎝ The number of electrons in this amount of charge is Q −8.4 × 10−17 C N = = = 5.3 × 102 electrons −19 e −1.60 × 10 C

†22-25.

Charge 1 is located at xy corredinate (2.0 m, 0 m) and charge 2 is located at (0 m, −3.0 m). Since the charges have the same sign, the force on charge 2 due to charge 1 will be directed away from charge 1 in the direction given by (−2.0 m)i + ( −3.0 m)j −2.0 −3.0 r= = i+ j 2 2 13 13 (−2.0 m) + (−3.0 m) The force exerted on charge 2 due to charge 1 is then qq F12 = F12 r = k 1 2 2 r r 2 −2.0 × 10−8 C )( −3.0 × 10−6 C ) ⎛ −2.0 ⎛ −3.0 ⎞ 9 Nim ⎞ ( = ⎜ 8.99 × 10 i+ j⎟ ⎟ ⎜ 2 2 2 C ⎠ (−2.0 m) + (−3.0 m) 13 ⎠ ⎝ 13 ⎝ = (−2.3 × 10−5 N)i + (−3.5 × 10−5 N)j Similarly, the force exerted on charge 1 due to charge 2 is qq F21 = F21 rˆ = k 1 2 2 rˆ r 2 −2.0 × 10−8 C )( −3.0 × 10−6 C ) ⎛ 2.0 ⎛ 3.0 ⎞ 9 Nim ⎞ ( = ⎜ 8.99 × 10 xˆ + yˆ ⎟ ⎟ ⎜ 2 2 2 C ⎠ (−2.0 m) + ( −3.0 m) 13 ⎠ ⎝ 13 ⎝ = (2.3 × 10−5 N)xˆ + (3.5 × 10−5 N)yˆ

22-26.

The magnitude of the electric force is to be equal to the weight of the plastic chip. q2 mg = k 2 r

mg (5.0 × 10−8 kg)(9.81 m/s 2 ) = (1.0 × 10−3 m) = 7.4 × 10−12 C 2 i N m k 8.99 × 109 C2 Let Q represent the amount of charge on the Earth. The charge on the moon will be 1.74 q= Q 6.38 If the magnitude of the electric force is to be equal to that of the gravitational force, then q=r

†22-27.


CHAPTER

22 Fe = Fg k

q1q2 mm = G 12 2 2 r r kQq = GMm

⎛ 1.74 ⎞ 2 k⎜ ⎟ Q = GMm ⎝ 6.38 ⎠

Q=

6.38GMm = 1.74k

⎛ N i m2 ⎞ 24 22 6.38 ⎜ 6.67 × 10−11 ⎟ ( 5.98 × 10 kg )( 7.35 × 10 kg ) 2 kg ⎝ ⎠ ⎛ N i m2 ⎞ 1.74 ⎜ 8.99 × 109 ⎟ C2 ⎠ ⎝

= 1.09 × 1014 C The number of electrons in this amount of charge is Q −1.09 × 1014 C N Earth = = = 6.81 × 1032 electrons e −1.60 × 10−19 C 1.74 1.74 q= Q= (1.09 × 1014 C ) = 2.97 × 1013 C 6.38 6.38 2.97 × 1013 C = 1.86 × 1032 electrons N moon = −19 1.60 × 10 C 22-28.

†22-29.

N =

Q +1.0 × 10−7 C = = 6.3 × 1011 ions e +1.60 × 10−19 C

There are 6.3 × 1011 ions per square meter. The fraction of surface atoms that must be ionized is 6.3 × 1011 ions/m 2 = 3.2 × 10−8 ions/atom 2.0 × 1019 atoms/m 2 q1q2 − r / r0 Fe_mod k r 2 e r = = e − r / r0 ≈ 1 − qq Fe r0 k 1 22 r The fractional deviation at r = 1.0 m is r 1.0 m = = 1.0 × 10−9 r0 1.0 × 109 m

The fractional deviation at r = 1.0 × 104 m is r 1.0 × 104 m = = 1.0 × 10−5 9 r0 1.0 × 10 m 22-30.

An electron is located at xy coordinate (4.0 × 10−11 m, 2.0 × 10−11 m) and a proton is located at the origin (0 m, 0 m). Since the charges have opposite signs, the force exerted on the electron due to the proton will be directed toward the proton in the direction (−4.0 × 10−11 m)i + (−2.0 × 10−11 m)j r= (−4.0 × 10−11 m) 2 + ( −2.0 × 10−11 m) 2

= −0.89i − 0.45 j


CHAPTER

22

The force exerted on the electron due to the proton is then q2 Fep = Fep r = −k 2 r r 2 1.60 × 10−19 C ) ( ⎛ 9 Nim ⎞ (−0.89i − 0.45 j) = − ⎜ 8.99 × 10 ⎟ C 2 ⎠ (−4.0 × 10−11 m) 2 + (−2.0 × 10−11 m) 2 ⎝ = (1.0 × 10−7 N)i + (5.2 × 10−8 N)j 2

Similarly, the force exerted on the electron due to the proton is q2 Fpe = − Fep r = −k 2 r r 2 1.60 × 10−19 C ) ( ⎛ 9 Nim ⎞ (−0.89i − 0.45 j) = ⎜ 8.99 × 10 ⎟ C 2 ⎠ (−4.0 × 10−11 m) 2 + (−2.0 × 10−11 m) 2 ⎝ = (−1.0 × 10−7 N)i + (−5.2 × 10−8 N)j 2

†22-31.

Under normal circumstances, an oxygen atom with 8 electrons, 8 protons, and 8 neutrons would be electrically neutral. However, if each of these particles has a slight differential charge, the maximum amount of charge the oxygen atom due to a charge differential between the 8 protons and electrons as well as the 8 neutrons is q = (8 + 8 + 8)(1.0 × 10−21)(1.60 × 10−19 C) = 3.8 × 10−39 C Assuming there are two such oxygen atoms, the ratio of the electric force to the graviational force is qq k 1 22 Fe kq 2 r = = Fg G m1m2 Gm 2 r2 where m is the mass of the electrons, protons, and neutrons in each oxygen atom. Inserting the values of the parameters gives: Fe kq 2 = Fg Gm 2 2 2 ⎛ 9 Nim ⎞ −39 ⎜ 8.99 × 10 ⎟ ( 3.8 × 10 C ) 2 C ⎠ ⎝

=

2 2 ⎛ −11 N i m ⎞ −31 −27 −27 ⎜ 6.67 × 10 ⎟ ⎣⎡8 ( 9.11 × 10 kg ) + 8 (1.673 × 10 kg ) + 8 (1.675 × 10 kg ) ⎤⎦ 2 kg ⎠ ⎝ −6 = 2.8 × 10 , The net force is attractive.

22-32.

The electric force provides the centripetal force to maintain the orbit of the electron in the classical orbital atomic model. mv 2 q2 =k 2 r r

v=

kq 2 = mr

2 2 ⎛ 9 Nim ⎞ −19 ⎜ 8.99 × 10 ⎟ (1.60 × 10 C ) 2 C ⎠ ⎝ = 2.2 × 106 m/s −31 −11 ( 9.11 × 10 kg )( 5.3 × 10 m )


CHAPTER

22

The orbital period is the circumference of the circle divided by the orbital speed. 2π r 2π (5.3 × 10−11 m) T = = = 1.5 × 10−16 s v 2.2 × 106 m/s †22-33.

22-34.

The situation is similar to example 6, +Q except that the negative charge on the y 0.5d axis has been replaced by a positive charge θ x Q. Study the drawing. The repulsive +q 0.5d electric force on the positive charge q due +Q to the upper charge Q is F1; and that due to the lower charge Q is F2. The net force is the sum of these two forces. The y components of these forces are equal in magnitude, but oppositely directed, so the y component of the net electric force on q is zero newtons. The x components of the two forces add together to give a net force F = F1 + F2 which is directed along the positive x axis with a magnitude Qq Qq x 2kqxQ F = 2k cos θ = 2k = 2 3/ 2 2 2 2 d 2 d 2 2 d d (2) + x ( 2 ) + x ( 2 ) + x ⎡⎣ 4 + x 2 ⎤⎦

F2 F1+F2

θ F1

The net force is directed in the +x direction. The three charges are positioned on the y axis, so the net force will be along the y axis. Let F1 be the force on the +10-µC charge due to the −40-µC charge and F2 be the force on the +10-µC charge due to the +40-µC charge. The magnitude of the net force on the +10 C charge is ⎡q q qq ⎤ F = F1 + F2 = k ⎢ 1 22 + 1 23 ⎥ r13 ⎦ ⎣ r12 ⎛ N i m2 = ⎜ 8.99 × 109 C2 ⎝

( +10 C )( +40 C ) ⎞ ⎡ ( +10 C )( −40 C ) + ⎟⎢ 3 2 (8.0 × 103 m) 2 ⎠ ⎢⎣ (3.0 × 10 m)

⎤ ⎥ ⎥⎦

= 3.4 × 105 N Because F1 has the larger magnitude and it is directed in the +y direction, the net force is also in 5 the +y direction. Therefore, the net force on the +10-C charge is F y = +3.4 × 10 N j

†22-35.

The two electric forces exerted on the electron due to the hydrogen nucleus (a proton) FH and the chlorine nucleus (17 protons) FCl are both attractive because the charges have opposite signs. Because each of these forces point in different directions, vector methods must be used to determine the net force on the electron. Consider the drawing. To determine the x and y components of FCl, the angle must be determined from the distances d1 and d2, which are given. ⎛d ⎞ ⎛ 5.0 × 10−11 m ⎞ θ = tan −1 ⎜ 1 ⎟ = tan −1 ⎜ ⎟ = 21.3° −10 ⎝ 1.28 × 10 m ⎠ ⎝ d2 ⎠


e

FCl

θ Cl

d2

F

FH d1 H

x

CHAPTER

22

The x component of the net force is the sum of the horizontal components of the forces FH and FCl. Fx = FH, x + FCl, x ⎛ N i m2 = ⎜ 8.99 × 109 C2 ⎝

⎤ ( −1.60 × 10−19 C ) (17 ) ( +1.60 × 10−19 C ) ⎞ ⎢⎡ ⎥ 0 cos 21.3 + ° ⎟ ⎥ (5.0 × 10−11 m) 2 + (1.28 × 10−10 m) 2 ⎠ ⎢⎣ ⎦

= 1.9 × 10−7 N

This component actually points in the −x direction as may be seen on the drawing. Similarly, the y component of the net force is Fy = FH, y + FCl, y ⎡ ( −1.60 × 10−19 C )( +1.60 × 10−19 C ) ⎤ ⎢ ⎥ −11 2 2 ⎢ ⎥ (5.0 × 10 m) ⎛ ⎞ i N m = ⎜ 8.99 × 109 ⎥ ⎟⎢ 2 19 19 − − C ⎠⎢ ⎝ ⎥ ( −1.60 × 10 C ) (17 ) ( +1.60 × 10 C ) + sin 21.3° ⎥ ⎢ 2 2 −11 −10 (5.0 × 10 m) + (1.28 × 10 m) ⎣⎢ ⎦⎥

= 1.7 × 10−7 N This component is directed in the −y direction. Therefore, the net electric force is F = −(1.9 × 10−7 N)i − (1.7 × 10−7 N)j

22-36.

For each of the five +Q charges, the contribution to the net electric force on the positive charge q is along a line between the two charges, Q and q, and directed away from the individual charge as shown. Each of the five contributions has the same magnitude. The five force vectors are equally distributed in angle around the center of the pentagon, so the angle between each vector is 72°. In the force diagram, the relevant angles with respect to the x and y axes are shown. We may now write down the x component of the net force on charge q as Fx = F1, x + F2, x + F3, x + F4, x + F5, x

+q +Q

+Q

+Q

Qq Qq Qq Qq ⎡ ⎤ = k ⎢ 0 − 2 cos18° − 2 sin18° + 2 sin18° + 2 cos18° ⎥ r r r r ⎣ ⎦ =0 Fy = F1, y + F2, y + F3, y + F4, y + F5, y Qq Qq Qq ⎡ Qq Qq = k ⎢ − 2 − 2 sin18° + 2 cos18° + 2 cos18° − 2 sin18° r r r r ⎣ r Qq Qq ⎡ Qq ⎤ = k ⎢ − 2 − 2 2 sin18° + 2 2 cos 36° ⎥ = 0 r r r ⎣ ⎦ where r is the distance from a vertex to the center of the pentagon. F = 0 N.


+Q

+Q y F3

36° F4

x 18° F 5 F2 72° F 18° 1

CHAPTER †22-37.

22

Each ball has three forces acting on it: the gravitational force of the Earth, the tension in the thread, and the electric force due to the other ball. Since each ball is in equilibrium, the sum of these forces is equal to zero newtons. The sum of forces in the vertical direction may be used to find the tension: ∑ Fy = T cos 20° − mg = 0

25 cm 10 cm

10 cm 20°

20° T

T

mg

mg

F21

mg cos 20° The sum of forces in the horizontal direction is Q2 F = T ° − k =0 sin 20 ∑ x r2 T =

F12

r 2T sin 20° r 2 mg sin 20° r 2 mg tan 20° = = k k cos 20° k The distance between the balls is r = (25 cm) + 2(10 cm)(sin 20°) = 31.84 cm = 0.3184 m. Q=

Q=

r 2 mg tan 20° = k

= 1.0 × 10 22-38.

−7

(0.3184 m) 2 (2.5 × 10−4 kg)(9.81 m/s 2 ) tan 20° 8.99 × 109

N i m2 C2

C

The x component of the net electric force is ⎡ Qq ⎤ 2Qq ⎥ θ θ Fx = k ⎢ cos cos − 2 2 ⎢⎣ ( d2 ) + x 2 ⎥⎦ ( d2 ) + x 2 x −Qq − kqQx =k = 2 2 3/ 2 2 2 d ( d2 ) + x 2 ( 2 ) + x ⎡⎣ d4 + x 2 ⎤⎦

+Q 0.5d 0.5d

−3Qq

( d2 )

2

+ x2

F1+F2

(d / 2)

( )

d 2 2

+x

θ

F1

F2

−2Q

The y component of the net electric force is ⎡ ⎤ Qq 2Qq ⎥=k − Fy = k ⎢ − sin sin θ θ 2 2 ⎢⎣ ( d2 ) + x 2 ⎥⎦ ( d2 ) + x 2

+q

θ

x

2

=

−3dkqQ 2 ⎡⎣ d4 + x 2 ⎤⎦ 2

3/ 2

Therefore,

F = F1 + F2 = − 22-39.

kqQx ⎡ d4 + x 2 ⎤ ⎣ ⎦ 2

3/ 2

i−

3dkqQ 2 ⎡⎣ d4 + x 2 ⎤⎦ 2

3/ 2

j

The x component of the net electric force is ⎡ Q2 Q2 2 ⎤ Fx = k ⎢ − 2 − 2 ⎥ 2L 2 ⎦ ⎣ L Q2 = −1.35k 2 L The y component of the net electric force is ⎡ Q2 Q2 2 ⎤ Q2 Fy = k ⎢ − 2 − 2 ⎥ = −1.35k 2 2L 2 ⎦ L ⎣ L


y F1

+Q L

+Q

−Q

F2

F3

L 2 L

+Q

x

CHAPTER

22

Therefore, Q2 Q2 i − 1.35 k j L2 L2 Due to the symmetry of the situation, the vertical components are equal in magnitude, but oppositely directed. Thus they cancel. The x components of each of these forces is directed parallel to the positive x axis. qQ 2 4 2kqQ = Fx = 4k 2 L 2 2 L2 F = F1 + F2 + F3 = −1.35k

22-40.

y +Q F2 L

F3

2

†22-41.

(

= 8.99 × 109

N i m2 C2

)

y

The magnitude of the y component of the net force is qe Fy = F1, y + F2, y = 0 + k 22 sin θ r2 −19 ⎞ ⎢⎡ ( −1.60 × 10 C ) ( −30 C ) ⎛ 6.0 km ⎞ ⎥⎤ ⎟ ⎟ 3 2 3 2 ⎜ ⎠ ⎣⎢ (4.0 × 10 m) + (6.0 × 10 m) ⎝ 52 km ⎠ ⎦⎥


F2 −e

F1

52 km

6.0 km

−30 C

x

−Q

F

+40 C

= −3.1 × 10−15 N

= 6.9 × 10−16 N Therefore, the net force on the electron is F = F1 + F2 = −(3.1 × 10−15 N)i + (6.9 × 10−16 N)j

F1

L

+Q

θ

4.0 km

⎡ ( −1.60 × 10−19 C ) ( +40 C ) ⎤ ⎢ ⎥ ⎢ ⎥ (4.0 × 103 m) 2 ⎢ ⎥ ⎢ ( −1.60 × 10−19 C ) ( −30 C ) ⎛ 4.0 km ⎞ ⎥ + ⎢ ⎥ (4.0 × 103 m) 2 + (6.0 × 103 m) 2 ⎜⎝ 52 km ⎟⎠ ⎥⎦ ⎢⎣

⎛ N i m2 = ⎜ 8.99 × 109 C2 ⎝

L 2 2

+q

( )

Therefore, the net force on the charge +q is 4 2kqQ F = F1 + F2 + F3 + F4 = i L2 Consider the drawing indicating the forces on the electron. The electric force F1 on the electron due to the positive charge is directed toward the positive charge, since it has an opposite sign. The force F2 exerted on the electron due to the negative charge is directed away from the negative charge, since both charges have the same sign. The net force on the electron is the vector sum of these two forces. The magnitude of the x component of the net force is qe qe Fx = F1, x + F2, x = k 12 + k 22 cos θ r1 r2

−Q

F4

x

CHAPTER 22-42.

22

Consider the drawing indicating the forces on the electron. The electric force F1 on the electron due to the positive charge is directed toward the positive charge, since it has an opposite sign. The force F2 exerted on the electron due to the negative charge is directed away from the negative charge, since both charges have the same sign. The net force on the electron is the vector sum of these two forces. The magnitude of the x component of the net force is qe qe ke(q1 + q2 ) Fx = F1, x + F2, x = k 12 cos θ + k 22 cos θ = cos θ r1 r2 r

y +40 C 3.0 km

−30 C

F F1

θ

θ

4.0 km

F2 −e 3.0 km x

−19 ⎛ N i m 2 ⎞ ( −1.60 × 10 C ) ( + 40 C − 30 C ) km ⎤ = ⎜ 8.99 × 109 cos ⎡⎣ tan −1 ( 3.0 ⎟ 4.0 km ) ⎦ 2 3 2 3 2 C ⎠ (4.0 × 10 m) + (3.0 × 10 m) ⎝ = 4.6 × 10−16 N The magnitude of the y component of the net force is qe qe ke(q1 + q2 ) Fy = F1, y + F2, y = k 12 sin θ + k 22 sin θ = sin θ r1 r2 r 2 ( −1.60 × 10−19 C ) ( +40 C + −30 C ) ⎡ −1 3.0 km ⎤ ⎛ 9 Nim ⎞ sin ⎣ tan ( 4.0 km ) ⎦ = ⎜ 8.99 × 10 ⎟ C2 ⎠ (4.0 × 103 m) 2 + (3.0 × 103 m) 2 ⎝ = 2.4 × 10−15 N Therefore, the net force on the electron is F = F1 + F2 = −(4.6 × 10−16 N) i + (2.4 × 10−15 N) j

22-43.

Consider the drawing. The ball on the left has a charge q1 = +2.0 × 25° 25° 10−7 C and the ball on the right has a charge q2 = +6.0 × 10−8 C. There are three forces acting on each ball: the gravitational force of the 10 cm 10 cm Earth, the tension in the thread, and the electric force due to the other ball. Since each ball is in equilibrium, the sum of these forces is equal T2 F12 to zero newtons. Because the angles are the same for both balls and F21 T1 the symmetry of the situation, the tension in both threads is identical. m1g m2g Because of this, the vertical component of the tension is the same for both balls. Thus, the mass of each ball is the same. Therefore, finding the mass of one ball gives the mass of the other. The distance between the two balls is r = 2(10 cm)sin 25°. The sum of forces in the horizontal direction for ball 1 is: qq ∑ Fx = T1 sin 25° − k r1 2 2 = 0 from which the tension in the left thread can be determined. 2 +2.0 × 10−7 C )( +6.0 × 10−8 C ) ⎛ q1q2 9 Nim ⎞ ( T1 = k 2 = ⎜ 8.99 × 10 = 1.28 × 10−3 N ⎟ 2 r sin 25° ⎝ C2 ⎠ ( 2(0.10 m) sin 25° ) The sum of the forces in the vertical direction for ball 1 is:


CHAPTER

∑F

y

22

= T1 cos 25° − m1 g = 0

−3 T1 cos 25° (1.28 × 10 N ) cos 25° m1 = = = 1.2 × 10−4 kg g 9.81 m/s 2

m1 = m2 = 1.2 × 10−4 kg 22-44.

Consider the three charges at the vertices of an equilateral triangle with sides of length a. The net force on the positive charge +q placed at the origin is to be determined. The x component of the net force on this charge is q2 q2 q2 q2 Fx = k 2 − k 2 cos 60° = k 2 = k 2 cos 60° 2a a a a and the y component of the net force on this charge is q2 q2 Fy = 0 − k 2 sin 60° = k 2 sin 60° a a The magnitude of the net force is 2

a 60° 60° a +q F−

F+

2 y

(

)

(

−q a

)

(

⎡ qq0 = k⎢ ⎢ 2 a 1 + 23 ⎣⎢

(

q = 2 2 1+ Q

(

3 2

)

−

) ( 2

2

)

(

Qq0 2 2

q0

)

) (

(

a

)

2

)

Qq0 2 − 2 2 2 a

(

a

60° 0.5a

+Q

In the y direction, The distance between the −q charge and q0 is a rqq0 = + a cos 30° = a 1 + 23 2 ⎡ ⎤ qq0 Qq0 Qq0 ⎥ Fy = k ⎢ − ° − ° sin 45 sin 45 2 2 2 ⎢ 2 ⎥ 3 2 2 2 a 2 a ⎢⎣ a 1 + 2 ⎥⎦

22-46.

−q

2

⎛ q2 ⎞ ⎛ q2 ⎞ q2 F = F + F = ⎜ k 2 cos 60° ⎟ + ⎜ k 2 sin 60° ⎟ = k 2 a ⎝ 2a ⎠ ⎝ a ⎠ The sum of electric forces acting on the charge q0 is zero newtons. Therefore, the x component of the net force must be equal to zero. Qq0 Qq0 cos 45° − k cos 45°= 0 Fx = k 2 2 2 2 2 a 2 a 2 x

†22-45.

+q

)

2

⎤ ⎡ qq0 2⎥ = k⎢ ⎥ ⎢ 2 3 2 ⎢⎣ a 1 + 2 ⎦⎥

(

)

2

⎤ 4Qq0 ⎥ − =0 2a 2 ⎥ ⎦⎥

= 9.85

All of the charges are alike, so all of the forces exerted on one of the corner charges by the others are directed away from the charge. Three of the charges are located a distance a away. Three charges are located a distance a 2 away; and one charge is a 3. Therefore, the net force on the corner charge is


x y

z

60° 0.5a

+Q

CHAPTER

22

⎡ q2 2q 2 q2 2 ⎤ kq 2 2 + Fx = Fy = Fz = k ⎢ 2 + ⎥= 3a 2 (a 2) 2 (a 3) 2 2 ⎦ ⎣a 2

F = 22-47.

22-48.

⎛ kq 2 2 ⎞ kq 2 3 2 kq 2 ⎛ 6 ⎞ F + F + F = 3 ⎜⎜ = = 2 ⎜⎜ ⎟⎟ ⎟ 2 3a 2 a ⎝ 3 ⎟⎠ ⎝ 3a ⎠ 2 x

2 y

2 z

⎡ −Qq ⎡1 ⎤ Qq ⎤ 1 F = k⎢ + 2 ⎥ = kQq ⎢ 2 − 2 2 ⎥ x ⎦ ( d + x) ⎦ ⎣ (d + x) ⎣x

−d/2

x

d/2

−Q

+Q

+q

As x becomes very large, the d term becomes negligible; and F goes to zero. Thus, it appears to the charge q that the other two charges, which are equal in magnitude, but opposite in sign, are really one neutral object. Both the charge +q and the line of charge are L d positive, so the net electric force on charge q will be dq′ directed to the left, the negative x direction. Consider q x an infinitesimally small portion of the rod with a charge dq′ = λdx. The magnitude of the force on charge q due to dq′ is dq′ λ dx dF = kq 2 = kq 2 r r The distance from the infinitesimal charge to charge q depends on how far it is from the end of the charged rod, x. From the drawing, we find r = d + L − x. Using this information, we may find the magnitude of the total force on charge q by integrating over the length of the rod. L λ dx F = kq ∫ 0 (d + L − x)2 Let u = d + L − x, then du = −dx and the limits of integration accordingly change. F = − kqλ ∫

22-49.

d

du 1 ⎞ ⎛1⎞ ⎛1 = + kqλ ⎜ ⎟ = + kqλ ⎜ − ⎟ 2 d +L u ⎝ u ⎠ d +l ⎝d d + L⎠ d

The tetrahedron is a four-sided pyramid with z each side an equilateral triangle. All of the y sides of the tetrahedron have length a; and x any two sides that meet at a vertex do so with a a 60° angle between them. The net force on a the charge shown on the lower left has components in the x, y, and z directions. Let F1 the base of the pyramid lie in the x-y plane, +Q then F2 2 F3 Q F1 = −k 2 j a Q2 Q2 F2 = k 2 (sin 60°)i − k 2 (cos 60°) j a a 2 Q Q2 Q2 F3 = k 2 (sin 30°)(cos 60°)i − k 2 (cos 30°)(cos 60°) j − k 2 (sin 60°)k a a a


+Q

+Q a +Q

CHAPTER

22-50.

Vector adding these three forces together gives, Q2 F = F1 + F2 + F3 = k 2 ( sin 60°+ ( sin 30° ) (cos 60°) ) i − a 2 Q Q2 k 2 (1+ cos 60°+ (sin 30°)(cos 60°) ) j − k 2 (cos 60°)k a a 2 kQ = 2 (1.116i − 1.75 j − 0.5k ) a Both lines of charge are positive, so the electric L force on the left rod will be directed to the left, the dq2 +++ negative x direction. The ends of the rod nearest x2 each other are separated by a constant distance x. Consider an infinitesimally small portion of the rod with a charge dq1 = λ dx1, where λ = Q/L. On the other rod, there is an identical infinitesimal amount of charge dq2 = λ dx2. The magnitude of the force on dq2 due to dq1 is dq dq λ 2 dx1dx2 dF = k 1 2 2 = k r ( x1 + x2 + x) 2

x

22

L dq1 x1

+++

To find the total repulsive force of one rod on the other, a double integral is required. F = kλ 2 ∫

L

0

∫

L

0

L

L dx1dx2 1 = kλ 2 ∫ − dx2 2 0 ( x1 + x2 + x) x1 + x2 + x 0

L⎡ L ⎤ 1 1 = kλ 2 ∫ ⎢ − dx2 = k λ 2 ⎡⎣ ln ( x2 + x ) − ln ( L + x2 + x ) ⎤⎦ ⎥ 0 0 ⎣ x2 + x L + x2 + x ⎦ = k λ 2 ⎡⎣ ln ( L + x ) − ln ( x ) − ln ( 2 L + x ) + ln ( L + x ) ⎦⎤

2

⎛Q⎞ ⎛ L+ x⎞ ⎛ L+x ⎞ = k ⎜ ⎟ ln ⎜ ⎟ − ln ⎜ ⎟ ⎝L⎠ ⎝ x ⎠ ⎝ 2L + x ⎠

22-51.

The x component of the net electric force is ⎡ 2Qq ⎤ ⎥ θ Fx = k ⎢ cos 2 ⎢⎣ ( d2 ) + x 2 ⎥⎦ x 2Qq 2kqQx =k = 2 2 3/ 2 2 2 d 2 ( d2 ) + x ( 2 ) + x ⎡⎣ d4 + x 2 ⎤⎦ The y component of the net electric force is ⎡ ⎤ Qq Qq ⎥=0 sin θ + sin θ Fy = k ⎢ − 2 2 ⎢⎣ ( d2 ) + x 2 ⎥⎦ ( d2 ) + x 2 Therefore, 2kqQx F = F1 + F2 = 2 i 3/ 2 ⎡ d4 + x 2 ⎤ ⎣ ⎦


+Q 0.5d 0.5d +Q

x

F2

θ +q

θ

F1

F1+F2

CHAPTER

22

The magnitude of F is x F = −2kqQ 2 3/ 2 ⎡ d4 + x 2 ⎤ ⎣ ⎦ To find the maximum value of F, we take the derivative and equate it to zero. ⎡ ⎤ ⎡ ⎤ −5 / 2 dF d ⎢ x 1 6 x2 d 2 2 ⎥ = −2kqQ ⎢ ⎥=0 ⎡ ⎤ = −2kqQ 2 − + x 3/ 2 ⎦ 2 ⎥ ⎢ ⎡ d 2 + x2 ⎤3 / 2 ⎥ d dx dx ⎢ 2 ⎣4 ⎡ ⎤ + x ⎢⎣ ⎢⎣ ⎣ 4 ⎥⎦ ⎣4 ⎦ ⎥⎦ ⎦ d2 = 0.35d 8 (a) The x component of the net electric force is ⎡ −2Qq ⎤ ⎥ Fx = k ⎢ cos θ 2 ⎢⎣ ( d2 ) + x 2 ⎥⎦ x −2Qq −2kqQx =k = 2 2 3/ 2 2 2 d 2 ( d2 ) + x ( 2 ) + x ⎡⎣ d4 + x 2 ⎤⎦ x=

22-52.

+Q

F1+F2

0.5d 0.5d

x

θ θ

F1 F2

−q

+Q

The y component of the net electric force is ⎡ ⎤ Qq Qq ⎥=0 sin θ + sin θ Fy = k ⎢ − 2 2 ⎢⎣ ( d2 ) + x 2 ⎥⎦ ( d2 ) + x 2 Therefore, −2kqQx F = F1 + F2 = 2 i 3/ 2 ⎡ d4 + x 2 ⎤ ⎣ ⎦ (b) The magnitude of F is x F = −2kqQ 2 3/ 2 ⎡ d4 + x 2 ⎤ ⎣ ⎦ If x is small, then as it is squared in the denominator, the x2 term becomes negligible and we have x −2kqQ F = −2kqQ 2 3 / 2 = x d 3 ⎡ d4 ⎤ ( ) 2 ⎣ ⎦ which is proportional to x. (c) A force that is proportional to x is similar to that of a spring force. Using the answer from part (b), we have −2kqQ F = Ax where A = 3

( d2 )

The period of the simple harmonic motion that results is

22-53.

−1 / 2

−1 / 2 ⎛ 2kqQ ⎞ 2π kqQ ⎞ ⎛ ⎟ T = = = 2π ⎜ = 4π ⎜ 3 ⎟ ⎜ m ( d )3 ⎟ ω A/ m ⎝ md ⎠ 2 ⎝ ⎠ Each of the following processes is impossible because charge is not conserved in them. p + p → n + n + π+ (e + e ≠ 0 + 0 + e) (e + e ≠ 0 + e + 0) p + p → n + p + π0 0 − p + p → n + p + π + π (e + e ≠ 0 + e + 0 + (− e))

2π


CHAPTER 22-54.

†22-55. 22-56. †22-57.

22-58. †22-59.

22

The net charge of the left hand side of the reaction is +2e, since the water molecule is electrically neutral. The right hand side of the reaction is (−2e) + 8e + ne. If the right hand side is to balance the left hand side, n = 4. The left hand side of the cobalt reaction has charges: 4e + (−Ne) or (4 − N)e. The right hand side shows that the Co atom with a charge +3e. Setting the two sides equal to each other gives N = 1. Qouter = Qtotal − Qinner = (−1.0 × 10−8 C) − (+2.0 × 10−8 C) = −3.0 × 10−8 C For every silver ion produced, there must also be an electron produced. To deposit a mole of silver would require a mole of electrons. For the deposition of 1.0 g of silver, the number of silver ions produced is ⎛ 1 mole ⎞ 23 ions n = (1.0 g ) ⎜ = 5.6 × 1021 ions ⎟ 6.02 × 10 107.9 g mole ⎝ ⎠

Therefore, 5.6 × 1021 electrons flow during the deposition. 6.0 × 10−11 C )( −6.0 × 10−11 C ) 2 ( q1q2 9 Nim F = k 2 = 8.99 × 10 C2 = 8.1 × 10−10 N 2 r ( 0.20 m )

(

)

If Fg = Fe, then mm qq G 1 2 2 = k 1 22 r r Because the masses of the two grains are equal to m and each has a charge q, we can multiply both sides by r2 and make the appropriate subtitutions to get Gm 2 = ke 2 2

8.99 × 109 NCi m2 ke 2 k −19 =e = 1.60 × 10 C = 1.9 × 10−9 kg −11 N i m 2 G G 6.67 × 10 kg 2

m= 22-60.

The torque due to the electric force between the probe and the end of the arm is

(

q2 τ = Fe l = k 2 l = 8.99 × 109 r †22-61.

N i m2 C2

)

( 2.0 × 10

−9

C)

(0.030 m) 2

2

( 0.15 m ) = 6.0 × 10−6

From the geometry of an equilateral triangle of sides of length a, the distance from each of the vertices to the center is a d = = 0.577 a 2 cos 30° Because all of the charges at the vertices are equal, each exerts an equal force (in magnitude) on the charge at the center. The sum of these forces must equal zero, if the system of charges is to remain in equilibrium. The sum of the x components and y components is ∑ Fx = F2 cos 30° − F1 cos 30° = 0

∑F

y

= F1 − F2 sin 30° − F3 sin 30° = 0


Nim

+Q

y x a

F1 F3

+Q

a

a

−q F2 +Q

CHAPTER

22

The magnitude of each of these three forces is Qq Qq F =k 2 =k 2 r ( 0.577a )

22-62.

The geometry of this situation is such that any negative charge q will satisfy the equilibrium requirement. Each ball has three forces acting on it: the gravitational force of 20 cm the Earth, the tension in the thread, and the electric force due to the other ball. Since each ball is in equilibrium, the sum of these forces is equal to zero newtons. θ 10 cm 10 cm θ The angle that each thread makes with the vertical is ⎛ 5 cm ⎞ θ = sin −1 ⎜ ⎟ = 30° T F12 F21 T ⎝ 10 cm ⎠ The sum of forces in the vertical direction may be used to find mg mg 10 cm the tension: ∑ Fy = T cos 30° − mg = 0 mg cos 30° The sum of forces in the horizontal direction is Q2 F T k = sin 30 ° − =0 ∑ x r2 T =

Q= =

22-63.

r 2T sin 30° = k

r 2 mg sin 30° k cos 30°

(0.10 m) 2 ( 2.0 × 10−4 kg )( 9.81 m/s 2 ) tan 30° 8.99 × 10

(

Q2 F = k 2 = 8.99 × 109 r

N i m2 C2

)

9 N i m2 C2

( 2.0 × 10

−11

C)

( 0.010 m )

= 3.5 × 10−8 C

2

2

= 3.6 × 10−8 N

To determine the acceleration, the mass must be determined. For each drop the mass is m = ρV = ρ ( 43 π R 3 ) a= 22-64.

F F = = 4 m ρ ( 3 π R3 )

(

3.6 × 10−8 N 1000 kg 1 m3

)

4 3

π (0.0005 m)

3

= 6.9 × 10−2 m/s 2

The total charge each hour is 3600 s Q = (1.0 × 10−4 C/s ) = 0.36 C/h 1h The total number of electrons flowing into the rod in the hour is Q 0.36 C n= = = 2.3 × 1018 electrons e 1.60 × 10−19 C


CHAPTER

22-65.

F =k

Q2 r2

Q=

Fr 2 = k

( 0.30 N )( 0.18 m ) 8.99 × 10

9 N i m2 C2

22

2

= 1.0 × 10−6 C

The number of electrons in excess on one ball an in deficit on the other is 1.0 × 10−6 C Q = = 6.5 × 1012 electrons n= e 1.60 × 10−19 C 22-66.

The distance between the centers of the balls is r = (0.20 m)sin(22.5°) = 0.077 m

45° 0.10 m

The sum of vertical forces acting on the elevated ball is ∑ Fy = F12 sin 22.5° − mg = 0 F12 =

mg Q2 =k 2 sin 22.5° r 2

Q= †22-67.

0.10 m

mgr = k sin 22.5°

T r

(1.5 × 10

−4

kg )( 9.81 m/s

8.99 × 10

9 Nim C2

2

2

) ( 0.077 m )

( sin 22.5° )

F21

F12 mg

2

= 5.0 × 10−8 C

n → p + e + ? On the left side the neutron is neutral, therefore, the total charge on the right of the arrow must add to zero. Because the proton and electron have charges +e and −e, respectively, the charge on the missing particle must be equal to zero. ∆ ++ → p + π 0 + ? The particle on the left has a charge of +2e. On the right, the proton is +e the π particle is neutral. The missing particle must have a charge of +e to give a total charge on the right equal to that on the left. ∆ + → n + ? The particle on the left has a charge of +e. On the right, the neutron has zero charge, so the missing particle must have a charge of +e to give a total charge on the right equal to that on the left. π − → µ − + ? The π and µ particles both have a charge −e, so the missing particle must have zero charge.


CHAPTER 23

THE ELECTRIC FIELD


23-2.

The electron has a charge q = −1.60 × 10−19 C. Because it has a charge, the electron experiences a force F = qE when it enters an electric field E. In the present situation, the magnitude of the force is F = qE = (1.60 × 10−19 C)( 3.4 × 105 N/C) = 5.44 × 10−14 N = 5.4 × 10−14 N The electric force on the electron causes it to have an acceleration F 5.44 × 10−14 N a= = = 6.0 × 1016 m/s 2 m 9.11 × 10−31 kg F = qE = ma −31 16 2 m a ( 9.11 × 10 kg )( +1.0 × 10 m/s i ) E= = = −5.7 × 104 N/C i −1.60 × 10−19 C q

†23-3.

23-4.

The direction of acceleration of the electron is opposite that of the electric field. E x xˆ Example 2 gives the components of the electric field: θ 3 Ex = 7.3 × 10 N/C Ey = −2.7 × 103 N/C E The components are drawn to scale. From the drawing, the direction of E may be determined using the components. E ⎛ −2.7 × 103 N/C ⎞ θ = tan −1 y = tan −1 ⎜ ⎟ = −20 ° or 20° below the horizontal axis 3 Ex ⎝ 7.3 × 10 N/C ⎠

E y yˆ

The sum of the vertical forces acting on the dust grain is Fy = Fe + Fg = qE − mg = ma −1.60 × 10−19 C ) ( −100 N/C ) ( qE a= −g = − 9.81 m/s 2 = 6.2 m/s 2 1.0 × 10−18 kg m

a = +6.2 m/s 2 j †23-5.

To balance the downward force of gravity on the drop, an upward electric force due to the electric field is required. Because Fe = qE and the charge an the electron is negative, the electric field must also point downward to produce an upward force on the electron. The sum of the vertical forces acting on the drop is Fy = Fe + Fg = qE − mg = 0

E=

mg ρVg = = q q

⎛4 ⎝3

⎞ ⎠

ρ ⎜ π R3 ⎟ g q

( 8.0 × 10 kg/m ) 43 π (1.0 × 10 2

=

3

−1.60 × 10

E = −2.1 × 105 N/C j


−19

−6

C

m ) ( 9.81 m/s 2 ) 3

= −2.1 × 105 N/C

CHAPTER 23-6.

†23-7.

23

F = qE = (1.60 × 10−19 C)( 8.0 × 105 N/C) = 1.28 × 10−13 N = 1.3 × 10−13 N The electric force on the electron causes it to have an acceleration F 1.28 × 10−13 N a= = = 1.4 × 1017 m/s 2 m 9.11 × 10−31 kg

The downward-directed gravitational force on the proton at the surface of the neutron star would be (1.4 × 1030 kg )(1.67 × 10−27 kg ) = 1.6 × 10−15 N Mm Fg = G 2 = ( 6.67 × 10−11 N i m 2 /kg 2 ) 2 r (1.0 × 104 m ) The upward-directed electric force on the proton is Fe = qE = (1.60 × 10−19 C)( 6.0 × 103 N/C) = 9.6 × 10−16 N Taking the ratio of the two forces, we find that Fe = 0.6 Fg

23-8.

†23-9.

Fy = Fe + Fg = qE − mg = 0 −6 2 mg (1.2 × 10 kg )( 9.81 m/s ) E= = = 9.1 × 103 N/C −9 q 1.3 × 10 C 1 The kinetic energy of the electron is K = mv 2 , so the speed of the electron with that energy is 2 given by 2K v= m The electric force that accelerates the electron is Fe = qE = ma, so the acceleration of the electron is qE a= m To determine the distance traveled by the electron as it achieves the kinetic energy indicated above may be determined by applying the kinematic equation, v 2 − v02 = 2a ( x − x0 )

x − x0 = 23-10.

v 2 − v02 2K / m K 3.0 × 10−19 J = = = = 6.3 × 10−7 m 6 −19 2a 2qE / m qE (1.60 × 10 C )( 3.0 × 10 N/C )

Fy = Fe + Fg = qE − mg = 0 −27 2 mg mg 56 (1.67 × 10 kg )( 9.81 m/s ) E= = = = 2.21 × 10−7 N/C q Ze 26 (1.60 × 10−19 C )

23-11.

23-12.

⎡ ⎤ −19 q C ⎥ 9 2 2 ⎢ 1.60 × 10 = 8.99 × 10 N i m /C = 5.1 × 1011 N/C E= 2 2 11 − ⎢ ( 5.3 × 10 m ) ⎥ 4πε 0 r ⎣ ⎦ ⎡ 92 (1.60 × 10−19 C ) ⎤ 1 q 9 2 2 ⎢ ⎥ = 2.4 × 1021 N/C = i E= 8.99 × 10 N m /C ⎢ ( 7.4 × 10−15 m )2 ⎥ 4πε 0 r 2 ⎣ ⎦ 1


CHAPTER

23

1 ⎛ q1 q2 ⎞ ⎜ − ⎟ 4πε 0 ⎝ r12 r22 ⎠ ⎡ 35 (1.60 × 10−19 C ) 19 (1.60 × 10−19 C ) ⎤ 9 2 2 ⎢ ⎥ = 5.1 × 1012 N/C = 8.99 × 10 N i m /C − 2 2 11 10 − − ⎢ ( 9.3 × 10 m ) (1.89 × 10 m ) ⎥⎦ ⎣

23-13.

E = E1 + E2 =

23-14.

E = E1 + E2 =

1 ⎛ q1 q2 ⎞ ⎜ + ⎟ 4πε 0 ⎝ r12 r22 ⎠

⎡ (1.60 × 10−19 C ) 1.60 × 10−19 C ) ⎤ ( ⎢ ⎥ = 8.99 × 10 N i m /C + ⎢ (1.05 × 10−10 m )2 (1.05 × 10−10 m )2 ⎥ ⎣ ⎦ 11 = 2.6 × 10 N/C, directed toward the electron 9

†23-15.

2

2

The electric field from a positive charge is directed radially outward from the charge in all directions. Only the electric field line that passes through the point P is shown, however. Using the Pythagorean theorem, the distance from the charge to point P is r=

( 0.15 m )

2

E P

r

+ ( 0.10 m ) = 0.18 m 2

The magnitude of the electric field at P is −10 ⎛ 1 q C⎞ 9 2 2 1.0 × 10 E= 8.99 × 10 N m /C i = ⎜ ⎟ = 28 N/C 2 2 ⎜ ( 0.18 m ) ⎟ 4πε 0 r ⎝ ⎠ 23-16.

E = E1 + E2 + E3 = E=

†23-17.

1 ⎛ q1 q2 q3 ⎞ + ⎟ ⎜ + 4πε 0 ⎝ r12 r22 r32 ⎠

⎞ 1 ⎛ Q 2Q Q Q + 2 − = ⎜− 2 2 ⎟ 4πε 0 ⎝ ( x + d ) x ( x − d ) ⎠ 4πε 0

⎛ ⎞ 1 2 1 + 2 − ⎜− 2 2 ⎟ x (x − d ) ⎠ ⎝ (x + d )

We’ll find the electric field at each of the mid-points labeled A, B, C, and D due to the contributions to the field from the four charges at the corners of the square. To calculate the magnitude of each of the contributions, the distance from each charge to a given point must be determined. For each of the midpoints, there are two charges at a distance of r1 = L/2 and two charges at a distance of

A D

B C

2

L 5 ⎛ L⎞ r2 = ⎜ ⎟ + L2 = 2 ⎝2⎠ The angle θ will also be useful in determining the components of the electric force contributions.

L

θ

r2

L/2


CHAPTER

23

⎛ L/2⎞ −1 ⎛ 1 ⎞ ⎟ = tan ⎜ ⎟ = 26.57° L ⎝ ⎠ ⎝ 2⎠

θ = tan −1 ⎜

L/2 L/2 1 5 = = = 1 r2 5 5 L 5 2 2 2 5 L L cos θ = = = = 1 5 r2 5 L 5 2 To determine the electric field, remember that at a given point the direction of the electric field due to an individual charge will point toward that charge, if the charge is negative, and away from the charge, if it is positive. In the following, take note of the signs, which take into account the above fact. For point A: EA = E1 + E 2 + E3 + E 4 sin θ =

and

=

=

1 4πε 0

Q 4πε 0

⎧ ⎡ ⎪ ⎢ Q ⎪ Q ⎢ −Q sin θ + + i i ⎨ 2 2 2 ⎢⎛ 1 ⎛1 ⎞ ⎞ ⎪⎛ 1 L⎞ ⎢⎜ L 5 ⎟ ⎟ ⎜ L⎟ ⎪⎜ ⎝2 ⎠ ⎠ ⎣⎢ ⎝ 2 ⎩⎝ 2 ⎠ ⎤ ⎡ ⎡ ⎢ ⎥ ⎢ ⎢ 2 − 2 sin θ ⎥ i = 1 Q ⎢ 2 − 2 ⎢ L2 ⎛ 1 4πε 0 L2 ⎢ ⎞ ⎥ L 5 ⎢ ⎥ ⎢ ⎜ ⎟ ⎢⎣ ⎢⎣ ⎝2 ⎠ ⎥⎦

For point B: ⎧ ⎡ ⎪ ⎢ Q ⎪ Q ⎢ −Q cos θ i + EB = k ⎨ j j + + 2 2 2 ⎢⎛ 1 ⎛1 ⎞ ⎞ ⎪⎛ 1 L⎞ ⎢⎜ L 5 ⎟ ⎜ L⎟ ⎪ ⎜⎝ 2 ⎟⎠ ⎝2 ⎠ ⎠ ⎣⎢ ⎝ 2 ⎩

i+

Q cos θ ⎛1 ⎞ ⎜ L 5⎟ ⎝2 ⎠

2

⎤ ⎡ ⎥ ⎢ −Q sin θ j⎥ + ⎢ i+ 2 ⎥ ⎢⎛ 1 ⎞ ⎥ ⎢⎜ L 5 ⎟ ⎠ ⎦⎥ ⎣⎢ ⎝ 2

⎤ 2 5 ⎥ 2 5 ⎥ i = ⎛⎜ 1.15 × 1010 N i m 2 C2 ⎛1 ⎞ ⎥ ⎝ 5 ⎥ ⎜ ⎟ ⎝2 ⎠ ⎥⎦

−Q sin θ ⎛1 ⎞ ⎜ L 5⎟ ⎝2 ⎠

2

⎤ ⎡ ⎥ ⎢ Q cos θ j⎥ + ⎢ i+ 2 ⎥ ⎢⎛ 1 ⎞ ⎥ ⎢⎜ L 5 ⎟ ⎠ ⎦⎥ ⎣⎢ ⎝ 2

www.elsolucionario.net

⎛1 ⎞ ⎜ L 5⎟ ⎝2 ⎠

2

⎞Q ⎟ 2i ⎠L

−Q sin θ ⎛1 ⎞ ⎜ L 5⎟ ⎝2 ⎠

⎡ ⎤ ⎡ ⎤ 2 ⎢ ⎥ ⎢ 5 ⎥ 2 2 2 sin θ ⎥ kQ ⎢ 5 ⎥ j = ⎛⎜ 1.15 × 1010 N i m ⎞⎟ Q j = kQ ⎢ 2 − j = 2 − 2 2 ⎢L L2 ⎢ C 2 ⎠ L2 ⎛1 ⎞ ⎥ ⎛1 ⎞ ⎥ ⎝ 5⎟ ⎥ ⎢ ⎢ ⎜ L 5⎟ ⎥ ⎜ ⎢⎣ ⎝2 ⎠ ⎦⎥ ⎝2 ⎠ ⎦⎥ ⎣⎢

483

−Q cos θ

2

⎤⎫ ⎥⎪ ⎪ j⎥ ⎬ ⎥⎪ ⎥⎪ ⎦⎥ ⎭

⎤⎫ ⎥⎪ ⎪ j⎥ ⎬ ⎥⎪ ⎥⎪ ⎦⎥ ⎭

CHAPTER

23

For point C: ⎧ ⎡ ⎪ ⎢ −Q ⎪ −Q ⎢ Q sin θ i + EC = k ⎨ i i + + 2 2 2 ⎢⎛ 1 ⎛1 ⎞ ⎞ ⎪⎛ 1 L⎞ ⎢⎜ L 5 ⎟ ⎟ ⎜ L⎟ ⎪⎜ ⎝2 ⎠ ⎠ ⎣⎢ ⎝ 2 ⎩⎝ 2 ⎠

Q cos θ ⎛1 ⎞ ⎜ L 5⎟ ⎝2 ⎠

2

⎤ ⎡ ⎥ ⎢ Q sin θ j⎥ + ⎢ i+ 2 ⎥ ⎢⎛ 1 ⎞ ⎥ ⎢⎜ L 5 ⎟ ⎠ ⎦⎥ ⎣⎢ ⎝ 2

−Q cos θ ⎛1 ⎞ ⎜ L 5⎟ ⎝2 ⎠

2

⎡ ⎤ ⎡ ⎤ 2 ⎢ ⎥ ⎢ 5 ⎥ 2 2 2 sin θ ⎥ kQ ⎢ 5 ⎥ i = − ⎜⎛ 1.15 × 1010 N i m ⎟⎞ Q i = kQ ⎢ − 2 + i = − 2 + 2 2 ⎢ L L2 ⎢ C2 ⎠ L2 ⎛1 ⎞ ⎥ ⎛1 ⎞ ⎥ ⎝ 5⎟ ⎥ ⎢ ⎢ ⎜ L 5⎟ ⎥ ⎜ ⎝2 ⎠ ⎦⎥ ⎝2 ⎠ ⎦⎥ ⎣⎢ ⎣⎢ For point D: ⎧ ⎡ ⎤ ⎡ ⎪ ⎢ ⎥ ⎢ −Q ⎪ −Q ⎢ −Q cos θ i + Q sin θ j⎥ + ⎢ Q cos θ i + Q sin θ ED = k ⎨ j j + + 2 2 2 2 2 2 ⎢⎛ 1 ⎛1 ⎞ ⎞ ⎛1 ⎞ ⎥ ⎢⎛ 1 ⎞ ⎛1 ⎞ ⎪⎛ 1 L⎞ L L 5 L 5 L 5 L 5 ⎢⎜ ⎜ ⎟ ⎟ ⎜ ⎟ ⎥ ⎢⎜ ⎟ ⎜ ⎟ ⎪ ⎜⎝ 2 ⎟⎠ ⎢⎣ ⎝ 2 ⎝2 ⎠ ⎠ ⎝2 ⎠ ⎥⎦ ⎢⎣ ⎝ 2 ⎠ ⎝2 ⎠ ⎩

23-18.

⎡ ⎤ ⎡ ⎢ ⎥ ⎢ 2 2 sin θ ⎥ kQ ⎢ = kQ ⎢ − 2 + = −2 + j 2 ⎢ L L2 ⎢ ⎛1 ⎞ ⎥ ⎢ ⎢ ⎜ L 5⎟ ⎥ ⎢⎣ ⎢⎣ ⎝2 ⎠ ⎥⎦ Consider the drawings. This construction is called a kite quadrilateral in geometry. To determine the electric field contributions, the angle θ must be determined. AB = 9.58 × 10−11 m A 1 1 ∠BAE = ∠BAD = (105°) = 52.5° 2 2 AE = AB sin 52.5°

⎤ 2 5 ⎥ 5 ⎥j= 2 ⎛1 ⎞ ⎥ 5⎟ ⎥ ⎜ ⎝2 ⎠ ⎥⎦

AC = AE + EC = 1.20 × 10−10 m

⎛ N i m2 − ⎜ 1.15 × 1010 C2 ⎝

EH2

θ E

θ θ C

D

= 1.20 × 10−10 m − ( 9.58 × 10−11 m ) sin 52.5° = 4.40 × 10−11 m

BE = DE = AB cos 52.5° = ( 9.58 × 10−11 m ) cos 52.5° = 5.83 × 10−11 m

⎛ 5.83 × 10−11 m ⎞ BE = tan −1 ⎜ ⎟ = 52.96° −11 EC ⎝ 4.40 × 10 m ⎠

BC = DC =

⎤⎫ ⎥⎪ ⎪ j⎥ ⎬ ⎥⎪ ⎥⎪ ⎥⎦ ⎭

⎞Q ⎟ 2j ⎠L

B

EC = AC − AE = AC − AB sin 52.5°

θ = tan −1

⎤⎫ ⎥⎪ ⎪ j⎥ ⎬ ⎥⎪ ⎥⎪ ⎦⎥ ⎭

BE 5.83 × 10−11 m = = 9.68 × 10−11 m cos θ cos 52.96°


θ θ EH1

EO

CHAPTER

23

The net electric field at point P will be directed toward the right because the vertical components of the electric fields due to the hydrogen ions are equal in magnitude and oppositely directed. Therefore, the net electric field is EP = EO + EH1cos θ + 2E H2 cos θ = EO + 2EH cos θ ⎡ 1 ⎢ QO = 4πε 0 ⎢ AC ⎢⎣

⎤ 2QH cos θ ⎥ i+ i 2 2 ⎥ BC ⎥⎦ 2 ⎡ 16 1.60 × 10 −19 C 2 (1.60 × 10−19 C ) cos 52.96° ⎤ ( ) 9 Nim ⎢ ⎥ i = 1.78 × 1012 N/C i = 8.99 × 10 + 2 −11 ⎥ C 2 ⎢ (1.20 × 10−10 m )2 ( 9.68 × 10 m ) ⎣ ⎦

( )

†23-19.

( )

At the point P, there are three contributions to the net electric field, one from each of the charges. The direction of each contribution is toward a charge, if the charge is negative, and away from a charge, if it is positive. Q1 and Q3 are positive and Q2 is positive. The directions of the electric field contributions are shown in the drawing. The distances and angles need to be determined before the net field can be calculated. For Q1, the distance to P is and angle θ1 are r1 =

( 2.0 km )

2

10.0 km

Q1

θ3

Q2

θ2 θ1

5.0 km

P

Q3 2.0 km

+ ( 3.0 km ) = 3.6 km 2

⎛ 3.0 km ⎞ ⎟ = 56.3° ⎝ 2.0 km ⎠ The other distances and angles are

θ1 = tan −1 ⎜ r2 =

( 3.0 km )

2

+ ( 3.0 km ) = 4.2 km

2

+ ( 3.0 km ) = 6.7 km

2

θ 2 = 45.0° r3 =

( 6.0 km )

2

⎛ 3.0 km ⎞

θ 3 = tan −1 ⎜ ⎟ = 26.6° ⎝ 6.0 km ⎠ The net electric field is then EP = E1 + E2 + E3 ⎧ ⎡ Q1 sin θ1 Q1 cos θ1 ⎤ ⎡ Q2 sin θ 2 Q2 cos θ 2 ⎤ ⎫ − + − − i j i j⎥ ⎪ ⎪⎢ ⎥ ⎢ 2 2 2 2 r1 r2 r2 1 ⎪ ⎣ r1 ⎦ ⎣ ⎦⎪ = ⎨ ⎬ 4πε 0 ⎪ ⎡ Q3 sin θ 3 Q3 cos θ 3 ⎤ ⎪ i+ j⎥ 2 ⎪+ ⎢ r 2 ⎪ r3 3 ⎦ ⎩ ⎣ ⎭ =

⎫⎪ Q3 Q2 1 ⎧⎪ Q1 ⎨ 2 [ sin θ1 i − cos θ1 j] + 2 [ −sin θ 2 i − cos θ 2 j] + 2 [ sin θ 3 i + cos θ 3 j] ⎬ 4πε 0 ⎩⎪ r1 r2 r3 ⎭⎪


8.0 km

CHAPTER

23

⎧ ⎫ 40 C ⎪ ⎪ i j sin 56.3 ° − cos 56.3 ° + [ ] ⎪ ( 3.6 × 103 m )2 ⎪ ⎪ ⎪ 2 ⎪ ⎪⎪ Nim ⎪ 40 C = 8.99 × 109 −sin 45.0° i − cos 45.0° j] + ⎬ [ ⎨ 2 2 C ⎪ ( 4.2 × 103 m ) ⎪ ⎪ ⎪ 10 C ⎪ sin 26.6° i + cos 26.6° j] ⎪ 2 [ ⎪ ⎪ 3 ⎪⎩ ( 6.7 × 10 m ) ⎪⎭ = 9.5 × 103 N/C i − 2.8 × 104 N/C j 23-20.

This problem is similar to Problem 23-19, except that the position at which the electric field is to be determined varies as the airplane flies. At any given instant, the airplane is at a point P located a distance x from the vertical line through the locations of the three charges.

Q1 x 10.0 km

Q2 5.0 km

r1

P

r2 r3

8.0 km

2.0 km

EP = E1 + E2 + E3 ⎧ ⎡ Q1 sin θ1 Q1 cos θ1 ⎤ ⎡ Q2 sin θ 2 Q2 cos θ 2 ⎤ ⎫ i j i j⎥ ⎪ − + − − ⎪⎢ ⎥ ⎢ 2 2 2 2 r1 r2 r2 1 ⎪ ⎣ r1 ⎦ ⎣ ⎦⎪ = ⎨ ⎬ 4πε 0 ⎪ ⎡ Q sin θ Q3 cos θ 3 ⎤ ⎪ 3 3 i+ j⎥ 2 ⎪+ ⎢ r 2 ⎪ r3 3 ⎦ ⎩ ⎣ ⎭ 40 C ⎧ ⎫ [ x i − 2000 j] + ⎪ 2 ⎪ 2 3/ 2 2 ⎪ ⎡ x + ( 2000 m ) ⎤ N m i ⎪⎣ ⎦ ⎪⎪ E = 8.99 × 109 ⎨ ⎬ 2 40 C 10 C C ⎪ ⎪ x x 3000 6000 − − + + i j i j [ ] [ ] 2 3/ 2 2 3/ 2 ⎪⎡ 2 ⎪ 2 ⎤ ⎡ x + ( 6000 m ) ⎤ ⎪⎩ ⎣ x + ( 3000 m ) ⎦ ⎪⎭ ⎣ ⎦

which has a magnitude 1/ 2

2 ⎧⎡ ⎫ ⎤ ⎪⎢ 10 C 40 C 40 C ⎥ 2⎪ + − ⎪⎢ 3 / 2 3 / 2 3 / 2 ⎥ x ⎪ 2 2 2 2 2 2 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎪ + 6000 m x x 2000 m x 3000 m + + ( )⎦ ( )⎦ ( ) ⎦ ⎦⎥ ⎪⎪ 2 ⎢ ⎣ ⎣ 9 N i m ⎪⎣ ⎣ E = 8.99 × 10 ⎨ 2 ⎬ C2 ⎪ ⎡ ⎤ ⎪ −80000 C i m 120000 C i m 60000 C i m ⎥ ⎪ ⎪ ⎢ − + 3 / 2 3 / 2 3 / 2 ⎪+ ⎢ 2 ⎥ ⎪ 2 ⎡ x 2 + ( 3000 m )2 ⎤ ⎡ x 2 + ( 6000 m )2 ⎤ ⎥ ⎪ ⎪ ⎢⎣ ⎡⎣ x + ( 2000 m ) ⎤⎦ ⎣ ⎦ ⎣ ⎦ ⎦ ⎭ ⎩


CHAPTER

23

Here is a graph of the electric field versus horizontal distance from the charges.

Electric Field, E (N/C)

200000

150000

100000

50000

0 0

500

1000

1500

2000

x (m)

The electric field rapidly goes to zero N/C. Following the results from Example 3, the magnitude of the vertical component of the electric field as a function of horizontal distance from the charges is ⎡ ⎤ 2 (30 C)(4000 m) ⎥ 9 N i m ⎢ ( −40 C)(10000 m) E = 2 × 8.99 × 10 + C 2 ⎢ ⎡ x 2 + (10000 m )2 ⎤ 3 / 2 ⎡ x 2 + ( 4000 m )2 ⎤ 3 / 2 ⎥ ⎢⎣ ⎣ ⎦ ⎣ ⎦ ⎥⎦ Because of the symmetry of the system, the electric field at the ground is always vertical. The factor of two in the electric field is due to the fact that the charge and the image charge both have the same vertical component. Here is a graph of the electric field as a function of horizontal distance. 30000

25000

Electric Field, E (N/C)

23-21.

20000

15000

10000

5000

0 0

2000

4000

6000

8000

x (m)


10000

CHAPTER 23-22.

23

z

Consider the drawing of the NaCl crystal structure. A chlorine ion is located at the origin of an xyz coordinate system; and the ions are at the corners of a cube with sides of length L = 2.82 × 10−10 m. Let’s find the net electric field acting at the origin due to the seven other ions. First, the electric field due to the four positive sodium ions is e 1 e 1 E Na = + ⎡ ( −i ) + ( − j ) + ( −k ) ⎦⎤ 2 2 4πε 0 L 4πε 0 L 3 3 ⎣

x

1 ⎞ ⎛ ⎜1 + ⎟ ⎡⎣ ( i ) + ( j ) + ( k ) ⎦⎤ 3 3⎠ ⎝ Then, the electric field due to the chlorine ions is 1 −e 1 −e 1 −e ECl = ⎡ ( −i ) + ( − j ) ⎦⎤ + ⎡ ( −i ) + ( −k ) ⎦⎤ + ⎡ ( − j ) + ( −k ) ⎦⎤ ⎣ ⎣ 2 2 2 4πε 0 L 2 2 4πε 0 L 2 2 4πε 0 L 2 2 ⎣ 1 e = ⎡ ( i ) + ( j ) + ( k ) ⎦⎤ 2 4πε 0 L 2 ⎣ The net electric field at the origin is e 1 e ⎛ 1 ⎞ 1 1+ E = E Na + ECl = − ⎟ ⎣⎡ ( i ) + ( j ) + ( k ) ⎦⎤ + 4πε 2 ⎣⎡ ( i ) + ( j ) + ( k ) ⎦⎤ 4πε 0 L2 ⎝⎜ 3 3⎠ 2 0 L =−

e 4πε 0 L2 1

⎛ 1 2⎞ + ⎜⎜ −1 − ⎟ ⎡ ( i ) + ( j ) + ( k ) ⎦⎤ 2 ⎟⎠ ⎣ 3 3 ⎝ The net force acting on the chlorine ion at the origin is the product of the chlorine ion’s charge and the net electric field at the origin where it is located. 1 e2 ⎛ 1 2⎞ 1 F = qE = ( − e ) E = − − − + ⎟ ⎡ ( i ) + ( j ) + ( k ) ⎤⎦ 2 ⎜ 4πε 0 L ⎜⎝ 2 ⎟⎠ ⎣ 3 3 =

e 4πε 0 L2 1

The magnitude of this force is 1 e2 ⎛ 1 2⎞ F = 1 + − ⎟ 3 2 ⎜ 4πε 0 L ⎜⎝ 2 ⎟⎠ 3 3 2 2 ⎛ 9 Nim ⎞ −19 8.99 × 10 ⎜ ⎟ (1.60 × 10 C ) 2 C ⎠ =⎝ 2 ( 2.82 × 10−10 m )

†23-23.

⎛ 1 ⎜⎜ 3 + − 3 ⎝

3⎞ −9 ⎟⎟ = 2.43 × 10 N 2⎠

From Problem 16, the electric field produced at a distance x from the central charge is ⎞ 1 ⎛ q1 q2 q3 ⎞ Q ⎛ 1 2 1 E = E1 + E2 + E3 = + 2 − ⎜ 2 + 2 + 2⎟= ⎜− 2 2 ⎟ 4πε 0 ⎝ r1 r2 r3 ⎠ 4πε 0 ⎝ ( x + d ) x (x − d ) ⎠ where d is the distance between the charges to the left and right of the central charge and the central charge. The above equation may be rewritten as ⎛ ⎞ ⎟ Q ⎜ 1 2 1 + 2 − E= ⎜− ⎟ d 4πε 0 ⎜ x 2 (1 + d ) 2 x x 2 (1 − ) 2 ⎟⎟ ⎜ x x ⎠ ⎝


y

CHAPTER

23-24.

†23-25.

23

When x >> d, we can apply the approximation given in the problem statement. ⎛ ⎞ ⎜ ⎟ Q Q 1 1 ⎜− ⎟≈ E= +2+ 2 d⎞ d ⎞ ⎟ 2πε 0 x 2 4πε 0 x ⎜ ⎛ ⎛ + − 1 2 1 2 ⎟ ⎜ ⎟ ⎜ ⎜ x⎠ x ⎠ ⎟⎠ ⎝ ⎝ ⎝ As x goes to infinity, E goes to zero. The electric field contribution from each of the positive charges is directed radially outward from each charge. Assume that charges are located on each corner of the cube with sides of length a, except for the front, lower right corner. There are three charges nearest to the empty corner at distance a, three charges at distance a(2)1/2, and one charge at distance a(3)1/2. The net electric field will be directed below, forward, and toward the right of the cube. The magnitude of the net electric field is ⎛ 1 1 1 1 1 1 1 1 ⎞ Q ⎛ 3 3 1 ⎞ E= Q⎜ 2 + 2 + 2 + 2 + 2 + 2 + 2 ⎟ = ⎜ 2 + 2 + 2⎟ 4πε 0 ⎝ r1 r2 r3 r4 r5 r6 r7 ⎠ 4πε 0 ⎝ a 2a 3a ⎠ (a) Equation 23.11 gives the electric field on the axis of a ring of radius R at a distance y from the center as 1 Qy Ey = 4πε 0 ( R 2 + y 2 )3 / 2 where a charge Q is distributed uniformly around the ring. To determine the location for the maximum electric field along the axis, take the derivative of the electric field with respect to y and equate it to zero. ⎡ ⎤ ⎡ ⎤ dE y 1 3 2 y2 d ⎢ 1 Qy ⎥= Q ⎢ ⎥=0 = − dy dy ⎢ 4πε 0 ( R 2 + y 2 )3 / 2 ⎥ 4πε 0 ⎢ ( R 2 + y 2 )3 / 2 2 ( R 2 + y 2 )5 / 2 ⎥ ⎣ ⎦ ⎣ ⎦ 1=

3y2 ( R2 + y2 )

3 y2 = R2 + y2 R 2 2 (b) The drawing shows the field distribution for two positive charges. The field distribution for the ring is the same as that for two positive charges rotated around the y axis as indicated. y=±


+

R

+

CHAPTER 23-26.

23

The electric field between two flat plates with a uniform charge distribution σ is given by Equation 23.13 σ (Q / A) E= =

ε0 Q = ε 0 EA

†23-27.

ε0

= ⎡⎣8.85 × 10−12 C2 / ( N i m 2 ) ⎤⎦ ( 2.0 × 105 N/C ) (0.0030 m)(0.0030 m) = 1.6 × 10−11 C This problem is exactly the same as Example 5. Note the comments that follow Example 5, that Equation 23.10, although solved for an infinitely long rod, approximates the electric field for a very long, straight rod as in this problem. Therefore, applying Equation 23.10 gives, 1 λ E0.50 m = 2πε 0 x

=

1

⎛ 2π ⎜ 8.85 × 10−12 ⎝ Likewise, 1 λ E1.0 m = 2πε 0 x =

E1.50 m

⎛ 2π ⎜ 8.85 × 10−12 ⎝ 1 λ = 2πε 0 x

⎛ 2.0 × 10−14 C/m ⎞ −4 ⎜ ⎟ = 3.6 × 10 N/m 1.0 m C2 ⎞ ⎝ ⎠ ⎟ N i m2 ⎠

⎛ 2.0 × 10−14 C/m ⎞ −4 ⎜ ⎟ = 2.4 × 10 N/m 2 1.50 m ⎛ ⎞ C ⎝ ⎠ 2π ⎜ 8.85 × 10−12 2 ⎟ i N m ⎝ ⎠ The net electric field is the vector sum of the y contributions from the charges along each axis. As in Example 5, the x component of the net field that is contributed by the rod on the x axis is 1 1 σdy Ex − x = − 2 4πε 0 ( x + y 2 )1 / 2 =

23-28.

1

⎛ 2.0 × 10−14 C/m ⎞ −4 ⎜ ⎟ = 7.2 × 10 N/m 0.50 m C2 ⎞ ⎝ ⎠ ⎟ N i m2 ⎠

1

and that contributed by the rod on the y axis is ⎞ 1 λ⎛ y Ex − y = − ⎜1 + 2 2 1/ 2 ⎟ 4πε 0 x ⎝ (x + y ) ⎠

Therefore, the x component of the net electric field at point (x, y) is


0.2 0

P

0.5

σdx

x

CHAPTER ⎡1 ⎛ ⎤ ⎞ y 1 − 2 ⎢ ⎜1 + 2 2 1/ 2 ⎟ 2 1/ 2 ⎥ (x + y ) ⎠ (x + y ) ⎦ ⎣x⎝ Subsituting the values given in the problem gives, ⎡ ⎛ 0.20 m ⎢ 1 ⎜1 + 2 2 1/ 2 2 ⎢ 0.50 m ⎜ + (0.50 m) (0.20 m) ⎛ ( ) 9 Nim ⎞ −12 C ⎞ ⎛ ⎝ Ex = − ⎜ 1.0 × 10 ⎟⎢ ⎟ ⎜ 8.99 × 10 2 m C ⎝ ⎠⎝ ⎠⎢ 1 ⎢− 1/ 2 ⎢ ( (0.50 m) 2 + (0.20 m 2 ) ) ⎣ Ex = −

23

λ 4πε 0

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎥ ⎥ ⎥ ⎥ ⎦

= −7.0 × 10−3 N/C Due to the symmetry of the situation, we can write the y component by simply switching x and y to get ⎤ ⎞ λ ⎡1 ⎛ x 1 Ey = − − 2 ⎢ ⎜1 + 2 2 1/ 2 ⎟ 2 1/ 2 ⎥ 4πε 0 ⎣ y ⎝ (x + y ) ⎠ (x + y ) ⎦

⎛ = − ⎜ 1.0 × 10−12 ⎝

⎡ ⎛ 0.50 m ⎢ 1 ⎜1 + 1/ 2 2 2 ⎢ 0.20 m ⎜ (0.50 m) + (0.20 m) 2 ) C ⎞⎛ ( 9 Nim ⎞ ⎝ ⎢ 8.99 × 10 ⎟ ⎟⎜ m ⎠⎝ C2 ⎠ ⎢ 1 ⎢− 1/ 2 2 ⎢ ( (0.50 m) + (0.20 m 2 ) ) ⎣

⎞⎤ ⎟⎥ ⎟⎥ ⎠ ⎥ = −0.11 N/C ⎥ ⎥ ⎥ ⎦

The electric field at point (0.5, 0.2) is E = −0.0070 N/C i − 0.11 N/C j †23-29.

The net electric field at point P is the vector sum of the contributions from each rod of length L with a uniform charge per unit length λ. Let’s look at the contribution from the rod on the y axis. We will be integrating the contributions from the infinitesimal charges λ dy from top to bottom of the rod. The maximum value of the angle θ is ⎛ L ⎞ −1 θ = tan −1 ⎜ ⎟ = tan ( 2 ) = 63.4° ⎝ L/2⎠ and its minimum value is 0°. Using the approach taken in Example 5, the x component will be λ 2 63.4° cosθ dθ Exy = 4πε 0 L ∫0

=

λ 1 sinθ 2πε 0 L

63.4° 0

=

λ ⎛ sin 63.4° ⎞ ⎜ ⎟ 2πε 0 ⎝ L ⎠


L/2

θ

λ dy L/2

P

L/2

CHAPTER

23

The y component contribution to the electric field at P due to the rod on the y axis is λ 2 63.4° sinθ dθ Eyy = 4πε 0 L ∫0 =

63.4° λ 1 λ ⎛ 1 − cos 63.4° ⎞ ( −cosθ ) 0 = ⎜ ⎟ 2πε 0 L 2πε 0 ⎝ L ⎠

The contributions from the rod on the x axis are similar. Consider the drawing. The integration will be from right to left as the angle φ changes from 0° to 63.4°. The contribution of the rod on the x axis to the x and y components of the net electric field at P are λ 2 63.4° sinφ dφ E xx = 4πε 0 L ∫0 =

E yx =

L/2

P

φ L/2 L/2

λ dx

63.4° λ 1 λ ⎛ 1 − cos 63.4° ⎞ ( −cosφ ) 0 = ⎜ ⎟ 2πε 0 L 2πε 0 ⎝ L ⎠

λ 2 63.4° λ 1 cosθ dθ = sinθ ∫ 4πε 0 L 0 2πε 0 L

63.4° 0

=

λ ⎛ sin 63.4° ⎞ ⎜ ⎟ 2πε 0 ⎝ L ⎠

Adding these contributions together gives

E= =

23-30.

λ 2πε 0

⎡ ⎛ sin 63.4° ⎞ ⎛ 1 − cos 63.4° ⎞ ⎤ ⎟+⎜ ⎟ ⎥ (i + j) ⎢⎜ L L ⎠ ⎝ ⎠⎦ ⎣⎝ 2 ⎛ 10 N i m ( ) 2.6 × 10 + = i j ⎜ C2 C2 ⎞ ⎝ L ⎟ N i m2 ⎠

0.230λ ⎛ −12 ⎜ 8.85 × 10 ⎝

⎞λ ⎟ (i + j) ⎠L

Consider an infinitesimal portion of a uniform line of charge. Regardless of the length, finite or infinite, there is essentially no distance between these infintesimal portions, so the Coulomb force becomes infinite. Thus, the tension in the line is infinite. In reality, charge is quantized and there is a finite distance between the charges. So, in a finite rod, the tension may be quite large, but not infinite. In the case of the infinite line, there is an infinite amount of charge on each side of a given infinitesimal element of charge, so the Coulomb force would be infinite for that reason as well.


CHAPTER †23-31.

Along a line that is parallel to both wires and located a distance 2d from each wire, the net electric field will be the sum of the contributions from each wire, E = Eleft + Eright. At each point along that line, the point P in the drawing, the horizontal components of the contributions from the two lines will be equal in magnitude, but oppositely directed, and cancel each other. This leaves a net, upwardly directed electric field as shown. The net electric field is, therefore, E = ⎡⎣ Eleft cosθ + Eright cosθ ⎤⎦ j = [ 2 Eleft cosθ ] j where θ is given by the drawing to the right. ⎛ d ⎞ ⎛1⎞ θ = sin −1 ⎜ ⎟ = sin −1 ⎜ ⎟ = 30° ⎝ 2d ⎠ ⎝2⎠ and the magnitude of the left and right contributions to the net electric field are identical. Example 5 gives the magntitude of the electric field from a charge distributed over an infintely long rod. We’ll apply that solution here giving 1 λ E = [ 2 Eleft cosθ ] j = ( cos 30° ) j πε 0 2d

23-32.

As in example 5, the magnitude of the electric field at a point located a perpendicular distance r from an infintely long wire with a charge distribution λ is 1 λ E= 2πε 0 r

23

E Eright

Eleft P

2d

2d

+

+ d

θ

2d

d P (x,y,z)

Let E = Ex + Ey be the net electric field due to the line charges that lie along the x and y axes. The electric field due to the x axis line charge is 1 λ 1 λ Ex = rˆ = ( yj + zk ) 2 2πε 0 r 2πε 0 ( y + z 2 ) Similarly, the electric field due to the y axis line charge is 1 λ Ey = (xi + zk) 2 2πε 0 ( x + z 2 ) The net electric field at point P at (x, y, z) is then 1 λ 1 λ E = Ex + Ey = ( y j + zk) + (xi + zk) 2 2 2 2πε 0 ( y + z ) 2πε 0 ( x + z 2 ) =

λ 2πε 0

⎧⎪ ⎡ ⎤ ⎫⎪ 1 1 xi yj ⎢ ⎥ zk⎬ + + + ⎨ 2 2 2 2 2 2 2 2 ⎪⎩ ( x + z ) ( y + z ) ⎢⎣ ( y + z ) ( x + z ) ⎥⎦ ⎪⎭


CHAPTER †23-33.

23

As in Example 5, components of the electric field in the y direction are given by π /2 1 ∞ λ cosθ λ cos θ dθ Ey = dz = 2 ∫ ∫ φ 0 4πε 0 4πε 0 y r

y

θ

where φ is the angle that the vector from the origin to point P makes with the line perpendicular to the x axis from point P. φ = −tan−1 (x/y) Ey =

π /2 λ λ cos θ dθ = sin θ ∫ φ 4πε 0 y 4πε 0 y

P (x,y)

x ∞

0

π /2 −1

− tan ( x / y )

=

λ ⎡ ⎢1 + 4πε 0 y ⎢ ⎣

⎤ ⎥ x 2 + y 2 ⎥⎦ x

Similarly, for the x component, π /2 1 ∞ λ ( −sin θ ) λ Ex = dz = ( −sin θ ) dθ −1 2 ∫ ∫ − 0 r 4πε 0 4πε 0 y tan ( x / y )

−λ

= 23-34.

4πε 0 y

π /2

cos θ

− tan −1 ( x / y )

=

⎤ 1 ⎡ −λ ⎢ ⎥ 4πε 0 ⎢ x 2 + y 2 ⎥ ⎣ ⎦

Using the results from Problem 22, with x = 0 for point P, the components of the electric field are ⎤ λ ⎡ x λ λ ⎢1 + ⎥= Ey = [1 + 0 ] = 2 2 4πε 0 y ⎢ 4πε 0 y x + y ⎥⎦ 4πε 0 y ⎣ ⎤ −λ 1 ⎡ −λ ⎢ ⎥= 2 2 4πε 0 ⎢ x + y ⎥ 4πε 0 y ⎣ ⎦ The electric field at P is Ex =

E = Ex i + E y j =

†23-35.

λ ( − i + j) 4πε 0 y

As in Example 7, the net electric field for a single sheet that is parallel to the x-z plane, is E=

σ j 2ε 0

(i)

In between two such sheets that have opposite charges, the electric field is given by Equation 23.13 E=

σ j ε0

(ii)

We note that the magnitude of the electric field in equation (ii) is twice that in equation (i), where the magnitude is equal to


CHAPTER σ 2.0 × 10−6 C/m 2 = = 1.13 × 105 N/C 2ε 0 2 8.85 × 10−12 C 2 / ( N i m 2 )

E=

(

A

)

In the space above or below two oppositely-charged parallel sheets, the individual electric fields add together resulting in zero electric field. Apply these results and the fact that electric fields superpose linearly to the situation shown in Figure 23.37, reproduced here, in the regions labeled A through D. In region A, only the uppermost sheet contributes to the electric field, giving EA = 1.13 × 105 N/C j.

23

B C D

In region B between the two positively charged sheets, there will be no field due to the positive sheets. There is a net field downward due to the negative sheet, EB = −1.13 × 105 N/C j. In region C, the electric field is the strongest as there are three downward contributions from the three sheets, EC = 3(−1.13 × 105 N/C)j = −3.39 × 105 N/C j. In region D, the contribution to the electric field due to the two lower sheets is zero, so the only contribution is due to the upper positively-charged sheet. Its contribution is directed downward, therefore, ED = −1.13 × 105 N/C j. 23-36.

As in Example 7, each sheet contributes the electric field components shown, with magnitude

σ E= 2ε 0

135°

45° The horizontal components of these electric fields are equal in magnitude, but opposite in direction; and therefore, they cancel. The vertical components are ⎛ ⎞ ⎛ σ ⎞ 3.0 × 10−4 C/m 2 ⎟ j = 3.1 × 107 N/C j E = 2⎜ ° cos 22.5° ⎟ j = ⎜ cos 22.5 −12 2 2 ⎜ ⎟ 2 ε i 8.85 × 10 C / N m ⎝ 0 ⎠ ( ) ⎝ ⎠ As in Example 7, each sheet contributes the electric field components shown, with magnitude

(

23-37.

Ebottom

Etop

E=

)

σ 2ε 0

The components from each sheet are at right angles with respect to each other. Therefore, the magnitude of the net electric field in each quadrant will be 2

2

⎛ σ ⎞ ⎛ σ ⎞ σ 2 E= ⎜ ⎟ +⎜ ⎟ = 2ε 0 ⎝ 2ε 0 ⎠ ⎝ 2ε 0 ⎠ =

(

( 3.0 × 10

2 8.85 × 10

−4

−12

C/m 2 ) 2

C /(Nim 2

2

))

= 2.4 × 107 N/C

The field is directed at an angle of 45° with respect to each sheet as shown.


CHAPTER 23-38.

†23-39.

23

From Problem 23-10, the electric field is 2.21 × 10−7 N/C. ⎛ C2 ⎞ 2.21 × 10−7 N/C ) = 1.96 × 10−18 C/m 2 σ = ε 0 E = ⎜ 8.85 × 10−12 2 ⎟( N m i ⎝ ⎠ The electric field at a point P on the z axis is that due to a full sheet with a uniform charge distribution minus the field due to similarly charged disc of radius R. For the charged disc, we calculated the electric field in similar manner as that shown in example 7, using concentric rings of charge. For a ring of radius r and width dr, the charge on the ring is dq = σdA = σ(2πr dr). The electric field at P due to this ring of charge is 1 (2π r ) zσ dE = dr 4πε 0 ( z 2 + r 2 )3 / 2 The electric field due to the entire disc of radius R is found by integrating R R ⎞ 1 (2π r ) zσ zσ ⎡ σ ⎛ z 2 2 −1 / 2 ⎤ = − + = Edisc = ∫ dr 2 z r 1− ( ) ⎜ ⎟ 3 / 2 0 4πε 2 2 ⎦⎥ 0 2ε 0 ⎝ 4ε 0 ⎣⎢ z 2 + R2 ⎠ 0 (z + r )

The net field at P is E = Esheet − Edisc =

23-40. †23-41.

σ σ ⎛ − ⎜1 − 2ε 0 2ε 0 ⎝

As in Example 7, E =

zσ 2ε 0

∫

2

r dr

R2

R1

⎞ σ ⎛ ⎟= ⎜1 − 1 − z + R ⎠ 2ε 0 ⎝ z

2

(z

2

+ r2 )

3/ 2

=

⎞ zσ ⎟=− z +R ⎠ 2ε 0 z 2 + R 2 z

2

zσ ⎡ 1 ⎢ − 2ε 0 ⎢ z 2 + R 2 2 ⎣

2

⎤ ⎥ 2 z 2 + R1 ⎥⎦ 1

(a) Consider an infinitesimally small element of charge dq = (Q/l)dl′ located on the rod. The contributions to the electric field at P due to all such charge elements are directed toward the positive x direction and the net electric field is l

1 Q l dl ′ 1 Q⎛ 1 ⎞ 1 Q⎛1 1 ⎞ = EP = ⎜1 − ⎟ = ⎜ − ⎟ 2 ∫ 0 ′ 4πε 0 l ( x + l ′ ) 4πε 0 l ⎝ x + l ⎠ 0 4πε 0 l ⎝ x x + l ⎠ EP =

1 Q⎛1 1 ⎞ ⎜ − ⎟i 4πε 0 l ⎝ x x + l ⎠

(b) Due to the symmetry of the situation, the x component of the electric field at the point P′ will be zero and the y component will be 1 Q l / 2 cosθ Ey = dl ′ 4πε 0 l ∫− l / 2 r 2

2y θ

Let l′= y tan θ, then dl′= y sec2 θ dθ and r2 = y2 sec2 θ Then, applying the relationships shown in the drawing, the electric field at P′ is 1 Q tan −1 ( l / 2 y ) 1 Q 1 Q 2l tan −1 ( l / 2 y ) Ey = cosθ dθ = sin θ − tan −1 ( l / 2 y ) = −1 ∫ 4πε 0 l − tan ( l / 2 y ) 4πε 0 l 4πε 0 ly l 2 + 4 y 2 E=

1 Q 1 j 2πε 0 y l 2 + 4 y 2


l2 + 4 y2 l

CHAPTER 23 23-42.

Due to the symmetry of the situation, the x and z components of the net electric field at the point P located at (x, 0, 0) will be equal to zero. The net field will be in the −y direction, since the x components of the electric fields due to the postively charged rod E+ and the negatively charged rod E− are equal in magnitude, but oppositely directed. In the x-y plane, the distance from each charged rod to the point P is

P

x

+λ d

y

−λ d

z

d 2 + x2

r=

Applying the result from Example 5, the net electric field will be ⎛ 1 ⎞ λ λ E = −2 ⎜ j ⎟j = 2 2 πε 0 d 2 + x 2 ⎝ 2πε 0 d + x ⎠ When x >> d, the result becomes that of Example 5, that the field is inversely proportional to x. As x goes to infinity, the electric field goes to zero. 23-43.

x P

E+ E−

y

θ +λ d

d

−λ

(a) The distance from the end of the rod to the point P is x To avoid confusion, points on the charged rod will be indicated by l′ where −l ≤ l′ ≤ 0. The total charge is 0

−λ0l ′ −λ ⎛ l ′2 ⎞ −λ ⎛ l 2 ⎞ 1 Q = ∫ λ dl ′ = ∫ dl ′ = 0 ⎜ ⎟ = 0 ⎜ ⎟ = λ0l −l −l l l ⎝ 2 ⎠ −l l ⎝2⎠ 2 from which, 2Q λ0 = l (b) Consider an infinitesimally small element of charge dQ = (−λ0l′/l)dl′=(−2Q l′/l2) dl′ located on the rod at l′. The contributions to the electric field at P due to all such charge elements are directed toward the negative x direction and the net electric field is −1 Q ⎡ ⎛ x + l ⎞ x ⎤ = − 1⎥ ln ⎜ ⎟+ 2 ⎢ 2πε 0 l ⎣ ⎝ x ⎠ x + l ⎦ 0

0

The net electric field at P is l x ⎞ ⎤ −1 2Q l l ′dl ′ −1 2Q ⎡ ⎛ ′ ln x l = + + ⎢⎜ ( ) ⎟ ⎥ 4πε 0 l 2 ∫0 ( x + l ′ )2 4πε 0 l 2 ⎢⎣ ⎝ x + l ′ ⎠ 0 ⎥⎦ −1 Q ⎡ x ⎤ ln ( x + l ) + = − ln ( x ) − 1⎥ 2 ⎢ 2πε 0 l ⎣ x+l ⎦

EP =

EP =

−1 Q ⎡ ⎛ x + l ⎞ x ⎤ ln ⎜ − 1⎥ i ⎟+ 2 ⎢ 2πε 0 l ⎣ ⎝ x ⎠ x + l ⎦

(c) When x >> l, the electric field resembles that of a point charge and goes to zero as x goes to infinity.


CHAPTER 23 23-44.

Due to the symmetry of the situation, the net electric field will only have a component in the j direction. An infinitesimal amount of positive charge dQ gives rise to an electric field at P as shown. The total electric field at P to all such charges on the semicircle is 1 Q π /2 E= cos θ dθ 4πε 0 π R ∫−π / 2 =

†23-45.

Q 4π 2ε 0 R

sin θ

π /2 −π / 2

=

j i

θ P

dQ

Q 2π 2ε 0 R

In this situation, there are two sides that have a positive charge distribution and two sides with a negative charge distribution. Each side will contribute an equal magnitude electric field to the net electric field at P. Consider the bottom side of the plate shown in the drawing, which has a negative charge distribution. Due to the symmetry of the situation, the x component of the electric field at the point P′ will be zero and the y component will be −λ l / 2 cosθ E= dl ′ 4πε 0 ∫− l / 2 r 2

P

θ

2 2

l

l/2 l/2

l/2

Let l′= (l/2) tan θ, then dl′= (l/2) sec2 θ dθ and r2 = (l/2)2 sec2 θ Then, applying the relationships shown in the drawing, the electric field at P due to the lower edge is −λ π / 4 −λ −λ 2 π /4 cosθ dθ = sin θ −π / 4 = Ey = ∫ π − / 4 2πε 0 l 2πε 0l 2πε 0l Ebottom = −

λ 2 j 2πε 0l

Similarly, the contribution from the charges on the right side will be of the same magnitude, but directed toward the right, λ 2 Eright = i 2πε 0l From the two positive charged sides, λ 2 Eleft = i 2πε 0l E top = −

λ 2 j 2πε 0l

Combining all of the contributions gives the net field at the point P λ 2 λ 2 E= i − j πε 0l πε 0l


CHAPTER 23 23-46.

Consider the cylinder to be made up of a series of ring dl′ elements of thickness dl′. Example 6 demonstrates the l′ determination of the electric field at a point at a distance y l above the center of a ring of charge. The surface area of a given ring element is dA = 2πR dl′; and the total surface area of the tube is A = 2πRl. The surface charge density is R Q σ = 2π Rl Let the central axis of the tube be parallel to the y axis. Each ring has a charge dq = σ dA =σ 2πR dl′. The magnitude of the net electric field at the end of the tube at the point on the central axis is 1 l l ′ dq 1 l σ (2π R ) l ′ dl ′ σ R l l ′ dl ′ E= = = 3 / 2 3 / 2 ∫ ∫ ∫ 4πε 0 0 ( l ′2 + R 2 ) 4πε 0 0 ( l ′2 + R 2 ) 2ε 0 0 ( l ′2 + R 2 )3 / 2 l

⎧ ⎫ ⎧ ⎫ 1 1 Q⎪1 1 σR⎪ ⎪ ⎪ = − = − ⎨ ⎬ ⎨ ⎬ 2ε 0 ⎪ ( l ′2 + R 2 )1 / 2 ⎪ 4πε 0 l ⎪ R ( l 2 + R 2 )1 / 2 ⎪ ⎩ ⎭0 ⎩ ⎭

23-47.

The net electric field is the vector sum of the contributions from the charges along each axis. As in Example 5, the x component of the net field that is contributed by the rod on the x axis is 1 λ Ex − x = 2 4πε 0 ( x + y 2 )1 / 2 and that contributed by the rod on the y axis is ⎞ 1 λ⎛ y Ex − y = ⎜1 + 2 2 1/ 2 ⎟ 4πε 0 x ⎝ (x + y ) ⎠

y

x

P

σdy y 0

σdx

x

Therefore, the x component of the net electric field at point (x, y) is Due to the symmetry of the situation, we can write the y component by simply switching x and y to get ⎤ ⎞ λ ⎡1 ⎛ x 1 Ey = + 2 ⎢ ⎜1 + 2 2 1/ 2 ⎟ 2 1/ 2 ⎥ 4πε 0 ⎣ y ⎝ (x + y ) ⎠ (x + y ) ⎦ The electric field at point (x, y) is ⎧⎡ 1 ⎛ ⎫ ⎤ ⎞ y 1 + 2 i ⎪ ⎪ ⎢ ⎜1 + 2 2 1/ 2 ⎟ 2 1/ 2 ⎥ (x + y ) ⎠ (x + y ) ⎦ λ ⎪⎣ x ⎝ ⎪ E= ⎨ ⎬ 4πε 0 ⎪ ⎡ 1 ⎛ ⎤ ⎪ ⎞ x 1 ⎪ + ⎢ y ⎜ 1 + ( x 2 + y 2 )1 / 2 ⎟ + ( x 2 + y 2 )1 / 2 ⎥ j ⎪ ⎠ ⎦ ⎭ ⎩ ⎣ ⎝ 23-48.

(a) Consider an infinitesimal portion of a uniform line of charge. Regardless of the length, finite or infinite, there is essentially no distance between these infintesimal portions, so the Coulomb force becomes infinite. Thus, the tension in the line is infinite. In reality, charge is quantized and there is a finite distance between the charges. So, in a finite rod, the tension may be quite large, but not infinite. In the case of the infinite line, there is an infinite amount of charge on each side of a given infinitesimal element of charge, so the Coulomb force would be infinite for that reason as well.


CHAPTER 23

†23-49.

(b) Divide the rod into three segments, two pieces of length L − 4R and one piece in the middle of the other two of length 4R. The two end pieces are semi-infinite rods. The force of two such semi-infinite rods on each other was found in Problem 22-50, ⎛L+ x⎞ ⎛ L+x ⎞ F = k λ 2 ln ⎜ ⎟ − ln ⎜ ⎟ ⎝ x ⎠ ⎝ 2L + x ⎠ ⎛ L + 4R ⎞ ⎛ L + 4R ⎞ = k λ 2 ln ⎜ ⎟ − ln ⎜ ⎟ ⎝ 4R ⎠ ⎝ 2L + 4R ⎠ ⎛ L + 4R ⎞ ⎜ ⎟ ⎛ 2L + 4R ⎞ ⎛ L ⎞ = k λ 2 ln ⎜ 4 R ⎟ = k λ 2 ln ⎜ = k λ 2 ln ⎜ + 1⎟ ⎟ ⎝ 4R ⎠ ⎝ 2R ⎠ ⎜⎜ L + 4 R ⎟⎟ ⎝ 2L + 4R ⎠ When L >> R, the tension is proportional to kλ2 and ln (L/R), where λ = 1.0 × 10−7 C/m. Therefore, the approximate tension is 1 mN. Because of the symmetry of these joined rods, the electric field will be directed toward the right (+i direction) as shown. The net electric field at P will be the sum of the contributions from the arc segment and the two straight segments. From the arc segment, 1 λ 67.5° 1 λ 67.5° E Earc = cos θ dθ = sin θ −67.5° P ∫ − ° 67.5 4πε 0 R 4πε 0 R =

λ sin 67.5° 2πε 0 R

For each linear segment, extending from l = 0 to ∞, the contribution to the electric field is that determined in Problem 23-33 with x = 0 and y = R. 1 λ E= 4πε 0 R

R 67.5°

Ex E

Ey 45.0°

The electric field vectors due to these linear segments are directed as shown with respect to the symmetry axis. The magnitude of the electric field due to these components is 2

2

⎛ 1 λ⎞ ⎛ 1 λ⎞ 2 λ E= E +E = ⎜ ⎟ +⎜ ⎟ = 4πε 0 R ⎝ 4πε 0 R ⎠ ⎝ 4πε 0 R ⎠ This electric field is directed at 67.5° + 45.0° = 112.5° The total electric field is then 2 x

Earc =

2 y

λ λ sin 67.5° + 2 cos 112.5° = 0.0609 ε0 R 2πε 0 R

(

)


CHAPTER 23 23-50.

Charge is uniformly distributed over the sheet, so one approach is to treat the sheet as being composed of strips of length L and width dy′. The integration will be a double integration in x and y Because of the symmetry about the y axis, only the x component of the net electric field at P at (0, y) will be zero. We need only integrate from 0 to +L/2 in x and then double the magnitude.

E=

L

dx L

where σ = Q/L2 and y − y′ cos θ = 2 ( y − y ′ ) + x′2 E=

(

⎡ 2 σ 2 ⎢ L = + j ln 2πε 0 L ⎢ y − y ′ ⎣⎢ ⎡ y+ ⎢( σ = ln ⎢ πε 0 L ⎢ ( y − ⎢⎣ 23-51.

L 2 L 2

(

⎤ ⎥ ′ dy 1/ 2 ⎥ ⎥⎦

)

+L / 2

) + ( y − y′ )2 ⎤⎥ ⎥ ( y − y′ ) ⎦⎥ − L / 2

L 2 2

( L + 2 ( ) + ( y − ) ) ⎤⎥⎥ j ) ( L + 2 ( ) + ( y + ) ) ⎥⎥ ⎦ )

L 2 2

L 2 2

L 2 2

L 2 2

The solution is a sketch of the electric field.

−3q

+2q


θ

x′

L/2 L/2 2σ cos θ j∫ dx′ dy ′ ∫ L − / 2 0 4πε 0 ( y − y ′) 2 + x′2

L/2 L/2 ( y − y′ ) 2σ dx′ dy ′ j∫ ∫ 4πε 0 − L / 2 0 ⎡ ( y − y ′) 2 + x′2 ⎤ 3 / 2 ⎣ ⎦ ⎡ L/2 ⎢ σ L/2 E= j∫ ⎢ − / 2 L 2 2 2 2πε 0 ⎢⎣ ( y − y ′ ) y + ( L2 ) − 2 yy ′ + y ′

dE

y′

i

P

j

CHAPTER 23 23-52.

+ + + + + + + + + + + + + + + + + + + +

−

23-53.


23-54.


+ +

+

+


CHAPTER 23 23-55.


+

+

−

23-56.


+ +

†23-57.

+ +

+ +

+ +

+ +

+ +

The magnitude of the electric field is F ma E= = q q From the equations of kinematics, the acceleration is v 2 − v02 = 2a∆x

v 2 − v02 2∆x Therefore, a=

7 9.11 × 10−31 kg ( 3.3 × 10 m/s ) F ma m v 2 − v02 = = = = 3.1 × 105 N/C E= −19 2(0.010 m) q q q 2∆x −1.60 × 10 C 2

23-58.

Assume the electron is sufficiently close to the large disk such that the local electric field is approximately that of an infinite sheet at the location of the charge, then according to Equation 23.12, the electric field is σ F ma Ey = = = q q 2ε 0 a=

σq 2ε 0 m


CHAPTER 23 Applying the kinematic equations, we can then solve for the time and final speed. 1 x = v0t − at 2 2 4 xε 0 m 2x t= = a σq =

4(0.0010 m) ⎡⎣8.85 × 10−12 C 2 / ( N i m 2 ) ⎤⎦ 9.11 × 10−31 kg (5.0 × 10−8 C/m 2 )(1.60 × 10−19 C)

= 2.0 × 10−9 s

v 2 − v02 = 2ax v=

2ax =

σ qx ε0m

(3.0 × 10−8 C/m 2 )(1.60 × 10−19 C)(0.0010 m) = 7.7 × 105 m/s ⎡8.85 × 10−12 C2 / ( N i m 2 ) ⎤ ( 9.11 × 10−31 kg ) ⎣ ⎦ Because it has a positive charge and it is directed toward the positively-charged sheet, the proton experiences an electric force of magnitude F in direction opposite to its initial velocity. Assume the electron is sufficiently close to the large disk such that the local electric field is approximately that of an infinite sheet at the location of the charge, then according to Equation 23.12, the electric field is σ F ma Ey = = = q q 2ε 0 =

†23-59.

a=

σq 2ε 0 m

Applying the relevant kinematic equation gives v 2 − v02 = 2a ( x − x0 ) v0 =

−2a ( x − x0 ) =

−σ q ( x − x0 )

ε0m

−(5.0 × 10−8 C/m 2 )(1.60 × 10−19 C)(0.0015 m − 0.040 m) = 1.4 × 105 m/s ⎡8.85 × 10−12 C2 / ( N i m 2 ) ⎤ (1.67 × 10−27 kg ) ⎣ ⎦ Taking the entry point as (x = 0, y = 0), 1 1 x = v0 x t + ax t 2 y = v0 y t + a y t 2 and 2 2 There is no acceleration in the x direction; and initially, there is no y component of the velocity. Therefore, the time the electron takes to travel from entering the parallel plates until it leaves the plates is L t1 = v0 x =

23-60.


CHAPTER 23 With this expression for the time, deflection y1 may be determined, where 2

1 1 ⎛ F ⎞⎛ L ⎞ 1 ⎛ eE ⎞ ⎛ L ⎞ y1 = a y t 2 = ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ ⎟⎜ 2 2 ⎝ m ⎠ ⎝ v0 x ⎠ 2 ⎝ m ⎠ ⎝ v0 x ⎠

2

2 −19 1 ⎛ (1.60 × 10 C ) ( 400 N/C ) ⎞ ⎛ 0.040 m ⎞ ⎟⎜ = ⎜ = 2.2 × 10−3 m ⎟ ⎝ 5.0 × 106 m/s ⎟⎠ 2⎜ 9.11 × 10−31 kg ⎝ ⎠ There is no vertical acceleration after it leaves the plates. To determine the vertical deflection, the y component of the velocity as it leaves the plates must be determined. 2eEL v y2 = v02 y + 2aL = 2aL = m 2eEL vy = m

y2 = v y t 2 = v y =

D D = v0 x v0 x

2eEL m

2 (1.60 × 10−19 C ) ( 400 N/m )( 0.040 m ) 0.120 m 5.0 × 106 m/s 9.11 × 10−31 kg

= 5.69 × 10−2 m y = y1 + y2 = 2.2 × 10−3 m + 5.69 × 10−2 m = 5.9 × 10−2 m †23-61.

Taking the entry point as (x = 0, y = 0), 1 1 x = v0 x t + ax t 2 y = v0 y t + a y t 2 and 2 2 There is no acceleration in the x direction; and initially, there is no y component of the velocity. Therefore, the time the electron takes to travel from entering the parallel plates until it leaves the plates is x L t= = v0 x v0 x With this expression for the time, the deflection y1 is 2

2

1 1 ⎛ F ⎞⎛ L ⎞ 1 ⎛ eE ⎞ ⎛ L ⎞ y1 = a y t 2 = ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ ⎟⎜ 2 2 ⎝ m ⎠ ⎝ v0 x ⎠ 2 ⎝ m ⎠ ⎝ v0 x ⎠ from which the initial velocity for which the electron would strike the top plate is 2

v0 x =

eEL = 2my1

⎛σ ⎞ e ⎜ ⎟ L2 ⎝ ε0 ⎠ = 2m( d / 2)

(1.60 × 10 C ) ⎛⎜⎝ 1.5 × 10 mC ⎞⎟⎠ ( 0.030 m ) ( 9.11 × 10 kg ) ( 0.0020 m ) −19

= 23-62.

eσ L2 md ε 0 −7

2

−31

2

= 110 m/s

We assume that x = 0 and y = 0 at the left edge of the lower, positively-charged plate. We’ll also ignore the force of gravity acting on the electrons. As in the previous two problems, a = (eE)/m, and the range (the distance at which the electron will strike the lower plate) is R = v0xt. To determine the time of flight t, consider the motion in the y direction.


CHAPTER 23 y = v0 y t + t=

2v0 y ay

=

1 2 ayt = 0 2 2mv0 y eE

Then, ⎛ 2mv0 y ⎞ 2mv02 cos θ sin θ R = v0 x t = v0 x ⎜ ⎟= eE ⎝ eE ⎠

2 ( 9.11 × 10−31 kg )( 4.0 × 106 m/s ) cos 35° sin 35° 2

= †23-63.

(1.60 × 10

−19

C ) ( 3000 N/C )

= 0.029 m

We assume that x = 0 and y = 0 at the left edge of the lower, positively-charged plate. The vertical acceleration of the beads will be a = (neE)/m, and the range (the distance at which the electron will strike the lower plate) is R = v0xt. To determine the time of flight t, consider the motion in the y direction. 1 y = v0 y t + a y t 2 = 0 2 2v0 y 2mv0 y t= = ay neE where n is the number of elementary charges (electrons), m is the mass of the bead (the mass of the additional electrons on the bead is negligible), and the effects of gravity are negligible. ⎛ 2mv0 y ⎞ 2mv02 cos θ sin θ R = v0 x t = v0 x ⎜ ⎟= neE ⎝ neE ⎠ The maximum number of charges that can be on the beads such that they will travel R = L is 2mv02 cos θ sin θ n= ReE 2 ( 2.0 × 10−14 kg ) ( 3.0 m/s ) cos 5.0° sin 5.0° 2

=

23-64. †23-65.

23-66.

23-67.

( 0.20 m ) (1.60 × 10−19

C ) ( 900 N/C )

= 1085 electrons (a) p = lQ = (0.0020 m)(1.0 × 10−14 C) = 2.0 × 10−17 C • m (b) τ = −pE sin θ = −(2.0 × 10−17 C • m)(6.0 × 105 N/C) sin 90° = −1.2 × 10−11 N • m In Figure 23.34, the distance between the two charges is 5.0 km or 5.0 × 103 m. The dipole moment is p = lQ = (5.0 × 103 m)(40 C) = 2.0 × 105 C • m (a) p = lQ = (5.3 × 10−11 m)(1.60 × 10−19 C) = 8.5 × 10−30 C • m (b) Since the electron has equal probability of being at any angle in its orbit, the time averaged value of the dipole moment vector is 0 C • m. (a) p = lQ = (1.0 × 10−10 m)(1.60 × 10−19 C) = 1.6 × 10−29 C • m (b) The assumption of the ions being pointlike is incorrect. While electrons are moving, they are confined to particular directions due to the bonds between the hydrogen and chlronine ions. Therefore, there is a net dipole moment, but it is reduced due to the motion of electrons.


CHAPTER 23 23-68. †23-69.

τ = −pE sin θ = −(3.4 × 10−30 C • m)(2.0 × 106 N/C) sin 45° = −4.8 × 10−24 N • m Summing the forces on the ball in each direction gives, T cosθ − mg = 0 and qE − T sin θ = 0 By solving the left equation for the tension T and substituting into the right equation above, the electric field is ⎛ mg ⎞ qE − ⎜ ⎟ sinθ = 0 ⎝ cosθ ⎠ E=

( 3.0 × 10−5 kg )( 9.81 m/s2 ) tan 35° = 520 N/C mg tan θ = 4.0 × 10−7 C q

⎛ mg ⎞ qE − ⎜ ⎟ sinθ = 0 ⎝ cosθ ⎠

E= 23-70.

( 3.0 × 10−5 kg )( 9.81 m/s2 ) tan 35° = 520 N/C mg tan θ = 4.0 × 10−7 C q

From the data for the proton, the electric field may be determined. Fproton = eE = mproton aproton E=

mproton aproton

e Then, the acceleration on an electron placed in the same field is Felectron = eE = melectron aelectron aelectron = =

eE melectron

=

e melectron

mproton aproton melectron

⎛ mproton aproton ⎞ ⎜ ⎟ e ⎝ ⎠

(1.67 × 10 =

−27

kg )( 2.0 × 1010 m/s 2 )

9.11 × 10−31 kg

= 3.7 × 1013 m/s 2

The acceleration of the electron is due west and that of a proton is due east. †23-71.

The net electric field is the sum of the contributions from the rod and the sheet separately. The drawing shows the location of the point P at which the field is to be determined. The point P lies above the sheet at 0.010 m from the rod. Let’s assume that the point is in the positive x direction with respect to the rod, which is parallel to the y axis. At P, the electric field due to the charged sheet is given by Equation 23.12: Esheet =

σ j 2ε 0

At P, the electric field due to the charged rod is given by Equation 23.10: 1 λ Erod = i 2πε 0 x


E θ P

CHAPTER 23 Summing these two contributions gives the net electric field at P. ⎛ 3.0 × 10−7 C/m ⎞ 1 λ σ 1 E= i+ j= ⎜ ⎟i + 2πε 0 x 2ε 0 0.010 m ⎛ C2 ⎞ ⎝ −12 ⎠ 2π ⎜ 8.85 × 10 2 ⎟ N m i ⎝ ⎠

2.0 × 10−4 C/m 2 ⎛ C2 2 ⎜ 8.85 × 10−12 N i m2 ⎝

⎞ ⎟ ⎠

j

= 5.40 × 105 N/C i + 1.13 × 107 N/C j E=

( 5.40 × 10

5

N/C ) + (1.13 × 107 N/C ) = 1.1 × 107 N/C 2

2

⎛ 5.40 × 105 N/C ⎞ Ex = tan −1 ⎜ ⎟ = 3° with respect to the y axis (parallel to the rod) 7 Ey ⎝ 1.13 × 10 N/C ⎠ σ 2.0 × 10−5 C/m 2 (a) E y = = = 2.26 × 106 N/C 2 C ε0 8.85 × 10−12 N i m2

θ = tan −1

23-72.

(b) Each sheet contributes an electric field E =

σn 2ε 0

where n = 1 for the top sheet and n = 2 for the middle sheet. The direction of the field contribution from each sheet is directed vertically away from a given sheet. In the space above the middle sheet, the field is 1 σ 1.0 × 10−5 C/m 2 Eabove = ⎡⎣ ( σ 1 − σ 1 ) + σ 2 ⎦⎤ j = 2 j = j = 5.6 × 105 N/C j 2 2ε 0 2ε 0 ⎛ ⎞ C 2 ⎜ 8.85 × 10−12 ⎟ N i m2 ⎠ ⎝ In the region below the sheet, all of the components are directed downward. Therefore, 1 2σ + σ 2 Ebelow = − ⎡⎣ ( σ 1 + σ 1 ) + σ 2 ⎦⎤ j = − 1 j 2ε 0 2ε 0 =−

23-73.

5.0 × 10−5 C/m 2 ⎛ C2 2 ⎜ 8.85 × 10−12 N i m2 ⎝

Ep = E1 + E2 =

⎞ ⎟ ⎠

j = −2.8 × 106 N/C j

1 ⎡ ⎛ Q1 Q2 sin θ 2 − ⎢⎜ 4πε 0 ⎣ ⎝ r12 r22

⎞ Q2 cos θ 2 ⎤ j⎥ ⎟i − r22 ⎠ ⎦

⎡⎛ ⎛ 5.0 km ⎢⎜ ( 30 C ) ⎜⎜ 2 2 ⎢⎜ ( 5.0 km ) + ( 6.0 km ) 40 C ⎢⎜ ⎝ − 2 2 2 ⎢⎜ 3 5.0 × 10 m ) ( 5.0 × 103 m ) + ( 6.0 × 103 m ) ⎢⎜ ( 1 ⎢⎜ = ⎢⎜ 4πε 0 ⎢ ⎝ ⎛ ⎞ ⎢ 6.0 km ⎟ ⎢ ( 30 C ) ⎜ ⎜ 5.0 km 2 + 6.0 km 2 ⎟ ⎢ ) ( ) ⎠ ⎝ ( ⎢− j 2 2 3 3 ⎢ 5.0 × 10 m + 6.0 × 10 m ( ) ( ) ⎣⎢ = 1.16 × 104 N/C i − 3.40 × 103 N/C j


P

+40

⎞ ⎞ ⎤ 10.0 km ⎟⎟ ⎥ ⎟⎟ ⎥ ⎠ ⎟ i⎥ ⎟ ⎥ 4.0 ⎟ ⎥ ⎟ ⎥ ⎟ ⎥ ⎠ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦⎥

E2 −30 5.0

θ2

E

θ

E1

CHAPTER 23 E=

(1.16 × 10

4

N/C ) + ( 3.40 × 103 N/C ) = 1.21 × 104 N/C 2

2

⎛ 1.16 × 104 N/C ⎞ ⎟ = 74° 3 ⎝ 3.40 × 10 N/C ⎠ The magnitude of the electric field at point P due to each charge 1 Q E= 4πε 0 r 2

θ = tan −1 ⎜

23-74.

Due to the symmetry of the situation, the net electric field will be directed toward the negative charge because the components parallel to the base of the triangle due to the positive charges are equal in magnitude and opposite in direction. The net field is then, 1 Q (1 + 2 sin θ ) j E= 4πε 0 r 2

30°

P

r

r/2

30° L/2

From the drawing, r 2 L2 L + which gives r = r2 = 4 4 3 3 Q 3 Q E= j (1 + 2 sin 30°) j = 4πε 0 L2 2πε 0 L2 23-75.

(a) E = =

1

∞

4πε 0

∫

λ 4πε 0

∫

∞

λ dl ′

y

( d + l ′) 2

0

0

1 ⎛λ⎞ dl ′ = ⎜ ⎟ 2 ( d + l ′) 4πε 0 ⎝ d ⎠

dq′ l′

q ⎛λ⎞ (b) F = qE = ⎜ ⎟ 4πε 0 ⎝ d ⎠

23-76.

P

d

x

(c) The magnitude of the force is the same as in (b), but its direction is opposite that of the rod on the charge. Since the point is on the bisector of the angle, it is an equal perpendicular distance d from both infinite rods. Also, since the linear charge distribution is the same for both rods, the magnitude of the electric field contributed by each rod is the same. The net field will α/2 d be directed along the bisector of angle α. Each rod d P contributes 1 λ E= 2πε 0 d to the net electric field at P in perpendicular directions relative to each electric field vector. Therefore, the magnitude of the net electric field is 2

2

⎛ 1 λ⎞ ⎛ 1 λ⎞ 2 λ E= ⎜ ⎟ +⎜ ⎟ = 2πε 0 d ⎝ 2πε 0 d ⎠ ⎝ 2πε 0 d ⎠


CHAPTER 23 †23-77.

The contributions to the net electric field at P are shown for the three charges on the x axis. Due to symmetry, the x componets are equal in magnitude and opposite in direction and cancel each other. The net electric field is ⎛ ⎞ Q ⎜ 1 1 ⎟j E= − 2πε 0 ⎜ y 2 ( y 2 + d 2 )3 / 2 ⎟ ⎝ ⎠

E2 y

E1

E3

d 2Q

d

θ θ

−Q

x

−Q

23-78.

The distance from each side to the central point of the equilateral triangle is d = (l/2) tan 30°. The magnitude of each contribution will be the same; and the directions will be as shown in the drawing. Due to symmetry, the vector sum of horizontal components of E1 andE2 cancel each other. The net electric field is directed toward the negatively charged side. The magnitude of the field is, see Problem 23-41(b), ⎞ 1 ⎛ λ E= ⎜ 2 ⎟ ( 2 cos 60° + 1) 2πε 0 ⎝ l + 4d 2 ⎠ ⎛ ⎞ +λ +λ ⎜ ⎟ ⎛ ⎞ λ ⎜ 1 λ ⎜ 1 ⎟ ⎟ E1 E2 = = ⎜ ⎟ 2 ⎜ ⎟ 2 πε 0 ⎜ πε l ⎜ ⎟ 0 E3 1 + tan 30° ) ⎛l ⎞ ⎟ 2 ⎝ ( ⎠ ⎜ l + 4 ⎜⎝ 2 tan 30° ⎟⎠ ⎟ ⎝ ⎠ −λ 0.276λ = ε 0l

23-79.

Following Example 6, the electric field on the y axis due to a ring of charge is 1 dq y Ea = 2 ∫ 4πε 0 a + y 2 where a is the radius of an infinitesimal ring of width da with a total charge dq. For this case, the disc will be viewed as a series of rings. The surface charge distribution is σ = Q/(πR2). Then, dq = (2πa)σ da 2Q dq = (2π a)σ da = 2 a da R The net electric field at P is then R 1 R 2Qay da Qy a da = E= 3 / 2 3/ 2 2 ∫ ∫ 0 0 4πε 0 2πε 0 R R2 ( y2 + a2 ) ( y2 + a2 ) =

R Qy ⎡ Qy 2 2 −1 / 2 ⎤ y a − + = ) 2 ⎢ ( ⎥ ⎣ ⎦ 2πε 0 R 2πε 0 R 2 0

⎡ ⎤ 1 ⎢1 − ⎥= Q ⎢ y ( y 2 + R 2 )1 / 2 ⎥ 2πε 0 R 2 ⎣ ⎦


R y

P

a

r

⎡ ⎤ y ⎢1 − ⎥ ⎢ ( y 2 + R 2 )1 / 2 ⎥ ⎣ ⎦

CHAPTER 23 Rewrite the above solution as ⎡ ⎤ Q ⎢ y Q ⎥ E= 1− = 1/ 2 2 2 2 ⎢ ⎥ 2πε 0 R 2πε 0 R 2 y R + ( ) ⎣ ⎦

−1 / 2 ⎡ ⎡ R2 ⎤ ⎤ ⎢1 − ⎢1 + 2 ⎥ ⎥ y ⎦ ⎥ ⎢⎣ ⎣ ⎦

⎡ R2 ⎤ If y >> R, then ( R /y ) << 1 and the quantity ⎢1 + 2 ⎥ y ⎦ ⎣ Therefore, Q ⎡ 1 R2 ⎤ 1 Q 1 1 − − E= ⎢ ⎥= 2πε 0 R 2 ⎣ 2 y 2 ⎦ 4πε 0 y 2 2

23-80.

−1 / 2

2

≈1−

1 R2 2 y2

which is the result for a point charge. See Section 15.4. When the dipole is at an angle θ with respect to the electric field, the restoring torque is τ = Iα = Iθ = − pE sin θ Applying the small angle approximation, sin θ ≈ θ, Iθ = − pEθ pE θ I Since this has the form of the equation for simple harmonic motion, the angular frequency is

θ =−

1 f = 2π

pE 1 = I 2π

( 6.1 × 10

−30

C i m )( 2.0 × 106 N/C )

1.93 × 10−47 kg i m 2

= 1.3 × 1011 Hz


CHAPTER 25

THE ELECTROSTATIC POTENTIAL


W = Q∆V = 2 × 105 × 12 = 2.4 × 106 J

25-2.

W = q∆V = 1.6 × 10–19 × 1.5 = 2.5 × 1019 J

25-3.

V ( p2 ) − V ( p1 ) = − ∫ E i dr = − ∫

p2

600 m

p1

0m

⎛ −100 V ⎜ ⎝ m

⎞ ⎟ dZ ⎠

= +60, 000 V (earth and plane) The Eiffel tower is grounded; therefore ∆V = 0. 25-4.

E-field strengths as shown. ∆V = V(p2) – V(p1) p2

= − ∫ E • dr p1

⎛ −3.4 × 105 V ⎞ =– ⎜ ⎟ (0.01 m) m ⎝ ⎠ 5 ⎛ −1.1 × 10 V ⎞ –⎜ ⎟ (0.01 m) m ⎝ ⎠ = 4500V †25-5.

25-6.

The potential energy of the proton is U = qp V. 1 The kinetic energy of the proton is: K = m p v 2 2 1 From conservation of energy: qp V = m p v 2 2 −19 2q pV 2 × 1.6 × 10 C × 2.5 × 105 V ⇒ v= = = 6.9 × 106 m/s mp 1.67 × 10−27 kg

Energy = q∆V Energy 20 × 109 eV ∆V = = = 20 × 109 V q e ∆E 20 × 109 V = = 1.25 × 107 V/m l 1600 m 1 (a) By conservation of energy, E = mv2 – eV(r) 2 –19 ∆E = e∆V(r) = 1.6 × 10 × 12 J = 1.92 × 10–18 J E=

25-7.

1 2 mv = ∆E ⇒ υ = 2

2∆E / m =

2 × 1.92 × 1018 J = 2.1 × 106 m/s 9.1 × 1031


CHAPTER

25

(b) The negative terminal is at –6V potential, and the positive is at +6V. Since infinity is at 0V (ground potential), ∆V = 6 and ∆E = e∆V. Then 1 e∆V = mv2 2 or 1 (1.6 × 10–19)6 = (9.1 × 10−31 ) v2 2 υ = 1.45 × 106 m/s 25-8.

Assume a constant field between electrodes ∆V = ∫ E i dl = El = (3 × 106 )(6.35 × 10−4 )

(0.025 in = 0.35 × 10–4 m) Therefore, V = 1900V

eV =

1 mp v 2 ⇒ v = 2

25-10.

V1 =

q 9 × 109 N m 2 /C 2 × 2 × 10−12 C = 0.30 m 4πε 0 r

†25-11.

25-12. †25-13.

2eV = me

2 × 1.6 × 10−19 C × 2000 V = 2.7 × 107 m/s 9.1 × 10−31 kg

†25-9.

1

At r = 0.6 m, 1 V2 = V1 = 3 × 10−2 V 2 Kinetic energy = Potential energy 1 ⇒ me v 2 = eV 2 me v 2 9.1 × 10−31 kg × (1.8 × 107 )2 m 2 /s 2 V ⇒ = = = 9.2 × 102 Volts −19 2e 2 × 1.6 × 10 C mv02 1 me v 2 = Potential energy = e E0 x ⇒ x = 2 2eE0

When the electron is a distance r from the proton, its total energy is: qp 1 Etotal = me ve2 − | e | V , where V = , | e | = q p = 1.6 × 10−19 C 2 4πε 0 r ⇒ 1 9 × 109 Nm 2 /C 2 Etotal = × 9.1 × 10−31 kg × (4.0 × 105 ) 2 − (1.6 × 10−19 C) 2 × = 1.92 × 10−20 J 2 4.3 × 10−9 m When the electron is very far from the proton, potential energy = 0, 1 Etotal = me ve2 2 1 From conservation of energy, me ve2 = 1.92 × 10−20 J 2 ⇒ v=

2 × 1.92 × 10−20 J = 2.05 × 105 m/s 9.1 × 10−31 kg


CHAPTER

25-14.

25 q

⇒ q = 4πε 0 rV 4πε 0 r Since q is not changing, 4πε 0 rV ⇒ rV 1 1 = 4πε 0 r2V2 1 1 = r2V2 ⇒ (12 cm)(50000 V) = (17 cm) V

V =

⇒ V = 3.53 × 104 V

†25-15.

VP , where VP is the potential at the point where the worm is located, r is the r distance from the center of the dome to the worm, which is 65 cm. q q , Vsphere = . If q is the charge on the dome, then VP = 4πε 0 r 4πε 0 R R 15 (75000 V) = 1.73 × 103 V Therefore, rVP = RVsphere ⇒ VP = Vsphere = 65 r 1.73 × 103 V = 2.66 × 104 V/m So at point P, the electric field is E = 0.65 m When the worm achieves equilibrium, mg = QE

Electric field E =

⇒ Q=

25-16.

25-17.

mg 0.2 × 10−3 kg × 9.8 m/s 2 = = 7.36 × 10−8 C 4 4 2.66 × 10 V/m 2.66 × 10 V/m

q 82 × 1.6 × 1019 = 9 × 109 × V/m = 1.7 × 107 V/m 15 7.1 × 10 4πε 0 r 1 3q V(r) at center is: = 2.5 × 107 V/m 4πε 0 2 r V(r) =

1

Potential energy at the surface [see Equation (25.33)] Qe 9 × 109 N m 2 /C2 × 78e 2 C2 U = qV(r) = = = 2.57 × 10–12 J −15 7.0 × 10 m 4πε 0 R 3 ⎛ 1 Qe ⎞ −12 ⎜ ⎟ = 3.86 × 10 J 2 ⎝ 4πε 0 r ⎠ Let r be distance from center of nucleus. Then, 1 Q 78 × 1.6 × 1019 1.12 × 107 V(r) = = 9 × 109 × V/m = V/m r r 4πε 0 r 1 E = mv2 + qV(r). At closest approach v = 0. Therefore, 2 ⎛ 1.12 × 107 ⎞ –12 E = qV(r0) = (2 × 1.6 × 10–19) ⎜ ⎟ J = 1.7 × 10 J r0 ⎝ ⎠ U(r = 0) = eV(0) =

25-18.

Therefore r0 = 25-19.

2 × 1.6 × 10−19 × 1.12 × 107 = 2.1 × 10−14 m 1.7 × 10−12

Potential at surface =

1 Q 94 × 1.6 × 10−19 = 9 × 109 × V = 1.80 × 107 V −15 7.5 × 10 4πε 0 r

Energy of α-particle at this distance E = qV(r) = (2 × 1.6 × 10–19) × 1.80 × 107 J = 5.8 × 10−12 J = min kinetic energy needed


CHAPTER

25-20.

2Q

At a distance x from the origin, V = V1 + V2 =

4πε 0 x 2 + †25-21.

d 4

2

Q

=

2πε 0 x 2 + 2

2

d2 4

3 ⎛a⎞ ⎛a⎞ ⎛a⎞ a At the center of the cube, V = 8 , where r = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = 2 4πε 0 r ⎝2⎠ ⎝ 2⎠ ⎝ 2⎠ Q 4Q ⇒ V =8 = 3 πε 0 3a 4πε 0 a 2 At the center of a face, Q

V1 = V2 = V3 = V4 =

V5 = V6 = V7 = V8 =

⇒

Q 2

⎛a⎞ ⎛a⎞ 4πε 0 ⎜ ⎟ + ⎜ ⎟ ⎝2⎠ ⎝ 2⎠

2

=

Q 2 2aπε 0

Q 2

2

⎛a⎞ ⎛a⎞ 4πε 0 ⎜ ⎟ + ⎜ ⎟ + a 2 ⎝2⎠ ⎝2⎠

V = 4V1 + 4V5 =

2Q

+

2aπε 0

2Q 3aπε 0

At the center of an edge, Q Q = V1 = V4 = ⎛ a ⎞ 2πε 0 a 4πε 0 × ⎜ ⎟ ⎝2⎠ Q V2 = V3 = V5 = V6 =

25-22.

2

2

=

=

=

2Q 4 3aπε 0

2Q ⎛ 1 ⎞ 1+ ⎜ ⎟ aπε 0 ⎝ 3⎠

Q 2 5πε 0 a

⎛a⎞ 4πε 0 a 2 + ⎜ ⎟ ⎝2⎠ Q Q = V7 = V8 = 2 6πε 0 a ⎛a⎞ 4πε 0 a 2 + ⎜ ⎟ + a 2 ⎝2⎠ Q ⎛ 2 1⎞ ⇒ V = 2V1 + 4V2 + 2V7 = 1+ + ⎟ ⎜ aπε 0 ⎝ 5 3⎠ Q 2Q − V = 2 2 4πε 0 x + D 4πε 0 ( x − D ) 2 + D 2

At x = D, V has the maximum value.


25

CHAPTER †25-23.

25

Divide the arc into infinitesimal pieces with length R dθ . Each piece has charge dq = λ Rdθ . The λ Rdθ dq = . potential from this piece is dV = 4πε 0 R 4πε 0 R

Integrate to find V. λ, R, are constants, so λR θ λθ V = dθ = ∫ 4πε 0 R 0 4πε 0 25-24.

z

σ

σ

-2σ 0

y

2d

d

x

The electric field from a sheet with surface charge density σ is given by: E =

σ . ε0

⎛ σ 2σ σ ⎞ 2σ For 0 ≤ y ≤ d , E1 = ⎜ + j − ⎟j= ε0 ⎝ ε0 ε0 ε0 ⎠ ⎛ σ 2σ σ ⎞ 2σ For d ≤ y ≤ 2d , E2 = ⎜ − j − ⎟j=− ε0 ⎝ ε0 ε0 ε0 ⎠ σ 2σ σ For y ≥ 2d , E3 = − + =0

ε0

ε0

ε0

r

Since V (r ) = − ∫ E i dr , so if we define y = 0 to be the reference point for electric potential, we ref

have: y

For 0 ≤ y ≤ d , V = − ∫ 0

2σ

ε0

dy = −

2σ y

ε0

y

d

For d ≤ y ≤ 2d , V = − ∫ E1dy − ∫ E 2 dy = − d

0

d

For y ≥ 2d , V = − ∫ E1dy − 0

25-25.

2d

∫ E2 dy − d

2σ d

ε0

y

∫ E dy = − 3

2d

+

2σ ( y − d )

ε0

2σ d

ε0

+

=

σ (2 y − 4d ) ε0

2σ (2d − d )

ε0

=0

1 Qq (2 × 1.6 × 10−19 ) × (88 × 1.6 × 10−19 ) = 9 × 109 × = 5.5 × 10−12 J 7.4 × 10−15 4πε 0 r 1 (b) All U has been converted to kinetic energy mv2. Therefore, 2

(a) U =


CHAPTER

v= 25-26.

2U = m

2 × 5.5 × 10−12 m/s = 4.0 × 107 m/s. 6.7 × 10−27

1 Q ∑r 4πε 0 −40 10 ⎞ ⎛ 40 = 9 × 109 ⎜⎝ 2000 + 3000 + 6000 ⎟⎠

(a) V =

= 7.5 × 107 V (b) V =

1 4πε 0

Q

∑r

⎛ 40 40 + + = 9 × 109 ⎜ 2 2 2 3000 + 50002 ⎝ 2000 + 5000

25-27.

⎞ ⎟ 6000 + 5000 ⎠ 10

3

2

⎛ 2 × 1.6 × 10−19 −1.6 × 10−19 −1.6 × 1019 ⎞ 1 Q 9⎜ ⎟ −10 V= ∑ = 9 × 10 0.3 × 10−10 0.632 × 10−10 ⎠ ⎝ 0.6 × 10 4πε 0 r 9 × 1.6 = −23 V 0.632 dq 1 λ ds , where the charge per unit length λ = Q / l dV = = 4πε 0 r 4πε 0 h 2 + s 2

=– 25-28.

⎛ ⎜ λ ds λ ⇒ V= ∫ = ln ⎜ 2 2 4πε 0 ⎜ − l / 2 4πε 0 h + s ⎜ ⎝ ⎛l 2 2 ⎞ ⎜ 2 + (l / 4) + h ⎟ Q = ln ⎜ ⎟ 4πε 0 l ⎜ l + (l 2 / 4) + h 2 ⎟ ⎜ ⎟ ⎝2 ⎠ l/2

l + (l 2 / 4) + h 2 2 l + (l 2 / 4) + h 2 2

⎞ ⎟ ⎟ ⎟⎟ ⎠


25

CHAPTER

25-29.

25-30.

25

l 1 Q h ;λ= . = ;h= l l 2 3 3 2 Therefore, V(0) ⎡ l ⎤ (l 2 /4) + l 2 + ⎢ ⎥ 4(3) Q⎛ 1 ⎞ ⎢ 2 ⎥ = ⎜ ⎟ ln l ⎝ 4πε 0 ⎠ ⎢ − l + (l 2 /4) + l 2 /r (3) ⎥ ⎢ ⎥ ⎢⎣ 2 ⎥⎦ 2 ⎞ ⎛ 1+ ⎜ Q⎛ 1 ⎞ 3 ⎟⎟ = ⎜ ⎟ ln ⎜ l ⎝ 4πε 0 ⎠ ⎜ −1 + 2 ⎟ ⎜ ⎟ 3⎠ ⎝ Q⎛ 1 ⎞ ⎛2+ 3⎞ = ⎜ ⎟ ⎟ ln ⎜ l ⎝ 4πε 0 ⎠ ⎝⎜ 2 − 3 ⎠⎟ Outside shell, E field is zero. (By symmetry all E-fields point radially outward): V(c) = V(∞) = 0. E-field inside shell given by Gauss’ Law—Charge enclosed in surface radius r, b < r < c is ⎡ r 3 − b3 ⎤ Q + ⎢Q 3 3 ⎥ ⎣ c −b ⎦ and Q E(r) 4πr2 = net ⇒ E(r) =

ε0

Qnet 4πε 0 r 2

⎡ 1(r 3 − b3 ) ⎤ Q = 2 ⎢ 3 3 ⎥ 4πε 0 r ⎣ (c − b ) ⎦ 4πε 0 ⎛ c3 ⎞ Q = − r⎟ 3 3 ⎜ 3 4πε 0 (c − b ) ⎝ r ⎠

E (r ) =

Q

⎡1 ⎢ r2 ⎣

r ⎤ ⎛ c ⎞ − 3 ⎜ 3 3 ⎟ 3 ⎥ ⎝c −b ⎠ c −b ⎦

b

⎛ c3 r 2 ⎞ Q V (b) − V (c) = − ∫ E ( r )dr = − ⎜− − ⎟ c 4πε 0 (c 3 − b3 ) ⎝ r 2 ⎠c ⎛ c3 b 2 c3 c 2 ⎞ ⎛ c3 b 2 3 2 ⎞ Q Q = + − − = − c ⎟ = V (b ) ⎜ ⎟ ⎜ + 4πε 0 (c 3 − b3 ) ⎝ b 2 2 ⎠ 4πε 0 (c 3 − b3 ) ⎝ b 2 2 ⎠ c b

From r = b to r = a, E =

Q

1 4πε 0 r 2

Therefore, a

Q

b

4πε 0

V (a ) − V (b) = − ∫ E (r )dr = −

−∫

a

b

a

⎛ Q ⎞ 1 Q dr = ⎜ ⎟ = 2 r ⎝ 4πε 0 ⎠b 4πε 0


⎛1 1⎞ ⎜ − ⎟ ⎝a b⎠

CHAPTER ⎛ c3 b 2 3 2 ⎞ + − c ⎟ 1 1 Q ⎛1 1⎞ Q ⎜ b ⎛ ⎞ 2 2 + − V (a ) = V (b) + ⎜ ⎟ ⎜ − ⎟= ⎜ ⎟ 4πε 0 ⎝ a b ⎠ 4πε 0 ⎜ c 3 − b3 ⎝ a b ⎠⎟ ⎜ ⎟ ⎝ ⎠ From r = 0 to r = a by Gauss’ Law Qr 3 Q r E ( r )4π r 2 = ⇒ E (r ) = ε 0 a3 4πε 0 a 3 Therefore, V(0) – V(a) = – ∫

0

a

a

⎛ ⎞ r Q Q r2 ⎟ = dr = ⎜ 3 3 4πε 0 a ⎝ 2(4πε 0 )a ⎠0 2(4πε 0 )a Q

Therefore, ⎛ c3 b 2 3 2 ⎞ + − c ⎜ 1 Q 1 Q ⎛ 3 1 1 ⎞⎟ = − ⎟⎟ V (0) = V (a ) + ⎜ b 3 2 32 + ⎜ 2 4πε 0 a 4πε 0 ⎜ c −b ⎝ 2 a b ⎠⎟ ⎜ ⎟ ⎝ ⎠ 25-31.

V = V(1) + V(2) + V(3) + V(4) Q l ds V (1) = 4πε 0 l ∫0 s + x ⎛l + x⎞ ln ⎜ ⎟ 4πε 0 l ⎝ x ⎠ x =l Q du V (2) = ∫ x 2 4πε 0l u + l2 x =e Q = ln υ + υ 2 + l 2 4πε 0l x =

Q

)

(

V(2) =

V(3) =

⎡ ( x + l ) + ( x + l )2 + l 2 ln ⎢ 4πε 0l ⎢ x + x2 + l 2 ⎣

Q

Q 4πε 0

∫

l

0

dy y 2 + x2

=

Q 4πε 0l

(

⎤ ⎥ ⎥⎦

ln y +

y 2 + x2

⎛ l + l 2 + ( x + l )2 ⎞ ⎟ ln ⎜ ⎟ x+l 4πε 0 ⎜ ⎝ ⎠ ⎛ l + l 2 + ( x + l )2 Q ⎡ ⎛l + x⎞ ⎢ ln ⎜ V = ⎟ + ln ⎜⎜ l+x 4πε 0l ⎢ ⎝ x ⎠ ⎝ ⎣

V(4) =

)

l

0

=

⎛ l + l 2 + x2 ⎞ ln ⎜ ⎟ ⎟ x 4πε 0l ⎜⎝ ⎠ Q

Q

⎛ ( x + l ) + ( x + l )2 + l 2 + ln ⎜ ⎜ x + x2 + l 2 ⎝

⎞ ⎛ l + l 2 + x2 ⎞ ⎟ + ln ⎜ ⎟ ⎜ ⎟ ⎟ x ⎝ ⎠ ⎠

⎞⎤ ⎟ ⎥ Volts ⎟⎥ ⎠⎦


25

CHAPTER 25-32.

25

For the straight rods, ⎛ (r − r ) + r1 ⎞ λ ln ⎜ 2 1 V= ⎟ 4πε 0 ⎝ r1 ⎠ =

⎛r ⎞ λ ln ⎜ 2 ⎟ for each. 4πε 0 ⎝ r1 ⎠

For a semicircle of radius r1 λ ds dV = 4πε 0 r1 π r1 ds λ λπ λ V= = = ∫ 0 r1 4πε 0 4πε 0 4ε 0 (independent of r1) Therefore, ⎤ ⎛r ⎞ λ λ λ ⎡ ⎛ r2 ⎞ V= en ⎜ 2 ⎟ + = ⎢ ln ⎜ ⎟ + π ⎥ Volt 2πε 0 ⎝ r1 ⎠ 2ε 0 2πε 0 ⎣ ⎝ r1 ⎠ ⎦ 25-33.

There is no electric field in cavity. Therefore, V(a) – V(0) = 0 By Gauss’ Law, E-field for a < r < b given by ⎛ r 2 − a2 ⎞ e E(2πre) = λ ⎜ 2 2 ⎟ ⎝ b − a ⎠ ε0

λ 1 ⎛ r 2 − a2 ⎞ ⎜ ⎟ 2πε 0 r ⎝ b 2 − a 2 ⎠ ⎛ a2 ⎞ λ r E(r) = − ⎜ ⎟ r ⎠ 2πε 0 (b 2 − a 2 ) ⎝ ⇒ E=

a

⎛ r2 ⎞ V(b) – V(a) = − ∫ E (r ) dr = − a 2 enr ⎟ ⎜ 2 2 a 2πε 0 (b − a ) ⎝ r ⎠b 2 2 2 λ λ ⎡ a en(b / a ) a +b ⎛b⎞ + a2 ln ⎜ ⎟ = = − ⎢ 2 2 2 2 2πε 0 (b − a ) 2 ⎝ a ⎠ 2πε 0 ⎣ b − a

λ

b

1⎤ ⎥ 2⎦

V (b) V (0) [since V(a) = V(0)] 25-34.

From Chapter 23, Problem 34, E =

⎛ Z ⎜1 2 2πε R ⎝ Z + R 2 Q

0

2

⎞ ⎟ along axis. ⎠

Taking V(∞) = 0 we have

∫

Z

2πε 0 R

2

V(Z) = =

∞

Q

E ( Z ′) dZ ′ +

(Z′

Q 2πε 0 R

Z ′2 + R 2

)

Z ∞

=

2

∫

∞

Z

⎛ ⎜1 − ⎝

Q 2πε 0 R 2

(

⎞ ⎟ dZ′ Z′ + R ⎠ Z′

2

2

Z 2 + R2 − Z

)


CHAPTER

25-35.

The electric field in the tube is given by

25

λ 1 and the potential difference between r = a, r = b, 2πε 0 r

given by a

∆V = − ∫ E (r ) dr = b

λ λ 2.5 λ ⎛b⎞ ln ⎜ ⎟ = ln = ln(1000) −3 2πε 0 ⎝ a ⎠ 2πε 0 2.5 × 10 2πε 0

λ 103 = 145 V = 2πε 0 ln1000 λ 1 145 Thus, E at the surface of wire = = 1.16 × 107 V/m = 2πε 0 r 1.25 × 105 Therefore,

1 1 145 = 1.16 × 104 v/m = 2πε 0 r 1.25 × 102 λ 1 The electric field in the tube given by and potential difference between r = a, r = b, given 2πε 0 r E at cylinder =

25-36.

a

by − ∫ E (r ) dr = b

λ ⎛b⎞ ln ⎜ ⎟ . 2πε 0 ⎝ a ⎠

λ λ 1 × 103V ⎛ 1.0 ⎞ 3 = 145 V = 1 × 10 V ⇒ = ln ⎜ ⎟ ln1000 2πε 0 ⎝ 0.001 ⎠ 2πε 0 λ 1 145 V E at surface of wire = = 5.7 × 106 V/m = 2πε 0 r 0.001 m λ 1 145V = 5.7 × 103 V/m E at cylinder = = 2πε 0 r 1.0m

Here ∆V =

25-37.

∫ E i ds

=

Then Q =

Q

ε0

∫

r

0

, where Q = charge inside a sphere of radius r, ρ = kr –5/2 C/m3 4π r 2 ρdr =

( 4π r ) E = 8πεk 2

∫

r

0

r

dr

0

r

4π r 2 ( kr −5 / 2 ) dr = 4π k ∫

= 8π k r

r

0

E=

2k ε0r 3 / 2

r r r⎛ 2k But V(r) = V(∞) − ∫ E i dr = − ∫ E i dr = − ∫ ⎜ ∞ ∞ ∞ ε r3 / 2 ⎝ 0 4k Volt V(r) = ε0 r

⎞ −2k ⎟ dr = ε0 ⎠

Take any point (x,y,z) on the surface of the sphere. Then


dr ∞ r3/ 2

∫

r

CHAPTER

25

x2 + y2 + z2 = R2 1 ⎡ Q −Q ( R / h) ⎤ V = ⎢ + ⎥ r2 4πε 0 ⎣ r1 ⎦ r1 and r2 as shown r1 = x 2 + y 2 + (hz ) 2 2

⎛ R2 ⎞ r2 = x + y + ⎜ − z⎟ ⎝ h ⎠ Therefore, ⎡ ⎤ ⎢ ⎥ ⎥ 1 1 Q ⎢ R V(x,y,z) = − ⎢ 2 ⎥ 2 4πε 0 ⎢ x + y 2 + (h − z ) 2 h ⎛ R2 ⎞ ⎥ − z⎟ ⎥ x+ y+⎜ ⎢ ⎢⎣ ⎝ h ⎠ ⎥⎦ ⎤ Q ⎡ 1 R 1 − = ⎢ 2 2 2 2 1/ 2 2 2 2 2 4 2 1/ 2 ⎥ 4πε 0 ⎣ ( x + y + z − 2 zh + h ) h ( x + y + z − 2 R z / h + R /h ) ⎦ 2

2

⎤ Q ⎡ 1 1 − ⎢ 2 2 1/ 2 2 2 4 2 1/ 2 ⎥ 4πε 0 ⎣ ( R − 2 zh + h ) h / r ( R − 2 R z /h + R /h ) ⎦ ⎤ Q ⎡ 1 1 − 2 =0 = ⎢ 2 2 1/ 2 2 1/ 2 ⎥ 4πε 0 ⎣ ( R − 2 zh + h ) (h − 2 zh + R ) ⎦ =

Therefore, potential constant is at 0 on entire surface. 25-38. d/2

+σ d/3

−σ +σ d/2

–σ

In order to make the uncharged conducting sheet in the middle to have zero electric field, the top of the middle sheet must have charge density −σ , its bottom must have charge density +σ . Take the negative charged sheet at the bottom as reference point. The top of the conducting sheet d d 1 2 d d 1 d is located at y = + × = d . Its bottom is located at y = − × = . 2 3 2 3 2 3 2 3 Based on the charge densities and their location, the electric fields at different regions are: 1 σ σ σ σ σ − + − =− For 0 ≤ y ≤ d , E1 = − ε0 3 2ε 0 2ε 0 2ε 0 2ε 0 1 2 For d ≤ y ≤ d , E 2 = 0 3 3


CHAPTER

For

⎛ σ σ σ σ ⎞ σ E3 = ⎜ − + − − ⎟j=− j ε0 ⎝ 2ε 0 2ε 0 2ε 0 2ε 0 ⎠ E4 = 0

2d ≤ y ≤ d, 3

For y ≥ d ,

r

Using V(r) = − ∫ E i dr ref

y

1 For 0 ≤ y ≤ d , 3

V1 = − ∫ E1 i dr = 0

for

1 2 d ≤ y ≤ d, 3 3

For

2d ≤ y ≤ d, 3 d /3

V3 = −

σd d V2 = constant = V1 ( ) = 3 3ε 0

y

2d / 3

∫ E dy − ∫

E 2 dy −

1

0

σ y ε0

d /3

∫

E 3 dy =

σd /3 σ (2d / 3 − y ) σ (d − y ) +0+ = ε0 ε0 ε0

E3 dy =

σd /3 σ (2d / 3 − d ) +0+ =0 ε0 ε0

2d / 3

for y ≥ d , d /3

V4 = −

∫

2d / 3

∫

E1dy −

0

†25-39.

d

E2 dy −

d /3

∫

2d / 3

0 < r < R, E1 = 0. R < r < R1, E2 =

Q 4πε 0 r 2

R1 < r < R2, E3 = 0. r > R2 , E4 = 0 Use the electric fields above to find electric potential r

using: V (r ) = −

∫

R2

E i dr

reference point

(a) r > R2, Va = 0 (b) R1 < r < R2, Vb = constant = Va(r = R2) = 0 R2 R r Q Q Q − (c) R < r < R1. Vc = − ∫ 0dr − ∫ 0dr − ∫ dr = 2 4πε 0 r 4πε 0 R1 4πε 0 r ∞ R2 R1 (d) 0 < r < R, Vd = constant = Vc(r = R) = 25-40.

Q 4πε 0 R1

−

Q 4πε 0 R

3Q − Q Q = 2 4πε 0 r 2πε 0 r 2 For a ≤ r ≤ b, E = 0 Q For r ≤ a, E = − 4πε 0 r 2

For r ≥ b, E =

Define V = at r = ∞ r

For r ≥ b, V1 (r ) = − ∫ ∞

Q 2πε 0 r

R

R1

2

dr =

Q 2πε 0 r


25

CHAPTER

25

a ≤ r ≤ b, V2 (r ) = V1 (r = b) = b

2πε 0 r

∞

†25-41.

2πε 0 b b

Q

r ≤ a, V3 (r ) = − ∫

Q

2

(−Q) Q Q 1 1 + dr = ( − ) 2 4πε 0 r 2πε 0 b 4πε 0 a r ∞

dr − ∫

First, find the electric fields in different regions using Gauss’ Law: q Qr 1 Q 4 3 Qr 3 πr = ⇒ E1 = r < a, 4π r 2 E1 = inside = 3 ε0 ε 0 4 π a3 3 ε0a 4πε 0 a 3 3 Q a ≤ r ≤ b , E2 = 4πε 0 r 2 b ≤ r ≤ c, E3 = 0 (It is a conductor) Q + 2Q 3Q = r ≥ c E4 = 2 4πε 0 r 4πε 0 r 2 r

Now use V (r ) = − ∫ Edr to find the potential at different regions: ∞

r

r

3Q 3Q dr = 2 4πε 0 r 4πε 0 r ∞

r ≥ c, V4 ( r ) = − ∫ E4 dr = − ∫ ∞

b ≤ r ≤ c, V3 = constant = V4(r = c) = c

a ≤ r ≤ b,

4πε 0 c

b

r

V2 (r ) = − ∫ E4 dr − ∫ E3 dr − ∫ E2 dr = ∞

r ≤ a,

3Q

c

b

c

b

a

r

∞

c

b

a 2

3Q 4πε 0 c

+

V1 (r ) = − ∫ E4 dr − ∫ E3 dr − ∫ E2 dr − ∫ E1dr =

3Q 4πε 0 c

+

Q 4πε 0 r

−

Q 4πε 0 b

−

Qr

8πε 0 a

3

|ra

Q ⎛3 1 1 r 1 ⎞ ⎜ + − − 3 + ⎟ 4πε 0 ⎝ c a b a 2a ⎠ 2

= 25-42.

†25-43.

∂V = –(2x + 2y) ∂x ∂V Ey = – = –2x ∂y ∂V Ez = – =0 ∂z Therefore, at x – 2, y – 2, E = −8î − 4ˆj V/m Ex = –

Example 9 gives: Q x Ex = 2 2 4πε 0 ( x + y + z 2 )3 / 2 Q y Ey = 2 2 4πε 0 ( x + y + z 2 )3 / 2


Q 4πε 0 r

−

Q 4πε 0 b

CHAPTER Ez =

Q

z

4πε 0 ( x + y + z 2 )3 / 2 2

2

2

⎛ Q ⎞ x2 + y 2 + z 2 Q 2 1 E +E +E =⎜ =( ) ⎟ 2 2 2 3 2 4πε 0 ( x + y 2 + z 2 ) 2 ⎝ 4πε 0 ⎠ ( x + y + z ) Q Q = ⇒ | E | = Ex2 + E y2 + Ez2 = 4πε 0 ( x 2 + y 2 + z 2 ) 4πε 0 r 2 1 Q 2 1 3Q for (r ≤ R ) V =− r + 3 4πε 0 2 R 4πε 0 2 R dV 1 Qr Qr =− =− 3 dr 2πε 0 2 R 4πε 0 R 3 dV Qr ⇒ Er = − = , which agrees with 24.18. 4πε 0 R 3 dr 2 x

†25-45.

25-46.

†25-47.

25-48.

2 y

2 z

∂V = −2 xy − 3 yz ∂x ∂V Ey = − = − x 2 − 3 xz − 2 zy ∂y ∂V Ez = − = −3 xy − y 2 ∂z ⇒ E = −(2 xy + 3 yz )i − ( x 2 + 3 xz + 2 zy ) j − (3 xy + y 2 )k Ex = −

∂V 2π 2π x 2π y 2π z =− sin cos cos a a b c ∂x ∂V 2π 2π x 2π y 2π z Ey = − =− cos sin cos ∂y b a b c ∂V 2π 2π x 2π y 2π z Ez = − =− cos cos sin ∂z c a b c Q 1 ∑ i (a) V(x,y,z) = ri 4πε 0 Ex = −

⎞ 1 ⎛ 6e 6e ⎜ ⎟ + 4πε 0 ⎜ ( x + 0.4 × 10−10 ) 2 + y 2 + z 2 ( x − 0.8 × 10−10 ) 2 + y 2 + z 2 ⎟⎠ ⎝ dV (b) Ex = − dx ⎫ 1 ⎧⎛ 1 ⎞ 6e × 2( x + 0.4 × 10−10 ) 2e × 2( x + 0.8 × 10−10 ) ⎛ 1⎞ − + − = ⎨⎜ ⎟ ⎜ ⎟ 2 2 3/ 2 2 2 3/ 2 ⎬ −10 2 −10 2 4πε 0 ⎩⎝ 2 ⎠ [( x + 0.4 × 10 ) + y + z ] ⎝ 2 ⎠ [( x + 0.8 × 10 ) + y + z ] ⎭ =

Ey = − =

dV dy

⎫ 1 ⎧⎛ 1 ⎞ 6e × 2 y 2e × 2 y ⎛ 1⎞ + ⎜− ⎟ ⎨⎜ − ⎟ −10 2 2 2 3/ 2 −10 2 2 2 3/ 2 ⎬ 4πε 0 ⎩⎝ 2 ⎠ [( x + 0.4 × 10 ) + y + z ] ⎝ 2 ⎠ [( x + 0.8 × 10 ) + y + z ] ⎭


25

CHAPTER

†25-49.

25 p

V =

z

4πε 0 ( x + y + z 2 )3 / 2 ∂V p 3 xz ⇒ Ex = − = 2 ∂x 4πε 0 ( x + y 2 + z 2 )5 / 2 2

2

Ey = −

∂V p 3 yz = 2 4πε 0 ( x + y 2 + z 2 )5 / 2 ∂y

Ez = −

⎤ ∂V p ⎡ 3z 2 1 = − 2 ⎢ 2 2 2 5/2 2 2 3/ 2 ⎥ 4πε 0 ⎣ ( x + y + z ) (x + y + z ) ⎦ ∂z

On z axis, x = y = 0

⇒

p

Ex = Ey = 0, | E | = Ez =

2πε 0 z 3

On x axis, y = z = 0,

⇒ 25-50.

Ex = Ey = 0, | E | = Ez = −

p 2πε 0 x 3

At point P, electric potential produced by positive charged rod is: r =l + x Q dr Q l+x V+ = = ln ∫ x 4πε 0l r = x r 4πε 0l Electric potential produced by negative rod is: y =l / 2 Q dy V− = − ∫ 2 4πε 0 l y =− l / 2 ⎛ l ⎞ 2 + x ⎜ ⎟ +y ⎝2 ⎠

⎡ =− ln ⎢ y + 4πε 0l ⎢ ⎣ Q

2 ⎛l ⎞ ⎤ y2 + ⎜ + x ⎟ ⎥ ⎝2 ⎠ ⎥⎦

y = l/ 2

| y =− l/ 2

2 ⎡ l2 ⎛ l Q ⎧⎪ ⎡ l 2 ⎛ l l⎤ l ⎤ ⎫⎪ ⎞ ⎞ ⎢ ⎥ ⎢ =− + ⎜ + x⎟ + − ln + ⎜ + x⎟ − ⎥⎬ ⎨ ln 4πε 0l ⎪ ⎢ 4 ⎝ 2 2⎥ 2⎥⎪ ⎢ 4 ⎝2 ⎠ ⎠ ⎦ ⎣ ⎦⎭ ⎩ ⎣ Vtotal = V+ + V− 2

=

E=−

Q 4πε 0l

ln

2 2 ⎡ l2 ⎛ l l+x Q ⎧⎪ ⎡ l 2 ⎛ l l⎤ l ⎤ ⎫⎪ ⎞ ⎞ − + ⎜ + x ⎟ + ⎥ − ln ⎢ + ⎜ + x⎟ − ⎥⎬ ⎨ ln ⎢ x 4πε 0l ⎪ ⎢ 4 ⎝ 2 2⎥ 2 ⎥⎪ ⎢ 4 ⎝2 ⎠ ⎠ ⎦ ⎣ ⎦⎭ ⎩ ⎣

dVtotal i dx


CHAPTER

=

=

=

=

25-51.

⎧ ⎡ ⎛l ⎞ ⎪ ⎢ −⎜ + x⎟ Qi ⎪ x x−l − x ⎢ 1 2 ⎝ ⎠ −⎢ × ⎨− 2 2 2 2 2 x 4πε 0l ⎪ l + x ⎢ l + ⎛ l + x⎞ + l l + ⎛ l + x⎞ ⎟ ⎜ ⎟ ⎪ ⎢⎣ 4 ⎝⎜ 2 2 4 ⎝2 ⎠ ⎠ ⎩ 2 2 ⎧ ⎡ l2 l ⎡ l2 ⎛ l l⎤ ⎛l ⎞ ⎞ ⎪ ⎢ + ⎜ + x⎟ − − ⎢ + ⎜ + x⎟ + ⎥ 2 ⎢ 4 ⎝2 2⎥ ⎢ 4 ⎝2 ⎠ ⎠ Q i ⎪⎪ l ⎣ ⎦ −⎢ ⎨ 2 4πε 0l ⎪ x(l + x) ⎢ ⎛l ⎞ ⎜ + x⎟ ⎢ ⎪ 2 ⎝ ⎠ ⎢ ⎪⎩ ⎣ ⎧ ⎫ ⎪ ⎪ Qi ⎪ l l 1 ⎪ − ⎨ ⎬ 2 4πε 0l ⎪ x(l + x) l + x l 2 ⎛ l ⎞ ⎪ + ⎜ + x⎟ ⎪ 2 ⎪ 4 ⎝2 ⎠ ⎭ ⎩ ⎧ ⎫ ⎪ ⎪ Qi ⎪ 1 1 ⎪ − ⎨ ⎬ 2 2 4πε 0 ⎪ x(l + x) ⎛ l ⎞ l ⎛l ⎞ ⎪ + ⎜ + x⎟ ⎪ ⎜ + x⎟ ⎪ ⎝2 ⎠ 4 ⎝2 ⎠ ⎭ ⎩

(a) From Equation 41, potential due to ring is 1 Q V= 2 4πε 0 ( z + R 2 )1 / 2 Consider the tube to be made up of many small rings. The potential due to a ring element is 1 dQ dV = 2 4πε 0 ( z + R 2 )1 / 2 Q But dQ = dx′. Therefore, l 1 Q dx′ dV = 2 4πε 0 l ( z + R 2 )1 / 2 Where z = distance from ring element to the point on the axis, x, 1 Q dx′ ⇒V therefore z = x – x ′. Therefore, dV = 4πε 0 l [( x − x′) 2 + R 2 ]1 / 2 1 Q x′= l / 2 dx′ = ∫ dv = ∫ / 2 x l ′= 4πε 0 l [( x − x′) 2 + R 2 ]1 / 2 Let x′ = x – y′ dx′ = –dy′ Then


−

2

l2 ⎛ l l ⎞ + ⎜ + x⎟ − 4 ⎝2 2 ⎠

⎤⎫ ⎥⎪ ⎥ ⎪⎪ ⎥⎬ 2 l2 ⎛ l ⎞ ⎥⎪ + ⎜ + x⎟ ⎥⎪ 4 ⎝2 ⎠ ⎥⎪ ⎦⎭

⎤⎫ ⎥⎪ ⎥⎪ ⎥⎬ 2 l2 ⎛ l ⎞ ⎥⎪ + ⎜ + x⎟ ⎪ 4 ⎝2 ⎠ ⎥⎦ ⎭ ⎛l ⎞ −⎜ + x⎟ ⎝2 ⎠

1

⎛l ⎞ −⎜ + x⎟ 2 ⎝ ⎠

25

CHAPTER

25

1 Q y ′= xl / 2 dy ′ 2 ∫ ′= / 2 y x l + 4πε 0 l ( y ′ + R 2 )1 / 2 y′ = x + l / 2 1 Q = ln y′ + y′2 + R 2 y′ = xl / 2 4πε 0 l

V =

(

=

)

2 2 1 Q ⎛ x + l / 2 + ( x + l / 2) + R ln ⎜ 4πε 0 l ⎜ xl / 2 + ( x + l / 2) 2 + R 2 ⎝

(Table of integrals) ⎞ ⎟ ⎟ ⎠

∂V ˆ (b) By symmetry, the electric field must only point in x-direction. Therefore, E = Ex î = – i ∂x 2 ⎛ ⎞ 1 Q ∂ l l⎞ ⎛ ⎜ = ln x + + ⎜ x + ⎟ + R 2 ⎟ ⎟ 4πε 0 l ∂x ⎜ 2 2⎠ ⎝ ⎝ ⎠ 2 ⎞ ∂V ⎛⎜ l ⎛ l⎞ – ln x − + ⎜ x ⎟ + R 2 ⎟ ⎟ ∂x ⎜ 2 ⎝ 2⎠ ⎝ ⎠ 1 Q ⎧⎪1 + ( x + l/2)[( x + l/2) 2 + R 2 ]1 / 2 =– ⎨ 4πε 0 l ⎪ x + l/2 + ( x + l/2) 2 + R 2 ⎩ 1 + ( x − l/2)[( x − l/2) 2 + R 2 ]−1 / 2 ⎫⎪ − ⎬ x − l/2 + ( x − l/2) 2 + R 2 ⎪⎭

25-52. 25-53.

1 ⎛ (40)(−40) (40)(10) (40 × 10) ⎞ 9 + + ⎜ ⎟ = 3.6 × 10 J 4πε 0 ⎝ 5000 8000 3000 ⎠ Potential energy ⎛ ⎞ ke 2 k (2e) U = 2⎜ + −10 −10 ⎟ ⎝ 0.958 × 10 ⎠ 2(0.958 × 10 cos37.5°) By (2) U =

where k = U=

1 4πε 0

ke 2 0.958 × 10−10

1 ⎛ ⎞ ⎜ −4 + ⎟ 2cos37.5° ⎠ ⎝

= −8.1 × 10−18 J

At a point outside the spherical charge distribution, the charge can be replaced by a point charge as was shown using Gauss’ Law in Chapter 24. ⎞ −2e 2 1 ⎛ −2e 2 e2 + + ⎜ −10 −10 −10 ⎟ 4πε 0 ⎝ 0.2 × 10 0.2 × 10 0.4 × 10 ⎠ ⎛ −2 × (1.60 × 10−19 ) 2 (1.60 × 10−19 ) 2 ⎞ −17 + = 9.0 × 109 ⎜ 2 ⎟ = −4.0 × 10 J(−250 eV) −10 −10 0.2 × 10 0.4 × 10 ⎝ ⎠

25-54.

By (2) U =

25-55.

U=

1 Q 2 d 1 (1.0 × 106 ) 2 0.001 = 6.3 × 104 J = 2 ε 0 A 2 (8.85 × 1012 )(0.09)


CHAPTER

25-56.

U=

1 ε0E2 × volume (see Equation 12) 2

−4 1 8.85 × 10–12 × 25 × 1010 × 0.2 × 5 × 10–4 = 1.1 × 10 J 2 −2 1 1 U = QV = 2 × 10–6 × 3 × 104 = 3 × 1 J 2 2 2 1 1 ⎛ C ⎞ 21 2 V U = ε0E2 = 8.85 × 10–12 ⎜ × (3.4 × 10 ) = 5.1 × 1031 J/m3 ⎟ m2 2 2 ⎝ Vm ⎠ −8 3 1 1 (a) u = ε0E 2 = (8.85 × 10−12 ) × 1002 J/m3 = 4.4 × 10 J/m 2 2 (b) Volume of air ≈ 4πR2 ∆R = 4π(6,380,000 m)2(10,000 m) = 5.12 × 1018 m3 Therefore, Energy U = uV = 4.4 × 10–8 J/m3 (5.12 × 1018)m3 = 2.3 × 1011 J

=

25-57. 25-58. 25-59.

25-60.

†25-61. 25-62.

The potential at point 1 is: Q Q ⎛5 2Q 2Q 1 ⎞ V1 = + − = − ⎜ ⎟ 4πε 0 2a 4πε 0 a 4πε 0 3a 2πε 0 a ⎝ 4 3⎠ So the energy of caused by charge 1 is: Q2 ⎛ 1 Q2 5⎞ U = −QV1 = 0.67 − = − 2πε 0 a ⎜⎝ 3 4 ⎟⎠ 2πε 0 a

2(-Q)

1(Q)

3(Q)

6(-Q) 5(Q)

Consider all pairs, using (25.48), we get: Q2 U(total) = –1.01 πε 0 a 1 1 u = ε 0 E 2 = × 8.85 × 10−12 C/(Nm 2 ) × (2.0 × 1010 )2 N 2 /C2 = 1.77 × 109 J/m3 2 2 (a) Consider a typical charge −e. −e ⎛ 2e 2e 2e 2e e2 ⎛ 1 1 1 ⎞ ⎞ − + − + ...... ⎟ = − U = ⎜ ⎜ 1 − + − + ... ⎟ 4πε 0 ⎝ a 2a 3a 4a 2πε 0 a ⎝ 2 3 4 ⎠ ⎠

1 2 1 3 1 4 x + x − x + ... 2 3 4 1 1 1 Let x = 1 ⇒ ln 2 = 1 − + − + ... 2 3 4 2 e ⇒ U =− ln 2 2πε 0 a (b) ln(1 + x) = x −

(c) If a = 3.0 × 10−10 m, U =−

(1.6 × 10−19 C) 2 ln 2 = 1.06 × 10−18 J = 6.6 eV 2π × 8.85 × 10−12 C/(Nm 2 ) × 3 × 10−10 m


4(-Q)

25

CHAPTER

25-63.

U=

=

25-64.

U =

25 1 ⎛ 4Q 2 Q2 ⎞ +2 ⎜ ⎟ 4πε 0 ⎝ d 2d ⎠ Q2

4πε 0 d

(4 + 2)

12kQ 2 − 2d

charge pairs along the free diagonals

U=

25-65.

4kQ 2 − 3d

12kQ 2 d

pairs along body diagonals

pairs along edges

4kQ 2 ⎛ 3 1 1 5.83kQ 2 ⎞ − − ; k = = 3 J ⎟ 4πε 0 d ⎜⎝ 2 d 3 ⎠

Let R be distance between α-particles. ⎛ 4e 2 4e 2 4e 2 ⎞ 1 U= =⎜ + + ⎟= R R ⎠ 4πε 0 ⎝ R 1 ⎛ 12e 2 ⎞ ⎜ ⎟ 4πε 0 ⎝ R ⎠ = 9.0 × 109 ×

12 × (1.6 × 10−19 ) 2 J 3.0 × 10−15

= 9.2 × 10–13 J = 5.8 × 106 eV 25-66.

Let distances between α-particles be R. By an extension of formula (2),

U=

2e 2 ×6 4πε 0 R 1

factor of 6 appearing because there are six lines joining the α-particles to each other. (2 × 1.6 × 10−19 ) 2 1 = 9.0 × 109 × 6 × eV 3.0 × 10−15 1.6 × 10−19 = 1.2 × 107 eV

25-67.

Energy U =

1 1 1 q1q2 Vcloud (r1)q1 + Vcloud (r2)q2 + 2 2 4πε 0 r12

Qr 3 1 3Q + 3 4πε 0 2 R 4πε 0 2 R R At position of electrons r = , 2 therefore, Vcloud (r) = –

1


CHAPTER 1 Q 1 3Q ⎛r⎞ Vcloud ⎜ ⎟ = – + 2 πε R πε 4 8 4 ⎝ ⎠ 0 0 2R 1 11Q = 4πε 0 8 R Therefore, 1 ⎛ 1 11Q 1 11Q ⎞ ⎛ 1 q1q2 ⎞ U= ⎜ q1 + q2 ⎟ + ⎜ ⎟ 2 ⎝ 4πε 0 8R 4πε 0 8R ⎠ ⎝ 4πε 0 r12 ⎠ ⎛ 11(2 × 1.6 × 10−19 )(1.6 × 1−19 ) ⎞ 9⎜ ⎟ = 9.0 × 10 8 × 0.5 × 1−10 ⎝ ⎠ (1.6 × 10−19 )(−1.6 × 10−19 ) J = –8.1 × 10–18 J = −50 eV −10 0.5 × 10

+ 9.0 × 109

1 1 1 Q2 1 (7.5 × 10−6 ) 2 QV = = (9.0 × 109 ) J = 1.7 J 2 2 4πε 0 R 2 0.15

25-68.

As in Example 1, U =

25-69.

Electrical energy in the space between R and 2R is given by

⎛1 ⎞ dU = µdv = ⎜ ε 0 E 2 ⎟ (4π r 2 dr ) ⎝2 ⎠ 2R 1 ⎛ ε Q ⎞ 2 2 0 U= ∫ ⎜ ⎟ (4πr )dr R 2 4πε r 2 0 ⎝ ⎠

=

Q2 8πε 0

=

Q2 1 R 8πε 0 2

Utotal =

25-70.

†25-71.

∫

2R

R

dr Q2 ⎛ 1 ⎞ = ⎜ ⎟ 8πε 0 ⎝ 2 R ⎠ r2

Q2 ⎛ 1 ⎞ Q2 ⎛ 1 ⎞ Q2 ⎛ 1 1 ⎞ Q2 ⎛ 7 ⎞ + = + = ⎜ ⎟ ⎜ ⎟J ⎜ ⎟ ⎜ ⎟ 8πε 0 ⎝ 2 R ⎠ 8πε 0 ⎝ 5R ⎠ 8πε 0 ⎝ 2 R 5R ⎠ 8πε 0 ⎝ 10 R ⎠

By (9), U =

1 ⎛ Q2 ⎜ 8πε 0 ⎝ R

⎞ 1 e2 = = mec2 ⎟ ⎠ 8πε 0 R

Therefore, ⎞ 1 ⎛ e 2 ⎞ 8.99 × 109 ⎛ (1.60 × 10−19 ) 2 R = = = 1.41 × 10−15 m × ⎜ ⎟ ⎜ 2 31 8 2 ⎟ 2 8πε 0 ⎝ me c ⎠ ⎝ 9.11 × 10 × (3.00 × 10 ) ⎠ First, use Gauss’ Law to find the electric field in regions r ≤ R and r ≥ R 4 3 Qr 3 Qr ⇒ E (r ≤ R) = πr = 3 4 4πε 0 R 3 ε0 ε 0 π R3 3 ε0 R 3 q Q Q 4π E (r ≥ R) = inside = ⇒ E (r ≥ R) = ε0 ε0 4πε 0 r 2 1 Since there is no charge outside the sphere, when we use U = ∫ V (r ≤ R )dQ to calculate the 2 energy, we only need the potential inside the sphere: V (r ≤ R ) , which can be found by: 4π E ( r ≤ R) =

qinside

=

1

Q


25

CHAPTER

25 R

r

V (r ≤ R ) = − ∫ E (r ≥ R)dr − ∫ E ( r ≤ R) dr = ∞

R

Q

4πε 0 R

−

Qr 2 Q 3Q Qr 2 + = − 3 8πε 0 R 8πε 0 R 8πε 0 R 8πε 0 R 3

R

⇒

U =

1 1 3Q Qr 2 V r R dQ ( ) ( )dQ ≤ = − 2∫ 2 ∫0 8πε 0 R 8πε 0 R 3

Since dQ = ρ dv =

Q

4π r 2 dr =

3Qr 2 dr R3

4 π R3 3 R R 1 3Q Qr 2 3Qr 2 3Q 2 3r 2 r 4 3Q 2 1 3 2 U dr ( ) ( So, = ∫ − = − 3 )dr = Q /R = 3 3 3 ∫ 2 0 8πε 0 R 8πε 0 R 16πε 0 R 0 R R R 20πε 0 R 4πε 0 5 25-72.

By Example 4, 1 3Q 2 1 3e 2 U= = n (eV) 4πε 0 5R 4πε 0 5 R where n = # of electric charge 9 Pu: n = 94 U = 1.03 × 10 eV

235

25-73. 25-74.

235

Np : n = 93

U = 1.01 × 109 eV

235

U : n = 92

U = 9.9 × 108 eV

235

Pa : n = 94

U = 9.7 × 108 eV

⎞ eV 1 3q 2 ⎛ 3e 2 = ⎜ 9 × 109 × = 8.6 × 105 eV 15 ⎟ 4πε 0 5R ⎝ 5(0.1 × 10 ) ⎠ e 1 U = ∫ u dV = ∫ ε0E2 dV. 2 The electric field E between the sphere and the shell is 1 Q 4πε 0 r 2

U =

where Q = charge on sphere. dV = 4πr 2 dr Therefore, 2

⎛ ⎞ Uspace = ∫ ε 0 ⎜ 1 Q2 ⎟ 4πr 2 dr R ⎝ 4πε 0 r ⎠ (R1,R2 = inner and outer radii) R2

Uspace =

1 Q2 2 4πε 0

Utotal = Uspace =

∫

R2

R1

r 1 Q2 = r 2 2 4πε 0

R

2 1 Q2 ⎡ 1⎤ − = ⎢ r⎥ ⎣ ⎦ R1 2 4πε 0

⎛ 1 1 ⎞ ⎜ − ⎟ ⎝ R1 R2 ⎠

Q2 ⎛ 1 1 ⎞ 1 ⎞ 9 –6 2 ⎛ 1 − ⎜ − ⎟ = (9.0 × 10 ) × (1.0 × 10 ) ⎜ ⎟ 8πε 0 ⎝ R1 R2 ⎠ ⎝ 0.1 0.2 ⎠

= 0.0225 J


CHAPTER

†25-75.

(a) The gravitational field g = By analogy E ≅ g;

ε0 E 2

Since u =

2

1 4πε 0

= 4πε0

GM 1 Q ; The electric field E = . 2 r 4πε 0 r 2

≅ G

E2 = 8π

E2 ⎛ 1 ⎞ 8⎜ ⎟ ⎝ 4πε 0 ⎠

π

g2 8π G 1 3 2 (b) From the result of Problem 71, U elec= Q /R 4πε 0 5 1 Use the analogy ≅ G , M ≅ Q, the gravitational energy of the moon is: 4πε 0

substituting analogous variables in gives u =

3 GM 2 = 1.24 × 1029 J 5 R U grav (3 / 5)(GM 2 /R ) 3 GM = = = 1.9 × 1011 U restmass Mc 2 5 Rc 2

Ugrav =

25-76.

(a) U =

1 3 Q2 , from Equation 25 for a uniform 4πε 0 5 R

sphere. Therefore, U = 9.0 × 109 3 (92 × 1.60 × 10−19 ) 2 J = 1.6 × 10−10 J = 9.9 × 108 eV 5 7.4 × 1015

(b) Let R1 be radius of palladium nucleus, R0 be radius of uranium. By constant volume 4 4 2 × π R13 = π R03 3 3 ⇒ 2 R13 = R03 ⇒ R1 = 2−1 / 3 R0 U = Uint + Uinter 1 3 (46e) 2 Uint = 2 × 4πε 0 5 R1 = 2 × 9.0 × 109 × Uinter =

3 (46 × 1.60 × 10−19 ) 2 1/3 2 = 1.0 × 10–10 5 7.4 × 1015

Q2 (46 × 1.6 × 10−19 ) 2 1/3 = 9.0 × 109 × 2 = 4.2 × 10–11 J 4πε 0 rinter 2 × 7.4 × 10−15

1

Therefore, U = 1.4 × 10−10 J


25

CHAPTER

25

(c) At this time Uinter → 0 (since rinter → ∞). Therefore, 1 3 Q2 (46 × 1.6 × 10−19 ) 2 1/3 U = Uint = 2 × = 2 × 9.0 × 109 × 0.6 × 2 4πε 0 5 R1 7.4 × 10−15 = 1.0 × 10−10 J

(d) ∆U = 1.6 × 10–10 J – 1.0 × 10–10 J = 6 × 10−11 J = 4 × 108 eV (e)

25-77.

288

U has atomic mass 238 g/mol = 0.238 kg/mol Therefore Mol. of U = 1/0.238 = 4.20 mol Number of U atoms = 4.20 mol × 6.02 × 1023/mol = 2.5 × 1024 atoms Each atom releases 6 × 10–11 J. Therefore, energy released = 6 × 10–11 × 2.5 × 1024 J = 1.5 × 1014 J

Let V2,V1 be potential at wire, shell, respectively. Charge Q on wire,

–Q on shell. Then 1 1 1 U= ∑ QV = QV2 + (−Q ) V1 2 2 2 1 = Q(V2 – V1) 2 V2 – V1 = 1.0 × 103 V. Since V2 – V1 = ∆V =

∫

E dr =

λ ⎛b⎞ ln ⎜ ⎟ 2πε 0 ⎝ a ⎠

⎛b⎞ ln ⎜ ⎟ where l = length of tube 2πε 0l ⎝ a ⎠ (see Question 29), 2πε 0l Q= (V2 − V1 ) ln(b / a )

=

Q

Therefore, 1 U = 2πε0l 2

25-78.

⎡ ⎛ b ⎞ ⎤ –1 2 ⎢ ln ⎜ a ⎟ ⎥ (V2 – V1) ⎣ ⎝ ⎠⎦

7 ⎡ ⎛ 1.0 ⎞ ⎤ –1 3 2 = 4.1 × 10 J = π(8.85 × 10–12) 0.10 ⎢ ln ⎜ ⎟ ⎥ (1.0 × 10 ) ⎣ ⎝ 0.001 ⎠ ⎦ 1 R = 1 × 10–15 m, µ = ε0E2 2 1 when U = U = ∫ u dv = ∫ ε 0 E 2 (4π r 2 dr ) 2 For r ≤ R/2, from Example 24.4, 2e 2 U1 = 8πε 0 5 R

For r ≥ R, E = 0; U = 0 ⎛ r 3 − R3 ⎞ ⎜ ⎟ R e ≤ r ≤ R (4π)r 2E = ⎜1 − 3 8 3 ⎟ R −R ⎟ 2 ε0 ⎜ ⎜ ⎟ 8 ⎝ ⎠


CHAPTER

E=

2e ( R3 − r 3 ) 7πε 0 r 2 R 3

R ≤ r ≤ R; 2 R 1 4e 2 1 3 U2 = ε0 ∫ ( R − r 3 ) 2 (4πr 2) dr 3 2 4 R / 2 2 (7πε 0 R ) r

=

⎛ R6 ⎞ − 2rR 3 + r 4 ⎟ dr 6 ∫⎜ 2 49πε 0 R ⎝ r ⎠

8e 2

After simplification, U = U1 + U2 = 25-79.

0.98e 2 = 7.1 × (105 ) eV 8πε 0 R

Electric field in the cylinder is (by Gauss’ Law) Q λ l π r 2 λ lr 2 (i) r < a, 2πrl E(r) = net = = ε0 π a2 ε0a2 ε0 Therefore, E(r) =

λ lr 2 λr = 2πε 0 rl 2πε 0 a 2

(ii) Between the rod and the cylinder, λ 1 a < r < b E(r) = 2πε 0 r (a) Therefore, between the rod and the cylinder, 2

1 1 ⎛ λ 1⎞ λ2 ε0E2 = ε0 ⎜ ⎟ = 2 2 ⎝ 2πε 0 r ⎠ 8π 2ε 0 r 2 (b) Inside the cylinder, the energy per unit length is: u=

2

1 1 ⎛ λr ⎞ λ 2r 2 = ε0 E 2 = ε0 ⎜ 2 ⎟ 2 2 ⎝ 2πε 0 a ⎠ 8π 2ε 0 a 4 Total energy of the system is in length l is: r =a r =b λ 2r 2 λ2 U = ∫ udv = ∫ dV + ∫r =a 8π 2ε 0 r 2 dV r = 0 8π 2ε a 4 0 u=

where U is energy in cylinder in length l. Take volume elements to be concentric cylindrical shells of volume 2πrl dr. Therefore, a b λ 2r 2 λ2 U= ∫ 2π rl dr + ∫a 8π 2ε 0 r 2 2πrl dr 0 8π 2ε a 4 0 =

b dr ⎞ l λ 2 ⎛ a r 3 dr lλ 2 ⎛ 1 b⎞ + = ⎜ ∫0 4 ⎟ ⎜ + ln ⎟ ∫ a a⎠ a r ⎠ 4πε 0 ⎝ 4 4πε 0 ⎝

Therefore, energy per unit length U′ = U/l = 25-80.

F =

λ2 ⎛ 1 b⎞ ⎜ + ln ⎟ 4πε 0 ⎝ 4 a⎠

qQ 4πε 0 r 2

From Newton’s Second Law, m

d 2x qQ = 2 4πε 0 r 2 dt


25

CHAPTER

25

Since m

d2x dv dv dx dv dv qQ =m =m = mv , Therefore, mv = dt 2 dt dx dt dt dt 4πε 0 x 2

v

m ∫ vdv =

⇒

0

⇒

†25-81.

3d

∫ d

dx x2

⇒

mv 2 qQ 1 1 qQ ( − )= = 2 4πε 0 d 3d 6πε 0 d

qQ 3πε 0 md

v=

a=

qQ 4πε 0

F qproton E 1.6 × 10−19 C × 2.5 × 106 V/m = = = 2.40 × 1014 m/s 2 m mproton 1.67 × 1027 kg

⇒

x=

v 2 − v02 0 − (6.9 × 106 m/s) 2 = = 9.96 × 10−2 m 2a 2 × 2.24 × 1014 m/s 2

V = Ex = 2.5 × 106 V/m × 9.96 × 10−2 m = 2.50 × 105 V

25-82.

U(1,0,0) =

qq′ 9 × 109 × (1.6 × 10−19 ) 2 = = –2.3 × 10–18 J 10−10 4πε 0 r

1

U(0.5,0.5,0) = –

9 × 109 × (1.6 × 10−19 ) 2 = –3.3 × 10–18 J 0.707 × 10−10

Work required ∆W = –U(1,0,0) + U(0.5,0.5,0) 1018 = –10–18 J = = −6.25 eV 1.6 × 1019 25-83.

(a) Evaluate

∫ E i dr

by taking the route of

straight line from (0,0,0) to (3,0,0) ∫ Ex dx + ∫ E y dy + ∫ Ez dx = 0 = 25-84.

∫

x =3

x =0

6 x 2 y dx =

∫

x =3

x =0

0 dx (since y = 0) = 0

⎛1 ⎞ (KE + PE) = ⎜ mv 2 + U ⎟ = 0 at r = ∞ ⎝2 ⎠ From conservation of energy, at any other point 2U KE = –PE or v = m −Qe −9 × 109 × 82e 2 = 2.66 × 10−12 J At the nuclear surface, U = = 15 7.1 × 10 4πε 0 R

v|r = R =

2 × 2.66 × 10−12 = 4.1 × 107 m/s −31 3490 × 9.1 × 10

At the center of the nucleus ⎛ 3 ⎞ ⎛ −2U ⎞ 7 v|r = 0 = ⎜ ⎟ i ⎜ ⎟ = 5.02 × 10 m/s ⎝2⎠ ⎝ m ⎠


CHAPTER 25-85.

25

Electric field pointing radially out given by 1 λ 2πε 0 r Therefore, V(b) – V(a) = − ∫

b

a

1 λ dr 2πε 0 r

⎛ λ ⎞ ⎛a⎞ =⎜ ⎟ ln ⎜ ⎟ ⎝ 2πε 0 ⎠ ⎝ b ⎠ Therefore, ⎛ λ ⎞ ⎛a⎞ ∆u = q∆V = –l ⎜ ⎟ ln ⎜ ⎟ ⎝ 2πε 0 ⎠ ⎝ b ⎠ ⎛ 12 mm ⎞ –16 = –1.6 × 10–19 × –5.5 × 10–8 × 2 × 9.0 × 109 ln ⎜ ⎟ = 4.29 × 10 J. mm 0.8 ⎝ ⎠ This energy converted to kinetic energy

1 2 mv = E = v ⇒ 2 25-86.

2 × 4.29 × 10−15 m/s = 3.1 × 107 m/s 9.1 × 1031

2E = m

(a) Potential from a point charge is Vpoint =

q

4πε 0 r

The rod can be divided into small segments with length dx. Each dx can be seen as a point charge Q with dq = dx l dq Q dx The potential from each segment is dV = = 2 4πε 0 r 4πε 0 l x + y 2 ⇒ V =

Q

4πε 0 l

(b) E y = − 25-87.

σ=

Q = A

x =l

dx

∫

x2 + y 2

x =0

=

Q

4πε 0 l

ln

l + l 2 + y2 y

l + l 2 + y2 ∂V Q ∂ Q =− (ln )= ∂y 4πε 0l ∂y y 4πε 0 y l 2 + y 2

Q

=

4Q C/m2 2 3π R

R ⎞ )⎟ 4 ⎠ ⎝ Potential at point P is given by 1 dQ 2π drσ dV = = 4πε 0 µ 4πε 0 r 2 + x 2 V= =

σ 2ε 0

⎛

π ⎜ (R2

2

R

rdr

R/ 2

2

∫

r + x2

⎞ R2 σ ⎛ 2 + x2 ⎟ ⎜ R + x2 ⎟ 2ε 0 ⎜⎝ 4 ⎠


CHAPTER

=

25

2Q 3πε 0 R 2

⎛ ⎜ R2 + x2 ⎜ ⎝

⎞ R2 + x2 ⎟ ⎟ 4 ⎠

⎛ ⎜ ∂V 2Q ⎜ x ˆ E = –i = 2 2 ∂x 3¹ε 0 R ⎜ R + x 2 ⎜ ⎝ 25-88.

⎞ ⎟ x ⎟ î ⎟ R2 + x2 ⎟ 4 ⎠

q2 ⎛ 1 1 1 1 1 1⎞ + + + + + ⎟ ⎜ 4πε 0 ⎝ L 2L L L 2L L ⎠ 2 q ⎛4 2 ⎞ = + ⎜ ⎟ 4πε 0 ⎝ L 2L ⎠ q2 = (4 + 2) 4πε 0 L U=

This is the kinetic energy, at large distances, of all four particles. ⎛1 ⎞ K.E. = 4 ⎜ mv 2 ⎟ when v = speed of each. ⎝2 ⎠ 2 q (4 + 2) 2mv2 = 4πε 0 ⎡ q2 ⎤ v2 = ⎢ (4 + 2) ⎥ m/sec ⎣ 8πε 0 mL ⎦ or 1/ 2

⎡ q2 ⎤ v2 = ⎢ (4 + 2) ⎥ ⎣ 8πε 0 mL ⎦

25-89.

m/sec

1 1 1 Vother (1) q1 + Vother (2) q(2) + . . . = Vother (i) q (i) 2 2 2 where Vother = potential at dv due to all charges du 1 du (Q / l ) Q du dvother = = 4πε 0 u + x + V 4πε 0l µ + x + V l Q l dµ Vother = ∫ dVother = ∫ 0 4πε 0l 0 µ + x + V V Q ∑i q(i) = ∫0 dV l Therefore, U=


CHAPTER

U= = = 25-90.

1⎡ Q ⎢ 2 ⎣ 4πε 0l Q2

4πε 0l

2

∫

l

0

∫

dv

l

0

du µ + x +V

∫

l

0

∫

V

0

Q ⎛ Q ⎞⎤ dv ⎜ ⎟ ⎥ = 4πε 0l 2 ⎝ l ⎠⎦ 2

du Q2 = µ + x + v 4πε 0l 2

∫

l

0

∫

l

0

V du dv ∫ µ+x+v 0

⎛l + x+v⎞ dv ln ⎜ ⎟ ⎝ x+v ⎠

⎡ ( x + 2l ) x x + 2l ⎤ x ln + 2l ln ⎢ 2 4πε 0 l ⎣ (x + l) x + l ⎥⎦ Q2

2

1 3Q 2 4πε 0 5R 1 3 ∆U = ( Q22Q12 ) = 3.6 × 106 eV 4πε 0 5 R

U=

Therefore, 1 3 l 2 (82 − 7 2 ) R= m = 3.5 × 1015 m 4πε 0 5 3.7 × 106 l 25-91.

There is no electric field in the cavity, hence its energy = 0. Q ⎛ r 3 − a3 ⎞ By Gauss’ Law, E(r) in the shell is given by: 4πr 2E(r) = ⎜ ⎟ ε 0 ⎝ b3 − a 3 ⎠ ⎛ Q 1 ⎛ r 3 − a3 ⎞ Q a3 ⎞ = − r Therefore, E(r) = ⎜ ⎟ ⎜ ⎟ 4πε 0 r 2 ⎝ b3 − a 3 ⎠ 4πε 0 (b3 − a 3 ) ⎝ r2 ⎠ 2

2

⎛ ⎞ ⎛ 1 a3 ⎞ Q 2 r − ε 0 ∫r =a 2 ⎜⎝ 4πε 0 (b3 − a3 ) ⎟⎠ ⎜⎝ r 2 ⎟⎠ 4πr dr r =b ⎛ 1 Q2 a6 ⎞ 4 3 = − + r 2 a r ⎜ ⎟ dr 2 4πε 0 (b3 − a 3 ) 2 ∫r = a ⎝ r2 ⎠

1 Therefore, Ushell = ∫ ε0E 2 dV. = 2

r =b

b

⎡ r5 1 Q2 a6 ⎤ 3 2 − − a r = ⎢ ⎥ 2 4πε 0 (b3 − a 3 ) 2 ⎣ 5 r ⎦a ⎛ b5 ⎞ 1 Q2 a6 a6 3 2 a b − − − + a5 + a5 ⎟ = ⎜ 3 3 2 2 4πε 0 (b − a ) ⎝ 5 b 5 ⎠ ⎛ b5 1 Q2 a6 9 5 ⎞ − a 3b 2 − + a ⎟ 3 3 2 ⎜ 2 4πε 0 (b − a ) ⎝ 5 b 5 ⎠ Q From Gauss’ Law, E (r > b) = , 4πε 0 r 2

Ushell =

2

1 ⎛ Q ⎞ 1 Q2 2 4πr So, Uoutside= ∫ ε0 ⎜ dr = ⎟ 2 ⎝ 4πε 0 r 2 ⎠ 8πε 0 b b therefore Q2 9 ⎛6 6 ⎞ U total = b − 3a 3b3 + a 5b ⎟ (after simplification) 3 3 2 ⎜ 8πε 0 b(b − a ) ⎝ 5 5 ⎠ ∞


25

CHAPTER

25-92.

25

Density of the original nucleus is: ρ0 =

4 π R3 3 This should be equal to the density of the final two nuclei. Therefore, both nuclei have density q charge of each nucleus q / 2 , where V is the volume of one nucleus, which is ρn = = = 4 volume of each nucleus V 3 πR 3 4 equal to π r 3 3 q/2 q R So: = ⇒ r = 1 / 3 = 0.794 R 4 3 4 2 πr π R3 3 3 Each of the final nuclei has energy: 1 3(q / 2) 2 1 ⎛ 3q 2 / 4 ⎞ 0.315 3q 2 Un = ( )= ⎜ ⎟= 4πε 0 5r 4πε 0 ⎝ 5 × 0.794 R ⎠ 4πε 0 5R

The two nuclei have energy: †25-93.

q

0.630 3q 2 , which is 0.630 times the energy of the original nucleus. 4πε 0 5R

Use the electric fields found in Problem 71: ρr 3Q E (r ≤ R) = , where ρ = 3ε 0 4π R 3 Q E (r ≥ R) = 4πε 0 r 2 ∞ ⎤ 2 2 ≤ + π | E ( r R ) | 4 r dr | E (r ≥ R) |2 4π r 2 dr ⎥ ⎢∫ ∫ 2 ⎣0 R ⎦ 2 R 2 ∞ ⎡ 4πρ 4π Q dr ⎤ r 4 dr + ⎢ 2 ∫ 2 2 ∫ 2 ⎥ 16π ε 0 R r ⎦ ⎣ 9ε 0 0 ⎡ 4πρ 2 R 5 Q2 ⎤ + ⎢ ⎥ 2 2 4πε 0 R ⎦ ⎣ 9ε 0 5 ⎡ Q2 Q2 ⎤ 3Q 2 + = ⎢ ⎥ 2 2 ⎣ 20πε 0 R 4πε 0 R ⎦ 20πε 0 R

⇒ U = =

ε0

=

ε0

=

ε0

2 2 2

ε0 ⎡ R


CHAPTER 26

CAPACITORS AND DIELECTRICS


The potential is defined as 1 Q V = 4πε 0 R

Since the potential is the same on the two spheres, if we increase the radius by a factor of three, the amount of charge on the sphere would have to also increase by a factor of three. Mathematically, 1 Q1 1 Q2 = V = 4πε 0 R 4πε 0 3R

Q1 Q2 = R 3R Q2 = 3Q1 When the charges are the same, the difference in potentials is Q = 4πε 0V1 R = 4πε 0V2 (3R) V1 = 3V2 26-2.

†26-3.

26-4.

(a) C = 4πε 0 R = 4π (8.85 × 10−12 F/m)(0.18 m) = 2.0 × 10−11 F

(b) Q = CV = (2.0 × 10−11 F)(2.0 × 105 V) = 4.0 × 10−6 C or 4.0 µC If your head can be modeled as a sphere of radius R = 0.10 m, then the capacitance of it would be C = 4πε 0 R = 4π ( 8.85 × 10−12 F/m ) (0.10 m) = 1.1 × 10−11 F according to Equation 26.3. Equation 26.2 gives the relationship of the amount of charge to the capacitance C and the potential V, Q = CV = (1.1 × 10−11 F)(1.0 × 105 V) = 1.1 × 10−6 C or 1.1 µC Let the radius of the solid, inner sphere be R1; and that of the outer, hollow sphere be R2. Assume the inner sphere has a net positive charge Q distributed equally over its surface. The capacitance is Q C= V To determine the capacitance, the potential between the two surfaces must be determined. 4πε 4πε 0 Q Q Q = = = R10 = C= R 1 1 Q dr 1 V − ∫ E dr − 1 − 2 ∫ R 4πε 0 2 r R1 R2 r =

4π ( 8.85 × 10−12 F/m ) 1 1 − 0.050 m 0.12 m

R2

= 9.5 × 10−12 F


CHAPTER †26-5.

26-6.

†26-7.

26-8.

26-9.

Equation 26.9 gives the capacitance of a parallel plate capacitor with a plate area A and distance between plates d as −12 2 ε A ε π r 2 ( 8.85 × 10 F/m ) π (0.20 m) = = 1.1 × 10−9 F C= 0 = 0 1.0 × 10−3 m d d

When connected to a 12-V battery, the amount of charge Q on each plate will be Q = CV = (1.1 × 10−9 F)(12 V) = 1.3 × 10−8 C ε A ε A C= 0 gives d = 0 d C V VC (4.4 V)(3.0 × 10−6 F) E= = = = 9.9 × 104 V/m d ε 0 A ( 8.85 × 10−12 F/m ) (15 m 2 ) The charge on each plate is Q = CV = (4.0 × 10−6 F)(9.0 V) = 3.6 × 10−5 C To reverse the polarity of the plates, you must move twice this amount of charge. If you just move an amount Q of charge, the plates would return to electrically neutral. An amount of charge 2Q is equal to −5 2Q 2 ( 3.6 × 10 C ) n= = = 4.5 × 1014 electrons −19 e 1.60 × 10 C As in Problem 26-4, 4πε 0 4πε 0 Q Q Q = = = = C= a a 1 1 dr V − ∫ E dr − Q 1 − 2 ∫ b 4πε 0 r a b rb ε A C= 0 d dC (5.0 × 10−4 m)(10.0 × 10−12 F) Amin = = = 5.6 × 10−4 m 2 ε0 8.85 × 10−12 F/m Amax =

26-10. 26-11. 26-12. †26-13.

26

dC

ε0

=

(5.0 × 10−4 m)(120 × 10−12 F) = 6.8 × 10−3 m 2 −12 8.85 × 10 F/m

Q = CV = (1.0 × 10−7 F)(5.0 V) = 5.0 × 10−7 C or 0.50 µC Q = CV = (50.0 F)(2.50 V) = 125 C Q = CV = (5.1 × 10−3 F)(250 V) = 1.3 C Equation 26.9 gives the capacitance of a parallel plate capacitor with a plate area A and distance between plates d as ε A (8.85 × 10−12 F/m)(2 × 10−7 m) 2 C= 0 = = 1.8 × 10−17 F d 2.0 × 10−8 m The charge on the plate is Q = CV = (1.8 × 10−17 F)(0.10 V) = 1.8 × 10−18 C


CHAPTER

26-14.

A=

d1C1

ε0

∆C = C2 − C1 =

†26-15.

⎛d ⎞ ε 0 d1C1 − C1 = C1 ⎜ 1 − 1 ⎟ d2 ε 0 ⎝ d2 ⎠

⎛ 5.0 mm ⎞ = 6.0 × 10−13 F ⎜ − 1 ⎟ = 1.44 × 10−11 F = 1.4 × 10−11 F 0.20 mm ⎝ ⎠ Q = CV = (1.44 × 10−11 F)(8.0 V) = 1.2 × 10−10 C From Problem 25-35, the thin wire at the center of the tube has a radius of 1.25 × 10−5 m and the tube radius is 1.25 × 10−2 m. The electric field inside the tube is 1 λ E= 2πε 0 r The potential difference between the tube and the wire is determined from the electric field through integration. r1 r λ r1 1 λ ln 2 ∆V = − ∫ E dr = − dr = ∫ r2 r 2 2πε 0 r 2πε 0 r1 The capacitance per unit length is

2π ( 8.85 × 10−12 F/m ) 2πε 0 = = 8.0 × 10−12 F/m −2 λ r2 r2 ∆V ⎛ ⎞ 1.25 × 10 m ln ln ln ⎜ ⎟ −5 2πε 0 r1 r1 ⎝ 1.25 × 10 m ⎠ Cparallel = C1 + C2 + C3 = 3.0 µ F + 5.0 µ F + 7.5 µ F = 15.5 µ F

C=

26-16.

1 Cseries

λ

=

=

λ

=

1 1 1 1 1 1 2 + + = + + = 3 µF C1 C2 C2 3.0 µ F 5.0 µ F 7.5 µ F

Cseries = 1.5 µ F †26-17.

To determine the net capacitance, begin by combining the two capacitors on the left in series, then combine that result with the capacitor on the right in parallel. 1 1 1 1 1 = + = + Cseries C1 C2 5.0 µ F 3.0 µ F Cseries = 1.88 µ F C = Cseries + C3 = 1.88 µ F + 8.0 µ F = 9.88 µ F = 9.9 µ F

26-18.

26

⎛ 1 1 ⎞ Q V = V1 + V2 = Q ⎜ + ⎟= ⎝ C1 C2 ⎠ Cseries


CHAPTER Q=

26 V

⎛ 1 1 ⎞ + ⎜ ⎟ ⎝ C1 C2 ⎠ Q V V1 = = C1 ⎛ 1 1 ⎞ C1 ⎜ + ⎟ ⎝ C1 C2 ⎠ 24 V = = 14.8 V ⎛ ⎞ 1 1 −6 5.0 × 10 F ⎜ + ⎟ −6 −6 ⎝ 5.0 × 10 F 8.0 × 10 F ⎠ V2 =

†26-19.

Q = C2

V

⎛ 1 1 ⎞ C2 ⎜ + ⎟ ⎝ C1 C2 ⎠ 24 V = = 9.23 V = 9.2 V ⎛ ⎞ 1 1 −6 8.0 × 10 F ⎜ + ⎟ −6 −6 ⎝ 5.0 × 10 F 8.0 × 10 F ⎠ The total charge on the three capacitors is the sum of the charge stored on each capacitor. Q = Q1 + Q2 + Q3 = C1V + C2V + C3V

= V (C1 + C2 + C3 ) = ( 30 V ) (1.0 × 10−5 F + 1.5 × 10−5 F + 3.0 × 10−5 F) = 0.00165 C = 1.7 × 10−3 C −1

26-20.

−1

⎛1 1⎞ ⎛1 1⎞ ⎛1 1⎞ Ceq = ⎜ + ⎟ + ⎜ + ⎟ + ⎜ + ⎟ C C C C ⎝ ⎠ ⎝ ⎠ ⎝C C⎠ −1

26-21. 26-22.

−1

=

C C C 3 + + = C 2 2 2 2

−1

C C 2 ⎛1 1 1⎞ ⎛1 1 1⎞ C =⎜ + + ⎟ +⎜ + + ⎟ = + = C 3 3 3 ⎝C C C⎠ ⎝C C C⎠ For the three capacitors on the right in series, the net capacitance is −1

⎛ 1 1 1 ⎞ + + Cright = ⎜ ⎟ = 3.0 µ F ⎝ 9.0 µ F 9.0 µ F 9.0 µ F ⎠ For the three capacitors on the left side, the net capacitance is −1

⎛ 1 1 ⎞ + Cleft = ⎜ ⎟ = 3.0 µ F ⎝ 3.0 µ F + 3.0 µ F 6.0 µ F ⎠ Therefore, combining the left and right sides in parallel with the capacitor at the top in series, the net capacitance for this combination is −1

†26-23.

−1

⎛ ⎛ 1 1 ⎞ 1 1 ⎞ C=⎜ + + ⎟⎟ = ⎜ ⎟ = 3.0 µ F ⎜C +C Ctop ⎠ ⎝ 3.0 µ F + 3.0 µ F 6.0 µ F ⎠ right ⎝ left When capacitors are connected in parallel, the total charge is divided among the capacitors, Q = Q1 + Q2 = C1V + C2V = (C1 + C2 )V = CparallelV Since the capacitors have equal capacitance values, the total charge is 12.0 µC. When capacitors are connected in series, the capacitors all have the same charge, regardless of the values of the capacitances. Therefore, the charge on the plates of the 6.0 µF capacitor must be 12.0 µC. The two 4.0 µF capacitors have a combined capacitance of


CHAPTER

26

C = 4.0 µ F + 4.0 µ F = 8.0 µ F The unknown voltage for the series combination of the above capacitance with the 6.0 µF capacitor is ⎛ 1 ⎛ ⎞ 1 ⎞ 1 1 −6 + + V = Q⎜ ⎟ = (12.0 × 10 C ) ⎜ ⎟ = 3.5 V −6 −6 ⎝ 8.0 × 10 F 6.0 × 10 F ⎠ ⎝ C1 C2 ⎠ −1

26-24.

⎛ 1 1 ⎞ Q = CseriesV = ⎜ + ⎟ V ⎝ C1 C2 ⎠ 1.5 V = = 2.14 × 10−6 C = 2.1 × 10−6 C or 2.1 µ C ⎛ ⎞ 1 1 + ⎜ ⎟ −6 −6 ⎝ 2.0 × 10 F 5.0 × 10 F ⎠ Q 2.14 × 10−6 C = = 1.07 V V2.0 µ F = C 2.0 × 10−6 F Q 2.14 × 10−6 C = = 0.428 V = 0.43 V C 5.0 × 10−6 F When the first capacitor is connected to the battery, its plates are charged to Q = CV = ( 2.5 × 10−6 F ) ( 9.0 V ) = 2.25 × 10−5 C V5.0 µ F =

†26-25.

Then, the battery is removed and the capacitor is connected in series with an uncharged 5.0-µF capacitor. When two capacitors are connected in series, the charge on each capacitor is the same, regardless of their capacitance. When the second capacitor is connected, electrons from the negative plate move through the connecting wire and collect on the plate at the other end of the wire. Charges will move until equilibrium is achieved. That will occur when the charge on all of the plates is equal to Q 2.25 × 10−5 C = = 1.13 × 10−5 C = 1.1 × 10−5 C Q′ = 2 2 The potential difference across each of the two capacitors is Q 1.13 × 10−5 C = = 4.5 V V2.5 µ F = C 2.5 × 10−6 F Q 1.13 × 10−5 C = = 2.3 V C 5.0 × 10−6 F This arrangement is equivalent to three capacitors connected in parallel. The two plates in the middle will have opposite charges on their opposite sides. Therefore, ⎛ε A⎞ C = 3⎜ 0 ⎟ ⎝ d ⎠ V5.0 µ F =

26-26.


CHAPTER †26-27.

26

If the capacitors are arranged like this: When you combine the two capacitors on either the left or the right in series, the result is 1 1 1 2 = + = Cseries C C C C C C 2 Then, when you combine these equivalent capacitances in parallel, you get C C Cparallel = Cseries + Cseries = + = C 2 2 Cseries =

C

26-28.

C

This is the only combination of four capacitors of equal capacitance C that will result in an equivalent capacitance of C. The capacitors are initially charged with charges Q1 = C1V and Q2 = C2V where C1 > C2

when they were individually connected to the battery of voltage V = 24 V. When they are subsequently connected together, the charges redistribute until equilibrium is established and both capacitors are at the same potential difference. Let Q′1 and Q′2 represent the final charge on the plates of each capacitor. The relationship is Q1′ Q2′ = C1 C2 We also know that charge is conserved in this process, where the total charge is Q1 + Q2 = ( C1 − C2 ) V = Q1′ + Q2′ The minus sign occurs because the negative plate of one capacitor is connected to the positive plate of the other. ⎛C ⎞ ( C1 − C2 ) V = Q2′ ⎜ 1 − 1⎟ ⎝ C2 ⎠ ⎛ C − C2 ⎞ ⎛ 6.0 × 10−6 F − 2.0 × 10−6 F ⎞ −6 Q2′ = ⎜ 1 C V = ⎟ 2 ⎜ ⎟ ( 2.0 × 10 F ) ( 24 V ) −6 −6 C C 6.0 × 10 F 2.0 × 10 F + + ⎝ ⎠ 2 ⎠ ⎝ 1 −5 = 2.4 × 10 C

and ⎛C ⎞ ⎛ 6.0 × 10−6 F ⎞ −5 Q1′ = ⎜ 1 ⎟ Q2′ = ⎜ ⎟ ( 2.4 × 10 C ) −6 C 2.0 × 10 F ⎝ ⎠ ⎝ 2⎠ −5 = 7.2 × 10 C

The final potential difference for each capacitor is Q′ 7.2 × 10−5 C V1′ = 1 = = 12.0 V C1 6.0 × 10−6 F V2′ =

Q2′ 2.4 × 10−5 C = = 12.0 V C2 2.0 × 10−6 F


CHAPTER †26-29.

26

The capacitors are initially charged with charges Q1 = C1V , Q2 = C2V , and Q3 = −C3V when they were individually connected to the battery of voltage V = 36 V. The convention has been used that as we follow a counterclockwise loop, the voltage increases when moving from a negative plate to a positive plate. When they are subsequently connected together, the charges redistribute until equilibrium is established and both capacitors are at the same potential difference. Let Q′1, Q′2, and Q′3 represent the final charge on the plates of each capacitor and apply the principle of the conservation of charge to each of the three segments (remembering that charge does not pass across the gap between plates). Q2′ − Q1′ = Q2 − Q1 = ( C2 − C1 ) V (1) = ( 5.0 × 10−6 F − 2.0 × 10−6 F ) (36 V) = 1.08 × 10−4 C

Q3′ − Q2′ = Q3 − Q2 = ( C3 − C2 ) V

(2)

Q1′ − Q3′ = Q1 − Q3 = ( C1 − C3 ) V

(3)

= ( 7.0 × 10−6 F − 5.0 × 10−6 F ) (36 V) = 7.2 × 10−5 C = ( 2.0 × 10−6 F − 7.0 × 10−6 F ) (36 V) = − 1.8 × 10−4 C

also, the sum of the voltages as the loop is made around the circuit is zero, therefore Q′ Q′ Q′ (4) V1′ + V2′ + V3′ = 1 + 2 + 3 = 0 C1 C2 C3 From (4), we can solve for Q′1 to find ⎛ Q′ Q′ ⎞ Q1′ = C1 ⎜ 2 + 3 ⎟ ⎝ C2 C3 ⎠ Substituting this result into (3) gives ⎛ Q′ Q′ ⎞ C1 ⎜ 2 + 3 ⎟ − Q3′ = − 1.8 × 10−4 C ⎝ C2 C3 ⎠

⎛C Q2′ = −1.8 × 10−4 C ⎜ 2 ⎝ C1

⎛ C1 ⎞⎛ C ⎞ ⎞ − 1⎟ ⎜ 2 ⎟ ⎟ − Q3′ ⎜ ⎠ ⎝ C3 ⎠ ⎝ C1 ⎠

⎛ 5.0 × 10−6 F ⎞ ⎛ 2.0 × 10−6 F ⎞ ⎛ 5.0 × 10−6 F ⎞ ′ = ( −1.8 × 10−4 C ) ⎜ − − Q 1 ⎟ ⎟ ⎟⎜ 3⎜ −6 −6 −6 ⎝ 2.0 × 10 F ⎠ ⎝ 7.0 × 10 F ⎠ ⎝ 2.0 × 10 F1 ⎠ = −4.50 × 10−4 C + 1.79Q3′ Then, substitute this result into Equation 2. Q3′ − Q2′ = = 7.2 × 10−5 C Q3′ + 4.50 × 10−4 C − 1.79Q3′ = 7.2 × 10−5 C Q3′ = 4.78 × 10−4 C = 4.8 × 10−4 C and Q2′ = −4.50 × 10−4 C + 1.79Q3′ = −4.50 × 10−4 C + 1.79(4.78 × 10−4 C) = 4.06 × 10−4 C = 4.1 × 10−4 C


CHAPTER

26

Q2′ − Q1′ = 1.08 × 10−4 C Q1′ = Q2′ − 1.08 × 10−4 C = 4.06 × 10−4 C − 1.08 × 10−4 C = 2.97 × 10−4 C = 3.0 × 10−4 C

26-30.

†26-31.

The voltage across the plates of the third capacitor is Q′ 4.78 × 10−4 C VPP′ = 3 = = 68 V C3 7.0 × 10−6 F ε A C =κ 0 d dC (5.0 × 10−5 m)(3.0 × 10−6 F) = = 7.4 m 2 A= −12 κε 0 (2.3) ( 8.85 × 10 F/m ) For the sphere in free space, the capacitance is C0 = Q/V where Q is the charge on the sphere and V is the potential, 1 Q V = 4πε 0 R Therefore, Q C0 = = 4πε 0 R 1 Q 4πε 0 R

When the sphere is in a dielectric medium, the capacitance increases by a factor of the dielectric constant κ, C = κ C0 = 4πε 0κ R 26-32.

(a) C = κ

ε0 A d

= ( 3.4 )

( 8.85 × 10 F/m )( 0.050 m ) ( 2.0 × 10 m ) −12

2

−4

= 7.52 × 10−9 F = 7.5 × 10−9 F or 7.5 nF (b) E = V =

1 Q 1 CV = κ ε0 A κ ε0 A

κε 0 AE C

=

(3.4) ( 8.85 × 10−12 F/m ) (0.050 m 2 ) ( 40 × 106 V/m ) 7.52 × 10−9 C

= 8.0 × 103 V (c) Q = CV = ( 7.52 × 10−9 F ) (8.0 × 103 V) = 6.0 × 10−5 C 26-33.

26-34.

C =κ

ε0 A d

d C ⎛ 5 × 10−8 m ⎞ ⎛ F ⎞ ⎛ 100 cm ⎞ −5 =⎜ κ = ⎟ ⎜ 9 × 10 ⎟⎜ ⎟ −12 cm 2 ⎠ ⎝ 1 m ⎠ ε 0 A ⎝ 8.85 × 10 F/m ⎠ ⎝ = 5100 1 Q 1 CV E= = κ ε0 A κ ε0 A

A=

2

1 CV κ ε0 E


CHAPTER

Apolystyrene = Anylon = †26-35.

26

(1.0 × 10−6 F ) ( 25 V ) 1 = 0.047 m 2 2.5 ( 8.85 = × 10−12 F/m )( 24 × 106 V/m )

(1.0 × 10−6 F ) ( 25 V ) 1 = 0.058 m 2 3.5 ( 8.85 × 10−12 F/m )(14 × 106 V/m )

The initial capacitance of the empty capacitor is C0. When the dielectric completely fills the space between the capacitor plates, the capacitance is C. One relationship between these two capacitances is due to the conservation of charge, Q = C0V0 = CV C V0 = C0 V According to Equation 26.21, C V0 3.0 × 105 V κ = = = = 1.7 1.8 × 105 V C0 V

26-36.

The capacitor with two dielectrics within is equivalent to two capacitors in series, each with a single dielectric inside of thickness d/2. The electric field within the each dielectric is 1 E = Efree

κ

and the potential difference is for the capacitor is d d ∆V = E1 + E2 2 2 The charge on each plate is of equal magnitude Q = C∆V, where the series capacitance is 1 1 1 d d d ⎛ κ1 + κ 2 ⎞ ∆V = + = = + = ⎜ ⎟ C C1 C2 Q 2ε 0κ 1 A 2ε 0κ 2 A 2ε 0 A ⎝ κ 1κ 2 ⎠ C= †26-37.

26-38.

2ε 0 A ⎛ κ 1κ 2 ⎞ ⎜ ⎟ d ⎝ κ1 + κ 2 ⎠

The capacitor contains two dielectrics of thickness d side by side. We can consider this in the same manner as two capacitors in parallel. In parallel, the capacitances add together, ⎛ε A⎞ ⎛ε A⎞ ⎛ε A⎞ C = κ1 ⎜ 0 ⎟ + κ 2 ⎜ 0 ⎟ = (κ1 + κ 2 ) ⎜ 0 ⎟ ⎝ 2d ⎠ ⎝ 2d ⎠ ⎝ 2d ⎠ Note that the area of each of the plates of the parallel capacitors is one half that of the one shown in Figure 26.34. The capacitor shown in Figure 26.35 can be modeled as two capacitors in series with a plate separation d/2 as in Problem 26-36, except that in this case one of the equivalent capacitors does not contain a dielectric. The capacitances are 2ε A 2ε κ A C1 = 0 and C2 = 0 d d The net capacitance is then found by summing these capacitances in series, 1 1 1 d d d ⎛ 1⎞ = + = + = ⎜1 + ⎟ κ⎠ C C1 C2 2ε 0 A 2ε 0κ A 2ε 0 A ⎝ 2ε A ⎛ κ ⎞ C= 0 ⎜ ⎟ d ⎝ κ + 1⎠


CHAPTER 26-39.

26

(a) The electric field in the dielectric is E = Efree/κ and the potential difference between the plates is d d E d⎛ 1⎞ ∆V = E + Efree = free ⎜ 1 + ⎟ κ⎠ 2 2 2 ⎝ 2∆V ⎛ κ ⎞ Efree = ⎜ ⎟ d ⎝1+ κ ⎠ (b) E =

Efree

κ

=

2∆V ⎛ 1 ⎞ ⎜ ⎟ d ⎝1+ κ ⎠

2ε ∆V ⎛ κ ⎞ Q = ε 0 Efree = 0 ⎜ ⎟ A d ⎝1+ κ ⎠ The density of the bound charge on the surface of the dielectric is related to the free charge density. 2∆V ⎛ 1 ⎞ ⎜ ⎟ σ bound E d ⎝1+ κ ⎠ 1 = = = 2∆V ⎛ κ ⎞ κ Efree σ free ⎜ ⎟ d ⎝1+ κ ⎠ σ 2ε ∆V ⎛ 1 ⎞ σ bound = free = 0 ⎜ ⎟ κ d ⎝1+ κ ⎠ ε A ε A C1.0 atm = κ 1.0 0 and C3.0 atm = κ 3.0 0 d d Therefore, C3.0 atm κ 3.0 Q3.0 = = C1.0 atm κ 1.0 Q1.0 (c) σ free =

26-40.

The dielectric constant is related to the pressure, κ −1 = α p where α is a proportionality constant. From Table 26.1, κ1.0 = 1.00054. Using this information, α can be found. κ −1 = α p κ − 1 1.00054 − 1 α = 1.0 = = 5.4 × 10−4 / atm p 1.0 atm −4 Q3.0 κ 3.0 α p2 + 1 ( 5.4 × 10 /atm ) (3.0) + 1 = = = = 1.00108 Q1.0 κ 1.0 1.00054 1.00054

†26-41.

Therefore, with a factor of 3 increase in pressure, the charge on the plates increases by about 0.11%. (a) The electric field in the space between the conductors is E=

λ 2πε 0 r

Assuming the charge per unit length on each conductor is the same, the potential difference between the conductors is


CHAPTER ∆V =

b

a

E dr =

∫

b

a

λ λ dr = 2πε 0 r 2πε 0

∫

b

a

1 dr r

λ

⎛b⎞ ln ⎜ ⎟ ⎝a⎠ The capacitance is then 2πε 0 L λL λL Q = = = C0 = λ ∆V ∆V ⎛b⎞ ⎛b⎞ ln ⎜ ⎟ ln ⎜ ⎟ 2πε 0 ⎝a⎠ ⎝a⎠ F⎞ ⎛ 2π ⎜ 8.85 × 10−12 ⎟ (0.50 m) m⎠ ⎝ = = 2.0 × 10−11 F 4.0 mm ⎛ ⎞ ln ⎜ ⎟ 1.0 mm ⎝ ⎠ (b) C = κC0 = (1.433)(2.0 × 10−11 F) = 2.9 × 10−11 F 1 1 1 2 (a) = + = C C1 C2 C0 =

26-42.

∫

2πε 0

⎛ 2.0 × 10−6 F ⎞ C0 −6 ∆V = ⎜ ⎟ (9.0 V) = 9.0 × 10 C 2 2 ⎝ ⎠ 1 1 1 1 ⎛ 1⎞ (b) = + = ⎜1 + ⎟ κ⎠ C C0 κ C0 C0 ⎝ ⎛ κ ⎞ ⎛ 2.5 ⎞ −6 −6 C = C0 ⎜ ⎟ = ( 2.0 × 10 F ) ⎜ ⎟ = 1.43 × 10 F ⎝ κ + 1⎠ ⎝ 3.5 ⎠ The charge on each capacitor is the same; and its value is Q = C ∆V = (1.43 × 10−6 F ) (9.0 V) =1.287 × 10−5 C = 1.3 × 10−5 C Q = C ∆V =

The potential difference across the empty capacitor is Q 1.287 × 10−5 C ∆V0 = = = 6.4 V C0 2.0 × 10−6 F The potential difference for the filled capacitor is then Q 1.287 × 10−5 C ∆V = = = 2.57 V = 2.6 V κ C0 2.5 ( 2.0 × 10−6 F ) 26-43.

1 1 1 2 = + = C C0 C0 C0 C0 2.0 × 10−6 F = = 1.0 × 10−6 F 2 2 Q = C ∆V = (1.0 × 10−6 F ) (9.0 V) = 9.0 × 10−6 C C=

∆Vfilled =

Q 9.0 × 10−6 C = = 1.8 V κ C0 2.5 ( 2.0 × 10−6 F )

∆Vempty =

Q 9.0 × 10−6 C = = 4.5 V C0 2.0 × 10−6 F


26

CHAPTER

26-44.

26

(a) σ free, R2 =

Q Q = A 4π R22

(b) The total electric field within the capacitor is the sum of the fields from bound and free surface charges: E = Efree + Ebound This is then equivalent to Efree = Efree + Ebound

κ

⎛1 ⎞ ⎛1− κ ⎞ Efree − Efree = Efree ⎜ − 1 ⎟ = Efree ⎜ ⎟ κ ⎝κ ⎠ ⎝ κ ⎠ This can then be expressed in terms of surface charge densities as ⎛ κ − 1⎞ σ bound = ⎜ ⎟ σ free ⎝ κ ⎠ For the problem at hand, we have κ −1 ⎛ κ −1⎞⎛ Q ⎞ σ bound, R1 = σ free = ⎜ ⎟⎜ 2 ⎟ κ ⎝ κ ⎠ ⎝ 4π R1 ⎠ Ebound =

1

(c) σ bound, R2 = −

26-45.

κ −1 ⎛ κ −1⎞⎛ Q ⎞ σ free = − ⎜ ⎟⎜ 2 ⎟ κ ⎝ κ ⎠ ⎝ 4π R2 ⎠

The minus sign indicates the opposite sign of the sign at each side of the dielectric. κ −1 ⎛ 2.8 − 1 ⎞ −6 2 −6 2 (a) σ bound, inside = − σ free = − ⎜ ⎟ 4.0 × 10 C/m = −2.6 × 10 C/m κ ⎝ 2.8 ⎠ The charge per unit length is the same for both surfaces, only the radii are different, so ⎛ κ − 1 ⎞ ⎛ Rinside ⎞ σ bound, outside = ⎜ ⎟ σ free ⎟⎜ ⎝ κ ⎠ ⎝ Routside ⎠

⎛ 2.8 − 1 ⎞ ⎛ 0.20 cm ⎞ 2 2 −6 −6 =⎜ ⎟⎜ ⎟ 4.0 × 10 C/m = 1.7 × 10 C/m 2.8 0.30 cm ⎝ ⎠⎝ ⎠ 1 1 ⎛ 1 λ ⎞ 1 ⎛ 1 Q ⎞ 1 ⎛ 1 ( 2π Rinside ) lσ free ⎞ (b) E = Efree = ⎜ ⎟ ⎟= ⎜ ⎟= ⎜ κ κ ⎝ 2πε 0 R ⎠ κ ⎝ 2πε 0 lR ⎠ κ ⎝ 2πε 0 lR ⎠ 1 ⎛ Rinside ⎞ = σ free κε 0 ⎜⎝ R ⎟⎠ At the inner surface, the radio of the radii is equal to 1. σ 4.0 × 10−6 C/m 2 = 1.6 × 105 V/m Einner = free = 2 κε 0 ⎛ ⎞ C 2.8 ⎜ 8.85 × 10−12 ⎟ N i m2 ⎠ ⎝ At the outer surface, σ ⎛R ⎞ Eouter = free ⎜ inside ⎟ = κε 0 ⎝ R ⎠

4.0 × 10−6 C/m 2 ⎛ C2 2.8 ⎜ 8.85 × 10−12 N i m2 ⎝

⎛ 0.20 cm ⎞ 5 ⎜ ⎟ = 1.1 × 10 V/m ⎞ ⎝ 0.30 cm ⎠ ⎟ ⎠


CHAPTER

(c) E = Efree =

26

σ ⎛R 4.0 × 10−6 C/m 2 ⎛ 0.20 cm ⎞ λ ⎞ 5 = free ⎜ inside ⎟ = ⎜ ⎟ = 3.0 × 10 V/m 2 C 2πε 0 R 0.30 cm ε0 ⎝ R ⎠ ⎝ ⎠ 8.85 × 10−12 2 Nim

26-46.

The electric field in either the dielectric or the empty space is 1 Q E1 = 4πε 0 r 2 which can be integrated to find the potential difference across the dielectric, 3 R / 2 dr Q Q ⎛1 2 ⎞ 1⎛ Q ⎞ ∆V1 = = ⎟ ⎜ − ⎟= ⎜ 2 ∫ R 4πκε 0 r 4πκε 0 ⎝ R 3R ⎠ 3 ⎝ 4πκε 0 R ⎠ an in the empty space, Q 2 R dr Q ⎛ 2 1 ⎞ 1⎛ Q ⎞ ∆V2 = = − ⎟ ⎜ ⎟= ⎜ 2 ∫ 3 / 2 R 4πε 0 r 4πε 0 ⎝ 3R 2 R ⎠ 6 ⎝ 4πε 0 R ⎠ The net potential difference is the sum of the two potential differences above. 1⎛ Q ⎞ 1⎛ Q ⎞ ∆V = ∆V1 + ∆V2 = ⎜ ⎟+ ⎜ ⎟ 3 ⎝ 4πκε 0 R ⎠ 6 ⎝ 4πε 0 R ⎠

†26-47.

The capacitance is 24πε 0 R Q Q C= = = ∆V 1 ⎛ Q ⎞ 1 ⎛ Q ⎞ ⎛ 2 + 1⎞ ⎟ ⎜ ⎟+ ⎜ ⎟ ⎜ ⎠ 3 ⎝ 4πκε 0 R ⎠ 6 ⎝ 4πε 0 R ⎠ ⎝ κ The electric force is given by F = qE, where the electric field within the oil is q qq 1 F = 1 Efree = 1 2 2 κ 4πε 0κ r

( 2.0 × 10

=

26-48.

−6

C )( 3.0 × 10−6 C )

= 0.050 N 4π ⎡⎣8.85×10−12 C 2 / ( N i m 2 ) ⎤⎦ (3.0)(0.60 m) 2 Let E0 be the magnitude of the electric field at the surface of the sphere. To a good approximation, the electric field will be directed radially outward from both sides of the sphere.

E0 =

σ κε 0

σ =

Q = κε 0 E0 A

Qtop Q

=

κ air ε 0 E0 A κ air 1.0 1 = = = κ airε 0 E0 A + κ oilε 0 E0 A κ air + κ oil 1.0 + 3.0 4

Therefore, the fraction of the total charge on the bottom of the sphere is †26-49.

The total charge on the inner sphere is Q = σ top A + σ bottom A = σ top 2π R12 + σ bottom 2π R12 = 2π R12 ( σ top + σ bottom ) The electric field at the surface of both dielectrics is


3 . 4 (i)

CHAPTER

E0 =

26 σ top σ = bottom κ topε 0 κ bottomε 0

σ top σ bottom = κ top κ bottom

(ii)

Substituting Equation (ii) into (i) gives ⎛ ⎞ κ Q = 2π R12σ top ⎜ 1 + bottom ⎟ ⎜ κ top ⎟⎠ ⎝ The electric field at a distance r from the center of the sphere is 2π R12σ top σ top R12 1 E= = 2π r 2κ topε 0 κ topε 0 r 2 The potential difference is then, σ top R12 R2 1 σ top R12 R2 dr ∆V = ∫ E dr = = R1 κ topε 0 ∫R1 r 2 κ topε 0

⎛ 1 1 ⎞ ⎜ − ⎟ ⎝ R1 R2 ⎠

The capacitance is

⎛ κ 2π R12σ top ⎜ 1 + bottom ⎜ κ top Q ⎝ C= = 2 σ top R1 ⎛ 1 ∆V 1 ⎞ ⎜ − ⎟ κ topε 0 ⎝ R1 R2 ⎠ 26-50.

26-51.

26-52.

U =

⎞ ⎟⎟ ⎠ =

2πε 0

⎛ 1 1 ⎞ ⎜ − ⎟ ⎝ R1 R2 ⎠

(κ

top

+ κ bottom )

1 1 2 2 C ( ∆V ) = ( 3.0 × 10−3 F ) (100 V ) = 15 J 2 2

(a) C =

ε0 A

( 8.85 × 10

2

−12

⎛ 1m ⎞ F/m ) 900 cm 2 ⎜ ⎟ ⎝ 100 cm ⎠ = 1.6 × 10−10 F 0.0050 m

= d Q 6.0 × 10−8 C (b) V = = = 377 V = 380 V C 1.59 × 10−10 F ∆V 377 V (c) E = = = 7.5 × 104 V/m d 0.0050 m 2 1 1 (d) u = ε 0 E 2 = ⎡⎣8.85 × 10−12 C 2 / ( N i m 2 ) ⎤⎦ ( 7.53 × 104 V/m ) = 0.025 J/m3 2 2 2 ⎛ 1m ⎞ −5 (e) U = uV = ( 0.025 J/m3 )( 900 cm 2 ) ⎜ ⎟ (0.0050 m) = 1.1 × 10 J ⎝ 100 cm ⎠ The dielectric constant of Plexiglas is κ = 3.4 (a) C =

ε 0κ A d

=

( 8.85 × 10

2

−12

⎛ 1m ⎞ F/m ) ( 3.4 ) ( 900 cm ) ⎜ ⎟ ⎝ 100 cm ⎠ = 5.4 × 10−10 F 0.0050 m 2

Q 6.0 × 10−8 C = = 111 V = 110 V C 5.42 × 10−10 F ∆V 111 V (c) E = = = 2.2 × 104 V/m d 0.0050 m

(b) V =


CHAPTER

26

2 1 1 ε 0 E 2 = ⎡⎣8.85 × 10−12 C2 / ( N i m 2 ) ⎤⎦ ( 2.21 × 104 V/m ) = 0.0022 J/m3 2 2 2 ⎛ 1m ⎞ −7 (e) U = uV = ( 0.00217 J/m3 )( 900 cm 2 ) ⎜ ⎟ (0.0050 m) = 9.8 × 10 J ⎝ 100 cm ⎠ The total charge on the capacitor plates is Q = CV = (10 × 10−6 F )( 2.0 × 104 V ) = 0.2 C

(d) u =

†26-53.

26-54.

26-55.

The electric energy stored in the capacitor when it is charged is 2 1 1 2 U = C ( ∆V ) = (10 × 10−6 F )( 2.0 × 104 V ) = 2000 J 2 2 (a) C = Q/V = (4.0 × 10−3 C)/ (2.0 × 105 V) = 2.0 × 10−8 F ε A (b) C = 0 d −12 2 ε A ( 8.85 × 10 F/m )( 0.50 m ) = 2.2 × 10−4 m d = 0 = C 2.0 × 10−8 F (c) E =

∆V 2.0 × 105 V = = 9.1 × 108 V/m −4 d 2.2 × 10 m

(d) U =

2 1 1 2 C ( ∆V ) = ( 2.0 × 10−8 F )( 2.0 × 105 V ) = 4.0 × 102 J 2 2

U =

1 2 C ( ∆V ) 2

∆V = 26-56.

†26-57.

2U = C

2 ( 40 J )

2 × 10−5 F

= 2000 V

Since the capacitance is given by C =

ε0 A

, increasing the distance d by a factor of three will d reduce the capacitance by a factor of three. The stored electrical energy for the capacitor is given 1 2 by U = C ( ∆V ) 2 Reducing the capacitance reduces the stored energy by a factor of three. The stored charge on the plates of a capacitor is given by Q = CV For the combination of capacitances and breakdown voltages given, the maximum amount of stored charge for the two types of capacitors is Qsuper = CsuperVsuper = ( 6.8 F )( 2.5 V ) = 17 C Qsupply = CsupplyVsupply = ( 8.20 × 10−4 F ) ( 400 V ) = 0.33 C

The supercapacitor stores 52 times more charge than the power supply capacitor. 1 The stored electrical energy for the capacitor is given by U = CV 2 2 Therefore, 1 1 2 2 U super = CsuperVsuper = (6.8 F) ( 2.5 V ) = 21 J 2 2 1 1 2 2 U supply = CsupplyVsupply = ( 8.20 × 10−4 F ) ( 400 V ) = 66 J 2 2


CHAPTER

26-58.

26

The stored electrical energy of the power supply capacitor is about three times greater than that of the supercapacitor. The equivalent capacitance for capacitors in series is 1 1 1 = + Cseries C1 C2 If a dielectric is inserted into one of the capacitors, C1 for example, its capacitance will decrease by a factor (κ + 1). 1 1 1 = + Cseries C1 C2 2

Cseries =

†26-59.

2

C1C2 C1 C1 C1 = = = C1 + C2 κ C1 + C1 C1 ( κ + 1) κ + 1

The result is that the stored energy, which is directly proportional to the capacitance, will decrease by a factor (κ + 1). −12 2 ε A ( 8.85 × 10 F/m )( 0.040 m ) (a) C0 = 0 = = 7.08 × 10−10 F = 7.1 × 10−10 F −4 d 5.0 × 10 m Q0 = C0V = ( 7.08 × 10−10 F ) (12 V ) = 8.496 × 10−9 C = 8.5 × 10−9 C 1 1 2 C0V 2 = (7.08 × 10−10 F) (12 V ) = 5.1 × 10−8 J 2 2 (b) Since the battery remains connected to the capacitor, the potential difference across the plates remains constant at 12 V. The capacitance will increase by a factor of κ = 3.0. C = κC0 = (3.0)(7.08 × 10−10 F) = 2.1 × 10−9 F The insertion of the dielectric allows more charge to be added to the plates Q = CV = ( 2.12 × 10−9 F ) (12 V ) = 2.54 × 10−8 C = 2.5 × 10−8 C W = ∆U =

This also increases the electric energy stored in the capacitor. To increase that energy, the battery does additional work to move those charges, which is equal to the change in the stored energy, 1 1 W = ∆U = ( ∆C ) V 2 = ( C − C0 ) V 2 2 2 1 2 = (2.12 × 10−9 F − 7.08 × 10−10 F) (12 V ) = 1.0 × 10−7 J 2 (c) In this case, the capacitance returns to that found in part (a), C0 = 7.1 × 10−10 F, and the charge on the plates remains that found in part (b), Q = 2.5 × 10−8 C The potential difference across the plates is Q 2.54 × 10−8 C V = = = 35.8 V = 36 V C0 7.08 × 10−10 F (d) In this case, the battery charges the capacitor until it reaches the value found in part (a), Q0 = 8.5 × 10−9 C and the capacitance is that found in part (b), C = 2.1 × 10−9 F The potential difference across the plates is Q 8.496 × 10−9 C V = 0 = = 4.0 V C 2.12 × 10−9 F


CHAPTER

26-60.

Cseries =

26

( 5.0 × 10−6 F )( 8.0 × 10−6 F ) = 3.08 × 10−6 F C1C2 = C1 + C2 ( 5.0 × 10−6 F ) + ( 8.0 × 10−6 F )

1 1 2 CV 2 = (3.08 × 10−6 F) ( 24 V ) = 8.9 × 10−4 J 2 2 1⎛ V⎞ 1 1 Q∆V ( CV ) Ed 2 ⎜⎝ κε 0 A d ⎟⎠ Ed U = 2 = 2 = u= volume volume volume volume 1 (κε 0 AE ) Ed 1 2 = = κε 0 E 2 volume 2 In this proof, the following relationships have been used. κε A Q = CV volume = Ad V = Ed C= 0 d U 1 u= = κε 0 E 2 volume 2 ⎛ 3.6 × 106 J ⎞ 1.0 × 106 kW i h ) ⎜ ( ⎟ U ⎝ 1 kW i h ⎠ = volume = 2 1 1 κε 0 E 2 ( 3.0 ) ⎡⎣8.85 × 10−12 C2 / ( N i m 2 ) ⎤⎦ ( 5.0 × 107 V/m ) 2 2 8 3 = 1.1 × 10 m U =

26-61.

26-62.

†26-63.

Let V1 represent the voltage across capacitor C1; and V2 represent the voltage across capacitors C2 and C3. The sum of these voltages must equal that of the battery, V = V1 + V2 = 200 V In charging, the plates of the capacitors acquire a charge Q. Therefore, Q Q Q V1 = and V2 = = C1 Cparallel C2 + C3 ⎛ 1 ⎞ Q Q 1 + = Q⎜ + ⎟ C1 C2 + C3 ⎝ C1 C2 + C3 ⎠ V 200 V Q= = 1 1 ⎛ ⎞ ⎛ ⎞ 1 1 + +⎜ ⎜ ⎟ − − − 6 6 6 C1 C2 + C3 ⎝ 2.0 × 10 F ⎠ ⎝ 6.0 × 10 F + 8.0 × 10 F ⎟⎠ = 3.5 × 10−4 C This is the charge on the plates of capacitor number 1. Then, the potential difference across the capacitors can be determined. Q 3.5 × 10−4 C V1 = = = 175 V C1 2.0 × 10−6 F

V =

V2 = V − V1 = 200 V − 175 V = 25 V The amount of charge on capacitors 2 and 3 is given by Q2 = C2V2 = ( 6.0 × 10−6 F ) ( 25 V ) = 1.5 × 10−4 C Q3 = C3V3 = ( 8.0 × 10−6 F ) ( 25 V ) = 2.0 × 10−4 C


CHAPTER

26-64.

†26-65.

26

The stored energy in each capacitor is 1 1 2 2 U1 = C1V1 = (2.0 × 10−6 F) (175 V ) = 3.1 × 10−2 J 2 2 1 1 2 2 U 2 = C2V2 = (6.0 × 10−6 F) ( 25 V ) = 1.9 × 10−3 J 2 2 1 1 2 2 U 3 = C3V2 = (8.0 × 10−6 F) ( 25 V ) = 2.5 × 10−3 J 2 2 Since they are initially connected in parallel the potential difference across each capacitor is the same, 240 V. Each capacitor will have a charge Q that will flow around the circuit when they are reconnected in series. This amount of charge is Q = CV = ( 5.0 × 10−6 F ) ( 240 V ) = 1.2 × 10−3 C The energy stored in the system when the capacitors are charged in parallel is 1 1 2 U = CV 2 = (10)(5.0 × 10−6 F) ( 240 V ) = 1.44 J 2 2 This is the same amount of energy that will be released when the capacitors are reconnected in series. The potential difference between the first and last terminals will be (10)(240 V) = 2400 V The capacitor contains two dielectrics of thickness d side by side. We can consider this in the same manner as two capacitors in parallel. In parallel, the capacitances add together, ⎛ε A⎞ ⎛ε A⎞ ⎛ε A⎞ ⎛ε A⎞ C = κ 1 ⎜ 0 ⎟ + κ 2 ⎜ 0 ⎟ = ( κ 1 + κ 2 ) ⎜ 0 ⎟ = (1 + κ ) ⎜ 0 ⎟ ⎝ 2d ⎠ ⎝ 2d ⎠ ⎝ 2d ⎠ ⎝ 2d ⎠ When the charge is added to the plates, the potential difference is Q Q V = = ε A C (1 + κ ) ⎛⎜ 0 ⎞⎟ ⎝ 2d ⎠ 2.0 × 10−6 C = = 1.3 × 104 V 2 − 12 ⎛ 8.85 × 10 F/m ( 0.10 m ) ⎞ ⎟ (1 + 2.5 ) ⎜⎜ −3 ⎟ 2 1.0 × 10 m ( ) ⎝ ⎠ The total electric energy stored in the capacitor when the dielectric is inserted halfway into the capacitor is 1 U = CV 2 2 ⎛ 8.85 × 10−12 F/m ( 0.10 m )2 ⎞ 2 1 ⎟ (1.29 × 104 V ) = (1 + 2.5 ) ⎜ − 3 ⎜ ⎟ 2 2 (1.0 × 10 m ) ⎝ ⎠ = 0.0129 J The empty capacitor with the same amount of charge would have a stored energy, 1 1⎛ε A⎞ U = C0V 2 = ⎜ 0 ⎟ V 2 2 2⎝ d ⎠ =

−12 1 ⎛ 8.85 × 10 F/m ( 0.10 m ) ⎜ 2 ⎜⎝ 1.0 × 10−3 m

2

⎞ 2 ⎟ (1.29 × 104 V ) ⎟ ⎠

= 0.0074 J


CHAPTER

26

The difference in the stored energy is equal to the magnitude of the work done in either inserting or removing the dielectric by a distance d/2. Using this approach, the force is W 2∆U 2(0.0129 J − 0.0073 J) F = = = = 11 N d /2 d 1.0 × 10−3 m 26-66.

⎛⎛ −12 F ⎞ 2 ⎟ 0.040 m ⎜ ⎜ 8.85 × 10 m ⎠ (a) Q = CV = V = ⎜⎝ d 2.0 × 10−4 m ⎜ ⎜ ⎝ V 12 V E= = = 6.0 × 104 V/m d 2.0 × 10−4 m

ε0 A

⎞ ⎟ ⎟ (12 V ) = 2.1 × 10−8 C ⎟ ⎟ ⎠

⎛ 1 1 ⎞ (b) ∆Q = ∆CV = ⎜ − ⎟ ε 0 AV ⎝ d1 d 2 ⎠ ⎛ ⎞⎛ 1 1 −12 F ⎞ 2 =⎜ − ⎟ ⎜ 8.85 × 10 ⎟ ( 0.040 m ) (12 V ) −4 −4 2.0 × 10 m 3.0 × 10 m m ⎠ ⎝ ⎠⎝ −9 = 7.1 × 10 C

V 12 V = = 4.0 × 104 V/m d 3.0 × 10−4 m

E= †26-67.

The equivalent capacitance for the two capacitors in series is 4.0 × 10−6 F )( 6.0 × 10−6 F ) ( C1C2 = C12 = C1 + C2 ( 4.0 × 10−6 F ) + ( 6.0 × 10−6 F )

= 2.4 × 10−6 F The charge on the plates of capacitors in series is the same for each capacitor, regardless of its capacitance. That charge is Q1 = Q2 = C12V = ( 2.4 × 10−6 F ) ( 400 V ) = 9.6 × 10−4 C The potential differences across these two capacitors is then Q 9.6 × 10−4 C V1 = = = 240 V C1 4.0 × 10−6 F V2 =

Q 9.6 × 10−4 C = = 160 V C2 6.0 × 10−6 F

Note that these voltages add to 400 V, which is that applied to the two free terminals. The charge on the plates of the third capacitor is Q3 = C3V = ( 3.0 × 10−6 F ) ( 400 V ) = 1.2 × 10−3 C The potential difference is the same as that applied to the two free terminals, 400 V. The stored energy in each capacitor may now be determined. 1 1 2 2 U1 = C1V1 = (4.0 × 10−6 F) ( 240 V ) = 0.115 J 2 2 1 1 2 2 U 2 = C2V2 = (6.0 × 10−6 F) (160 V ) = 0.077 J 2 2 1 1 2 2 U 3 = C3V2 = (3.0 × 10−6 F) ( 400 V ) = 0.24 J 2 2


CHAPTER 26-68.

26

In series and parallel, we have −1

⎛ ⎞ 1 1 1 −7 Cseries = ⎜ + + ⎟ = 5.5 × 10 F −6 −6 −6 ⎝ 1.0 × 10 F 2.0 × 10 F 3.0 × 10 F ⎠ Cparallel = 1.0 × 10−6 F + 2.0 × 10−6 F + 3.0 × 10−6 F = 6.0 × 10−6 F

For a combination of series and parallel, there are three possibilities, (1.0 × 10−6 F )( 2.0 × 10−6 F ) + 3.0 × 10−6 F C1C2 + C3 = Ccombo1 = C1 + C2 (1.0 × 10−6 F ) + ( 2.0 × 10−6 F ) = 3.7 × 10−6 F Ccombo2

2.0 × 10−6 F )( 3.0 × 10−6 F ) ( C2C3 = + C1 = + 1.0 × 10−6 F −6 −6 C2 + C3 ( 2.0 × 10 F ) + ( 3.0 × 10 F ) = 2.2 × 10−6 F

Ccombo3 =

1.0 × 10−6 F )( 3.0 × 10−6 F ) ( C1C3 C + 2 = + 2.0 × 10−6 F −6 −6 C1 + C3 (1.0 × 10 F ) + ( 3.0 × 10 F )

= 2.8 × 10−6 F 26-69.

CC ⎛ CC ⎞ ⎛ CC ⎞ (a) Cequivalent = ⎜ + C = 2C ⎟+C +⎜ ⎟=2 2C ⎝C +C⎠ ⎝C +C⎠ = 2 ( 4.0 × 10−6 F ) = 8.0 × 10−6 F

26-70.

(b) This circuit appears to be a series circuit consisting of five equivalent capacitances −1 1 ⎛1 1 1 1 1⎞ Cequivalent = ⎜ + + + + ⎟ = C 5 ⎝C C C C C⎠ 1 = ( 4.0 × 10−6 F ) = 8.0 × 10−7 F 5 The 5.0 µF capacitor is initially charged to Q1 = C1V1 = ( 5.0 × 10−6 F ) 40 V = 2.0 × 10−4 C and the 8.0 µF capacitor is initially charged to Q2 = C2V2 = ( 8.0 × 10−6 F ) 60 V = 4.8 × 10−4 C When the two capacitors are connected to each other, the charge will be redistributed so that each capacitor has the same amount of charge on each plate since they would then be connected in series to each other. The final charge on the plates will be Q = Q1 + Q2 = 6.8 × 10−4 C The potential difference across each capacitor will be Q 6.8 × 10−4 C V5.0 = = 140 V = C1 5.0 × 10−6 F V8.0 =

Q 6.8 × 10−4 C = 85 V = C2 8.0 × 10−6 F


CHAPTER †26-71.

The initial capacitance is ε A C0 = 0 d0 After the dielectric is added and the plate separation is increased, the new capacitance is ε κA C= 0 d The ratio of these two capacitances is 1.5. Therefore, the value of κ may be determined. ε0 A C0 d0 d = = = 1.5 ε κ A κ d0 C 0 d 3d 0 d = = 2.0 κ = 1.5d 0 1.5d 0

26-72.

(a) Q = C0V = (2.5 × 10−5 F)(50 V) = 1.25 × 10−3 C = 1.3 × 10−3 C (b) C = κ C0 = (2.3)(2.5 × 10−5 F) = 5.75 × 10−5 F = 5.8 × 10−5 F (c) The minimum distance between the plates of the empty capacitor is V 50 V = = 1.67 × 10−5 m d = 6 Efree 3.0 × 10 V/m The maximum voltage that may be applied to the plates with the dielectric present is then ∆V = Ed = (1.8 × 107 V/m )(1.67 × 10−5 m ) = 3.0 × 102 V The maximum charge that can be placed on this capacitor with the dielectric in place is Q = CV = κ C0V = (2.3)(2.5 × 10−5 F)(3.0 × 102 V) = 1.7 × 10−2 C

26-73.

U =

1 1 2 2 CV = (2.0 × 10−4 F) ( 360 V ) = 13 J 2 2 2 ⎛ U 1 1 C2 ⎞ = κε 0 E 2 = ( 2.2 ) ⎜ 8.85 × 10−12 u= 7.0 × 107 V/m ) 2 ⎟( volume 2 2 Nim ⎠ ⎝ 4 3 4 3 = 4.77 × 10 J/m = 4.8 × 10 J/m

U 13 J = = 2.7 × 10−4 m3 4 3 u 4.77 × 10 J/m (a) As in Example 9, the capacitance of a cylindrical capacitor of length l with a dielectric κ is l C = 2πκε 0 ⎛b⎞ ln ⎜ ⎟ ⎝a⎠ where a and b are the radii of the inner and outer conductors, respectfully. For the Leyden jar given, l 0.15 m = 2π (6.0) ( 8.85 × 10−12 F/m ) = 1.9 × 10−9 F Ccylinder = 2πκε 0 b 7.5 cm ⎛ ⎞ ⎛ ⎞ ln ⎜ ⎟ ln ⎜ ⎟ a 7.3 cm ⎝ ⎠ ⎝ ⎠ volume =

26-74.


26

CHAPTER

26

(b) If the jar is treated as a parallel plate capacitor, then −12 2 ε 0κ A ε 0κπ R 2l ( 8.85 × 10 F/m ) (6.0)π (0.075 m )(0.15 m) = = Cplate = d d 0.020 m −12 = 7.0 × 10 F 26-75.

V = V1 + V2 = V1 + 3.0 V = 9.0 V V1 =

Q = 6.0 V C1

Q = C1V1

26-76.

26-77.

−6 Q C1V1 ( 4.0 × 10 F ) ( 6.0 V ) C2 = = = = 8.0 × 10−6 F V2 V2 3.0 V 1 1 2 2 U1 = C1V1 = (4.0 × 10−6 F) ( 6.0 V ) = 7.2 × 10−5 J 2 2 1 1 2 2 U 2 = C2V2 = (8.0 × 10−6 F) ( 3.0 V ) = 3.6 × 10−5 J 2 2 The initial equivalent capacitance for the two identical capacitors is −1 C ⎛1 1⎞ Ceq = ⎜ + ⎟ = 2 ⎝C C⎠ When one has a dielectric inserted, the equivalent capacitance changes to −1 1 ⎞ κ ⎛1 Ceq = ⎜ + C ⎟ = κ +1 ⎝ C κC ⎠ The change in potential energy is then given by 1 1⎛ κ 1 ⎞ 1 ⎛ κ 1⎞ 1 ⎛ C ⎞⎛ κ −1 ⎞ 2 ∆U = ( ∆C ) V 2 = ⎜ − ⎟V 2 = ⎜ ⎟ ⎜ C − C ⎟V 2 = C ⎜ ⎟V 2 2⎝κ +1 2 ⎠ 2 ⎝κ +1 2⎠ 2 ⎝ 2 ⎠ ⎝ (κ + 1) ⎠

The underlined factor above is the factor by which the stored energy is reduced by the insertion of the dielectric. As in Problem 26-36, we can treat a layered capacitor as two capacitors in series the equivalent capacitance given by C ⎛ l l ⎞ =⎜ + ⎟ l ⎝ C1 C2 ⎠

−1

⎛ ⎛c⎞ ⎛b⎞ ⎜ ln ⎜ a ⎟ ln ⎜ c ⎟ =⎜ ⎝ ⎠+ ⎝ ⎠ ⎜ 2πε 0κ 1 2πε 0κ 2 ⎜ ⎝

⎛ ⎛c⎞ ⎛b⎞⎞ ⎜ ln ⎜ a ⎟ ln ⎜ c ⎟ ⎟ = 2πε 0 ⎜ ⎝ ⎠ + ⎝ ⎠ ⎟ κ2 ⎟ ⎜ κ1 ⎜ ⎟ ⎝ ⎠

−1

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

−1

⎛ ⎞ ⎜ ⎟ κ 1κ 2 ⎟ = 2πε 0 ⎜ ⎛c⎞ ⎛b⎞⎟ ⎜ ⎜ κ 2 ln ⎜ a ⎟ + κ 1ln ⎜ c ⎟ ⎟ ⎝ ⎠ ⎝ ⎠⎠ ⎝


CHAPTER 26-78.

26

As in Problems 26-36 and 37, this situation may be modeled as three capacitors, with two in series in parallel with the third. The area of each equivalent capacitor is one-half that of the original capacitor. The series capacitance is 1 1 d d ∆V 1 1 1 d ⎛ κ2 + κ3 ⎞ 2 2 = + = = + = ⎜ ⎟ 1 1 C C1 C2 Q ε 0κ 2 A ε 0κ 3 A ε 0 A ⎝ κ 2κ 3 ⎠ 2 2 ε A⎛ κ κ ⎞ C23 = 0 ⎜ 2 3 ⎟ d ⎝ κ2 + κ3 ⎠ Using the above result, the capacitor can now be viewed as containing two dielectrics of thickness d side by side. In parallel, the capacitances add together, ε A⎛ κ κ ⎞ ε A⎛κ κκ ⎞ ⎛ε A⎞ C = C1 + C23 = ⎜ 0 ⎟ κ 1 + 0 ⎜ 2 3 ⎟ = 0 ⎜ 1 + 2 3 ⎟ d ⎝ κ2 + κ3 ⎠ d ⎝ 2 κ2 + κ3 ⎠ ⎝ 2d ⎠

26-79.

The charge on the plates is Q = CV. Since the charge remains constant, the effect of removing the carbon dioxide is C1V1 = C2V2 V2 =

C1V1 κ 1C0V1 κ 1 V1 = = κ 2 C0 κ2 C2

∆V = 26-80.

⎛κ ⎞ κ1 ⎛ 1.000 98 ⎞ V1 − V1 = V1 ⎜ 1 − 1 ⎟ = 48 V ⎜ − 1 ⎟ = 4.7 × 10−2 V κ2 ⎝ 1.000 00 ⎠ ⎝ κ2 ⎠

(a) For the empty capacitor, we can use Gauss’ law to find the potential difference, R2 R2 Q Q ⎛ 1 1 ⎞ ∆V = − ∫ E i d l = ∫ dr = ⎜ − ⎟ R1 R1 4πκε r 2 4πκε 0 ⎝ R1 R2 ⎠ 0 (b) C =

26-81.

⎛ RR ⎞ Q = 4πκε 0 ⎜ 1 2 ⎟ ∆V ⎝ R2 − R1 ⎠

⎛1 σ ⎞ ⎛1 Q ⎞ Q2 (a) F = QE = Q ⎜ ⎟ = Q⎜ ⎟= ⎝ 2 ε0 ⎠ ⎝ 2 Aε 0 ⎠ 2 Aε 0 2 Q (b) W = F i s = − Fs = − ∆d 2 Aε 0 ⎛ 1 1 ⎞ Q2 ⎛ d2 d1 ⎞ Q2 Q2 d d − = − = − = ∆d ( ) ⎜ ⎟ ⎜ ⎟ 2 1 2 ⎝ ε 0 A ε 0 A ⎠ 2ε 0 A 2ε 0 A ⎝ C2 C1 ⎠ (d) Comparing the answers from parts (a) and (c), one can easily see that ∆U Q2 =− F =− ∆d 2 Aε 0 (c) ∆U =

Q2 2

The minus sign arises because the force is directed oppositely to that due to the electric field.


CHAPTER 27

CURRENTS AND OHM’S LAW


27-2.

†27-3.

27-4. †27-5.

27-6. 27-7.

To determine the number of electrons, we must first determine the total amount of charge that flows through the circuit in one hour. Since current is defined as the amount of charge passing through part of the circuit per unit time (1 A = 1 C/s), we can write ⎛ 1 C/s ⎞ ⎛ 3600 s ⎞ 0.5 A ) ⎜ ( ⎟⎜ ⎟ Q It ⎝ 1 A ⎠ ⎝ 1 h ⎠ = 1.1 × 1022 electrons/hour = = n= e e 1.60 × 10−19 C

Q = It = ( 20 000 A ) 1.0 × 10−4 s = 2 C The conventional current is a flow of positive charge from ground to the cloud. However, it is electrons that are moving from cloud to ground. Therefore, Qground = −2C The total charge on each capacitor plate after the battery is connected is Q = CV. This resulted from the movement of n = Q/e electrons to one of the plates, leaving behind an equal amount of positive charge on the opposite plate. The discharge current is I = Q/t. Therefore, the required time for discharge will be −6 Q CV ( 40 × 10 F ) ( 9.0 V ) = = = 1.2 × 10−4 s t= I I 3.0 A After the discharge, the plates return to a neutral, or uncharged, state. −2 −7 Q It ( 2.5 × 10 A )( 5.0 × 10 s ) C= = = = 8.9 × 10−9 F V V 1.4 V Ohm’s law gives the relationship between the potential difference V and the current I within a circuit with a resistance R. 12 V V I = = = 4.0 A R 3.0 Ω The electric field within the wire is the potential difference per unit length, 12 V V E= = = 6.0 V/m 2.0 m l V 1.5 V 7.5 V V I2 = 2 = = 2.5 A R= = = 3.0 Ω 0.50 A R 3.0 Ω I At time t = 5.0 s, we can find the current by simply entering the given values into the given equation, I (t ) = At + Bt 2 = ( 0.50 C/s 2 ) ( 5.0 s ) + ( 0.20 C/s3 ) ( 5.0 s ) = 7.5 A 2

To find the total charge, integrate the current equation for the time interval, 0 ≤ t ≤ 5.0 s. 5.0 s 5.0 s 1 1 Q = ∫ ( At + Bt 2 ) dt = At 2 + Bt 3 0 2 3 0

1 1 2 3 0.50 C/s 2 ) ( 5.0 s ) + ( 0.20 C/s3 ) ( 5.0 s ) ( 2 3 = 14.6 C = 15 C =


CHAPTER 27-8.

27

Q (t ) = 5.0t − 2.0t 2 Q (1.0 s) = 5.0 (1.0 s ) − 2.0 (1.0 s ) = 3.0 C 2

Q (2.0 s) = 5.0 ( 2.0 s ) − 2.0 ( 2.0 s ) = 2.0 C 2

Q (3.0 s) = 5.0 ( 3.0 s ) − 2.0 ( 3.0 s ) = −3.0 C 2

The change in sign is due to the reversal of the current in the circuit. The current at each of these times is found by taking the derivative of the total charge equation since dQ I (t ) = = 5.0 − 4.0t dt I (1.0 s) = 5.0 − 4.0(1.0 s) = 1.0 A I (2.0 s) = 5.0 − 4.0(2.0 s) = −3.0 A

†27-9.

I (3.0 s) = 5.0 − 4.0(3.0 s) = −7.0 A Again, the change in sign reflects the reversal of current. The current is instantaneously zero at t = 1.2 s. The current at an elapsed time t is given by I (t ) = A(t − 2.0 s) 2 = ( 0.25 C/s 2 ) (t − 2.0 s) 2 The current at times t = 0.0 s and 1.0 s are then, I (0.0 s) = ( 0.25 C/s 2 ) (0.0 − 2.0 s) 2 = 1.0 A I (1.0 s) = ( 0.25 C/s 2 ) (1.0 − 2.0 s) 2 = 0.25 A

The total charge that accumulates on the capacitor plates in an elapsed time of 2.0 s is found by integrating the equation for the current. 2.0 s 2.0 s 1 2 3 Q = ∫ A(t − 2.0) dt = A(t − 2.0) 0 3 0

= 27-10.

27-11.

1 1 2 3 3 0.25 C/s 2 ) ( 0.0 s ) − ( 0.25 C/s3 ) ( −2.0 s ) = C ( 3 3 3

The number of free electrons for sodium is given by electrons ⎞ ⎛ kg ⎞ ⎛ 6.02 × 1023 ⎟ ⎜ 968 3 ⎟ N A d ⎜⎝ mole ⎠ ⎝ m ⎠ = = 2.53 × 1028 electrons/m3 n= kg ⎞ A ⎛ ⎜ 0.0230 ⎟ mole ⎠ ⎝ According to Equation 27.12, m ρ = 2e ne τ 9.11 × 10−31 kg = = 1.6 × 10−7 Ω i m 2 electrons ⎛ ⎞ −19 −15 28 ⎜ 2.53 × 10 ⎟ (1.60 × 10 C ) ( 8.8 × 10 s ) 3 m ⎝ ⎠ me 9.11 × 10−31 kg (a) τ = 2 = ne ρ ( 5.8 × 1028 / m3 )(1.60 × 10−19 C )2 (1.6 × 10−8 Ω i m )

= 3.8 × 10−14 s


CHAPTER

27

(b) vd =

27-12.

− (1.60 × 10−19 C ) ( 8.0 V/m ) ( 3.8 × 10−14 s ) −eEτ = = 0.054 m/s me 9.11 × 10−31 kg

The speed of the electrons v is given as twelve times the speed given by Equation 19-25, which gives the rms speed for molecules in an ideal gas as 3kBT vrms = me The mean free path is given by l = vτ where m τ = 2e ne ρ and electrons ⎞ ⎛ kg ⎞ ⎛ 6.02 × 1023 10 490 3 ⎟ ⎜ ⎟ ⎜ N d mole ⎠ ⎝ m ⎠ = 5.85 × 1028 electrons/m3 n= A =⎝ kg ⎞ A ⎛ ⎜ 0.1079 ⎟ mole ⎠ ⎝ where ρ is the electrical resistivity, d is the density, NA is Avogadro’s number, and A is the atomic mass. 3kBT me l = 12vrmsτ = 12 me ne 2 ρ

J⎞ ⎛ 3 ⎜ 1.38 × 10−23 ⎟ ( 293 K ) ( 9.11 × 10−31 kg ) K⎠ = 12 ⎝ 2 9.11 × 10−31 kg ⎛ 28 electrons ⎞ −19 −8 ⎜ 5.85 × 10 ⎟ (1.60 × 10 C ) (1.6 × 10 Ω i m ) 3 m ⎝ ⎠ −8 = 5.3 × 10 m †27-13.

27-14. †27-15.

The resistance of a wire is defined in Equation 27.14 as l R=ρ A where ρ is the resistivity, l is the length of the wire, and A is its cross-sectional area. For this constantan wire, the resistivity is given in Table 27.2; and the resistance is l 0.30 m R = ρ = ( 49 × 10−8 Ω i m ) = 3.0 Ω 2 A ⎛ 2.5 × 10−4 m ⎞ π⎜ ⎟ 2 ⎝ ⎠ −14 6 V = IR = (4.0 × 10 A)(1.0 × 10 Ω) = 4.0 × 10−8 V Ohm’s law gives the relationship between the current and the potential difference in a circuit element as V = IR. In the case of the light bulb, the filament has a resistance of 88 Ω when the filament is hot. When the potential difference of 115 V is applied to the bulb, it will have a current of V 115 V I = = = 1.307 A = 1.3 A R 88 Ω The current is defined as the amount of charge passing through a section of wire per unit time, I = Q/t. The number of electrons, then, passing through the filament each second is


CHAPTER

27

Q It (1.307 A ) (1.0 s) = = = 8.2 × 1018 electrons −19 e e 1.60 × 10 C V 12 V I = = = 4.0 × 102 A R 0.030 Ω When resistors are connected in series, such as the eight pieces that formed the original wire, the resistances add together to give an equivalent resistance R. Let R′ represent the resistance of each of the eight pieces. Then, when connected in series, their resistance is RS = R′ + R′ + R′ + R′ + R′ + R′ + R′ + R′ = 8 R′ Therefore, R R′ = 8 When resistors are connected in series, the equivalent resistance is found using 1 1 1 1 1 4 = + + + = RP R′ R′ R′ R′ R′ n=

27-16. †27-17.

27-18.

Taking the inverse, gives the resistance when four of the pieces are connected in parallel. R′ R / 8 R RP = = = 4 4 32 The orbital angular momentum of the electrons moving around the loop of radius r is l L = ∑ mvr r

where m is the total mass of the electrons and v is their drift velocity. From the information given concerning the current, each second there are

Q It = e e electrons passing a given point on the loop. The total number of electrons is expected to be uniformly distributed around the ring. If these electrons are moving at the drift velocity, then each one will move a distance l in a time t′, where l = vt′. The number within the portion of the loop of arc length l is n n′ = 2π r l The angular momentum is then 2π r It 2π r L = ∑ mvr = n′me vr = n me vr = me vr vt e vt n=

−31 2π me IR 2 2π ( 9.11 × 10 kg ) ( 4.0 A )( 0.020 m ) = = = 5.7 × 10−14 kg i m 2 /s e 1.60 × 10−19 C

2

†27-19.

We assume that by drawing the wire that the total volume of the wire remains unchanged. Therefore, if the initial radius is r1 and the final radius r2, then lA1 = 2lA2 2

lπ r1 = 2lπ r2 r2 =

2

2 r1 2


CHAPTER

27

The resistance is given by l l =ρ 2 R1 = ρ π r1 A1

R2 = ρ

27-20.

2l 2l =ρ 2 =ρ A2 π r2

2l ⎛ 2 ⎞ π⎜ r1 ⎟ ⎝ 2 ⎠

2

⎛ l ⎞ = 4 ⎜ ρ 2 ⎟ = 4 R1 = 4(0.10 Ω) = 0.40 Ω ⎝ π r1 ⎠ The electrons in the wire move along the wire at the drift velocity, eEτ vd = me

where the electric field E in the wire is given by I E = ρj = ρ A and the mean free time τ is m τ = 2e ne ρ and electrons ⎞ ⎛ kg ⎞ ⎛ 6.02 × 1023 8920 3 ⎟ ⎜ ⎟ ⎜ N d mole ⎠ ⎝ m ⎠ n= A =⎝ = 8.45 × 1028 electrons/m3 kg A ⎛ ⎞ ⎜ 0.063546 ⎟ mole ⎝ ⎠ Therefore, me l me lA l l nlAe nlπ r 2 e t= = = = = = eEτ m ρI vd I I e τ e ρ I 2e me A ne ρ 2 ⎛ 28 electrons ⎞ −19 ⎜ 8.45 × 10 ⎟ ( 2.0 m ) π ( 0.0020 m ) (1.60 × 10 C ) 3 m ⎠ =⎝ 1.5 A 5 5 = 2.263 × 10 s = 2.3 × 10 s or 63 hours

27-21.

E = ρj = ρ

I 80 A = (1.7 × 10−8 Ω i m ) = 0.069 V/m 2 A π ( 0.0025 m )

27-22.

E = ρj = ρ

I 750 A = (1.7 × 10−8 Ω i m ) = 0.018 V/m 2 A π ( 0.015 m )

†27-23.

The total length l of the wire is equal to the number of turns N around the spool times the circumference of the spool. The resistance R of the wire of cross-sectional area A is given by


CHAPTER

R=ρ =ρ

27-24.

R=ρ

N (2π rspool ) l l =ρ 2 =ρ πr π r2 A N (2rspool ) r2

= (1.7 × 10 Ω i m ) −8

l l =ρ 2 A πr

R8 = (1.7 × 10−8 Ω i m ) R9 = (1.7 × 10−8 Ω i m )

⎛1 ⎞ −4 ⎜ (4.5 × 10 m) ⎟ 2 ⎝ ⎠

100 m ⎛1 ⎞ π ⎜ (3.264 × 10−3 m) ⎟ ⎝2 ⎠ 100 m

2

= 0.20 Ω

⎛1 ⎞ π ⎜ (2.906 × 10−3 m) ⎟ ⎝2 ⎠ 100 m

2

= 0.26 Ω

R10 = (1.7 × 10−8 Ω i m ) R11 = (1.7 × 10−8 Ω i m ) R12 = (1.7 × 10−8 Ω i m )

†27-25.

( 260 )( 2 ) ( 5.0 × 10−3 m )

2

= 0.32 Ω

2

= 0.41 Ω

⎛1 ⎞ π ⎜ (2.588 × 10−3 m) ⎟ ⎝2 ⎠ 100 m ⎛1 ⎞ π ⎜ (2.305 × 10−3 m) ⎟ ⎝2 ⎠ 100 m

2

= 0.87 Ω

= 0.51 Ω 2 ⎛1 ⎞ −3 π ⎜ (2.053 × 10 m) ⎟ ⎝2 ⎠ Ohm’s law can be applied to find the resistance of the wire, which in turn is related to the resistivity of the metal as follows. ∆V l l =ρ =ρ 2 R= I A πr −4 ∆V π r 2 (12.0 V ) π ( 2.5 × 10 m ) = = 5.7 × 10−7 Ω ⋅ m ρ= 3.75 A 1.10 m Il ( )( ) 2

27-26.

†27-27.

27-28.

l l 200 × 103 m = ρ 2 = ( 2.8 × 10−8 Ω i m ) = 7.9 Ω 2 πr A ⎛1 ⎞ π ⎜ (0.030 m) ⎟ ⎝2 ⎠ The resistance of metals varies with temperature as described by Equation 27.22: ∆R = αR0∆T, where R0 is an initial resistance at a temperature T0 and α is a constant called the temperature coefficient of resistance (see Table 27.2). From Problem 27-26, the initial resistance is 7.9 Ω at 20°C. When the temperature increases to 50°C, the resistance increases by ∆R = αR0∆T = (3.9 × 10−3/°C)(7.9 Ω)(50 – 20°C) = 0.92 Ω ∆R = αR0∆T ∆R (0.60 − 0.50 Ω) ∆T = = = 33 °C α R0 ( 6.0 × 10−3 / o C ) ( 0.50 Ω ) R=ρ


27

CHAPTER †27-29.

27-30.

†27-31.

27-32.

27

The current density j is defined as the current I passing through a cross-sectional area A within a circuit. I I 30 A = = 5.69 × 106 A/m 2 = 5.7 × 106 A/m 2 j= = 2 2 A πR ⎛1 ⎞ π ⎜ (2.59 × 10−3 m) ⎟ 2 ⎝ ⎠ The electric field E is the product of the current density j and the resistivity ρ, E = ρ j = (1.7 × 10−8 Ω i m )( 5.69 × 106 A/m 2 ) = 0.097 V/m j=

I I = = A π R2

= 1.4 × 107 A/m 2

π ⎜ (3.0 × 10−10 m) ⎟ 2

(109.8 Ω − 100.0 Ω ) = 25°C ∆R = 0.0°C + α R0 ( 3.9 × 10−3 /°C ) (100.0 Ω )

The temperature coefficient of resistance for silicon is given in Table 27.4. ∆R ∆T = T − T0 = α R0 T = T0 +

27-34.

2

⎛1 ⎞ ⎝ ⎠ Comparing Equation 27.21 with the given alternative equation, we see that 1 1 −1 σ = = = 5.9 × 107 ( Ω i m ) −8 ρ 1.7 × 10 Ω i m ∆R ∆T = T − T0 = α R0 T = T0 +

27-33.

1.0 × 10−12 A

R=ρ

(10.0 Ω − 12.0 Ω ) = 22°C ∆R = 20°C + α R0 ( −8 × 10−2 /°C ) (12.0 Ω )

l l =ρ 2 A πr 2

⎛1 ⎞ π ⎜ (1.0 × 10−3 m) ⎟ (1.0 Ω ) 2 AR π r R 2 ⎠ l= = = ⎝ = 0.022 m 3.5 × 10−5 Ω i m ρ ρ

†27-35.

27-36.

To apply Ohm’s Law, we must first determine the resistance of the rod. The resistance is given by l l R=ρ =ρ 2 πr A where ρ is the resistivity of iron, r is the radius of the rod, and l is its length. This is then substituted into Ohm’s law to find the potential difference. (1.0 × 104 A ) ( 0.5 m ) = 9.9 V Il ∆V = IR = ρ 2 = (1.0 × 10−7 Ω i m ) 2 πr π ( 0.0040 m ) (a) As in Problem 27-35, (12 A )( 25 m ) = Il ∆V = IR = ρ 2 = (1.7 × 10−8 Ω i m ) 0.97 V 2 πr ⎛1 ⎞ π ⎜ (0.00259 m) ⎟ ⎝2 ⎠ There are two wires that carry current to and from the air conditioner, each with a potential difference of 0.97 V. Therefore, the voltage delivered to the air conditioner is


CHAPTER

†27-37.

27

Vair cond. = V − 2∆V = 115 V − 2(0.97 V) = 113 V (12 A )( 25 m ) = Il (b) ∆V = IR = ρ 2 = (1.7 × 10−8 Ω i m ) 1.5 V 2 πr ⎛1 ⎞ π ⎜ (0.00205 m) ⎟ ⎝2 ⎠ Vair cond. = V − 2∆V = 115 V − 2(1.54 V) = 112 V The length and resistance of each of the two wires is the same. Therefore, RCu = RAl

ρCu

l l = ρ Al 2 2 π rCu π rAl

The ratio of the radii of the two wire is l l ρCu 2 = ρ Al 2 π rCu π rAl

ρCu rAl ρ Al

rCu =

The mass m of each wire is the product of its density D and its volume V = Al = πr2l. The mass of the aluminum wire is kg ⎞ 2 ⎛ mAl = Dπ rAl2 l = ⎜ 2.7 × 103 3 ⎟ π ( 0.015 m ) (100 m ) = 190 kg m ⎝ ⎠ The mass of the copper wire is 2

mCu

27-38.

⎛ ρ Cu ⎞ rAl ⎟⎟ l = Dπ r l = Dπ ⎜⎜ ⎝ ρ Al ⎠ 2 Cu

kg ⎞ ⎛ 1.7 × 10−8 Ω i m ⎞ 2 ⎛ = ⎜ 8.9 × 103 3 ⎟ π ⎜ ⎟ ( 0.015 m ) (100 m ) = 380 kg −8 m ⎠ ⎝ 2.8 × 10 Ω i m ⎠ ⎝ The mass of the copper wire is twice that of the aluminum wire. ( 25 A )(1.0 m ) = 0.13 V Il (a) ∆V = IR = ρ 2 = (1.7 × 10−8 Ω i m ) 2 πr ⎛1 ⎞ π ⎜ (0.00205 m) ⎟ ⎝2 ⎠ (b) E = ρ j = ρ

27-39.

∆V = IR = ρ r2 = ρ r=

(1.7 × 10−8 Ω i m ) ( 25 A ) = 0.13 V/m I = 2 A ⎛1 ⎞ π ⎜ (0.00205 m) ⎟ ⎝2 ⎠

Il π r2

Il π∆V

ρ

Il = π∆V

( 2.8 × 10

−8

Ω i m)

( 25 A )(15 m ) π ( 5.0 V )

= 8.176 × 10−4 m

Diameter = 2r = 2 ( 8.176 × 10−4 m ) = 1.6 × 10−3 m or 0.16 cm


CHAPTER

27-40.

27

∆V = IR = ρ r1 =

ρ

Il π r2

I1l = π∆V

(1.7 × 10

−8

Ω i m)

( 25 A )( 9.0 m ) π (1.2 V )

= 0.0010 m

diameter = 2r1 = 0.00200 m ≈ gauge #12 r2 =

ρ

I 2l = π∆V

(1.7 × 10

−8

Ω i m)

( 35 A )( 9.0 m ) π (1.2 V )

= 0.00119 m

diameter = 2r2 = 0.00238 m ≈ gauge #10 r3 =

†27-41.

27-42.

ρ

I 3l = π∆V

(1.7 × 10

−8

Ω i m)

( 45 A )( 9.0 m ) π (1.2 V )

= 0.00135 m

diameter = 2r3 = 0.00270 m ≈ gauge #9 The electric field between the plates of the capacitor is ∆V E= d The current density within the polyethylene is j = I/A = E/ρ, where ρ is given for polyethylene in Table 27.3. Therefore, 8.0 × 10−2 m 2 )( 2.0 × 104 V ) ( EA A∆V = = = 8.0 × 10−5 A I = −4 11 ρ dρ Ω i 1.0 × 10 m 2.0 × 10 m ( )( )

R=ρ

l l =ρ 2 A π router − rinner 2

(

= (1.7 × 10−8 Ω i m )

)

30 m = 0.0081 Ω π ( (0.0060 m) 2 − (0.0040 m) 2 )

∆V = IR = ( 600 A )( 0.0081 Ω ) = 4.9 V †27-43.

Let the inner radius of the polyethylene cylinder be a and its outer radius be b. The resistance for a thin cylinder of radius r is ρL L ρ 1 R= =ρ = A 2π rL 2π r Treating the polyethylene cylinder as a series of thin cylinders connected in series, each of surface area equal to the circumference times the length, means that we need to integrate the resistances for the thin cylinders to obtain the total resistance of the entire polyethylene cylinder. 11 dr ρ ⎛ b ⎞ 2.0 × 10 Ω i m ⎛ 4.0 mm ⎞ 10 = = ln ln ⎜ ⎟ = 4.4 × 10 Ω ∫a r 2π ⎜⎝ a ⎟⎠ 2π ⎝ 1.0 mm ⎠ The current between the two conductors is then given by Ohm’s Law, ∆V 300 V I = = = 6.8 × 10−9 A R 4.4 × 1010 Ω

R=

ρ 2π

b


CHAPTER 27-44.

27

Req = R1 + R2 + R3 = 4.0 Ω + 6.0 Ω + 8.0 Ω = 18.0 Ω I =

12 V ∆V = = 0.67 A Req 18.0 Ω

∆V1 = IR1 = ( 0.67 A )( 4.0 Ω ) = 2.7 V

∆V2 = IR2 = ( 0.67 A )( 6.0 Ω ) = 4.0 V ∆V3 = IR3 = ( 0.67 A )( 8.0 Ω ) = 5.3 V

†27-45.

When resistors are connected in parallel, the equivalent resistance is −1

−1

⎛ 1 1 1 ⎞ 1 1 ⎞ ⎛ 1 + + + Req = ⎜ + ⎟ =⎜ ⎟ = 2.2 Ω ⎝ 5.0 Ω 7.0 Ω 9.0 Ω ⎠ ⎝ R1 R2 R3 ⎠ The total current in the circuit is found using Ohm’s law with the value for the equivalent resistance. ∆V 12 V I = I1 + I 2 + I 3 = = = 5.4 A Req 2.2 Ω I1 =

12 V ∆V = = 2.4 A 5.0 Ω R1

I2 =

12 V ∆V = = 1.7 A 7.0 Ω R2

12 V ∆V = = 1.3 A 9.0 Ω R3 ∆V ∆V ∆V ∆V I = I brass + I iron = + = + Rbrass Riron ρ brass L / A ρiron L / A I3 =

27-46.

The ratio of the current carried by the brass wire to that carried by the iron wire is ∆V I brass ρ L / A ρiron = brass = ∆V I iron ρ brass ρiron L / A I brass =

ρiron I iron ρ brass

⎛ ρ ⎞ I = I iron ⎜ 1 + iron ⎟ = 6.0 A ρ brass ⎠ ⎝ 6.0 A = 2.5 A I iron = 1.0 × 10−7 1+ 7 × 10−8 I brass = I − I iron = 6.0 A − 2.5 A = 3.5 A 27-47.

The resistance of each wire will increase as the temperature is increased to 90°C. Let R′brass and R′iron represent the new resistances for brass and iron, respectively. From Equation 27.22, the new resistances will be given by ′ − Rbrass = α brass Rbrass ∆T ∆Rbrass = Rbrass ′ = Rbrass + α brass Rbrass ∆T Rbrass Similarly,


CHAPTER

27

′ = Riron + α iron Riron ∆T Riron Then, the total current is ∆V ∆V ′ + I iron ′ = I = I brass + = 6.0 A ′ ′ Rbrass Riron The ratio of the current carried by the brass wire to that carried by the iron wire is ∆V ′ ′ ρ ′ L / A ρiron ρ + αρiron ∆T I brass = brass = = iron ∆V ′ ′ ρ brass ρ brass + αρ brass ∆T I iron ′ L/ A ρiron

′ = I brass

27-48.

ρiron + αρiron ∆T ′ I iron ρ brass + αρ brass ∆T

⎛ ρ + αρiron ∆T ⎞ ′ ⎜ 1 + iron I = I iron ⎟ = 6.0 A ρ brass + αρ brass ∆T ⎠ ⎝ 6.0 A ′ = = 1.7 A I iron −7 ⎛ 1.0 × 10 Ω i m + ( 5 × 10−3 / °C ) ( 70 °C ) ⎞ ⎟ 1+ ⎜ ⎜ 7 × 10−8 Ω i m + ( 2 × 10−3 / °C ) ( 70 °C ) ⎟ ⎝ ⎠ ′ = I − I iron ′ = 6.0 A − 1.7 A = 4.3 A I brass For resistor one: The first two stripes are digits: brown = 1, red = 2 The third stripe is the multiplier: orange = 103 R1 = 12 × 103 Ω or 12 kΩ For resistor two: yellow = 4, violet = 7 red = 102 R2 = 47 × 102 Ω or 0.47 kΩ When connected in parallel, the equivalent resistance is −1

†27-49.

−1

⎛ 1 ⎛ ⎞ 1 ⎞ 1 1 2 Req = ⎜ + + ⎟ =⎜ ⎟ = 34 × 10 Ω 4 3 Ω Ω R R 1.2 × 10 4.7 × 10 ⎝ ⎠ 2 ⎠ ⎝ 1 The corresponding color code is orange-yellow-red Connecting the three resistors in series gives: Req = R1 + R2 + R3 = 2.0 Ω + 3.0 Ω + 4.0 Ω = 9.0 Ω When resistors are connected in parallel, the equivalent resistance is −1

−1

⎛ 1 1 1 ⎞ 1 1 ⎞ ⎛ 1 Req = ⎜ + + + + ⎟ =⎜ ⎟ = 0.92 Ω ⎝ 2.0 Ω 3.0 Ω 4.0 Ω ⎠ ⎝ R1 R2 R3 ⎠ Then, we can do combinations of series and parallel as follows:


CHAPTER

27-50.

⎛ 1 1 ⎞ Req = R1 + ⎜ + ⎟ ⎝ R2 R3 ⎠

−1

⎛ 1 1 ⎞ Req = R2 + ⎜ + ⎟ ⎝ R1 R3 ⎠

−1

⎛ 1 1 ⎞ Req = R3 + ⎜ + ⎟ ⎝ R1 R2 ⎠

−1

⎛ 1 1 ⎞ Req = ⎜ + ⎟ ⎝ R1 R2 + R3 ⎠

−1

⎛ 1 1 ⎞ Req = ⎜ + ⎟ ⎝ R2 R1 + R3 ⎠

−1

⎛ 1 1 ⎞ + Req = ⎜ ⎟ ⎝ R3 R1 + R2 ⎠ In series, we have Req = R1 + R2 = 80 Ω

−1

27

−1

1 ⎞ ⎛ 1 = 2.0 Ω + ⎜ + ⎟ = 3.7 Ω 4.0 Ω ⎠ ⎝ 3.0 Ω −1

1 ⎞ ⎛ 1 = 3.0 Ω + ⎜ + ⎟ = 4.3 Ω 4.0 Ω ⎠ ⎝ 2.0 Ω −1

1 ⎞ ⎛ 1 = 4.0 Ω + ⎜ + ⎟ = 5.2 Ω 3.0 Ω ⎠ ⎝ 2.0 Ω −1

1 ⎛ 1 ⎞ + =⎜ ⎟ = 1.6 Ω 3.0 Ω + 4.0 Ω ⎠ ⎝ 2.0 Ω −1

1 ⎛ 1 ⎞ =⎜ + ⎟ = 2.0 Ω 2.0 Ω + 4.0 Ω ⎠ ⎝ 3.0 Ω −1

1 ⎛ 1 ⎞ =⎜ + ⎟ = 2.2 Ω 2.0 Ω + 3.0 Ω ⎠ ⎝ 4.0 Ω

R1 = 80 Ω − R2 Then, in parallel, we can substitute the above result. −1

⎛ 1 ⎛ 1 1 ⎞ 1 ⎞ Req = ⎜ + + ⎟ =⎜ ⎟ ⎝ R1 R2 ⎠ ⎝ 80 − R2 R2 ⎠ R2 ( 80 − R2 ) R2 ( 80 − R2 ) = = 15 Ω 80 − R2 + R2 80

−1

= 15 Ω

R22 − 80 R2 + 1200 = 0 R2 = 20 Ω or 60 Ω

†27-51.

Then, R1 = 60 Ω or 20 Ω Therefore, the values of the two resistors are 20 Ω and 60 Ω. The three appliances are connected in parallel to the outlet. Therefore the equivalent resistance will be, −1

⎛ 1 1 1 ⎞ 1 1 ⎞ ⎛ 1 Req = ⎜ + + + + ⎟ =⎜ ⎟ ⎝ 11 Ω 16 Ω 12 Ω ⎠ ⎝ R1 R2 R3 ⎠ The total current will then be ∆V 115 V I = = = 27 A Req 4.2 Ω

27-52.

−1

= 4.2 Ω

There may be a circuit breaker in the circuit that is rated for less than this current, perhaps 20 A. If that is the case, the circuit breaker switch would open and current would stop. ∆V 115 V R1 = = = 130 Ω I1 0.87 A In series, we have


CHAPTER

27

2 R2 =

27-53.

∆V 115 V = = 167 Ω I2 0.69 A

R2 = 83 Ω The resistance is reduced because the temperature of the filament wire is less because the current through each bulb is less than when one bulb is connected. 2 6.0 V ) π ( 0.0015 m ) ( ∆V ∆V ∆V π r 2 = = = = 210 A I Cu = L ρL R (1.7 × 10−8 Ω i m ) (12.0 m ) ρ A 2 2 2 2 ∆V π router − rinner ( 6.0 V ) π ⎡⎣ ( 0.0020 m ) − ( 0.0015 m ) ⎤⎦ I rubber = = ρL (1 × 1013 Ω i m ) (12.0 m )

(

)

= 2.7 × 10−19 A 27-54.

†27-55.

The cross-sectional area at the cut is reduced, which d would increase the resistance at the cut. The result is three sections of wire connected in series as shown in l the drawing. The equivalent resistance is ⎛ ⎞ ⎜l −d d ⎟ ⎛ l + 3d ⎞ + = ρ⎜ Req = ρ ⎜ ⎟ ⎟ 1 ⎟ ⎝ A ⎠ ⎜⎜ A A⎟ 4 ⎠ ⎝ The ratio giving the percent increase in resistance is then, ⎛ l + 3d ⎞ ρ⎜ ⎟ Req A ⎠ l + 3d 0.50 m + 3(0.0040 m) = ⎝ = = = 1.024 l 0.50 m R l ρ A Therefore, the resistance increases by 2.4%. The distance between points A and C is l and let the distance from point A to point P where the short is be d. When the resistance is measured between points A and B, 2d = 30 Ω RAB = ρ A Similarly, between points C and D, 2(l − d ) RCD = ρ = 70 Ω A Taking the ratio of these two resistances gives 2d ρ RAB d 30 Ω A = = = 2 ( l − d ) l − d 70 Ω RCD ρ A 3 3 d = l = ( 5.0 km ) = 1.5 km from point A 10 10


CHAPTER

27-56.

(a) j =

ρ= (b) R = †27-57.

27

E

ρ 100 V/m E = = 2.5 × 1013 Ω i m j 4.0 × 10−12 A/m 2 ∆V ∆V ∆V 4.0 × 105 V = = = = 2.0 × 102 Ω I jA j 4π r 2 ( 4.0 × 10−12 A/m 2 ) 4π ( 6.38 × 106 m )2

By twisting the wires together to form one equivalent wire, the cross-sectional area is substantially increased and the resistance is then decreased. l l 1.0 m = (1.7 × 10−8 Ω i m ) R=ρ =ρ 2 2 A 24π r 24π ( 2.65 × 10−4 m ) = 3.2 × 10−3 Ω

⎛ ⎜ 1 ⎛ 1 ⎞ 1 1 ⎜ = + R=⎜ + ⎟ l l R R ⎜ρ ⎝ 1 2 ⎠ ρ ⎜ A A2 1 ⎝ ∆V ( A1 + ∆V ∆V I = = = ρl R ρl A1 + A2 −1

27-58.

−1

⎞ ⎟ ρl ⎟ = A ⎟ 1 + A2 ⎟ ⎠ A2 )

ρI ρI ∆V = = 2 l A1 + A2 π ( r1 + r22 ) I1 =

(18 A ) (0.013 m) 2 r2 ∆V ∆V ∆V A1 = = =I 2 1 2 = ρl R1 l ρ r1 + r2 ( (0.013 m)2 + (0.0105 m)2 ) A1

= 10.89 A = 11 A I 2 = I − I1 = 18 − 11 A = 7 A †27-59.

The resistance of the matrix can be determined from the formula for the resistance in terms of the resistivity. The area of the copper is the cross sectional area of the matrix minus the area of the 896 filaments. l l l =ρ R=ρ =ρ 2 A ACu − ASC π ( rCu − rSC2 ) 1 R = (1.7 × 10−8 Ω i m ) 2 2 l π ⎡⎢ ( 3.5 × 10−4 m ) − 896 ( 5.0 × 10−6 m ) ⎤⎥ ⎣ ⎦ −2 = 5.4 × 10 Ω/m

27-60.

The equivalent resistance is found in steps. The three resistors at the top of the circuit are combined as two resistors in parallel in series with another equivalent resistance. −1

⎛1 1⎞ Req1 = R + ⎜ + ⎟ ⎝R R⎠ This resistance is in parallel with another resistor R.


CHAPTER

27 −1

†27-61.

−1 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ −1 1 1 ⎟ = ⎜⎜ 1 + 1 ⎟⎟ = ⎛ 1 + 2 ⎞ Req = ⎜ + ⎜ ⎟ −1 ⎜R ⎝ R 3R ⎠ ⎛1 1⎞ ⎟ ⎜⎜ R R + R ⎟⎟ R + ⎜ + ⎟ ⎟⎟ ⎜⎜ 2⎠ ⎝ ⎝R R⎠ ⎠ ⎝ 3 3 = R = ( 3.0 Ω ) = 1.8 Ω 5 5 (a) R1 is in series with the other two resistors that are in parallel configuration. −1

−1

⎛ 1 1 ⎞ 1 ⎞ ⎛ 1 Req = R1 + ⎜ + + ⎟ = 2.0 Ω + ⎜ ⎟ = 4.4 Ω Ω Ω⎠ R R 4.0 6.0 ⎝ 3 ⎠ ⎝ 2 (b) The total current through this network of resistors is ∆V 8.0 V I = = = 1.8 A Req 4.4 Ω (c) The potential difference for resistors in series is ∆V = ∆V1 + ∆V2 = IR1 + IR23 = 8.0 V

The same current passes through the first resistor as that found in part (b), I1 = 1.8 A The potential drop for this resistor is ∆V1 = I1R1 = (1.8 A)(2.0 Ω) = 3.6 V For the resistors in parallel, the potential difference across the second and third resistors is ⎛ 1 1 ⎞ ∆V23 = I ⎜ + ⎟ ⎝ R2 R3 ⎠

I2 =

∆V23 R2

−1

⎛ 1 1 ⎞ I⎜ + ⎟ R2 R3 ⎠ ⎝ = R2

−1

1 ⎞ ⎛ 1 + 1.8 A ⎜ ⎟ ⎝ 4.0 Ω 6.0 Ω ⎠ = 4.0 Ω

−1

= 1.1 A

Then, ∆V2 = I2R2 = (1.1 A)(4.0 Ω) = 4.4 V I 3 = I − I 2 = 1.8 A − 1.1 A = 0.7 A 27-62.

and ∆V3 = ∆V2 = 4.4 V (a) In this problem, resistors 2 and 3 are in parallel with each other; and, in turn, with resistor 1. We may write a parallel combination of these resistors in one step as follows: −1

−1

⎛ 1 1 1 ⎞ 1 1 ⎞ ⎛ 1 Req = ⎜ + + + + ⎟ =⎜ ⎟ = 1.84 Ω = 1.8 Ω ⎝ 4.0 Ω 6.0 Ω 8.0 Ω ⎠ ⎝ R1 R2 R3 ⎠ (b) The total current through this network of resistors is ∆V 12.0 V I = = = 6.5 A Req 1.84 Ω

(c) Since the resistors are all in parallel, the potential difference across each resistor is that of the battery, ∆V1 = ∆V2 = ∆V3 = 12.0 V The currents through these resistors is then, ∆V 12.0 V I1 = = = 3.0 A 4.0 Ω R1 I2 =

∆V 12.0 V = = 2.0 A 6.0 Ω R2


CHAPTER I3 =

†27-63.

27-64.

27-65.

∆V 12.0 V = = 1.5 A 8.0 Ω R3

Since the resistors are all connected in parallel, the potential difference across each one is the same. The current that passes through the resistor is equal to the potential difference divided by the value of the resistor. ∆V = IR2 = ( 6.0 A )( 6.0 Ω ) = 36 V Begin by labeling the corners of the cube as shown in the simplified drawing to the right. The edges of the cube all have the same resistance R. If a voltage is applied between points A and H, then a current will flow through the network. Since the current must pass through three resistors along any path taken from A to H, each path is equivalent. The total current comes into node A and is split into three currents, each of value I/3. Then, at the next node encountered by a given current, point B for example, the current is again split in half. The current passing through the third leg, for point C to H for example, is the result of two currents entering the node at C. Following the path A-B-C-H (or equivalently A-D-CH, A-F-E-H, A-F-G-H), we may write the potential drops along this path as 1 1 1 5 ∆V = IR + IR + IR = IR 3 6 3 6 The equivalent resistance for this network is then, ∆V 5 Req = = R 6 I Let R be the resistance across the end-points AB of the infinite network. Then, the resistance across points C and D, with the three resistances between it and AB removed (as in Figure 27.33b) is also R. Hence our circuit is equivalent to the circuit shown below. The equivalent resistance of resistors in parallel in the diagram is −1

−1

⎛ 1 1 ⎞ 1⎞ ⎛ 1 Req = ⎜ + + ⎟ ⎟ =⎜ ⎝ 1.0 Ω R ⎠ ⎝ R1 R2 ⎠ R (1.0 Ω ) = R + 1.0 Ω Therefore, the series resistance across points AB is R (1.0 Ω ) RAB = 1.0 Ω + + 1.0 Ω = R R + 1.0 Ω which may be rewritten, R2 − 2R − 2 = 0

(

)

(

)

R = 1 ± 3 Ω = 1 + 3 Ω = 2.73 Ω


E

H

F

G D

C

A

A

B

1Ω

C R

1Ω B

1Ω

D

27

CHAPTER 27-66.

27

The potential difference across each resistor is that of the battery, 12.0 V. Therefore, the current through each will be proportional to its resistance. ∆V 12.0 V = = 4.0 A I1 = R1 3.0 Ω I2 =

∆V 12.0 V = = 2.4 A R2 5.0 Ω

∆V 12.0 V = = 1.5 A R3 8.0 Ω The total current is the sum of these three currents, I = 7.9 A. Q I = t ⎛ 3600 s ⎞ 5 5 Q = It = ( 40 A )(1 .0 h ) ⎜ ⎟ = 1.44 × 10 C = 1.4 × 10 C 1 h ⎝ ⎠ 5 Q 1.44 × 10 C n= = = 9.0 × 1023 electrons −19 e 1.60 × 10 C I3 =

27-67.

27-68.

(a) R = ρ

( 250 000 m ) = 6.0 Ω l = (1.7 × 10−8 Ω i m ) 2 A π ( 0.015 m )

(b) E = ρ j = ρ 27-69.

E=

27-70.

RAg

(1500 A ) = 0.036 V/m I = (1.7 × 10−8 Ω i m ) 2 A π ( 0.015 m )

115 V ∆V = = 2300 V/m 0.050 m d = R0 (1 + α∆T ) = RCu

RAg = ρ Ag

L L (1 + α Ag ∆T ) = ρ Cu A A

1 + α Ag ∆T =

ρCu ρ Ag

⎞ ⎛ 1.7 × 10−6 Ω i m ⎞ 1 ⎛ ρ Cu 1 − 1⎟ = − 1 ⎟ = 16°C ⎜⎜ ⎜ −3 −6 ⎟ α Ag ⎝ ρ Ag ⎠ ⎠ 3.8 × 10 /°C ⎝ 1.6 × 10 Ω i m Electrical resistance for most materials increases with temperature as indicated by Equation 27.22 and the values of α given in Tables 27.2, 27.3, and 27.4. For this material, α must be determined. ∆R1 = α∆T1 = 0.0445 R0 ∆T =

†27-71.

0.0445 0.0445 = = 8.9 × 10−4 /°C ∆T1 (70.0 − 20.0)°C The temperature coefficient of electrical resistance allows one to determine the resistance change for any temperature change. Reapplying the formula above gives, ∆R2 = α∆T2 = ( 8.9 × 10−4 /°C ) (120.0 °C − 20.0°C) = 0.089 or 8.9% R0

α =

27-72.

R=ρ

( 8.2 m ) l = (1.7 × 10−8 Ω i m ) = 0.876 Ω = 0.88 Ω 2 A ⎛1 ⎞ π ⎜ (0.00045 m) ⎟ ⎝2 ⎠ 602 www.elsolucionario.net

CHAPTER

27

12 V ∆V = = 14 A 0.876 Ω R ( 0.60 m ) l (a) R = ρ = (1.7 × 10−8 Ω i m ) 2 A ⎛1 ⎞ π ⎜ (0.0050 m) ⎟ ⎝2 ⎠ −4 −4 = 5.19 × 10 Ω = 5.2 × 10 Ω I =

27-73.

(b) ∆V = IR = ( 600 A ) ( 5.19 × 10−4 Ω ) = 0.31 V 27-74.

The ten wires carry current in parallel to give an equivalent resistance R. −1

⎛ 10 ⎞ 1 R=⎜ ⎟ = R1 10 ⎝ R1 ⎠ Therefore, each resistor has resistance R1 = 10R. Then, when six of these wires are connected in series, the equivalent resistance is Req = R1 + R1 + R1 + R1 + R1 + R1 = 6 R1 = 6(10 R ) = 60 R 27-75.

27-76.

27-77.

Connecting two sets of two resistors in parallel gives, −1 −1 R R ⎛1 1⎞ ⎛1 1⎞ = + + + Req ⎜ ⎟ ⎜ ⎟ = + = R = 1.0 Ω 2 2 ⎝R R⎠ ⎝R R⎠ ∆V1 3.0 V I1 = = = 0.38 A R 8.0 Ω ∆V2 1.5 V I2 = = = 0.19 A R 8.0 Ω ∆V ∆V 12 V 12 V Req = R1 + R2 = + = + = 4.0 × 101 Ω I1 I2 0.50 A 0.75 A The current through the resistors will be ∆V 12 V I = = = 0.30 A Req 4.0 × 101 Ω

27-78.

⎛ ⎛ R ⎞ ρ Fe L / AFe ⎞ I = I water + I Fe = 20 A = I Fe ⎜ 1 + Fe ⎟ = I Fe ⎜ 1 + ⎟ Rwater ⎠ ρ water L / Awater ⎠ ⎝ ⎝ ⎛ ⎞ ⎛ ρ A ⎞ ρ Feπ r12 ⎟ = I Fe ⎜ 1 + Fe water ⎟ = I Fe ⎜ 1 + 2 2 ⎟ ⎜ ρ A ρ π r r − ( ) water Fe ⎠ ⎝ water 2 1 ⎝ ⎠ ⎛ ⎞ ρ Fe r12 ⎟ = I Fe ⎜ 1 + 2 2 ⎜ ⎟ ρ r r − ( ) water 2 1 ⎝ ⎠ ⎞ I Fe ⎛ ρ Fe r12 ⎟ = ⎜1 + 2 2 ⎜ ⎟ I − r r ρ ( ) water 2 1 ⎝ ⎠

−1

2 ⎛ 10 × 10−8 Ω i m ) ( 0.010 m ) ( ⎜ = 1+ 2 2 ⎜⎜ 1.0 × 10−2 Ω i m ) ( 0.0125 m ) − ( 0.010 m ) ( ⎝

(

= 1 − (1.8 × 10−5 ) = 99.998%

)

⎞ ⎟ ⎟⎟ ⎠


−1

CHAPTER

27-79.

27-80.

27-81.

27

Therefore, the percentage of current in the water is 1.8 × 10−5 or 0.0018% I I 110 A = = 3.27 × 106 A/m 2 = 3.3 × 106 A/m 2 j= = 2 2 A πr ⎛1 ⎞ π ⎜ (0.00654 m) ⎟ ⎝2 ⎠ 3.27 × 106 A/m 2 j = vd = kg ⎞ ne ⎛ electron ⎞ ⎛ 23 atoms ⎞ ⎛ ⎜1 ⎟ ⎜ 6.02 × 10 ⎟ ⎜ 8920 3 ⎟ atom mole m ⎝ ⎠⎝ ⎠⎝ ⎠ 1.60 × 10−19 C ( ) kg 0.063546 mole = 2.4 × 10−4 m/s The equivalent resistance for the series combination is 5.0 Ω + 3.0 Ω = 8.0 Ω The current through both resistors is the same, I = 0.75 A. Therefore, the potential difference across these resistors is ∆V = IReq = ( 0.75 A )( 8.0 Ω ) = 6.0 V ⎛R⎞ ⎛R⎞ (a) Req = 2 A ⎜ ⎟ = 2(2π rL) ⎜ ⎟ A ⎝ ⎠ ⎝ A⎠ ⎛ 100 cm ⎞ = 4π ( 0.0013 m )( 0.080 m ) (1.0 × 105 Ω / cm 2 ) ⎜ ⎟ ⎝ 1m ⎠ = 1.3 × 106 Ω

12 V = 9.2 µ A, harmless 1.3 × 106 Ω 115 V = 88 µ A, harmless I2 = 1.3 × 106 Ω

(b) I1 =

I3 =

240 × 103 V = 0.18 A, fatal 1.3 × 106 Ω −1

27-82.

−1

⎛ 1 1 ⎞ 1 ⎞ 8 ⎛ 1 + (a) Req = ⎜ + ⎟ =⎜ ⎟ = Ω = 2.7 Ω 3 ⎝ 4.0 Ω 8.0 Ω ⎠ ⎝ R1 R2 + R3 ⎠ ∆V 1.5 V (b) I = = = 0.5625 A = 0.56 A 8 Req Ω 3 ∆V1 1.5 V (c) ∆V1 = 1.5 V and I1 = = = 0.38 A R1 4.0 Ω I 2 = I 3 = I − I1 = 0.5625 A − 0.375 A = 0.1875 A = 0.19 A ∆V2 = I 2 R2 = ( 0.1875 A )( 6.0 Ω ) = 1.1 V

∆V3 = I 3 R3 = ( 0.1875 A )( 2.0 Ω ) = 0.375 V = 0.38 V


2

CHAPTER 28

DIRECT CURRENT CIRCUITS


The kilowatt-hour (kW⋅h) is a unit of energy equal to 3.600 × 106 J. To determine the energy per kilogram of battery, simply divide the maximum energy storage of the battery by its mass. Energy 5.0 × 10−6 kW i h kW i h J Smallest battery: = = 3.3 × 10−3 = 1.2 × 104 Mass 0.0015 kg kg kg

Energy 5.0 × 103 kW i h kW i h = = 1.85 × 10−2 Mass kg ⎛ 1000 kg ⎞ 270 t ⎜ ⎟ ⎝ 1t ⎠ kW i h J = 1.9 × 10−2 = 6.7 × 104 kg kg

Largest Battery:

28-2.

(a) E = Pt = IVtζ ⎛ 3600 s ⎞ ED cell = (1.2 A )(1.5 V )(1.0 h ) ⎜ ⎟ = 6480 J = 6500 J ⎝ 1.0 h ⎠ ⎛ 3600 s ⎞ 6 6 Eauto battery = ( 55 A )(12 V )(1.0 h ) ⎜ ⎟ = 2.376 × 10 J = 2.4 × 10 J ⎝ 1.0 h ⎠ (b) The energy stored per unit volume is the energy density, U. E E 6480 J = = 135 J/cm 2 = 140 J/cm 2 U D cell = = 2 2 V πr l 1 ⎛ ⎞ π ⎜ (3.3 cm) ⎟ ( 5.6 cm ) ⎝2 ⎠ 6 2.376 × 10 J U auto battery = = 203 J/cm 2 = 2.0 × 102 J/cm 2 30 cm 17 cm 23 cm ( )( )( )

(c) We can also write an energy density as the energy stored per unit mass, u. E 6480 J uD cell = = = 7.5 × 104 J/kg m 0.086 kg uauto battery = 28-3. 28-4.

2.376 × 106 J = 1.0 × 105 J/kg 23 kg

⎛ 3600 s ⎞ 6 E = Pt = IVt = (1.0 A )(12 V )(160 h ) ⎜ ⎟ = 6.9 × 10 J ⎝ 1.0 h ⎠ ⎛ 1 C/s ⎞ ⎛ 3600 s ⎞ QAAA = It = ( 0.22 A ) ⎜ ⎟ (1.0 h ) ⎜ ⎟ = 792 C = 790 C ⎝ 1A ⎠ ⎝ 1h ⎠ ⎛ 1 C/s ⎞ ⎛ 3600 s ⎞ QAA = It = ( 0.70 A ) ⎜ ⎟ (1.0 h ) ⎜ ⎟ = 2520 C = 2500 C ⎝ 1A ⎠ ⎝ 1h ⎠

⎛ 1 C/s ⎞ ⎛ 3600 s ⎞ QC = It = ( 2.2 A ) ⎜ ⎟ (1.0 h ) ⎜ ⎟ = 7920 C = 7900 C ⎝ 1A ⎠ ⎝ 1h ⎠ ⎛ 1 C/s ⎞ ⎛ 3600 s ⎞ 4 4 QD = It = ( 4.4 A ) ⎜ ⎟ (1.0 h ) ⎜ ⎟ = 1.58 × 10 C = 1.6 × 10 C ⎝ 1A ⎠ ⎝ 1h ⎠


CHAPTER

†28-5.

28-6.

28

⎛ 3600 s ⎞ 2 EAAA = Pt = IVt = ( 0.22 A )(1.2 V )(1.0 h ) ⎜ ⎟ = 9.5 × 10 J 1.0 h ⎝ ⎠ ⎛ 3600 s ⎞ 3 EAA = Pt = IVt = ( 0.70 A )(1.2 V )(1.0 h ) ⎜ ⎟ = 3.0 × 10 J 1.0 h ⎝ ⎠ ⎛ 3600 s ⎞ 3 EC = Pt = IVt = ( 2.2 A )(1.2 V )(1.0 h ) ⎜ ⎟ = 9.5 × 10 J ⎝ 1.0 h ⎠ ⎛ 3600 s ⎞ 4 EAA = Pt = IVt = ( 4.4 A )(1.2 V )(1.0 h ) ⎜ ⎟ = 1.9 × 10 J ⎝ 1.0 h ⎠ The total energy supplied by the battery is equal to the product of the electrical power and the elapsed time. The power is the product of the current and voltage. Therefore, ⎛ 3600 s ⎞ 5 E = Pt = IVt = ( 0.65 A )(14.4 V )( 4.0 h ) ⎜ ⎟ = 1.3 × 10 J ⎝ 1.0 h ⎠ (a) E = Pt = IVt = ( 80 A )(12 V )( 3.0 s ) = 2.9 × 103 J

(b) t = †28-7.

28-8.

†28-9.

E 2880 J = = 48 s VI (12 V )( 5.0 A )

The voltages of the batteries add together since they are connected in series. The current is determined using Ohm’s Law. ∆V 3.0 V I = = = 0.375 A = 0.38 A 8.0 Ω R The total energy supplied by the battery is equal to the product of the electrical power and the elapsed time. The power is the product of the current and voltage. Therefore, ⎛ 3600 s ⎞ 3 E = Pt = IVt = ( 0.375 A )( 3.0 V )(1.0 h ) ⎜ ⎟ = 4.0 × 10 J ⎝ 1.0 h ⎠ Use Figure 29.10(b) with the polarity of the battery reversed and ε1 = 12.0 V, ε2 = 15.0 V, R1 = 4.0 Ω, and R2 = 2.0 Ω. Then, apply the Kirchoff voltage rule, starting and ending at the lower left corner. ε1 − IR1 + ε2 − IR2 = 0 ε + ε 2 12.0 V + 15.0 V = = 4.5 A I = 1 R1 + R2 4.0 Ω + 2.0 Ω The battery is connected to two lightbulbs, resistors, wired in parallel. When the currents are passing through the lightbulbs, the filaments are hot and R = 2.0 Ω. We can determine the equivalent resistance for the two resistors. −1 R 2.0 Ω ⎛1 1⎞ = 1.0 Ω Req = ⎜ + ⎟ = = 2 2 ⎝R R⎠ The total current I is then, ε 6.0 V = = 6.0 A I = Req 1.0 Ω

I1

R

I2

I R

The total current is also equal to the sum of the two currents, which are equal to each other because the resistances are equal, I = I1 + I2 = 2 I1 = 6.0 A I1 = I2 = 3.0 A


ε +

CHAPTER

28-10. †28-11.

28-12.

28

12.0 V = 20 cells 0.60 V/cell When R2 is reduced, the voltage across R1 remains the same, so its current does not change. However, since the equivalent series resistance of R2 + R3 is reduced, the current through those resistors increases. Let R represent the resistance of the ammeter and Rsh represent that of the shunt resistor. Since they are connected in parallel, the equivalent resistance is n=

−1

⎛ ⎞ ⎜1 ⎛1 1 ⎞ 1 ⎟ R Req = ⎜ + ⎟ =⎜ + 1 ⎟ = 11 ⎝ R Rsh ⎠ ⎜⎜ R R ⎟⎟ 10 ⎠ ⎝ The effective resistance is a factor of 11 smaller. The physicist must recalibrate the ammeter by a factor of 1/11 = 0.091. (a) The internal resistance of the battery, Ri, is in series with the internal resistance of the voltmeter R. The total equivalent resistance is Req = R + Ri The current flowing through the battery when the voltmeter is connected across it is ε ε 12.0 V I = = = = 2.4 × 10−4 A Req R + Ri 5.0 × 104 Ω + 0.020 Ω −1

28-13.

(b) ∆V = IRi = ( 2.4 × 10−4 A ) ( 0.020 Ω ) = 4.8 × 10−6 V 28-14.

As in Problem 28-12 above, I =

Req

ε

ε

=

R + Ri

9.00 V − 47.0 Ω = 1.13 Ω 0.187 A I The holiday lights are connected in series, so the equivalent resistance is Req = 25R, where R is the resistance of an individual bulb. The equivalent resistance can be found from the voltage and current information given. ε = IReq = 25IR Ri =

†28-15.

ε

−R=

ε

115 V = 62 Ω 25I 25 ( 0.074 A ) When the bulb burns out, the circuit is broken and the current stops flowing. Since there is no current, the potential difference across the other resistors is zero The potential difference across the burned out bulb is the same as the outlet, 115 V. This high potential difference produces a short, essentially removing that bulb from the circuit. This leaves 24 bulbs with an equivalent resistance of 24R. The current in the circuit then increases to ε = 24 I 2 R R=

I2 =

ε 24 R

=

=

115 V = 0.077 A 24 ( 62 Ω )

The potential difference across each bulb is ∆V = IR = ( 0.077 A ) (62 Ω) = 4.8 V


CHAPTER

28-16.

I1 =

28 ε Req

=

ε R1 + Ri

⎛ R1 ⎞ ∆V1 = I1 R1 = ε ⎜ ⎟ ⎝ R1 + Ri ⎠ ⎛ ε ⎞ Ri = R1 ⎜ − 1⎟ ⎝ ∆V1 ⎠ ∆V2 = I 2 R2 =

ε R2 ⎛ ε

⎞ R2 + R1 ⎜ − 1⎟ ⎝ ∆V1 ⎠ ∆V2 ( R2 − R1 ) ( 6.19 V )( 47.0 Ω − 22.0 Ω ) = 6.65 V = 6.7 V = ε = ⎛ ∆V2 ⎞ 47.0 Ω − 22.0 Ω ⎛ 6.19 V ⎞ ( )⎜ ⎟ ⎜ R2 − R1 ⎟ ⎝ 5.74 V ⎠ ∆V1 ⎠ ⎝

†28-17.

⎛ ε ⎞ ⎛ 6.65 V ⎞ − 1 ⎟ = ( 22.0 Ω ) ⎜ − 1 ⎟ = 3.5 Ω Ri = R1 ⎜ ⎝ 5.74 V ⎠ ⎝ ∆V1 ⎠ (a) The equivalent resistance for the four resistors is found by combining R3 and R4 in series and then combining them with R2 in parallel. This equivalent resistance is then added to R1 to give the equivalent for all four. −1

⎛ 1 1 ⎞ Req = R1 + ⎜ + ⎟ ⎝ R2 R3 + R4 ⎠ The current through the voltage source and the first resistor is then ε ε 12 V I = = = = 1.71 A = 1.7 A −1 −1 Req 1 ⎛ 1 ⎛ 1 ⎞ 1 ⎞ + 3.0 Ω + ⎜ R1 + ⎜ + ⎟ ⎟ ⎝ 8.0 Ω 3.0 Ω + 5.0 Ω ⎠ ⎝ R2 R3 + R4 ⎠ (b) The voltage drop across R2 is found by subtracting the voltage drop across R1 from ε. ∆V2 = ε − IR1 = 12 V − (1.71 A )( 3.0 Ω ) = 6.87 V ∆V2 6.87 V = = 0.86 A R2 8.0 Ω Since the equivalent resistance for R3 and R4 in series is equal to R2 and the potential drop across them is the same as that for R2, I2 = I3 = I4 = 0.86 A Equation 28.10 gives the current for the loop shown in Figure 28.10(b). ε − ε 2 12.6 V − 11.1 V I = 1 = = 12.5 A R1 + R2 0.12 Ω I2 =

28-18.

The energy delivered to the weak battery in the first 30 seconds is the total energy delivered by the strong battery minus the energy dissipated by the resistance in the circuit. Thus, E = ( Pb − PR ) t = ( ε b I − I 2 R ) t =

( (12.6 V )(12.5 A ) − (12.5 A ) ( 0.12 Ω ) ) (30 s) = 4160 J 2

If the terminals were connected postive to negative by mistake, the situation would be as in Problem 28-8,


CHAPTER I =

28-19.

ε1 + ε 2

=

R1 + R2 ε = I ( R + Ri )

12.6 V + 11.1 V = 198 A = 2.0 × 102 A 0.12 A

( 0.40 A ) (1.0 Ω + Ri ) = ( 0.22 A ) (2.0 Ω + Ri ) 0.40 V + ( 0.40 A ) Ri = 0.44 V + ( 0.22 A ) Ri

R

ε

I

+ Ri

0.44 V − 0.40 V = 0.22 Ω 0.40 A − 0.22 A ε = I ( R + Ri ) = ( 0.40 A ) (1.0 Ω + 0.22 Ω) = 0.49 V Ri =

28-20.

28

Combining the resistors connected in parallel gives −1

⎛ 1 1 1 1 ⎞ + + Req = ⎜ + ⎟ ⎝ R1 R2 R3 R4 ⎠ The total current passing through the batteries is then ⎛ 1 ε 1 1 1 ⎞ =ε⎜ + + + I = ⎟ Req ⎝ R1 R2 R3 R4 ⎠

†28-21.

⎛ 1 1 1 1 ⎞ 1 1 1 ⎞ ⎛ 1 = I1 R1 ⎜ + + + + + + ⎟ = ( 5.0 A )( 25 Ω ) ⎜ ⎟ ⎝ 25 Ω 15 Ω 12 Ω 6.0 Ω ⎠ ⎝ R1 R2 R3 R4 ⎠ = 44.6 A = 45 A To determine the current through the 200-Ω resistor, we can determine the potential difference across it by subtracting the potential difference across the 20-Ω resistor from the battery voltage. To carry this out, first determine the total current passing through the battery. −1

⎛ 1 1 1 ⎞ + Req = ⎜ + ⎟ + R4 ⎝ R1 R2 R3 ⎠ I =

ε Req

=

ε −1

⎛ 1 1 1 ⎞ + ⎜ + ⎟ + R4 ⎝ R1 R2 R3 ⎠ ∆V1 = ε − IR4 I1 =

=

45 V −1

1 1 ⎞ ⎛ 1 + + ⎜ ⎟ + 20 Ω ⎝ 200 Ω 67 Ω 33 Ω ⎠

∆V1 ε − IR4 45 V − (1.2 A )( 20 Ω ) = = = 0.105 A = 0.11 A 200 Ω R1 R1 −1

28-22.

= 1.12 A

−1

⎛ 1 1 ⎞ 1 ⎞ ⎛ 1 + + (a) Req = R1 + ⎜ ⎟ + R2 = 25 Ω + ⎜ ⎟ + 15 Ω = 53.33 Ω = 53 Ω ⎝ 40 Ω 20 Ω ⎠ ⎝ R3 R4 ⎠ (b) The current through resistors one and two is given by ε 12 V I = I1 = I 2 = = = 0.23 A Req 53.33 Ω The current through R3 is R4 20 Ω ⎛ 12 V ⎞ I3 = I =⎜ = 0.075 A ⎟ R3 + R4 ⎝ 53.33 Ω ⎠ 40 Ω + 20 Ω ⎛ 12 V ⎞ I 4 = I − I3 = ⎜ ⎟ − 0.075 A = 0.15 A ⎝ 53.33 Ω ⎠


CHAPTER

28

†28-23.

Since three of the wires are carrying current into the junction, the fourth wire must be carrying all of that current out of the junction. I 4 = I1 + I 2 + I 3 = 10.0 A + 8.5 A + 12.5 A = 31.0 A

28-24.

Taking a counterclockwise loop around the left portion of the circuit beginning and ending at the upper left corner, the voltage drops are ε1 − I1 R1 = 0 (i) A second loop around the right portion of the circuit, starting at the lower right corner gives, ε 2 − I 2 R2 + I1 R1 = 0 (ii) Adding equations (i) and (ii) together gives, ε 2 − I 2 R2 + I1 R1 + ε1 − I1 R1 = 0

ε 2 − I 2 R2 + ε1 = 0 ε1 + ε 2 6.0 V + 4.0V I2 =

R2

=

5.0 A

= 2.0 A

Substituting this result into equation (i) gives ε1 − I1 R1 = 0

†28-25.

ε1

6.0 V = 2.0 A 3.0 Ω R1 Finally, at the junction at the top, middle point in the circuit, I3 = I1 + I2 = 4.0 A where I3 is the current through the source ε1. Taking a counterclockwise loop around the left portion of the circuit beginning and ending at the upper left corner, the voltage drops are ε1 − I1 R1 = 0 (i) A second loop around the right portion of the circuit, starting at the lower right corner gives, ε 2 − I 2 R2 + I1 R1 = 0 (ii) Adding equations (i) and (ii) together gives, ε 2 − I 2 R2 + I1 R1 + ε1 − I1 R1 = 0 I1 =

=

ε 2 − I 2 R2 + ε1 = 0 ε1 + ε 2 10.0 V + 6.0V R2 =

28-26.

I2

=

2.0 A

= 8.0 Ω

Because of the symmetry of the circuit, a current loop that is clockwise in the left portion of the circuit should be equal in magnitude, but oppositely directed in the right portion of the circuit. The circuit loop around the left portion is ε1 − I1 R1 − I 2 R3 = 0

ε1 − I1 R1 − 2 I1 R3 = 0 ε1 − I1 ( R1 + 2 R3 ) = 0 I1 =

ε1 R1 + 2 R3

=

4.0 V = 1.0 A 2.0 Ω + 2(1.0 Ω)


CHAPTER †28-27.

Begin by drawing the five currents on the circuit diagram as shown. There are six unknowns, the five currents and R5, so at least six equations must be written and solved simultaneously. Here is one set of such equations: Starting at the lower left junction and following a clockwise loop: ε1 − R1 I1 − R3 I 3 = 0 10.0 − 2.0 I1 − 4.0 I 3 = 0

I1

I2 I4

I3 I5

(1)

I1 = 2.0 I 3 − 5.0 Starting at the lower right junction and following a counterclockwise loop: ε 2 − R2 I 2 − R4 I 4 = 0 10.0 − 2.0 I 2 − 4.0 I 4 = 0

(2)

I 2 = 2.0 I 4 − 5.0 Starting at the lower left junction and following a counterclockwise loop: − R5 I 5 + R4 I 4 − R3 I 3 = 0 − R5 I 5 − 4.0 I 2 − 4.0 I 3 = 0

(3)

R5 I 5 = −4.0( I 2 + I 3 ) Starting at the lower left junction and following a clockwise loop: ε1 − R1 I1 + R2 I 2 − ε 2 + R5 I 5 = 0 10.0 − 2.0 I1 + 2.0 I 2 − 10.0 + R5 I 5 = 0 −2.0 I1 + 2.0 I 2 + R5 I 5 = 0

(4)

R5 I 5 = 2.0( I1 − I 2 ) Then, some junction equations I1 + I 2 = I 3 + I 4 (5) I 3 = I1 + I 5 (6) I 2 = I 4 + I5 (7) Adding Equations (1) and (2) together and making use of Equation (5) gives I1 + I 2 = 2.0 I 3 − 5.0 + 2.0 I 4 − 5.0 = 2.0( I 3 + I 4 ) − 10 = I 3 + I 4 I1 + I 2 = I 3 + I 4 = 10 A Setting Equations (3) and (4) equal to each other, we have −4.0( I 2 + I 3 ) = 2.0( I1 − I 2 ) 1 I 3 = − ( I1 + I 2 ) = −5 A 2 1 I 4 = ( I1 + I 2 ) = 15 A 2 I 2 = 2.0 I 4 − 5.0 = 2.0 (15 ) − 5.0 = 25 A

I1 = 2.0 I 3 − 5.0 = 2.0 ( −5 ) − 5.0 = −15 A

I 5 = I 3 − I1 = −5 A − 15 A = −20 A 2.0( I1 − I 2 ) 2.0(−15 − 25) R5 = = = 4.0 Ω I5 20


28

CHAPTER 28-28.

28

The figure shown is the same as Figure 28-17(b), except that the second battery has been reversed. As a result, Equations 28.14 and 28.15 become −4 I 2 + 12 − 4 I1 = 0 (1) 8 − 2I3 + 4I 2 = 0

(2)

and we have the junction rule for point A: I1 = I2 + I3 (3) Substituting equation 3 into Equation 1 gives −4 I 2 + 12 − 4 ( I 2 + I 3 ) = 0

8I 2 + 4 I 3 = 12 3 − I3 2 Substituting this result into Equation 2, we find ⎛ 3 − I3 ⎞ 8 − 2I3 + 4 ⎜ ⎟=0 ⎝ 2 ⎠ 8 − 2I3 + 6 − 2I3 = 0 I2 =

I 3 = 3.5 A 3 − I 3 3 − 3.5 A = = −0.25 A= 0.25 A 2 2 I1 = 3.25 A I2 =

†28-29.

The minus sign indicates that choice of current direction for I2 was incorrect. Using Figure 28.44 for the two loops shown, (1) −4 I 2 + 12 − 4 I1 = 0 −8 − 2 I 3 + 12 − 4 I1 = 0

(2)

I1 = I 2 + I 3

(3)

Substituting Equation (3) into Equation (1) gives −4 I 2 + 12 − 4 ( I 2 + I 3 ) = 0 8I 2 + 4 I 3 = 12 I3 = 3 − 2I2 Substituting this result into Equation (2), we find −8 − 2 ( 3 − 2 I 2 ) + 12 − 4 ( I 2 + I 3 ) = 0 −8 − 6 + 4 I 2 + 12 − 4 I 2 − 4 I 3 = 0 1 A = −0.50 A 2 Then, this result can be used to find I2: I3 = 3 − 2I 2 I3 = −

−0.50 = 3 − 2 I 2 I 2 = 1.75 A I1 = I 2 + I 3 = 1.75 A − 0.50 A = 1.25 A which is the same result as found using the two alternative loops.


CHAPTER 28-30.

5 − 4 I1 − 2 I 3 = 0

(i)

3 − 3I 2 − 2 I 3 = 0

(ii)

I 3 = I1 + I 2

(iii)

Substituting Equation (iii) into Equation (i) gives 5 − 4 ( I3 − I 2 ) − 2I3 = 0 5 + 4I 2 − 6I3 = 0 6I3 − 5 4 Substituting this result into Equation (ii) gives ⎛ 6I − 5 ⎞ 3 − 3⎜ 3 ⎟ − 2I3 = 0 ⎝ 4 ⎠ 18I 3 15 + − 2I3 = 0 3− 4 4 27 I3 = A = 1.04 A 26 ⎛ 27 ⎞ 6⎜ ⎟ − 5 6I − 5 26 = ⎝ ⎠ = 0.31 A I2 = 3 4 4 I1 = I 3 − I 2 = 1.04 A − 0.31 A = 0.73 A I2 =

The potential difference between P and P′ is equal to the sum of the potential drops, VPP′ = − I1 R1 − I 2 R2 = − ( 0.73 A )( 4.0 Ω ) − ( 0.31 A )( 3.0 Ω ) = −2.0 V 28-31.

The loop and junction equations are ε1 − 4 I1 − 4 I 2 = 0 (1) −8 − 2 I 3 + 4 I 2 = 0

(2)

I1 = I 2 + I 3

(3)

Equation 2 gives −8 − 2 I 3 + 4 I 2 = 0 I3 = 4 − 2I 2 Substituting Equation 3 into 1 and applying the above result, ε1 − 4 ( I 2 + I 3 ) − 4 I 2 = 0

ε1 − 4 I 2 − 4 ( 4 − 2 I 2 ) − 4 I 2 = 0 ε1 − 4 I 2 − 16 + 8I 2 − 4 I 2 = 0 ε1 = 16 V

28-32.

Assume that I1 is directed upward through resistor R1 and that I2 passes through R2. Taking two loops around this circuit and substituting the given values, we find ε ′ − Ri′ ( I1 − I 2 ) − R1 I1 = 0 (1) 6.0 − 0.020 I1 + 0.020 I 2 − 0.50 I1 = 0 6.0 − 0.52 I1 + 0.020 I 2 = 0 I2 =

0.52 I1 − 6.0 0.020


28

CHAPTER

28

ε − Ri I 2 + Ri′ ( I1 − I 2 ) − ε ′ − R2 I 2 = 0

(2)

ε − ε ′ − ( Ri + R2 − Ri′ ) I 2 + Ri′I1 = 0 12.0 − 6.0 − 0.245 I 2 − 0.020 I1 = 0 6.0 − 0.245 I 2 − 0.020 I1 = 0

Now, substitute the result found from Equation 1 ⎛ 0.52 I1 − 6.0 ⎞ 6.0 − 0.245 ⎜ ⎟ − 0.020 I1 = 0 ⎝ 0.020 ⎠ −6.39 I1 + 79.5 = 0 I1 = 12.4 A

†28-33.

Then, 0.52 I1 − 6.0 0.52 (12.4 A ) − 6.0 I2 = = = 23.5 A 0.020 0.020 We can select three loops within the circuit shown and write the loop equations as follows. ε − R3 I 3 − R4 I 6 = 0 (1)

ε − R1 I 2 − R2 I 4 = 0

(2)

− R1 I 2 − R5 I 5 + R3 I 3 = 0

(3)

I2

(6)

I6 = I3 + I5

(7)

R1

I4 R2

I5 R5

− R2 I 4 + R4 I 6 + R5 I 5 = 0 (4) Here are some junction equations: I1 = I 2 + I 3 = I 4 + I 6 (5) I 2 = I 4 + I5

P′

I3 R3 I1

R4 P

I6

Substitute the values given for the resistors and the battery, suppressing the units, gives 12 − 6.0 I 3 − 2.0 I 6 = 0 (1) 1 I6 3 12 − 2.0 I 2 − 4.0 I 4 = 0 I 3 = 2.0 −

(2)

I 2 = 6.0 − 2.0 I 4 −2.0 I 2 − 3.0 I 5 + 6.0 I 3 = 0

(3)

−4.0 I 4 + 2.0 I 6 + 3.0 I 5 = 0

(4)

Adding together the results from Equations 1 and 2 and applying Equation 5 gives I 2 + I3 = I 4 + I6 6.0 − 2.0 I 4 + 2.0 −

1 I6 = I4 + I6 3

4 I6 = 0 3 6.0 − 2.25I 4 − I 6 = 0

8.0 − 3.0 I 4 −

I 6 = 6.0 − 2.25 I 4


CHAPTER This result can be used along with Equation 6 in Equation 3: −2.0 I 2 − 3.0 I 5 + 6.0 I 3 = 0

1 ⎞ ⎛ −2.0 ( 6.0 − 2.0 I 4 ) − 3.0 ( I 2 − I 4 ) + 6.0 ⎜ 2.0 − I 6 ⎟ = 0 3 ⎠ ⎝ 1 ⎛ ⎞ −2.0 ( 6.0 − 2.0 I 4 ) − 3.0 ( 6.0 − 2.0 I 4 − I 4 ) + 6.0 ⎜ 2.0 − ( 6.0 − 2.25I 4 ) ⎟ = 0 3 ⎝ ⎠ −12 + 4.0 I 4 − 18 + 6.0 I 4 + 3.0 I 4 + 4.5I 4 = 0 −30 + 17.5I 4 = 0 I 4 = 1.71 A I 6 = 6.0 − 2.25 I 4 = 6.0 − 2.25 (1.71 A ) = 2.15 A I 2 = 6.0 − 2.0 I 4 = 2.58 A 1 1 I 6 = 2.0 − ( 2.15 A ) = 1.28 A 3 3 −4.0 I 4 + 2.0 I 6 + 3.0 I 5 = 0 I 3 = 2.0 −

4.0 I 4 − 2.0 I 6 4.0 (1.71) − 2.0 ( 2.15 ) = = 0.85 A 3.0 3.0 The currents in the resistors are: R1: I2 = 2.58 A R2: I4 = 1.71 A R3: I3 = 1.28 A R4: I6 = 2.15 A R5: I5 = 0.85 A The potential difference between points P and P′ is ∆V = IR = (0.85 A)(3.0 Ω) = 2.6 V I5 =

28-34.

Start at the lower mid-point junction and make a clockwise loop as follows: − R1 I1 + R3 I 3 + R2 I 2 = 0 (1) Start at the upper mid-point junction and make a counterclockwise loop as follows: ε1 + ε 2 − I 3 R3 = 0 (2) Start at the lower right junction and make a counterclockwise loop as follows:

ε 2 − I 3 R3 + I1 R1 = 0

(3)

From Equation 2, ε + ε 2 6.0 V + 3.0 V = = 4.5 A I3 = 1 R3 2.0 Ω From Equation 3, I R − ε 2 ( 4.5 A )( 2.0 Ω ) − 3.0 V I1 = 3 3 = = 1.0 A 6.0 Ω R1 From Equation 1,


I3 I2

I4 I1

I5

28

CHAPTER

I2 =

†28-35. 28-36.

28-37.

28-38.

†28-39.

28 R1 I1 − R3 I 3 ( 6.0 Ω )(1.0 A ) − ( 2.0 Ω )( 4.5 A ) = R2 4.0 Ω

= −0.75 A Note that I2 was chosen to point in the wrong direction. The currents in the resistors are: R1: I1 = 1.0 A R2: I2 = 0.75 A R3: I3 = 4.5 A The currents in the batteries are: ε1: I4 = I1 + I3 = 5.5 A ε1: I5 = − I2 + I3 = 5.25 A The power used by the device is the product of the current and voltage. P = IV = (2.0 × 10−4 A)(3.0 V) = 6.0 × 10−4 W Cost = ( energy used )( rate ) = Pt ( rate ) = I 2 Rt ( rate ) 2 ⎛ 1 kW ⎞ ⎛ $0.15 ⎞ = (15 A ) ( 8.0 Ω )( 4.0 h ) ⎜ ⎟⎜ ⎟ = $1.08 ⎝ 1000 W ⎠ ⎝ 1 kW ⋅ h ⎠ Cost = ( energy used )( rate )

⎛ 1 kW ⎞ ⎛ $0.15 ⎞ = (100 W )( 24 h ) ⎜ ⎟⎜ ⎟ = $0.36 ⎝ 1000 W ⎠ ⎝ 1 kW ⋅ h ⎠ P = IV = I 2 R P 1500 W I = = = 13 A V 115 V P 1500 W R= 2 = = 8.9 Ω 2 I (13 A ) As current passes through a resistor, energy is dissipated in the resistor that becomes heat. In the case of this refrigerator, the maximum current that one could pass through such a resistor is P 350 × 10−6 W = = 3.7 × 10−3 A R 25 Ω P = IV = I 2 R

I = 28-40.

P 1.0 × 1012 W = = 8.696 × 109 A = 8.7 × 109 A V 115 V P 1.0 × 1012 W R= 2 = = 1.3 × 10−8 Ω 2 9 I ( 8.696 × 10 A )

I =

E = Pt = (1.0 × 1012 W )(1.5 × 10−14 s ) = 0.015 J/pulse E = ( 0.015 J/pulse )( 20 pulses/s ) = 3.0 × 10−1 W t P 3.0 × 10−1 W I ′ = ave = = 2.6 × 10−3 A V 115 V The heater is connected across a potential difference of 115 V, so the power is given by V2 P= R eq Pave =

†28-41.


CHAPTER

28

where Req is the equivalent resistance of the two equivalent resistors connected either in series or in parallel, depending on how the switch is set. V2 V2 V2 Plow = = = R eq R + R 2 R V2 V2 2V 2 = = R eq ⎛ 2 ⎞ −1 R ⎜ ⎟ ⎝R⎠ Since the low power value is given, we can find the high power value by taking a ratio of these two equations. 2V 2 Phigh = R2 = 4 V Plow 2R Phigh = 4 Plow = 4(275 W) = 1100 W Phigh =

28-42.

⎛ 0.5 A ⎞ ∆V = IR = ⎜ ⎟ (0.10 W) ( 5.0 Ω ) = 0.25 V ⎝ 1W ⎠ 2

Pincident

⎡ ⎛ 0.5 A ⎞ ⎤ ⎜ ⎟ (0.10 W) ⎥ ( 5.0 Ω ) ⎢ 2 I R ⎝ 1W ⎠ ⎦ = = ⎣ = 0.125 = 0.13 (0.10 W) Pincident

P V 1 P= 2 P I = V

⎛ 746 W ⎞ 0.75 hp ⎜ ⎟ ⎝ 1 hp ⎠ = 4.9 A = 115 V m 1 gh = ( 2800 m 3 /s )(1000 kg/m 3 )( 9.81 m/s 2 ) ( 51 m ) = 7.0 × 108 W t 2 7.0 × 108 W = = 2.9 × 103 A 3 240 × 10 V

PR

28-43. 28-44.

†28-45.

I =

Current is defined as the amount of charge passing a given point per unit time. In this case, the current is given, so, Q = It The number of protons generated is equal to this charge divided by the unit charge. −6 Q It (1.0 × 10 C/s ) n= = = = 6.25 × 1012 protons/s = 6.3 × 1012 protons/s −19 e e 1.60 × 10 C

28-46.

Each proton has an energy of ⎛ 1.60 × 10−19 J ⎞ −10 Ep = 7.0 × 108 eV ⎜ ⎟ = 1.12 × 10 J/proton 1 eV ⎝ ⎠ So, the total power of the beam is P = nE=(6.25 × 1012 protons/s)(1.12 × 10−10 J/proton) = 7.0 × 102 W P 1200W I = = = 10.4 A = 1.0 × 101 A 115 V V V 2 (115 V ) = = 11 Ω 1200 W P 2

R=


CHAPTER

28

28-48.

E 5.0 × 106 W i h = = 6.7 h P ⎛ 746 W ⎞ 1000 hp ⎜ ⎟ ⎝ 1hp ⎠ P = IV = (80 A)(12 V) = 960 W ⎛ 1 hp ⎞ P = 960 W ⎜ ⎟ = 1.3 hp ⎝ 746 W ⎠ E = Pt = ( 960 W )( 2.5 s ) = 2400 J

28-49.

t=

28-50.

Adding all of the values given gives: P = ( 40 )( 75 W ) + 9000 W + 1400 W + 1000 W + 700 W + 5000 W + 700 W

28-47.

t=

E E 2.5 × 106 J ⎛ 1 h ⎞ = = ⎜ ⎟ = 58 h P IV (1.0 A )(12 V ) ⎝ 3600 s ⎠

+ 1100 W + (2)1200 W + (2)200 W + 1100 W + 1400 W + 250 W + 200 W + 150 W

†28-51.

28-52.

= 27 800 W or 27.8 kW P 27 800 W I = = = 240 A V 115 V The power lost due to the resistance of the wire is determined by the resistance of the wire. The resistance of a wire is given L R=ρ A Given this relationship and the information given in the problem and in Table 27.2, the power loss is L 60 × 103 m 2 P = I 2 R = I 2 ρ = ( 600 A ) ( 2.8 × 10−8 Ω i m ) = 1.2 × 106 W or 1.2 MW 2 A ⎛1 ⎞ π ⎜ (0.025 m) ⎟ ⎝2 ⎠ The power dissipated by the initial resistor is P=

ε2

R The emf is increased by 25% to a value ε′ = 1.25ε. To maintain the ratio, the resistance would have to be increased to 2 R′ = R (1.25 ) = 1.56 R 28-53.

The resistor should be increased by 56% The power dissipated by the initial resistor is P=

28-54.

ε2

R The resistance is increased by 20% to a value R′ = 1.2R. To maintain the ratio, the emf would have to be increased to ε ′ = ε 1.2 = 1.1ε The voltage should be increased by 10% ⎛ 1.000 kg ⎞ 1 liter ⎜ ⎟ ( 4180 J/(kg i °C) )( 50.0°C ) E mC ∆T ⎝ 1 liter ⎠ = = = 1.02 × 102 s t= 2 2 P I R ( 32.0 A ) ( 2.00 Ω )


CHAPTER †28-55.

28

The easiest way to produce the same power requirement, even though the applied potential differences are changed, is to adjust the equivalent resistance in the circuit. Since there is a switch, the two equal heater coils will be added to the circuit in different ways depending on how the switch is closed to create different paths. For 115 V operation and 230 V operation, the required resistances are V2 P= Req

V 2 (115 V ) = = 13.225 Ω P 1000 W 2 V 2 ( 230 V ) R230 = = = 52.90 Ω P 1000 W Note that R230 = 4R115. Therefore, for two equal resistors to produce the same power for the two different voltages, the resistors must be arranged as follows: 2

R115 =

R R

This part of the circuit may slide either left or right until contact is made.

V

28-56.

When the switch slides to the left, the result is two equal resistors in parallel. When the switch slides to the right, the result is two equal resistors in series. In the latter case, the equivalent resistance is 2R = 52.90 Ω for the 230 V case. Therefore, R = 26.45 Ω. When they are then connected in parallel, the equivalent resistance is R/2 = 13.225 Ω as calculated above for the 115 V case. cost = ( energy used )( rate )

28-57.

⎛ 4.0 min ⎞ ⎛ 1 h ⎞ ⎛ 1 kW ⎞ ⎛ $0.15 ⎞ = ( 7.0 W ) ⎜ ⎟ = $0.026/year ⎟ ( 365 days ) ⎜ ⎟⎜ ⎟⎜ ⎝ 60 min ⎠ ⎝ 1000 W ⎠ ⎝ 1 kW i h ⎠ ⎝ 1 day ⎠ P = IV = ( 2.5 × 10−6 C/s )( 2.0 × 105 V ) = 0.50 W

28-58. †28-59.

( 2.7 A )(14 V ) Pout IV = = = 0.12 or 12% 3 Pin SA (1.0 × 10 W/m 2 ) ( 0.58 m )( 0.53 m )

When the battery with an internal resistance Ri is connected to a resistor of resistance R, the equivalent resistance for this series circuit is the sum of the two resistances.

I =

ε Req

=

ε R + Ri

and the power dissipated by R in the circuit is 2

⎛ ε ⎞ P=I R=⎜ ⎟ R ⎝ R + Ri ⎠ To determine when the power has a maximum, we can take the derivative of the power with respect to R and set that equal to zero. 2


CHAPTER

28

2 ⎡⎛ 1 ⎞2 ⎤ ⎡ 2R ⎤ 2 dP d ⎡⎛ ε ⎞ ⎤ 1 2 d ⎢⎜ ⎢⎜ = + ⎥ε = 0 ⎟ R⎥ = ε ⎟ R⎥ = − ⎢ 3 2 dR dR ⎢ ⎝ R + Ri ⎠ ⎥ dR ⎢ ⎝ R + Ri ⎠ ⎥ + + R R R R ⎢ ⎥ ( ) ( ) i i ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 2R 1 = 3 2 ( R + Ri ) ( R + Ri )

2 R ( R + Ri ) = ( R + Ri ) 2

2

3

3

R 3 + R 2 Ri − RRi − Ri = 0

( R − Ri )( R + Ri )

2

=0

R = ± Ri

28-60.

Of course, resistance must be positive. Therefore, the maximum power dissipated by the resistor occurs when R = Ri. The current in the circuit is I =

ε Req

ε

=

R + Ri

So, the potential difference across the light bulb is ( 3.0 V )( 6.0 Ω ) = 2.11 V = 2.1 V εR V = IR = = R + Ri 6.0 Ω + 2.5 Ω The power delivered to the lightbulb is ( 3.0 V )( 2.11 V ) = 0.74 W εV P = IV = = R + Ri 6.0 Ω + 2.5 Ω The power lost to the internal resistance is 2

⎛ ⎛ ε ⎞ ( 3.0 V ) ⎞ Pi = I 2 Ri = ⎜ ⎟ 2.5 Ω = 0.31 W ⎟ Ri = ⎜ ⎝ R + Ri ⎠ ⎝ 6.0 Ω + 2.5 Ω ⎠ (a) Without the internal resistance, the power delivered to the circuit would be P = Iε. The power lost to the internal resistance is Pi = I2Ri Taking the ratio of these two quantities gives the fraction of the power lost to the internal resistance. (1.0 A )( 0.40 Ω ) = 0.033 or 3.3% Pi I 2 Ri IR = = i = P Iε 12 V ε 2 (10.0 A )( 0.40 Ω ) = 0.33 or 33% P I Ri IR (b) i = = i = ε P Iε 12 V Clearly, using the battery at lower current is more efficient. 2

†28-61.

(115 V ) V2 V2 = = = = 3310 W −1 −1 Req ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ + ⎜ ⎟ ⎜ + ⎟ ⎝ 12.0 Ω 6.0 Ω ⎠ ⎝ R1 R2 ⎠ 2

28-62.

(a) Pparallel

(115 V ) V2 V2 = = = 735 W Req R1 + R2 12.0 Ω + 6.0 Ω 2

(b) Pseries = 28-63.

(a) V = IR = I ρ

L 40 m = (1.0 × 104 A )(1.7 × 10−8 Ω i m ) = 177 V = 180 V 2 A π ( 0.0035 m )


CHAPTER

28

(b) P = IV = (1.0 × 104 A ) (177 V ) = 1.8 × 106 W 28-64.

P = I 2R = I 2ρ

L A

( 30 A ) P I2 =ρ = (1.7 × 10−8 Ω i m ) = 2.9 W/m 2 L A ⎛1 ⎞ π ⎜ (0.00259 m) ⎟ ⎝2 ⎠ V P / L 2.9 W/m = = = 0.097 V/m L I 30 A Assume all of the power consumed goes into Joule heating. The cooling water must carry away this heat to prevent the temperature of the coils from becoming too large. As water flows through the coils, its temperature is to have a maximum temperature increase of 60.0°C. Each second, the heating rate is ∆m P = IV = c∆T ∆t ( 200 A )( 400 V ) IV ∆m ⎛ 60 s ⎞ ⎛ 1 liter ⎞ = = ⎟ = 19 liters/min ⎜ ⎟⎜ c∆T ( 4180 J/(kg ⋅ °C) )( 60 °C ) ⎝ 1 min ⎠ ⎝ 1.0 kg ⎠ ∆t 2

†28-65.

28-66.

P40 mA = I 2 R = ( 0.040 A ) (11 Ω) = 0.018 W 2

P40 mA = I 2 R = ( 3.0 A ) (0.030 Ω) = 0.27 W 2

Pacc = I 2 R = ( 3.0 A ) (11 Ω) = 99 W 2

28-67.

28-68.

According to Equation 28-25, the value of the unknown resistor is R 2.0 × 103 Ω Rx = 3 R1 = ( 350 Ω ) = 7.0 × 102 Ω R2 1.0 × 103 Ω V = I ( R + Ri ) I =

V R + Ri

ε = V + IRi ⎛ Ri ⎞ 0.20 Ω V ⎛ ⎞ Ri = V ⎜ 1 + ⎟ = (1.4993 V ) ⎜ 1 + ⎟ = 1.4994 V 5000 Ω + 0.20 Ω ⎠ R + Ri R + Ri ⎠ ⎝ ⎝ ε = I ( R + Ri ) = I ′R =V +

28-69.

I′ =

28-70.

I ( R + Ri )

( 3.955 A )( 3.002 Ω )

= 3.958 A 3.000 Ω ⎛ Ri ⎞ As in Problem 28-68, ε = V ⎜ 1 + ⎟ R + Ri ⎠ ⎝ We want the ratio of the emf to the measured voltage to be within 1.0%. Therefore, the maximum resistance that gives a 1.0% difference is R

=


CHAPTER ε V

28

=1+

Ri = 1.010 R + Ri

Ri = 0.010 R + Ri Ri ⎛ 1 ⎞ − Ri = 0.020 Ω ⎜ − 1 ⎟ = 1.98 = 2.0 Ω 0.010 0.010 ⎝ ⎠ The resistance of the two lead wires is combined in a series connection with the specimen. The total resistance is L RCu = Rsample + 2 Rleads = Rsample + 2 ρ A 3.0 m = 5.0 Ω +2 (1.7×10−8 Ω i m ) = 5.021 Ω = 5.0 Ω 2 π ( 0.00125 m ) R=

†28-71.

Rconstantan = Rsample + 2 ρ

L A

= 5.0 Ω + 2 ( 49 × 10−8 Ω i m )

28-72.

28-73. 28-74. 28-75.

28-76.

3.0 m

π ( 0.00125 m )

2

= 5.6 Ω

Using low resistance leads has minimal effect on the measurement. However, some materials have significant resistance that does affect the measurement. τ = RC τ 15 s = = 7.5 × 105 Ω R= −5 C 2.0 × 10 F τ = RC = (100 Ω)(3.2 × 10−6 F) = 3.2 × 10−4 s τ 0.0027 s = 3.6 × 10−6 F = 36 µ F C= = R 75 Ω 0.37 = e − t / RC t −0.994 = − RC t = 0.994 RC = 0.994 (100 Ω ) ( 4.0 × 10−6 F ) = 4.0 × 10−4 s 1 I 0 = I 0 e − t / RC 10 et / RC = 10 t1 / 10 = RC ln(10) = 2.3RC I (t ) =

t1 / 100 = RC ln(100) = 4.6 RC †28-77.

For two capacitors in series, the equivalent capacitance is −1

⎛ 1 1 ⎞ Ceq = ⎜ + ⎟ ⎝ C1 C2 ⎠ The characteristic time is calculated using Equation 28-34. ⎛ 1 1 ⎞ τ = RCeq = R ⎜ + ⎟ ⎝ C1 C2 ⎠

−1

⎛ ⎞ 1 1 = ( 8.0 × 10 Ω ) ⎜ + ⎟ −6 −6 ⎝ 2.0 × 10 F 4.0 × 10 F ⎠ 3


−1

= 0.011 s

CHAPTER 28-78.

28

Equation 28.31 is Q = Cε (1 − e − t / RC ) At two time constants, t = 2RC, Q = Cε (1 − e −2 ) = Cε ( 0.8647 ) Therefore, at two time constants, the charge reaches approximately 86.5% of its final value. At three time constants, t = 3RC, Q = Cε (1 − e −3 ) = Cε ( 0.9502 ) At three time constants, the charge reaches approximately 95.0% of its final value. At five time constants, t = 5RC, Q = Cε (1 − e −5 ) = Cε ( 0.9933 )

†28-79.

Therefore, at five time constants, the charge reaches approximately 99.3% of its final value. At t = 0, the charge on the capacitor is zero coulombs because no current is in the circuit. When the switch is moved to position one, a current exists in the circuit that charges the capacitor. There are two resistors in series with an equivalent resistance R = R1 + R2. Therefore, applying Equation 28-31, the charge on the capacitor as a function of time is Q = Cε (1 − e − t / RC ) = Cε 1 − e − t /( R1 + R2 ) C

(

)

The voltage drop across R1 is determined by applying Equation 28.33 I =

ε

(e R

− t / RC

1

∆V = IR1 = 28-80.

) = (R

ε

(e +R)

− t /( R1 + R2 ) C

2

ε R1 ( R1 + R2 )

(e

− t /( R1 + R2 ) C

)

)

As in the previous problem, I =

ε

(e R

− t / RC

) = (R

1

ε

(e +R)

− t /( R2 + R3 ) C

2

) 2

⎡ ⎤ ε The power dissipated in resistor R2 is given by P = I R2 = ⎢ e − t /( R2 + R3 ) C ⎥ R2 ⎣ ( R1 + R2 ) ⎦ 2

(

)

The total energy dissipated in this resistor is ∞ ∞ R2ε 2 R2ε 2 ⎛ ( R2 + R3 )C ⎞ R2ε 2C −2 t /( R2 + R3 ) C E = ∫ P dt = e dt = = ⎟ 2( R + R ) 0 ( R1 + R2 ) 2 ∫0 ( R1 + R2 ) 2 ⎜⎝ 2 ⎠ 2 3

(

†28-81.

)

(a) The amount of charge on the capacitor plates is given by Q (t ) = A + Be − t / τ At time t = 0, the capacitor begins fully charged with an initial value Q0 = A + B = ε1C. Prior to that time the capacitor had become fully charged by a current passing through R1 from the battery ε1. After the switch is moved to position 2, depending on whether ε2 is greater than, equal to, or less than ε1, the capacitor will either further charge, remain the same, or discharge, respectively. At time t = ∞, the capacitor plates will be charged to Q(∞) = ε2C. Therefore, A = ε 2C Then, from the earlier relationship, A + B = ε1C B = ε1C − A = ε1C − ε 2 C = C (ε1 − ε 2 )


CHAPTER

28

Since the switch is in position 2, any current that flows in the circuit is through the resistor R2, so the time constant for this circuit is τ = R2C The result is that the charging of the capacitor is completely described as Q (t ) = ε 2 C + ( ε1 − ε 2 ) Ce − t / R2C (b) This case is simply the reverse of (a). At time t = 0, the capacitor begins fully charged with an initial value Q0 = A + B = ε2C. After the switch is moved to position 1, at time t = ∞, the capacitor plates will be charged to Q(∞) = ε1C. Therefore, A = ε1C Then, B = ε 2C − A = ε 2C − ε1C = C (ε 2 − ε1 ) Since the switch is in position 1, any current that flows in the circuit is through the resistor R1, so the time constant for this circuit is τ = R1C The result is that the charging of the capacitor is completely described as Q (t ) = ε1C + ( ε 2 − ε1 ) Ce − t / R1C 28-82.

Q (t ) = V0 e − t / RC C ⎛ V (t ) ⎞ ln ⎜ ⎟ = −t / RC ⎝ V0 ⎠

V (t ) =

†28-83.

t

5.0 × 10−3 s

= 1.2 × 104 Ω 60 ⎛ V0 ⎞ ⎛ ⎞ −6 C ⎜ ln ⎟ ( 0.25 × 10 F ) ln ⎜ 1.2 ⎟ V ( t ) ⎝ ⎠ ⎝ ⎠ Applying Equation 28-31, the charge on the capacitor as a function of time is Q = Cε (1 − e − t / RC ) R=

=

The time at which the capacitor reaches one-half of its final charge is Cε = Cε (1 − e − t / RC ) 2 1 1 e − t / RC = 1 − = 2 2 Take the natural logarithm of both sides. ⎛1⎞ −t / RC = ln ⎜ ⎟ ⎝2⎠ Moving the minus sign inside the logarithm inverts the fraction to give t = RC ln(2) = 0.693(8.0 × 103 Ω)(2.0 × 10−6 F) = 0.011 s The final energy stored in the capacitor is Q02 ( ε C ) 1 = = ε 2C 2C 2C 2 The time at which one-half of this energy is achieved is 1 2 − t / RC 2 2 ε C ) Q 2 (t ) ε C (1 − e U (t ) = = = 2 2C 2 2 2 (1 − e−t / RC ) = 12 2

U =


CHAPTER

1 − e − t / RC =

1 2

e − t / RC = 1 −

1 2

⎛ 1⎞ −t / RC = ln ⎜⎜ 1 − ⎟ 2 ⎟⎠ ⎝

t = 1.228 RC = 1.228 ( 8.0 × 103 Ω )( 2.0 × 10−6 F ) = 0.020 s 28-84.

Equation 28.27 is dQ Q ε− R− =0 dt C Substituting Equation 28.31 into this equation gives d Q ε − R ⎡⎣ Cε (1 − e − t / RC ) ⎤⎦ − = 0 dt C Cε (1 − e − t / RC ) d − t / RC ε − C ε R (1 − e =0 )− dt C ⎛ 1 − t / RC ⎞ − t / RC ε − Cε R ⎜ e )=0 ⎟ − ε (1 − e ⎝ RC ⎠

ε − ε e− t / RC − ε + ε e −t / RC = 0 28-85.

0=0 The final voltage on the capacitor is the sum of the voltage due to the initial charging plus that due to the discharging that occurs when the battery is replaced. ∆V = ε 2 (1 − e − t / RC ) − ε1 (e − t / RC ) = 0

ε 2 − ( ε1 + ε 2 ) e− t / RC = 0 e − t / RC =

ε2

ε1 + ε 2 ε2 −t/RC = ln ε1 + ε 2

t = RC ln

ε1 + ε 2 ⎛ 8.0 V ⎞ = ( 6.0 × 105 Ω )( 9.0 × 10−6 F ) ln ⎜ ⎟ ε2 ⎝ 3.0 V ⎠

= 5.296 s = 5.3 s The current is given by Equation 28.35, ε 3.0 V = 5.0 × 10−6 A I = 2 = 5 R 6.0 × 10 Ω 28-86.

U =

∫

∞

Pdt =

∫

∞

I 2 Rdt = ∫

∞

ε2 2

e −2t / RC Rdt =

ε2

R∫

∞

e −2t / RC dt

R 1 ε ⎛ − RC ⎞ = = Cε 2 ⎜ ⎟e 0 2 R⎝ 2 ⎠ The energy required to charge the capacitor with a total charge Q is equal to the work done, 0

2

0

0

0

− t / RC ∞


28

CHAPTER

28

q Q2 1 2 dq = = Cε ∫0 C 2C 2 Since an equal amount of energy is lost to Joule heating, the battery must supply twice this energy to charge the plates. Therefore, the total energy is ⎛1 ⎞ E = U + W = 2 ⎜ Cε 2 ⎟ = Cε 2 ⎝2 ⎠ (a) Taking the shortest path, from right hand to right foot, gives a total resistance of Req = R1 + R2 + R3 + R4 + R5 + R6 + R7 W =

28-87.

Q

= 365 Ω + 150 Ω + 137 Ω + 18 Ω + 70 Ω + 195 Ω + 445 Ω = 1380 Ω V 600 × 103 V = = 430 A R 1380 Ω (c) Req = R3 + R4 + R5 (b) I =

= 137 Ω + 18 Ω + 70 Ω

28-88. †28-89.

= 225 Ω V 600 × 103 V I = = = 2670 A R 225 Ω R ρ L / A ρCu 1.7 × 10−8 Ω PAl V2 /R = 2 Al = Cu = Cu = = = 0.61 PCu V / RCu RAl ρ Al L / A ρ Al 2.8 × 10−8 Ω The resistance of the first bulb may be determined from the power and the potential difference across the bulb. P1 = R1 =

( ∆V )

2

R1

( ∆V )

2

P1

Similarly, R2

( ∆V ) =

2

P2

When these bulbs are connected in series, the two bulbs have an equivalent resistance Req = R1 + R2 Using this information, the current through the two bulbs connected in series may then be determined. P1 P2 ∆V ∆V ∆V = = = I = 2 2 Req R1 + R2 ( ∆V ) + ( ∆V ) ∆V ( P1 + P2 ) P1 P2 The power delivered by the first bulb is now, 2

⎛ ⎞ ⎛ ⎞ P1 P2 P1 P2 P1′ = I R1 = ⎜⎜ ⎟⎟ R1 = ⎜⎜ ⎟⎟ ⎝ ∆V ( P1 + P2 ) ⎠ ⎝ ∆V ( P1 + P2 ) ⎠

2

2

( 75 W )(130 W ) = = 2 2 ( P1 + P2 ) ( 75 W + 130 W ) P1 P22

( ∆V )

2

P1

2

= 3.0 × 101 W

For the second bulb,


CHAPTER 2

⎛ ⎞ ⎛ ⎞ P1 P2 P1 P2 P2′ = I R2 = ⎜⎜ ⎟⎟ R2 = ⎜⎜ ⎟⎟ ⎝ ∆V ( P1 + P2 ) ⎠ ⎝ ∆V ( P1 + P2 ) ⎠

2

2

=

28-90.

2

P2

2

P12 P2

( P1 + P2 )

( 75 W ) (130 W ) 2 ( 75 W + 130 W )

( ∆V )

28

2

=

= 1.7 × 101 W

The bulb that was dimmer when connected in parallel is now the brighter bulb. Consider the circuit shown to the right. Two I3 possible loops are shown and three currents have been labeled. At the junction point at the I2 middle of the upper portion of the circuit, 2 1 I 3 = I1 + I 2

I1

Then, for the first loop, we have ε1 + I 2 R2 − ε 2 = 0 10 + 8.0 I 2 − 12 = 0 2.0 V = 0.25 A 8.0 Ω For the second loop, ε 2 − I 2 R2 − I 3 R1 = 0 I2 =

10 − ( 8.0 Ω )( 0.25 A ) − 10.0 I 3 = 0 8V = 0.8 A 10.0 Ω The power delivered by the 12 V battery is P = I 2V = (0.25 A)(12 V) = 3.0 W I3 =

28-91.

Consider the circuit shown to the right. Two possible loops are shown and three currents have been labeled. At the junction point at the middle of the lower portion of the circuit, I 3 = I1 + I 2 (1) Then, for the first loop, we have ε − I1 Ri + I 2 Ri′ − ε = 0 (2) I1 Ri = I 2 Ri′ For the second loop, ε − I 2 Ri′ − I 3 R = 0

ε

ε

R 2 I3

Ri′ I2

1

Ri I1

(3)

Using the results from Equations (1) and (2) in equation (3) gives the current through resistor Ri′ ε − I 2 Ri′ − ( I1 + I 2 ) R = 0 ⎛

ε − I 2 Ri′ − ⎜ I 2 ⎝

I2 =

R′ ⎞ ⎟ R − I2 R = 0 Ri ⎠

ε ⎛ R′ ⎞ Ri′ + R ⎜ − 1 ⎟ ⎝ Ri ⎠

Then, for Equation (2),


CHAPTER

28

Ri′ Ri R′ I1 = I 2 i = Ri

I1 = I 2

28-92.

Ri′ ε Ri′ = Ri R′ + RR′ − RRi ⎛ R′ ⎞R Ri′ + R ⎜ i − 1 ⎟ i ⎝ Ri ⎠

ε

This problem is similar to that of Problem 28-91. We simply relabel the resistors as shown. Two possible loops are shown and three currents have been labeled. At the junction point at the middle of the lower portion of the circuit, I 3 = I1 + I 2 (1) Then, for the first loop, we have ε 2 − I1 R2 + I 2 R1 − ε1 = 0 12.0 − 0.20 I1 + 0.25I 2 − 6.0 = 0

2 I2

I3

(2)

1 I1

I 2 = 0.80 I1 − 24.0 For the second loop, ε1 − I 2 R1 − I 3 R3 = 0

(3)

6.0 − 0.25 I 2 − 0.50 I 3 = 0

Using the results from Equations 1 and 2 in Equation 3 gives the current through resistor Ri′ 6.0 − 0.25 ( 0.80 I1 − 24.0 ) − 0.50 ( I1 + 0.80 I1 − 24.0 ) = 0 6.0 + 6.0 − 0.20 I1 − 0.50 I1 − 0.40 I1 + 12.0 = 0 24 − 1.10 I1 = 0 I1 = 21.8 A The current through the resistor R1 is I 2 = 0.80 ( 21.8 A ) − 24.0 A= − 6.6 A It is 6.6 A directed from the bottom toward the top. The current through the resistor R3 is I 3 = 21.8 A − 6.6 A = 15.2 A 28-93.

The resistors R2 and R3 are connected in parallel, so they be combined into a single equivalent resistor −1

−1

⎛ 1 1 ⎞ 1 ⎞ ⎛ 1 + + R23 = ⎜ ⎟ =⎜ ⎟ = 5.71 Ω ⎝ 8.0 Ω 20.0 Ω ⎠ ⎝ R2 R3 ⎠ The circuit may then be redrawn as shown using the equivalent resistor. At the junction point at the middle of the lower portion of the circuit, I 2 = I1 + I 3 (1) Around the first loop: ε1 − I1 R1 − I 2 R23 = 0

1

12 − 10.0 I1 − 5.71I 2 = 0 I2 =

I1

(2)

12 − 10.0 I1 = 2.10 − 1.75I1 5.71


I2 R2

2 I3

CHAPTER

28

Around the second loop: ε 2 − I 2 R23 − I 3 R4 = 0 16 − 5.71I 2 − 10.0 I 3 = 0 16 − 5.71I 2 − 10.0 ( I 2 − I1 ) = 0

16 − 5.71 ( 2.10 − 1.75I1 ) − 10.0 ( 2.10 − 1.75I1 − I1 ) = 0 I1 = 0.45 A I 2 = 2.10 − 1.75 I1 = 2.10 − 1.75(0.453) = 1.31 A I 3 = I 2 − I1 = 1.31 A − 0.453 A = 0.85 A The batteries deliver P1 = I1ε1 = (0.453 A)(12 V) = 5.4 W P2 = I 3ε 2 = (0.854 A)(16 V) = 14 W 28-94.

28-95.

(a) Let ε represent the external emf in connected to the battery. A current of 6.0 A flows through the internal resistance Ri = 0.20 Ω. Therefore, the potential drop across the internal resistance is ∆V = (0.20 Ω)(6.0 A) = 1.2 V. The emf must be at least, ε = 12.0 V + 1.2 V = 13.2 V (b) P = I2R = (6.0 A)2(0.20 Ω) = 7.2 W L L (a) R = ρ = ρ 2 πr A r=

ρL = πR

( 2.8 × 10

−8

Ω i m )(1.0 × 105 m )

π ( 6.0 Ω )

= 0.0122 m

D = 2r = 0.024 m or 2.4 cm

28-96.

(b) P = I2R = (750 A)2(6.0 Ω) = 3.4 × 106 W or 3.4 MW J ⎞ ⎛ 1.0 kg ⎞ ⎛ 150 liters ⎜ ⎟ ⎜ 4180 ⎟ 50°C kg ⋅ °C ⎠ Q mc∆T ⎝ 1 liter ⎠ ⎝ P= = = = 8700 W t t ⎛ 3600 s ⎞ 1 h⎜ ⎟ ⎝ 1h ⎠

( ∆V )

P=

R

( ∆V )

R=

28-97.

2

2

(115 V )

2

= = 1.5 Ω 8700 W P (a) P = I ∆V ⎛ 745.7 W ⎞ 12 hp ⎜ ⎟ 1 hp ⎠ P ⎝ I = = = 46.6 A = 47 A ∆V 16(12 V) (b) The total energy U to be used is equal to the product of power delivered by the batteries and the elapsed time. U = P∆t U P In the time ∆t, the vehicle can travel a distance ∆t =


CHAPTER

28

6 km ⎞ ⎛ 1 h ⎞ ⎛ 1000 m ⎞ 16 ( 2.2 × 10 J ) U ⎛ = ⎜ 65 = 7.1 × 103 m or 71 km ⎟⎜ ⎟⎜ ⎟ h ⎠ ⎝ 3600 s ⎠ ⎝ 1 km ⎠ P ⎝ ⎛ 745.7 W ⎞ 12 hp ⎜ ⎟ ⎝ 1 hp ⎠ Q mLV = P= t t ( 6.0 kg − 3.7 kg ) ( 2.26 × 106 J/kg ) mLV mLV t= = = = 1130 s or about 19 min P I ∆V ( 20 A )( 230 V )

x = v∆t = v

28-98.

28-99.

28-100.

P = I 2R = I 2ρ

L 0.60 m 2 = ( 500 A ) (1.7 × 10−8 Ω i m ) = 130 W 2 A π ( 2.5 × 10−3 m )

The power delivered by the battery is equal to Iε, so the fraction of the delivered power that is lost to Joule heating is P 130 W = = 0.022 or 2.2% I ε (500 A)(12 V) 1 Q = Q0 e − t / RC = Q0 2 1 e − t / RC = 2 ⎛1⎞ −t/RC = ln ⎜ ⎟ ⎝2⎠ t = RC ln ( 2 ) = ( 5.0 × 108 Ω )( 8.0 × 10−6 F ) ln ( 2 ) = 2770 s

28-101.

(a) Assuming the plates initially had no charge, then at time t = 0 s, the charge on the plates is Q0 = 0 C. The capacitor is then charging, so that at time t = 0.0010 s, the charge on the plates is −0.0010 s / (100 Ω )( 2.0×10−5 F ) ⎞ Q0.0010 = ε C (1 − e − t / RC ) = ( 6.0 V ) ( 2.0 × 10−5 F ) ⎛⎜ 1 − e ⎟ ⎝ ⎠ −5 = 4.7 × 10 C −0.0020 s / (100 Ω )( 2.0×10−5 F ) ⎞ Q0.0020 = ε C (1 − e − t / RC ) = ( 6.0 V ) ( 2.0 × 10−5 F ) ⎛⎜ 1 − e ⎟ ⎝ ⎠ −5 = 7.6 × 10 C

(b) The final value of the charge is Q∞ = ε C = ( 6.0 V ) ( 2.0 × 10−5 F ) = 1.2 × 10−4 C (c)

dQ ⎛ 1 ⎞ ε 6.0 V = ⎜ Cε e − t / RC = = = 6.0 × 10−2 A ⎟ Ω 100 dt RC R ⎝ ⎠ t =0


CHAPTER 29

MAGNETIC FORCE AND FIELDS


The magnetic field produced by the current is: µI 4π × 10−7 × 50A B = 0 into the page ⇒ B = = 3.33 × 10−4 T. 2π × 0.03 m 2π r

| F | = q | v × B | = (1.6 × 10−19 C) × (2 × 106 m/s) × (3.33 × 10−4 T) = 1.07 × 10−6 N

29-2.

Since the magnetic field is pointing into the paper and the velocity is perpendicular to and away from the current, the right hand rule says the direction of v × B is in the direction of the current. Because the electron has a negative charge, the force’s direction opposite the direction of the current. µI 4π × (10−7 Ns 2 /C2 ) × 30A F = qvB = qv 0 = (1.6 × 10−19 C) × (5.0 × 106 m/s) × 2π r 2π × 0.1m = 4.8 × 10−17 N a=

†29-3.

F 4.8 × 10−17 N = = 2.87 × 1010 m/s 2 m 1.67 × 10−27 kg

Energy = ⇒

v=

1 me v 2 = U , where me is the mass of the electron. 2 2U = me

2 × 3 × 10−15 J = 8.12 × 107 m/s −31 9.1 × 10 kg

Since the magnetic field B =

µ 0 I 4π × (10−7 Ns 2 /C2 ) × 12A = = 8 × 10−6 T 2π r 2π × 0.3m

Therefore, the magnetic force F = qvB = (1.6 × 10−19 C) × (8.12 × 107 m/s) × (8 × 10−6 T) = 1.04 × 10−16 N ⇒

29-4.

B=

a=

F 1.04 × 10−16 N = = 1.14 × 1014 m/s 2 −31 me 9.1 × 10 kg

µ 0 I 4π × 10−7 Ns 2 /C2 × 25A = = 3.33 × 10−4 T 2π r 2π × 1.5 × 10−2 m

The magnetic force on one electron is:


CHAPTER

29

Fe = evB = (1.6 × 10−19 C) × (7.0 × 103 m/s) × (3.33 × 10−4 T) = 3.73 × 10−19 N We assume the copper atom (electrons and the nucleus) move with the speed 7.0 × 103 m/s, since copper nucleus has 29 protons, the magnetic force on the copper nucleus is: Fnucleus = 29evB = 29(1.6 × 10−19 C) × (7.0 × 103 m/s) × (3.33 × 10−4 T) = 1.08 × 10−17 N †29-5.

Since the whole atom is neutral, so the force acting on the whole atom is zero. µI 4π × (10−7 Ns 2 /C2 ) × 30A B= 0 k= k = 2.40 × 10−6 Tk 2π r 2π × 2.5m F = qv × B = 1.6 × 10−19 C (2.0i − 3.0 j + 4.0k ) × 2.4 × 10−6 k T = −1.15 × 10−24 i − 7.68 × 10−25 j The magnitude of the force is therefore:

| F | = (7.68) 2 + 11.522 × 10−25 N = 1.38 × 10−24 N 29-6. †29-7.

29-8.

1

ε 0 µ0

=

1 (8.85 × 10

−12

C /Nm ) × 4π × (10−7 Ns 2 /C 2 ) 2

2

= 3.0 × 108 m/s

The magnitude of the magnetic force is: F = qvB sin θ ⇒ Fmax = qvB

For F to be 1/3 of Fmax, sin θ = 1 3 F = qv × B

⇒

θ = 19.5°

⇒ | F | = qvB sin120° = (1.6 × 10−19 C ) × ( 2.0 × 105 m/s ) × ( 0.33 T ) × sin120o = 9.15 × 10−15 N

†29-9.

The direction of the force is going into the paper. Suppose the magnitude of the magnetic field is B, then the magnetic force is qpvB, where qp is the charge of the proton. F qp vB , where mp is the mass of the proton. = a= m mp ⇒

B=

mp a qp v

(1.67 × 10 (1.6 × 10

kg ) × ( 3.2 × 1014 m 2 /s )

−27

=

−19

C ) × ( 8.0 × 105 m/s )

= 4.18 T

In order t make a force in the west direction, the direction of the magnetic field should point downward. 29-10.

29-11. 29-12.

qp vB = mp g

⇒

B=

mp g qp v

=

(1.67 × 10

(1.6 × 10

−19

− 27

)

kg × 9.8 m/s 2

C ) × ( 6.0 × 10 m/s ) 4

= 1.70 × 10−12 T

The direction of the magnetic field is north. µ I 2 × 104 A B = 0 = ( 2.0 × 10−7 Ns 2 /C 2 ) × = 4.0 × 10−3 T. 2π r 1.0 m (a) B =

µ0 I 1.8 × 103 A = ( 2.0 × 10−7 Ns 2 /C 2 ) × = 1.44 × 10−5 T. 2π r 25m

(b) Earth’s field is greater by factor 0.60 × 10−4 = 4.2 times At the ground under the power line. 1.44 × 10−5


CHAPTER

29-13.

1 2 1.6 × 10−19 J mv = 3 × 104 eV = 4.8 × 10−15 J 2 1eV Therefore, K =

v=

29-14. †29-15.

2K = me

2 × ( 4.8 × 10−15 J )

( 9.11 × 10

−31

= 1.03 × 108 m/s

kg )

F = qv × B Since direction is perpendicular, |v × B| = vB Therefore, |F| = qvB = (1.6 × 10–19 C) × (1.03 × 108 m/s) × (5 × 10–5T) = 8.2 × 10–16 N F = qv × B = 2 × (1.6 × 10−19 C ) ( 5.0i − 3.0j ) m/s × ( −4.0i + 2.5 j ) T = 1.6 × 10−19 k N

F = qv × B = ( −1.6 × 10−19 C ) ( −3.0i + 4.0 j − 3.5k ) m/s × ( 2.5i + 3.6 j + 1.5k )

= ( − 2.98 × 10−18 i + 6.8 × 10−19 j + 3.33 × 10−18 k ) N 29-16.

mp g = qvB

⇒ †29-17.

v=

mp g qB

=

(1.67 × 10

(1.6 × 10

−19

− 27

)

kg × 9.8 m/s 2

C ) × ( 4.2 × 10−5 T )

= 2.44 × 10−3 m/s moving East

Let x, y, z axes point East, North, vertically up, respectively. B = 0.17 × 10−4 ˆj − 0.60 10−4 kˆ v = −1.0 × 106 î

Then, F = qv × B ˆ = (−1.6 × 10−19 )(−1.0 × 106 )iˆ × (0.17 × 10−4 ˆj − 0.60 × 10−4 k) ˆ N F = 1.6 × 10−19 (60 ˆj + 17 k)

29-18.

Therefore, the force has a magnitude 9.98 × 10–18 N and points in direction tan–1 (17/60) = 16º above the horizontal in North direction 1 e2 (1.6 × 10−19 ) 2 9 The electric force will be Fe = = = 8.2 × 10−8 N 9.0 × 10 × ( ) 4πε 0 r 2 (0.53 × 10−10 ) 2 The magnetic force will be Fm = qvB = (1.6 × 10–19 C) × (2.2 × 106m/s) × (108 T) = 3.5 × 10−5 N( Fm > 400 Fe )

†29-19.

29-20.

Yes, it will be deformed. dq d dl (a) I = = λl = λ = λv dt dt dt µ I µ λv (b) B = 0 = 0 2π r 2π r 1 λ , so λ = 2πε 0 Er Since E = 2πε 0 r µI µv Therefore, B = 0 = 0 2πε 0 Er = ε 0 µ 0 vE 2π r 2π r s From Gauss’ Law, E =

ε0

Therefore, B = µ 0ε 0 vE = µ 0ε 0 v

s

ε0

= µ 0 vs


29

CHAPTER

†29-21.

B=

29-22.

B=

29 −7 2 2 µ 0 I 4π × (10 Ns /C ) × 1900A = = 0.19 T 2π r 2π × ( 0.2 × 10−2 m )

−7 2 2 µ 0 I 4π × (10 Ns /C ) × 20A = = 4 × 10−3 T −3 2π r 2π × (1.0 × 10 m )

F = qvB = (1.6 × 10−19 C ) × (1.0 × 106 m/s ) × ( 4 × 10−3 T ) = 6.4 × 10−16 N a=

†29-23.

F 6.4 × 10−15 N = = 7.03 × 1014 m/s 2 −31 m 9.1 × 10 kg

F and a are pointing toward the current. −7 2 2 −3 µ I 4π × (10 Ns /C ) × ( 2 × 10 A ) Bedge = 0 = = 4 × 10−8 T −2 2π r 2π × (1 × 10 m )

F = qvB = (1.6 × 10−19 C ) × ( 3 × 108 m/s ) × ( 4 × 10−8 T ) = 1.92 × 10−18 N

29-24. †29-25.

B=

µ0 I 2r

=

4π × (10−7 Ns 2 /C 2 ) × 2.0A 2 × (1.5 × 10 m ) −2

= 8.38 × 10−5 T

The magnetic field produced by one turn is: −7 2 2 µ 0 I 4π × (10 Ns /C ) × 2A B1 = = = 8.38 × 10−5 T 2r 2 × (1.5 × 10−2 m ) The total magnetic field is: B = 60 B1 = 60 × 8.38 × 10−5 T = 5.03 × 10−3 T

29-26. †29-27.

−3 Bmax 2π r ( 9.5 T ) × 2π × (1.5 × 10 m ) µ0 I B= ⇒ I max = = = 7.13 × 104 A 2π r µ0 4π × (10−7 Ns 2 /C2 )

B =

⇒

µ0 I 2π R cos θ 4π r 2 µ0 R 2 I B =

∫ B ds = 29-28.

where r =

z2 + R2 ,

R

cos θ =

2

z + R2

2( z 2 + R 2 )3 / 2

µ0 R 2 I 2

∞

µ0 R 2 I dz z = ∫−∞ ( z 2 + R 2 )3 / 2 2 R2 z2 + R2

∞

= µ0 I / R −∞

The first wire does not produce any magnetic field at its own center. So B1 = 0. The magnetic field produced by the second wire at the center of the first one is: µI B2 = 0 2π d µI µI Similarly, B3 = 0 = 0 2π 2d 4π d µ0 I µ0 I B4 = = 2π 3d 6π d µI µI B5 = 0 = 0 2π 4d 8π d µI µI B6 = 0 = 0 2π 5d 10π d


CHAPTER

29-29.

29-30.

The magnetic fields above are all in the same direction. 6 µI ⎛ 1 1 1 1 ⎞ 0.36 µ 0 I Therefore, Btotal = ∑ Bi = 0 ⎜ 1 + + + + ⎟ = 2π d ⎝ 2 3 4 5⎠ d i =1 µI At the center of the ring, Bz = 0 (see Equation 29.26) 2R −7 2 2 4π × (10 Ns /C ) × 35A Bz = = 7.3 × 10−5 T 2 × 0.3 m F = qv × B Since v and B are perpendicular |F| = qvB. Acceleration a = F/m = qvB/me. Therefore, a = (1.6 × 10–19C) × (1.2 × 106m/s) × (7.3 × 10–5T)/(9.1 ×10–31kg) = 1.5 × 1013 m/s2 µ I ⎛ kˆ kˆ ⎞ µ I ( x − y ) ˆ B (P ) = 0 ⎜ − ⎟ = 0 k xy 2π ⎜⎝ y x ⎟⎠ 2π

µ0 I 20A = ( 2.0 × 10−7 Ns 2 /C 2 ) × = 5.0 × 10−6 T 2π r 0.80m

29-31.

(a) B =

29-32.

⎛ 5.0 × 10−6 T ⎞ (b) Deviation is given by tan −1 ⎜ ⎟ = 16° −4 ⎝ 0.18 × 10 T ⎠ Magnitude of field from each wire is µ0 I µI B= = 0 2π (2d ) 4π d The horizontal components of the field cancel. The vertical components add. The net field at P is ⎛ d /2⎞ 2 B sin θ = 2 B ⎜ ⎟ ⎝ 2d ⎠ 2µ 0 I 1 µI = = 0 downward 4π d 4 8π d

29-33.

B contributed by wires 1, 3 are

B 2d

I (into the page)

I(out of the page)

d

µ0 I each, in direction of 2π d

dotted line. Therefore µ I Net B = 0 2, in the direction shown. Current from 2 2π d contributes π0 1 µ I ⎛ 2⎞ = 0 ⎜⎜ ⎟ also in same direction. Therefore, 2π 2d 2π d ⎝ 2 ⎟⎠ magnetic field is B =

P

B3

I1

P

B2

µ0 I 3 2 2π d 2

B1

The direction has to be at 45° to the two equal components, pointing downward and to the left, which is in the same direction as B2 shown in the figure.


29

CHAPTER

29-34.

B1 = B1 =

29 µ0 I 2r

µ0 I 2r

⇒ B =

†29-35.

29-36.

i=

4π × (10−7 Ns 2 /C 2 ) × 10A

j=

4π × (10−7 Ns 2 /C 2 ) × 10A

2 × 0.2 m 2 × 0.2 m

i = 3.14 × 10−5 Ti

B j = 3.14 × 10−5 Tj

B12 + B22 = 4.44 × 10−5 T

B1

The magnetic field at the center of the circle is produced by the straight current and circular µI µI current. Bstraight = 0 , Bcircle = 0 2π R 2R The direction of Bstraight goes out of the page, the direction of Bcircle goes into the page. µI µI Therefore, |B| = Bcircle – Bstraight = 0 – 0 2 R 2π R The direction of B points into the page. In between the wires, the fields add. µ I µ I 1 ⎞ ⎛ 1 −4 + 0 = ( 2.0 × 10−7 ) × 8.0 ⎜ + B= 0 ⎟ = 6.4 × 10 T. 2π d1 2π d 2 ⎝ 0.005 0.005 ⎠ Outside wires, fields cancel; therefore, µ I µ I 1 ⎞ ⎛ 1 −5 B= 0 + 0 = ( 2.0 × 10−7 ) × 8.0 ⎜ − ⎟ = 4.3 × 10 T. 2π d1 2π d 2 ⎝ 0.015 0.025 ⎠

†29-37.

r < r1: From Ampere’s law, 2π rB = µ 0i

π r 2 Ir 2 = 2 π r12 r1 µi µ Ir 2 µ Ir B= 0 = 0 2 = 0 2 2π r 2π rr1 2π r1

where i = I ⇒

r1< r < r2: i = I

⇒

B=

B2

µ0 I 2π r


CHAPTER

29-38.

⎛ r2 − r2 µ0i µ 0 I r32 − r 2 r2 − r2 ⎞ r2 < r < r3: i = I ⎜ 1 − 2 22 ⎟ = I 32 ⇒ = B= r3 − r2 ⎠ r3 − r22 2π r 2π r r32 − r22 ⎝ r > r3, i = 0 ⇒ B = 0. µ dI µ 0σ dx dB = 0 = , where z is the distance from the plate. 2π r 2π z 2 + x 2 B = ∫ dB =

µ 0σ +∞ dx cos θ ∫ 2 2π −∞ z + x 2

where cos θ =

z 2

z + x2

. θ

Therefore, µ σ z +∞ dx µ σ z 1 −1 x x =+∞ µ 0σπ µ 0σ tan |x =−∞ = = 0 = B= 0 2 2 ∫ 2π −∞ z + x 2π z 2π 2 z

z

r I

B is in the i direction.

†29-39.

29

x

In the quadrant where y > 0, z > 0, – ∞ < x < + ∞: from the currents in the x-y plane, B xy = ⇒

µ 0σ 2

i , from the currents in the x-z plane, B xz = −

µ 0σ 2

i

B = B xy + B yz = 0

In the quadrant where y < 0, z > 0, – ∞ < x < + ∞: from the currents in the x-y plane, B xy = ⇒

µ 0σ

B = µ 0σ i

2

i , from the currents in the y-z plane, B yz =

µ 0σ 2

i

In the quadrant where y < 0, z < 0, – ∞ < x < + ∞: from the currents in the x-y plane, B xy = − B yz =

µ 0σ

µ 0σ 2

i , from the currents in the y-z plane,

i 2 ⇒ B=0 In the quadrant where y > 0, z < 0, − ∞ < x < + ∞ :

from the currents in the x-y plane, B xy = −

B yz = − ⇒

29-40.

µ 0σ

µ 0σ 2

i , from the currents in the y-z plane,

i 2 B = − µ 0σ i

Assume the current is uniformly distributed in the cross section in the conductor. Treat the I conductor as a solid cylinder with radius R1, current density σ 1 = , and a solid cylinder with π R12 I radius R2, current density σ 2 = − . π R12


CHAPTER

29

The magnetic field at any point equals the superposition of the two magnetic fields produced by these two currents. (a) Below the center, r ≤ R1 , see Figure a.

B1 =

µ 0i1 µ Iπ r 2 µ Ir = 0 = 0 2 2 2π r 2π r π R1 2π R1

B2 = − ⇒

µ 0 i2 2π ( R1 − 2 R2 + R2 + r )

=−

Fig. a µ0 I

2

R2 2π ( R1 − R2 + r ) R12

R1-2R2

2 ⎞ µ I⎛ r R2 B= 0 ⎜ 2 − 2 ⎟ 2π ⎝ R1 R1 ( R1 − R2 + r ) ⎠

r

(b) Below the conductor, r > R1 , see Figure b. µI B1 = 0 2π r

B2 = − ⇒

µ 0 i2 2π (r + R1 − 2 R2 + R2 )

B=

=−

µ0 I

Fig. b 2

R2 2π (r + R1 − R2 ) R12

R1-2R2

2 µ0 I ⎛ 1 R2 ⎞ 1 ⎜ − ⎟ 2π ⎝ r (r + R1 − R2 ) R12 ⎠

r

(c) Above the center, for r ≤ R1 – 2R2, see Figure c. µ Ir Ir 2 2π rB1 = µ 0i1 = µ 0 2 ⇒ B1 = 0 2 , R1 2π R1

B2 = − ⇒

µ 0i2 2π ( R1 − 2 R2 + R2 − r )

B=

=−

µ0 I

R22 2π ( R1 − R2 − r ) R12

µ0 I ⎛ r R22 ⎞ 1 − ⎜ ⎟ 2π ⎝ R12 R1 − R2 − r R12 ⎠


Fig. c r

R1-2R2

CHAPTER (d) Above the center, R1 ≥ r ≥ R1 – 2R2,, see Figure d. µ Ir B1 = 0 2 2π R1 µ 0 i2 B2 = − 2π ( R1 − 2 R2 + R2 − r )

=−

29

R1-2R2+R2-r =R1-R2-r

Fig. d

µ0 I

( R1 − R2 − r ) 2 2 2π ( R1 − R2 − r ) R1

r

R1-2R2

µ 0 I R1 − R2 − r 2 2π R1 µI r µI R − R2 − r ⇒ B= 0 ( 2 − 1 ) = 0 2 ( R1 − R2 + 2r ) 2 2π R1 2π R1 R1

= =−

(e) Above the conductor, r ≥ R1, see Figure e. µI B1 = 0 2π r 2 µ0 I ⎛ 1 R2 ⎞ 1 B ⇒ = ⎜ − ⎟ 2π ⎝ r (r − R1 + R2 ) R12 ⎠

29-41.

Fig. e

r

R1-2R2

By symmetry the B-field must point in direction shown (since any component down or up from one segment of current on one side will be canceled by current on other side). Break current into little strips, dx. I Current through strip is dx. Horizontal component of B due to this strip is: b µ0 µ0 µ dI dI Z z I cos θ = = 0 2 dx 2 2π x 2 + z 2 2π x 2 + z 2 x 2 + z 2 2π x + z b Therefore, µ I x =b / 2 z B= 0 ∫ dx 2 x =− b / 2 2π b x + z2 µ Iz x =b / 2 2 dx = 0 2 ∫ 2π b 0 x + z2 µ I 1 ⎛ x ⎞ b/2 = 0 2 tan −1 ⎜ ⎟]0 2π b z ⎝z⎠ µ I = 0 tan −1 ( b / 2 z ) π b


CHAPTER

29-42.

29

Let Bup be the magnetic field from the upper wire, Blow be the magnetic field from the lower wire. Bup =

For z > 2R,

µ0 I =− 2π rdown 2π ( z + R) µI µ0 I z 1 1 B1 = − 0 ( )=− + 2π z − R z + R π (z 2 − R2 )

Blow = ⇒

µ0 I µ0 I =− 2π rup 2π ( z − R) µ0 I

For R < z < 2R, Bup = − Blow = − ⇒

2π ( z + R)

µ0 I ⎛ z − R 1 ⎞ ⎜ 2 + ⎟ z + R⎠ 2π ⎝ R

For 0 < z < R, Bup =

⇒

µ0 I

π ( R − z )2 µ0 I ( R − z ) = 2π ( R − z ) π R 2 2π R 2

µ0 I 2π ( z + R)

B3 =

µ0 I ⎛ R − z µ0 I 1 ⎞ z2 − = − ⎜ ⎟ z + R⎠ 2π ⎝ R 2 2π R 2 ( z + R )

For − R < z < 0 ,

Bup =

µ0 I 2π ( R + | z |)

=

µ0 I 2π ( R − z )

(since z < 0)

µ I ( R − | z |) µ I (R + z) π ( R − | z |) 2 =− 0 =− 0 2 2 2π ( R − | z |) 2π R 2π R 2 πR µI µ0 I R+z z2 1 ) B4 = 0 ( − = 2π R − z 2π R 2 ( R − z ) R2

Blow = −

=>

µ I ( z − R) π ( z − R)2 =− 0 2 2π ( z − R) π R 2π R 2

µ0 I

B2 = −

Blow = −

µ0 I

µ0 I

For −2 R < z < − R ,

Bup =

µ0 I 2π ( R + | z |)

=

µ0 I 2π ( R − z )


CHAPTER

29

µ0 I

µ I (| z | − R) µ 0 I ( z + R) π (| z | − R) 2 =− 0 = 2 2π (| z | − R) 2π R 2 2π R 2 πR µI z+R 1 + ⇒ B5 = 0 ( ) R2 2π R − z µ0 I µ0 I For z < −2 R , Bup = = 2π ( R + | z |) 2π ( R − z ) µ0 I µ0 I Blow = − = 2π (| z | − R) 2π ( z + R ) µI 1 1 ⇒ B6 = 0 ( + ) 2π R − z R + z Blow =

†29-43.

Bmax happens at z = 2R or z = – 2R. 2µ0 I Bmax = 3π R Let Bup be the magnetic field from the upper wire, Blow be the magnetic field from the lower wire. Assume the current in the upper wire is in the negative x-direction, the current in the lower wire is in the positive x-direction. µI µ0 I Bup = 0 = For z > 2R, 2π rup 2π ( z − R )

µ0 I µ0 I =− 2π rdown 2π ( z + R ) µI µ0 I R 1 1 − )= B1 = 0 ( π (z 2 − R2 ) 2π z − R z + R

Blow = −

⇒

For R < z < 2R, Bup = Blow = −

⇒

π ( z − R)2 µ0 I ( z − R) = 2π ( z − R) π R 2 2π R 2

µ0 I 2π ( z + R)

B2 =

µ0 I ⎛ z − R 1 ⎞ ⎜ 2 − ⎟ 2π ⎝ R z + R⎠

For 0 < z < R, Bup = − Blow = −

µ0 I

µ0 I

µ I (R − z) π ( R − z )2 =− 0 2 2π ( R − z ) π R 2π R 2

µ0 I 2π ( z + R)

µ0 I ⎛ R − z 1 ⎞ ⎜ 2 + ⎟ 2π ⎝ R z + R⎠ µ0 I µ0 I For − R < z < 0 , Bup = − =− (since z < 0) 2π ( R + | z |) 2π ( R − z ) µ0 I µ I (R + z) π ( R − | z |) 2 − µ 0 I ( R − | z |) Blow = − = =− 0 2 2 2π ( R − | z |) 2π R 2π R 2 πR µI 1 R+z − ⇒ B4 = 0 ( − ) R−z R2 2π ⇒

B3 = −


CHAPTER

29

For −2 R < z < − R ,

Bup = −

µ0 I

=−

2π ( R + | z |)

µ0 I

µ0 I 2π ( R − z )

µ I (| z | − R ) µ 0 I ( z + R ) π (| z | − R ) = 0 = 2 2π ( R − | z |) 2π R 2 2π R 2 πR µI z+R 1 − ⇒ B5 = 0 ( ) R2 2π R − z µ0 I µ0 I For z < −2 R , Bup = − =− 2π ( R + | z |) 2π ( R − z ) µ0 I − µ0 I Blow = = 2π (| z | − R ) 2π ( z + R) µI µ0 I R 1 1 ⇒ B6 = − 0 ( + )= π ( z2 − R2 ) 2π R − z R + z Blow =

2

The max. B happens at z = 0. |Bmax | = |B3 (z = 0)| = 29-44.

µ0 I πR

The disc can be thought of as a collection of rings each of radius r with current dI = Qrωdr/πR2. The field due to each of these rings is given by µ µ Qω r dB = 0 dI = 0 dr (see equation 29.19) 2r 2r π R 2

Rotational axis

Disk

†29-45. 29-46.

Therefore, the field in the center of the disk will be the sum of these contributions: µ Qω R µ Qω B = ∫ dB = 0 2 ∫ dr = 0 = r 0 2π R 2π R B 5 × 10−2 T B = µ 0 nI ⇒ I = = = 26.5 A µ 0 n ( 4π × 10−7 Ns 2 /C2 ) × (15 × 102 turns/m ) n=

260 turns = 1.3 × 104 turns/m 2 × 10−2 m

B = µ 0 nI = ( 4π × 10−7 Ns 2 /C2 ) × (1.3 × 104 turns/m ) × ( 8.0A ) = 0.131 T

29-47.

n=

1turn = 5 turns/m 20 × 10−2 m

B = µ 0 nI = ( 4π × 10−7 Ns 2 /C2 ) × ( 5 turns/m ) × ( 2 × 103 A ) = 1.26 × 10−2 T

29-48.

The magnetic field of the first solenoid has a magnitude B = µ 0 nI . The magnitude of the magnetic field of the second solenoid is twice as big. The magnetic fields are at right angles to each other, and the magnitude of their vector sum is B=

( B1 )

2

+ ( B2 ) = 2

( µ 0 nI )

2

+ ( 2µ 0 nI ) = µ 0 nI 5 2

This magnetic field makes an angle of tan–1(1/2) = 27° with the axis of the solenoid with the stronger current. The field lines are as shown by the figure.


CHAPTER

B2

29

B

B1

†29-49.

µ 0 IN (4π × 10−7 T i m/A)(125 × 103 A)(80 turns) = = 3.34 T 2π rmin 2π (0.60 m) µ IN (4π × 10−7 T i m/A)(125 × 103 A)(80 turns) = 0 = = 1.43T 2π rmax 2π (1.4 m)

Bmax = Bmin

rmin r rmax

29-50.

B solenoid = µ 0 nI , r ≤ R , the direction pointing along the axis. µ I′ B wire = 0 , the direction pointing perpendicular to the wire, along the concentric circles around 2π r the wire. The fields will always be perpendicular to each other and the resultant field’s magnitude inside the solenoid will be given by 2

B net =

µ I′ 2 ( µ 0 nI ) + ⎛⎜ 0 ⎞⎟ , the field lines will form something like an upward spiral. ⎝ 2π r ⎠ Bsolenoid Bwire

Bsolenoid

⇒ Bwire

†29-51.

Btotal

For r < r1, construct an Amperian loop that encloses all the coils. The total current enclosed by the loop is I enclosed = nn′(r2 − r1 ) IL. As with the solenoid with a single layer of turns, the magnetic field outside is very small, so integrating B around the loop gives BL, and we get B = µ 0 nn′(r2 − r1 ) I , r < r1 For r1 < r < r2, the Amperian loop is within the coils and the current enclosed is I enclosed = nn′(r2 − r ) IL , giving B = µ 0 nn′(r2 − r ) I , r1 < r < r2


CHAPTER 29-52.

29

The spinning rod produces a current that looks like the current through the coils in Problem 51. Construct an Amperian loop as shown in the diagram. The rotating charge within the cylindrical shell of inner radius r and outer radius R rotating with angular speed ω constitutes a current I enclosed = ωρπ ( R 2 − r 2 ) L.

R

ω

Then integrating B around the loop gives

r

L

BL = µ 0ωρπ ( R 2 − r 2 ) L B = µ 0ωρπ ( R 2 − r 2 ) †29-53.

29-54.

The magnetic field from each of the four segments has the same magnitude. Each field has a magnitude of I µ0 I (cos 45° − cos135°) Bsegment = ⎛ L⎞ 4π ⎜ ⎟ ⎝2⎠ The fields from all the segments point in the same direction, so 45º µ0 I 4µ0 I Btotal = 4 Bsegment = (cos 45° − cos135°) = ⎛ L⎞ 2π L π⎜ ⎟ ⎝2⎠ µ0 I L 2 (2 cos θ1 ) , where cos θ1 = = Btop = Bbottom = . Thus 2 2 L ⎛ ⎞ 5 + L ( L / 2) 4π ⎜ ⎟ ⎝2⎠ 2µ0 I Btop = Bbottom = . π 5L µI ( L / 2) 1 = Bleft = Bright = 0 (2 cos θ 2 ) , where cos θ 2 = . Thus 2 2 4π L 5 L + ( L / 2)

Bleft = Bright =

135º

µ0 I . All the fields are in the same direction, so the magnitude of the total field 2 5π L

is B = Btotal = 2 Btop + 2 Bleft = †29-55.

µ0 I 5µ 0 I (4 + 1) = πL 5π L

The straight parts do not produce any magnetic field at the point, because r is parallel to the current and from Biot Savart’s Law, (29.23), sin θ = sin 0 = 0. Thus using (29.27), µ I ∆θ µ 0 Iπ µ 0 I (4π × 10−7 T i m/A)(5A) = = = = 7.9 × 10−5 T , directed into B = Bsemi-circle = 0 4π r 4π r 4r 4(0.02 m) the paper.


CHAPTER 29-56.

B = B1 − B2 =

µ0 I ⎛ 1

1 ⎞ ⎜ − ⎟ into the paper. 4 ⎝ R1 R2 ⎠

†29-57.

From the top straight wire, (semi-infinitely long,) B1 =

µ0 I , 4π R

From the vertical straight wire, (also semi-infinitely long,) B2 =

µ0 I , 4π R

From the quarter-circle, use (29.27), noticing that ∆θ = π / 2 , B3 = =

29-58.

µ 0 I ∆θ µ 0 I . = 4π R 8R

B1, B2, and B3 are in the same direction, therefore, µI µI B = B1 + B2 + B3 = 0 + 0 , pointing into the page. 2π R 8 R µi For a current in a circle with radius r, B = 0 . The electron motion constitutes a current 2r ev . Angular momentum is L = mvr = 1.05 × 10−34 J i s , so i= 2π r 1.05 × 10−34 J ⋅ s L = = 2.18 × 106 m/s. Thus v= mr (9.1 × 10−31 kg)(5.29 × 10−11 m) i=

(1.6 × 10−19 C)(2.18 × 106 m/s) = 1.05 × 10−3 A , and 2π (5.29 × 10−11 m)

B=

µ0i 2r

=

(4π × 10−7 T i m/A)(1.05 × 10−3 A) = 12 T 2(5.29 × 10−11 m)


29

CHAPTER †29-59.

29

Use equation (29.28). From the right segment, (see Fig. 1 on the right). θ1 = 45°, θ 2 = 135°, r = L / 2 µ0 I ⇒ Bright = (cos 45° − cos135°) 4π ( L / 2) µI = 0.225 0 . The direction is into the paper. L

Fig. 1

From the left segment, (see Fig. 2.) 3L / 2 θ1 = tan −1 ( ) = tan −1 (3) = 71.6°, L/2 θ 2 = 180° − 71.6° = 108.4°, r = 3L / 2 µ0 I ⇒ Bleft = (cos 71.6° − cos108.4°) 4π (3L / 2) µI = 3.35 × 10−2 0 . The direction is out of the paper. L

Fig. 2

From the top segment, (see Fig. 3). 1 L/2 θ1 = tan −1 ( ) = tan −1 ( ) = 18.4°, 3L / 2 3 θ 2 = 45°, r = L / 2 µ0 I µI (cos18.4° − cos 45°) = 3.85 × 10−2 0 . ⇒ Btop = 4π ( L / 2) L

Fig. 3

The direction is out of the paper. µI Bbottom = Btop = 3.85 × 10−2 0 . The direction is out of the L paper. Therefore, the magnitude of the magnetic field is: |B| = |Bleft + Btop + Bbottom – Bright| µI µI µI µI = | 3.35 × 10−2 0 + 3.85 × 10−2 0 + 3.85 × 10−2 0 − 0.225 0 | L L L L µ0 I µ0 I = | −0.11 | = 0.11 L L The sum of the four fields is negative. That means the total B is in the direction of Bright, i.e., the direction of B is into the paper


CHAPTER 29-60.

Pick a small segment from the loop. At point P, it produces magnetic field dB: µ I ds sin θ dB = 0 ( Biot-Savart Law.) 4π r2 In this case, the angle between r and ds (the current) is 90o, so sinθ = 1. Consider the whole loop. The only non-zero component of the magnetic field is Bz. µ I ds cos φ R dBz = 0 where cos φ = sin φ ' = , and r 2 = R 2 + z 2 2 2 r 4π R + z2 Notice that both φ and r are constants, we have: µ I cos φ Bz = ∫ dBz = 0 2π R 4π r 2 µI µ0 I R 1 R2 2 R π = 0 = 4π R 2 + z 2 R 2 + z 2 2 ( R 2 + z 2 )3 / 2

†29-61.

φ

φ’

r R

(a) The point P is at z = R/2 since the separation between the coils is equal to their radius. The top and bottom coils produce magnetic fields at P that have the same magnitude and direction. Thus the magnitude of the total field at P is µI µ 0 IR 2 µ 0 IR 2 8µ 0 I R2 = = ⇒ BP = BP = 2 0 3 / 2 3 / 2 2 2 3/ 2 2 ( z + R ) z = R / 2 ⎡ ( R/2) 2 + R 2 ⎤ 125R ⎡⎣ 5R 2 /4 ⎤⎦ ⎣ ⎦ (b) At any point z, the fields produced by the coils are µI µ0 I R2 R2 , = Bbottom = 0 B top 2 ( z 2 + R 2 )3 / 2 2 ⎡ ( R − z)2 + R 2 ⎤3 / 2 ⎣ ⎦ Their first derivatives are 3µ I dBbottom zR 2 , =− 0 2 ( z 2 + R 2 )5 / 2 dz

At z = R/2,

dBtop dz

=−

dBtop dz

dBtop dBbottom , so dz dz

=

3µ 0 I (R − z)R2 2 ⎡ ( R − z )2 + R 2 ⎤ 5 / 2 ⎣ ⎦ +

z=R / 2

dBbottom dz

=0 z=R / 2

The second derivatives are ⎤ d 2 Bbottom 3µ 0 IR 2 d 3µ 0 IR 2 ⎡ z 1 5 z = − = − − ⎢ 2 2 2 5/2 2 2 5/ 2 2 2 7/2 ⎥ dz 2 dz ( z + R ) 2 ⎣ (z + R ) 2 (z + R ) ⎦ 2 3µ 0 IR ( z 2 + R2 − 5z 2 ) =− 2( z 2 + R 2 )7 / 2 d 2 Bbottom 3µ IR 2 ( R 2 − 4 z 2 ) =− 0 2 2 dz 2( z + R 2 )7 / 2 Similarly, ⇒

d 2 Btop dz 2

3µ 0 IR 2 ⎡⎣ R 2 − 4( R − z ) 2 ⎤⎦ = 7/2 2 ⎡⎣ ( R − z ) 2 + R 2 ⎤⎦

Both of these derivatives cancel each other at z = R/2.


29

CHAPTER

29-62.

29

Use the result of 29.60, dB = ⇒

dB =

⇒

B=

R 2 NI / L dz 2 ( z 2 + R 2 )3 / 2

z0 − L

†29-63.

µ 0 R 2 NI 2L

R2 NI dI , where dI = dz 2 ( z 2 + R 2 )3 / 2 L

µ0

∫

z0

=

µ0

µ0

µ 0 R 2 NI R 2 NI / L dz = 2 ( z 2 + R 2 )3 / 2 2L

z0 − L

∫

z0

1 dz ( z 2 + R 2 )3 / 2

⎫⎪ µ 0 NI ⎧⎪ z0 z0 − L z z = z0 − L = − | ⎨ ⎬ z = z0 R 2 ( z 2 + R 2 )1 / 2 2L ⎪ z 2 + R2 ( z0 − L) 2 + R 2 ⎭⎪ ⎩ 0

The magnetic field produced by the current is: µ I 4π × 10−7 Ns 2 /C2 × 30A = 6 × 10−5 T B= 0 = 2π r 2π × 0.1m

| F | = | qv × B | = (1.6 × 10−19 C ) × ( 5 × 106 m/s ) × ( 6 × 10−5 T ) = 4.8 × 10−17 N

29-64.

a=

F 4.8 × 10−17 N = = 2.87 × 1010 m/s 2 m 1.67 × 10− 27 kg

B=

−7 2 2 µ 0 I ( 4π × 10 Ns /C ) × (15A ) = = 6 × 10−5 T 2π r 2π × 0.05 m

⇒ †29-65.

F = evB = (1.6 × 10−19 C ) × ( 3 × 108 m/s ) × ( 6 × 10−5 T ) =2.88 × 10−15 N , parallel to the

current. The magnetic field from the small circle is: 4π × 10−7 Ns 2 /C2 ) × ( 4A ) ( µ0 I = = 4.19 × 10−5 T Bsmall = 2rsmall 2 × ( 6 × 10−2 m ) Bbig =

µ0 I 2rbig

( 4π × 10 =

−7

Ns 2 /C 2 ) × ( 4A )

2 × ( 8 × 10−2 m )

= 3.14 × 10−5 T

If two currents flow in the same direction, they produce magnetic fields in the same direction. ⇒ B = 4.19 × 10−5 T + 3.14 × 10−5 T = 7.33 × 10−5 T If two currents flow in the opposite direction, they produce magnetic fields in opposite directions. ⇒ B = 4.19 × 10−5 T − 3.14 × 10−5 T = 1.05 × 10−5 T 29-66.

B= V =

µ0 N i µ0 N V = 2π r 2π r R

−4 B 2π rR ( 8 × 10 Τ ) × 2π × ( 0.2 m ) × ( 20 Ω ) = µ0 N ( 4π × 10−7 Ns2 /C2 ) × (150 turns )

= 106.7 V


CHAPTER †29-67.

The magnetic field at the center of the circle is the superposition of the field from the circle and the field from the straight wire. µI µI B circle = 0 into the paper, Bstraight = 0 to the right. 2R 2π R 2

2

µI µI 1 ⎛µ I⎞ ⎛ µ I ⎞ | B | = ⎜ 0 ⎟ + ⎜ 0 ⎟ = 0 1 + 2 = 1.05 0 π R 2R ⎝ 2 R ⎠ ⎝ 2π R ⎠ The angle between B and the straight wire is: 1 ⎛ µ I µ I⎞ θ = tan −1 ⎜ 0 / 0 ⎟ = tan −1 = 17.7° π π 2 R 2 R ⎝ ⎠

29-68.

I = 2π fq = 2π × (15 / s ) × ( 2.0 × 10−6 C ) = 1.88 × 10−4 A B=

†29-69.

µ0 I

=

( 4π × 10

−7

Ns 2 /C2 ) × (1.88 × 10−4 A )

2r 2 × 0.08 m r < R1 , B = µ 0 nI + µ 0 nI = 2µ 0 nI

= 1.48 × 10−9 T

R1 < r < R2 , because it is outside the small solenoid, so the B from the small one is zero. Therefore, B = µ 0 nI r > R2 , it is outside both solenoids, therefore, B = 0. 29-70.

Inside the solenoid, | B total | =

B0

µ 0 2 n 2 I 2 + B02

The angle between Btotal and the axis of the solenoid is: | B0 | B θ = tan −1 = tan −1 0 µ 0 nI | Bsolenoid |

Bintotal

µ0nI

Outside the solenoid, B total = B 0 29-71.

Make an Amperian loop as shown by the figure. (Top view.) The current enclosed in the loop is: i = rnI From Ampere’s law, 2lB = µ 0i = µ 0 lnI

29-72.

29

⇒

B=

µ 0 nI

2 This magnetic field is different from the expression for the magnetic field of a solenoid, because if you imagine the current form a solenoid, it is not a thin cylinder. Instead, it is a flat array of wires, there is a field on both sides of the Amperian loop. The direction of B is mostly parallel to the plane of the wires, along the Amperian loop in the figure shown, and perpendicular to the current. The only exception is along the edges. For R < r < 2R, Iπ (r 2 − R 2 ) I (r 2 − R 2 ) 2π rB = µ 0i , where i = = 2 3R 2 π ( 2R ) − R2

(

⇒

B=

)

µ 0 I (r − R ) 6π R 2 r 2

2


CHAPTER

⇒

29

at r =

µ I [(9 R 2 / 4) − R 2 ] 5µ 0 I 3R = , B= 0 2 6π R 2 (3R / 2) 36π R

For r > 2R, i = I, 2π rB = µ 0i

⇒ 29-73.

29-74.

29-75.

⇒

B=

at r = 3R, B =

µ0 I 2π r

µ0 I 6π R

The ends of the wire contribute nothing since de × rˆ = 0. Contribution of the top segment: µ I µ I 2 L = 0 Btop = 0 (into paper) 2 2 1/ 2 4π L ( L + L / 4) 4π L 5 Contributions of the side segments II, III (also into paper) given by: µ 2I µ 2I 2 µ I 4 BII = 0 = 0 = BIII (cos α ) = 0 4π L 4π L 5 4π L 5 Since they all point the same way, they add. B at point P is µ I⎛ 2 µI 5 4 4 ⎞ µ 0 I 10 B = Btop + BII + BIII = 0 ⎜ + + = = 0 ⎟ 4π L ⎝ 5 2π L 5 5 ⎠ 4π L 5 µI From the half loop standing up, B1 = 0 i B2 4R µI From the half loop lying flat, B 2 = 0 j 4R µ0 I 2µ 0 I ⇒ B= (i + j) ⇒ | B | = 4R 4R

B

B1

Contributions all add in same direction (here out of paper). The magnetic field due to the 135º arc is ⎛ 3π ⎞ ⎟ µ I ⎛ 3⎞ µ I⎜ BII = 0 ⎜ 4 ⎟ = 0 ⎜ ⎟ 2 R ⎜ 2π ⎟ 2 R ⎝ 8 ⎠ ⎜ ⎟ ⎝ ⎠ µI Contributions from I, III are the same, and are given by: BI = BIII = 0 4π R Therefore, the total B-field is µ I ⎛ 3 ⎞ 2µ I µ I ⎛ 1 3⎞ B = BI + BII + BIII = 0 ⎜ ⎟ + 0 = 0 ⎜ + ⎟ 2 R ⎝ 8 ⎠ 4π R 2R ⎝ π 8 ⎠


CHAPTER 30

CHARGES AND CURRENTS IN MAGNETIC FIELDS


Equation 30.6 gives the radius of the circular orbit as a function of the particle momentum, its charge, and the strength of the magnetic field. Since the particle is a proton, its charge is 1.60 × 10−19 C. p r= qB

p = qBr = (1.60 × 10−19 C ) ( 20 T )( 3.5 m ) = 1.1 × 10−17 kg i m/s 30-2.

The kinetic energy is related to the momentum of the particle as 1 p2 K = mv 2 = 2 2m p = 2mK

Applying Equation 30.6, the radius of the circular orbit is

30-3.

r=

p = qB

r=

p qB

2mK = qB

⎛ 1.60 × 10−19 J ⎞ 2 (1.67 × 10−27 kg )(1.0 × 107 eV ) ⎜ ⎟ 1 eV ⎝ ⎠ = 9100 m −19 −5 (1.60 × 10 C )( 5.0 × 10 T )

p = qBr = (1.60 × 10−19 C )( 3.3 × 10−5 T )( 6.5 × 106 m ) = 3.4 × 10−17 kg i m/s 30-4.

r=

p 1 × 10−16 kg i m/s = = 6 × 1010 m −19 −8 qB (1.60 × 10 C )(1 × 10 T )

30-5.

r=

p qB

B=

p 5.3 × 10−16 kg i m/s = = 3.3 T qr (1.60 × 10−19 C )(1.0 × 103 m )

30-6.

30-7.

qB 2π m q 2π f 3.51 × 1010 rad/s = = = 1.76 × 1011 C/kg (it's an electron) m B ( 0.200 T ) f =

r=

p qB

p = qBr = (1.60 × 10−19 C ) ( 2.0 T )(1.2 m ) = 3.8 × 10−19 kg i m/s


CHAPTER

30-8.

r=

30 mv qB

⎛ 0.332 m ⎞ C ) ( 0.050 T ) ⎜ ⎟ qBr 2 ⎝ ⎠ = 3.3 × 10−26 kg = m= 4.0 × 104 m/s v The electrons are following circular orbits with a frequency called the cyclotron frequency, which is given by Equation 30.7. qB f = 2π m 9 −31 2π fm 2π (1 × 10 Hz )( 9.11 × 10 kg ) = = 0.036 T B= 1.60 × 10−19 C q

(1.60 × 10

†30-9.

30-10.

−19

For the particle on the left, the radius is approximately 1.0 cm, so the actual radius is 0.1 m. The momentum is pleft = qBr = (1.60 × 10−19 C ) (1.0 T )( 0.1 m ) = 1.6 × 10−20 kg i m/s For the particle on the right, the radius is approximately 0.5 cm, so the actual radius is 0.05 m; and the momentum is pright = qBr = (1.60 × 10−19 C ) (1.0 T )( 0.05 m ) = 8.0 × 10−21 kg i m/s

†30-11.

The track on the right is the antielectron because it has a positive charge and the magnetic field is directed into the paper. The force causes it to spiral inward in a counterclockwise orbit. The track on the left is that for the electron. Since the particles are the nucleii of the two carbon isotopes, each nucleus contains the same number of protons. Therefore, the charge on each particle is the same, +6e. The difference in the two types of nucleii is in their masses. However, the problem specifies that the momentum, p = mv, of the two particles is identical. The radii of the circular orbits are p p and r12 = 12 r14 = 14 qB qB Taking the ratio of these two radii gives p12 r12 p qB = = 12 = 1 p r14 p14 14 qB

30-12.

(a) p = qBr = (1.60 × 10−19 C ) (1.3 T )(128 m ) = 2.7 × 10−17 kg i m/s (b) f =

†30-13.

(1.60 × 10−19 C ) (1.3 T ) = 2.0 × 107 Hz qB = 2π m 2π (1.67 × 10−27 kg )

The radius of the circular orbit of the electron of kinetic energy K is given by p 2mK r= = qB qB

The magnetic field of a solenoid is directed parallel to its axis. The magnetic field strength inside the solenoid is B = µ 0 nI


CHAPTER

30

The radius of the circular orbit of the electron is to be less than 0.10 m, so the minimum current in the solenoid is found from the radius equation, 2mK 2mK r= = qB q µ 0 nI

30-14.

30-15.

⎛ 1.60 × 10−19 J ⎞ 2 ( 9.11 × 10−31 kg )( 3.0 × 104 eV ) ⎜ ⎟ 1 eV 2mK ⎝ ⎠ I = = = 0.39 A i T m turns 100 cm q µ 0 nr ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ −19 −7 (1.60 × 10 C ) ⎜⎝ 4π × 10 A ⎟⎠ ⎜⎝ 120 cm ⎟⎠ ⎜⎝ 1 m ⎟⎠ ( 0.10 m ) The electric force on the electron is Fe = qE. Since the electric field is directed downward, the electric force on the electron is directed upward. To cancel the effect of this field, the magnetic force on the electron must be directed downward. The magnetic force on the electron is given by Fm = qv × B. Since B is directed due north, if the velocity of the electron is directed due east, then the magnetic force would be directed downward. From the magnitudes of the forces, the magnitude of the velocity can be determined. qE = qvB E 100 V/m = 2 × 106 m/s v= = B 5.0 × 10−5 T (a) A magnetic force acts on the electron F = qv × B that is directed perpendicular to both the velocity and the magnetic field. This force causes the electron to follow a circular path that spirals in the direction of the magnetic field. 1.60 × 10−19 C )( 5.0 × 10−4 T ) ( qB (b) f = = = 1.4 × 107 Hz 2π m 2π ( 9.11 × 10−31 kg ) 1 1 = = 7.2 × 10−8 s f 1.4 × 107 Hz (c) The distance traveled in the direction of the magnetic field during this period is x = v//T = ( 4.0 × 106 m/s )( 7.2 × 10−8 s ) = 0.29 m T =

30-16.

The electric force on the electron is Fe = qE. The electric field is directed due north, the electric force on the electron is directed due north. To cancel the effect of this field, the magnetic force on the electron must be directed due south. The magnetic force on the electron is given by Fm = qv × B. Since B is directed vertically upward, if the velocity of the electron must be directed due east, then the magnetic force would be directed upward for a positively-charged particle. From the magnitudes of the forces, the magnitude of the velocity can be determined. qE = qvB E B The velocity selector allows particles with a particular speed to pass undeflected through the crossed electric and magnetic fields. That particular speed is found by equating the magnitudes of the electric and magnetic forces on the charged particles. qE = qvB v=

†30-17.

B=

E v


CHAPTER

30

The selected kinetic energy of the alpha particles is given as K = (1/2)mv2 = 2.0 × 103 eV. This corresponds to a speed of 2K v= m Therefore, E E 1.0 × 106 V/m B= = = = 3.2 T v 2K 2 ( 2.0 × 103 eV )(1.60 × 10−19 J/eV ) m 4.0 u (1.6605 × 10−27 kg/u )

30-18.

(a) l = 2r, where p mv = r= ′ qB qB′ ⎛l⎞ qB′ ⎜ ⎟ ⎝ 2 ⎠ = qB′Bl E 2E B (b) Since m is directly proportional to l, the ratio of the masses is munknown lunknown = m16.0 l16.0

qB′r = m= v

munknown = m16.0 †30-19.

lunknown ⎛ 32.86 cm ⎞ = (16.0 u ) ⎜ ⎟ = 18.0 u l16.0 ⎝ 29.20 cm ⎠

(a) The radius of the circular orbit is given by p mv r= = qB qB Since the magnitude of the charge of the proton is the same as that for the electron and they are in the same magnetic field, the momentum of both particles must be the same if they are on the same circle. Therefore, the magnitude of the velocity of the proton can be determined. me ve = mp vp −31 6 me ve ( 9.11 × 10 kg )( 3. × 10 m/s ) vp = = = 1.6 × 103 m/s −27 mp 1.67 × 10 kg

(b) The radius of the circle is p mv r= = qB qB The electron travels a distance se = rθe = vet while the proton travels a distance sp = rθp = vpt during the same elapsed time t. The relationship between the two angles is θe = 2π − θp.


CHAPTER t=

se ve

rθ p = vp t = vp

θp =

vp ve

θe =

se rθ = vp e ve ve

vp ve

( 2π − θ ) p

⎛ 1.6 × 103 2π ⎜ 6 ve ⎝ 3.0 × 10 θp = = vp ⎛ 1.6 × 103 +1 1+ ⎜ 6 ve ⎝ 3.0 × 10 F = IL × B 2π

30-20.

vp

m/s ⎞ ⎟ m/s ⎠ m/s ⎞ ⎟ m/s ⎠

= 3.3 × 10−3 rad

B

F = ILB sin α = ( 6.0 A )( 0.10 m )( 0.40 T ) sin 30° = 0.12 N

†30-21.

30

The direction of the force F is out of the plane of the paper.

I

The magnetic force of one wire on a segment of the other wire that is parallel to it is given by F = IL × B F = ILB sin α = ILB In this case, α = 90°, since the wires are parallel to each other. The magnetic field produced by a wire carrying a current I is at a distance r,where the segment of wire is in this case, is given by µI B= 0 2π r Therefore, the magnetic force on the 0.010 m segment of wire due to the other wire is Tim ⎞ 2 ⎛ 4π × 10−7 20 A ) ( 0.010 m ) ( ⎜ ⎟ 2 µI µI L ⎝ A ⎠ F = I1 LB = I1 L 0 2 = 0 = = 6.7 × 10−5 N 2π r 2π r 2π ( 0.012 m ) Tim ⎞ 2 ⎛ 4π × 10−7 1800 A ) ( 0.50 m ) ( ⎜ ⎟ µI µI L ⎝ A ⎠ F = I1 LB = I1 L 0 2 = 0 = = 8.1 × 10−2 N 2π r 2π r 2π ( 4.0 m ) 2

30-22. †30-23.

30-24.

Consider an element dl of the long, straight wire. As the current I flows through the loop, the magnetic field B produced by the loop exerts a magnetic force F on the element dl given by dF = I dl×B This force is perpendicular to element dl. Applying Newton’s third law of motion, the element dl exerts a force that is directed oppositely to dF, which is also perpendicular to element dl. F = IL × B = IL × B = I [r2 − r1 ] × B

= (15 A ) [( 0.50 m − 0.35 m ) i + ( −0.25 m − 0.50 m ) j + ( 0.40 m − 0 ) k ] × ( 2.0 T ) k = (15 A ) [( 0.15 m ) i + ( −0.75 m ) j + ( 0.40 m ) k ] × ( 2.0 T ) k = (15 A ) ⎡⎣ ( −0.75 m )( 2.0 T ) i − ( 0.15 m )( 2.0 T ) j⎦⎤ = − ( 22.5 N ) i −

( 4.5 N ) j


CHAPTER †30-25.

30

Since the magnetic force on a segment of current carrying wire in a uniform magnetic field is given by dF = I dl×B There is no force on segments that are parallel to the magnetic field, including the incoming and outgoing portions as well as the segment labeled B in the drawing. For the segments labeled A and D in the drawing, the magnetic force is equal in magnitude and opposite in direction, so the net force on those two segments is zero. Therefore, the net force on the wire is equal to the magnetic force on segment C in the drawing. The direction of the force is in the −j direction. Its magnitude is F = ILB sin α = ILB = ( 25 A )( 0.25 m )( 2.0 T ) = 12.5 N = 13 N

C

B A

D

F = −13 N j 30-26.

F = I ′ l × B( I ) F = I ′lB = I ′l

30-27.

30-28.

µ0 I µ II ′l = 0 2π d 2π d

The direction of the force is toward the long, straight wire and perpendicular to it. F ILB = a= m m 5 2 ma ( 0.20 kg ) (1.0 × 10 m/s ) I = = = 1.0 × 106 A LB ( 0.10 m )( 0.20 T ) (a) Consider a small element dl of the wire that is oriented perpendicular to the magnetic field and carrying a current I. The magnetic force on the wire is d F = I d l × B = IB dl (in the direction shown on the drawing)

dF T

××× ××× dθ

T r

The element makes an angle dθ with the center of curvature. The net tensile force is in the direction opposite to the direction of the magnetic force and is of equal magnitude. Therefore, d F = I d l × B = IB dl = IB rdθ IB rdθ = T sin dθ = Tdθ T r= BI p (b) Since rorbit = qB we can substitute the value given for the momentum to show that p qT T rorbit = = = =r qB qBI BI †30-29.

The torque on a current-carrying coil of wire in a magnetic field is given by Equation 30.17, τ = µ B sin θ Since the plane of the coil is parallel to the magnetic field, the angle θ is 90°, so the torque is found by applying Equation 30.16.


CHAPTER

30

τ = µ B sin θ = µ B = NIAB

= (120 ) (1.0 × 10−3 A ) ( 0.010 m )( 0.020 m )( 0.010 T ) = 2.4 × 10−7 N i m

30-30.

τ = µ B sin θ = IAB sin θ

= ( 25 A ) π ( 0.20 m ) ( 3.9 × 10−5 T ) sin 16° = 3.4 × 10−5 N i m 2

†30-31.

(a) The torque is given by Equation 30.17, τ = µ B sin θ = (1.41 × 10−26 A i m 2 ) ( 0.80 T ) sin θ = 1.1 × 10−26 sin θ N i m

Torque, τ (N-m)

If this torque is plotted as a function of 0 ≤ θ ≤ 360°, we find the sinusoidal behavior of the torque. (b) The energy required to flip the proton is equal to the work W done by the magnetic force on the proton as it rotates through 180°.

1.0x10

-26

5.0x10

-27

0.0

-5.0x10

-27

-1.0x10

-26

0

30

60

90

120 150 180 210 240 270 300 330 360 Angle, θ

W = 30-32.

∫

π 0

π

π

0

0

τ dθ = µ B ∫ sin θ dθ = µ B cos θ

= 2 µ B = 2.2 × 10−26 J

The torque is τ = µ B sin θ = ( 3.2 N i m ) sin θ Since there is a commutator on this motor, the torque is a rectified sine wave; and the average torque over a full revolution is equal to the average torque over a half-revolution. ( 3.2 N i m ) π sin θ dθ = 1.0 N i m cos θ π = 2.0 N i m 1 π τ average = ∫ τ dθ = ( ) ∫ 0

π

0

π

0

The power of the motor is the work done per unit time. W τ averageθ rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 hp ⎞ ⎛ = = τ averageω = ( 2.0 N i m ) ⎜ 50 Paverage = ⎟⎜ ⎟⎜ ⎟ = 0.84 hp t s ⎠ ⎝ 1 rev ⎠ ⎝ 745.7 W ⎠ t ⎝ 30-33. 30-34. †30-35.

µ = NIA = ( 60 )( 0.30 A ) π ( 0.020 m ) = 0.023 A i m 2 2

µ = IA = I (2π rl ) = ( 4.0 × 103 A ) 2π ( 0.040 m )( 0.30 m ) = 3.0 × 102 A i m 2 The charge is uniformly distributed over the paper disc and is assumed to remain fixed with respect to the paper, which is an insulator. As the disc rotates, it may be viewed as concentric rings of charge. For a ring at distance r with width dr, the amount of charge is Q 2Qr dq = π (r + dr ) 2 − π r 2 ) = 2 dr 2 ( R πR


ω

r dq

CHAPTER

30

During one period of rotation, T = 2π/ω, the charge dq passes a given point. The current represented by the rotation of dq about the axis is 2Qr dr 2 dq Qω dI = rdr = R = 2π T π R2

ω

The magnetic moment dµ for this ring is then, Qω Qω d µ = dIA = dI (π r 2 ) = rdr (π r 2 ) = 2 r 3 dr π R2 R The total magnetic dipole moment is found by integrating this result for 0 ≤ r ≤ R. R Qω R Qω 1 µ = 2 ∫ r 3 dr = 2 r 4 0 = Qω R 2 0 R 4R 4 The magnetic and electric fields surrounding the spinning paper disk are illustrated as follows:

E, B

30-36.

(a) Considering the symmetry of the problem, the magnetic forces on the two horizontal wires of the loop are equal in magnitude, but opposite in direction, so there is no 6.0 cm net magnetic force in the direction parallel to the wire. There is also no net torque on the loop. There are vertical 8.0 cm components of the magnetic forces that add together to give a net force in a direction perpendicular to the wire as shown in the side view drawing. The net force is then µ I′ F = 2 ILB sin θ = 2 IL 0 sin θ 2π r Tim ⎛ ⎞ (4π × 10−7 ) (95 A) ⎜ ⎟ 0.060 m A = 2(70 A) (0.25 m) ⎜ ⎟ 1 / 2 2 2 2 2 1/ 2 ⎜⎜ 2π ( (0.060 m) + (0.080 m) ) ⎟⎟ ( (0.060 m) + (0.080 m) ) ⎝ ⎠ Tim 0.060 m = (70 A) (0.25 m) (4 × 10−7 ) (95 A) A ( (0.060 m)2 + (0.080 m)2 ) = 4.0 × 10–3 N (b) The lever arm is l = 0.060 m. Therefore, the torque is


θ θ

CHAPTER τ = 2 ILBl cos θ = 2 IL

30

µ0 I ′ l cos θ 2π r

⎛ ⎞ ⎛ −7 T i m ⎞ ⎜ 4π × 10 ⎟ ( 95 A ) ⎜ ⎟ ( 0.060 m )( 0.080 m ) A ⎠ ⎝ ⎟ = 2 ( 70 A )( 0.25 m ) ⎜ 1 / 2 ⎜ 2π ( 0.060 m )2 + ( 0.080 m )2 ⎟ ( 0.060 m )2 + ( 0.080 m )2 ⎜ ⎟ ⎝ ⎠ ( 0.060 m )( 0.080 m ) Tim ⎞ ⎛ = ( 70 A )( 0.25 m ) ⎜ 4 × 10−7 ⎟ ( 95 A ) 2 2 A ⎠ ⎝ ( 0.060 m ) + ( 0.080 m )

(

) (

(

)

1/ 2

)

= 3.2 × 10−4 N i m †30-37.

(a) The torque on the needle due to the horizontal component of the Earth’s magnetic field is given by Equation 30.17, τ = µ B sin θ = ( 9.0 × 10−3 A i m 2 )(1.9 × 10−5 T ) sin θ = 1.71 × 10−7 N i m sin θ

= 1.7 × 10−7 N i m sin θ (b) The rotational equation of motion is τ = Iθ = −1.7 × 10−7 N i m sin θ

( 2.0 × 10

−8

kg i m 2 ) θ + 1.7 × 10−7 N i m sin θ = 0

Note, the torque has a minus sign because the torque acts in the direction opposite to the angular displacement. For small oscillations, sin θ ≈ θ. So, ⎛ 1.71 × 10−7 N i m ⎞ θ +⎜ θ =0 −8 2 ⎟ ⎝ 2.0 × 10 kg i m ⎠

θ + ( 8.55 rad/s 2 ) θ = 0

This equation is of the form θ + ω 2θ = 0 , so the frequency of small oscillations is f =

ω 8.55 rad/s 2 = = 0.47 Hz 2π 2π

30-38.

µ = 1 + χ = 1 + 7.9 × 10−4 = 1.00079 µ0

30-39.

(a) M =

30-40.

(a) B = Bexternal + Bmatter = (1 + χ ) Bexternal =

30-41.

Ti m ⎞⎛ turns ⎞ ⎛ = ( 200 ) ⎜ 4π × 10−7 ⎟ ⎜ 320 ⎟ ( 5.00 A ) = 0.40 T A ⎠⎝ m ⎠ ⎝ µ Ti m ⎞⎛ turns ⎞ ⎛ µ 0 nI = (1500 ) ⎜ 4π × 10−7 (b) B = ⎟ ⎜ 320 ⎟ ( 5.00 A ) = 3.0 T µ0 A ⎠⎝ m ⎠ ⎝ (a) B = Bexternal + Bmatter = 0 = (1 + χ ) Bexternal

Bmatter

2.19 T = 1.74 × 106 A/m = 1.7 × 106 A/m µ0 −7 T i m 4π × 10 A 6 µ M 1.74 × 10 A/m = = = 2.0 × 10−23 A i m 2 (b) −3 28 V n 8.5 × 10 m =

µ µ µ 0 nI Bexternal = µ0 µ0

χ = −1


CHAPTER

30

(b) µ = µ 0 (1 + χ ) = µ 0 (1 − 1) = 0 30-42.

†30-43.

The current must move around in a shallow region near the surface of the cylinder in such a way as to create a magnetic field that is equal in magnitude and in the opposite direction to the applied field. The result is zero magnetic field within the cylinder. This is similar to the magnetic field produced within a solenoid. Here the magnetic field created by the surface currents is given by an equation similar to 29.20, I B = µ0 L The currents are uniformly distributed over the surface to cancel the applied field, The total current is I B = µ0 L −2 BL ( 2.7 × 10 T ) ( 0.10 m ) I = = = 2149 A = 2.1 × 103 A Tim µ0 4π × 10−7 A This corresponds to a current density of ( 2149 A ) I I = = 4.3 × 1011 A/m 2 J = = −8 A Lλ ( 0.10 m ) ( 5.0 × 10 m )

The Hall potential difference is given by Equation 30.30, ( 50 A )( 2.0 T )( 0.0030 m ) IBd ∆VH = = C ⎞ 2 neA ⎛ −19 28 electrons ⎞ ⎛ ⎜ 8.5 × 10 ⎟ ⎜ 1.60 × 10 ⎟ π ( 0.0015 m ) 3 m electron ⎠ ⎝ ⎠⎝ = 3.1 × 10−6 V or 3.1 µ V

30-44.

Consider the drawing. The relevant component of the magnetic field is the one perpendicular to the wire.

θ

B

( 50 A )( 2.0 T )( 0.0030 m ) sin 60° IBd sin 60° = C ⎞ 2 neA ⎛ −19 28 electrons ⎞ ⎛ ⎜ 8.5 × 10 ⎟ ⎜ 1.60 × 10 ⎟ π ( 0.0015 m ) 3 m electron ⎠ ⎝ ⎠⎝ −6 = 2.7 × 10 V or 2.7 µ V

∆VH =

30-45.

IB neh neh∆VH B= I C ⎞ ⎛ −19 −4 −1 19 electrons ⎞ ⎛ ⎜ 2.0 × 10 ⎟ ⎜ 1.60 × 10 ⎟ ( 3.0 × 10 m )(1.3 × 10 V ) 3 m electron ⎠ ⎠⎝ =⎝ 5.0 × 10−3 A

∆VH =

= 2.5 × 10−2 T


CHAPTER

30-46.

n=

(1.0 × 10−1 A ) ( 8.0 T ) IB = = 9.26 × 1027 /m3 ∆VH eh ( 5.4 × 10−6 V )(1.60 × 10−19 C )(1.0 × 10−4 m )

Number electrons per atom = 30-47.

(a) vd = =

30

9.26 × 1027 electrons/m3 = 0.40 electrons/atom 2.3 × 1028 atoms/m3

I neA

⎛ electron ⎞ ⎛ 23 ⎜1 ⎟ ⎜ 6.02 × 10 atom ⎠ ⎝ ⎝

0.30 A atoms ⎞ ⎛ 1 mole ⎞ ⎛ 1532 kg ⎞ −19 −5 −3 ⎟⎜ ⎟ (1.60 × 10 C )( 6.5 × 10 m )(1.0 × 10 m ) ⎟⎜ mole ⎠ ⎝ 0.0855 kg ⎠ ⎝ 1 m3 ⎠

= 2.67 × 10−3 m/s = 2.7 × 10−3 m/s (b) ∆VH =

IB neh

( 0.30 A )( 6.0 T )

=

⎞ ⎛ 1532 kg ⎞ ⎛ electron ⎞ ⎛ 23 atoms ⎞ ⎛ 1 mole −19 −5 ⎜1 ⎟ ⎜ 6.02 × 10 ⎟⎜ ⎟ (1.60 × 10 C )( 6.5 × 10 m ) ⎟⎜ 3 atom ⎠ ⎝ mole ⎠ ⎝ 0.0855 kg ⎠ ⎝ 1 m ⎠ ⎝ −5 = 1.6 × 10 V 30-48.

(1.5 A )( 2.0 T ) IB = = 8.3 × 1026 /m3 −4 ∆VH eh (1.5 × 10 V )(1.60 × 10−19 C )(1.5 × 10−4 m )

(a) n = (b) vd =

30-49.

RH = − =

I 1.5 A = = 3.8 × 10−2 m/s 26 3 −19 neA ( 8.33 × 10 /m )(1.60 × 10 C )(1.5 × 10−4 m )( 2.0 × 10−3 m )

1 ne

⎛ electron ⎞ ⎛ 23 ⎜1 ⎟ ⎜ 6.02 × 10 atom ⎠ ⎝ ⎝

−1 atoms ⎞ ⎛ 1 mole ⎞ ⎛ 860 kg ⎞ −19 ⎟⎜ ⎟ (1.60 × 10 C ) ⎟⎜ mole ⎠ ⎝ 0.0391 kg ⎠ ⎝ 1 m3 ⎠

= −4.7 × 10−10 m3 / C 30-50.

30-51.

(a) Since Fm = qv × B and v and B are in the same direction, the magnetic force on the electron is 0 N. (b) The direction is due south; and the magnitude is F = evB = (1.60 × 10−19 C)(6.0 × 106 m/s)(0.60 T) = 5.8 × 10−13 N (c) The direction is due west; and the magnitude is F = evB = (1.60 × 10−19 C)(6.0 × 106 m/s)(0.60 T) = 5.8 × 10−13 N (d) The direction is due west; and the magnitude is F = evB cos 30° = (1.60 × 10−19 C)(6.0 × 106 m/s)(0.60 T) cos 30° = 5.0 × 10−13 N (a) Since Fm = Il × B and l and B are parallel, the magnetic force on the electron is 0 N. (b) The direction is due north; and the magnitude is F= IL×B F = ILB sin α = ILB = ( 3.0 A )( 0.10 m )( 0.60 T ) = 0.18 N (c) The direction is due east; and the magnitude is F = ILB sin α = ILB = ( 3.0 A )( 0.10 m )( 0.60 T ) = 0.18 N


CHAPTER

30

(d) The direction is due east; and the magnitude is F = ILB sin α = ( 3.0 A )( 0.10 m )( 0.60 T ) sin 60° = 0.16 N 30-52.

†30-53.

The magnetic force is only exerted on the segment of wire that follows the y axis. The direction is +k; and the magnitude is F = ILB sin α = ILB = ( 5.0 A )( 0.25 m )(1.5 T ) = 1.9 N The radius of the circular orbit for this proton is given by Equation 30.5 as mv r= qB which can be rearranged to solve for the magnitude of the magnetic field, −27 5 mv (1.67 × 10 kg )( 4.0 × 10 m/s ) B= = = 0.010 T qr (1.60 × 10−19 C ) ( 0.40 m ) The clyclotron frequency for this proton is given by Equation 30.7. (1.60 × 10−19 C ) ( 0.010 T ) = 1.6 × 105 Hz qB = f = 2π m 2π (1.67 × 10−27 kg )

30-54.

The charge is uniformly distributed over the ring and is assumed to remain fixed with respect to the ring, which is an insulator. During one period of rotation, T = 2π/ω, the charge Q passes a given point. The current represented by the rotation of Q about the axis is Q Q Qω = = I = 2 π 2π T

ω

The magnetic moment for this ring is then, Qω 1 1 2 µ = IA = (π r 2 ) = Qω r 2 = ( 5.0 × 10−6 C )( 6.3 × 104 rad/s ) ( 0.030 m ) 2π 2 2 2 −4 = 1.4 × 10 A i m †30-55.

(a) Taking the entry point as (x = 0, y = 0), 1 x = v0 x t + ax t 2 2 1 and y = v0 y t + a y t 2 2 There is no acceleration in the x direction; and initially, there is no y component of the velocity. Therefore, the time the electron takes to travel from entering the parallel plates until it leaves the plates is L t1 = v0 x

y

With this expression for the time, deflection y may be determined, where 2

2

1 1 ⎛ F ⎞⎛ L ⎞ 1 ⎛ eE ⎞ ⎛ L ⎞ eEL2 y = ayt 2 = ⎜ ⎟ ⎜ = = ⎟ ⎟ ⎜ ⎟⎜ 2 2 ⎝ m ⎠ ⎝ v0 x ⎠ 2 ⎝ m ⎠ ⎝ v0 x ⎠ 2mv 2 (b) If the electric and magnetic fields are crossed, both being perpendicular to the electron velocity, and the magnitudes of their forces on the electron are equal, then the electron will pass through the plates undeflected. Setting the forces equal to each other, we have


CHAPTER

30

evB = eE E B Substituting this result into the equation derived in part (a), we have, eEL2 y= 2mv v=

e 2 yv 2 2 y ( E/B ) 2 yE = = = 2 2 m EL2 EL2 B L (a) The electron is accelerated both in the direction opposite to that of the electric field, but also in the direction perpendicular to it by the magnetic field. The result is that the electron follows a helical path around the electric field lines. 1.60 × 10−19 C )( 5.0 × 10−4 T ) ( qB = = 1.398 × 107 Hz = 1.4 × 107 Hz (b) f = 2π m 2π ( 9.11 × 10−31 kg ) 2

30-56.

The time taken by the electron to make a single loop on the helix is equal to the cyclotron period, 1 1 T = = = 7.15 × 10−8 s f 1.398 × 107 Hz (c) The electron is decelerated along the x-axis in the electric field for a time −31 6 mv0 // ( 9.11 × 10 kg )( 4.0 × 10 m/s ) = = 1.139 × 10−8 s t= 3 −19 eE (1.60 × 10 C )( 2.0 × 10 V/m ) At that time the parallel component of the velocity goes to zero. After that, it starts increasing in direction opposite to the electric field. For the full period, the change in velocity is: 2.0 × 103 V/m ) ( eE eE 2π m E T = ∆v = axT = = 2π = 2π = 2.5 × 107 m/s −4 m m eB B ( 5.0 × 10 T ) The translation along the x-axis is initially in the positive direction, then the electron spirals backward to: 1 x = voxT − axT 2 2 −19 3 2 1 (1.6 × 10 C )( 2 × 10 V/m ) −8 6 = ( 4.0 × 10 m/s )( 7.15 × 10 s ) − 7.15 × 10−8 s ) ( −31 2 9.11 × 10 kg

†30-57.

= −0.61 m The negative sign means that the electron is positioned to the left of its original position. For the particles to collide at the mid-point, they must be forced to follow a circulr path of radius 0.25d. The radius of the circular path is given by by Equation 30.5 as mv d r= = qB 4 From which, the magnetic field is 4mv B= qd


CHAPTER 30-58.

30

Consider the drawing at the right, which shows the current I2 following a counterclockwise path around the rectangular loop; and the current in the long, straight wire moving from left to right. The current in the wire produces a magnetic field through the loop that is directed upward, out of the page. As a result, the magnetic forces on the two 18 cm segments will be equal in magnitude, but oppositely directed. There will, however, be a net force on the two 12 cm sides, since the magnetic field at the side closest to the wire has a larger magnitude than the side further away. The force on the nearest side is µI F1 = I 2 LB1 = I 2 L 0 1 2π Ra

I2

I1

Similarly, the force on the farthest side of the loop is µI F1 = I 2 LB1 = I 2 L 0 1 2π Rb

†30-59.

The net force on the loop will then be µI ⎛ 1 1 ⎞ F = F1 − F2 = F1 = I 2 LB1 = I 2 L 0 1 ⎜ − ⎟ 2π ⎝ Ra Rb ⎠ ⎛ −7 T i m ⎞ ⎜ 4π × 10 ⎟ ( 40 A ) 1 1 ⎞ A ⎠ ⎛ −4 = ( 60 A )( 0.12 m ) ⎝ − ⎜ ⎟ = 7.2 × 10 N 2π 0.060 m 0.24 m ⎝ ⎠ In the problem, the directions of the currents are not specified, so the net force will be directed perpendicularly to the wire. The magnetic field exerts a torque τ on the hoop, which is τ = Imα, where Im is the moment of inertia for the hoop of mass m and radius R. Since the plane of the hoop is parallel to the magnetic field, there will be a maximum torque on the hoop as it is released. τ µ B sin θ NIAB sin θ 2 NI π R 2 B sin θ = = = α = 1 1 2mR 2 Im mR 2 mR 2 2 2 2 NI π B sin θ 2 ( 25 )( 5.0 A ) π ( 0.20 T ) sin 90° = = = 3.1 × 103 rad/s 2 0.050 kg m


CHAPTER 31

ELECTROMAGNETIC INDUCTION


ε = vBl v = 88 km/h = 24.4 m/s l = 2.1 m ε = vBl = 24.4 × 0.58 × 10–4 × 2.1 V = 3.0 × 10−3 V

31-2.

Left side is positive, right is negative. 5 ms–1 = 267 ms–1; therefore, ε = vBl v = 960 km/h = 960 × 18 ε = 267 × 0.60 × 10–4 × 47 V = 0.75 V

31-3.

ε = vBl; therefore, v = v=

ε Bl

B = 0.70 gauss = 0.70 × 10–4 T

7.0 × 103V = 0.5 m/s ( 0.70 × 104 × 200 ) T

31-4.

ε = lvB = (0.5 m)(9.0 m/s)(6 × 10−5 T) = 2.7 × 10−4 V

31-5.

ε = vBl = 1.5 ms–1 × 1.5 × 10–2 T × 0.10 m = 2.2 × 10−3 V

31-6.

E = v × B; v ⊥ B ⇒ E = vB

†31-7.

E = (1.2 × 107 m/s)(5.5 × 10−5 T) = 660 V/m The induced emf is ε = lvB. From Ohm’s Law ε = IR ⇒ IR = lvB if the rails and the rod have negligible resistance. Then IR (1.5 A)(2.2 Ω) = = 0.48 m/s v= lB (0.86 m)(8.0 T)

As the rod moves to the left, there is an increasing flux caused by an external magnetic field pointing into the page. Lenz’ Law says the induced current must produce a magnetic field that opposes this change in flux, so the field caused by the induced current must point in the opposite direction from the external field, or the induced field must point out of the page. By the right hand rule, this means the induced current must be counter-clockwise. If the current is counter-clockwise, it is going down through the rod in the figure. The magnetic force due to the external field is Fm = I l × B. l points down and B points into the page, so by the

right-hand rule the magnetic force points to the right. Thus an external force with the same magnitude but pointing to the left must be applied to keep the rod moving at a constant velocity. Since l and B are perpendicular, the magnitude of the magnetic force is IlB, and the magnitude of the external force is Fext = (1.5 A)(0.86 m)(8.0 T) = 10 N Thus Fext = 10 N to the left.


CHAPTER 31-8.

†31-9.

31

The component of the induced electric field along the length of the antenna is vB cos 52°. By the right hand rule, the direction of this field is up if the antenna is moving east. Thus the induced emf is ε = vBl cos 52°, where l is the length of the antenna. The speed is 80 km/h = 22.2 m/s, so ε = vBl cos 52° = (22.2 m/s)(0.7 × 10−4 T)(0.75 m) cos 52° = 7.2 × 10−4 V with the top of the antenna positive relative to the bottom. As the rotor turns, the end of each blade sweeps out an area A = πl2, where l is the length of the ⎛ π l2 ⎞ dΦB dA blade. The rate of change of magnetic flux is =B = B⎜ ⎟ , where T is the time for dt dt ⎝ T ⎠

one complete revolution of the blade. At a rate of 3.0 rev/s, T = 1/3 s. Substituting the other data gives ε = (6.5 × 10−5 T)(π )(4.0 m)2 (3.0 s −1 ) = 9.8 × 10−3 V 31-10.

The coil can be rotated through an angle of 180º ± θ, where θ is given by BA cos θ = 0.30BA. ⇒ cos–10.30 = 73°. So the coil can be rotated through 253° counterclockwise or it could be rotated through 107° clockwise.

B

B

A

30º 31-11.

31-12. 31-13.

A

The area vector points straight up and the magnetic field points 69° below the horizontal, so the angle between B and A is 159°. The flux is Φ B = B i A = BA cos θ = (5.9 × 10−5 T)(1.0 m 2 ) cos(159°) = −5.5 × 10−5 T i m 2 1 min ⎛ ⎞ × 2π radian/rev ⎟ (2.0 × 10−4 m 2 ) = 0.25 V. 60 s ⎝ ⎠ 2 See problem 9. |ε| = πr Bf , where r is the radius of the disk and f = 1/T is the rotation frequency. Then ε 6.0 V f = = = 22 rev/s 2 π r B π (1.2 m) 2 (0.060 T)

ε max = NBω A = (300)(0.020 T) ⎜ 2000 rev/min ×

31-14.

See Problem 9. |ε| = πr2Bf , where r is the radius of the star and f = 1/T is the rotation frequency. Then ε = (1.0 × 108 T)(π )(10 × 103 m)2 (30 s −1 ) = 9.4 × 1017 V

31-15.

The field is along the axis of the solenoid, so it’s perpendicular to the patient’s cross section and the flux is BA. There is a change in flux caused by the change in area from 0 to A as the patient is dΦB dA moved into the field. The induced emf is ε = =B . Assuming a roughly circular cross dt dt section, the area is related to the circumference: C 2 (0.90 m) 2 = = 0.0645 m 2 . C = 2π r ; A = π r 2 ⇒ A = 4π 4π dA (0.0645 − 0) m 2 The average rate of change of A is = = 0.00645 m 2 /s, so dt 10 s dA ε =B = (1.5 T)(0.00645 m 2 /s) = 9.7 × 10−3 V. dt This is a very low voltage, so it’s not necessary to push the patient in more slowly.


CHAPTER 31-16.

†31-17.

31

View the coil from the left end. The N pole of the magnet is being moved toward the coil, so B always points into the page in our diagram. (a) As the magnet approaches, B is increasing into the page. The induced current must produce a magnetic field that opposes this change, so the induced field must point out of the page. By the right hand rule, this requires a counter-clockwise current. (b) As the middle section of the magnet passes through, B is relatively uniform so there is no change in the flux as that part of the magnet moves through. The induced current goes to zero. (c) As the magnet leaves the coil on the other side, B still points into the page but is decreasing. The induced current must produce a magnetic field that opposes this change, so the induced field must point into the page. By the right hand rule, this requires a clockwise current. Φ B = NB i A = NBA cos θ . A ∆B ε = NA cos θ ∆t

B

5.5 T ∆B = = 0.157 T/s 35 s ∆t

30º

ε = (250)(0.35 m 2 )(cos 60°)(0.157 T/s) = 6.88 V ρL

The resistance of the wire is R = Awire

. The area of the wire is A = π r 2 = π (0.50 × 10−3 m) 2 = 7.85 × 10−7 m 2 .

To find the length, use the circumference of the coil which is related to the area: 2 A = π rcoil ; C = 2π rcoil ⇒ C = 2 π A = 2 π (0.35 m 2 ) = 2.10 m. There are 250 turns each with

a circumference of 2.10 m, so the total length of wire is (250)(2.10 m) = 524 m. The resistance is (1.7 × 10−8 Ω i m)(524 m) R= = 11.4 Ω 7.85 × 10−7 m 2

ε

6.88 V = 0.606 A R 11.4 Ω The power is P = I ε = (0.606 A)(6.88 V) = 4.17 W, so the total energy is I =

=

W = P∆t = (4.17 W)(35 s) = 146 J. 31-18.

ε

0.15 V = = 0.010 A R 1.5 Ω As the area decreases, the flux due to the inward field decreases. The induced current must flow clockwise to produce an additional inward flux to oppose this change. Thus the current flows from right to left through the resistor. The motional electric field at any point on the CD is E = v × B. The velocity v is always I =

†31-19.

⎛ 100 × 10−4 m 2 ⎞ dΦB dA =B = (0.30 T) ⎜ ⎟ = 0.15 V dt dt ⎝ 0.020 s ⎠

ε =

perpendicular to B, so the magnitude of the electric field is vB. At a distance r from the center of the CD, the speed is v = ωr, so E = ωrB and the direction of E is radial. (Whether E is radial inward or outward depends on the direction of rotation and the direction of B.) The magnitude of the emf between points at r1 and r2 from the center is


CHAPTER

ε =

31-20.

31

∫

r2

r2

Edr = ω B ∫ rdr =

ω B (r22 − r12 )

2 2π (210 rev/min)(1 min / 60 s)(1.5 T)[(0.058 m) 2 − (0.023 m) 2 ] = 2 = 0.047 V dΦB dh = wB = wBv. Φ B = whB ⇒ ε = dt dt The flux due to the inward field is decreasing, so the induced current must be clockwise to induce a field also pointing inward. The current through the top segment is left to right, producing an upward force on the loop. The magnitude of this force is ε ( wB ) 2 v = Fm = IwB = wB R R The induced current increases with increasing speed, increasing the magnitude of the upward force. The downward force on the loop is mg, which is constant. Thus the acceleration decreases, reaching zero when the magnitude of the upward magnetic force equals the weight. This determines the terminal speed: r1

r1

I

h

( wB ) 2 vt = mg R (0.012 kg)(9.81 m/s 2 )(0.17 Ω) mgR vt = = ( wB ) 2 (0.25 m)2 (1.8 T) 2 = 0.099 m/s †31-21.

The flux through the strip shown is d Φ B = B(r )hdr

=

w

d

µ 0 Ihdr 2π r

The total flux is found by integrating: µ Ih d + w dr µ 0 Ih d + w ln ΦB = 0 ∫ = d 2π d r 2π Substituting d = 0.10 m, w =0.30 m, h = 0.50 m, I = 2.5 A gives (4π × 10−7 T i m/A)(2.5 A)(0.50 m) 0.4 ΦB = = 3.5 × 10−7 T i m 2 ln 2π 0.1

I h r dr


CHAPTER 31-22.

A strip through one side of the solenoid is shown in the diagram. From Equation 29.22 the field a distance r from the center of the torus is µ NI B(r ) = 0 2π r The flux through the strip shown is d Φ B = B(r )hdr =

†31-23.

w

R

µ 0 NIhdr 2π r

h

The total flux is found by integrating: r µ 0 NIh R + w dr µ 0 NIh R + w ΦB = = ln R 2π ∫R r 2π dr −7 (4π × 10 T i m/A)(150)(2.0 A)(0.040 m) 13.5 2 −6 = = 1.6 × 10 T i m ln 2π 7 (a) Inside the solenoid, the magnetic field is B = µ 0 ( N / L) I where N/L is the number of turns per unit length. For a circle of radius r < a, where a is the radius of the solenoid, the flux is Φ B = BA = µ 0 ( N / L) I π r 2 If the current is changing with time, the magnitude of the induced emf is dΦB dI = µ 0 ( N / L)π r 2 . ε = dt dt The magnitude of the induced electric field is ε µ r ( N / L) dI E = = 0 2π r 2 dt At r = 2.0 cm, this is (4π × 10−7 T i m/A)(0.020 m)(350 m −1 ) E = (1.5 A/s) = 6.6 × 10−6 V/m 2 (b) Outside the solenoid, the magnetic field is essentially zero. For a circle of radius r > a, the total flux will be Φ B = BA = µ 0 ( N / L) Iπ a 2 , no matter what r is. Then the induced field will be E = E =

21-24.

31

ε µ (π a 2 )( N / L) dI µ 0 a 2 ( N / L) dI = 0 = dt dt 2π r 2π r 2r (4π × 10−7 T i m/A)(0.030 m) 2 (350 m −1 ) 2(0.04 m)

(1.5 A/s) = 7.4 × 10−6 V/m

Φ B = BA cos θ dΦB dθ ε = = BA sin θ dt dt I =

B

ε dq BA dθ sin θ = = dt R R dt

A

B

A

BA sin θ dθ ⇒ dq = R If the coil is rotated from θ1 to θ2, the total charge that flows is θ2 BA q = ∫ dq = cos θ 2 − cos θ1 θ1 R


CHAPTER

†31-25.

31

For θ1 = 0 and θ2 = 90º, this is BA (8.2 T)(π )(0.035 m)2 = = 0.13 C q = 0.25 Ω R From Problem 19, ω B (r22 − r12 ) ε = 2 Here r1 = 0, r2 = r = 6.0 cm. The current through the resistor is ε 2π (300 rev/s)(5.5 T)(0.060 m) 2 I = = = 0.565 A R 2(33 Ω) Assuming the disk itself has no resistance and rotates without friction, the electrical power is all absorbed by the resistor. This must be equal to the mechanical power required to turn the disk. Thus I ε = τω

=

I 2R

ω

(0.565 A) 2 (33 Ω) = 5.59 × 10−3 N i m ω ω 2π (300 rev/s) (a) l = 4 ft 9 in = 4.75 ft × 0.3048 m/ft = 1.45 m 5 v = 80 mi/hr = 80 × 1.6 km/h = 80 × 1.6 × ms–1 = 35.6 ms–1 18 ∴ε = vBl = ( 35.6 × 0.62 × 10−4 × 1.45 ) V = 3.2 × 10−3 V

τ =

31-26.

I ε

B

=

(b) Resistance of axle R =

ρl

=

ρl π r2

A r = 1.5 in = 0.125 × 0.3048 m = 0.0381 m; therefore, 10 × 10−8 Ωm × 1.45m R= = 3.2 × 10–5 Ω; therefore, 2 π ( 0.0381m )

I=

V 3.2 × 10−3 V = = 100Α R 3.2 × 105 Ω

(c) P = I ε = (100 A)(3.2 × 10−3 V) = 0.32 W per axle. The magnitude of the magnetic retarding force is given by FM = IlB = (100 A)(1.45 m)(6.2 × 10−5 T) = 9.0 × 10−3 N per axle. 31-27.

The length of the wire is 4 × 8.0 cm = 32.0 cm = 0.32 m. The radius is 1.0 mm = 1 × 10–3 m; ρl therefore, resistance of wire =R A 1.7 × 108 × 0.32 = Ω = 1.73 × 10–3 Ω. The area of the loop is −3 2 π (10 ) A = 8 cm × 8 cm = 64 cm2 = 6.4 × 10–3 m2; therefore, ε = – = – ( 6.4 × 103 m 2 ) 80 Ts–1 = – 0.512 V; therefore, I =

31-28.

dF dt

ε 0.512 V = = 300 A R 1.73 × 10−3Ω

The length of the wire is 4 × 8.0 cm = 32.0 cm = 0.32 m. The radius is 1.0 mm = 1 × 10–3 m; therefore, resistance of wire ρ l 1.7 × 10−8 × 0.32 R= = Ω = 1.73 ×10−3 Ω. 2 − 3 A π (10 )


CHAPTER

31

The area of the loop is A = 8 cm × 8 cm = 64 cm2 = 6.4 × 10–3 m2; therefore, dΦ dΦ ; therefore, = ( 6.4 × 10−3 m 2 ) (80 T/s) = 0.512 V ε = dt dt ε 0.512V I= = − 300A. R 1.73 × 103 Ω

†31-29.

The direction of the current depends on how the field is oriented relative to the loop. By Lenz’ Law, the induced current must flow in a direction that opposes the change in flux causing it. Thus if the loop is in the plane of the page and the magnetic field is increasing out of the page, the induced current will go clockwise around the loop. If the magnetic field is increasing into the page, the induced current will be counterclockwise. Resistance, R, of the loop l R=ρ A (1.7 × 10−8 Ω i m)(2.0 m) = 2 π (1.3 × 10−3 m ) = 6.40 × 10–3 Ω

dΦ d dx = ( BA) = hB = hBv = (0.20 m) (0.050 T)(0.40 m/s) = ε ind = 4.0 × 10−3 V dt dt dt dΦ d dx = ( BA) = hB = hBv = (0.20 m) dt dt dt (0.050 T)(0.40 m/s) = ε ind = 4.0 × 10−3 V Iind =

31-30.

ε ind R

=

4.0 × 10−3 V = 0.63 Α 6.40 × 10−3 Ω

Since there is an increasing flux caused by an inward magnetic field, the induced current must produce an outward field to oppose that change. Thus the current will be counterclockwise as shown. The cause of the interaction between the washer and the solenoid is the increasing flux through the washer. Therefore, from Lenz’s law, the washer will move in such a manner as to oppose it by reducing the flux linkage. This will happen if the washer moves away from the solenoid where the density of field lines is less than at the end of the solenoid.


CHAPTER †31-31.

31

B = B0 sin ωt = B0 sin 2π ft = (0.9 T) sin 2π(60)t= (0.9 T) sin 377t dΦΒ dB = –πr2 At a radius r < Rbetatron, ε = – ; ε = –πr2[(377)(0.90)] cos 377t dt dt = –1.10 × 103 r2 cos 377t = –ε0 cos 377t, where the amplitude ε0 = 1.10 × 103 r2 V. But ε0 = ∫ E0 i ds = 2πr E0; therefore, E0 = At r = 0.80 m, ε 0 = 704 V, E0 = 140 V/m

ε0 = 175r V/m 2π r

dB = −π (1.0) 2 (377)(0.9) cos 377t , dt ε B2 (.0236 T) 2 175 = = 221 J/m3 Then E0 = 0 = V/m. At r = 1.5 m, so u = −7 2π r 2µ 0 2(4π × 10 T i m/A) r E0 = 117 V/m

2 2 For r > Rbetatron, the total flux is π Rbetatron B, so ε = −π Rbetatron

31-32.

†31-33.

See Problem 21. µ Il d + l Φ B = 0 ln d 2π µ Il d µ Il dΦB 1 1 ⎞ d (d ) ε =− =− 0 − ⎟ [ ln(d + l ) − ln d ] = − 0 ⎛⎜ dt 2π dt 2π ⎝ d + l d ⎠ dt I µ 0 Il 2 v ε = 2π d (d + l )

Φ B = NBA cos θ dΦB dθ ε = = NBA sin θ dt dt I =

d v

B

A

ε dq NBA dθ sin θ = = dt R R dt

NBA sin θ dθ R If the coil is rotated from θ1 to θ2, the total charge that flows is θ2 NBA q = ∫ dq = cos θ 2 − cos θ1 θ1 R For θ1 = 0 and θ2 = 180º, this is 2 NBA 2(60)(0.050 T)(π )(0.090 m) 2 = = 0.010 C q = R 15 Ω

⇒ dq =

B

A


l

CHAPTER

31-34.

I1 = 12 sin (120πt); Φ2 = L21I1 (a) Since Φ2 = L21I1, Φ2 = 0 at t = 0 dI1 = ( 2 × 102 ) × 12 × 120π = 90 V dt (c) At t = 0 flux in (2) is increasing to the right; induced emf will oppose this increase; therefore, direction of induced current (see fig.) (opposite to direction of current in coil 1). Φ = L21I1 F (60)(0.1) = L21 = = 0.4 H 15 I1 (b) ε2 = – L21

†31-35.

31-36.

†31-37.

31-38.

2 Φ 30Bπ r 2 30 µ 0 nI1π ( 0.5 × 10 ) = = (a) L21 = I1 Ι1 I1

2

where n = 400 turns/m, πr2 = cross-sectional area of the coil L21 = 30 × 4π × 10–7 × 400 × π(0.01) 2 = 4.74 × 10–6 H dI (b) |ε1| = L21 = 4.74 × 10–6 × 200 = 9.47 × 10–4 V dt Φ 50 Wb (a) Φ = LI ⇒ L = = = 0.5 H I 100 A dI = – 0.5 H(– 20 A/s) = 10 V (b) Induced emf ε = –L dt (a) By (32), L = µ0n2eπR2 = (1.26 × 106 ) × ( 2000 ) 2 × 1 × π ( 0.02 ) 2 H = 6.3 × 103 H dI = – ( 6.3 × 10−3 H ) × 3.0 × 102 A/s = 1.9 V dt By Kirchhoff’s laws, the applied voltages around loop must equal zero. The emf of battery = ε, dI . (back) emf of inductance = – L dt dI dI ε 24 V = = Hence ε – L =0 ⇒ dt dt L 2.2 × 10−3 H (b) ε = – L

†31-39.

= 1.1 × 104 A / s. 31-40.

dI d = − L (C1t − C2 t 2 ) = − L(C1 − 2C2 t ) dt dt ε (1.0 s) = −(7.5 × 10−3 H)[65 A/s − 2(25 A/s 2 )(1.0 s)] = −0.11 V

ε (t ) = − L

ε (2.0 s) = −(7.5 × 10−3 H)[65 A/s − 2(25 A/s 2 )(2.0 s)] = 0.26 V 31-41.

ε = −L

dI = −(25 H)(0.075 A/s) = −1.9 V dt


31

CHAPTER 31-42.

31

The voltage across the inductance is constant with a magnitude ε = 5.0 V. For an inductor,

ε ε dI dt ⇒ I − I 0 = t. If the initial current I0 = 0, then for the data given in , so dI = L L dt LI (2.5 × 10−6 H)(2.0 × 10−3 A) the problem t = = = 1.0 × 10−9 s. ε 5.0 V ε =L

†31-43.

31-44.

†31-45.

31-46.

The magnetic field inside the solenoid B1= µ0nI1. The flux due to this field through each loop of the second coil is µ0nI1A, where A is the cross-sectional area of the solenoid (because there is no field outside the solenoid). Thus the total flux through the loop caused by the solenoid is ΦB1 = Φ 200 µ0nI1A, and the mutual inductance is M = B1 = 200 µ 0 nA = 200µ 0 nπ R 2 . The shape of coil I1 doesn’t matter because there is no magnetic field outside of the solenoid. Let current I run through coil radius r1. Then field generated in coil equals B = µ0n1I. Then flux through larger coil, length l, is n2l (area of field) × B L = n2l ( π r12 ) µ0n1I. Hence mutual inductance, per unit length = 12 e Φ = = µ 0π r12 n1n2 . Il µ Ih d+w = 3.47 × 10−7 T i m 2 , where three significant ΦB1 = MI1. From problem 21, Φ B1 = 0 1 ln 2π d figures have been kept because this result will be used for another calculation. Φ 3.47 × 10−7 T i m 2 = 1.4 × 10−7 H. M = B1 = I1 2.5 A dI d = − L ( I max sin ω t ) = −ω LI max cos ω t. The amplitude of the induced emf is dt dt ε 0 = ω LI max . For a solenoid,

ε = −L

µ0 N 2 A

(4π × 10−7 T i m/A)(15) 2 (π )(0.030 m) 2 = 3.20 × 10−6 H. l 0.25 m ε 0 = (1.2 × 107 s −1 )(3.20 × 10−6 H)(2.5 A) = 96 V L = µ 0 n 2 Al =

31-47.

=

Let ε be back emf across both—by Kirchhoff’s law, these must be equal. Let I1,I2 be currents through L1,L2, respectively. Then dI dI dI I1 + I2 = I, 1 + 2 = dt dt dt dI dI Now ε = – L1 1 = – L2 2 ; dt dt dI dI1 dI 2 therefore, = + = dt dt dt ⎛ 1 ε ε 1 ⎞ ε dI – = – ε⎜ + = – , this gives – ⎟ . Since L dt L2 L2 ⎝ L1 L2 ⎠ 1 1 1 = + L L1 L2


CHAPTER

31-48.

31-49. 31-50.

dI . Then dt dI dI back emf across both equals – L1 – L2 =– dt dt dI ( L1 + L2 ) = ε dt ε this gave L1 + L2 = L. Since L = – dI / dt 1 1 By (37) U = LI2 = ( 4.0 × 10−8 H ) (25 A) 2 = 1.2 × 10−5 J 2 2 2 B (1.0 T) 2 = = 4.0 × 105 J/m3 uB = −7 2 µ 0 2(4π × 10 T i m/A) Let current I through both be changing at

†31-51.

ε0 E 2

(8.85 × 10−12 C2 /N i m 2 )(1.0 V/m) 2 = 4.4 × 10−12 J/m3 2 2 2 B0 µ0 I 2 1 ⎛ µ0 I ⎞ = = u= 2µ 0 2µ 0 ⎜⎝ 2π r ⎟⎠ 8π 2 r 2

uE =

=

u=

(4π × 10−7 T i m/A)(24 A) 2 = 1.0 J/m3 8π 2 (0.0030 m) 2

31-52.

u=

B0 (103 T) 2 = = 4.0×1011 J/m3 2µ 0 2(4π × 10−7 T i m/A)

31-53.

u=

31-54.

31-55.

B2 = 3.98 × 105 B 2 J/m3 2µ0

Use a spreadsheet to do the repetitive calculations. B (T) u (J/m3) 1.00E + 08 4.0E + 21 1000 4.0E + 11 45 8.1E + 08 8 2.5E + 07 2 1.6E + 06 1.5 9.0E + 05 4π 3 (r2 − r13 ). The volume of a spherical shell: V = 3 B 2 ⎡ 4π 3 (5.0 × 10−5 ) 2 ⎤ 3 − = U = uV = ( r r ) (12.43 − 6.383 ) × 1018 J = 6.9 × 1018 J 2 1 ⎥ 2µ 0 ⎢⎣ 3 6 × 10−7 ⎦ LI 2 35 H = (55 A) 2 = 5.29 × 104 J 2 2 B2 (9.0 T) 2 u= = = 3.22 × 107 J/m3 2µ 0 2(4π × 10−7 T i m/A)

U =

V =

U 5.29 × 104 J = = 1.64 × 10−3 m3 u 3.22 × 107 J/m3


31

CHAPTER

31-56.

U =

31 LI 2 2 2

⎛ 60 ⎞ U (60 A) = U (30 A) ⎜ ⎟ = 8.0 × 10−3 J ⎝ 30 ⎠ 2

31-57. 31-58.

⎛ 90 ⎞ U (90 A) = U (30 A) ⎜ ⎟ = 1.8 × 10−2 J ⎝ 30 ⎠ 2 LI (6.6 × 10−4 H/km)(800 A) 2 U = = = 211 J/km 2 2 v 2 1 2 B where V = volume of toroid; u= B ; therefore, U = uV = 2µ 0 2µ 0 therefore, 2m0U 2 × 1.26 × 106 H/m × 1.0 × 105 × 103 × 3600 J = 9100 m3 = V= 102 T 2 B2 Assume outer radius is twice inner radius, inner radius = R. Using the hint, 2 3 ⎛3 ⎞ ⎛R⎞ Volume = 2π ⎜ R ⎟ π ⎜ ⎟ = π2R2; 4 ⎝2 ⎠ ⎝ 2⎠ 3 therefore, π2R3 = 9100 m3 ⇒ R 4 ( 4 × 9100 ) m = 11m (i.e., size ≈ 40 m across). = 3 3π 2

†31-59.

By Ampere’s law, we see that there is no magnetic field either inside the smaller tube or outside the bigger one since the net current inside these is zero. The only field is in between the tubes, and by Ampere’s law again this is given by µ I B = 0 . Therefore, magnetic 2π r µ I2 1 2 energy density u(r) = B = 02 2 . 8π R 2m0 Therefore, U = ∫ u(r)dV. Take dV = thin concentric cylinders, thickness dr; therefore, r2 1 µ µ 2πlr dr = 0 lI2 dV – l 2πr dr; therefore U = 02 I2 ∫ 2 = r r 1 r 8π 4π ⎛r ⎞ ln ⎜ 2 ⎟ i U ⎝ r1 ⎠ = 1.0 × 10–7

H ⎛ 3 ⎞ −3 × 1 m × (120 A ) 2 × ln ⎜ ⎟ = 1.0 × 10 J m 1.5 ⎝ ⎠


CHAPTER

31-60.

ε (1 − e− tR/L ) R (a) I t = os = 0 A I =

(see Equation 45)

6 (1 − e−0.01×100 / 2 ) = 0.024 A 100 6 I t = 0.02 s = (1 − e−0.02 ×100 / 2 ) = 0.038 A 100 6 1 − e −∞ ) = 0.06 A (b) I t =∞ = ( 100 dI ε R ε (c) = = 3 A/s = • e − tr/L 0 t = t=0 L R L dt Applying Kirchhoff’s voltage rule gives dI − IR − L =0 dt dI R + I =0 dt L ε dI ε −t / τ =− e . Substituting these expressions into the differential equation If I = e − t / τ , then R τR dt gives I

†31-61.

31

−

t = 0.01s

=

ε ε + = 0. τR L

L , the expression for I satisfies the differential equation. R Throwing the switching from 1 to 2 will feed the current from L to R. L The time constant is τ = = 10 s therefore R L 0.2 R= = = 0.02 Ω 10 10

If τ = 31-62.

†31-63.

31-64.

L L 45 H ⇒R= = = 3.75 Ω. R 12 s τ I (0) 2 (b) The initial power is P (0) = . The initial current I(0) is the current that was flowing in R the circuit at the instant the switch was opened, which is given as 65 A. Thus P (0) = I (0) 2 R = (65 A) 2 (3.75 Ω) = 1.58 × 104 W

(a) τ =

(c) The total energy dissipated is equal to the energy initially stored in the inductor: LI (0) 2 (45 H)(65 A) 2 U = = = 9.51 × 104 J 2 2 (a) There was no current through the inductor before the switch was closed, so the current through the inductor must still be zero immediately after the switch is closed, and all the current from the battery must flow through the two resistors. There is the same current through both resistors because they are connected in series. The current is


CHAPTER

31

I (0) =

ε R1 + R2

(b) After the switch has been closed for a long time, the voltage across the inductor will be zero. Since L is in parallel with R1, that means the voltage across R1 must also be zero and there is no current through R1. The total current flows through L and R2, giving I (∞ ) = †31-65.

ε

R2

.

(a) The voltage across L is zero, the current through the circuit is ε 12 V I = = = 2.0 A R2 6Ω (b) When the switch is moved to the lower position, L is now in series with R1 and R2. At the instant the switch is moved, the current through the inductor must be the same as just before the change, so the current through L and the two resistors must still be 2.0 A. According to Kirchhoff’s voltage law, the sum of voltages changes around the circuit must be zero, giving I ( R1 + R2 ) + VL = 0 VL = − I ( R1 + R2 ) = −(2.0 A)(1506 Ω) = −3.0 × 103 V

31-66.

After the switch has been closed for a long time, the voltage across L will be zero. The voltage across each resistor will be ε, so the current through R2 is ε/R2. Since L is in series with R2, this is also the initial current through L: I L (0) =

ε

R2

. When the switch is opened, the current decays

exponentially from this initial value. So for t ≥ 0, ε −t / τ L IL = e , where τ = because the two resistors are connected in series. R1 + R2 R2 The voltage across each resistor is given by Ohm’s Law: εR V1 = IR1 = 1 e − t / τ R2 V2 = IR2 = ε e − t / τ 31-67.

(a) When the switch is in the upper position, L is in series with R2: L τ upper = R2 (b) When the switch is in the lower position, L is in series with both R1 and R2: L τ lower = R1 + R2

31-68.

I = I 0 e−t / τ I = e−t / τ I0 t

τ

= − ln

τ =−

I I0

t 0.060 s =− = 0.0866 s ln( I / I 0 ) ln(1/ 2)


CHAPTER Thus L = Rτ = 0.866 H 31-69.

From Kirchhoff’s rule, di (1) ε = L1 1 – iR = 0 dt di (2) ε – L2 2 – iR = 0 dt Multiply (1) by L2 and (2) by L1, add the resulting equations to get

( L2 + L1 ) ε – L1L2

L L di d – iR = 0 ( i1 + i2 ) – iR ( L1 + L2 ) = 0 or ε – 1 2 dt L2 + L1 dt

di – iR = 0 dt 1 1 1 + and i = i1 + i2 where = L L1 L2 ε–L

(a)

di dt

= I0 t =0

d (1 − e − tR / L ) dt

t=0

=

ε L

=

3 = 2.25 A/s 4/3

di1 L2 di2 2 di2 1 di2 = = = dt 4 dt 2 dt L1 dt di1 di2 di = = 1.5 2 = 2.25 A/s dt dt dt di di 2.25 Therefore 2 = = 1.5 A/s and 1 = 2.25 – 1.5 = 0.75 A/s dt dt 1.5 ε 3 (b) Final steady current in R = = = 0.5 A R 6 Final steady-state current through L = 0.5 A. The steady-state current through L1 and L2 obey d d ( L1i1 ) = (l2i2 ) = 0 or L1i1 = i2L2 = constant dt dt Since i1 = 0 if i2 = 0, the constant term must be zero. Thus i1 L 1 = 2= i2 L1 2 2i1 = i2 Since i1 + i2 = 0.5 A, 3i1 = 0.5 A 0.5 = 0.17 A and i2 = 0.33 A Thus i1 = 3 31-70.

Joule heat =

=

ε 2L 2R

2

=

∫

∞

0

I 2 dt =

ε2 R

∫

∞

0

e −2tR / L dt =

∞

ε2

⎛ L ⎞ e −2tR / L ⎜ − ⎟ R ⎝ 2R ⎠ 0

1 2 L I 0 = The initial magnetic energy stored in L 2


31

CHAPTER

†31-71.

31-72. 31-73.

31-74.

31 ε

3 = 0.125 Ω I 24 From the short-circuiting experiment, I (t ) 12 (0.22) I (t ) = I (0)e − t / τ ⇒ s = 0.317 s. = e−t / τ ⇒ = e − (0.22 / τ ) ⇒ τ = − I (0) 24 ln(0.5) L τ = ⇒ L = τ R = (0.317 s)(0.125 Ω) = 0.0396 H R ε = lvB = (20 × 103 m)(7.7 × 103 m/s)(2.7 × 10−5 T) = 4.2 × 103 V

Internal resistance R =

=

dΦB d = − ( BA cos ω t ) = ω BA sin ω t. Here A is the dt dt area of a semicircle of radius r and the angle ωt is measured from the vertical when the wire is horizontal. Then the amplitude of the emf is ε 0 = ω BA = (2π )(1.0 rev/s)(5.0 × 10−5 T)(π )(0.70 m) 2 /2 = 2.4 × 10−4 V

ε =−

r

31-75.

ε = BAω sin ωt (12); therefore, amplitude of the voltage is BAω Ampl ω Ampl ⇒ f= BAω = Ampl ⇒ ω = BA 2π BA 1 12.0 V 2π (2.0 × 10−2 T) (0.02 m 2 ) 4.8 × 103 s −1 = = = 40 Hz 120 120 Equation. 29.26 gives the magnitude of the magnetic field at the center of a circular loop of radius a: µI B= 0 2a The energy density is 2 ⎛ 1 ⎞ ⎛ µ0 I ⎞ µ 0 I 2 (4π × 10−7 T i m/A)(3.0 A) 2 B2 u= =⎜ = = = 0.014 J/m3 ⎟⎜ ⎟ 2 2 2 µ 0 ⎝ 2 µ 0 ⎠ ⎝ 2a ⎠ 8a 8(0.010 m)

31-76.

Average radius r =

0.40 + 1.50 m = .95 m. The circular cross section has a diameter D = 1.10 2 m, giving a cross sectional area A = 0.950 m2. Then the volume is V = 2π rA = (2π )(0.95 m)(0.950 m 2 ) = 5.67 m3 . U = uV =

B 2V (4.0 T) 2 (5.67 m3 ) = = 3.6 × 107 J 2µ0 2(4π × 10−7 T i m/A)


CHAPTER †31-77.

31-78.

The only component of B that matters is the component perpendicular to the velocity of the shark. When traveling at v = 25 km/h = 6.94 m/s north, the component of B that is perpendicular to v is B cos 40°. Then ε = vlB⊥ = (6.94 m/s)(0.30 m)(4.7 × 10−5 T)(cos 40°) = 7.5 × 10−5 V.

Force on the rod = I ∫ dl i B) = B0Il, since B is

ε , and by R dΦ dB dB = – x0l ; = – (Area) Faraday’s law ε = – dt dt dt therefore, dB x0l ε dt I= =– R R Therefore, acceleration a = F/m B Il B l ⎛ dB 1 ⎞ a = 0 = 0 ⎜ − x0 l ⎟ m m ⎝ dt R ⎠ B l 2 x dB =− 0 0 mR dt (Minus sign indicates that force is in direction shown.) perpendicular to l, and constant, and I =

31-79.

31

ε = vBl 5 ms–1 = 16.7 ms–1 18 (a) ε = 16.7 × 0.62 × 10–4 × 2.5 V = 2.6 × 10−3 V v = 60 km/h = 60 ×

(b) E due to induced emf = 2.6 × 10–3/2.5 = 1.0 × 10–3 v/m. This field is offset by accumulation of static charge on the walls of the boxcar. Therefore, net field = 0. (c) E =

σ ε0

; therefore,

σ = ε0E = 8.85 × 10–12

N c2 1.0 × 10–3 = 9.2 × 10−15 c/m 2 2 c Nm


CHAPTER

31-80.

†31-81.

31-82.

†31-83.

I =

31 ε

=

NA dB / dt

R R (5.0 × 10−3 A)(15 Ω) dB IR = = = 0.75 T/s dt NA (25)(4.0 × 10−3 m3 ) dB dI (a) = µ0 n = (4π × 10−7 T i m/A)(300 m −1 )(50 A/s) = 0.0188 T/s dt dt (b) The total flux in the secondary coil is Φ B1 = N 2 BA1 , where N2 is the number of turns in the secondary coil and A1 is the area of the primary solenoid. The actual area of the secondary doesn’t matter because the field of the primary solenoid is essentially zero outside of it. B is the magnitude of the field inside the primary solenoid. Thus the induced emf is d Φ2 dB ε2 = = N 2 A1 1 = (120)(π )(0.030 m) 2 (0.0188 T/s) = 6.4 × 10−3 V dt dt (c) Because the total flux in the secondary coil only depends on the cross sectional area of the primary solenoid, changing the radius of the secondary coil will not change the value of the induced emf. The result is still 6.4 × 10-3 V. Vapplied − ε back 115 V − ε back = = 0.10 A. To three significant figures, this gives 2.0 Ω R ε back = 115 V. The resistance of the coil in the motor is found by measuring the voltage and current when the motor isn’t turning: Vapplied 115 V Rcoil = = = 3.19 Ω 36 A I stationary Then the current when the motor is running is Vapplied − ε back I run = Rcoil

ε back = Vapplied − I run Rcoil = 115 V − (4 A)(3.19 Ω) = 102 V 31-84.

µ 0 NI

(4π × 10−7 T i m/A)(1500)(5.0 A) = 0.0236 T l 0.40 m B2 (.0236 T) 2 (b) u = = = 221 J/m3 2µ 0 2(4π × 10−7 T i m/A)

(a) B =

=

(c) U = uV = u (π r 2l ) = (221 J/m3 )(π )(0.015 m) 2 (0.40 m) = 0.0624 J †31-85.

(a)

The maximum current is ε 1.2 V I max = = = 48 A R 0.025 Ω The time constant for the circuit is L 0.50 H τ = = = 20 s R 0.025 Ω

The time for the current to decay to half its maximum value is given by


CHAPTER I (t ) = I max e − t / τ ⇒ e − t / τ =

31-86.

I I max

=

1 2

t = −τ ln(1/ 2) = −(20 s)(−0.693) = 14 s LI 2 (0.50 H)(48 A) 2 = 576 J (b) U 0 = max = 2 2 When I = Imax/2, U = U0/4, so 25% of the energy remains. µI µ ⎛ dI ⎞ dB (a) At a distance r, B = 0 ⇒ = 0 ⎜ ⎟. 2π r 2π r ⎝ dt ⎠ dt (b) The total flux through the loop is Φ B = Bal , so the induced emf is

ε =−

µ al ⎛ dI dΦB dB = −al =− 0 ⎜ 2π r ⎝ dt dt dt

⎞ ⎟. ⎠


31

CHAPTER 32

ALTERNATING CURRENT CIRCUITS

Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions Manual, available for purchase. Answers to all solutions below are underscored.

32-1.

(a) P =

2 ε RMS

R

= ε RMS I RMS ; therefore, I RMS =

ε RMS

=

1200 W = 10.4 A 115 V

2 ε RMS = ε max );

2 I RMS = I max (same as

And

P

therefore, I max = 2 I RMS = 14.8 A (b) Power = ε (t ) I (t ) = ε max cos ω t

ε max cos ω t R

=

2 ε max

R

cos 2 ω t

2 2ε RMS cos 2 ω t = 2 P cos 2 ω t, where P = average power = 1200 W. Therefore R Max power when cos2 ωt = 1 ⇒ Power = 2400 W Min power when cos2 ωt = 0 ⇒ Power = 0 W

=

32-2.

PAC =

2 ε RMS

R

; therefore R =

2 ε RMS

PAC

where R = resistance, PAC = average power on AC. Then for

V2 V2 = 2 PAC . But V = 115 V and εRMS = 115 V; therefore, ε RMS R

DC voltage of V, Power PDC =

PDC = PAC . I.e., power consumed = 400 W. †32-3.

(a) Max instant voltage ε max =

Max instant current I max =

2ε rms =

2 I rms =

2 × (230 × 103 V) = 3.25 × 105 V

2 × 740 A = 1.05 × 103 A

(b) Max power = ε max I max = (325 × 103 V)(1.05 × 103 A) = 3.41 × 108 W

Average power = ε rms I rms = (230 × 103 V)(740 A) = 1.70 × 108 W 32-4.

32-5.

P = 4600 hp × 746 W/hp = 3.43 × 106 W P 3.43 × 106 W (a) I RMS = = = 3.1 × 103 A ( RMS value) ε RMS 1100 V (b) High voltage implies low current, which keeps Joule heating in wires (hence energy loss) to a minimum (remember P = I2R). P 1200 W = = 10.4 A (a) I rms = ε rms 115 V (b) I max = 2 I rms = 14.8 A P (c) R = 2 = 11.1 Ω I rms

32-6.

P=

2 ε rms

I rms =

R

⇒ ε rms =

P

ε rms

=

PR = (50 Ω)(2.0 W) = 10 V

2.0 W = 0.20 A 10 V


CHAPTER ε max

32

2.2 kV = 1.56 kV 2 2 ε 1.56 × 103 V I rms = rms = = 31.1 A R 50 Ω P = I ε = (31.1 A)(1.56 × 103 V) = 48.3 × 103 W or 48.3 kW

32-7.

ε rms =

32-8.

ε rms =

ε rms 2

=

=

ω BA

2 ε ω B A2 4π 2 f 2 B 2 A2 = = P= 2R 2R R (2)(1.0 W)(11 Ω) 2 PR = = 2.5 rev/s f = 2π BA (2π )(2.0 T)(0.15 m 2 ) 2 rms

2

2

To get 100 W, the rotation rate must be increased by a factor of 100 to 25 rev/s. †32-9.

I rms =

ε rms XC

= ω Cε rms

Cε rms = (2.2 × 10−6 F)(12 V) = 2.64 × 10−5 F i V. At ω = 1.2 × 103 rad/s,

32-10.

†32-11.

Irms = ωCεrms = (1.2 × 103 rad/s)( 2.64 × 10–5 FV) = 0.032 A. At ω = 2.4 × 103 rad/s, the current will double to 0.064 A. 1 1 XC = = = 1.1 × 103 Ω. At ω = 18 × 103 rad/s, the reactance 3 ω C (6.0 × 10 rad/s)(0.15 × 10−6 F) will decrease by a factor of 3 to 3.7 × 102 Ω. X ω ω f 1 XC = ⇒ C ,1 = 2 ⇒ X C ,2 = X C ,1 1 = X C ,1 1 . Given: XC,1 = 3.0 × 106 Ω at f1 = 2.0 kHz. ωC ω2 X C ,2 ω1 f2 Then at f2 = 3.0 kHz,

32-12. †32-13.

⎛ 2.0 ⎞ 6 X C ,2 = (3.0 × 106 Ω) ⎜ ⎟ = 2.0 × 10 Ω. ⎝ 3.0 ⎠ At f2 = 4.0 kHz, ⎛ 2.0 ⎞ 6 X C ,2 = (3.0 × 106 Ω) ⎜ ⎟ = 1.5 × 10 Ω. 4.0 ⎝ ⎠ 1 1 1 XC = ⇒ f = = = 1.3 × 103 (1.3 kHz) 5 2π fC 2π X C C 2π (2.0 × 10 Ω)(600 × 10−12 F) 1 1 (a) X C = = = 208 Ω 3 ω C (6.0 × 10 rad/s)(8.0 × 10−7 F) ε 0.20 V = 9.62 × 10−4 A (b) I max = max = 208 Ω XC (c) I = I max cos(ω t + π / 2). At t = 0, I = I max cos(π / 2) = 0. At t = π/(4ω),

I = I max cos(π / 4 + π / 2) = I max cos(3π / 4) = −6.80 × 10−4 A 32-14.

The current leads the voltage by 90º or π/2 radians, so when the voltage is a maximum the current is zero. Thus when ε = 12 V, I = 0. One quarter of a cycle later, the current will have its ε 1 1 maximum value I = max . X C = = = 833 Ω. Then 3 XC ω C (3.0 × 10 rad/s)(0.40 × 10−6 F)


CHAPTER

32

12 V = 0.0144 A. One half cycle after the voltage is a maximum, the current will again 833 Ω have a value of zero. Three quarters of a cycle after the voltage is a maximum, the current will have its maximum value but it will have the opposite sign: I = −0.0144 A. I =

32-15.

ε max

I max =

XC

⇒ XC =

ε max I max

=

4.0 × 10−3 V = 5.0 × 105 Ω 8.0 × 10−9 A

1 1 = = 1.0 × 10−9 F 3 ω X C (2.0 × 10 rad/s)(5.0 × 105 Ω) 1 1 XC = = = 1.25 × 107 Ω. 4 ω C (4.0 × 10 rad/s)(2.0 × 10−12 F)

C= 32-16.

ε max = I max X = (5.0 × 10−5 A)(1.25 × 108 Ω) = 625 V 32-17.

XC = P=

32-18.

1 1 1 = = Ω. The instantaneous power is 2 −6 ω C (6.0 × 10 rad/s)(2.0 × 10 F) 0.0012

ε2 XC

2 ε max cos 2 (ω t )

XC

2 = (0.0012)ε max cos 2 (600t ). A spreadsheet will be used to do the

repetitive calculations assuming an amplitude of 1.00 V. (If a calculator is used, make sure it’s in radian mode.) t (s) 0.0012cos2(600t) (W) 0.001 8.17E–04 0.002 1.58E–04 0.003 6.19E–05 0.004 6.52E–04 1 1 X empty = = = 5.56 × 104 Ω ω Cempty (6.0 × 104 rad/s)(0.30 × 10−9 F)

X full = I empty = I full = †32-19.

=

1 1 = = 3.88 × 104 Ω −9 4 ω C full (6.0 × 10 rad/s)(0.43 × 10 F)

ε X empty

ε X full

=

=

15 V = 2.7 × 10−4 A 5.56 × 104 Ω

15 V = 3.9 × 10−4 A 3.88 × 104 Ω

ω = 2πf = 2π(60) = 377 rad/s The capacitor and resistor have the same voltage across them because they are in parallel. The current through the resistor is I R =

ε R

current through the capacitor is I C = ⇒ XC = R ⇒

1

ωC

=R⇒C=

and the

1

ωR

ε XC =

6.2 V 60 Hz

C

47 Ω

. IR = IC

1 = 5.6 × 10−5 F. Note that the actual (377 rad/s)(47 Ω)

amplitude of the voltage does not matter.


CHAPTER 32-20.

Ceq = 3(0.80 µF) = 2.40 µF. 1 1 X Ceq = = = 231 Ω 3 ω Ceq (1.8 × 10 rad/s)(2.4 × 10−6 F)

ε rms

I rms =

X Ceq

=

24 V = 0.10 A 231 Ω

dI dI ε 25 V ⇒ = = = 1.0 × 105 A/s dt dt L 250 × 10−6 H

32-21.

ε =L

32-22.

XL = ωL X L = (6.0 × 103 rad/s)(0.25 H) = 1.5 × 103 Ω X L = (1.8 × 104 rad/s)(0.25 H) = 4.5 × 103 Ω XL 2.0 × 103 Ω = = 1.6 × 103 Hz 2π L 2π (0.20 H)

32-23.

X L = 2π fL ⇒ f =

32-24.

ω = 2πf = 2π(60) = 377 rad/s. 1 1 XC = = = 5.6 × 104 Ω ω C (377 rad/s)(4.7 × 10−8 F) X L = ω L = (377 rad/s)(6.6 × 10−4 H) = 0.25 Ω Since R = 7.3 × 10–2 Ω, the ranking is X C > X L > R.

32-25.

I =

ε (ε / L) 40 = = A ωL ω ω

40 = 0.033 A 1.2 × 103 rad/s 40 I = = 0.017 A 2.4 × 103 rad/s εmax = ImaxXL. XL = ωL = (4.0 × 104 rad/s)(0.01 H) = 400 Ω. εmax = (5.0 × 10–5 A)( 400 Ω) = 0.020 V Assume f = 60 Hz so ω = 2πf = 2π(60) = 377 rad/s. Then X L = ω L = (377 rad/s)(6.2 H) = 2.33 × 103 Ω I =

32-26. 32-27.

I = 32-28.

XL

XL = L=

†32-29.

ε

=

ε max I max

XL

ω

=

115 V = 0.049 A 2.33 × 103 Ω =

8.0 × 10−3 V = 4.0 × 104 Ω −7 2.0 × 10 A

4.0 × 104 Ω = 13 H 3.0 × 103 rad/s

ε dI 12 V ⎛ dI ⎞ ⇒ ⎜ ⎟ = max = = 0.343 A/s dt L 35 H ⎝ dt ⎠ max ⎛ dI ⎞ ε max = I max X L ⇒ L ⎜ ⎟ = I max X L = 2π fLI max ⎝ dt ⎠ max (dI / dt ) max 0.343 A/s = = 0.00218 Hz f = 2π I max 2π (25 A)

V =L


32

CHAPTER

32-30.

32

⎛ dI ⎞ 6 −9 ⎜ ⎟ = 2π fI max = 2π (1.2 × 10 Hz)(5.2 × 10 A) = 0.0392 A/s dt ⎝ ⎠ max ⎛ dI ⎞ ε max = L ⎜ ⎟ = I max X L = (0.0392 A/s)(0.90 × 10−3 H) = 3.5 × 10−5 V ⎝ dt ⎠ max

(See Problem 29.) An alternative solution is X L = ω L = 2π fL = 2π (1.2 × 106 Hz)(0.9 × 10−3 H) = 6.79 × 103 Ω

ε max = I max X L = (5.2 × 10−9 A)(6.79 × 103 Ω) = 3.5 × 10−5 V 32-31.

X L = 2π fL = 2π (47 × 10−6 H) f = (2.95 × 10−4 ) f Ω

At 60 Hz, X L = (2.95 × 10−4 Ω i s)(60 Hz) = 0.018 Ω At 100 MHz, X L = (2.95 × 10−4 Ω i s)(100 × 106 Hz) = 3.0 × 104 Ω 32-32.

(a) XL = ωL = 6.0 × 103 s–1 × 4.0 × 10–2 H = 240 Ω ε 0.20 V (b) Max current I max = max = = 8.33 × 10−4 A XL 240 Ω ε sin ω t (c) By Eq. 32.19, I (t ) = max = 8.33 × 10−4 sin ω t ωL At t = 0, I(t) = 0 At t =

32-33.

π π , I (t ) = 8.33 × 10−4 sin = 5.89 × 10−4 A 4ω 4

X L = 2π fL, X C =

1 2π fC

X L = 3 X C ⇒ 2π fL = f = 32-34.

1 2π

3 1 = LC 2π

3 2π fC 3 = 3.6 × 103 Hz (3.0 × 10 H)(2.0 × 10−6 F) −3

Because they are connected in parallel, the voltage across both L and C must be ε0 cos ωt. Therefore, by Kirchhoff’s Law, dI dI ε (t ) ε (t ) − L L = 0 ⇒ L = dt dt L ε ε 1 ⇒ I L (t ) = ∫ ε (t )dt = max ∫ cos ω t dt = max sin ω t (through L) ωL L L Q dQ dε = ε (t ) ⇒ = I C (t ) = C = −ω Cε max sin ω t (through C ) C dt dt Instantaneous power to L: 2 ε2 ⎛ε ⎞ ε PL = ε (t ) I L (t ) = ε max cos ω t ⎜ max sin ω t ⎟ = max cos ω t sin ω t = max sin 2ω t 2ω L ⎝ ωL ⎠ ωL Instantaneous power to C: ε 2 ωC 2 PC = ε (t ) I C (t ) = ε max cos ω t ( −ω Cε max sin ω t ) = −ε max ω C cos ω t sin ω t = max sin 2ω t 2 Total power from source:


CHAPTER

32

2 ⎛ 1 ⎞ ε sin 2ω t − ω C ⎟ max P = PL + PC = ⎜ 2 ⎝ ωL ⎠

†32-35.

(a) The amplitude of the voltage remains the same, so the amplitude of the current through the resistor is unchanged. I max = 5.0 × 10−6 A. 1 and XL = ωL, decreasing ω by a factor of 2 requires doubling C and L to ωC keep the reactances equal to the resistance. Thus C = 2 × (8.3 × 10–7 F) = 1.66 × 10–6 F L = 2 × (3.3 × 10–2 H) = 6.6 × 10–2 H L and C have been chosen to make the reactances equal to the resistance. The currents through L and C then have equal magnitudes and are 180º out of phase with each other, so they cancel. The total current in the circuit is just the current through the resistor: 1.0 × 10−3 V ε ε I = = max cos ω t = cos ω t = (5.0 × 10−6 A) cos ω t R R 200 Ω At t = 0, I = (5.0 × 10−6 A) cos 0 = 5.0 × 10−6 A (b) Since X C =

32-36.

π π , I = (5.0 × 10−6 A) cos = 0 2ω 2 π At t = , I = (5.0 × 10−6 A) cos π = −5.0 × 10−6 A ω At t =

32-37.

IC = I L ⇒ X C = X L ⇒

ω= 32-38.

1 = ωL ⇒ ω = ωC

1 −6

(0.045 H)(0.25 × 10 F)

1

LC

= 9.4 × 103 rad/s

(a) Eq. 31.23 gives the inductance per unit length of a long solenoid in terms of the number of turns per unit length (n). Here we have the total number of turns (N) and the length. Using n =

N l

µ N2A L = µ0 ( N / l )2 A ⇒ L = 0 . For a circular cross section, the inductance is l l πµ N 2 r 2 π (4π × 10−7 T i m/A)(103 ) 2 (2.0 × 10−3 m) 2 = = 3.95 × 10−4 H L= 0 0.040 m l ε 1.2 × 10−4 V (b) XL = ωL = (9.0 × 103 rad/s)(3.95 × 10–4 H) = 3.55 Ω. I = = = 3.38 × 10−5 A XL 3.55 Ω gives

LI 2 (3.95 × 10−4 H)(3.38 × 10−5 A) 2 = = 2.25 × 10−13 J 2 2 1 1 By Eq. 32.26, ω 0 = ⇒ f0 = Hz = 379 Hz −6 LC 2π 2.2 × 10 × 8.0 × 10−2 (c) U =

32-39. 32-40. 32-41.

1 1 = = 2.0 × 10−3 H 2 2 3 4π f C 4π (8.0 × 10 Hz) 2 (0.20 × 10−6 F) 2π LC 1 1 = Hz = 663 Hz Capacitors in parallel add, so f 0 = −3 2π L(2C ) 2π 1.2 × 10 × 2 × 2.4 × 10−5 f =

1

⇒L=

2


CHAPTER

32-42.

32 1

ω=

LC

Cmax =

1 2 4π Lf min 2

1 1 1 = = 6.6 × 10−10 F 2 2 −6 4π L f max 4π (15 × 10 ) (1600 × 103 ) 2 Q = Qmax cos ω t ⇒ I = −ω Qmax sin ω t I max = ω Qmax = ω Cε max = 2π fCε max

Cmin = 32-43.

1 1 1 ⇒C= 2 = 2 2 ω L 4π f L LC 1 1 = = 6.0 × 10−19 F 2 −6 4π (15 × 10 ) (530 × 103 ) 2

; therefore, ω 2 =

2 2

I max 3.5 × 10−5 A = = 2.11 Hz 2π Cε max 2π (22 × 10−6 F)(12 V)

f =

For the inductor, I max = 32-44.

[Quality factor] =

ω0 =

1 LC

=

ε max ε max 12 V ⇒L= = = 259 H 2π fL 2π fI max 2π (2.11 Hz)(3.5 × 10−3 A)

ω0 L R

1 −6

−9

(25 × 10 H)(100 × 10 F)

= 6.32 × 105 rad/s

(6.32 × 105 rad/s)(25 × 10−6 H) = 21 0.75 Ω [Quality factor] Number of oscillations = N = =6

[Quality factor]=

†32-45.

[Quality factor] =

ω0 L

π

R 1 1 ω 02 = ⇒L= 2 ω0 C LC

[Quality factor] = C=

1 ω 0 RC

1 1 1 = = 6 ω 0 R × [Quality factor] 2π fR × [Quality factor] 2π (3.5 × 10 Hz)(2.2 × 10−3 Ω)(750)

= 2.76 × 10−8 F L= 32-46.

1 1 = = 7.50 × 10−8 H 2 2 6 4π f C 4π (3.5 × 10 Hz) 2 (2.76 × 10−8 F) 2

(a) When the switch is opened, the capacitance in the circuit changes to the series equivalent of C1 and C2. This is less than C1, so the resonant frequency will increase.


CHAPTER

2 1.5 1 0.5 -1

0 -0.5 0

-0.5

0.5

1

1.5

2

2.5

3

-1 -1.5 -2

(b) When the switch is closed, the resonant frequency will return to its original value. 2 1.5 1 0.5 0 -1

-0.5 0 -1

1

2

3

4

-1.5 -2

†32-47.

I max =

ε max Z

⇒Z =

ε max I max

R 2 + X C2 ⇒ X C =

Z =

=

163 V = 326 Ω. 0.50 A Z 2 − R 2 = (326 Ω) 2 − (50 Ω) 2 = 322 Ω

ω = 2π f = 377 rad/s C=

1

ω XC

=

1 = 8.23 × 10−6 F (377 rad/s)(322 Ω)

2 I max R (0.50 A) 2 (50 Ω) = = 6.25 W 2 2 ε 115 V Z = rms = = 46.0 Ω I rms 2.5 A

P=

32-48.

XL =

Z 2 − R 2 = (46.0 Ω) 2 − (16 Ω) 2 = 43.1 Ω

ω = 2π f = 377 rad/s 43.1 Ω = 0.114 H 377 rad/s ω I 2 R I (ε / Z ) R 1 ⎛R⎞ (a) P = max = max max = I max ε max ⎜ ⎟ 2 2 2 ⎝Z⎠ (b) Assume the reactance is inductive. From the impedance triangle, R cos φ = Z 1 P = I max ε max cos φ 2 L=

†32-49.

XL

=


Z

φ R

XL

32

CHAPTER

32

(c) cos φ =

2P 2(750 W) = = 0.642 I max ε max (14 A)(167 V)

φ = cos −1 (0.642) = ±0.874 radian = ±50° Even though the answer was derived assuming there is an inductor in the circuit, it is not possible to tell from the information given whether the circuit contains a capacitor or an inductor or both because cos φ = cos(−φ ). The same answer would have come from assuming a capacitor or a 32-50.

combination of capacitance and inductance in the circuit. 1 1 (a) ω 0 = = = 8.23 × 103 rad/s −3 −6 LC (36 × 10 H)(0.40 × 10 F)

ε

100 V = 19.2 A. R 5.2 Ω X L = ω L = (8.23 × 103 rad/s)(36 × 10−3 H) = 296 Ω

(b) At resonance Z = R, so I =

=

ε L = IX L = 5.68 × 103 V (c) The impedance of the RC part of the circuit is Z RC =

R 2 + X C2 = (5.2) 2 + (296) 2 Ω = 296 Ω , where we have used XC = XL at resonance.

Thus ε12 = 5.68 × 103 V. (d) At ω = 4.7 × 105 rad/s, XC = 5.19 Ω, XL = 1.69 × 104 Ω. The total impedance of the circuit is now Z =

R 2 + ( X L − X C ) 2 = 1.69 × 104 Ω , and I =

now is Z RC = †32-51.

ε

Z

= 5.91 × 10−3 A. The RC impedance

R 2 + X C2 = (5.2) 2 + (5.19) 2 Ω = 7.35 Ω. ε12 = IZ RC = 4.4 × 10−2 V.

(a) At resonance, XC = XL ⇒ Z = R. Then the current through the circuit is ε 12 V I max = max = = 4.0 A. ε L = ε C = IX C = (4.0 A)(468 Ω) = 1.9 × 103 V R 3.0 Ω (b) The voltages across C and L cancel because they are 180º out of phase with each other. Then Kirchhoff’s voltage law requires ε R = ε max = 12 V. 2 I max R (4.0 A) 2 (3 Ω) = = 24 W 2 2 ω L X L X C 468 (d) [quality factor] = = = = = 1.6 × 102 3 R R R ε ε max ω Cε max I max = max = = . 2 2 Z ω 1 + ( CR ) 1 ⎛ ⎞ R2 + ⎜ ⎟ ⎝ ωC ⎠ ω CRε max . ε R , max = IR = 1 + (ω CR) 2

(c) P =

32-52.

When ωRC = 1/10, (1/10)ε max ε = max = ε max /10 ε R , max = 101 1 + (1/10) 2 When ωRC = 10, ε R , max =

10ε max 1 + 102

= 0.995ε max .


CHAPTER ε R , max = ⇒ω =

†32-53.

I max =

ε max

1

⇒

2

=

2

ω CR 1 + (ω CR )

1 RC

ε max Z

ε max

=

ε R , max = IR =

R + (ω L ) 2

1 + (ω L / R ) 2

ε max

ε R , max =

2

Rε max

When ωL/R = 1/10, 1 + (1/10)

2

=

⇒ω = 32-54.

ε max 2

⇒

1 2

=

ε max

=

R 1 + (ω L / R ) 2

.

.

10ε max 101

When ωL/R = 10, ε R , max =

ε R , max =

⇒ 1 + (ω CR) 2 = 2(ω CR ) 2

2

= 0.995ε max

ε max 1 + (10)

=

2

1 1 + (ω L / R)

2

ε max 101

= ε max /10

⇒ 1 + (ω L / R) 2 = 2(ω L / R ) 2

R L

At resonance, Z = R and the maximum power is Pmaximum =

2 ε max

I max ε max cos φ . 2 ε R I max = max , cos φ = Z Z 2 2 ε R ε max R ⇒ P = max 2 = 2 2Z 2[ R + ( X L − X C ) 2 ]

2R

. At any frequency the average

power is P =

P = Pmaximum/2 means 2 2 ε max R ε max = 2[ R 2 + ( X L − X C ) 2 ] 4R

2 R 2 = R 2 + ( X L − X C )2 ⇒ X L − XC = ±R Below resonance, XL < XC, so to find the lower half power frequency we must solve X L − X C = − R. Above resonance, XL > XC, so to find the upper half power frequency we must solve X L − X C = + R. Lower frequency:


32

CHAPTER

32 1 = − R ⇒ ω12 LC + ω1 RC − 1 = 0 ω1C

ω1 L −

− RC ± ( RC ) 2 + 4 LC 2 LC Choose the + sign to get a positive frequency:

ω1 =

− RC + ( RC ) 2 + 4 LC 2 LC Upper frequency: 1 ω2 L − = R ⇒ ω 22 LC − ω 2 RC − 1 = 0 ω 2C

ω1 =

RC ± ( RC ) 2 + 4 LC 2 LC Again choose the + sign to get a positive frequency:

ω2 =

ω2 =

RC + ( RC ) 2 + 4 LC 2 LC

The bandwidth is ω 2 − ω1 =

RC + ( RC ) 2 + 4 LC − RC + ( RC ) 2 + 4 LC R − = 2 LC 2 LC L

†32-55.

IC

IR

IL 2

2

I max = I max =

ε ⎞ ⎛ε ⎞ ⎛ε I R2 + ( I C − I L ) 2 = ⎜ max ⎟ + ⎜ max − max ⎟ . XL ⎠ ⎝ R ⎠ ⎝ XC ε max Z

⇒ 2

2

1 1 ⎞ ⎛1⎞ ⎛ 1 = ⎜ ⎟ +⎜ − ⎟ ⇒ Z ⎝ R ⎠ ⎝ XC X L ⎠ 2 ⎡ ⎛ 1 ⎞2 ⎛ 1 1 ⎞ ⎤ Z = ⎢⎜ ⎟ + ⎜ − ⎟ ⎥ ⎢⎣ ⎝ R ⎠ ⎝ X C X L ⎠ ⎥⎦

−1 / 2


CHAPTER

32-56.

V 1 1 15 V − = 0 ⇒ Z = R. Then I max = max = = 3.0 A. 5.0 Ω R XC X L ω 1 1 = = 1.15 × 106 rad/s ⇒ 0 = 5.77 × 105 rad/s. −6 −9 2 LC (30 × 10 H)(25 × 10 F)

(a) At resonance, (b) ω 0 =

X L = ω L = 17.3 Ω, X C =

1 = 69.3 Ω ωC −1 / 2

32-57. 32-58. †32-59.

32-60.

32-61.

−1 / 2

2 2 ⎡ ⎛ 1 ⎞2 ⎛ 1 ⎡⎛ 1 ⎞2 ⎛ 1 1 ⎞ ⎤ 1 ⎞ ⎤ Z = ⎢⎜ ⎟ + ⎜ − = ⎢⎜ ⎟ + ⎜ − Ω = 12.6 Ω ⎟ ⎥ ⎟ ⎥ ⎢⎣ ⎝ R ⎠ ⎝ X C X L ⎠ ⎥⎦ ⎣⎢ ⎝ 15 ⎠ ⎝ 69.3 17.3 ⎠ ⎦⎥ V 15 V I max = max = = 1.2 A Z 12.6 Ω N2 ε 2 ε 5000 = ; therefore, N 2 = 2 N1 = × 100 turns = 4348 turns ε1 N1 ε1 115

N2 ⎛ 80 ⎞ = (115 V) ⎜ ⎟ = 7.67 V N1 ⎝ 1200 ⎠ N 22 = 0.044 Transformer 1: 22 kV to 500 kV 1 = N 2 500 N 500 Transformer 2: 500 kV to 66 kV 1 = = 7.6 N2 66 N 66 = 17 Transformer 3: 66 kV to 4.0 kV 1 = N 2 4.0 N 4000 Transformer 4: 4.0 kV to 115 V 1 = = 35 N2 115 N 2 ε 2 1200 = = = 100 N1 ε1 12 I2 3.0 A I2 = = = 0.030 A N 2 / N1 100

ε 2 = ε1

Primary power = secondary power. P = I1ε1 ⇒ I1 = I2 =

P

ε2

P

ε1

=

2000 × 106 W = 9.09 × 104 A 3 22 × 10 V

6

=

2000 × 10 W = 5.00 × 103 A 400 × 103 V

32-62.

Since no power is lost, ε1 I1 = ε 2 I 2 ⇒ I1 =

32-63.

ε 2 = ε1

32-64.

I1 =

ε2 6V I2 = × 3.0 A = 0.16 A ε1 115 V

N2 ⎛ 1 ⎞ = (120 V) ⎜ ⎟ = 12 V N1 ⎝ 10 ⎠ P = I 2ε 2 = (3.0 A)(12 V) = 36 W

I2 =

P

ε1 P

ε2

=

1.50 × 109 W = 1.96 × 103 A 3 765 × 10 V

=

1.50 × 109 W = 4.35 × 103 A 3 345 × 10 V


32

CHAPTER

32-65. 32-66.

†32-67.

Total power = 16.0 W. I1 =

†32-69.

P

=

ε1

16.0 W = 0.139 A 115 V

N 2,max

⎛ 250 ⎞ = (115 V) ⎜ ⎟ = 144 V. The smallest change in N2 is one turn, which is N1 ⎝ 200 ⎠ 144 V 1/250 of the total. So ∆ε min = = 0.58 V. 250

ε 2,max = ε1

For a resistor, the voltage and current are in phase with each other. Thus I =

ε max

ε R

at all times.

163 V = 1.63 A. In a sinusoidal wave form, a zero comes one quarter cycle after a R 100 Ω maximum or minimum, and a maximum (minimum) is followed one-half cycle later by a minimum (maximum). Thus one quarter cycle after the maximum of 1.63 A, ε = 0, so I = 0. One half cycle later, ε = –163 V, so I = –1.63 A. Note that no trigonometric calculations are needed for this problem. ε A (8.85 × 10−12 C2 /N i m 2 )(0.30 m 2 ) = 1.33 × 10−7 F (a) C = 0 = d 2.0 × 10−5 m 1 1 (b) X C = = = 837 Ω ω C (9.0 × 103 rad/s)(1.33 × 10−7 F) ε 12 V I = = = 0.014 A X C 837 Ω I max =

32-68.

32

=

(c) Q = CV = (1.33 × 10–7 F)(12 V) = 1.6 × 10–6 C CV 2 (1.33 × 10−7 F)(12 V) 2 (d) U = = = 9.6 × 10−6 J 2 2 (a) Flux through loop = φ = BA = (0.15)(.04) sin ωt = 0.006 sin 360t. dφ Induced emf: ε = − = −(360)(0.006) cos 360t = −2.16 cos 360t V. The induced emf is added dt to the battery voltage, so Kirchhoff’s Voltage Law gives V + ε = IR ⇒ 12 − 2.16 cos 360t = 5 I (t ). The instantaneous current is I(t) = (2.4 – 0.432 cos 360 t) A.

(b) The average power is P = I 2 R , where I 2 is the average of the square of the current. 2π / ω 1 T 1 2 I 2 = ∫ I 2 (t ) dt = ( 2.4 − 2.16 cos ω t ) dt. When the integrand is expanded and ∫ 0 0 T (2π / ω ) the three terms are integrated, there will be a constant term multiplied by (2π/ω), which cancels the same factor in front of the integral. There will be a cos term that averages to zero. The cos2 term just gives the rms value of the induced current, so the final result is result is 2 ⎡ ⎛ 0.432 ⎞ ⎤ 2 P = ⎢ (2.4 A) + ⎜ A ⎟ ⎥ (5 Ω) = 29.3 W (Without the magnetic field, the power delivered ⎝ 2 ⎠ ⎦⎥ ⎣⎢ by the battery alone is 28.8 W.)


CHAPTER

32-70.

1 1 = = 208 Ω −3 −1 ω C 6.0 × 10 s × 8.0 × 10−7 F ε 0.20 V (b) Max current = max = = 9.6 × 10−4 A XC 208 Ω ε cos ω t 0.20 = (c) I = max cos ω t = (9.6 × 10−4 A) cos ω t. XC 208

(a) Reactance X C =

At t = 0, I = 9.6 × 10–4 A At t = 32-71.

π π , I = 9.6 × 10−4 cos = 6.8 × 10−4 A 4ω 4

The charge Q is the same on both capacitors C1 and C2. The circuit equation is Q Q ε max sin ω t − 1 − 2 = 0. Since Q1 = Q2 = Q C1 C2 ⎡1 1 ⎤ + ⎥=0 ⎣ C1 C2 ⎦ ⎡ CC ⎤ (a) Q = ⎢ 1 2 ⎥ ε max sin ω t = (3.6 × 10−8 )(1.8) sin 120π t ⎣ C1 + C2 ⎦

ε max sin ω t − Q ⎢

Q = 6.48 × 10−8 sin 120π t C (b) Maximum when sin 120 πt = 1, or whenever 120π t =

π 3π 2

,

2

,… Minimum whenever

120π t = 0, π , 2π , … First maximum occurs when t = 1/240 sec First minimum occurs when t = 0. 2 1 Q2 1 Qmax ; U max = where C = 3.6 × 10–8 F. (c) Energy U = 2 C 2 C 1 (6.48 × 10−8 ) 2 U max = = 5.83 × 10−8 J. 2 3.6 × 10−8 U =

32-72.

1 1 Q2 1 (6.48 × 10−8 ) 2 U dt dt sin 2 ω t dt = = T∫ T ∫ 2C T ∫ 2(3.6 × 10−8 )

⎡1 T ⎤ U U = U max ⎢ ∫ sin 2 ω t dt ⎥ = max = 2.92 × 10−8 J. 0 2 ⎣T ⎦ (a) ω = 120π rad/s = 377 rad/s. Find Ceq to find the total current: −1

⎛ 1 C2C3 1 ⎞ (2.0)(1.0) ⎤ ⎡ −8 −8 + = ⎢ 5.0 + Ceq = C1 + ⎜ ⎟ = C1 + ⎥ × 10 F = 5.67 × 10 F 2.0 1.0 + + C C C C ⎣ ⎦ 3 ⎠ 2 3 ⎝ 2 1 1 = = 4.68 × 104 Ω X Ceq = ω Ceq (377 rad/s)(5.67 × 10−8 F) I Ceq =

ε max X Ceq

π⎞ π⎞ 2.0 V ⎛ ⎛ cos ⎜ 120π t + ⎟ = cos ⎜ 120π t + ⎟ = −(4.27 × 10−5 ) sin (120π t ) A 4 2 ⎠ 4.68 × 10 Ω 2⎠ ⎝ ⎝

I max = 4.27 × 10−5 A


32

CHAPTER

32

(b) The voltage across C1 is VC1 = ε = 2.0 cos(120π t ) V. X C1 =

1 1 = = 5.31 × 104 Ω. −8 ω C1 (377 rad/s)(5.0 × 10 F)

ε max

2.0 V π⎞ π⎞ ⎛ ⎛ cos ⎜ 120π t + ⎟ = cos ⎜ 120π t + ⎟ = −(3.77 × 10−5 ) sin (120π t ) A 4 2 ⎠ 5.31 × 10 Ω 2⎠ X C1 ⎝ ⎝ To find the current through C2 and C3, find the current the equivalent series capacitance: C2C3 (2.0)(1.0) × 10−8 F = 6.67 × 10−9 F = C23 = 2.0 + 1.0 C2 + C3 I C1 =

X C 23 = I 23 =

1 1 = = 3.98 × 105 Ω ω C23 (377 rad/s)(6.67 × 10−9 F)

2.0 V π⎞ ⎛ cos ⎜ 120π t + ⎟ = −(5.03 × 10−6 ) sin (120π t ) A 5 3.98 × 10 Ω 2⎠ ⎝

The voltages across the two capacitors must add to 2.0 V. Since C2 = 2C3, VC2 = VC3/2. Thus VC 2 = 0.67 cos(120π t ) V VC 3 = 1.33cos(120π t ) V 32-73.

32-74.

X = ω L = (3.0 × 103 rad/s)(3.0 × 10−3 H) = 9.0 Ω ε 8.0 V I max = max = = 0.89 A 9.0 Ω XL One quarter cycle later, I = 0. One half cycle later, I = –0.89 A. Three-quarters of a cycle later, I = 0. dI (a) ε = L = ω LI 0 cos ω t = (120π )(1.6 × 10−3 )(180) cos120π t = 109 cos120π t V dt at t = 0, ε = 109 V

1 120π π = cos = 0 ⇒ ε = 0 sec, cos 240 240 2 1 1 (1.6 × 10−3 )(180) 2 sin 2 120π t (b) U = LI 2 = LI 02 sin 2 ω t = = 25.9 sin 2 120π t J 2 2 2 At t = 0, U = 0. 1 π At t = s , U = 25.9 sin 2 = 25.9 J 2 240 dU dI (c) P = = LI = L ( I 0 cos ω t ) (ω I 0 sin ω t) = ω LI 02 sin ω t cos ω t dt dt 1 P = 0 at t = 0 and t = s. 240 (a) The total current through each circuit element is the superposition of the currents from the individual sources. For the resistor, ε battery + ε AC 3.0 V + (1.5 V) cos(6000π t ) I R (t ) = = = 1.5 × 10−3 A + (7.5 × 10−4 A) cos(6000π t ) R 2.0 × 103 Ω at t =

†32-75.

To find the current through the inductor, use the fact that the initial current is zero and must change with time after the sources are connected. Write a Kirchhoff loop equation including the sources and the inductor:


CHAPTER

32

dI dI ε battery + ε AC 3.0 + 1.5 cos(6000π t ) =0⇒ = = = 60 + 30 cos(6000π t ). dt dt L 0.050 Integrate to get I(t): 30 I (t ) = ∫ [ 60 + 30 cos(6000π t ) ]dt + I (0) = 60t + sin(6000π t ) + I (0). 6000π The initial current is zero, so the final result is I (t ) = 60t + (1.6 × 10−3 ) sin(6000π t ) A.

ε battery + ε AC − L

(b) The instantaneous power absorbed by the inductor is PL = I (t )(ε battery + ε AC ) = [60t + (1.6 × 10−3 ) sin(6000π t )][3.0 + 1.5 cos(6000π t )] = 180t + 90t cos(6000π t ) + (4.8 × 10−3 ) sin(6000π t ) + (2.4 × 10−3 ) sin(6000π t ) cos(6000π t ) W The instantaneous power dissipated by the resistor is PR =

2 ε battery

R

+

2 ε AC

R

=

(3.0 V) 2 + (1.5 V) 2 cos 2 (6000π t ) = 4.5 × 10−3 +1.1 × 10−3 cos 2 (6000π t ) W 2 × 103 Ω −1

32-76.

32-77.

32-78.

⎛ 1 C1C2 1 ⎞ (20 µ F)(10 µ F) + = = 6.67 µ F. The The equivalent capacitance is Ceq = ⎜ ⎟ = C1 + C2 30 µ F ⎝ C1 C2 ⎠ 1 1 = = 50 Hz resonant frequency is f = 2π LCeq 2π (1.5 H)(6.67 × 10−6 F) 1 Q 2 (t ) 1 2 + LI (t ), at t = 0, I (t ) = 0, Q(t ) = 1.2 × 10−4 C ; therefore, 2 C 2 1 (1.2 × 10−4 ) 2 U (0) = J = 1.4 × 10−3 J 2 5.0 × 10−6 1 The current will slosh around at angular frequency ω = . When current is maximum, there LC is no charge on capacitor, and the energy is fully magnetic. Since ⎛ t ⎞ I (t ) = I max sin ω t = I max sin ⎜ ⎟ , this first occurs when ⎝ LC ⎠ t π π π = ⇒t= LC = 5.0 × 10−6 × 5.0 × 10−2 s = 7.9 × 10–4 s (fully magnetic). The 2 2 2 LC energy is fully electric when I(t) = 0, occurring at t = 0 and next when t = π ⇒ t = π LC = 1.6 × 10−3 s LC U (t ) =

I max =

ε max R 2 + X L2

=

ε max R 2 + (ω L) 2

=

ε max R 1 + (ω L / R ) 2

⇒R=

ω = 2π f = 2π (1.2 × 103 Hz) = 7.54 × 103 rad/s. R=

24 V (0.45 A) 1 + [(7.54 × 103 rad/s)(5.0 × 10−3 s −1 )]2

= 1.4 Ω


ε max I max 1 + (ω L / R) 2

CHAPTER

†32-79.

32

(a) f 0 =

1 2π LC

=

1 2π (55 × 10 H)(0.25 × 10 F) −3

(b) At resonance, Z = R, so I max =

ε max

−6

=

= 1.36 × 103 Hz

15 V = 2.5 A. 6Ω

R 1 1 XC = = = 469 Ω 2π f 0 C 2π (1.36 × 103 Hz)(0.25 × 10−6 F)

ε C ,max = I max X C = (2.5 A)(469 Ω) = 1.2 × 103 V (c) [quality factor] =

X XL 469 = C = = 78 , where the fact that at resonance XL = XC has been 6 R R

used. 32-80.

⎛ N2 ⎞ ⎟ ⎝ N1 ⎠1

ε1,2 = ε 2,1 = ε1,1 ⎜

2

⇒ ε 2,2

⎛N ⎞ ⎛N ⎞ ⎛N ⎞ ⎛N ⎞ = ε1,2 ⎜ 2 ⎟ = ε1,1 ⎜ 2 ⎟ ⎜ 2 ⎟ = ε1,1 ⎜ 2 ⎟ ⎝ N1 ⎠1 ⎝ N1 ⎠1 ⎝ N1 ⎠ 2 ⎝ N1 ⎠1 2

†32-81.

⎛ 2200 ⎞ ε 2,2 = (115 V) ⎜ ⎟ = 28.4 kV ⎝ 140 ⎠ P is the same at all points in the generation and transmission process. P = (115 V)(5.0 A) = 575 W. At the generator, P 575 W = = 0.026 A I generator = ε generator 22 × 103 V In the transmission line, P 575 W = = 0.0012 A I line = ε line 500 × 103 V

32-82.

Amount lost in transmission = 10% of 1.0 × 108 W = 1.0 × 107 W. Let R be resistance of transmission line. Power transmitted = VI. Therefore, for same power to be transmitted, ⎛V I ⎞ V1I1 = V2I2 ⇒ I2 = ⎜ 1 1 ⎟ . But power dissipated for first current is I12 R = 1.0 × 107 W and ⎝ V2 ⎠ 2

2

⎛V ⎞ ⎛V ⎞ power dissipated for second current is I 22 R = ⎜ 1 ⎟ I12 R = ⎜ 1 ⎟ 1.0 × 107 W.V1 = 760000 V ; ⎝ V2 ⎠ ⎝ V2 ⎠ 2

32-83.

⎛ 760000 ⎞ 7 7 V2 = 340000 V; therefore, I 22 R = ⎜ ⎟ × 1.0 × 10 W = 5 × 10 W ⎝ 340000 ⎠ 7 Then 5.0 × 10 W, or 50% of the power, would be lost at 340 kV. N 200 ε L = 2 ε1 = × 60 cos (2π × 1500t ) = 15 cos(3000π t ) V N1 800


CHAPTER 33

ELECTROMAGNETIC WAVES


(a) The potential difference is V =

C=

ε0 A d

, so E =

Q

ε0 A

.

(b) Φ E = EA ⇒ Φ E = (c) I displacement = ε 0

Q V Q , so the electric field is E = = . The capacitance is C d Cd

Q

ε0

d Φ E dQ = dt dt

dQ is the current charging the capacitor, so I displacement = I . dt The current intercepted by the surface is I, the current in the wire. The opening in the surface is A/4, where A the total surface area of one of the capacitor plates. If we assume that the displacement current is evenly distributed across A, then the displacement current intercepted is 3I/4 because the total displacement current must be equal to I. To find the total current intercepted by the surface, we must recognize that the current fluxes have opposite senses for the wire and the plate–current is entering the surface where it intersects the wire and displacement current is exiting the surface between the plates. Thus the net current intercepted by the surface is I/4, which is just the total displacement current contained within the loop of radius R/2 formed by the hole in the surface between the plates. (a) The displacement current between the plates is equal to the conduction current in the wires. So I displacement = 4.0 A. (d)

33-2.

†33-3.

(b) Idisplacement = ε0

I displacement dΦE dΦE = = ; therefore, ε0 dt dt

4.0 A = 4.5 × 1011 V i m/s 8.85 × 10−12 C2 /N i m 2


CHAPTER

33

dV dE = 2 × 103 V/s. But E = V/d, where d is the plate separation of 0.2 cm. Then = dt dt ⎛ 1 ⎞ dV = 1.0 × 106 V/m i s ⎜ ⎟ ⎝ d ⎠ dt dΦE (b) I displacement = ε 0 dt dΦE dΦE dE = 2.78 × 10−5 r 2 = π r2 = π × 106 r 2 ⇒ I displacement = ε 0 dt dt dt At r = 0.15 m, Idisplacement = 6.3 × 10–7 A; At r = 0.30 m, Idisplacement = 2.5 × 10–6 A. µε r ⎛ dE ⎞ dE −13 = 2⎜ (c) B = 0 0 r ⎟ = 8.33 × 10 T at r = 0.15 m, and dt 2 2c ⎝ dt ⎠ B = 1.67 × 10−12 T at r = 0.30 m

33-4.

(a)

33-5.

(a) Displacement current = current into capacitor = 0.10 A. (b) Uniform field between plates. Therefore, flux through a specified area is proportional to the π 52 1 of plate of capacitor. Since Id = ε0 d Φ ; area. Area radius 5 cm as described takes up = 2 π 20 16 dt 0.10 A ⇒ Id ∝ Φ ∝ Area; therefore, displacement current through that area = = 6.3 × 10−3 A 16 dΦ ∫ B ds = µ0ε 0 dt E . For r < R, Φ E = EA = π r 2Ct 2

33-6.

dΦE = 2π r 2 Ct dt B (2π r ) = µ 0ε 0 ( 2π r 2Ct ) B = µ 0ε 0 rCt = (4π × 10−7 T i m/A)(8.85 × 10−12 C2 /N i m 2 )(5 × 104 V/m i s 2 )rt = (5.56 × 10−13 T/m i s)rt At t = 0.50 s, r = 1.0 cm, B = 2.8 × 10–15 T For r > R, Φ E = EA = π R 2Ct 2

dΦE = 2π R 2 Ct dt B (2π r ) = µ 0ε 0 ( 2π R 2Ct )

µ 0ε 0 R 2Ct

(4π × 10−7 T i m/A)(8.85 × 10−12 C2 /N i m 2 )(5 × 104 V/m i s 2 )(0.05 m) 2 t r r −15 (1.39 × 10 T i m/s)t = r At t = 2.0 s, r = 6.0 cm, B = 4.6 × 10−14 T B=

=


CHAPTER 33-7.

33

The displacement current between the capacitor plates must equal the current I coming in, i.e., I = Id. The flux through the dielectric is given by Q 1 Φ= (since E = Efree );

κε 0

κ

dΦ 1 dQ I = = ; dt κε 0 dt κε 0 dΦ therefore, I d = I = κε 0 . Substituting this into Eq. 33.3 gives dt dΦ ∫ B i ds = µ0 (I + I d ) = µ0 I + κ µ0ε 0 dt dQ dε dΦE = I = Id = ε0 =C = Cεmaxω cos ωt dt dt dt = (2 × 10–12 F)(0.5 V)(4 × 103 rad/s)cos (ωt) = 4 × 10−9 cos(ω t ) Α

therefore,

33-8.

33-9.

(a) Id = ε0 d Φ . But Φ = EA, where E is electric field through the space in the capacitor and A = dt V where V = voltage, d = distance across area of capacitor; therefore, Id = ε0A dE . But E = d dt ε A dV capacitor; therefore, Id = ε0 A dV . But C = 0 ; therefore, Id = C = 2.0 × 10–6 F × 1.0 × 103 d dt d dt V/s = 2.0 ×10–3 A. ∆V (b) Real current I = . We want ∆V = IR = (2.0 × 10–3 A × 5.0 × 105 Ω) R = 1000 V. Since voltage rises at 1000 V/s, this occurs in 1.0 s. ε (π r 2 dV ε A dV dV (c) From (a), Id = 0 = 0 = 1 × 103 V/s , , r = 0.20 m, d dt dt d dt 4V 4 (∆V / ∆t )∆t (4)103 = = ∆t. and I = 9R 9 (9) R R R = 5 × 105 Ω. Then B =

B= C=

ε π r2 3 ⎤ µ0 1 ⎡ (4)103 ∆t + 0 (10 ) ⎥ ⎢ 2π r ⎣ (9) R d ⎦

(4)103 µ0 ∆t µε + 0 0 (0.2 × 103 ). We determine d through (9)0.4π R 2d

ε0 A

, d=

ε0 A

so that B =

(4)103 µ0 ∆t 200µ0C , A = π(.3)2. + (9)0.4π R 2A

C 200µ0C = 8.9 × 1010 T at t = 0, B = 2A at t = 1 s, B = 8.9 × 10–10 + 8.9 × 10–10 T = 1.78 × 10−9 T d

at t = 2.0 s, B = (8.9 × 10–10)2 + 8.9 × 10–10 = 2.7 × 10−9 T


CHAPTER

33-10.

33

µ0ε 0 rωE0 cos ωt (since r < radius of capacitor). 2 ∆V 5000 V = = 3.33 × 106 V/m and ω = 2πf Here, E0 = d 1.5 × 10−3 m

By Example 1, B =

= 2π × 60 rad/s = 377 rad/s, and r = 20 cm = 0.20 m. Therefore, 1.26 × 10−6 × 8.85 × 10−12 s 2 1 V B= × 0.20 m × 377 × 3.33 × 106 cos ωt 2 s 2 m m = 1.4 × 10–9 T cos ωt. Amplitude = 1.4 × 10−9 T †33-11.

(a) J =

ρI ρ Ct I E = ⇒E= = . 2 2 πR ρ πR π R2

(b) For r ≤ R, the Ampere-Maxwell law gives dΦE ⎞ ⎛ B (2π r ) = µ 0 ⎜ I + ε 0 ⎟ dt ⎠ ⎝ Ir 2 Ctr 2 The current within radius r is I = J (π r 2 ) = 2 = 2 . R R 2 2 ρ Ct (π r ) ρ Ctr Φ E = EA = = R2 π R2 dΦE ρ Cr 2 = dt R2 ⎛ π r 2C ρ Cr 2 ⎞ B (2π r ) = µ 0 ⎜ 2 + ε 0 ⎟ R2 ⎠ ⎝ R

µ 0Cr (t + ε 0 ρ ) 2π R 2 µ ε ρ Cr At t = 0, B = 0 0 2 2π R (c) For r ≥ R, the current within r is I = Ct, and the Ampere-Maxwell Law gives ρ Ct (π R 2 ) Φ E = EA = = ρ Ct π R2 dΦE = ρC dt B (2π r ) = µ 0 ( Ct + ε 0 ρ C ) B=

µ 0C (t + ε 0 ρ ) 2π r µ ε ρC At t = 0, B = 0 0 2π r B=

33-12.

As a check, note that the answers to (b) and (c) are the same at r = R. Ct dΦE C ⎛J⎞ ⎛ I ⎞ The electric flux through one turn is Φ E = EA = ⎜ ⎟ A = ⎜ ⇒ = . An ⎟A= σ σ dt ⎝σ ⎠ ⎝σ A⎠ Amperian loop around the coils encloses N turns, so the Ampere-Maxwell Law gives ε ⎞ µ NC ⎛ ε 0 ⎞ µ ε NC C⎞ ⎛ ⎛ . Bl = µ 0 N ⎜ I + ε 0 ⎟ = µ 0 NC ⎜ t + 0 ⎟ ⇒ B = 0 t + ⎟ . At t = 0, B = 0 0 ⎜ l ⎝ σ⎠ σ ⎠ σ ⎠ σl ⎝ ⎝


CHAPTER

33-13.

For a medium with dielectric constant κ, Gauss’ Law becomes

∫

E i dA =

Q

κε 0

33

and

Maxwell-Ampere’s Law becomes dΦ ∫ B i ds = µ0 ( I + κε 0 dt E ) so that Maxwell’s equations become: Q ∫ E i dA = ∫ B i dA = 0

κε 0

33-14.

∫ E i ds = − ∫ Β i dA = q

m

dΦE dΦ B i ds = µ 0 ( I + κε 0 ) ∫ dt dt (1), similar to Gauss’ Law for

electricity. Applying this to a point charge by constructing a surface with radius r around the q charge gives B = m 2 , as required. As with 4π r electric charge, if we sent a flow of magnetic charge into a capacitor, there must be a magnetic

dΦB dt

dqm dt

dqm . The magnetic dt displacement current will be proportional to the rate of change of the magnetic flux, by analogy dΦB . This is verified with electric displacement current. Then magnetic current Im must equal dt dq dΦB by (1): ∫ Β i dA = Φ B = qm ⇒ I m = m = . The version of Ampere’s Law containing dt dt displacement current says ∫ B i ds = µ 0 ( I + I displacement ), so Faraday’s Law must be modified to

displacement current that is equal to the magnetic conduction current

include magnetic current: Therefore,

†33-15.

⎛ dΦB ⎞ + I m ⎟ (2). The other two laws remain dt ⎠

∫ E i ds = − ⎜⎝

the same. As implied in the text, an accelerated positive charge produces a radiation (transverse) electric field that points in the opposite direction from the component of the acceleration parallel to the radiation field. For points on the line perpendicular to the acceleration, the direction of the radiation field is exactly opposite the direction of the acceleration. For an accelerated electron, the radiation electric field on the line perpendicular to the acceleration will point in the same direction as the acceleration because of the electron’s negative charge. Suppose the electron is initially moving to the right. When it stops, its acceleration will point to the left, so the transverse electric will also point left. In the diagram, the pulse is propagating “up” in the top half of the figure and “down” in the bottom half. The electric component of the radiation field along a line perpendicular to the acceleration will point left everywhere. To satisfy the right-hand rule relation between E, B, and the propagation direction, the magnetic field must point out of the page in the top half of the figure and into the page in the bottom half.


B

E

a

E

× B

CHAPTER 33-16.

†33-17.

33

The diagram is a mirror image of Figure 33.5. In the diagram for this problem, the charge is initially stationary at point 1, which is at time t = 0. It undergoes a leftward acceleration a until time τ (indicated by the arrow pointing from 1 to 2), which is at point 2, and then travels to the left at constant speed v = aτ. At time t it has reached point 3, which is a aτ 2 distance + v(t − τ ) from point 1. + + + 2 The large circle has radius ct and is 3 2 1 centered on point 1. The smaller circle has radius c(t – τ) and is centered on point 2 to show how far the field had reached after the disturbance ended. The field lines at the larger circle are directed from point 1. Lines are drawn from point 3 to show the direction of the field at that location. By extending those lines to the smaller circle centered on point 2, we can determine the change in the field caused by the acceleration. The “kinks” connecting the field lines from the smaller circle to the ones at the larger circle represent the average change in the field caused by the acceleration. Note that the component of the kink connecting the two parts of the field has a transverse component that points opposite the component of the acceleration that is perpendicular to the Coulomb field direction (parallel to the radiation field direction). To set this problem up, we use the nontechnical interpretation of the word “acceleration” to mean “an increase in velocity” because “deceleration” is a colloquial expression that means “a decrease in velocity” (referring to an acceleration that points in the opposite direction from the initial velocity). Assume there is a positive charge initially at point 1 in the diagram. A circle of radius 7 shows how far the field from the charge has reached from that position at time t. At t = 0, the charge moves to the right with constant acceleration a to the right for time τ to point 2. It then moves with constant acceleration –a (the “deceleration”) for time τ, finally coming to rest again at point 3. The largest circle has radius ct and is centered on point 1. The next smaller circle has radius c(t – τ) and is centered on point 2. The smallest circle has radius c(t –2τ) and is centered on point 3.


CHAPTER 33-18.

33-19.

33

The diagram will look like Figure 33.8 except that if the electric field vectors point up as in 33.8, the magnetic field vectors will point in the –z direction so that the direction of E × B will point in the –x direction. To satisfy the requirement that E × B points up when E points north, B must point west.

E

B 33-20.

†33-21.

33-22. 33-23.

If the charge on the sheet is positive, there will be a Coulomb field pointing away from the sheet on each side. So on the –x side, the Coulomb field will point in the –x direction. The acceleration in the +y direction will cause a radiation field in the –y direction. To satisfy the requirement that E × B points in the –x direction, B must point in the +z direction. Consider the small rectangle of length l shown. The gray area c represents the region containing the electric field, which is pointing out of the paper. If the pulse has crossed a length x of the rectangle, the flux is Φ E = Elx , and the rate of change of the flux is dΦE d dx E = ( Elx) = El = Elc, since the speed of the front edge of l dt dt dt the field is c. Then the displacement current/length along the loop is Id = ε 0 Ec = (8.85 × 10−12 C 2 /N i m 2 )(4.0 × 10−3 V/m)(3.00 × 108 m/s) = l 1.1 × 10−5 A/m.

1 ly = (3.00 × 108 m/s)(365 days)(24 h/day)(3600 s/h) = 9.46 × 1015 m ∆x 0.01 m = = 3.3 × 10−11 s ∆x = c∆t ⇒ ∆t = 8 c 3.00 × 10 m/s

33-24.

t=

d 2(3.8 × 108 m) = = 2.5 s c 3.00 × 108 m/s

33-25.

t=

d 1.50 × 1011 m = = 5.00 × 103 s (8.33 min) c 3.00 × 108 m/s

33-26. 33-27.

33-28.

d 2.0 × 106 m = = 6.7 × 10−3 s c 3.00 × 108 m/s ∆x 1.0 m ∆t = = = 3.3 × 10−9 s c 3.00 × 108 m/s 1 f = = 3.0 × 108 Hz (0.30 GHz) ∆t These limitations are based on fundamental properties of nature and cannot be overcome. (a) vertical. (b) To maximize the flux (and thus the change in flux), the vector that defines the plane of the circle must be parallel to the magnetic field. This means that the plane of the circle must include the direction of propagation of the wave. t=


CHAPTER 33-29.

33-30.

33

By right-hand rule, B points north. E Magnitude = 0 = 2.0 × 109 T c

(direction of propagation) If the original beam is unpolarized, the first polarizer transmits half of the incident intensity. The second polarizer then transmits a fraction cos2φ of that intensity. The final intensity is ⎛I ⎞ I final = ⎜ 0 ⎟ cos 2 φ ⎝ 2⎠ I final cos 2 φ = = 0.30 I0 2

φ = cos −1 0.60 = 39° †33-31.

33-32.

The incident polarization is at 45º to the preferential direction of the first polarizer, so the intensity after the first polarizer is I I1 = I 0 cos 2 45° = 0 2 This light is now polarized at 45º to the preferred direction of the second polarizer, so I I I 2 = I1 cos 2 45° = 1 = 0 2 4 The first Polaroid transmits half of the incident intensity. The second transmits a fraction cos2 φ I cos 2 φ of that, so transmitted = 2 I0 30° :

I transmitted cos 2 30° = = 0.38 I0 2

45° :

I transmitted = 0.25 I0

I transmitted = 0.13 I0 If the original beam is unpolarized, the first polarizer transmits half of the incident intensity. The second polarizer then transmits a fraction cos2 φ of that intensity. The final intensity is ⎛I ⎞ ⎛I ⎞ I final = ⎜ 0 ⎟ cos 2 φ = ⎜ 0 ⎟ cos 2 80° = 0.015I 0 ⎝ 2⎠ ⎝ 2⎠ 1.5% of the incident light is transmitted. If the original beam is unpolarized, the first polarizer transmits vertically polarized light with half the incident intensity. The second polarizer then transmits a fraction cos2 60º of that intensity: I I ⎛I ⎞ I 2 = ⎜ 1 ⎟ cos 2 60° = 1 = 0 4 8 ⎝2⎠ The third polarizer is oriented 20º from the vertical, which is 40º from the polarization direction after the second polarizer. The final intensity is ⎛I ⎞ I final = I 2 cos 2 40° = ⎜ 0 ⎟ cos 2 40° = 0.073I 0 ⎝8⎠ 60° :

†33-33.

33-34.


CHAPTER

†33-35.

The intensity transmitted by the first polarizer is I1 = I 0 cos 2 45° =

I0 . The intensity transmitted 2

by the second polarizer is ⎛I ⎞ I final = I1 cos 2 φ = ⎜ 0 ⎟ cos 2 φ ⎝ 2⎠ 2 I final cos φ = = 0.30 I0 2

φ = cos −1 0.60 = 39° 33-36.

(a) See Problem 31 for the solution to this part. (b) After the first polarizer, I1 = I 0 cos 2 30°. After the second polarizer, I 2 = I1 cos 2 30° = I 0 ( cos 2 30° ) . After the third polarizer, 2

I 3 = I 2 cos 2 30° = I 0 ( cos 2 30° ) = 0.42 I 0 3

90° . After the second polarizer, N 2 N 90° 90° ⎞ ⎛ ⎛ 2 90° ⎞ I 2 = I1 cos 2 = I 0 ⎜ cos 2 After N polarizers, I = I . cos 0 ⎜ N ⎟ ⎟ . If N = 90, N N ⎠ N ⎠ ⎝ ⎝

(c) After the first polarizer, I1 = I 0 cos 2

I 90 = I 0 ( cos 2 1° )

As N → ∞, cos 2

33-37.

90

= 0.97.

90° → cos 2 0 = 1 , so lim I N = I 0 . N →∞ N

ωx ⎞ ωx ⎞ ⎛ 2 ⎛ (a) E = E0 sin 2 ⎜ ω t − ⎟ + cos ⎜ ω t − ⎟ c ⎠ c ⎠ ⎝ ⎝

z

= E0 1 = E0 (b)At x = 0, E(t ) = E0 ⎡⎣ j sin ( ω t ) + k cos ( ω t ) ⎦⎤ , so E(0) = E0 k ⎛ π ⎞ E⎜ ⎟ = E0 j ⎝ 2ω ⎠

t = π/(2ω)

33-38.

y

t = π/ω t=0

⎛π ⎞ E ⎜ ⎟ = − E0 k ⎝ω ⎠ ⎛ 3π ⎞ E⎜ ⎟ = − E0 j ⎝ 2ω ⎠

t = 3π/(2ω)

So the electric field rotates counterclockwise with angular frequency ω. ⎛π ⎞ E0′′ = E0 cos φ cos ⎜ − φ ⎟ . The intensity will be a maximum when the magnitude of E0′′ is a ⎝2 ⎠ maximum. dE0′′ d ⎛π ⎞ ⇒ = E0 cos φ cos ⎜ − φ ⎟ = 0 dφ dφ ⎝2 ⎠ ⎛π ⎞ ⎛π ⎞ ⇒ − sin φ cos ⎜ − φ ⎟ + cos φ sin ⎜ − φ ⎟ = 0 ⎝2 ⎠ ⎝2 ⎠


33

CHAPTER

33

⎛π ⎞ ⎛π ⎞ cos ⎜ − φ ⎟ = sin φ ; sin ⎜ − φ ⎟ = cos φ 2 2 ⎝ ⎠ ⎝ ⎠ 2 2 ⇒ − sin φ + cos φ = 0 ⇒ 2 cos 2 φ = 1 1 ⇒ cos φ = 2 ⇒ φ = 45° 33-39.

We will make the simplifying assumption that the second polarizer acts as an ideal polarizing filter, i.e., the transmitted intensity is determined by the projected angle that the incoming wave sees when approaching the second Polaroid. The effect of oblique incidence on the transmitted intensity will be neglected. If I0 is the intensity of light incident on the second polarizer, then the transmitted intensity is I = I0 cos2 θ

where cos θ =

33-40. 33-41.

cos ø

cos 2 ø + sin 2 ø cos 2 α as can be seen from part (a). Parts (b) and (c) are views along the y-, and the z-axis respectively I 0 cos ø shown for clarity. Thus I = 2 cos ø + sin 2 ø cos 2 α (Note: Angle ø has been shown opposite to that shown in the book, for clarity.) c 3.00 × 108 m/s λ= = = 1.85 m f 162.6 × 106 Hz f =

c

λ

=

3.00 × 108 m/s = 3.0 × 1011 Hz 10−3 m


CHAPTER

33-42.

λ=

c 3.00 × 108 m/s 3.00 × 108 m/s = = = 120 m f f 2.5 × 106 Hz

Use a spreadsheet to make a table of all the wavelengths.

33-43.

f (MHz) λ (m) 2.5 120 5.0 60 10 30 15 20 20 15 c By λ = we have: f AM 530 kHz ⇒ λ =

3.00 × 108 m/s 3.00 × 108 m/s 566 m; = 1600 kHz ⇒ λ = = 187 m 530 × 103 Hz 1600 × 103 Hz

3.00 × 108 m/s 3.00 × 108 m/s = 3.41 m; 108 MHz ⇒ λ = = 2.78 m 6 88 × 10 Hz 108 × 106 Hz qB The cyclotron frequency is Eq. 30.7 f = . For protons, 2π m (1.6 × 10−19 C)(8.0 T) f = = 1.22 × 108 Hz 2π (1.67 × 10−27 kg) FM 88 MHz ⇒ λ =

33-44.

λ= †33-45.

c 3.00 × 108 m/s = = 2.5 m f 1.22 × 108 Hz

(a) λ =

c 3.00 × 108 m/s 3.00 × 108 m/s = = = 1.5 × 10−10 m (X-ray) 18 f f 2.0 × 10 Hz

(b) λ =

c 3.00 × 108 m/s 3.00 × 108 m/s = = = 1.0 × 10−2 m, or 1.0 cm (microwave radio) 10 f f 3.0 × 10 Hz

(c) λ =

c 3.00 × 108 m/s 3.00 × 108 m/s = = = 5.0 × 106 m (“electric” wave) f f 60 Hz

c

f =

33-47.

λ=

33-48.

λmax = 550 nm, which is yellow-green, according to the caption in Figure 33-24. λ1/2 ≈ 510 nm (aqua) and 610 nm (yellow-orange). λ1/4 ≈ 490 nm (blue) and 640 nm (orange). ε E 2 (8.85 × 10−12 C2 /N i m 2 )(80 V/m) 2 u= 0 = = 2.8 × 10−8 J/m3 2 2 ε 0 E02 (8.85 × 10−12 C2 /N i m 2 )(0.12 V/m) 2 uE = uB = = = 6.4 × 10−14 J/m3 2 2 E2 (cB ) 2 cB 2 (3.00 × 108 m/s)(2.0 × 10−10 T) 2 S = = = = = 9.5 × 10−6 W/m 2 −7 µ0c µ0c µ0 4π × 10 T i m/A

33-49. 33-50. 33-51.

λ

=

3.00 × 108 m/s = 1.4 × 109 Hz 21 × 10−2 m

33-46.

c 3.00 × 108 m/s 3.00 × 108 m/s = = = 0.027 m f f 1.1 × 1010 Hz


33

CHAPTER

33

S =

E02 2µ 0 c

33-52.

⇒ E0 =

33-53.

2µ 0 cS =

2(4π × 10−7 T i m/A)(3.00 × 108 m/s)(1.4 × 103 W/m 2 ) = 1.03 × 103 V/m

B0 =

E0 1.03 × 103 V/m = = 3.42 × 10−6 T c 3.00 × 108 m/s

B=

E0 0.50 T = 1.7 × 109 T . = c 3 × 108

By right-hand rule, B points vertically downwards.Direction of Poynting vector = direction of propagation = NORTH. 1 EB Magnitude S = µ0 =

1 × 0.50 × 1.7 × 10–9 1.26 × 10−6

= 6.6 × 104 W/m 2 33-54.

S =

E02 2µ 0 c

⇒ E0 = B0 = †33-55.

33-56.

2µ 0 cS =

2(4π × 10−7 T i m/A)(3.00 × 108 m/s)(1.0 × 109 W/m 2 ) = 8.68 × 105 V/m

E0 8.68 × 105 V/m = = 2.89 × 10−3 T c 3.00 × 108 m/s

The average intensity S at a distance r from a source emitting power P equally in all directions is P S = . The power is 4π r 2 P = 4π r 2 S = (4π )(4.3 × 1017 m) 2 (1.2 × 10−8 W/m 2 ) = 2.8 × 1028 W. S =

P 1.2 W = = 2.44 × 105 W/m 2 2 −3 A π (1.25 × 10 m)

S =

E02 ⇒ E0 = 2µ 0 c

2µ 0 cS =

2(4π × 10−7 T i m/A)(3.00 × 108 m/s)(2.44 × 105 W/m 2 )

= 1.4 × 104 V/m 33-57.

P = 4π r 2 S = (4π )(1.5 × 1011 m) 2 (1.4 × 103 W/m 2 ) = 4.0 × 1026 W.

33-58.

S =

Power 6.0 × 103 W = = 4.8 × 108 W/m 2 2 − 3 Area π ( 2.0 × 10 m )

Erms = Brms =

Eo2 / 2 ⇒ Eo =

2 Erms ⇒ S =

Eo2 E2 = rms ⇒ E = 2µ o c µo c

Erms 1.332 × 106 V/m = = 4.4 × 10−3 T 8 c 3.00 × 10 m/s


µ o cS = 1.3 × 106 V/m

CHAPTER 33-59.

33-60.

†33-61.

33-62.

Magnitude of average energy flux at 5 km is given by E2 (0.22 V/m) 2 S = 0 = = 6.42 × 10−5 W/m 2 . 2µ 0 c 2(4π × 10−7 T i m/A)(3.00 × 108 m/s) Power of transmitter = flux × area = S × 4πr2 = 6.4 × 10–5 W/m2 × 4π(5000 m)2 = 2.02 × 104 W W × 13 cm2 = 1.3 W. Power incident on cell = 0.10 cm 2 Power out of cell = VI = 0.45 V × 0.20 A = 0.09 W. Therefore, Power out 0.09 = 0.069 ( 7%) Efficiency = = 1.3 Power in The electric field from the transmitter is produced by accelerated charges, so the magnitude is ⎛ r ⎞ 1 proportional to 1/r. Since E ∝ , we have E1 = ⎜ E0 0 ⎟ ; therefore, at 12.0 km, r ⎝ r1 ⎠ E1 = (0.13 V/m)(6/12) = 0.065 V/m. At 18.0 km, E1 = (0.13 V/m)(6/18) = 0.043 V/m. E02 P P S S = . = = 2 2µ 0 c 4π r 2 4π r ⇒r=

†33-63.

S =

1 E0

µ 0 cP 1 (4π × 10−7 T i m/A)(3.00 × 108 m/s)(100 W) = = 39 m 2π 2.0V/m 2π

P 5 × 10−3 W = = 6.37 × 103 W/m 2 A π (5 × 10−4 m) 2

E02 S = ⇒ E0 = 2µ 0 c

2µ 0 cS =

2( µ 0 )(3.00 × 108 m/s)(6.37 × 103 W/m 2 ) = 2.19 × 103 V/m

The total energy contained within a volume V of the field is W = uV =

33-64.

†33-65.

33

ε 0 E02 2

V . The volume of a

cylinder of length l and radius r is πr2l, so (8.85 × 10−12 C 2 /N i m 2 )(2.19 × 103 V/m)(π )(5 × 10−4 m)(0.1 m) W = = 1.67 × 10−10 J 2 ∆W 0.50 J P= = = 2.5 × 107 W ∆t 2.0 × 10−8 s S =

P 5 × 10−3 W = = 7.96 × 1012 W/m 2 A π (1.0 × 10−3 m) 2

S =

E02 ⇒ E0 = 2µ 0 c

2µ 0 cS =

2( µ 0 )(3.00 × 108 m/s)(7.96 × 1012 W/m 2 ) = 7.75 × 107 V/m

The average intensity S at a distance r from a source emitting power P equally in all directions is P 75 W 2S S = , since the = = 1.49 W/m 2 . The radiation pressure is [ pressure] = 2 2 c 4π r 4π (2.0 m) mirror reflects the light. The magnitude of the force on the mirror is 2 2S (π Rmirror ) F = [ pressure] × [area of mirror ] = c 2 2 2(1.49 W/m )(π )(0.05 m) = = 7.81 × 10−11 N 8 3.00 × 10 m/s


CHAPTER

33-66.

S =

33 P 50 × 103 W = = 1.59 × 10−6 W/m 2 4π r 2 4π (50 × 103 m) 2

E02 ⇒ E0 = S = 2µ 0 c 33-67.

33-68.

33-69.

33-70.

2µ 0 cS =

2( µ 0 )(3.00 × 108 m/s)(1.59 × 10−6 W/m 2 ) = 0.0346 V/m

The total power output is P = eSA, where e is the efficiency. Then P 1.0 × 106 A= = = 8.0 × 103 m 2 . eS (0.25)(500) W π energy × (10 cm) 2 ; therefore, Energy flux at spot = = (a) Energy through lens = 0.10 2 cm area 4 0.10 W/cm 2 × (π /4)(10 cm) 2 = 40 W/cm 2 (π /4)(0.50) 2 cm 2 (b) = Yes! Let S be flux at the binoculars. With naked eye, energy through eye per second = 7.02 × S. Energy through binoculars = 502 × S. All this goes into the eye. Therefore, Factor increase = 502 × S = 51 times 7.02 × S S = r=

E02 E02 P P P = . By Eq. 33.24, = . Therefore, = . Then S 2 A 4π r 2µ 0 c 2 µ 0 c 4π r 2

µ 0 cP = 2π E02

(4π × 10−7 T i m/A)(3.00 × 108 m/s)(104 W) = 2π (2 × 10−4 V/m) 2

(2 × 10−7 )(3.00 × 108 )(104 ) 2 × 10−4

m = 3.9 × 106 m †33-71.

The radiation pressure is [ pressure] =

S , since the radiation is absorbed. The magnitude of the c

radiation force is Fr = [ pressure] × [cross sectional area ] =

2 S (π Rearth ) (1400 W/m 2 )(π )(6.38 × 106 m) 2 = c 3.00 × 108 m/s

= 6.0 × 108 N. The magnitude of the gravitational force exerted by the sun is Gmearth msun (6.67 × 10−11 N i m 2 /kg 2 )(5.98 × 1024 kg)(1.99 × 1030 kg) = = 3.5 × 1022 N. Fg = (1.50 × 1011 m) 2 r2 33-72.

The effect of the radiation pressure is negligible. The potential difference from one end of the wire to the other end is ∆V = IR, so the magnitude of the electric field within the wire is E = ∆V / l. The field points in the direction of I. At the surface µI of the wire, the magnitude of the magnetic field is B = 0 . The field lines form a circular 2π r pattern around the wire with the direction at any point given by the right hand rule. The Poynting E×B . E and B are perpendicular to each other, so the magnitude of S is vector is given by S =

µ0

S =

EB

µ0

=

VI . The direction of the Poynting vector is radially inward toward the center of the 2π rl


CHAPTER

†33-73.

33

wire. Note that the magnitude of the Poynting vector is equal to the electrical power dissipated by the wire divided by its surface area. This means the power dissipated comes from the electric and magnetic fields around and within the wire. ε E 2 (8.85 × 10−12 C2 /N i m 2 )(2.0 × 10−3 V/m) = = 1.77 × 10−17 J/m3 . For each pulse, (a) uelectric = 0 2 2 uelectric = umagnetic = 1.77 × 10–17 J/m3. (b) Suppose the waves are propagating along the x direction (one to the right, the other to the left), and at some instant both electric fields point in the +y direction. Then, since they are traveling

E

E

B

B in opposite directions, their magnetic fields must cancel as shown in the diagram. If at some other instant, their electric fields point in opposite directions along the y axis, then their magnetic fields point in the same direction but the electric fields cancel. Thus at every instant the superposition of the two waves has one of the net field components equal to zero while the other component has twice the magnitude of the individual pulses. For the specific case shown, Etotal = 4.0 × 10−3 V/m, and Btotal = 0. If the electric fields cancel, then ⎛ 2.0 × 10−3 V/m ⎞ = 2⎜ = 1.33 × 10−11 T 8 2 ⎟ c ⎝ 3.0 × 10 m/s ⎠ (c) The energy density for each field is proportional to the square of the amplitude of that field. Since one of the net fields in the superposition is always zero and the nonzero field has twice the magnitude of the field in each individual pulse, the nonzero field will have four times the energy density of one of the individual pulses. To give a concrete example, suppose that the electric fields are aligned as shown in the diagram. Then uelectric = 7.08 × 10-17 J/m3 , and umagnetic = 0.

Btotal = 2

Esingle

The other possibility is umagnetic = 7.08 × 10−17 J/m3 , and uelectric = 0. Note that in either case the

33-74.

energy density is equal to the total energy density in the individual fields as calculated in (a). ∂2 E ∂2 E = µ ε . If E y = E0 sin(kx − ω t ) , then Eq. 33.33 says 0 0 ∂x 2 ∂t 2 ∂E y ∂2 E = kE0 cos(kx − ω t ), 2 = −k 2 E0 sin( kx − ω t ) ∂x ∂x ∂E y ∂2 E = −ω E0 cos( kx − ω t ), 2 = −ω 2 E0 sin(kx − ω t ) ∂t ∂t Substituting into the wave equation gives −k 2 E0 sin(kx − ω t ) = − µ 0ε 0ω 2 E0 sin(kx − ω t ).


CHAPTER

33

The sine factors cancel and we’re left with 2

†33-75.

1 ⎛ω ⎞ , ⎜ ⎟ = ε k ⎝ ⎠ 0 µ0 which means the wave is propagating with speed c. So the equation given does describe an electromagnetic wave. The wave is moving in the +x direction because of the minus sign in the argument of the sine function. By the right hand rule, B must point in the +z direction when E points in the +y E direction. So the equation for B is Bz = 0 sin(kx − ω t ). c 8 c 3.00 × 10 m/s λ= = = 80.0 m f 3.75 × 106 Hz 2π k= = 0.0785 m −1

λ ω = 2π f = 2.56 × 107 s −1

The wave is traveling in the –z direction because of the plus sign in the argument of the cosine function. Since the directions of the fields and propagation directions are related by the right hand rule, the electric field must point in the +y direction when the magnetic field points in the +x direction to give the correct propagation direction. Thus the equation for the electric field is E y = cB0 cos(kx + ω t ) 33-76.

Define a new variable u = x – ct. Then

∂f ∂f ∂ 2 f ∂2 f = and ; 2 = ∂x ∂u ∂x ∂u 2

∂f ∂f ∂u ∂f ∂ 2 f ∂2 f = = −c ; 2 = c 2 2 . Substituting into the wave equation gives ∂t ∂u ∂t ∂u ∂x ∂u 2 2 2 1 ∂ f ∂ f ∂ f ∂2 f 2 − = µ ε 0 ⇒ − c µ ε = 0 because c 2 = 0 . Thus f(x – ct) satisfies the 0 0 0 0 2 2 2 2 ∂x ∂t ∂u ∂u µ 0ε wave equation. For the specific case of E = E0 exp[−k 2 ( x − ct ) 2 ] = E0 exp[−k 2u 2 ] , we find ∂E ∂E = = −2k 2uE0 exp[−k 2u 2 ] ∂x ∂u ∂2 E ∂2 E = = −2k 2 E0 exp[−k 2u 2 ] + (2k 2 u ) 2 E0 exp[ −k 2 u 2 ] = 2k 2 E0 exp[ − k 2 u 2 ] ( 2k 2u 2 − 1) ∂x 2 ∂u 2 ∂E ∂E = −c = 2ck 2uE0 exp[− k 2u 2 ] ∂t ∂u 2 2 ∂ E 2 ∂ E = = −2c 2 k 2 E0 exp[ −k 2 u 2 ] − (2ck 2u ) 2 E0 exp[− k 2u 2 ] = 2c 2 k 2 E0 exp[− k 2u 2 ] ( 2k 2u 2 − 1) c 2 2 ∂t ∂u Substituting this into c2 =

10

µ 0ε

∂2 E ∂2 E ∂2 f ∂2 f − µ 0ε 0 2 = 0 − c 2 µ 0ε 0 2 = 0 completes the proof because 2 2 ∂x ∂t ∂u ∂u

, causing all the terms to cancel to give 0 = 0.


CHAPTER 33-77.

Applying the line integral in the Ampere-Maxwell Law around the rectangle in the xz plane in Figure 33.27 gives ∫ B ds = Bdz − ( B + dB)dz = −dB dz. The electric flux through the rectangle is d Φ E = Edx dz , which gives

∂Φ E ∂E = dxdz. Then the Ampere-Maxwell Law gives ∂t ∂t

∂E dx dz , or ∂t ∂B ∂E − = µ 0ε 0 , ∂x ∂t which is Eq. 33.32. dΦE dE dE I displacement = ε 0 = ε0 A = ε 0 (π r 2 ) dt dt dt = (8.85 × 10−12 C2 /N i m 2 )(π )(0.15 m)2 (3.8 × 1013 V/m i s) = 24 A ∆V V0 sinω t = (a) The current in the thin wire is given by I = . R R (b) Displacement current dΦE d ( EA) d (V / d ) A ⎛ dV ⎞ A = ε0 = ε0 A = ε0 ⎜ Id = ε0 ⎟ = ε 0 ωV0 cosω t dt dt dt d ⎝ dt ⎠ d (c) Current arriving at the outside terminal must equal current through the thin wire plus V sinω t A + ε 0 V0ω cosω t displacement current: I outside = 0 R d −dB dz = µ 0ε 0

33-78.

†33-79.

33

(d)

∫

B i ds = µ 0 ( I + I d ). As in Example 1, we have

µ0 ( I + I d )in V sinω t . Here, I = 0 . 2 r R But Id is not the entire displacement current, since r < radius of plate. ⎛ I π r 2 ⎞ ε 0π r 2 dΦE V0ω cos ωt; therefore, ∝ area, this gives Idin = ⎜ d Since Id∝ ⎟= d dt ⎝ Α ⎠

2πrB = µ0(I + Id)in, B =

B=

⎞ µ ⎛ V sinω t ε 0π r µ0 1 ⎛ V0 sinω t ε 0π r 2 ⎞ V0ω cosω t ⎟ = 0 ⎜ 0 V0ω cosω t ⎟ + + ⎜ 2π r ⎝ R d d ⎠ ⎠ 2π ⎝ rR


CHAPTER

33

33-80.

Assume a positively charged particle is initially moving to the right. Let t = 0 be the instant the particle comes to rest. At – τ it begins to slow down with a constant acceleration pointing to the left as shown. At some time –t before it stopped it was at the point shown. The large circle centered on the point labeled with –t has radius ct and shows how far the field produced at that instant has propagated by the time the particle stops. The smaller circle is centered on the point labeled –τ and has radius cτ. Field lines are drawn from the point at t = 0 to show the “kinks” produced by the acceleration from t = –τ to t = 0.

33-81.

∆t =

33-82.

33-83.

πr c

=

3.141 × 6.4 × 106 ≅ 0.067 s 3 × 108

I transmitted = cos 2 φ I0 I transmitted = 0.88 20° : I0 40° :


60° :


After the first polarizer, I1 = I 0 cos 2 60. The second polarizer is inclined 60º to the first, so I 2 = I1 cos 2 60 = I 0 cos 4 60° = (0.50 W/m 2 ) cos 4 60° = 0.31 W/m 2

33-84.

ωz ⎞⎤ ⎡ ⎛ (a) In minus z-direction ⎢ froms plus sign in ⎜ ωt + ⎟ , since c ⎠ ⎥⎦ ⎝ ⎣ waves in form f(z + vt) travel in minus direction.

⎛ 2⎞ (b) From diagram, E makes angle θ = tan–1 ⎜ ⎟ ⎝1⎠ = 63° with x − direction,

27° with y − axis, and 90° with z − axis.


CHAPTER

33

E0 in magnitude, and B-field is related to c E by right-hand rule, and since wave is propagating in minus-z-direction, we have E ωz ⎞ ⎛ ˆ E0 sin ⎛ ω t + ω z ⎞ B = ˆj 0 sin ⎜ ω t + ⎟ + 2i ⎜ ⎟ c c c ⎠ c ⎠ ⎝ ⎝ (c) Since B =

†33-85.

E = 140 V/m east, and the propagation direction is up. By the right-hand rule, B must point north so the direction of E × B is up. The magnitude of B is E 150 V/m B= = = 5.00 × 10−7 T. Thus the c 3.00 × 108 m/s

B E

magnetic field vector is B = 5.00 ×10–7 T north. 33-86. †33-87.

Assuming the wavelength is specified in air, f = S =

P 1.2 W = = 2.44 × 105 W/m 2 A π (1.25 × 10−3 m) 2

S =

E02 E2 = rms 2µ 0 c µ0 c

⇒ Erms =

c

λ

=

3.00 × 108 m/s = 75 Hz 4 × 106 m

µ 0 cS = (4π × 10−7 T ⋅ m/A)(3.00 × 108 m/s)(2.44 × 105 W/m 2 ) = 9.60 × 103 V/m

Erms = 3.2 × 10−5 T c Power 5 × 109 W = = 50 W/m 2 (a) S = Energy flux = 4 2 Area (10 m) Brms =

33-88.

(b) E0 =

2µ0 cS =

2(4π × 10−7 T i m/A)(3.00 × 108 m/s)(50 W/m 2 ) = 194 V/m V/m

E0 = 6.5 × 10−7 T c Power Flux at 10 km = . Area = 4πr2 = 4π(10,000 m)2 = 1.26 × 109 m2; therefore, Flux = \f(5000 Area E W, 1.26 × 109 m2) = 4.0 × 10–6 W/m2 = S. By (36), E0 = 2 µ0 cS = 0.055 V/m and B = 0 = c 0.055 V/m 10 = 1.8 × 10 T 3 × 108 ms −1 = 195 V/m and B =

33-89.

33-90.

−8 I ρ (12 A ) (1.7 × 10 Ω/m ) = = 3.8 × 10−2 V/m (a) E = 2 −3 A π (1.3 × 10 m )

(b) B =

(c) S =

−7 µ 0 I ( 4π × 10 T ⋅ m/A ) (12 A ) = = 1.8 × 10−3 T −3 2π r 2π (1.3 × 10 m )

EB

µ0

=

( 3.8 × 10 ( 4π

−2

V/m )(1.8 × 10−3 T )

× 10−7 T i m/A )

= 56.5 W/m 2


CHAPTER

33

(d) P = S ( 2π rl ) = ( 56.5 W/m 2 ) 2π (1.3 × 10−3 m ) (1 m ) = 0.46 W −8 I 2 ρ l (12 A ) (1.7 × 10 Ω/m ) (1 m ) = = 0.46 W 2 A π (1.3 × 10−3 m ) 2

(e) P = I 2 R = 33-91.

Power received by the Arecibo telescope = Flux × Area = 4.1 × 10–25 × π(150)2 = 2.9 × 10−20 W Total power emitted by the quasar = Flux × 4πR2 (R is the earth-quasar distance) = 4.1 × 10–25 × 4π(2.8 × 109 × 9.46 × 1015)2 = 3.6 × 1027 W

33-92.

mW W cm 2 W × 10−3 × 104 2 = 100 2 . Then 2 cm mW m m E02 S = ⇒ E0 = 2µ0 cS = 2 ( 4π × 10−7 T i m/A )( 3 × 108 m/s )(100 W/m 2 ) = 270 V/m 2µ0 c E 274.5 V/m B0 = 0 = = 9.2 × 10−7 T c 3 × 108 m/s

(a) 10

(b) P = SA = (100 W/m 2 )(1 m 2 ) = (100 J/s)/(4.186 J/cal) = 24 cal/s 33-93.

(a) Power = flux × area. Area = πr2 = π × (3.5 × 10–3)2 m2 = 3.85 × 10–5 m2. Therefore, P = 8.8 × 10–11 W/m2 × 3.85 × 10–5 m2 = 3.4 × 1015 W (b) Energy flux = Flux = r=

Power , and at a distance of r, area = 4πr2. Therefore, Area

P . Minimum flux = 8.8 × 10–11 W/m2. Therefore, max distance 2 4π r

P =r= 4π flux

3.9 × 1026 = 5.9 × 1017 m, which is the distance of faintest possible 4π (8.8 × 1011 )

star. Volume that distance covers around us =

4 3 πr 3

4 π(5.9 × 1017 m)3 = 8.8 × 1053 m3. One cubic light year = (9.47 × 1015 m)3 3 = 8.5 × 1047 m3, contains 3.5 × 10–3 stars. Therefore, number of visible stars 3.5 × 103 stars = 8.8 × 1053 m3 × = 3600 stars. 8.5 × 1047 m3 =


CHAPTER 34

REFLECTION, REFRACTION, AND OPTICS

Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions Manual, available for purchase. Answers to all solutions below are underscored. †34-1. 34-2.

†34-3.

Ideally, each plane mirror will reflect the same energy flux of 0.10 W/cm2. To start the canvas on fire requires forty times this flux, 4.0 W/cm2. Therefore, 40 mirrors are required. Light is reflected from the cat in all directions. Some of those rays will reflect from the mirrors in such a way that they may enter a viewers eye and be seen. Note that the reflected image to the right is left-right inverted such that each point of the cat is equally distant from the mirror as each corresponding point on the image. The image on the top is similarly left-right inverted. The viewer sees light coming directly from the cat to his eye and also sees the virtual light rays coming from the two images.

Consider the drawing showing the incident ray of light approaching a point on the surface of the mirror at an angle θ with respect to a line drawn perpendicular to the surface at that point. From the law of reflection, the reflected ray leaves the point in a direction that is an angle θ with respect to the perpendicular line. Since the angle between the plane of the surface and the perpendicular line is 90°, the angles β are equal to 90° − θ.


θ β

θ β

CHAPTER 34-4. †34-5.

34-6.

34-7.

34-8.

34

Since the image distance is equal to the object distance, the camera should be focused at a distance of 2(1.2 m) = 2.4 m. The drawing at the right shows Figure 34.73 with some added features. The line normal to the surface is rotated by 10° as the mirror is rotated by 10° and 10° shown as a dashed line. The net result is that the reflected ray makes an angle φ with respect to the original normal line. After the mirror is rotated, the φ incident ray makes an angle of θ + 10° with respect to the rotated normal line. Since the angle of reflection is equal to the angle of incidence, φ = θ + 10° + 10° = θ + 20°. Therefore, with respect to the original reflected ray, we have rotated it by 20°. For two mirrors placed at an angle θ with respect to each other. The number of images formed by the mirrors is 360° 360° N = −1 = −1 = 5 θ 60° For two mirrors placed at an angle θ with respect to each other. The number of images formed by the mirrors is 360° 360° N = −1 = −1 = 7 θ 45° 50° All light rays coming from the sun are assumed to be parallel to one another. The angle of incidence for the rays is 50°; and they are reflected from the mirror at an angle of 50°. Consider a ray that strikes the mirror at the very top. It is reflected to the left-most point of the patch of sunlight on the floor. The width of the patch H will equal the width of the mirror. Therefore, the area of the patch will be ( 0.50 m )( 0.50 m ) = 0.21 m 2 ⎛ H ⎞ A = hW = ⎜ ⎟W = tan 50° ⎝ tan 50° ⎠

50° h


CHAPTER †34-9.

34-10.

34

Let’s define the following lengths that will aid us in determining the necessary height of the mirror H. h′′ h : your height (1.80 m) h′ : the height of your eyes above the floor H h′′ : the length between your eyes and the top of your head = h − h′ θ h The drawing shows the minimum height of the mirror θ such that a ray of light from the top of your head will h′ strike the mirror in such a way that its reflection will enter your eye. Note that the top of the mirror is located one-half the distance between your eyes and the top of your head. Similarly, a ray coming from the bottom of your foot will reflect from the bottom-most point on the mirror and enter your eye. The bottom of the mirror is located exactly halfway between your eyes and the floor. Thus, the height of the mirror is 1 1 1 1 1 H = h′ + h′′ = ( h′ + h′′ ) = h = (1.80 m ) = 0.90 m 2 2 2 2 2 Similarly, for the width of the mirror, we can draw a view from above as shown. Assume that you are standing directly in front of the mirror, so that W your midline is directly across from the midline of the mirror. As before, a ray from each shoulder reflects from the corresponding left or right edge of the mirror and into your eyes as shown. From the diagram, we see that the minimum width of the mirror is 1 1⎛1 ⎞ W = ⎜ w − w′ ⎟ + w′ 2 2⎝2 ⎠ 1 ⎛1 ⎞ W = ⎜ w − w′ ⎟ + 2w′ = w + 2 w′ w 2 ⎝2 ⎠ w′ 1 = ( 0.50 m ) + 0.06 m = 0.31 m 2 The speed of light in a material is given by Equation 34.4 c v= n where the indices of refraction n are given in Table 34.1. Therefore, for crown glass, ncg = 1.52 and vcg is c 3.00 × 108 m/s vcg = = = 1.97 × 108 m/s ncg 1.52

Similarly, for light flint glass and heavy flint glass, the speeds are c 3.00 × 108 m/s = = 1.90 × 108 m/s vlf = nlf 1.58 vhf = †34-11.

c 3.00 × 108 m/s = = 1.81 × 108 m/s nhf 1.66

The relationship between speed v, wavelength λ, and frequency f is v = λf, so the frequency of the light is


CHAPTER

f =

34 c

λ

=

3.00 × 108 m/s = 4.74 × 1014 Hz 633 × 10−9 m

When the light is passing through the crown glass (n = 1.52), the frequency is not changed, but its speed and wavelength do change. The speed of light in the crown glass is c 3.00 × 108 m/s vcg = = = 1.97 × 108 m/s ncg 1.52 The wavelength of the light in crown glass is v c 3.00 × 108 m/s λ= = = = 4.16 × 10−7 m or 416 nm 14 f nf (1.52) (4.74 × 10 Hz) 34-12. †34-13.

34-14.

n=

c 3.00 × 108 m/s = = 2.42 v 1.24 × 108 m/s

The index of refraction for castor oil is 1.48, but it is only 1.36 in ethyl alcohol. To find the factor of the increase in speed, take the ratio of the speeds. c ve n n 1.48 = e = co = = 1.09 c vco ne 1.36 nco

n1 sin θ1 = n2 sin θ 2 ⎛ n1 ⎞ ⎛ 1 ⎞ sin θ1 ⎟ = sin −1 ⎜ sin 40° ⎟ = 29° ⎝ 1.33 ⎠ ⎝ n2 ⎠

θ 2 = sin −1 ⎜ †34-15.

Since the index of refraction is inversely proportional to the speed of the waves in the media, we can write a version of Snell’s Law for these sound waves as n1 sin θ1 = n2 sin θ 2 1 1 sin θ1 = sin θ 2 v1 v2 ⎛ v2 ⎞ ⎛ 1500 m/s ⎞ sin θ1 ⎟ = sin −1 ⎜ sin 10° ⎟ = 50° ⎝ 340 m/s ⎠ ⎝ v1 ⎠

θ 2 = sin −1 ⎜

The maximum angle is 49°. 50

40

Refracted Angle

34-16.

30

20

10

0 0

15

30

45

60

Incident Angle


75

90

CHAPTER †34-17.

34-18. 34-19.

34-20. †34-21.

The Sun’s light passes through the space (n1 = 1.0000) between it and the Earth’s upper atmosphere making an angle of incidence θ1 on the atmosphere (n2 = 1.0003). The angle of refraction is θ2 = 39.000°. Snell’s Law gives the angle of incidence ⎛n ⎞ ⎛ 1.0003 ⎞ θ1 = sin −1 ⎜ 2 sin θ 2 ⎟ = sin −1 ⎜ sin 39.000° ⎟ = 39.014° ⎝ 1.0000 ⎠ ⎝ n1 ⎠ d 2.0 m d′ = = = 1.5 m n 1.33 Let’s use an index of refraction that falls in the middle of the two indices given. sin θ1 sin 30° = (1.00 ) = 1.46, fused quartz n2 = n1 sin θ 2 sin 20.1° d 5.0 cm d′ = = = 3.2 cm n 1.58 The light rays are incident on the water-glass interface at an angle θ1 = 30.0° where it refracts at an angle θ2 as shown. Given the three indices of refraction, the angle θ3 can be found by applying Snell’s Law at the two θ1 interfaces. n1 n sin θ 2 = 1 sin θ1 n2 n2 θ2 ⎞ n2 n ⎛ n sin θ 2 = 2 ⎜ 1 sin θ1 ⎟ n3 n3 ⎝ n2 ⎠ ⎛ n ⎞ ⎛ 1.33 ⎞ sin ( 30.0° ) ⎟ = 41.7° θ 3 = sin −1 ⎜ 1 sin θ1 ⎟ = sin −1 ⎜ ⎝ 1.00 ⎠ ⎝ n3 ⎠ The wavelength λ′ in the unknown clear material is sin θ 3 =

34-22.

34

λ′ =

θ3

n3

λ

n2

where the index of refraction is sin θ1 n2 = n1 sin θ 2 Therefore,

λ′ =

34-23. 34-24.

λ n2

=

λ

sin θ1 n1 sin θ 2

=

550 nm = 365 nm sin 45° (1.00 ) sin 28°

⎛ n2 ⎞ ⎛ 1.33 ⎞ sin θ 2 ⎟ = sin −1 ⎜ sin 30.0° ⎟ = 42° ⎝ 1.00 ⎠ ⎝ n1 ⎠ c d The speed of light in the crown glass is v = = where d is the thickness of the glass, n is the n t index of refraction, and t is the transit time for the light in the glass. The difference in the transit times for red and blue light is d ∆t = tblue − tred = ( nblue − nred ) c −3 ⎛ 1.00 × 10 m ⎞ −14 =⎜ ⎟ (1.530 − 1.510 ) = 6.67 × 10 s 8 3.00 × 10 m ⎝ ⎠

θ1 = sin −1 ⎜


CHAPTER

34-25. 34-26.

34

⎛1⎞ ⎛ 1 ⎞ ⎝ ⎠ ⎝ ⎠ ⎛ n ⎞ θ 2 = sin −1 ⎜ 1 sin θ1 ⎟ ⎝ n2 ⎠

θ C = sin −1 ⎜ ⎟ = sin −1 ⎜ ⎟ = 24.4° n 2.42

θ1

n sin θ 3 = 2 sin ( 90° − θ 2 ) n3 ⎛ ⎛ n ⎞⎞ n = 2 sin ⎜⎜ 90° − sin −1 ⎜ 1 sin θ1 ⎟ ⎟⎟ n3 ⎝ n2 ⎠⎠ ⎝

θ2

θ3

90° − θ2

⎛ 1.58 ⎛ 1.00 ⎞⎞ sin ⎜ 90° − sin −1 ⎜ sin 70° ⎟ ⎟ = 1.27 1.00 ⎝ 1.58 ⎠⎠ ⎝ This indicates that θ3 does not exist and total internal reflection occurs. The ray does not exit the second side of the prism. ⎛ n ⎞ ⎛ n ⎞ ⎛ n ⎞ θ 2 = sin −1 ⎜ 1 sin θ1 ⎟ = sin −1 ⎜ 1 sin 30° ⎟ = sin −1 ⎜ 1 ⎟ ⎝ n2 ⎠ ⎝ n2 ⎠ ⎝ 2n2 ⎠ =

34-27.

34-28. †34-29.

For red light, n1 = 1.650, θ2 = 55.6° For green light, n1 = 1.660, θ2 = 56.1° For red light, n1 = 1.690, θ2 = 57.7° ⎛1⎞ ⎛ 1 ⎞ θ C = sin −1 ⎜ ⎟ = sin −1 ⎜ ⎟ = 41.8° ⎝n⎠ ⎝ 1.50 ⎠ (a) Consider light that is incident on an interface where n1 > n2. Depending on the angle of incidence, some of the light will be reflected at the interface and some will be transmitted at a refracted angle θ2. Snell’s Law gives n1 sin θ1 = n2 sin θ 2

sin θ1 =

n2 sin θ 2 n1

As the incident angle is varied with respect to the normal to the interface, a critical angle θc is reached where total internal reflection occurs and there is no transmitted light. The refracted angle is then 90°. n n sin θ c = 2 sin 90° = 2 n1 n1 n (b) sin θ c = 2 n1 ⎛ n2 ⎞ −1 ⎛ 1.20 ⎞ ⎟ = sin ⎜ ⎟ = 64.5° ⎝ 1.33 ⎠ ⎝ n1 ⎠

θ c = sin −1 ⎜ 34-30.

sin θ C = n=

1 n 1 1 = = 1.2 sin θ C sin 56°


CHAPTER

34-31.

34-32.

⎛1⎞ ⎛ 1 ⎞ ⎝ ⎠ ⎝ ⎠ If the glass is immersed in water, then ⎛n ⎞ ⎛ 1.33 ⎞ θ C = sin −1 ⎜ 2 ⎟ = sin −1 ⎜ ⎟ = 61.0° ⎝ 1.52 ⎠ ⎝ n1 ⎠

θ C = sin −1 ⎜ ⎟ = sin −1 ⎜ ⎟ = 41.1° n 1.52

⎛ n2 ⎞ −1 ⎛ 1.58 ⎞ ⎟ = sin ⎜ ⎟ ⎝ 1.52 ⎠ ⎝ n1 ⎠ The critical angle is undefined for this combination of materials. The critical angle for the flint glass-air interface is ⎛1⎞ ⎛ 1 ⎞ θ C = sin −1 ⎜ ⎟ = sin −1 ⎜ ⎟ = 39.3° n ⎝ ⎠ ⎝ 1.58 ⎠ The incidence angle within the crown glass is found via Snell’s Law. (1.52) sin θ = (1.58) sin (39.3°)

θ C = sin −1 ⎜

⎛ 1.58 ⎞ sin(39.3°) ⎟ = 41.1° ⎝ 1.52 ⎠ So, for light that leaks from the crown glass at angles > 41.1°, the light will be totally internally reflected. n sin θ C = 2 n1

θ = sin −1 ⎜

34-33.

n2 = n1 sin θ C = (1.000 26 ) sin 89.4° = 1.000 21 34-34.

Brewster’s Law (Equation 34.17) gives the relationship between the angle at which total polarization occurs and the index of refraction. n = tan θ p = tan (60.5°) = 1.77

†34-35.

Brewster’s Law (Equation 34.17) gives the relationship between the angle at which total polarization occurs and the index of refraction. tan θ p = n

θ p = tan −1 n = tan −1 (1.33 ) = 53.1° θ horizon = 90° − θ p = 90° − 53.1° = 36.9° 34-36.

34

The index of refraction of cubic zirconia may be found from the critical angle for total internal reflection. 1 sinθ C = n 1 n= sinθ C Then, Brewster’s law gives the angle for total polarization, ⎛ 1 ⎞ 1 ⎞ −1 ⎛ θ p = tan −1 n = tan −1 ⎜ ⎟ = tan ⎜ ⎟ = 65.6° ⎝ sin 27.0° ⎠ ⎝ sinθ C ⎠


CHAPTER †34-37.

34

The incident light is refracted at the air-glass interface to an angle θ2 given by ⎛ n ⎞ ⎛ 1.0 ⎞ sin 50° ⎟ = 30.7° θ 2 = sin −1 ⎜ 1 sin θ1 ⎟ = sin −1 ⎜ ⎝ 1.5 ⎠ ⎝ n2 ⎠

θ1 t

θ2

The hypotenuse of the triangle that has the side of length d shown in the drawing is

L

θ3

d

L = (2.00 × 10−3 m) ⎡⎣ tan ( 50° ) − tan ( 30.7° ) ⎦⎤ = 1.196 × 10−3 m To find d, the angle θ3 is found by applying Snell’s Law again. ⎛ n ⎞ ⎛ 1.5 ⎞ θ 3 = sin −1 ⎜ 2 sin θ 2 ⎟ = sin −1 ⎜ sin 30.7° ⎟ = 50° ⎝ 1.0 ⎠ ⎝ n3 ⎠ As expected, the final refracted ray is parallel to the initial incident ray. The distance d is then, d sin (90° − θ 3 ) = L d = L sin (90° − θ 3 ) = (1.196 × 10−3 m ) sin 40° = 7.7 × 10−4 m or 0.77 mm 34-38.

The first image occurs a distance l below the top surface of the glass (just as it does for a plane mirror). Let a represent the distance between the reflected ray and the second parallel ray leaving the top surface as shown. The distance below the first image to the second image is represented by x. Here is a magnified image of the rays within the glass: The incident ray approaches the surface at angle θ and is refracted at angle θ′. The distance a/2 = d tan θ′ Then, a = 2d tan θ′ The distance x is then given by a 2d tan θ ′ = x= tan θ tan θ

a

a/2

θ′ θ′

a/2 d

x

From Snell’s Law, there is a relationship between θ and θ′ and these angle are quite small, so the small angle approximation applies, i.e. tan θ ≈ θ and tan θ′ ≈ θ′. Therefore, 2dθ ′ 2dθ ′ 2d = = x= θ nθ ′ n 2d The distance from the top surface to the second image is l + x = l + n 4d ⎛ 2d ⎞ Similarly, the distance from the top surface to the third image is l + 2 x = l + 2 ⎜ ⎟=l+ n ⎝ n ⎠


CHAPTER †34-39.

34-40.

Since the incident ray is polarized in the plane of incidence, no reflected ray will occur at the Brewster angle. n 1.33 θ B = tan −1 2 = tan −1 = 53.1° n1 1.00 (a) The lateral displacement of the ray that originates from behind the glass is θ1 t d = sin (θ1 − θ 2 ) θ2 cos θ 2

t

and ∆l =

L

⎛ t ⎞ sin (θ1 − θ 2 ) d =⎜ ⎟ sin θ1 ⎝ cos θ 2 ⎠ sin θ1

θ3

d

When the angles are very small, cos θ ≈ 1 and sin θ ≈ θ Then, ⎛ d ⎛ θ1 − θ 2 ⎞ θ2 ⎞ ∆l = =⎜ ⎟ t = ⎜1 − ⎟ t θ1 ⎝ θ1 ⎠ θ1 ⎠ ⎝ Applying Snell’s Law for small angles, θ1 = nθ2, 1⎞ ⎛ ∆l = ⎜ 1 − ⎟ t n⎠ ⎝

34-41.

34-42.

1⎞ 1 ⎞ ⎛ ⎛ (b) ∆l = ⎜ 1 − ⎟ t = ⎜ 1 − ⎟ 2.0 mm = 0.67 mm n⎠ ⎝ ⎝ 1.5 ⎠ 1⎞ 1 ⎞ ⎛ ⎛ (c) ∆l = ⎜ 1 − ⎟ t = ⎜ 1 − ⎟ 8.0 mm = 2.7 mm n⎠ ⎝ ⎝ 1.5 ⎠ The maximum reflection at the back of the sphere occurs when a light ray enters at the top of the sphere as shown to a refracted angle θ2 = 45°, which is deduced from the geometry shown. In this case, θ1 = 90° and sin θ1 sin 90° 2 n= = = = 2 sin θ 2 sin 45° 2 The index of refraction must be at least 1, therefore 1≤ n ≤ 2 We can apply Snell’s Law to sound waves as we do for light, with the index of refraction inversely proportional to the speed of the wave in the given medium. Therefore, n v sin θ C = 2 = 1 n1 v2

θ2

h

Consider the drawing of this situation. The distance d from the ship to the point at which the sound wave will undergo total reflection is


θC d

∆l

34

CHAPTER

†34-43.

34

⎡ ⎛ v ⎞⎤ ⎡ ⎛ 340 m/s ⎞ ⎤ d = h tan θ C = h tan ⎢ sin −1 ⎜ 1 ⎟ ⎥ = (200 m) tan ⎢ sin −1 ⎜ ⎟ ⎥ = 46.5 m = 47 m ⎝ 1500 m/s ⎠ ⎦ ⎣ ⎝ v2 ⎠ ⎦ ⎣ Consider the drawing of the situation. A ray of light originating in water (n = 1.33) is incident on the water-oil θC θ′ interface. It is refracted at an angle θ′. If the ray is to be totally reflected at the oil-air interface, then θ′ ≤ θC. The θ maximum value of θ occurs when θ′ = θC. Applying Snell’s Law gives, n sin θ = n′ sin θ ′

n′ n

⎛1⎞ n sin θ max = n′ sin θ C = n′ ⎜ ⎟ = 1 ⎝ n′ ⎠ ⎛1⎞ ⎛ 1 ⎞ θ max = sin −1 ⎜ ⎟ = sin −1 ⎜ ⎟ = 48.8° ⎝n⎠ ⎝ 1.33 ⎠ The maximum angle has no dependence on n′. 34-44.

†34-45.

34-46.

The shark can be seen when the ray of light from the shark approaches the water-air interface at an angle less than the critical angle. This maximum angle was found in the previous problem to be θC = 48.8°. Therefore, using the right triangle shown, the horizontal distance is x = (5.0 m) tan θ C = (5.0 m) tan 48.8° = 5.7 m (a) Since the critical angle is given by ⎛ n′ ⎞ θ C = sin −1 ⎜ ⎟ ⎝n⎠ the index of refraction n′ of the solution is n′ = n sinθ C = (1.6640 ) sin ( 57.295° ) = 1.4002

5.0 m

θC x

(b) From the table, we see that the index increases by 0.0002 for each 0.10% increase in concentration. Therefore, this is a linear relationship between index and concentrations (at least for the range of values given). The value found in part (a) is halfway between the index values given for concentrations 40.20 and 40.30%, therefore, after interpolating, the unknown concentration is 40.25%. Consider the angles shown in the drawing. For total internal θ 90° reflection at the glass-air interface, the angle β = 90.0°. From θ′ the drawing, the following relationships are present: ′ θ′′ α + 45.0° + 90.0° − θ = 180° 45° α 45° α = 45.0° + θ ′ (i) β θ ′′= 90.0° − α = 45.0° − θ ′ (ii) The critical value of θ′′ is given by ⎛ n′ ⎞ ⎛ 1 ⎞ θ C′′ = sin −1 ⎜ ⎟ = sin −1 ⎜ ⎟ = 41.81° ⎝n⎠ ⎝ 1.50 ⎠ Applying equation (ii) gives θ ′= 45.0° − θ C′′ = 45.0° − 41.81° = 3.19°. At the air-glass interface,


CHAPTER

34

n1 sin θ = n2 sin θ ′ ⎛ n2 ⎞ ⎛ 1.50 ⎞ sin θ ′ ⎟ = sin −1 ⎜ sin ( 3.19° ) ⎟ = 4.79° ⎝ 1.00 ⎠ ⎝ n1 ⎠ (a) At the first air-water interface, the ray is refracted at the angle θ′. Note that all normals to the surface of a sphere intersect at the center of the sphere, D. Because of this, as the drawing shows, there are two equal, isosceles triangles. The reflected ray at B has equal angles θ′ about the normal. Because the incident angle at the final water-air interface is θ′, Snell’s Law gives the final refracted angle θ, which is equal to the initial incident angle. (b) The drawing shows the angle of deflection at each of the three points. The angles are evident from the drawing. (c) At the critical angle θC, d∆ dθ ′ = 2−4 =0 dθ dθ dθ ′ 1 = dθ 2 However, from Snell’s Law at the air-water interface, n1 sin θ1 = n2 sin θ 2 sin θ = n sin θ ′ Taking the derivative of this, gives cos θ dθ = n cos θ ′ dθ ′ dθ ′ cos θ 1 = = dθ n cos θ ′ 2

θ = sin −1 ⎜

†34-47.

⎛ 1 ⎞ Also, cos θ C′ = 1 − sin 2θ C′ = 1 − ⎜ 2 ⎟ sin 2θ C ⎝n ⎠ Therefore, cos θ C cos θ C 1 dθ ′ = = = 2 dθ n cos θ C′ ⎛ 1 ⎞ n 1 − ⎜ 2 ⎟ sin 2θ C ⎝n ⎠ cos θ C =

1 ⎛ 1 ⎞ n 1 − ⎜ 2 ⎟ sin 2θ C 2 ⎝n ⎠

1 2 ⎛ ⎛ 1 ⎞ 2 ⎞ 1 2 n ⎜ 1 − ⎜ 2 ⎟ sin θ C ⎟ = ( n − sin 2θ C ) 4 ⎝ ⎝n ⎠ ⎠ 4 1 1 cos 2 θ C + sin 2θ C = n 2 4 4 3 1 1 cos 2 θ C + ( sin 2θ C + cos 2 θ C ) = n 2 4 4 4

cos 2 θ C =


θ

A

θ′

θ′ θ′

D

θ′

B

θ C

θ

θ − θ′

A

θ′ D

θ′ C

θ′ θ

θ′ θ′

B

π − 2θ′

CHAPTER

34

3 1 1 cos 2 θ C = n 2 − 4 4 4 1 cos 2 θ C = ( n 2 − 1) 3 (d) Using the result from part (c) 1 2 θ C = cos −1 ( n − 1) = cos−1 13 ( (1.330)2 − 1) 3 = 59.585° = 59°35′ Then, ∆ = 2(θ C − θ ′) + (180° − 2θ ′) where θ′ is found from Snell’s Law at the air–water interface, sin θ C = n sin θ ′

⎛1 ⎞ n ⎝ ⎠ Therefore, ⎡ ⎛1 ⎞⎤ ⎛1 ⎞ ∆ = 2 ⎢θ C − sin −1 ⎜ sin θ C ⎟ ⎥ + 180° − 2 sin −1 ⎜ sin θ C ⎟ n n ⎝ ⎠⎦ ⎝ ⎠ ⎣ ⎛1 ⎞ = 2θ C + 180° − 4 sin −1 ⎜ sin θ C ⎟ ⎝n ⎠

θ ′ = sin −1 ⎜ sin θ C ⎟

⎛ 1 ⎞ sin 59.585° ⎟ = 137.483° = 137°29′ = 2 ( 59.585° ) + 180° − 4 sin −1 ⎜ ⎝ 1.33 ⎠ (e) As shown in the drawing, violet light exits the drop and follows a trajectory that never enters the eye, if the red light does reach the eye. The violet ray that the observer sees must come from a drop that is lower, so red appears higher in the sky than violet. The critical angle for violet light is 1 2 1 θ C = cos −1 n − 1) = cos −1 ( ( (1.342)2 − 1) 3 3 = 59.888° = 59°53′ Then,

⎡ ⎛1 ⎞⎤ ⎛1 ⎞ ∆ = 2 ⎢θ C − sin −1 ⎜ sin θ C ⎟ ⎥ + 180° − 2 sin −1 ⎜ sin θ C ⎟ ⎝n ⎠⎦ ⎝n ⎠ ⎣ ⎛1 ⎞ = 2θ C + 180° − 4 sin −1 ⎜ sin θ C ⎟ ⎝n ⎠

34-48.

⎛ 1 ⎞ sin 59.888° ⎟ = 139.235° = 139°14′ = 2 ( 59.888° ) + 180° − 4 sin −1 ⎜ 1.342 ⎝ ⎠ The radius of curvature is equal to twice the focal length, which may be determined using Equation 34.21. 1 1 1 + = s s′ f


CHAPTER

†34-49.

1 1 1 + = f 4.0 cm ∞ 1 f = R = 4.0 cm 2 R = 8.0 cm Three rays are drawn from the top of the object, represented by the tip of an arrow. Ray 1 is directed parallel to the principal axis and reflected along a line that if extended behind the mirror, passes through the focal point f of the convex mirror. Ray 2 is directed toward the focal point. That ray reflects and is directed parallel to the principal axis. Ray 3 is directed toward the center of curvature at a distance 2f behind the mirror. That ray reflects back on itself. The three reflected rays do not intersect on the object side of the mirror. If those reflected rays are extended behind the mirror, they intersect and form a virtual, upright, and reduced image at a point between the mirror and the focal point. Applying the mirror equation gives 1 1 1 + = s s′ f 1 1 1 + = 15.0 cm s′ −40.0 cm −1

34-50.

†34-51.

34

1 1 ⎛ ⎞ − s′ = ⎜ ⎟ = −10.9 cm ⎝ −40.0 cm 15.0 cm ⎠ The virtual image is formed 10.9 cm behind the mirror. Note that when applying the mirror equation to a convex mirror, the focal length is negative. 1 1 1 + = s s′ f 1 1 1 + = s s f s = 2f = R The object should be placed at the radius of curvature for the concave mirror. Since f = R/2, the mirror equation may be written, 1 1 2 + = s s′ R If a real image is to be formed, then s′ must be positive. 1 2 1 = − >0 ′ s R s 1 s> R 2


CHAPTER

34

For a virtual image, s′ is negative. 1 2 1 = − <0 ′ s R s 1 s< R 2 34-52.

†34-53.

34-54.

The object (the candle) is placed within the focal length of the concave mirror and forms an enlarged, upright, virtual image. 1 1 1 2f f s s′ = − s′ f s 1 1 1 = − s′ 20.0 cm 15.0 cm s′ = −60.0 cm Since the radius of the ornament (a sphere) is 6.0 cm, the focal length is f = −3.0 cm. The negative sign is used because the focal length is on the opposite side of the reflecting surface than the object (the candy cane). The mirror equation is used to find the image distance. 1 1 1 = − s′ f s 1 1 1 = − ′ −3.0 cm 15.0 cm s s′ = −2.5 cm The moon is so far away that its object distance is essentially infinite. We’ll apply the mirror equation for the two instances, recognizing that the focal length of the mirror does not change. Therefore, 1 1 1 + = s s′ f 1 1 1 1 + = + s1 s1′ s2 s2′ 1 1 1 1 + − = ∞ s1′ s2 s2′ 0+

1 1 1 − = 6.0 cm 8.0 cm s2′

s2′ = 24 cm 34-55.

34-56.

1 1 1 = − s f s′ 1 1 1 = − s −25.0 cm −10.0 cm s = 16.7 cm The only case of the image forming on the same side of the spoon as the object is when the woman is facing the concave side of the spoon and her eye is placed within the focal distance of the mirror. The relevant mirror equation is


CHAPTER

†34-57.

34-58.

34-59.

1 1 2 + = s s′ R For the two cases, we have 1 1 2 + = Concave: s 7.0 cm R 1 1 2 =− Convex: + s −2.0 cm R For these two equations with two unknowns, we find 1 1 1 1 + =− − s −2.0 cm s 7.0 cm 2 1 1 =− + 7.0 cm 2.0 cm s s = 3.11 cm = 3.1 cm and 1 1 2 + =− R 3.11 −2.0 cm R = 11.2 cm The magnification M of the mirror gives s′ M = − = 1.5 s ′ s = −1.5s = −1.5(20.0 cm) = − 30.0 cm Therefore, by knowing both the image and object distances the mirror equation can be used to find the radius of curvature. 1 1 2 + = s s′ R 1 1 2 + = 20.0 cm −30.0 cm R R = 120.0 cm The mirror is concave. 1 1 2 + = ′ s s R 1 2 1 2 1 = − = − s′ R s −4.5 cm 15 cm s′ = −1.956 cm = −1.96 cm −1.956 cm M =− = −0.13 15 cm The image size is 1/M = 7.6 times smaller than your face. (a) Consider the first mirror. 1 1 1 + = ′ s s f


34

CHAPTER

34

1 1 1 1 1 = − = − s′ f s 15.0 cm 20.0 cm s′ = 60.0 cm (b) The image formed by the first mirror becomes the object for the second mirror. The object distance is s = 80.0 cm − 60.0 cm = 20.0 cm. The final image is formed at 1 1 1 1 1 = − = − s′ f s 12.0 cm 20.0 cm s′ = 30.0 cm 34-60.

†34-61.

For the concave mirror, the image distance is found from the mirror equation. 1 1 1 + = ′ s s f 1 1 1 1 1 = − = − ′ s f s 30.0 cm 40.0 cm s′ = 120.0 cm This image then becomes the object for the convex mirror. 1 1 1 1 1 = − = − ′ s f s −30.0 cm 70.0 cm s′ = −21.0 cm Applying the thin lens formula Equation 34.27, we find ⎛ 1 1 1 ⎞ 1 ⎛ 1 ⎞ = ( n − 1) ⎜ + + ⎟ = (1.45 − 1) ⎜ ⎟ f ⎝ 10 mm 6.0 mm ⎠ ⎝ R1 R2 ⎠ f = 8.3 mm

34-62.

34-63.

34-64.

⎛ 1 1 1 ⎞ 1⎞ ⎛ 1 = ( n − 1) ⎜ + + ⎟ ⎟ = (1.58 − 1) ⎜ ∞ f R R 15 cm ⎝ ⎠ ⎝ 1 2 ⎠ f = 25.9 cm = 26 cm 1 1 1 (b) + = ′ s s f 1 1 1 1 1 = − = − ′ s f s 25.9 cm 40.0 cm s′ = 73 cm

(a)

⎛ 1 1 1 ⎞ ⎛ 2 ⎞ 2 ( n − 1) = ( n − 1) ⎜ + ⎟ = ( n − 1) ⎜ ⎟ = f R ⎝R⎠ ⎝ R1 R2 ⎠ R = 2 ( n − 1) f = 2 (1.52 − 1) (20 cm) = 20.8 cm = 21 cm ⎛ 1 1 1 ⎞ = ( n − 1) ⎜ + ⎟ f ⎝ R1 R2 ⎠ 1 1 1 1 1 = − = − R2 f ( n − 1) R1 16 cm (1.66 − 1) 27 cm R2 = 17 cm


CHAPTER

34-65.

34-66.

†34-67.

34-68.

34-69.

34-70.

34

⎛ 1 1 1 ⎞ 1 ⎞ ⎛ 1 = ( n − 1) ⎜ + + ⎟ = (1.58 − 1) ⎜ ⎟ f ⎝ 16 cm 12 cm ⎠ ⎝ R1 R2 ⎠ f = 12 cm ⎛ 1 1 1 ⎞ 1 ⎛ 1 ⎞ = ( n − 1) ⎜ + + ⎟ = (1.43 − 1) ⎜ ⎟ − f R R 2.5 cm 3.0 cm ⎝ ⎠ ⎝ 1 2 ⎠ f = 35 cm The ray diagram for this situation is similar to that shown in Figure 34.51. Apply Equation 34.28 using the sign conventions illustrated in Figure 34.48. Assuming the object is placed 15 cm to the left of the concave lens with f = −20 cm, 1 1 1 + = s s′ f 1 1 1 1 1 = − = − s′ f s −20 cm 15 cm s′ = −8.57 cm = −8.6 cm The image forms 8.6 cm to the left of the lens. The image is virtual since it is formed by the intersection of virtual rays on the left side of the lens. The magnification is given by Equation 34.30, s′ −8.57 cm M =− =− = 0.57 s 15 cm The image is smaller and upright. h If the object distance can be considered infinite; and the image has θ ′ = , then f ⎛ π rad ⎞ h = f θ ′ = (150 mm )( 0.52° ) ⎜ ⎟ = 1.36 mm ⎝ 180° ⎠ 1 1 1 + = s s′ f 1 1 1 1 1 = − = − s f s′ 0.130 m 2.0 m s′ = 0.139 m = 14 cm s′ 2.0 m M =− =− = 14 s 0.139 m s′ M = − = 2.0 s s′ = −2.0s 1 1 1 + = s −2.0 s f 2 1 2 − = s s f 1 1 s = f = ( 20 cm ) = 10 cm 2 2


CHAPTER †34-71.

34-72.

34

The ray diagram drawn for this situation shows that the image will be inverted, real, and enlarged. Applying the lens formula gives, 1 1 1 + = s s′ f 1 1 1 1 1 = − = − s′ f s 18 cm 30 cm s′ = 45 cm

f

f

The Gaussian lens makers formula is derived in most optics texts and may also be found on the internet. The problem asks for an argument for the modification of Equation 34.27 when the lens is surrounding by a medium other than air. When Snell’s Law applied to the interface between the surrounding medium and the lens gives n sin θ = n0 sin θ 0 . This gives, n sin θ 0 = sin θ n0 If the lens is in air, n0 = 1; and you have the situation that gives Equation 34.27. Since we have replaced n by n/n0 in Snell’s Law and Snell’s Law is the basis for Equation 34.27, it makes sense to simply make the same substitution to give: ⎞⎛ 1 1 ⎛ n 1 ⎞ = ⎜ − 1⎟ ⎜ + ⎟ f ⎝ n0 ⎠ ⎝ R1 R2 ⎠

†34-73.

Figure 34.81 shows two converging lenses in contact. Assume an object is placed a distance s to the left of the first lens, the image will be formed at a distance s1 to the right of the first lens. The distances are related by the lens formula. 1 1 1 = + f1 s s1 This image then becomes the object for the second lens. The final image is formed at a distance s2, which may again be found using the lens formula with the proper sign convention. 1 1 1 1 1 = + = − f 2 − s1 s2 s2 s1 Since both equations contain s1, we can further relate both equations. 1 1 1 = − s1 f1 s

1 1 1 1 1 1 = − = − + f 2 s2 s1 s2 f1 s 1 1 1 1 1 + = + = f1 f 2 s1 s2 f 34-74.

(a)

1 1 1 1 1 = − = − s′ f s 25 cm 80 cm s′ = 36.36 cm = 36 cm

(b) The image formed in part (a) becomes the object in part (b). The object distance for the mirror will be 60 cm − 36.36 cm = 23.64 cm. Then, using the mirror equation, we can find where the second image is formed.


CHAPTER

34

1 1 1 1 1 = − = − s′′ f m s′ 20 cm 23.64 cm s′′ = 130 cm (c) The image formed from the mirror becomes an object for the lens again. The object distance is negative since it is located 129.89 cm − 60 cm = 69.89 cm to the left of the lens, which in this case is a negative distance with regard to the lens formula. 1 1 1 1 1 = − = + ′′′ ′′ s f s 25 cm 69.89 cm s′′′ = 18.41 cm = 18 cm 34-75.

34-76.

†34-77.

The final image forms 18 cm to the right of the lens. 1 1 1 1 1 = − = − (a) s′ f s 30 cm 20 cm s′ = −60 cm 1 1 1 1 1 = − =− − (b) s′′ f s′ 30 cm 70 cm s′′ = −21 cm 1 1 1 1 = − = 3 s′ f 3f f 2 s′ = 3 f This image distance is 2f to the right of the second lens. 1 1 1 3 = + = s′′ f 2f 2f 2 s′′ = f 3 The final image is formed at a distance 2f/3 to the left of the left lens. Consider Figure 34.84. The usual image and object distances can be written: s = x + f and s′ = x′ + f. Using these expressions in the lens formula gives, 1 1 1 1 1 x + f + x′ + f x + x′ + 2 f = + = + = = ′ ′ ′ ′ f s s x+ f x + f ( x + f )( x + f ) xx + xf + x′f + f 2 1 x + x′ + 2 f = f xx′ + f ( x + x′ ) + f 2 f ( x + x′ ) + 2 f 2 = xx′ + f ( x + x′ ) + f 2 xx′ = f 2

34-78.

From Problem 34-73, 1 1 1 + = f1 f 2 f 1 1 1 1 1 = − = − f2 f f1 22.0 cm 12.0 cm f 2 = −26.4 cm


CHAPTER †34-79.

34-80.

†34-81.

34

First find the image for the mirror with the object placed at 15 cm in front of the mirror. The image is formed at 1 1 1 2 1 2 1 = − = − = − s′ f s R s −10.0 cm 15.0 cm s′ = −3.75 cm This indicates that the image is formed 3.75 cm behind the mirror, which is 23.75 cm from the lens. This image becomes the object for the lens. We then apply the lens formula to find the distance to the final image. 1 1 1 1 1 = − = − s′ f s 25.0 cm 23.75 cm s′ = −475 cm The final image is 475 cm behind the lens, which is also behind the mirror. Assume the lens with a focal length f1 = 40 cm is to the left of the lens with a focal length f2 = 60 cm. If parallel rays of light approach from the left, the first lens focuses the light at a point 40 cm to its right, which is 30 cm to the right of the right lens. That image becomes the object for the lens on the right, which focuses the light at a distance given by the lens formula. 1 1 1 1 1 = − = − s′ f s 60 cm −30 cm s′ = 20 cm This is located 20 cm to the right of the right lens. In this case of parallel rays of light approaching from the left, the effective focal length measured from the center of the two lens system is located at fleft = 20 cm − 5 cm = 15 cm to the right of the system center If parallel rays of light approach from the right, the right lens focuses the light at a point 60 cm to its left, which is 50 cm to the left of the left lens. That image becomes the object for the lens on the left, which focuses the light at a distance given by the lens formula. 1 1 1 1 1 = − = − s′ f1 s 40 cm −50 cm s′ = 22.22 cm = 22 cm This is located 22 cm to the left of the left lens. In this case of parallel rays of light approaching from the right, the effective focal length measured from the center of the two lens system is located at fright = 22 cm − 5 cm = 17 cm to the left of the system center From the drawing, the following relationships can be shown: θ1 = α + β (i) θ2 =β − α′ (ii) Snell’s Law applied at the interface is n1 sin θ1 = n2 sin θ 2 The angles are generally quite small; and the small angle approximation yields, n1θ1 = n2θ 2 . Multiplying equation (i) by n1 gives n1θ1 = n1α + n1β; and equation (ii) multiplied n2 gives n2θ2 = n2β − n2α′. These two equations are equal to each other, so


CHAPTER

θ2

θ1 α

34

h

β

α′

n1α + n1 β = n2 β − n2α ′ h h h + ( n1 − n2 ) = − n2 s R s′ n1 n2 n2 − n1 + = s s′ R In the second line above, we have used the small angle approximation that tan α = α = h/s and similar relations for the other angles. At the first interface, the result of Problem 34-81 gives, n1 n2 n2 − n1 + = (i) s1 s1′ R1 n1

34-82.

Similarly, at the second interface, n2 n1 n1 − n2 + = (ii) s2 s2′ R2 Assume the thickness of the lens is d, then the object distance at the second interface is properly related to the image distance resulting from refraction at the first interface as s2 = d − s1′. Substitute this relationship into equation (ii) and add equations (i) and (ii) together. n1 n2 n2 n n − n1 n1 − n2 + + + 1 = 2 + s1 s1′ d − s1′ s2′ R1 R2 In the thin lens approximation d is approximately zero, so n1 n2 n2 n1 n2 − n1 n1 − n2 + − + = + s1 s1′ s1′ s2′ R1 R2 ⎛ 1 n1 n1 1 ⎞ (iii) + = ( n2 − n1 ) ⎜ − ⎟ s1 s2′ ⎝ R1 R2 ⎠ Since n1 is 1 for air and s1 = s, the original object distance, and s2′ = s′, the final image distance, we can simplify equation (iii) by letting n2 = n to give ⎛ 1 1 1 1 ⎞ + = ( n − 1) ⎜ − ⎟ ′ s s ⎝ R1 R2 ⎠

The focal length is defined for an object placed at infinity or for an image at infinity, so


CHAPTER

34

1 1 1 1 1 = + = + f ∞ s′ s ∞

†34-83.

34-84.

Therefore, ⎛ 1 1 1 ⎞ = ( n − 1) ⎜ − ⎟ (34.27) f ⎝ R1 R2 ⎠ and 1 1 1 = + (34.28) f s s′ Apply the equation from Problem 34-81, n1 n2 n2 − n1 + = s s′ R at the first air-glass interface, with n1 = 1 (for air) and s = ∝ (since the Sun is very far away and the light comes in as parallel rays). n2 n2 − n1 n1 = − s′ R s 1.60 1.60 − 1.00 1 = − 7.5 cm s′ ∞ s′ = 20.0 cm The image distance then becomes the object distance for the second interface. n2 n2 − n1 n1 = − s′ R s 1.00 1.00 − 1.60 1.00 = − 7.5 cm 20.0 cm − 15.0 cm s′ s′ = −3.57 cm The final image is observed at 7.5 cm − 3.57 cm = 3.9 cm to the right of the center of the ball. The small segment is considered to be part of a lens with a radius of curvature R and focal length f. The lens makers’ formula (Equation 34.27) is applied to this lens segment. θ ⎛ 1 ⎛ 1 1 1 ⎞ 1 ⎞ n −1 = ( n − 1) ⎜ + ⎟ = ( n − 1) ⎜ + ⎟ = R f R ⎝ R1 R2 ⎠ ⎝ R1 ∞ ⎠ r θ R = f ( n − 1) f

Consider the drawing. r sin θ = R ⎛ ⎞ r ⎛r⎞ θ = sin −1 ⎜ ⎟ = sin −1 ⎜ ⎟ ⎝R⎠ ⎝ f (n − 1) ⎠ †34-85.

The camera is designed to focus the image at the plane of the film in the camera, so the image distance is adjustable so that a range of object distances can be accommodated. When the lens is 50 mm or 5.0 cm from the film, the object distance is 1 1 1 1 1 = − = − =0 s f s′ 5.0 cm 5.0 cm s=∞


CHAPTER

34-86.

34

When the lens is 62 mm or 6.2 cm from the film, the object distance is 1 1 1 1 1 = − = − ′ s f s 5.0 cm 6.2 cm s = 25.8 cm So, the camera can form a sharp image for object distances ranging from 25.8 cm to infinity. When the object is 20 cm from the lens, 1 1 1 1 1 = − = − s′ f s 1.5 cm 20 cm s′ = 1.6 cm When the object is at infinity, 1 1 1 1 1 = − = − ′ s f s 1.5 cm ∞ s′ = 1.5 cm

†34-87.

The f number is defined as the focal length of the lens divided by its diameter, f = f/d. The iris changes the effective diameter of the lens, but not its focal length. Since we want to relate the f number to an exposure time and the f number is related to the effective diameter, let’s define a new parameter called the exposure E, which is the product of the amount of light incident on the film (which is inversely proportional to the area) and the exposure time, E = At. For the first f number, 1.7, the exposure time is 1/250 s, therefore the exposure is 2 1 1

E = A1t1 = π R t =

π d12

π ( f/f1 )

2

t1 = t1 4 4 Similarly, for the second f number, the same exposure results when the proper exposure time is used. Therefore, the exposure equations for both f numbers are set equal to each other. We then solve for the second exposure time for an f number of 8.

π ( f/f1 )

2

t1 =

π ( f/f 2 )

4 2 2 f 2 t1 = f1 t2 2

t2 = 34-88.

34-89.

f 2 t1 = 2 f1

(8)

4

2

t2

⎛ 1 ⎞ s⎟ ⎜ ⎝ 250 ⎠ = 8.9 × 10−2 s (1.7) 2

2

(a) Since the image must be focused on the retina, the focal length of the eye is equal to the image distance 2.2 cm for an object placed at infinity. (b) For an object placed at 25 cm from the eye, the focal length must be 1 1 1 1 1 = + = + f s s′ 25 cm 2.2 cm f = 2.0 cm 1 1 1 (a) + = f1 f 2 f 1 1 1 1 1 = − = − f2 f f1 2.2 cm 2.0 cm f 2 = −22 cm This is a diverging lens.


CHAPTER

(b)

34 ⎛ 1 1 1 ⎞ n −1 n −1 = ( n − 1) ⎜ + + ⎟= f R1 R2 ⎝ R1 R2 ⎠ n −1 1 n −1 = − R2 f R1 1 1 1 1 1 = − = − R2 f ( n − 1) R1 −22 cm (1.33 − 1) −0.80 cm R2 = −0.72 cm

34-90.

†34-91.

The focal length of the contact lens in Problem 34-66 is 35 cm. The lens is convex and converging. This lens is designed for a far-sighted person. The number of diopters is the inverse of the focal length in meters, so 1/0.35 m = +2.9 diopters. The angular magnification is given by θ ′ 25 cm = f θ where f can be determined from the lens formula from the case when the object distance is infinite, then f = s and θ′ = h/f = h/s. For an image distance s′ = −25 cm, s is given by 1 1 1 = + s f 25 cm s=

( 25 cm ) f 25 cm + f

Then, θ′ h = = sθ θ

34-92.

h ( 25 cm ) 25 cm + f 25 cm = = +1 25 cm f f f ( ) h 25 cm + f The net angular magnification is, according to Equation 34.33, θ ′ s′ 25 cm s′ = = 550 M = θ s f ocular s where s is found from the lens formula. 1 1 1 = − s 0.40 cm 22.4 cm s = 0.407 cm The ocular focal length is then 25 cm s′ = 550 f ocular s 25 cm s′ 25 cm ⎛ 22.4 cm ⎞ = ⎜ ⎟ = 2.5 cm 550 s 550 ⎝ 0.407 cm ⎠ Then, then the ocular magnification is 25 cm M ocular = = 10 2.5 cm The magnification is given by Equation 34.35, except that the objective and ocular are reversed. f θ′ 2.5 cm 1 = ocular = = M = θ f objective 160 cm 64 f ocular =

†34-93.


CHAPTER

34

34-94.

M =

θ ′ f objective 90.0 cm = = = 72 θ f ocular 1.25 cm

34-95.

M =

θ ′ f objective 1680 cm = = = 1344 = 1340 θ f ocular 1.25 cm

34-96.

Since the light collected is proportional to the area of the mirror, we simply take the ratio of the new proposed mirror to that of the existing six mirrors. (6.5 m) 2 enhancement = = 2.2 2 6 (1.8 m )

†34-97.

The telescope designed by Galileo has the focal lengths of the objective and ocular lenses aligned at the same point to the right of the lenses as shown. A image formed by the converging ocular lens becomes a virtual object for the diverging objective lens.

virtual object

θ′ θ fobj foc

The objective lens then refracts the rays from the virtual object, producing an upright image that appears to be at infinity. The original rays come from infinity at an angle θ and exit the system at an angle θ′. Therefore, the angular magnification is f θ′ = − ocular M = θ f objective 34-98.

34-99.

The negative sign indicates that the image is erect, since focular is negative. The angles are shown in the drawing. The first angle of reflection is 60°. The ray forms one side of a triangle that is opposite the 145° angle. The angle on the lower left must be 30° 85° since it must be added to the angle of reflection 60° to give 90°. Since the three angle of a triangle must add to 180°, the third angle of the triangle is 5°. This means that the second angle of 30° incidence must be 85°. Since the angle of incidence is equal to the angle of reflection, the final reflected angle is also 85°. n (a) sin θ 2 = 1 sin θ1 n2 ⎛ 1.0 ⎞ sin 50° ⎟ = 30.7° = 31° 1.5 ⎝ ⎠ ⎛ 1.5 ⎞ (b) θ 3 = sin −1 ⎜ sin 30.7° ⎟ = 50° 1.0 ⎝ ⎠

θ 2 = sin −1 ⎜


5°

CHAPTER 34-100.

34

The situation is shown in the drawing. The refracted ray exits the water at an angle θ2 given by n sin θ 2 = 1 sin θ1 n2 ⎛ 1.33 ⎞ sin 40° ⎟ = 58.7° = 59° 1.0 ⎝ ⎠ The distance from the corner to the point where the ray exits the water is L, which can be found from L tan θ1 = d L = d tan θ1

θ 2 = sin −1 ⎜

θ2 d′

θ1 d L

The apparent depth d′ can then be determined. L tan θ 2 = d′ L d tan θ1 ( 2.0 m ) tan 40° = = = 1.02 m = 1.0 m d′ = tan θ 2 tan θ 2 tan 58.7° †34-101.

Consider the drawing. The refracted ray is parallel to the base of the equilateral triangle, so θ2 + 60° = 90°, so θ2 = 30°. Apply Snell’s Law to find the incident angle. n sin θ1 = 2 sin θ 2 n1

θ1

θ2 60°

⎛ 1.66 ⎞ sin 30° ⎟ = 56° ⎝ 1.0 ⎠ sin θ 2 sin 72° = (1.0 ) = 1.34 n1 = n2 sin θ1 sin 45°

θ1 = sin −1 ⎜ 34-102. †34-103.

This is similar to Example 4. A ray of light leaving the bottom of the beaker undergoes two refractions, one at the water-oil interface, and one at the oil-air interface as shown in the drawing. The thickness of each layer is 10 cm. If one were in the oil looking at the refracted ray, the apparent depth would be n 1.48 d ′ = d 2 = (10.0 cm ) = 11.13 cm n1 1.33 So, the apparent depth would be greater than the actual depth. To find the apparent depth one sees looking from above the container in air, we simply reapply the above equation. n 1.00 d ′′ = d ′ 3 = (11.13 cm ) = 7.5 cm 1.48 n2

34-104.

There are two refractions. The first is at the water–glass interface. n sin θ 2 = 1 sin θ1 n2 The second is at the glass–air interface.


θ3

d′′

θ2

θ1 d

d′

CHAPTER ⎛n ⎞ n2 n ⎛n ⎞ sin θ 2 = 2 ⎜ 1 ⎟ sin θ1 = ⎜ 1 ⎟ sin θ1 n3 n3 ⎝ n2 ⎠ ⎝ n3 ⎠ ⎛ 1.33 ⎞ sin 45.0° ⎟ = 70° θ 3 = sin −1 ⎜ 1.00 ⎝ ⎠ The drawing shows the relevant angles for the situation. The angle α is related to the angle of incidence θ as follows. 180° − 90° − α − (90° − θ ) = 0

34

sin θ 3 =

†34-105.

34-106.

90° − θ

θ

α =θ The requirement for total internal reflection is that 1 sin θ critical = n ⎛1⎞ ⎛ 1 ⎞ θ critical = sin −1 ⎜ ⎟ = sin −1 ⎜ ⎟ = 38.68° = 38.7° ⎝n⎠ ⎝ 1.60 ⎠ Total internal reflection will occur when θ ≥ θcritical. Therefore, the requirement is that α ≥ 38.7°. If we assume that the light emits light in all directions, then the dark portions of the water are produced by total internal reflection (because the light does not escape the pool). The critical angle for the water–air interface is ⎛1⎞ ⎛ 1 ⎞ θ critical = sin −1 ⎜ ⎟ = sin −1 ⎜ ⎟ = 48.75°, n ⎝ ⎠ ⎝ 1.33 ⎠ which determines whether light is reflected or transmitted. The radius of the circle 2.0 meters above the light is then r tanθ critical = . 2.0 m r = (2.0 m) tanθ critical = (2.0 m) tan (48.75°) = 2.3 m

34-107. Mirror Concave

34-108. †34-109.

Object distance Image characteristics s 2f Real, inverted, and reduced Convex all Virtual, upright, and reduced If a distant object (s ≈ ∞) is placed in front of a convex mirror, the virtual image appears at the focal point of the mirror. Therefore, since the radius of curvature R = 2f, R = 2(1.0 m) = 2.0 m. A convex mirror always produces virtual, upright, and reduced images, so the image is behind the mirror. To find its location, use the mirror formula. The focal distance is equal to one-half the radius of curvature, f = 0.5R = (0.5)(−4.0 m) = −2.0 m 1 1 1 1 1 = − = − s′ f s −2.0 m 12 m s′ = −1.7 m The magnification of the mirror is given by Equation 34.30.


CHAPTER

34

s′ −1.7 m =− = 0.142 s 12 m The magnification is positive, which indicates that the image is upright. The height of the image is ⎛ −1.7 m ⎞ h′ = Mh = − ⎜ ⎟ 1.7 m = 0.24 m ⎝ 12 m ⎠ The focal length of each mirror is one-half of the radius, or 30 cm. Light from the candle reflects from the concave mirror to form an image at 1 1 1 1 1 = − = − s′ f s 30 cm 40 cm s′ = 120 cm This image then becomes the object, located at 120 cm − 50 cm = 70 cm, behind the convex mirror, which forms the final image at 1 1 1 1 1 = − = − s′ f s −30 cm −70 cm s′ = −52.5 cm M =−

34-110.

The final image is located 52.5 cm behind the convex mirror. 34-111. Lens Concave Convex

34-112.

Object distance sf s 2f

Image characteristics Virtual, upright, and reduced Virtual, upright, and reduced Virtual, upright, and magnified Real, inverted, magnified Real, inverted, reduced

The image is formed by the intersection of virtual rays as shown in the diagram. The first ray leaves the top of the candle and travels parallel to the principle axis. The ray is refracted by the lens along a line that extends backward through the focal point. The second ray passes directly through the center of the lens and it is not refracted. The third ray is directed toward the focal point on the opposite side of the lens. This ray is refracted and travels parallel to the principle axis. The intersection of the three rays produces a virtual image between the lens and the focal point that is upright and reduced.


CHAPTER 34-113.

The image is formed by the intersection of virtual rays as shown in the diagram. The first ray leaves the top of the light bulb and travels parallel to the principle axis. The ray is refracted by the lens along a line that passes through the focal point on the opposite side of the lens. The second ray passes directly through the center of the lens and it is not refracted. The intersection of the two rays produces a virtual image between the lens and the focal point that is upright and enlarged. The image distance is 1 1 1 1 1 = − = − s′ f s 30 cm 15 cm s′ = −30 cm

34-114.

Apply the lens equation three times. Assume the light bulb is to the left of the first lens. For the first lens, the image is produced at 1 1 1 1 1 = − = − s1′ f s1 60 cm 20 cm s1′ = −30 cm This becomes the object for the second lens with an object distance of 30 cm + 12 cm = 42 cm. 1 1 1 1 1 = − = − s2′ f s2 60 cm 42 cm s2′ = −140 cm This image becomes the object for the third lens. The object distance is 140 cm + 12 cm = 152 cm. 1 1 1 1 1 = − = − s3′ f s3 60 cm 152 cm s3′ = 99 cm The final image is located 99 cm to the right of the right lens. The magnification is the product of the magnifications of each lens. ⎛ s′ ⎞ ⎛ s′ ⎞ ⎛ s′ ⎞ ⎛ −30 cm ⎞ ⎛ −140 ⎞ ⎛ 99 ⎞ M = M1M 2 M 3 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ ⎜ − 3 ⎟ = ⎜ − ⎟⎜ − ⎟⎜ − ⎟ = −3.3 ⎝ s1 ⎠ ⎝ s2 ⎠ ⎝ s3 ⎠ ⎝ 20 cm ⎠ ⎝ 42 ⎠ ⎝ 162 ⎠

34-115.

(a) For an image to be formed at infinity, the object must be placed at the focal point for the ocular lens. This distance is s′ =180 mm + 25 mm = 205 mm (b) The objective forms an image at a distance s′ given by 1 1 1 1 1 = − = − ′ s f s 1.9 mm 205 mm s = 1.9 mm which is the focal point of the objective. The angular magnification is 25 s′ 25 205 mm M = = = 108 f oc s 25 mm 1.9 mm


34

CHAPTER 35

INTERFERENCE AND DIFFRACTION


35-2.

†35-3.

Distance between node/antinode =

λ

. Hence, λ of wave = 4 × 2 m = 8 m 4 Since c = fλ , c = (4 × 107 Hz) × (8 m) = 3.2 × 108 m/s Since only one ray undergoes a phase change, the light interferes constructively when 1 3 5 2d = λ , λ , λ , ... 2 2 2 4 4 ⇒ λ = 4d , d , d , ... 3 5 Now d = 400 nm. Therefore, λ = 1600 nm, 533 nm, 320 nm, 229 nm, . . . The wavelength in air is found by multiplying by n = 1.35 (λair = nλsoap). The wavelengths in visible region are 530 nm × 1.35 = 720 nm (red) 320 nm × 1.35 = 432 nm (violet) Since there is a 180º phase shift upon both reflections, constructive interference requires that the thickness of the film be a integer number of wavelengths in the film. Thus the thinnest possible layer will have 2d = λ = λair/n: λ 350 nm ⇒ d = air = = 127 nm 2n 2(1.38)

For destructive interference, twice the thickness of the layer must be an odd number of halfwavelengths, so the thinnest layer will have 2d = λ/2 = λair/(2n): λ 700 nm ⇒ d = air = = 127 nm , 4n 4(1.38) 35-4.

which conveniently turns out to be the same as the layer that strongly reflects 350 nm UV. Because there is air between the glass plates, the light ray reflected from the bottom of the top plate does not undergo a 180º phase change, but the ray reflected from the top of the bottom plate does undergo a phase change. Thus the rays reflected from the left end where the plates are dark fringe touching (d = 0) are out of phase and there is a dark fringe there. At the second dark fringe, 2d = λ. At the third dark fringe, 2d = 2λ, and so on. At the 76th dark fringe, 2d = 75λ. The problem does not state exactly where the last fringe is relative to the hair, but the error caused by assuming the fringe is at the location of the hair is negligible. Thus the thickness of the hair is


CHAPTER d =

35 75λ 75(546 nm) = = 2.05 × 104 nm = 2.05 × 10−5 m (20.5 µm) 2 2

m2λ 2 4r 2 + m 2 λ 2 ⇒R= . Because r is five orders of 4 4mλ magnitude larger than λ, the m2λ2 term is negligible compared to the r2 term and we get r2 (0.060 m) 2 R≈ = = 244 m mλ 25(590 × 10−9 m) mλ R −

35-5.

From Problem 35.13, r =

35-6.

Using the equation from Problem 35.13, rair =

mλ R −

m2λ 2 4

mλ R m 2 λ 2 − n 4n 2 Because λ is small, λR >> λ2 and the λ2 term can be neglected. Then rair ≈ mλ R rliquid =

rliquid ≈

mλ R n

⎛ r ⇒ n = ⎜ air ⎜r ⎝ liquid 35-7. 35-8.

†35-9.

2

2 ⎞ ⎛ 2 ⎞ ⎟⎟ = ⎜ ⎟ = 1.78 ⎝ 1.5 ⎠ ⎠ λ λ 500 nm ⇒d = = = 78 nm 2d = 2n 4n 4(1.6)

(a) In order for standing waves to have nodes at mirrors, there must be an integral number of half-wavelengths in the tube. Therefore, ⎛ m + 1⎞ d = ( m + 1)(λ / 2) = ⎜ ⎟λ ⎝ 2 ⎠ 2d 2 × 0.3000 m (b) m = −1 = − 1 = 9.48 × 105 −9 λ 633 × 10 m Since m = number of nodes, there are 9.48 × 105 maxima and minima. The diagram shows an incident ray passing through all the layers with reflections within the film. Assume n1 = 1 (air) and n2 < n3. Then there is a 180º phase change on the first reflection within the film but no phase change on the second. (All the rays are actually normal to the layers. The reflected rays have been drawn at an angle for clarity.) Then the condition for constructive interference between the direct ray and the ray that has been reflected twice is λ 3λ 5λ 2d = , , , 2n2 2n2 2n2

n1 n2

d

n3

which is the condition for destructive interference between the incident and reflected rays in air. Now assume n2 > n3. Then there is a 180º phase change on both reflections in the film. Now the condition for constructive interference between the direct ray and the ray that has been reflected twice is


CHAPTER 2d =

35-10.

λ 2λ 3λ n2

,

n2

,

n2

35

,

which again is the condition for destructive interference between the direct and reflected rays in air. In both cases the wavelengths whose intensity are reduced in air (n1) are enhanced in the rays transmitted into the medium with index n3. mλ mλ ⇒d = . The order m must be Since noil < nwater, constructive interference requires 2d = 2noil noil different for the two wavelengths. The ratio of the wavelengths is 675/450 = 3/2, so that suggests m = 3 for 450 nm and m = 2 for 675 nm. If this guess is correct, d should have the same value for both wavelengths: 2(675 nm) d = = 540 nm 2(1.25) 3(450 nm) d = = 540 nm 2(1.25)

35-11.

35-12.

The slick must be 540 nm thick. The extra distance that light must travel to reach bottom plate and return = 2d. The ray reflected from the inner surface of the top plate does not suffer a phase reversal but the ray reflected from the top surface of the bottom plate does suffer a phase reversal. Therefore Equation 35.2 gives the condition for constructive interference: 1 3 5 2d = λ , λ , λ , ... (λ = wavelength in AIR) 2 2 2 Both reflected rays in air undergo a phase change on reflection, so the overall phase difference between the two rays in air is caused by the passage through the two films. The total phase change is ⎛ 2d n + 2d water nwater ⎞ φ = φoil + φwater = 2π ⎜ oil oil ⎟ The λ ⎝ ⎠ minimum phase change for destructive interference is π, so ⎛ 2d n + 2d water nwater ⎞ π = 2π ⎜ oil oil ⎟ λ ⎝ ⎠ ⇒ λ = 4(d oil noil + d water nwater )

oil

200 nm

water

400 nm

metal

= 4[(200 nm)(1.24) + (600 nm)(1.33)] = 3120 nm Other wavelengths that undergo destructive interference follow the pattern φ = mπ , where m is any odd integer. m = 5 and 7 give destructive interference at wavelengths of 624 nm and 446 nm, respectively, which are in the visible range.


CHAPTER †35-13.

35

The light that reflects within the lens does not change phase (because this is a glass/air interface, and nglass > nair), whereas light reflecting off the mirror does change phase. Therefore, at the center of the lens where distance between lens and flat glass is infinitesimal, the waves cancel immediately ⇒ dark spot. (b) Note that this is the formula for the nth dark ring. At radius r, let d be height of the curved part of lens above the mirror. Then d = R − R2 − r 2 For destructive interference to occur, 2d = mλ (where m = integer). Therefore, mλ d = = R − R2 − r 2 2 2

mλ ⎞ m2λ 2 ⎛ 2 R m R λ ⇒ R2 − r 2 = ⎜ R − = − + ⎟ 2 ⎠ 4 ⎝ ⇒ r 2 = mλ R −

m2λ 2 ⇒ r= 4

mλ R −

m2λ 2 4

(c) By (b), with m = 1, ⎛ λ2 ⎞ r = λR − ⎜ ⎟ ⎝ 4 ⎠ = 500 × 10−9 m × 3.0 m − [(500 × 10−9 m) 2 / 4] = 1.22 × 10−3 m (1.22 mm)

35-14.

35-15.

To go from one dark fringe to the next dark fringe, the mirror must move a distance λ/2. If m fringes are observed, the mirror moved a total distance d = mλ/2. Thus 2d 2 × 0.450 × 10−3 m λ= = = 6.33 × 10−7 m = 633 nm m 1422 mλ (930)(546 × 10−9 m) d = = = 2.57 × 10−4 m (0.257 mm) 2 2


CHAPTER 35-16.

Two adjacent rays are shown in the diagram. The upper ray has been reflected twice between the mirrors before exiting. The path difference between the two rays is 2 AB − AD. From the diagram, d AB = cos θ

C θ

B

A

CA = 2d tanθ AD = CA sinθ = 2d tan θ sinθ

35

D

θ

path difference = 2 AB − AD

d

⎛ 1 − sin θ ⎞ 2d − 2d tanθ sinθ = 2d ⎜ ⎟ cos θ ⎝ cos θ ⎠ ⇒ path difference = 2d cosθ 2

=

Since the reflected ray undergoes a 180º phase change upon each reflection, the requirement for constructive interference is 2d cosθ = mλ , m = 0,1, 2, 3, … 35-17. 35-18. 35-19.

35-20.

35-21.

mλ λ dm 605.8 × 10−9 m (19 × 104 s −1 ) = 5.76 × 10−3 m/s (5.76 mm/s) ⇒v= = 2 2 dt 2 Movement through 1 maximum ⇒ movement of λ/2, therefore distance = (605.7802 nm/2) × 36484.8 = 1.10509 × 10–2 m A change of ∆L in the arm length MM2 (see Fig. 35.9) introduces a path length change of 2∆L. If this change is required to maintain the number of fringes constant, the change must equal the difference in the optical path lengths in the two arms which equals 2Ln – 2L. ∆L Therefore, 2L(n – 1) = 2∆L or n − 1 = . For λ = 580 nm, ∆L = 0.01106 cm and L = 40 cm, L ∆L 0.01106 n =1+ =1+ = 1.000277 L 40 From Equation 35.7, the maxima are located at λ 2λ mλ sin θ = 0, , , ... d d d m λ ⎛ ⎞ or θ m = sin −1⎜ ⎟ ⎝ d ⎠ for λ = 2 cm and d = 5 cm, θm = sin–1(0.4 m) θ0 = 0 θ1 = 23.6° θ2 = 53.1° Everything same, except λ replaced by λwater = λ/n = λ/1.33. Therefore, θ ≈ sin θ = λ/nd, and ∆y ≈ rθ = rλ/nd 2.00 m × 589 × 10−9 m = = 7.4 × 10−3 m. 1.33 × 1.2 × 10−4 m Distance =


CHAPTER

35

35-22.

From Example 2, the distance between fringes is ∆y ≈ rθ for small angles, where r is the distance to the screen and θ is the angular location of the first maximum. Thus r λ (4 m)(633 × 10−9 m) ∆y = = = 0.017 m (1.7 cm) d 0.15 × 10−3 m

†35-23.

From Example 2, the distance between fringes is ∆y ≈ rθ, where r is the distance to the screen and θ is the angular location of the first maximum. The distance from the central maximum to the first minimum is half this value: ∆y λ = sin θ ≈ tan θ ⇒ tan θ = r d ∆y r λ (1.5 m)(546 × 10−9 m) ⇒ = = = 1.9 × 10−3 m (1.9 mm) −3 2 2d 2(0.22 × 10 m) d sin θ = mλ with d = λ/2 ⇒ sin θ = 2m. Because d < λ, the only maximum is the central maximum (m = 0), and the intensity has its lowest value at 90°. However, the intensity never reaches zero, even at 90°. I y ⎛πd ⎞ sinθ ⎟ . For small angles, tanθ ≈ tanθ ⇒ tanθ = , = cos 2 ⎜ From Equation 35.11, I max r ⎝ λ ⎠

35-24.

†35-25.

where y is the distance from the central maximum to the point of interest on the screen and r is the slit-to-screen distance. Thus I ⎛ π dy ⎞ = cos 2 ⎜ ⎟ I max ⎝ λr ⎠

35-26.

35-27.

35-28.

⎡ π (0.1 × 10−3 m)(1.5 × 10−2 m) ⎤ = cos 2 ⎢ ⎥ = 0.994 (633 × 10−9 m)(1.2 m) ⎣ ⎦ (Make sure your calculator is in radians mode for calculations like this!) This means ∆θ (between fringes) = ∆y/r = 1.4 cm/300 cm = 4.67 × 10–3 rad Since d sin θ ≈ θd (small θ) = nλ (n = integer), this gives ∆θd = λ ⇒ d = λ/∆θ = (6.943 × 10–7 m)/(4.67 × 10–3)= 1.49 × 10–4 m By measurement, d ≅ 0.9 cm Length of 13 waves = 2.5 cm λ ≅ 0.192 cm, λ / d = 0.213 Minima are not uniformly positioned, but occur at angles of +6 + 24 + 40 + 57 ⎫ ⎬ −5 − 17 − 31 − 46 ⎭ relative to center. Use average value: 5.5º 20.5º 35.5º 51.5º From Equation 35.9 d sin θ = λ/2, 3λ/2, 5λ/2, 7λ/2 We predict minima at 6.1º, 18.6º, 32.2º, 48.2º There is reasonable agreement (to within a few degrees) The angular positions that cause maximum intensity in the receiver correspond to the angular positions of the maxima that would be produced if the antennas acted as sources rather than


CHAPTER

35

receivers. So the first maximum away from the central maximum occurs when d sin θ = λ

⇒ θ = sin −1 35-29.

35-30.

35-31.

λ d

≈

λ d

=

21 × 10−2 m = 4.2 × 10−8 radian (0.0087 arc sec) 5 × 106 m

Frequency of 2300 MHz means λ = c/f = 0.13 m. Maximum amplitude occurs when constructive interference occurs. d sin θ = 0, λ, 2λ, 3λ, . . . where d = 1000 m. sin θ = 0, ± 1.3 × 10–4, ± 2.6 × 10–4, ± 3.9 × 10–4 . . . θ = 0, ± 0.0074º, ± 0.0149º, ± 0.0223º, . . . ⎛πd ⎞ From Equation 35.11, I = I max cos 2 ⎜ sin θ ⎟ . ⎝ λ ⎠ –4 d = 0.11 mm = 1.1 × 10 m λ = 694.3 nm = 6.943 × 10–7 m θ ≈ 0.012 m/1.5 m = 8 × 10–3 rad ⎛ π × 1.1 × 10−4 ⎞ I = I max cos 2 ⎜ sin 8 × 10−3 ⎟ = 0.45I max −7 ⎝ 6.943 × 10 ⎠

Therefore, intensity of that fringe is 0.45 of intensity of center fringe. I ⎛πd ⎞ sin θ ⎟ = cos 2 (12π sin θ ) ≈ cos 2 (12πθ ) for small angles. In a spreadsheet, = cos 2 ⎜ I max ⎝ λ ⎠ convert each of the angles to radians by multiplying by π/180 and then evaluate the trig function.

35-32.

θ (degrees) θ (radians) 1.0 0.0175 2.0 0.0349 3.0 0.0524 [path difference] = d sin θ ⎛ π [path difference] ⎞ I = I max cos 2 ⎜ ⎟ λ ⎝ ⎠

[path difference] =

†35-33.

I/Imax 0.626 0.064 0.154

λ

3 ⎛π ⎞ I ⇒ I = I max cos 2 ⎜ ⎟ = max 4 ⎝3⎠ Without the polystyrene, the phase difference between the waves from the two slits would be 2π d sinθ δ = . However, the light from the bottom slit passes through a thickness t of

λ

polystyrene, which adds an additional phase shift of

λ

. Thus the total phase shift is

2π (n − 1)t + . For constructive interference, δ must be an integer multiple of 2π: λ λ 2π d sinθ 2π (n − 1)t + = 2mπ ⇒ d sinθ + (n − 1)t = mλ λ λ

δ =

2π d sinθ

2π (n − 1)t


CHAPTER

35

(n − 1)t , where the minus d ∆y , sign means the central maximum is shifted downwards. For small angles, sinθ ≈ tanθ = − r where the minus sign indicates the downward shift of the central maximum. Thus ∆y (n − 1)t − =− r d d ∆y (0.20 × 10−3 m)(4.0 × 10−3 m) ⇒t= = = 5.4 × 10−7 m (n − 1)r (0.49)(3.0 m) For m = 0, the location of the central maximum is given by sin θ = −

35-34.

The first minimum is located at an angle given by d sinθ =

λ

2 ⎡ ⎤ v 340 m/s θ = sin −1 = sin −1 = sin −1 ⎢ ⎥ = 4.88° 2d 2 fd ⎣ 2(1000 Hz)(2.0 m) ⎦

λ

∆y = r tanθ = (5.0 m) tan 4.88° = 0.43 m (43 cm) 35-35.

(a) There is a difference in pathlength for the approaching rays of d sinφ with the top ray traveling farther than the bottom ray. There is a path difference of d sinθ from the slits to the final point where the waves interfere. Therefore, the net difference in pathlength is d sinθ − d sinφ (minus because on opposite side, bottom ray has to travel farther.) For constructive interference, the net path difference must be a whole number of wavelengths: d sinθ − d sinφ = 0, λ , 2λ , … For destructive interference the net path difference must be an odd number of half wavelengths: λ 3λ 5λ d sinθ − d sinφ = , , , … 2 2 2 (b) For the first two bright fringes at θ1 and θ2, d sinθ1 − d sinφ = λ d sinθ 2 − d sinφ = 2λ Use the small angle approximation sinθ ≈ θ and subtract the top equation from the bottom. The d sinφ terms cancel, giving λ = d (θ 2 − θ1 ) = d ∆θ , and the angular separation is independent of

35-36.

φ. For constructive interference, difference in pathlength 1 3 = λ , λ , ... 2 2 (because of phase reversal for ray 2), and for destructive it is 0, λ, 2λ. Pathlength for ray 1 = ( z − z0 ) 2 + l 2 . Length for longer route is exactly the same as length of ray passing straight through mirror and striking target at a distance z0 from mirror on opposite side. Therefore, Pathlength for ray 2 = ( z + z0 ) 2 + l 2


1

z – z0 2 z + z0

CHAPTER Therefore, difference = 35-37.

( z + z0 ) 2 + l 2 − ( z − z0 ) 2 + l 2

By first two lines, QED! Constructive interference occurs when difference in pathlengths = nλ where n is an integer. From (–x0, 0) to (x, y) is a distance ( x + x0 ) 2 + y 2 from (x0, 0) to (x, y) is distance ( x − x0 ) 2 + y 2 Therefore, difference in pathlengths is ( x + x0 ) 2 + y 2 − ( x − x0 ) 2 + y 2 , and for constructive interference, this must equal = nλ, where n = 0, ±1, ±2, ±3, . . . . QED (Hyperbola is defined as the locus of points with constant difference between two foci—in this case (–x0, 0) and (x0, 0)—and this equation describes that).

35-38.

†35-39.

Difference in pathlength |AC| – |BC| = ∆d. AC = (25 m/sin θ). We note that angle ACB is 2θ (see diagram), and therefore, BC = AC cos 2θ = (25 m/sin θ) cos 2θ. Therefore, AC – BC = ∆d 25 m = (1 − cos 2θ ). sin θ But 1 – cos 2θ = 2 sin2 θ. 25 m Therefore, ∆d = (2 sin 2 θ ) = 50 m sin θ sin θ For destructive interference, since a reflection (change of phase) is involved, minimum occurs when ∆d = λ. Here 3 × 108 m/s λ = c/v = = 5.0 m 6.0 × 107 Hz Therefore, ∆d = 50 m sin θ = λ = 5.0 m ⇒ sin θ = 1/10 ⇒ θ = 5.7° 1 d between slits = cm = 2 × 10−6 m 5000 λ = 650 nm = 6.50 × 10−7 m For principal maxima

⎛ 6.5 × 10−7 ⎞ d sinθ = mλ ⇒ θ = sin −1 (mλ / d ) = sin −1 ⎜ m = sin −1 (0.325 m) −6 ⎟ 2 × 10 ⎝ ⎠


35

CHAPTER

35

where m = 0, 1, 2, 3, . . . . For m = 1, θ1 = 19.0°, m = 2, θ 2 = 40.5°, m = 3, θ 3 = 77.2°

35-40.

Note that these angles are too large to use the small angle approximation that was suitable for other double slit problems. 1 cm = 1.818 × 10−4 cm = 1.818 × 10−6 m Distance d between slits = 5500 mλ ⎞ ⎛ d sinθ = mλ ⇒ θ = sin −1 ⎜ ⎟ ⎝ d ⎠ (a) For first order spectrum n = 1, ⎛ 5.8899 × 10−7 ⎞ = 18.904° θ1 = sin −1 (λ / d ) = sin −1 ⎜ −6 ⎟ ⎝ 1.818 × 10 ⎠

⎛ 5.8959 × 10−7 ⎞ = 18.924° −6 ⎟ ⎝ 1.818 × 10 ⎠ 0.020° × π Therefore, ∆θ = 0.020º; ∆y = r ∆θ = 3.0 m × rad = 1.0 × 10−3 m (1.0 mm) 180° (b) For second order spectrum n = 2, ⎛ 2 × 5.8899 × 10−7 ⎞ ⎛ λ⎞ θ 2 = sin −1 ⎜ 2 ⎟ = sin −1 ⎜ ⎟ = 40.388° −6 ⎝ d⎠ ⎝ 1.818 × 10 ⎠

θ1′ = sin −1 (λ ′ / d ) = sin −1 ⎜

⎛ 2 × 5.8959 × 10−7 ⎞ ⎟ = 40.437° −6 ⎝ 1.818 × 10 ⎠

θ 2′ = sin −1 (2λ ′ / d ) = sin −1 ⎜ Therefore, ∆θ = 0.049°

Therefore, ∆y = r ∆θ = 3.0 m × †35-41.

35-42.

Here d =

0.049° × π rad = 2.6 × 10−3 m (2.6 mm) 180°

1 1 cm = m. 5900 5.9 × 105

d sin θ = mλ ⇒ θ = sin–1(mλ/d) The angles extend from those corresponding to 400 nm = 4.00 × 10–7 m, to 700 nm = 7.00 × 10–7 m. First order. θ400 = sin–1(4.00 × 10–7 × 5.9 × 105) = 13.7° θ700 = sin–1(7.0 × 10–7 × 5.9 × 105) = 24.4° Second order. θ400 = sin–1(2 × 0.236) = 28.2° θ700 = sin–1(2 × 0.413) = 55.7° Third order. θ400 = sin–1(3 × 0.236) = 45.1° θ700 = sin–1(3 × 0.413). The argument here is greater than 1, so the diffraction angle reaches 90° at a wavelength shorter than 700 nm and 700 nm is not visible. Second and third orders overlap (second ends at 56°, third begins at 45°), third is incomplete. λ λ 633 × 10−9 From Equation 35,16, d sinθ = ⇒ sinθ = = = 4.22 × 10−4 radian. This is −3 N dN (0.25 × 10 )(6) half the width of the central maximum so [angular width] = 8.4 × 10–4 radian.


CHAPTER

†35-43.

From Equation 35.16, d sinθ =

λ N

⇒ sinθ ≈ θ =

λ dN

35

.

N = 40000, λ = 5.50 × 10−7 m 8.0 cm d = = 2.0 × 10−6 m 40000 Therefore, θ = (1/N)(λ/d) =

35-44.

1 5.50 × 10−7 × = 6.88 × 10−6 radian 40000 2 × 10−6

[angular width] = 2θ = 1.38 × 10–5 rad. I E2 I ∝ E 2 , so = where 3E0 is the amplitude of the total electric field for δ = 0 and E is I max (3E0 ) 2 the amplitude of the total field for other values of δ. For each of the diagrams, the x and y components of the fields are given by 2 ⎛ ⎞ Ex = E0 ⎜ 1 + ∑ cos mδ ⎟ m =1 ⎝ ⎠ 2

E y = E0 ∑ sin mδ m =1

from which the total field and thus the relative intensity can be found. A spreadsheet was used to tabulate the results. All the fields are given as multiples of E0.

†35-45.

35-46.

I/Imax delta delta Ex Ey E 0 0.000 3.000 0.000 3.000 1.000 π/4 0.785 1.707 1.707 2.414 0.648 π/2 1.571 0.000 1.000 1.000 0.111 2π/3 2.094 0.000 0.000 0.000 0.000 4π/5 2.513 0.500 –0.363 0.618 0.042 3.142 1.000 0.000 1.000 0.111 π For multiple slits, d sin θ = mλ. Adjacent images means the m’s are different by 1. d sin θm = mλ; d sin θm+1 = (m + 1) λ. Therefore, d(sin θm+1 – sin θm) = λ for small angles (sin θ ≈ θ). Therefore, d(θ2 – θ1) = d∆θ = λ ⇒ d = λ/∆θ Here λ = 6.70 × 10–7 m, ∆θ = 2.0 × 10–3 6.70 × 10−7 Therefore, d = = 3.4 × 10−4 m 2.0 × 10−3 I ∝ E 2 , so

I I max

=

E2 where 4E0 is the amplitude of the total electric field for δ = 0 and E is (4 E0 ) 2

the amplitude of the total field for other values of δ. For each of the diagrams, the x and y components of the fields are given by 3 ⎛ ⎞ Ex = E0 ⎜ 1 + ∑ cos mδ ⎟ m =1 ⎝ ⎠ 3

E y = E0 ∑ sin mδ m =1


CHAPTER

35

from which the total field and thus the relative intensity can be found. A spreadsheet was used to tabulate the results. All the fields are given as multiples of E0. Sketches and a plot follow the table. delta 0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π

delta 0.000 0.524 0.785 1.047 1.571 2.094 2.356 2.618 3.142

Ey 0.000 2.366 2.414 1.732 0.000 0.000 0.414 0.634 0.000

Ex 4.000 2.366 1.000 0.000 0.000 1.000 1.000 0.634 0.000

I/Imax 1.000 0.700 0.427 0.188 0.000 0.063 0.073 0.050 0.000

E 4.000 3.346 2.613 1.732 0.000 1.000 1.082 0.897 0.000

I /I max

1.0

0.5

0.0 0.0

1.0

2.0

3.0

Delta (radians)

35-47.

The electric field from the middle slit has twice the amplitude of the fields from the other slits because the intensity from the middle slit is twice the intensity from the others. The phasors then are added according to this rule: Ex = E0 + 2 E0 cosδ + E0 cos2δ E y = 2 E0 sinδ + E0 sin2δ


CHAPTER

35

The calculations and a plot were done using a spreadsheet. All the fields are given as multiples of E0. Sketches of the phasors follow the table and plot from the spreadsheet. delta 0 π/4 π/2 3π/4 π

delta 0.000 0.785 1.571 2.356 3.142

Ex 4.000 2.414 0.000 –0.414 0.000

Ey 0.000 2.414 2.000 0.414 0.000

E 4.000 3.414 2.000 0.586 0.000

I/Imax 1.000 0.729 0.250 0.021 0.000

I /I max

1.0

0.5

0.0 0.0

1.0

2.0

3.0

Delta (radians)

35-48.

The phase shift of the electric field from the bottom slit has twice the shift from the middle slit. The phasors then are added according to this rule: Ex = E0 + E0 cosδ + E0 cos 3δ E y = E0 sinδ + E0 sin3δ


CHAPTER

35

The calculations and a plot were done using a spreadsheet. All the fields are given as multiples of E0. Sketches of the phasors follow the table and plot from the spreadsheet. delta 0 π/4 π/3 π/2 2π/3 3π/4 π

delta 0.000 0.785 1.047 1.571 2.094 2.356 3.142

Ex 3.000 1.000 0.500 1.000 1.500 1.000 –1.000

Ey 0.000 1.414 0.866 0.000 0.866 1.414 0.000

I /I max

1.0

0.5

0.0 0.0

1.0

2.0

3.0

Delta (radians)


E 3.000 1.732 1.000 1.000 1.732 1.732 1.000

I/Imax 1.000 0.333 0.111 0.111 0.333 0.333 0.111

CHAPTER

35

35-49.

(a) The optical path length difference between 1 and 2 is |(d sin θ – d sin ø)|. This difference must equal an integral number of wavelengths for 1 and 2 to interfere constructively. Hence, the condition |d sin θ – d sin ø| = 0, λ, 2λ, . . . nλ (b) For θ = ø = 0º, the incident and the emerging lights will overlap. (c) For θ = ø, the condition |(d sin θ – d sin ø)| = 0 can always be satisfied independent of λ.

35-50.

The locations of the minima are given by a sinθ = mλ . The wavelength decreases as the index

†35-51.

35-52.

of refraction increases, so immersing the entire apparatus in water will cause all the angles to decrease. ⎛ 6.328 × 10−7 m ⎞ ⎛λ⎞ By Equation 35-17, the first minimum is at θ = sin–1 ⎜ ⎟ = sin–1 ⎜ ⎟= −3 ⎝a⎠ ⎝ 0.10 × 10 m ⎠ 6.33 × 10–3 rad The total angular width = 2 × 6.33 × 10–3 rad = 0.0127 rad. Therefore, width = r∆θ = 2.0 m × 0.0127 rad = 0.025 m (2.5 cm). Wavelength λ ≈ 2.1 mm. a ≈ 5 mm. Therefore, ⎛ nλ ⎞ –1 θ = sin–1 ⎜ ⎟ = sin (n 0.42). a ⎝ ⎠ For n = 1, θ ≈ 25°; n = 2, θ ≈ 57°. (Kind of hard to confirm—look at photos!)

†35-53.

(See Problem 51.) Full width = 3.0 cm ⇒ 2θ =

θ = 0.015 radian aθ = λ ⇒ a = 35-54.

0.03 m = 0.03 radian. Therefore 1.0 m

λ 633 × 10−9 m = = 4.2 × 10−5 m (0.042 mm) θ 0.015 radian

f = 820 Hz v 331 m/s v 331 = 0.4036 m. λ= = = 0.404 m λ = = υ 820 f 820 Hz Angular directions of minima are given by ⎛λ⎞ ⎛ 2λ ⎞ ⎛ 3λ ⎞ θ = ± sin −1 ⎜ ⎟ , ± sin −1 ⎜ ⎟ , ± sin −1 ⎜ ⎟ ,… a a ⎝ ⎠ ⎝ ⎠ ⎝ a ⎠ a = 1.0 m ⇒ θ = ±23.8°, ± 53.8°


CHAPTER

†35-55.

35-56.

35-57.

35

⎛ 20 m ⎞ ⎛λ⎞ –1 By Equation 35-17, θ = sin–1 ⎜ ⎟ = sin–1 ⎜ ⎟ = sin (0.4) = 0.41 radian. Then the angular ⎝a⎠ ⎝ 50 m ⎠ width is 2θ = 0.82 radian (47º). Note that the angle here is too large to use the small angle approximation. λ 0.25 The central maximum will have an angular width of 2θ, so sin θ = ≈ tan θ = . Therefore a 200 200 200 = 577 × 10–9 × = 4.6 × 10–4 m (0.46 mm). the slit width is a = λ × 0.25 0.25 (a) The half-angular width of the first minimum is ⎛ 550 × 10−9 m ⎞ ⎛λ⎞ ∆θ = sin–1 ⎜ ⎟ = sin–1 ⎜ ⎟ = 0.32° −3 ⎝a⎠ ⎝ 0.1 × 10 m ⎠ Since the angular diameter of the sun is 0.5°, the fringe pattern will not be resolved. ⎛ 550 × 10−9 m ⎞ ⎛λ⎞ (b) ∆θ = sin–1 ⎜ ⎟ = sin–1 ⎜ ⎟ = 3.2° −3 ⎝a⎠ ⎝ 0.01 × 10 m ⎠ The fringe pattern can be resolved. (c) If the filter is removed, the λ-values will have an approximate visible range of 400–700 nm. This range gives ∆θ values for a slit with a = 0.01 mm as ⎛ 400 × 10−9 m ⎞ ∆θ|400 nm = sin–1 ⎜ ⎟ = 2.3° −3 ⎝ 0.01 × 10 m ⎠ ⎛ 700 × 10−9 m ⎞ ∆θ|700 nm = sin–1 ⎜ ⎟ = 4.0° −3 ⎝ 0.01 × 10 m ⎠ The spread in fringe positions is ~1.7°, which will make the diffraction pattern so smeared as to be not observable.

35-58.

2

4 ⎡ sin(π / 2) ⎤ =⎢ = 2 = 0.41 ⎥ π I max ⎣ π / 2 ⎦ The angular location of the second order maximum is approximately halfway between the first θ − θ1 θ1 + θ 2 = . For small angles, the minima are at and second order minima, or θ = θ1 + 2 2 2 mλ 3λ , so the second order maximum is at θ = θm ≈ . From Equation 35.26 the phase angle is a 2a 2π 2π a ⎛ 3λ ⎞ a sinθ = φ= = 3π . Thus the relative intensity is λ λ ⎜⎝ 2a ⎟⎠ I

†35-59.

2

⎡ sin(φ / 2) ⎤ 2π a sinθ λ =⎢ , where φ = . The first minimum is at θ = for small angles, so ⎥ a I max ⎣ φ / 2 ⎦ λ θ λ 2π a ⎛ λ ⎞ = and φ = = π . Therefore 2 2a λ ⎜⎝ 2a ⎟⎠ I

I I max

2

⎡ sin(φ / 2) ⎤ 4 sin 2 (3π / 2) =⎢ = = 0.045 ⎥ 9π 2 ⎣ (φ / 2) ⎦


CHAPTER

35

35-60.

The figure shows closely spaced bright spots that decrease in intensity from a bright central maximum. The intensity of these closely spaced spots reaches a minimum and then increases again, reaching a secondary maximum at about 1°. The closely spaced spots are most likely an interference pattern from multiple slits, and the decrease and subsequent increase in overall intensity is probably a diffraction pattern from the individual slits. Comparing this diagram with Figure 35.38 shows that this is probably a two-slit interference pattern superimposed on the single slit diffraction pattern. (If there were more than two slits, we would expect more bright maxima. See Problem 62.) From the figure, it is apparent that the third interference maximum coincides with the first diffraction minimum because the third peak in the interference pattern is missing. Since the equation for the third interference maximum is d sin θ = 3λ and the equation for the first diffraction minimum is a sin θ = λ, we have d sinθ = a sinθ . This gives 3 1 1 a = d = (0.12 mm) = 0.04 mm. 3 3 The wavelength of the light then is λ = a sinθ = (0.04 × 10−3 m) sin 0.75° = 5.24 × 10−7 m (524 nm)

35-61.

(a) Interference maxima occur at angles given by d sinθ = pλ , p = 1, 2, 3,… Diffraction minima occur at angles given by a sinθ = qλ , q = 1, 2, 3,… Interference maxima are suppressed

35-62.

when they occur at the same angles as the diffraction minima, i.e., when d sinθ = pλ a sinθ = qλ d p ⇒ = a q If d = ma, the interference maxima are suppressed when p = mq, p = 1,2,3,… (b) Since p = qm, ∆p = m∆q. To go from one diffraction minimum to the next is a change ∆q = 1, so the change in p is ∆p = m. There are m interference maxima between diffraction minima. To find the combined relative effect of the interference and diffraction effects, multiply the relative intensities: ⎛ I ⎞ ⎛ I ⎞ ⎛ I ⎞ =⎜ ⎜ ⎟ ⎟ ⎜ ⎟ ⎝ I max ⎠total ⎝ I max ⎠interference ⎝ I max ⎠ diffraction The intensities for the three narrow slits were calculated using the method described in Problem 1 2π (24λ sinθ ) ≈ 24πθ , since the 44 where the phase shift from one slit to the next is δ = × 2 λ total distance from the top slit to the bottom is 24λ. The phase for the single slit diffraction pattern 2π is φ = a sinθ ≈ 12πθ for a = 12λ. Graphs for each effect are shown, and the result of

λ

multiplying the two patterns is shown in the third graph.


CHAPTER

35 Three-Slit Interference Pattern 1

0 -10

-5

0

5

10

Single Slit Diffraction Pattern 1

0 -10

-8

-6

-4

-2

0

2

4

6

8

10

Combined Interference and Diffraction Pattern 1

0 -10

-8

-6

-4

-2

0

2


4

6

8

10

CHAPTER

†35-63.

2

35

2

⎡ sin(φ / 2) ⎤ ⎛ sin x ⎞ =⎢ =⎜ ⎟ where x = φ/2. ⎥ I max ⎣ (φ / 2) ⎦ ⎝ x ⎠ d ⎛ sin 2 x ⎞ 2 x sin 2 x − 2 x 2 sin x cos x = 0. Only the numerator must be zero: ⎜ ⎟= dx ⎝ x 2 ⎠ x4 x sin x( x cos x − sin x) = 0. I

x = 0 makes the first two factors zero and gives the location of the central maximum. Other extrema occur whenever x = mπ, which identify the minima. (Remember that φ = 2x, so these correspond to φ = 2mπ.) To find the minima, we must solve ( x cos x − sin x) = 0 ⇒ tan x = x. Such a transcendental equation can be solved using a spreadsheet or tools like MathCAD or Maple. The problem asks for a trial and error solution, and the results are shown in the table below. We choose a starting value near φ = 3π, which is x = 1.5π. Choose x = 4.65 as the actual initial guess. The table shows how the trial values for x produce values of tan x that are either greater than or less than x. We stop when x and tan x are the same to within four significant figures. (This is more than the problem requires, but it allows us to express the final answer to three figures.) x tan x 4.65 16 4.60 8.9 4.55 6.1 4.50 4.6 4.45 3.7 4.49 4.4 4.492 4.464 4.494 4.506 4.493 4.485 The error for x = 4.493 is 0.008, which is smaller than the error for 4.494. So the result is φ = 2x = 8.986. The problem requires the answer to be given as a multiple of 3π: φ = 0.952(3π ). (There is 35-64.

only a small error caused by using 3π for the secondary maximum.) (a) If the waves arrive normally at the slit, they are all in phase in the slit and there is a path difference a sin φ a sin θ between the top and the middle for the waves 2 θ arriving at a distant screen. If the waves arrive at an incidence angle φ, the waves at the top are out of φ phase with the waves at the bottom. For the oblique waves to interfere destructively, we have a sin θ a λ a sin θ = + sin φ, 2 2 2 where the extra term is added because the waves at the top are “late” by that amount. Therefore, a sin θ – a sin φ = λ. Continuing we have a sin θ – a sin φ = λ, 2λ, 3λ . . . . (b) a sin θn – a sin φ = nλ; a sin θn + 1 – a sin φ = (n + 1)λ. Therefore, ∆θ = θn + 1 – θn = sin θn + 1 – sin θn


CHAPTER

35

1 [(a sin θn + 1 – a sin φ) – (a sin θn – a sin φ)] a 1 λ = [(n + 1)λ – nλ] = (independent of φ ) a a If the dots touch, their separation is equal to their diameter. Thus Rayleigh’s criterion says 1.22λ 1.22(550 × 10−9 m) θ = = = 1.12 × 10−3 radian. At a viewing distance of 2.0 m, this a 6.0 × 10−3 m =

†35-65.

corresponds to a separation (diameter) of ∆x = (2.0 m)(1.12 × 10−4 radian) = 2.2 × 10−4 m (0.22 mm). These are very tiny dots. 35-66.

35-67.

35-68.

⎛ λ⎞ 550 × 10−9 m ⎞ ⎛ –5 (a) By Equation, 35.27, θ = sin–1 ⎜ 1.22 ⎟ = sin −1 ⎜ 1.22 × ⎟ = 9.6 × 10 rad −3 a⎠ 7 × 10 m ⎠ ⎝ ⎝ –4 Therefore, angular size = 2θ = 1.92 × 10 radian (b) Diameter = r(2θ) = 1.92 × 10–4 × 23 mm = 4.4 × 10–3 mm (c) Area of rod = πR2 = π(2.2 × 10–3)2 mm2 = 1.53 × 10–5 mm2 Number of rods = 150,000 mm–2 × 1.53 × 10–5 mm2 = 2 rods λ⎞ 15 ⎞ ⎛ ⎛ By Equation 35.27, θ = sin–1 ⎜ 1.22 ⎟ = sin–1 ⎜ 1.22 ⎟ = 47° a⎠ 25 ⎠ ⎝ ⎝ Angular width = 2 × θ = 94°

λ=

c 3.00 × 108 m/s = = 0.02 m. By Equation 35.27, f 1.5 × 1010 Hz ⎛ 1.22λ ⎞ −1 ⎡ 1.22(0.02 m) ⎤ ⎟ = sin ⎢ ⎥ = 1.4°. Therefore, angular width = 2θ = 2.8º (0.0488 a 1m ⎝ ⎠ ⎣ ⎦

θ = sin −1 ⎜

†35-69.

35-70. †35-71.

35-72.

rad). Linear width = r∆θ = 5.0 km × 0.0488 rad = 0.24 km. ⎛λ⎞ Optical resolution ∆θ = 1.22 ⎜ ⎟ . λ = 550 nm, a = 2.4 m. Therefore, ∆θ = ⎝a⎠ −9 1.22 × 550 × 10 m 180° ⎞ ⎛ –5 = 2.8 × 10−7 radian. This equals ⎜ 2.8 × 10−7 × ⎟ = (1.6 × 10 )° = 0.06 2.4 m π ⎝ ⎠ seconds of arc, eight times better than earth. 0.21 m ⎛λ⎞ ∆θ = 1.22 × ⎜ ⎟ = 1.22 × = 3.4 × 103 rad = 0.19° = 12′ or arc 76 m ⎝a⎠ 1.8 m = 6.7 × 10−5 radian (a) Size of man ~ 1.8 m. Therefore angular resolution = 3 27 × 10 m (b) Assume a wavelength of 550 nm to find the lens diameter. Then Rayleigh’s criterion says −9 λ ⎛ λ ⎞ 1.22 × 550 × 10 m = 0.010 m (1.0 cm) θ = 1.22 or a = 1.22 ⎜ ⎟ = 6.67 × 10−5 radian a ⎝θ ⎠ λ⎞ 1.22 × 550 × 10−9 m ⎛ –1 ⎜ 1.22 –1 ⎟ The lens acts like an aperture. θ = sin ⎝ = 6.71 × 10–6 radian. 0.10 m a ⎠ = sin Angular width = 2θ = 1.34 ×10–5 radian. Therefore, linear width = r(2θ) = 18 cm × 1.34 × 10–5 radian = 2.4 × 10–4 cm


CHAPTER 35-73.

Minimum distance between moons required for resolution dmin = distance to Jupiter × ∆θ where ∆θ = 1.22 Therefore dmin =

35-74.

35

λ a

.

6.3 × 1011 × 1.22 × 550 × 10−9 = 6 × 107 m 7 × 10−3

The distance between any two moons is larger than dmin, hence it is possible to resolve the four moons with the naked eye. The limit on resolution does not prevent one from seeing the moons. However, if the diameter of the moon is less than dmin, the moon would not appear as a disc but as a flickering object to the naked eye. 6800 km Angular diameter of Mars: θ = = 8.72 × 10−5 radian. Minimum diameter of the 7.8 × 107 km objective lens required: 1.22 × 550 × 10−9 m a= = 7.7 × 10−3 m (7.7 mm) −5 8.72 × 10 radian

35-75.

35-76.

λ

550 × 10−9 m = 9.6 × 10−5 radian 7.0 × 10−3 m a 20 30 (b) 20′ to 30′ of arc = degree to degree = 9.7 × 10−5 radian to 1.5 × 10−4 radian 3600 3600 Elimination of tremor would somewhat improve resolution. λ 550 × 10−9 m = 1.34 × 10–4 radian = minimum angle where headlights θ = 1.22 = 1.22 × 5.0 × 10−3 m a

(a) θ = 1.22

= 1.22

resolved. d d θ = (d = distance between head lights, r = distance to other truck). Therefore, r = = r θ 1.8 m = 1.34 × 104 m = 13.4 km 1.34 × 10−4 †35-77.

35-78.

(a) θ = 1.22

λ

= 1.22 ×

550 × 10−9 m = 2.24 × 10−6 radian 0.30 m

a d (b) θ = (d = distance between points, r = distance to points). r Therefore, d = rθ. d = 150 × 103 m × 2.24 × 10–6 = 0.34 m (c) The image on the film will subtend an angle θ at a distance f from the lens, where f is the focal length. Thus distance on film = ∆x = fθ = 2.4 m × 2.24 × 10–6 = 5.4 × 10–6 m (0.0054 mm) λ 550 × 10−9 m (a) θ = 1.22 = 1.22 = 1.3 × 10 −5 radian 0.05 m a 550 × 10−9 m = 9.6 × 10−5 radian (b) θeye = 1.22 0.007 m Actually, it should be θeye DIVIDED by seven, since binoculars increase angles by seven. 9.6 × 10−5 radian Therefore, with binoculars, θ = = 1.4 × 10−5 radian. Eye determines angular 7 resolution.


CHAPTER

35-79.

35

∆θ = 1.22

λ a

= 1.22 ×

550 × 10−9 m 0.05 m = 2.5 × = 2.2 × 10−3 radian. Mice at 20 m give θ = 0.3 × 10−3 m 20 m

–3

35-80.

10 . Therefore, cat barely distinguishes the mice. nfilm < nlens, so there is a 180° phase change on reflection from the top and bottom of the film. Then destructive interference requires 2d = (a) d =

λ 4n film

=

2n film d 2(1.38)(99.6 nm) 275 nm mλ . The ⇒λ= = = n film m m m

longest wavelength that would experience constructive interference is 275 nm, which is not visible. The film has air on both sides, so only the ray reflected from the outer surface of the bubble undergoes a phase change. The ray reflected within the film (from the rear surface) does not suffer a phase change because the index of refraction of the air is less than the index of refraction of the soapy water. Then the requirement for constructive interference is mλ 2d = , m = 1, 3, 5, … 2n film ⇒d =

mλ 4n film

For m = 1, d = 35-82.

for the minimum thickness.

550 nm = 99.6 nm 4(1.38)

(b) For constructive interference, 2d =

†35-81.

λ

2n film

650 nm = 120 nm. We conclude the film must be at least 120 nm thick. 4(1.35)

The volume of the soap film is fixed, so as the bubble inflates and increases the surface area, the film thickness decreases. The volume of the film is V film = π r 2 d , where r the radius of the bubble and d is the thickness of the film. For two radii r1 and r2, r12 d1 = r22 d 2 , so

r2 = r1

d1 . The d2

thicknesses are proportional to the wavelengths for strong reflections (see Problem 81), so r2 λ1 650 = = = 1.2 430 λ2 r1 †35-83.

For double slit interference, d sinθ m = mλ . If θ is small, this becomes dθ m = mλ dθ m m Here m = 3 and θ3 = 0.52º = 9.08 × 10–3 radian. Thus (0.15 × 10−3 m)(9.08 × 10−3 radian) λ= = 4.54 × 10−7 m (454 nm) 3 1 1 1 cm = m ⇒ = 6 × 105 m −1 . Distance d between slits = 5 6000 6 × 10 d

⇒λ=

35-84.

⎛ λ⎞ Since d sinθ = nλ ⇒ θ = sin −1 ⎜ n ⎟ ⎝ d⎠ –9 λred = 656.3 × 10 m λblue = 434.2 × 10–9 m


CHAPTER

35-85.

35

First order spectrum ⇒ n = 1. Therefore, ⎛λ ⎞ θ red = sin −1 ⎜ red ⎟ = sin −1 (656.3 × 10−9 × 6 × 105 ) = 23.2° ⎝ d ⎠ ⎛λ ⎞ θ blue = sin −1 ⎜ blue ⎟ = sin −1 (434.2 × 10−9 × 6 × 105 ) = 15.1° ⎝ d ⎠ Therefore, ∆θ = 8.1º This is exactly like interference from two slits. Here wavelength 3 × 108 m/s λ = c/f = = 3 × 103 m = 3.0 km. 5 1 × 10 / s Then as in Example 2, λ 3λ 5λ sinθ = , , , ... 2d 2d 2d

λ

35-86.

= 3.0 km/6.0 km = 0.5 d θ = sin–1(0.25, 0.75). Therefore, θ = ± 14.5°, ± 48.6° From the diagram, half the path difference is d sinθ . For constructive interference, the total path difference must be an integer number of wavelengths. Hence the requirement 2d sinθ = mλ , m = 1, 2, 3,…

θ θ

d θ

†35-87.

(a) Let the distance between the buildings and their height be d and h, respectively. The reflected rays must travel a distance of 2 h 2 + ( d / 2) 2 . Therefore, the path difference between a direct ray and a reflected ray is 4h 2 + d 2 − d = 4h 2 + d 2 − d . 4 Because the reflected ray undergoes a phase change, destructive interference will occur when the path difference is an integral number of wavelengths: ∆d = 2 h 2 + (d / 2) 2 − d = 2

4h 2 + d 2 − d = mλ , m = 1, 2, 3,… (b) This condition can only be met for wavelengths that satisfy the condition 4(60 m) 2 + (2300 m) 2 − 2300 m 3.13 4h 2 + d 2 − d = = m m m m The lowest frequency corresponds to the longest wavelength, which has m = 1, or 3.13 m. Thus 3 × 108 m/s c = = 9.59 × 107 Hz (95.9 MHz) f min = λmax 3.13 m

λ=


CHAPTER 35-88.

35

The maxima locations are given by d sin θ = nλ;

λn =

n = 0, 1, 3 .... for θ = 30°, sin 30° =

1 , therefore 2

d 10 = cm n 2n

Thus, the dominant wavelengths are 10 cm, 5 cm, The corresponding frequencies are f n =

c

λn

=

331

λn

10 cm, etc. 3 Hz

f1 = 3310 Hz, f 2 = 2f1 ; f3 = 3f1 , etc. 35-89.

θ =

1.22λ 1.22(580 × 10−9 m) = = 1.01 × 10−4 radian. ∆x = rθ , where r = distance to headlights. −3 a 7.0 × 10 m

Thus ∆x r= =

θ

35-90.

1.3 m = 13 km 1.01 × 10−4 radian

Difference in pathlengths is still d sin θ, but now we want path difference to be equal to nλ – (λ/4), the λ/4 factor to compensate for the fact that south beacon emits at 1/4 cycle later than N. Therefore, for constructive interference d sin θ = nλ −

λ 4

Here λ = c / v =

. 3 × 108 m/s = 1500 m 2 × 105 / s

d = 3000 m. Therefore, λ⎛ 1⎞ 1⎞ ⎛ n = 0, ±1, ±2, . . . sin θ = ⎜ n − ⎟ = 0.5 ⎜ n − ⎟ d⎝ 4⎠ 4⎠ ⎝ θ = 22°, 61°, –7°, –39° (minus means south of EW line) †35-91.

At the first order minimum, a sinθ1 = λ ⇒ sinθ1 =

λ a

=

693 × 10−9 m = 0.0139. For an angle 0.050 × 10−3 m

this small, sin θ ≈ θ, so θ1 ≈ 0.0139 radian. The distance from the central maximum to the first minimum is ∆x1 = rθ1 , where r is the distance to the screen. Thus ∆x1 = (2.5 m)(0.0139 radian) = 3.5 cm The higher order minima are given by aθ m = mλ for small angles, so θ2 = 2 θ1, and ∆x2 = 2(2.5 m)(0.0139 radian) = 6.9 cm 35-92.

(a) a sinθ1 = λ ⇒ sinθ1 =

λ a

=

577 × 10−9 m = 5.25 × 10−3 ⇒ θ1 ≈ 5.25 × 10−3 radian. This is −3 0.11 × 10 m

half the angular width of the central maximum, so the full width is ∆x1 = 2rθ1 = 2(4.0 m)(5.25 × 10−3 radian) = 4.2 cm (b) The second minimum is at aθ 2 = 2λ ⇒ θ 2 = 2θ1 ⇒ θ 2 − θ1 = 2θ1 − θ1 = θ1. The width of the second maximum is half the width of the central maximum: ∆x2 = r (θ 2 − θ1 ) = rθ1 = (4.0 m)(5.25 × 10−3 radian) = 2.1 cm


CHAPTER 35-93.

35

Treat antenna as circular aperture. Then by Equation 35.27 0.10 m ⎛λ⎞ sin θ = 1.22 ⎜ ⎟ = 1.22 × = 8.13 × 10–5 ⇒ θ = 8.13 × 10–5 rad. 1500 m ⎝a⎠ Angular width = 2θ = 1.63 × 10−4 rad

35-94.

Diameter of beam when it hits earth = r∆θ = 35,000 km × 1.63 × 10–4 = 5.7 km 0.05 m (a) Angular resolution = θ = = 3.13 × 10−7 rad 3 160 × 10 m (b) Diameter of the aperture of the camera, 1.22λ 1.22(550 × 10−9 m) a= = = 2.15 m θ 3.13 × 10−7 rad

†35-95.

(a) The beam produced by the transmitter matches the diffraction pattern that would be formed by waves entering the aperture. Thus 1.22λ ⎛ 1.22λ ⎞ −1 ⎡ 1.22(0.04 m) ⎤ sinθ = ⇒ θ = sin −1 ⎜ ⎟ = sin ⎢ ⎥ = 0.0325 radian a ⎝ a ⎠ ⎣ 1.5 m ⎦ Angular width = 2θ = 0.0650 radian. Therefore, the linear width at 30 km = r∆θ = 30 km × 0.065 = 1.95 km Power . In front of the antenna, area = πR2 = π(0.75)2 m2 = 1.77 m2 (b) Flux = area 1.5 × 103 W Therefore, flux = = 847 W/m 2 2 1.77 m At 30 km, radius R′ = 1.95 km/2 = 975 m. Therefore, area = π(975 m)2 = 2.99 × 106 m2 1.5 × 103 W Therefore, flux = = 5.02 × 10−4 W/m 2 6 2 2.99 × 10 m


CHAPTER 36

THE THEORY OF SPECIAL RELATIVITY


When the Earth and Sun are moving in the same direction, their speeds relative to the ether add to give the maximum relative speed. When they are moving in opposite directions, their speeds subtract to give the minimum relative speed. Vmax = vsun + vearth = vsun + 30 km/s

vEarth

vSun

Vmin = vsun − vearth = vsun − 30 km/s (a) vsun = 30 km/s ⇒ Vmax = 60 km/s; Vmin = 0

(b)vsun = 60 km/s ⇒ Vmax = 90 km/s; Vmin = 30 km/s

vEarth 36-2.

L L , t2 = are the travel times for the light each way when the arm is parallel to c +V c −V the direction of V. The round trip time is L L L (c − V ) + L (c + V ) 2cL 2L t = t1 + t2 = + = = 2 = . c +V c −V (c − V )(c + V ) (c − V 2 ) c ⎡⎣1 − (V / c) 2 ⎤⎦

(a) t1 =

(b) When the arm is perpendicular to V, the times are equal: t1 = t2 = time is t⊥ = t1 + t2 =

2L 2

c −V

2

=

2L c 1 − (V / c) 2

L 2

c −V2

. The round trip

.

(c) The difference is

∆t = t − t⊥ = (d) T =

2 ⎛ 2L 2L 2L ⎡ V V 2 ⎞ ⎤ LV 2 1 1 − ≈ + − + ⎢ ⎜ ⎟⎥ = 3 c ⎣ c2 2c 2 ⎠ ⎦ c c ⎡⎣1 − (V / c) 2 ⎤⎦ c 1 − (V / c) 2 ⎝

1 λ = f c

∆t LV 2 ⎛ c ⎞ ⎛ L ⎞ ⎛ V ⎞ = 3 ⎜ ⎟ = ⎜ ⎟⎜ ⎟ T c ⎝ λ ⎠ ⎝ λ ⎠⎝ c ⎠

2

2

4 11 ⎛ 11 ⎞ ⎛ 3 × 10 ⎞ = ⎜ × 107 ⎟ ⎜ = × 10−1 = 0.22, or 22% 8 ⎟ 5 ⎝5 ⎠ ⎝ 3 × 10 ⎠ If the effect existed, it should have been easy to detect.


CHAPTER 36-3.

36

Let L1 be the parallel length and L2 be the perpendicular length. Then 2 L1 2 L1 ⎛ V2 ⎞ ≈ + t = 1 ⎜ ⎟ c ⎝ c2 ⎠ c ⎡⎣1 − (V / c) 2 ⎤⎦ t⊥ =

2 L2 c 1 − (V / c) 2

≈

2 L2 c

⎛ V2 ⎞ 1 + ⎜ ⎟ 2c 2 ⎠ ⎝

In the first orientation, 2 ⎛ 2 L1 2 L2 2⎡ ⎛ V ⎞ V 2 ⎞⎤ ∆t1 = − ≈ + − + L 1 L 1 ⎢ ⎜ ⎟ ⎜ ⎟⎥ 1 2 c ⎢⎣ ⎝ c2 ⎠ 2c 2 ⎠ ⎥⎦ c ⎣⎡1 − (V / c) 2 ⎦⎤ c 1 − (V / c) 2 ⎝ 2 2⎡ L2 ⎞ ⎛ V ⎞ ⎤ ⎛ − + − ( L L ) L ⎢ 1 2 ⎜ 1 ⎟⎜ ⎟ ⎥ c ⎢⎣ 2 ⎠ ⎝ c ⎠ ⎥⎦ ⎝ Rotate by 90° so 2 L2 2 L2 ⎛ V2 ⎞ ≈ + 1 t = ⎜ ⎟ c ⎝ c2 ⎠ c ⎡⎣1 − (V / c) 2 ⎤⎦

∆t1 =

t⊥ =

2 L1 c 1 − (V / c) 2

≈

2 L1 ⎛ V2 ⎞ + 1 ⎜ ⎟ 2c 2 ⎠ c ⎝

Now 2 L1 ⎞ ⎛ V ⎞ ⎤ 2⎡ ⎛ ∆t2 = ⎢ ( L2 − L1 ) + ⎜ L2 − ⎟ ⎜ ⎟ ⎥ 2 ⎠ ⎝ c ⎠ ⎥⎦ c ⎢⎣ ⎝

∆t1 − ∆t2 = = 36-4.

2 L2 ⎞ ⎛ L1 ⎞ ⎤ ⎛ V ⎞ ⎫⎪ ⎡⎛ 2 ⎧⎪ 2( ) L L L L − + − − − ⎨ 1 2 ⎢⎜ 1 2 ⎟ ⎜ 2 2 ⎟ ⎥ ⎜ c ⎟ ⎬ c ⎩⎪ ⎠ ⎝ ⎠ ⎦ ⎝ ⎠ ⎭⎪ ⎣⎝

2 2( L1 − L2 ) ⎡ 3⎛V ⎞ ⎤ ⎢2 − ⎜ ⎟ ⎥ 2 ⎝ c ⎠ ⎦⎥ c ⎣⎢

All the measured values for c are the same. Suppose that all the light pulses were emitted at the same time (i.e., the pulse from the pod was emitted at the instant the pod was ejected and the pulse from the ship was also emitted at that same instant). No direction of motion of the ship/pod relative to earth is given, so the question of arrival time is hard to answer exactly. If the pulses were emitted when the observers were aligned perpendicular to Earth, they see no length contraction of the ship–Earth distance and both observers will see the pulses arrive simultaneously. However, if the ship and pod measured some longitudinal component of displacement from earth when the pulses were emitted, both will see contraction of that component of displacement. Since the pod has a greater speed relative to earth than the ship, the contraction will be greater for the pod observer than for the ship observer. Thus the pod observer sees a shorter distance for the pod pulse to travel and will see the pod’s pulse arrive earlier than the ship’s pulse.


CHAPTER

†36-5.

∆t = time measured in the lab and ∆t′ = time in the neutron’s frame.

36

∆t 1 = ∆t ′ 1 − (V / c) 2

⎡ ⎛ V ⎞2 ⎤ 10 = ⇒ 100 ⎢1 − ⎜ ⎟ ⎥ = 1 1 − (V / c) 2 ⎢⎣ ⎝ c ⎠ ⎥⎦ 1

36-6.

1 V = 1− = 0.995 100 c V = 0.995c ∆t ′ ∆t = 1 − (V / c) 2 For V = 0.25c, ∆t =

15 1 − (0.25) 2

15 = 17.32 min 1 − 0.25

Similarly, for V0 = 0.5c, ∆t =

and for V0 = 0.9c, ∆t = 36-7.

38-8. †38-9.

We want ∆t =

= 15.49 min

15 1 − (0.9) 2

∆t ′ 1 − (V / c) 2

= 34.41 min

= 2∆t ′. Solving for V/c, we obtain V = 0.866c. That is, the pion must

have a speed of 0.866c or 2.60 × 108 m/s to make its lifetime, in the laboratory frame of reference, 5.2 × 10–8 s. ∆t 1 1 = = = 1.0000002 2 ∆t ′ 1 − (V / c) 1 − (200 km/s / 3 × 105 km/s) 2 ∆t =

2 ⎡ 1 ⎛V ⎞ ⎤ ≈ ∆t ′ ⎢1 + ⎜ ⎟ ⎥ , using the approximation (1 − x) n ≈ 1 − nx with 2 ⎝ c ⎠ ⎥⎦ 1 − (V / c) 2 ⎣⎢

∆t ′

2

1 ⎛V ⎞ x = ⎜ ⎟ and n = − as given in the hint. The difference between the time ∆t in the sun’s 2 ⎝c⎠ 2

⎛V ⎞ frame ⎜ ⎟ and the time ∆t′ in the earth’s frame is ⎝c⎠ 2 2 ⎡ 1 ⎛V ⎞ ⎤ ∆t ′ ⎛ V ⎞ ∆t − ∆t ′ = ∆t ′ ⎢1 + ⎜ ⎟ ⎥ − ∆t ′ = ⎜ ⎟ . (∆t > ∆t′ because the moving clock on the Earth 2 ⎝ c ⎠ ⎦⎥ 2 ⎝c⎠ ⎣⎢ appears to run slow compared to the clock at rest in the Sun’s frame.) To three significant figures, both times are equal to 1 year, or 3.15 × 107 s. The difference between the readings is 2

∆t − ∆t ′ = 36-10.

3.15 × 107 s ⎛ 30 km/s ⎞ ⎜ ⎟ = 0.16 s. 5 2 ⎝ 3 × 10 km/s ⎠

2 ⎡ 1 ⎛V ⎞ ⎤ 2 ′ ∆t = ∆t 1 − (V / c) ≈ ∆t ⎢1 − ⎜ ⎟ ⎥ . V = 2.6 × 10–5 c and 2 ⎝ c ⎠ ⎥⎦ ⎢⎣ ∆t = 25 hr = 90,000 s. Therefore,


CHAPTER

36-11.

36

⎡ (2.6 × 10−5 ) 2 ⎤ –5 ∆t ′ ≈ ⎢1 − ⎥ 90,000 s = 90,000 s – 3.0 × 10 s 2 ⎣ ⎦ His body clock fell behind by 3.0 × 10–5 s. 1 1 ⎛ ∆t ⎞ ⎛ ∆t ⎞ ⎛ ∆t ⎞ = 2⎜ = ;⎜ ⎜ ′⎟ = ⎟ ⎟ 2 ⎝ ∆t ⎠1 ⎝ ∆t ′ ⎠1 1 − (V / c) ⎝ ∆t ′ ⎠ 2 1 − (2V / c) 2

⎛ ∆t ⎞ ⎜ ′⎟ 1 − (V / c) 2 ⎝ ∆t ⎠ 2 = =2 1 − (2V / c) 2 ⎛ ∆t ⎞ ⎜ ′⎟ ⎝ ∆t ⎠1 ⎡ ⎛ 2V ⎞ 2 ⎤ ⎡ ⎛ V ⎞ 2 ⎤ ⇒ 4 ⎢1 − ⎜ ⎟ ⎥ = ⎢1 − ⎜ ⎟ ⎥ ⎣⎢ ⎝ c ⎠ ⎥⎦ ⎢⎣ ⎝ c ⎠ ⎦⎥ 2

⎛V ⎞ 15 ⎜ ⎟ = 3 ⎝c⎠ c V = = 0.447c 5 36-12.

The ratio of the periods is

Tearth f 1 = ⇒ earth = 1 − (V / c) 2 . Thus 2 Tship f ship 1 − (V / c)

f earth = f ship 1 − (V / c) 2 = (3 Hz) 1 − (0.80) 2 = 1.8 Hz †36-13.

Follow the method in Problem 9 with ∆t − ∆t ′ = 1.0s, ∆t ≈ ∆t ′ = 1 day = 86,400 s. Then

V = c 36-14.

2(∆t − ∆t ′) = ∆t

2(1.0 s) = 4.8 × 10−3 86, 400 s

V = 4.8 × 10–3c = 1.4 × 106 m/s. (a) In the Earth’s frame, the round trip would take 8.8 years at the speed of light. The ship is traveling at 0.10c, so the round trip takes 88 yr for the ship. (b) On the ship, ∆t ′ = ∆t 1 − (V / c) 2 = (88) 1 − (0.1) 2 = 87.6 yr

†36-15.

For a receding emitter, Equation 36.13 says

V = 0.50c:

f′ 1 − 0.50 = = 0.577 f 1 + 0.50

V = 0.70c:

f′ 1 − 0.70 = = 0.420 f 1 + 0.70

f′ 1 − (V / c) = f 1 + (V / c)

f′ 1 − 0.90 = = 0.229 f 1 + 0.90 f′ 1 − (V / c) = 0.20 ⇒ (0.20) 2 = f 1 + (V / c) ⇒ 0.04 + 0.04(V / c) = 1 − (V / c) 0.96 ⇒ V /c = = 0.923 1.04

V = 0.90c: 36-16.


CHAPTER 36-17.

36

Speed = rω = 2πrf f = 35,000/min = 583.3 Hz; r = 0.1 m Therefore, speed = 2π(583.3)(0.1)m/s = 366 m/s. 1 By Equation 36.5, the time dilation factor is ; 1 − (V / c) 2

v = 1.22 × 10–6 c. Then 2

1 1 − (V / c) 2 36-18.

≈1+

1 ⎛V ⎞ 1 −6 2 −13 ⎜ ⎟ = 1 + (1.22 × 10 ) = 1 + 7.4 × 10 = 1.00000000000074 2⎝ c ⎠ 2

∆t ′ = ∆t 1 − (V / c) 2 , and the time taken (in the Earth frame) for the ship to reach Andromeda is ⎛d⎞ d/V. Therefore ∆t ′ = ⎜ ⎟ 1 − (V / c) 2 . If c is given as 1 light-yr/yr rather than in m/s, all ⎝V ⎠ calculations can be done in terms of light-yr and years. Then d/c = 2.2 × 106 yr. Solve the equation for V/c: ⎛ d /c ⎞ 2 ∆t ′ = ⎜ ⎟ 1 − (V / c) V c / ⎝ ⎠ 2

1 − (V / c) 2 ⎛ ∆t ′ ⎞ ⎜ ⎟ = (V / c) 2 ⎝ d /c⎠

36-19.

2

2

⎞ 1 ⎛ ∆t ′ ⎞ 1 ⎛ 10 yr −11 ≈1− ⎜ ⎟ = 1 − 1.0 × 10 ⎟ =1− ⎜ 6 2 d c 2 / 2 2.2 × 10 yr ⎝ ⎠ ⎝ ⎠ ⎛ ∆t ′ ⎞ 1+ ⎜ ⎟ ⎝ d /c⎠ V = 0.999999999999c The time elapsed on Earth will be d/V. Since V is so close to c, the time will be unmeasurably different from the travel time for light, or ∆t = 2.2 × 106 yr. 2 1 1 ⎛V ⎞ 460 The difference factor is ≈ 1 + ⎜ ⎟ . Here V/c = = 1.53 × 10–6. The 8 2 c 2 3 × 10 ⎝ ⎠ 1 − (V / c)

V = c

1

(1.53 × 10 ) factor by which the clocks differ is 1 +

−6 2

36-20.

= 1 + 1.17 × 10–12 = 1.00000000000117.

2 ⎡ (1.53 × 10−6 ) 2 ⎤ –12 7 Clocks will differ by ⎢ ⎥ × 1 yr = 1.17 × 10 × 3.156 × 10 s 2 ⎣ ⎦ –5 = 3.7 × 10 s. The clock at north pole (stationary in Earth frame) will be ahead. (a) By symmetry, the time taken for the round trip is twice the time taken to get there, and the time taken to get there is twice the time taken to reach the midpoint.

x – x0 =

at 2 , so that t = 2

2( x − x0 ) = a

(2 × 2.2 ly × 9.46 × 1015 m/ly) = 2.06 × 108 s. (0.1 × 9.81 m/s 2 )

Therefore, the round trip takes 4 × 2.06 × 108 s = 8.24 × 108 s. (b) Instantaneously, dt ′ = 1 − (V / c) 2 dt, where dt′ is for the spaceship and dt is for the Earth.

dt′ = 1 − (at / c) 2 dt = 1 − (3.27 × 10−9 t ) 2 dt Let 3.27 × 10–9 t = sin θ. Then dt =

cos θ dθ . 3.27 × 10−9


CHAPTER

36

The limits are 0 and sin θ = 3.27 × 10–9 × 2.06 × 108 = 0.674. 0.674 cos θ dθ Then t ′ = ∫ (1 − sin 2 θ )1 / 2 0 3.27 × 10−9 1 = 3.27 × 10−9

t′ =

0.674

⎡ sin θ 1 − sin 2 θ 1 −1 ⎤ + sin (sin θ ) ⎥ ⎢ 2 2 ⎢⎣ ⎥⎦ 0

1 (0.249 + 0.370) = 1.89 × 108 s 3.27 × 10−9

Total time = 4t′ = 7.57 × 108 s (24 yr). 36-21. 36-22.

36-23.

L = L′ 1 − (V / c) 2 = (1.00 m) 1 − (0.5) 2 = 0.866 m ∆x . Here, ∆x = 100 m, and ∆x′ 2 1 1 3 3c ⎛V ⎞ = 0.866c ∆x ′ = 200 m. Thus 1 − (V / c) 2 = ⇒ 1 − (V / c) 2 = ⇒ ⎜ ⎟ = ⇒ V = 2 4 4 2 ⎝c⎠ A moving object appears contracted along the direction of motion. The transverse diameter is the “true” diameter (D′ = 0.20 m), so its apparent longitudinal diameter is ∆x = ∆x′ 1 − (V / c) 2 so 1 − (V / c) 2 =

D = D′ 1 − (V / c) 2 = (0.20 m) 1 − (0.3) 2 = 0.19 m. 36-24.

1 − (V / c) 2 =

∆x 26.7 . 96 km/hr = 26.7 m/s = c = 8.89 × 10–8 c ∆x′ 3 × 108 2

36-25. 36-26.

∆x 1 ⎛V ⎞ 1 ≈ 1 − ⎜ ⎟ = 1 − (8.89 × 10−8 ) 2 = 1 − 4.0 × 10−15 so the length contracts by a factor of 2⎝ c ⎠ 2 ∆x′ –15 –13 4.0 × 10 or 4.0 × 10 %. L 1 1 3 = 1 − (V / c) 2 = ⇒ 1 − (V / c) 2 = ⇒ V = c = 0.866c ′ 2 4 2 L y = y′ = 0.20 m. The x-side will be shortened to

x = x′ 1 − (V / c) 2 = (0.20 m) 1 − (0.8) 2 = 0.12 m. hypotenuse = ( x′) 2 + ( y ′) 2 = (0.12) 2 + (0.20) 2 cm = 0.233 cm. y′ 0.20 θ ′ = tan −1 = tan −1 = 59.0°. The remaining angle is 90° – 59.0° = 31.0°. 0.12 x′ †36-27.

L1 = L′ 1 − (V1 / c) 2 , L2 = L′ 1 − (V2 / c) 2 ⇒

θ′

L1 1 − (V1 / c) 2 = . L2 = 1.50L1, V1 = 0.80c. 1 − (V2 / c) 2 L2

2

1 1 − 0.82 1 − 0.82 ⎛2⎞ ⇒ = ⇒ = ⎜ ⎟ 1.5 1 − (V2 / c) 2 1 − (V2 / c) 2 ⎝3⎠ 4 − 4(V2 / c) 2 = 3.24 V2 4 − 3.24 = c 4 V2 = 0.44c


CHAPTER 36-28.

36

In the rest frame of the meter stick, x′ = 1.0 cos 30 = 0.866 m; y′ = 1.0 sin 30 = 0.5 m. In a new reference frame moving at V = 0.7c in the x-direction, y = y′= 0.5 m, and x = x′ 1 − (V / c) 2 = (0.866 m) 1 − (0.7) 2 = 0.618 m. In the new reference frame, the meter stick makes an angle with the x-axis given by tan θ ′ = 0.50/0.618, or θ ′ = 39.0°.

†36-29.

The sphere will appear compressed along the direction of motion, so it will look like an oblate spheroid with major semiaxis equal to the radius r′ in the rest frame of the sphere and minor semiaxis equal to the Lorentz contracted radius: a = r ′ 1 − (V / c) 2 . The volume of an oblate 4π r 2 a 4π ( r ′)3 . The sphere’s volume in its own rest frame is V ′ = . The charge 3 3 ρ V′ = , where ρ′ is the density in the density is inversely proportional to the volume, so ρ′ V spheroid is V =

sphere’s rest frame. Then ρ ( r ′)3 1 ρ′ 2.0 × 10−6 C/m3 ρ = = ⇒ = = = 3.3 × 10−6 C/m3 ρ ′ (r ′) 2 a 1 − (V / c) 2 1 − (V / c) 2 1 − (0.8) 2 36-30.

Draw the field pattern for the charge at rest and copy it into a “paint” program. Then use that program to change the scale so that the horizontal axis is shortened by the factor 1 − (V / c) 2 but the vertical axis remains unchanged. For V/c = 0.8, 1 − (V / c) 2 = 0.6 and the horizontal axis must be scaled to 0.6 of its original length. The results are shown in the two figures.

At rest

Moving right at 0.8c

(Note that even the spherical “charge” has changed shape.)


CHAPTER †36-31.

36-32.

In the frame of reference of the base, the belt is shortened according to ∆x = ∆x′ 1 − (V / c) 2 . The belt should therefore be tightened about the flywheels. However, if we are riding on the upper part of the belt, which is moving with a speed V, the base should appear shortened by the same factor as the upper belt appears shortened in the frame of the base. However, relative to the upper part of the belt the lower belt segment is moving with a speed greater than V and appears shortened even more than the base. Therefore the belt is tightened in both reference frames, and there is no paradox. From Equations 36.26 and 36.28 with V = 0.5c = 1.5 × 108 m/s, 6.0 × 108 m − (1.5 × 108 m/s)(4.0 s) x − Vt x′ = = =0 1 − (V / c) 2 1 − (0.5) 2 t′ =

†36-33.

36

t − xV / c 2 1 − (V / c) 2

=

4.0 s − (6.0 × 108 m)(0.5) /(3.00 × 108 m/s) 1 − (0.5) 2

= 3.5 s

The Galilean transformation would give (x′, t′) = (0, 4.0 s). Equations 36.26 and 36.28 are x − Vt x′ = (1) 1 − (V / c) 2 t′ =

t − xV / c 2

(2)

1 − (V / c) 2

Solve (2) for x: c 2 t − c 2 t ′ 1 − (V / c) 2 x= (3) V Solve (1) for t and substitute into (3): x − x′ 1 − (V / c) 2 t= V

c 2 x − c 2 x′ 1 − (V / c) 2 − c 2 t ′ 1 − (V / c) 2 V x= V c 2 x − c 2 x′ 1 − (V / c) 2 − c 2t ′ 1 − (V / c) 2 V ⎛ c2 ⎞ ⎛ x′ + Vt ′ ⎞ 2 2 x ⎜V − ⎟ = − ⎜ ⎟ c 1 − (V / c) V V ⎝ ⎠ ⎝ ⎠ xV =

x(V 2 − c 2 ) = −( x′ + Vt ′)c 2 1 − (V / c) 2 − xc 2 [1 − (V / c) 2 ] = −( x′ + Vt ′)c 2 1 − (V / c) 2 Divide through by c 2 [1 − (V / c) 2 ] and change the signs to get the final result: x′ + Vt ′ x= 1 − (V / c) 2 Solve (1) for t: x − x′ 1 − (V / c) 2 t= V

(4)


CHAPTER

36

Solve (2) for x and substitute into (4):

c 2 t c 2 t ′ 1 − (V / c) 2 x= − V V 2 2 2 ⎡ ⎛ c ⎞ 2 x′ ⎤ x′ 1 − (V / c) 2 c 2 t c t ′ 1 − (V / c) ⎛c⎞ 2 1 ( / ) − − = − − t V c ⎢t′ ⎜ ⎟ + ⎥ ⎜ ⎟ V2 V2 V V ⎥⎦ ⎝V ⎠ ⎣⎢ ⎝ V ⎠ ⎡ ⎛ c ⎞2 ⎤ ⎡ ⎛ c ⎞ 2 x′ ⎤ 2 t ⎢ ⎜ ⎟ − 1⎥ = 1 − (V / c) ⎢ t ′ ⎜ ⎟ + ⎥ V ⎦⎥ ⎣⎢ ⎝ V ⎠ ⎦⎥ ⎣⎢ ⎝ V ⎠

t=

⎛ c 2 − V 2 ⎞ c 2 1 − (V / c) 2 t⎜ ⎟= 2 V2 ⎝ V ⎠

⎡ ′ x′V ⎤ ⎢t + c2 ⎥ ⎣ ⎦ x′V ⎤ ⎡ tc 2 [1 − (V / c) 2 ] = c 2 1 − (V / c) 2 ⎢ t ′ + 2 ⎥ c ⎦ ⎣ 2 2 Divide through by c [1 − (V / c) ] to get the final result: t ′ + x′V / c 2 t= 1 − (V / c) 2 36-34.

Take the origin of the reference frame of y the ship to be at (x′, y′) = (0, 0). At t = t′, let this origin coincide with the origin of the reference frame (x, y) on the Earth, and at this instant the flash of light is emitted from the nose. Some time later the light x arrives at the rear of the ship, located at x′ = –300 m. (a) In the frame of the ship, the light travels 300 m, so it reaches the rear at 300 m d′ = = 1.0 × 10−6 s t′ = 3.00 × 108 m/s c (b) Using the results from Problem 33, the Earth time coordinate is t = Substituting gives t =

1.0 × 10−6 + (−300 m)(0.8) /(3.00 × 108 m/s) 1 − (0.8)

†36-35.

2

y′ V

x′

t ′ + x′V / c 2 1 − (V / c) 2

.

= 3.3 × 10−7 s

y

y′ V = 0.6c vx = 0.9c

x′

x

Both ships are approaching earth (x, y). Let the (x′, y′) system be the first ship. No direction is explicitly given for the motion of the second ship, but the term “following” means it is moving in the same direction as the first. The speed of the second ship measured from the earth frame is vx = 0.9c. Use Equation 36.36 to find the velocity relative to the first:


CHAPTER

36

vx − V (0.9 − 0.6)c = = 0.65c 2 1 − vxV / c 1 − (0.9)(0.6) vx − V v′x = 1 − vxV / c 2 vV ⎞ v′V ⎞ ⎛ ⎛ vx′ ⎜ 1 − x 2 ⎟ = vx − V ⇒ vx ⎜ 1 + x 2 ⎟ = v′x + V c ⎠ c ⎠ ⎝ ⎝ vx′ + V vx = 1 + vx′V / c 2

v′x = 36-36.

†36-37.

(a) Since the coordinates of the nova are given in the reference frame of the moving spaceship, use the inverse transformation from Problem 33 to find the coordinates in the Earth frame. V = 0.8c = 2.4 × 108 m/s.

y

y′ V

x′

x x′ + Vt ′

x=

1 − (V / c)

2

=

1.9 × 1017 m + (2.4 × 108 m/s)(−6 × 108 s) 1 − (0.8)

2

☼ Nova

= 7.67 × 1016 m

y = y ′ = 1.2 × 1017 m t=

36-38.

x′ + Vt ′ / c 2 1 − (V / c)

2

=

−6.0 × 108 s + (1.9 × 1017 m)(0.8) /(3.00 × 108 m/s) 1 − (0.8)

2

= −1.56 × 108 s

(b) According to the results of (a), the distance to the nova is shorter in the Earth’s frame than in the frame of the spaceship. Because Light travels at the same speed in both frames and the radio message also travels at the speed of light, the light from the nova will arrive at Earth before the radio message. d′ (a) In the spaceship frame the time for the light to travel to the ship is ∆t ′ = , where d′ is the c distance to the nova in ship’s frame: d ′ = ( x′) 2 + ( y ′) 2 = 1.92 + 1.22 × 1017 m = 2.25 × 1017 m. Then ∆t ′ =

2.25 × 1017 m = 7.49 × 108 s. The nova occurred at t′0 = –6.0 × 108 s in the ship’s frame, so 8 3.00 × 10 m/s

the light reaches the ship at t ′ = t0′ + ∆t ′ = (−6.0 + 7.49) × 108 s = 1.49 × 108 s.

(b) The captain’s coordinates when he sends the message are (x′, y′) = (0, 0). In the Earth frame, 0 + (2.4 × 108 m)(1.49 × 108 s) x′ + Vt ′ this is x = = = 5.96 × 1016 m , y = 0. The 2 2 1 − (V / c) 1 − (0.8) corresponding time is t =

t ′ + x′V / c 2 1 − (V / c)

2

=

1.49 × 108 s + 0 1 − (0.8)

2

message to travel from the ship to the Earth is ∆tmessage =

= 2.48 × 108 s. The time for the x 5.96 × 1016 m = = 1.96 × 108 s. Thus c 3.00 × 108 m/s

the message arrives at tmessage = tsent + ∆tmessage = 4.47 × 108 s.


CHAPTER

36

(c) In the Earth frame, the nova occurred at ( x, y, t0 ) = (7.67 × 1016 m, 1.2 × 1017 m, −1.56 × 108 s). The distance to the nova is d = 0.767 2 + 1.22 × 1017 m = 1.42 × 1017 m. The travel time for the light is ∆t =

d 1.42 × 1017 m = = 4.75 × 108 s. The light arrives at t = t0 + ∆t = (−1.56 + 4.75) × 108 s c 3.00 × 108 m/s

= 3.19 × 108 s. Note that this is earlier than the arrival of the message, as predicted in Problem 37. †36-39.

Let New York be the origin, so the explosion there occurs at t = 0. Also, let the origin (time and distance) in the primed (spaceship) frame be the same as for the unprimed frame. Then t − xV / c 2 . For the explosion in the spaceship frame, the explosion in Camden occurs at t ′ = 1 − (V / c) 2 in Camden to occur before New York, t ′ < 0, which means t < Vx/c2 or V > (c2t/x). t = 0.0003 s, so ⎛ 3.00 × 108 m/s × 0.0003 s ⎞ ⎛ ct ⎞ c⎜ ⎟ = c⎜ ⎟ = 0.6c , or V > 0.6c. 150 × 103 m ⎝ x⎠ ⎝ ⎠

36-40.

†36-41.

Let the atom be the primed system and the lab the unprimed system. Then V = 0.8c and the speed of the ejected particle in the atom’s frame is v′x = 0.5c. From Problem 36, the velocity of the v′x + V . If the particle is ejected in the forward direction, its particle in the lab frame is vx = 1 + v′xV / c 2 0.5c + 0.8c = 0.93c. If the particle is ejected in the backward velocity in the lab frame is vx = 1 + (0.5)(0.8) −0.5c + 0.8c direction, its velocity in the lab frame is vx = = 0.50c. (Its speed in the lab frame is 1 − (0.5)(0.8) the same as its speed in the atom’s frame, but the two velocities point in opposite directions.) dy ′ . v′y = dt ′ dy ′ = dy

dt ′ = v′y =

dt − Vdx / c 2 1 − (V / c) 2 v y 1 − (V / c) 2 (dy / dt ) 1 − (V / c) 2 dy = = dt − Vdx / c 2 1 − V (dx / dt ) / c 2 1 − vxV / c 2 1 − (V / c) 2

36-42.

The greatest possible combination of velocities occurs when the vectors point in opposite directions. For example, suppose V points right and vx points left. Then in the primed system, vx − V v′x = and the speeds will add in the numerator. The denominator will be greater than 1 − vxV / c 2 1 because the product term will be positive. Suppose the speeds are equal and differ from c by a small amount fraction δ. Then the relative speed will be c(1 − δ ) + c(1 − δ ) 2(1 − δ ) v′x = = c. This will always be less than c because the 2 1 + (1 − δ ) 2(1 − δ ) + δ 2 denominator is always larger than the numerator, if only by a tiny amount. If δ = 0, then the relative speed is c. But it can never be larger than c.


CHAPTER

†36-43.

36

The speed of light is measured relative to the water, so vx′ =

c . From Problem 36, n

c +V vx′ + V V ⎞ c V V2 ⎛c ⎞⎛ n vx = = ≈ ⎜ + V ⎟ ⎜1 − = − 2 +V − ⎟ 2 V 1 + vx′V / c nc ⎠ n n nc ⎝n ⎠⎝ 1+ nc V2 can be neglected because it is much smaller than all the other terms. The result is The term nc c 1 ⎞ ⎛ vx ≈ + V ⎜ 1 − 2 ⎟ n n ⎠ ⎝ 36-44.

Follow the method used to derive v′x: dv′ ax′ = x dt ′ dvx′ = d dt ′ =

2 (1 − vxV / c 2 )dvx − (vx − V )(Vdvx / c 2 ) dvx ⎡⎣1 − (V / c) ⎤⎦ vx − V = = 1 − vxV / c 2 (1 − vxV / c 2 ) 2 (1 − v′xV / c 2 ) 2

dt − Vdx / c 2 1 − (V / c) 2

2 2 (dvx / dt ) ⎡⎣1 − (V / c) 2 ⎤⎦ dvx′ dvx ⎡⎣1 − (V / c) ⎤⎦ 1 − (V / c) = = (1 − vxV / c 2 ) 2 (dt − Vdx / c 2 ) (1 − vxV / c 2 )3 dt ′

36-45.

36-46.

3/ 2

=

ax ⎡⎣1 − (V / c) 2 ⎤⎦

3/ 2

(1 − vxV / c 2 )3

⎡ ⎤ ⎡ ⎤ 1 1 K rel = mc 2 ⎢ − 1⎥ = mc 2 ⎢ − 1⎥ = 5.04 × 10−3 mc 2 ⎢⎣ 1 − (v / c) 2 ⎥⎦ ⎢⎣ 1 − (0.1) 2 ⎥⎦ 2 2 2 2 2 mv mc (v / c) mc (0.1) KN = = = = 5.00 × 10−3 mc 2 2 2 2 K rel 5.04 = = 1.008 KN 5.00 The deviation is about 0.80%. 2 2 ⎡ ⎤ mc 2 ⎛ v ⎞ 1 1 1⎛v⎞ 2 − 1 ≈ ⎜ ⎟ and K ≈ K = mc ⎢ − 1⎥ . For v/c << 1, ⎜ ⎟ . (Note 2⎝c⎠ 2 ⎝c⎠ ⎢⎣ 1 − (v / c) 2 ⎥⎦ 1 − (v / c ) 2 mv 2 that this is just , but here it is easier to work with the ratio v/c.) For v/c = 0.01, 2 K = 5 × 10–5mc2. Then K = (5 × 10–5)(0.01 kg)(3 × 108 m/s)2 = 4.5 × 1010 J.


CHAPTER

36-47.

36

⎛ ⎞ ⎜ ⎟ 2 mc ⎟ ⎜ K= – mc2 = 8 × 1019 J 2 ⎟ ⎜ v ⎜⎜ 1 − 2 ⎟⎟ c ⎠ ⎝ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎟ 2 ⎜ 1 m 1 − 1 ⎟ = 1000 kg (3 × 108 ) 2 2 ⎜ − 1 ⎟ = 8 × 1019 J mc2 ⎜ ⎜ ⎟ ⎟ s ⎜ v2 v2 ⎜⎜ 1 − 2 ⎟⎟ ⎜⎜ 1 − 2 ⎟⎟ c c ⎝ ⎠ ⎝ ⎠ 19 1 8 × 10 −1 = = 0.89, so that Then 2 1000 (3 × 108 ) 2 ⎛v⎞ 1− ⎜ ⎟ ⎝c⎠ 2

2

2

1 ⎛v⎞ ⎛v⎞ ⎛ 1 ⎞ ⇒1– ⎜ ⎟ = ⎜ 1− ⎜ ⎟ = ⎟ = 0.280 1.89 ⎝c⎠ ⎝ 1.89 ⎠ ⎝c⎠ v = 1 − 0.280 c = 0.85c 36-48.

†36-49.

2.6 × 106 m/s = 8.6667 × 10–3c. mv 2 (9.11 × 10−31 kg)(2.6 × 106 m/s) 2 KN = = = 3.07918 × 10−18 J 2 2 ⎡ ⎤ ⎡ ⎤ 1 1 K = mc 2 ⎢ − 1⎥ = (9.11 × 10−31 )(3.00 × 108 m/s) 2 ⎢ − 1⎥ 2 ⎢⎣ 1 − (8.6667 × 10−3 ) 2 ⎥⎦ ⎣⎢ 1 − (v / c) ⎦⎥ = 3.07938 × 10−18 J The calculator shows a difference if energy is computed to at least five significant figures. Rearrange Equation 36.44: −2

K K K ⎞ 1 1 ⎛ = −1 ⇒ =1+ ⇒ 1 − (V / c) 2 = ⎜ 1 + ⎟ . Solve this for 2 2 2 2 mc mc mc 2 ⎠ ⎝ 1 − (v / c ) 1 − (v / c ) ⎡ ⎛ v: v = ⎢ 1 − ⎜ 1 + ⎢ ⎝ ⎣ K 14 J. Then = mc 2 36-50.

−2 K ⎞ ⎤ 2 –31 16 2 2 – ⎟ ⎥ c. For an electron, mc = (9.11 × 10 kg)(9 × 10 m /s ) = 8.20 × 10 mc 2 ⎠ ⎥ ⎦ 1.6 × 10−13 J −2 = 1.95 and v = ⎡ 1 − ( 2.95 ) ⎤ c = 0.94c −14 ⎢ ⎥⎦ ⎣ 8.20 × 10 J

For an electron moving at v = 0.5c, or 1.5 × 108 m/s, the magnitude of the momentum is mv (9.11 × 10−31 kg)(1.5 × 108 m/s) p= = = 1.58 × 10−22 kg i m/s 2 2 1 − (v / c ) 1 − (0.5) ⎡ ⎤ ⎛ ⎞ 1 1 K = mc 2 ⎢ − 1⎥ = (9.11 × 10−31 kg)(3.00 × 108 m/s) 2 ⎜ − 1 ⎟ = 1.27 × 10−14 J 2 2 ⎝ 1 − 0.5 ⎠ ⎣⎢ 1 − (v / c) ⎦⎥


CHAPTER

36-51.

36 mv

p=

1 − (v / c ) 2 mc 2

E=

⇒ 1 − (v / c ) 2 =

1 − (v / c ) 2

mc 2 E

mvE vE = 2 mc 2 c Newtonian: pN = mv

⇒p=

36-52.

Relativistic: pR =

mv

1 − (v / c ) 2 2

†36-53.

⎛ 72 × 103 m/s ⎞ p Therefore, N = 1 − (v / c) 2 = 1 − ⎜ ⎟ = 0.99999997 8 pR ⎝ 3.00 × 10 m/s ⎠ The percentage difference is about 0.000003%. 2 2 ⎡ ⎤ ⎛ mc 2 ⎞ 1 ⎛v⎞ 2 2 K = mc ⎢ − 1⎥ ⇒ 1 − ⎜ ⎟ = ⎜ ⎟ . Multiply through by c : K ⎝c⎠ ⎢⎣ 1 − (v / c) 2 ⎥⎦ ⎝ ⎠ 2

2

⎛ mc 2 ⎞ 2 ⎛ mc 2 ⎞ 2 2 2 ⇒c −v =⎜ ⎟ c . Factor c − v : ⇒ (c − v)(c + v) = ⎜ ⎟ c . The speeds will all be ⎝ K ⎠ ⎝ K ⎠ very close to c, so c + v ≈ 2c. Substitute this and divide through by 2c to get an equation for c – v: 2

2

2

c ⎛ mc 2 ⎞ 2 c−v = ⎜ ⎟ . This can now be used to find c – v. We’ll need mc for protons, electrons, and 2⎝ K ⎠ gold. Proton: mc2 = 9.38 × 108 eV Electron: mc2 = 5.11 × 105 eV Gold nucleus: mc2 = 197 × 9.38 × 108 eV = 1.84 × 1011 eV 2

7 TeV protons: c − v =

c ⎛ 9.38 × 108 eV ⎞ −9 ⎜ ⎟ = 8.98 × 10 c = 2.7 m/s 2 ⎝ 7 × 1012 eV ⎠ 2

50 GeV electrons: c − v =

c ⎛ 5.11 × 105 eV ⎞ −11 ⎜ ⎟ = 8.98 × 10 c = 0.016 m/s 9 2 ⎝ 50 × 10 eV ⎠ 2

c ⎛ 1.84 × 1011 eV ⎞ −5 4 20 TeV gold nuclei: c − v = ⎜ ⎟ = 4.23 × 10 c = 1.3 × 10 m/s 2 ⎝ 20 × 1012 eV ⎠ 36-54.

mc2 for a proton is 1.50 × 10–10 J. Using the result from Problem 53, 2

c ⎛ 1.50 × 10−10 J ⎞ −24 −15 c−v = ⎜ ⎟ = 4.5 × 10 c = 1.4 × 10 m/s. 2⎝ 50 J ⎠ †36-55.

p=

mv 1 − (v / c ) 2

. For a speed as close to c as the one in this problem, many calculators will not

be able to evaluate 1 − (v / c) 2 –a value of zero will be returned. To get a value, follow the method described in Problem 53: Factor the expression in the square root and use c/v ≈ 1 in the


CHAPTER sum factor: 1 − (v / c) 2 = [1 − (v / c)][1 + (v / c)] ≈ calculator: p=

36

2[1 − (v / c)]. This can be evaluated with a

2(1 − 0.99999999967) = 2.57 × 10−5. To find p, use v ≈ c in the equation:

mv 1 − (v / c ) 2

=

(9.11 × 10−31 kg)(3.00 × 108 m/s) = 1.06 × 10−17 kg i m/s 2.57 × 10−5 2

36-56.

c ⎛ 1.50 × 10−10 J ⎞ −7 Follow the method in Problem 53. c − v = ⎜ ⎟ = 4.39 × 10 c = 132 m/s 2 ⎝ 1.6 × 10−7 J ⎠ For an ultrarelativistic particle, p =

36-57.

E 1.6 × 10−7 J = = 5.33 × 10−16 kg i m/s 8 c 3.00 × 10 m/s

Assume the particle is initially at rest. Then the total momentum after the disintegration must be zero with each piece carrying momentum of magnitude p in opposite directions. The total energy of each piece is given by E12 = ( pc) 2 + (m1c 2 ) 2 E22 = ( pc) 2 + (m2 c 2 ) 2 The initial energy was Mc2, so energy conservation gives E1 + E2 = Mc 2 ⇒ E1 = Mc 2 − E2 Square the expression for E1 and substitute the expression for E2: ( pc) 2 + (m1c 2 ) 2 = ( Mc 2 ) 2 − 2 E2 Mc 2 + E22 = ( Mc 2 ) 2 − 2 Mc 2 ( pc) 2 + (m2 c 2 ) 2 + ( pc) 2 + (m2 c 2 ) 2 Rearrange into a form that can be squared: 2Mc3 p 2 + (m2 c) 2 = M 2 c 4 + m22 c 4 − m12 c 4 2M p 2 + (m2 c) 2 = [ M 2 + (m2 − m1 )(m1 + m2 )]c Square both sides: 4M 2 [ p 2 + (m2 c) 2 ] = [ M 4 + 2M 2 (m2 − m1 )(m1 + m2 ) + (m2 − m1 ) 2 (m1 + m2 ) 2 ]c 2 The 4M 2 m22 c 2 term on the left can be subtracted from the 2M 2 m22 c 2 term on the left to give 4M 2 p 2 = [ M 4 − 2M 2 (m12 + m22 ) + (m2 − m1 ) 2 (m1 + m2 ) 2 ]c 2 It turns out that (m12 + m22 ) = ( m2 − m1 ) 2 + (m1 + m2 ) 2 , so 4M 2 p 2 = {M 4 − 2 M 2 [(m1 + m2 ) 2 + (m2 − m1 ) 2 ] + ( m2 − m1 ) 2 (m1 + m2 ) 2 }c 2 The expression in the curly brackets can be factored, and we’ll reverse the order of the terms in the mass difference factor to make it look like the desired result: 4M 2 p 2 = [ M 2 − (m1 + m2 ) 2 ][ M 2 + (m2 − m1 ) 2 ]c 2 Divide through by 4M 2 and take the square root: p=

36-58.

c M 2 − ( m1 + m2 ) 2 M 2 + (m1 − m2 ) 2

2M The total initial energy of m1 is E1 = m1c 2 , and the total initial energy of m2 is E2 = m2 c 2 + K = ( pc) 2 + (m2 c 2 ) 2 . The momentum of the final composite particle must also have magnitude p and the total energy of M must be equal to the total initial energy. Thus m1c 2 + m2 c 2 + K = ( pc) 2 + ( Mc) 2 . Square both sides and square the initial energy for m2:


CHAPTER

36

( pc) 2 + ( Mc 2 ) 2 = [(m1 + m2 )c 2 ]2 + 2(m1 + m2 )c 2 K + K 2 ( pc) 2 + (m2 c 2 ) 2 = (m2 c 2 ) 2 + 2m2 c 2 K + K 2 ⇒ ( pc) 2 = 2m2 c 2 K + K 2 Substitute the expression for (pc)2: 2m2 c 2 K + K 2 + ( Mc 2 ) 2 = [(m1 + m2 )c 2 ]2 + 2(m1 + m2 )c 2 K + K 2 Make cancellations: ( Mc 2 ) 2 = [(m1 + m2 )c 2 ]2 + 2m1c 2 K Finally, solve for M: M = (m1 + m2 ) 2 + 36-59.

2m1 K c2

From Problem 51, p =

vE . From Equation 36.51, E = ( pc) 2 + (mc 2 ) 2 . Substitute into p: 2 c

v ( pc) 2 + (mc 2 ) 2 c2 c 2p ⇒v= = ( pc) 2 + ( mc 2 ) 2 p=

36-60.

cp 2

p + (mc) 2

−2 ⎡ K ⎞ ⎤ ⎛ ⎢ ⎥ c. K = 5.3 × 10–9 J; mc2 = 1.50 × 10–10 J (see (a) From problem 49, v = 1 − ⎜ 1 + 2 ⎟ mc ⎠ ⎥ ⎢ ⎝ ⎣ ⎦ −2 ⎤ ⎡ ⎛ 5.3 × 10−9 J ⎞ ⎥ ⎢ Problem 54), so the proton’s speed in the lab is v = 1 − ⎜ 1 + ⎟ c = 0.99962c ⎢ 1.50 × 10−10 J ⎠ ⎥ ⎝ ⎣ ⎦ (b) Let this frame move at V relative to the lab. Let the moving proton have velocities vx1 = v (from (a)) relative to the lab and vx′1 relative to the moving frame, and let vx′ 2 be the velocity of the other particle in the moving frame. Since the target proton is at rest in the lab, its velocity in the moving frame must be vx′ 2 = −V . The velocity of the moving proton in the moving frame is v −V vx′1 = V . Thus = V ⇒ v − V = V − vV 2 / c 2 ⇒ vV 2 / c 2 − 2V + v = 0. Solving for V 2 1 − vV / c

gives V =

2 ± 4 − 4(v / c) 2 1 ± 1 − (0.99962) 2 c= c = 0.973c (choosing the root < 1). 2(v / c) 0.99962

(c) Each proton is traveling at 0.973 c in the moving frame, so the total energy of each is 1.50 × 10−10 J mc 2 = = 6.5 × 10−10 J E= 2 2 1 − (V / c) 1 − (0.973) †36-61. 36-62.

E = 2me c 2 =

2(9.11 × 10−31 kg)(9.00 × 1016 m 2 /s 2 ) = 1.02 × 106 eV (1.02 MeV) −19 1.6 × 10 J/eV

E = (∆m)c 2 ⇒ ∆m =

8.4 × 1013 J E = = 0.00093 kg (0.93 g) c 2 9.00 × 1016 m 2 /s 2


CHAPTER

36-63.

mthermal E 2.0 × 1041 J = thermal2 = × 100% = 1.1 × 10−4 % msun msun c (2.0 × 1030 kg)(9.00 × 1016 m 2 /s 2 )

36-64.

E = (∆m)c 2 ⇒ (∆m) =

†36-65.

36

1.3 × 108 J E = = 1.4 × 10−9 kg. This is 5 × 10–10 of the mass of 2 16 2 2 9.00 × 10 m /s c

one gallon of gasoline. The mass lost is mn − (m p + me ) = 1.674929 × 10−27 kg − (1.672623 × 10−27 kg + 9.11 × 10−31 kg) = 1.39 × 10−30 kg. E = (∆m)c 2 = 1.26 × 10−13 J. The rest mass energy of the electron is 8.20 × 10–14 J, so this is

36-66.

about 1.5 × the rest mass energy of the electron. mv mc(v / c) mc 2 = . Rearrange and square both: E= ; p= 2 2 1 − (v / c ) 1 − (v / c ) 1 − (v / c ) 2 2

1 ⎛ E ⎞ ⎜ 2⎟ = 1 − (v / c ) 2 ⎝ mc ⎠ 2

(v / c ) 2 ⎛ p ⎞ = ⎜ ⎟ 1 − (v / c ) 2 ⎝ mc ⎠ Subtract the bottom from the top 2

2

1 (v / c ) 2 ⎛ E ⎞ ⎛ p ⎞ − = − =1 ⎜ 2⎟ ⎜ ⎟ 1 − (v / c ) 2 1 − (v / c ) 2 ⎝ mc ⎠ ⎝ mc ⎠ ⇒ E 2 − ( pc) 2 = (mc 2 ) 2 This gives the final result: E 2 = ( pc) 2 + (mc 2 ) 2 †36-67.

The mass lost is ∆m = mK − 2mπ = [8.87 − 2(2.49)] × 10−28 kg = 3.89 × 10−28 kg. This corresponds to E = (∆m)c 2 = 3.50 × 10−11 J. This becomes kinetic energy of the pions. Since the

36-68.

pions have equal masses and their total momentum must be zero, they will move off in opposite directions with equal speeds. The energy released in the decay is equally divided between the −2 ⎡ K ⎞ ⎤ ⎛ pions, so each has kinetic energy K = E/2. From Problem 49, v = ⎢ 1 − ⎜ 1 + ⎟ ⎥ c. The rest mc 2 ⎠ ⎥ ⎢ ⎝ ⎣ ⎦ − 2 ⎡ 1.75 ⎞ ⎤ ⎛ mass energy is easily calculated to be 2.24 × 10–11 J, so v = ⎢ 1 − ⎜ 1 + ⎟ ⎥ c = 0.828c 2.24 ⎠ ⎥ ⎢ ⎝ ⎣ ⎦ 2.2 × 10−6 s ∆t ′ (a) The lifetime of the moving muon in the earth frame is ∆t = = 1 − (V / c) 2 1 − (0.99) 2 =

2.2 × 10−6 s 1 − (0.99)

2

= 1.56 × 10−5 s. Distance traveled = v∆t = 2.97 × 108 m/s × 1.56 × 10–5 s = 4.63

km. The muon decays at (20 – 4.63) km = 15.4 km. (b) Without time dilation, ∆d = v∆t = 2.97 × 108 m/s × 2.2 × 10–6 s = 653 m, so the muon decays at an altitude of (20 – 0.653) km= 19.3 km.


CHAPTER †36-69.

36-70.

36

Use c = 1 light-year/year for this problem. The lifetime of the cat in its own rest frame is 7 years ∆t ′ ∆t ′ = 7years. As measured on Earth, this time is ∆t = = = 11.7 years. 2 1 − (V / c) 1 − (0.8) 2 The ship is at a distance V∆t = (0.8 c)(11.7 years) = 9.33 light years from Earth. The radio signal travels at the speed of light (1 light-year/year), so the travel time for the signal to reach us is ∆T = 9.33 years. Thus, the message reaches us after ∆t + ∆T = (11.7 + 9.33) yr = 21 years (a) In the reference frame of proton, the earth speeds by at 0.8 c. The earth appears contracted along its polar diameter. Therefore, D = D′ 1 − (V / c) 2 = (12760 km) 1 − (.8) 2 = 7656 km. The equatorial diameter is the same as in the Earth frame.

(b) According to the proton, the time taken = it takes †36-71.

D 7.656 × 106 m = = 0.0319 s. According to Earth V 2.4 × 108 m/s

D′ 1.276 × 104 m = = 0.0532 s V 2.4 × 108 m/s

(a) Let t1 = 0, t2 = 1.00 µs, x1 = 0, x2 = 960 m. From Equation 36.28, t − xV / c 2 t′ = 1 − (V / c) 2 t1′ = 0 t2′ =

1.00 × 10−6 s − (960 m)(0.8) /(3.00 × 108 m/s) 1 − (0.8) 2

= −2.60 µ s

To the ship observer, the 600 nm pulse is emitted before the 400 nm pulse (b) Equations 36.26 and 36.28 were derived assuming that the origins of the coordinate systems coincided at t = t′ = 0. That means the 400-nm pulse is moving away from the earth and the 600 nm pulse is moving toward the Earth. The 400-nm light is red shifted to a longer wavelength, which means λearth > λship, or λship < 400 nm. The 600 nm light is blue shifted to a shorter wavelength, so λearth < λship, or λship > 600 nm. Great care must be used in applying the Dopplershift Equations 36.12, because the meanings of the primed and unprimed frequencies are opposite from the meaning of primed and unprimed variables in the Lorentz transformation equations. To clarify the situation, let’s give the equations in this form: 1−V /c Receding emitter: f received = f emitted 1+V /c 1+V /c f emitted 1−V /c where we understand that the Earth is the receiver and the ship is the emitter. Since this problem gives wavelengths and λ = c/f, rewrite the equations in terms of wavelength: Approaching emitter: f received =


CHAPTER

Receding emitter: λreceived =

36

1+V /c λemitted 1−V /c

1−V /c λemitted 1+V /c Use the equation for the receding emitter to find the emitted wavelength of the 400-nm light: 1+V /c 1−V /c 1 − 0.8 (400 nm) = 133 nm λreceived = λemitted ⇒ λemitted = λreceived = 1−V /c 1+V /c 1 + 0.8 Use the equation for the approaching emitter to find the emitted wavelength of the 600-nm light: 1−V /c 1+V /c 1 + 0.8 (600 nm) = 1800 nm λreceived = λemitted ⇒ λemitted = λreceived = 1+V /c 1−V /c 1 − 0.8 (c) The length of the ship corresponds to the position x′2 in the ship’s frame where the second pulse is released. Using Equation 36.26, x2 − Vt 960 − (2.4 × 108 m/s)(1 × 10−6 s) x2′ = = = 1.20 × 103 m. 2 2 1 − (V / c) 1 − (0.8) Approaching emitter: λreceived =

36-72.

The ship is 1.2 km long. Let the primed system represent the ship going east at V = 0.7c. The velocity of the westbound ship as measured on Earth is vx = –0.4c. Both emitters are receding, so use the equation for a receding emitter from Problem 71.

y

y′

vx = –0.4c

Eastbound: f received = Westbound: f received =

x

V = 0.7c

x′

1−V /c 1 − 0.7 f emitted = (2.00 GHz) = 0.840 GHz 1+V /c 1 + 0.7 1 − vx / c 1 + vx / c

1 − 0.4 (2.00 GHz) = 1.31 GHz 1 + 0.4

f emitted =

The velocity of the westbound ship relative to the eastbound ship is vx − V −0.4c − 0.7c = − 0.859c v′x = 2 1 − vxV / c 1 − (−0.4)(0.7) The frequency received by either ship will be 1 − vx / c 1 − 0.859 f received = f emitted = (2.00 GHz) = 0.551 GHz 1 + vx / c 1 + 0.859 *36-73.

The cube appears contracted along its direction of motion, so one side will be shortened. The other two sides will remain unchanged. The shortened side will have a length

V

L = L′ 1 − (V / c) 2 = (1.0 m) 1 − (0.6) 2 =

At rest


CHAPTER

36

0.80 m. The apparent dimensions will be 1.0 m × 1.0 m × 1.0 m. The area of the two faces perpendicular to the direction of motion will be 1.0 m2. The four faces whose planes are parallel to the direction of motion will have area = 0.80 m2. The volume of the cube as seen from Earth will be 0.80 m3. 36-74.

(a)

∆t ′ =

L′ 200 m = = 6.67 × 10−7 s. Here the 8 c 3.00 × 10 m/s

prime is used in the context of the Lorentz transformation: The ship is the primed coordinate system and the Earth is the unprimed system, so L′ is the length of the ship in its own rest frame. (b) To calculate the interval as measured in the Earth frame by using the motions of the light and the ship, look at the diagram. The ship appears contracted to a length L = L′ 1 − (V / c) 2 in the Earth frame. The light must travel that distance plus the distance V∆t that the nose travels while the light is reaching the nose. The total distance traveled by the light is c∆t. This must be equal to L+ V∆t, which gives an equation for ∆t: L′ 1 − (V / c) 2 + V ∆t = c∆t

L′ 1 − (V / c) 2 ⎛ L′ ⎞ 1 − (V / c) 2 ∆t = =⎜ ⎟ c −V ⎝ c ⎠ (1 − V / c)

L′ Ship frame

L

V∆t V

c∆t Earth frame

1+V /c ⎛ L′ ⎞ 1 + V / c =⎜ ⎟ = ∆t ′ 1−V /c ⎝ c ⎠ 1−V /c To calculate this using the Lorentz transformation, use the inverse time transform from Problem t ′ + Vx′ / c 2 ∆t ′ + V ∆x′ / c 2 . Taking intervals, ∆t = , where ∆t′ is the interval from (a) 33: t = 1 − (V / c) 2 1 − (V / c) 2 and ∆x′ = L′. Substituting gives L′ + V ( L′ / c ) / c 1+V /c ⎛ L′ ⎞ 1 + V / c ∆t = c =⎜ ⎟ = ( ∆t ′) , 2 2 1−V /c ⎝ c ⎠ 1 − (V / c) 1 − (V / c) which is the same result obtained by direct analysis of the motions. The numerical value is 1+V /c 1 + 0.95 ∆t = ( ∆t ′ ) = (6.67 × 10−7 s) = 4.16 × 10−6 s. 1−V /c 1 − 0.95 Note that time dilation cannot be used here because the two events (emission of the light from the tail and its arrival at the nose) happen at different locations in the rest frame of the ship.


CHAPTER 36-75.

36

Both the ship and the meteoroid are moving in the same direction, so the velocity of the vx − V 0.10c − 0.20c = = −0.102c. The speed is the meteoroid relative to the ship is vx′ = 2 1 − vxV / c 1 − (0.10)(0.2) magnitude of the velocity, so vx′ = 0.102c. The Galilean speed would be vx′ = vx − V = 0.1c − 0.2c = 0.10c. The deviation between the two values is 2%.

36-76.

Velocity of electron = V = – 0.95 c. Velocity of antielectron = Vx = 0.95 c Therefore, 0.95 − (−0.95) Vx′ = = 0.999 c 1 − 0.95(−0.95)

36-77.

36-78.

The speed of electron in the antielectron’s frame, and vice versa, is 0.999 c. v′ + Vx Use the inverse transform from Problem 36: vx = 1 + v′xV / c 2 v′ + Vx 0.9c + 0.7c Forward direction v′x = +0.90c vx = = = 0.98c 2 1 + v′xV / c 1 + (0.9)(0.7) v′ + Vx −0.9c + 0.7c Backward direction v′x = –0.90c vx = = = −0.54c 2 1 + v′xV / c 1 − (0.9)(0.7)

y1

y

x

v′12 = 0.6c

y2

v′23 = 0.8c

x1

x2

y3

V = 0.5c

x3

Only speeds are given, so let’s assume all the particles are moving in the same direction. Let V be the velocity of particle 3 in the lab frame (x, y). Let v′23 be the velocity of particle 2 relative to particle 3. Then from Problem 36 the velocity of particle 2 in the lab frame is ′ +V v23 0.8c + 0.5c v2 = = = 0.929c 2 ′ 1 + v23V / c 1 + (0.8)(0.5) Let v′12 be the velocity of particle 1 relative to particle 2. Since the velocity of particle 2 in the lab frame is 0.929c, the velocity of particle 1 in the lab frame is v12′ + v2 0.6c + 0.929c v1 = = = 0.982c 2 ′ 1 + v23v2 / c 1 + (0.6)(0.929) †36-79.

⎡ ⎤ ⎡ ⎤ 1 1 K = mc 2 ⎢ − 1⎥ = (50 × 103 kg)(9.00 × 1016 m 2 /s 2 ) ⎢ − 1⎥ = 6.96 × 1020 J ⎢⎣ 1 − (v / c) 2 ⎥⎦ ⎢⎣ 1 − (0.5) 2 ⎥⎦ The mass that would have to be converted to produce this much energy is 6.96 × 1020 J K ∆m = 2 = = 7.74 × 103 kg = 7.74 metric tons. 9.00 × 1016 m 2 /s 2 c


CHAPTER

36-80.

36

⎡ ⎤ 1 K = mc 2 ⎢ − 1⎥ = mc 2 . Therefore, ⎢⎣ 1 − (v / c) 2 ⎥⎦

1 1 − (v / c )

2

−1 = 1 ⇒

1 1 − (v / c ) 2

= 2. Solving

2

for v gives 36-81.

36-82.

1 3 3 v ⎛v⎞ =4⇒⎜ ⎟ = ⇒ = . Thus v = 0.866c. 2 1 − (v / c ) 4 c 2 ⎝c⎠

No speed is given for the asteroid, so let’s assume it’s at rest in the Solar System frame. Then its speed relative to the spaceship is 0.8c. ⎡ ⎤ ⎡ ⎤ 1 1 (a) K = mc 2 ⎢ − 1⎥ = (2.0)(9.00 × 1016 m 2 /s 2 ) ⎢ − 1⎥ = 1.2 × 1017 J ⎢⎣ 1 − (v / c) 2 ⎥⎦ ⎢⎣ 1 − (0.8) 2 ⎥⎦ 1.2 × 1017 J = 2.9 × 107 tons (29 MT) (b) E (tons of TNT) = 9 4.2 × 10 J/ton −2 ⎤ ⎡ ⎛ 5.3 × 10−9 ⎞ ⎥ ⎢ = − + v 1 1 (a) ⎜ ⎟ c = 0.99962c ⎢ 1.50 × 10−10 ⎠ ⎥ ⎝ ⎣ ⎦

(b) For an ultrarelativistic particle, E = K and p = 2

5.3 × 10−9 J E = = 1.8 × 10−17 kg i m/s. c 3.00 × 108 m/s 2

c ⎛ mc 2 ⎞ c ⎛ 1.50 × 10−10 J ⎞ −9 = ⎜ ⎟ ⎜ ⎟ = 1.1 × 10 c = 0.33 m/s. 2⎝ K ⎠ 2 ⎝ 3.2 × 10−6 J ⎠

36-83.

From Problem 53, c − v =

36-84.

(a) ∆E = (∆m)c2 = ⎡⎣1.6749 × 10−27 kg − (1.6726 × 10−27 kg + 9.11 × 10−31 kg ) ⎤⎦ c2 = 1.389 × 10–30 kg × 9 × 1016 m2/s2 = 1.25 × 10–13 J (a) In the rest frame of the K0, its energy is mK c 2 . The excess energy after the decay is

†36-85.

∆E = (mK − 2mµ )c 2 = [8.87 − 2(1.88)] × 10−28 kg × 9.00 × 1016 m 2 /s 2 = 4.60 × 10−11 J. Since the muons have equal masses they share this energy equally, with each having K = 2.30 × 10–11 J. The rest mass energy of each muon is 1.69 × 10–11 J. The speed of each in the K0 frame then is −2 ⎤ −2 ⎡ −11 ⎡ ⎛ ⎞ K ⎞ ⎤ 2.3 × 10 ⎛ v = ⎢ 1 − ⎜1 + ⎟ ⎥ c = 0.906c ⎟ ⎥ c = ⎢⎢ 1 − ⎜ 1 + 1.69 × 10−11 ⎠ ⎥ mc 2 ⎠ ⎥ ⎢ ⎝ ⎝ ⎣ ⎦ ⎣ ⎦ 0 (b) Let the K be moving to the right at 0.6c. Suppose the muon is also moving to the right in the same direction as the K0 and the antimuon is moving to the left. Then V = 0.60c and v′x = 0.906c. v′x + V 0.906c + 0.6c The velocity of the muon in the lab frame is vmuon = = = 0.98c. The 2 1 + vx′V / c 1 + (0.906)(0.6) velocity of the antimuon in the K0 frame is v′x = –0.906c, so the velocity of the antimuon in the v′x + V −0.906c + 0.6c lab frame is vantimuon = = = −0.67c. 2 1 + v′xV / c 1 − (0.906)(0.6)


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