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Solucionario Capitulo 14 - Paul

anschp15

propiedades termicas.doc

Physics, 6th Edition

Chapter 10. Uniform Circular Motion

Chapter 10. Uniform Circular Motion Centripetal Acceleration 10-1. A ball is attached to the end of a 1.5 m string and it swings in a circle with a constant speed of 8 m/s. What is the centripetal acceleration? ac

=

v2 R

=

(8 m/s) 2

ac = 42.7 m/s2

1.5 m

10-2. What are the period and frequency of rotation for the the ball in Problem 10-1? v = 2π fR;

v=

2π R

;

T=

=

1

T

f =

1 T

1.18 s

2π R v ;

=

2π (1.5 m) 8 m/s

;

T = 1.18 s

f = 0.849 rev/s

10-3. A drive pulley 6-cm 6-cm in diameter is set to rotate at 9 rev/s. What is the the centripetal acceleration of a point on the edge of the pulley? What would be the linear speed of a belt around the pulley? [ R = (0.06 m/2) = 0.03 m ] ac

= 4π 2 f 2 R = 4π 2 (9 rev/s)2 (0.03 m) ;

v = 2π fR = 2π (9 rev/s)(0.03 m) ;

ac = 95.9 m/s2 v = 1.70 m/s

10-4. An object revolves in a circle of diameter 3 m at a frequency of 6 rev/s. rev/s. What is the the period of revolution, the linear speed, and the centripetal acceleration? [ R = (3 m/2) = 1.5 m ] T =

1

=

1

;

T = 0.167 s ;


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10-5. A car moves around a curve 50 m in radius and undergoes a centripetal acceleration of 2 m/s2. What is its constant speed? a

v2 =

R

;

v

=

aR

(2 m/s) /s) 2 (50 (50 m) ;

=

v = 10.0 m/s

10-6. A 1500-kg 1500-kg car moves at a constant speed of 22 m/s along a circular track. The centripetal acceleration is 6 m/s2. What is the radius of the track and the centripetal force on the car? a=

F =

v2 R

mv 2 R

; R=

=

v2 a

=

(22 m/s) 2 6 m/s 2

(1500 kg)(22 m/s) 2 (80.7 m)

;

R = 80.7 m

;

Fc = 9000 N

10-7. An airplane dives dives along a curved path of radius R and velocity v. The centripetal acceleration is 20 m/s2. If both the velocity and the radius are doubled, what will will be the new acceleration? a1 =

v2 R

;

a2

=

(2 v ) 2 2R

=

4v2 2R

a2 = 2a1 = 2(20 m/s2;

;

a2

=

2v 2 R

;

a = 40 m/s2

Centripetal Force 10-8. A 20-kg child riding a loop-the-loop at the Fair moves at 16 m/s through a track of radius 16 m. What is the resultant force on the child? 2

2


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10-9. A 3-kg rock, attached to a 2-m cord, swings in a horizontal circle so that it makes one revolution in 0.3 s. What is the centripetal force on the rock? Is there an outward force on the rock? 2

Fc

= 4π

2

f mR = 4π 2

2

 1  (3 kg)(2 m) ;  0.3 s   

F c = 2630 N, No

10-10. An 8-lb object swings in horizontal circle with a speed of 95 ft/s. What is the radius of the path, if the centripetal force is 2000 lb? m=

R =

mv 2 F c

8 lb 32 ft/s

=

2

= 0.25 slug;

(0.25 slug)(95 ft/s) 2 2000 lb

mv 2

F c =

;

R

R = 1.13 ft

*10-11. Two 8-kg masses are attached to the end of a tin rod 400 mm long. The rod is supported in the middle and whirled in a circle. The rod can support a maximum tension of only 800 N. What is the maximum frequency of revolution? [ R = (400 mm/2) = 200 mm ] Fc

=

mv 2 R

; v

=

v = 2π fR; f

Fc R m

=

(800 N)(0.20 m)

=

v

2π R

8 kg

=

4.47 m/s 2π (0.20 m)

;

;

v = 4.47 m/s

f = 3.56 rev/s

*10-12. A 500-g damp shirt rotates against the wall of a washer at 300 rpm. The diameter of the

rotating drum is 70 cm. What is the magnitude and direction of the resultant force on the


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*10-13. A 70-kg runner rounds a track of radius 25 m at a speed of 8.8 m/s. What is the central force causing the runner to turn and what exerts the force? F c =

mv 2 R

=

(70 kg)(8.8 m/s) 2

;

25 m

F c = 217 N, friction

*10-14. In Olympic bobsled competition, a team takes a turn of radius 24 ft at a speed of 60 mi/h.

What is the acceleration? How many g’s do passengers experience? ac

=

v2 R

=

(88 ft/s) 2 24 ft

(60 mi/h = 88 ft/s)

ac = 323 ft/s2 or 10.1 g’s

;

Flat Curves and Banked Curves 10-15. On a rainy day the coefficient of static friction between tires and the roadway is only 0.4. What is the maximum speed at which a car can negotiate a turn of radius 80 m? mv 2 R

= µ s mg;

vc

=

µ s gR

=

vc = 17.7 m/s or

(0.4)(9.8 m/s 2 )(80 m) ; 63.8 km/h

10-16. A bus negotiates a turn of radius 400 ft while traveling at a speed of 60 mi/h. If slipping just begins at this speed, what is the coefficient of static friction between the tires and the road?

(60 mi/h = 88 ft/s) mv 2 R

= µ s mg ;

µ s

=

v2 gR

=

(88 ft/s) 2 2

(32 ft/s )(400 ft)

;

µs = 0.605


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*10-18. A 20-kg child sits 3 m from the center of a rotating platform. If µs = 0.4, what is the maximum number of revolutions per minute that can be achieved without slipping? (Slipping occurs when the centripetal force equals the maximum force of static friction.)

Fc

= 4π

2

f mR = µ s mg ; 2

µ s g

f =

=

4π 2 R

f = 0.182 rev/s (60 s/min);

(0.4)(9.8 m/s2 ) 4π 2 (3 m)

;

f = 10.9 rpm

*10-19. A platform rotates freely at 100 rpm. If the coefficient of static friction is 0.5, how far from the center of the platform can a bolt be placed to without slipping? f = 100 rev/min (1 min/60 s) = 1.67 rev/s;

Fc

= 4π

2

f mR = µ s mg ;

R=

2

µ s g

4π 2 f 2

=

µs = 0.5; R = ?

(0.5)(9.8 m/s 2 ) 4π 2 (1.67 rev/s)2

;

R = 21.1 cm

10-20. Find the required banking angle to negotiate the curve of Prob.10-15 without slipping. tan θ =

v2 gR

=

(17.7 m/s)2 2

(9.8 m/s )(80 m)

;

θ = 21.80

10-21. Find the required banking angle for Problem 10-16 to prevent slipping? tan θ =

v2 gR

=

(88 ft/s) 2 2

(32 ft/s )(400 ft)

;

θ = 31.20

10-22. The optimum banking angle for a curve of radius 20 m is found to be 280. For what speed was this angle designed?


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*10-23. A curve in a road 9 m wide has a radius of 96 m. How much higher than the inside edge should the outside edge be for an automobile to travel at the optimum speed of 40 km/h? v = 40 km/h = 11.1 m/s;

tan θ =

v2 gR

=

(11.1 m/s)2 2

(9.8 m/s )(96 m)

;

h = (9 m) sin θ ;

θ = 7.480;

tan θ =

v2 gR

h = (9 m) sin 7.480;

h = 1.17 m

The Conical Pendulum 10-24. A conical pendulum swings in a horizontal circle of radius 30 cm. What angle does the supporting cord make with the vertical when the liner speed of the mass is 12 m/s? tan θ =

v2 gR

=

(12 m/s)2 2

(9.8 m/s )(0.30 m)

;

θ = 88.80

10-25. What is the linear speed of the flyweights in Fig. 10-16 if L = 20 cm and θ = 600? What is L

the frequency of revolution? L = 20 cm = 0.20 m; R = L sin θ = (0.2 m) sin 600; R = 0.173 m tan θ

=

v2 gR

; v=

gR tan θ = (9.8 m/s2 )(0.173 m)( tan 600 ) ;

v = 2π fR; f

=

v

2π R

=

1.71 m/s 2π (0.173 m)

;

θ R

v = 1.71 m/s

f = 1.58 rev/s

10-26. If the length of the L in Fig. 10-16 is 60 cm. What velocity is required to cause the


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10-27. Each of the flyweights in Fig. 10-16 has a mass of 2 kg. The length L = 40 cm and the shaft rotates at 80 rpm. What is the tension in each arm? What is the angle θ ? What is the height h?

T cos

θ

80 rpm = 1.33 rev/s; T sinθ = F c = mv2 /R sin θ =

R L

=

R 0.4 m

;

cosθ =

mg T

=

θ

h

R

mg

T = 56.1 N

(2 kg)(9.8 m/s 2 ) 56.1 N

h = L cos θ = (0.4 m) cos 69.60 ;

L

T sin θ

 R  = 4π 2 f 2 mR T   0.4 m 

T = (0.4 m)(4 π2)(1.33 rev/s)2(2 kg); T cos θ = mg;

θ

;

h = 0.14 m

θ = 69.60

or

h = 14.0 cm

10-28. In Fig. 10-16, assume that L = 6 in., each flyweight is 1.5 lb, and the shaft is rotating at 100 rpm. What is the tension in each arm? What is the angle θ ? What is the distance h? 100 rpm = 1.67 rev/s; T sinθ = F c = mv2 /R m=

1.5 lb 32 ft/s

sin θ =

R L

=

T cos

θ

= 0.0469 slug ;

L = 6 in. = 0.50 ft

θ T sin θ

R 0.5 ft

;

 R  = 4π 2 f 2 mR T   0.5 ft 

T = (0.5 ft)(4 π2)(1.67 rev/s)2(0.0469 slug); T cos θ = mg;

cosθ =

mg

=

mg

T = 2.75 lb

(0.0469 slug)(32 ft/s2 )

;

θ = 54.30

L

θ R

h


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10-29. Consider the rotating swings in Fig. 10-17. The length L = 10 m and the distancea = 3 m. What must be the linear velocity of the seat if the rope is to make an angle of 300 with the vertical?

[ L = 10 m; a = 3 m,

θ = 300 ]

a

b = L sin θ = (10 m) sin 30 = 5 m; R = a + b = 7 m; 0

tan θ

v2 =

gR

;

v

=

θ

L b

gR tan θ

R = a + b

v = (9.8 m/s 2 )(7 m) tan 300

v = 6.73 m/s

10-30. What must be the frequency of revolution for the swing in Problem 10-17 if the angle θ is to be equal to 250? b = L sin 280 = (10 m) sin 280; b = 4.695 m; R = a + b = 7.695 m tan θ

=

v2 gR

;

v = 2π fR;

v=

f

gR tan θ ;

=

v

2π R

=

v = (9.8 m/s 2 )(7.695 m) tan 280 ; v = 6.33 m/s 6.33 m/s

2π (7.695 m)

;

f = 0.131 rev/s or 7.86 rpm

Motion in a Vertical Circle 10-31. A rock rests on the bottom of a bucket moving in a vertical circle of radius 70 cm. What is the least speed the bucket must have as it rounds the top of the circle if it is to remain in the bucket? [ Resultant force to center = mv2 /R ]

T


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10-32. A 1.2-kg rock is tied to the end of a 90-cm length of string. The rock is then whirled in a vertical circle at a constant speed. What is the critical velocity at the top of the path if the

v

string is not to become slack? 0

T

vc

+ mg = =

gR

mv 2

mg

; Critical speed vc is when T = 0

R

=

T

2

(9.8 m/s )(0.7 m) ;

R

vc = 2.97 m/s

*10-33. Assume that the rock of Problem 10-32 moves in a vertical circle at a constant speed of 8 m/s? What are the tensions in the rope at the top and bottom of the circle. At Top: T

T =

+ mg =

R

(1.2 kg)(8 m/s) 2 (0.9 m)

At Bottom: T

T =

mv 2

(0.9 m)

=

R

− mg

v T

− (1.2 kg)(9.8 m/s

2

− mg =

(1.2 kg)(8 m/s) 2

and T

mv 2

mv 2 R

and T

=

);

mv 2 R

T = 73.6 N

+ mg

mg

R

T

v mg

+ (1.2 kg)(9.8 m/s2 ) ;

T = 97.1 N

*10-34. A test pilot in Fig. 10-18 goes into a dive at 620 ft/s and pulls out in a curve of radius 2800 ft. If the pilot weighs 160 lb, what acceleration will be experienced at the lowest point? What is the force exerted by the seat on the pilot? W = 160 lb; m =160 lb/32 ft/s2 = 5 slugs; v = 620 ft/s;


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*10-34. (Cont.)

=

N


(5 slugs)(620 ft/s) 2 (2800 ft)

+ 160 lb ;

N

= 846 lb

*10-35. If it is desired that the pilot in Problem 10-34 not experience an acceleration greater than 7 times gravity (7g), what is the maximum velocity for pulling out of a dive of radius 1 km? a=

v2 R

= 7 g;

v=

7 gR

=

7(9.8 m/s2 )(1000 m) ;

v = 262 m/s or 943 km/h Note: The pilot actually “feels” a force that is eight times W: N–

mg = m(7g);

N

= 8mg

*10-36. A 3-kg ball swings in a vertical circle at the end of an 8-m cord. When it reaches the top of its path, its velocity is 16 m/s. What is the tension in the cord? What is the critical speed at the top? At Top: T

T =

[ R = 8 m; m = 3 kg; v = 16 m/s ]

+ mg =

(3 kg)(16 m/s) 2 (8 m)

When T = 0, vc

=

mv 2

and T

R

=

mv 2 R

− (3 kg)(9.8 m/s2 ) ; gR

=

v T

− mg

mg

R

T = 66.6 N

(9.8 m/s2 )(8 m) ;

vc = 8.85 m/s

*10-37. A 36-kg girl rides on the seat of a swing attached to two chains that are each 20 m long. If she is released from a position 8 m below the top of the swing, what force does the swing exert on the girl as she passes the lowest point? θ

8m


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*10-37. (Cont.)

At Bottom: T

T =

− mg =

(36 kg)(15.34 m/s) 2 (20 m)

mv2 R

and T

=

mv2 R

+ mg

+ (36 kg)(9.8 m/s2 ) ;

T = 776 N

Gravitation 10-38. How far apart should a 2-ton weight be from a 3-ton weight if their mutual force of attraction is equal to 0.0004 lb? m1

F

=

(2 ton)(2000 lb/ton)

=

Gm1m2

32 ft/s 2

R 2

R=

;

=

125 slugs; m2

Gm1m2 F

( G = 3.44 x 10-8 lb ft2/slug2 )

=

=

(3 ton)(2000 lb/ton) 32 ft/s 2

=

187.5 slugs

(3.44 x 10-8 lb ft2 /slug 2 )(125 slug)(187.5 slug) 0.0004 lb R = 1.42 ft

10-39. A 4-kg mass is separated from a 2 kg mass by a distance of 8 cm. Compute the gravitational force of attraction between the two masses. F =

Gm1m2 R

2

=

(6.67 x 10-11 N m2 /kg 2 )(4 kg)(2 kg) 0.08 m

; F = 8.34 x 10-8 N

*10-40. A 3-kg mass is located 10 cm away from a 6-kg mass. What is the resultant gravitational force on a 2-kg mass located at the midpoint of a line joining the first two masses? F 3 =

Gm3 m2 R 2

=

(6.67 x 10-11 N m2 /kg2 )(3 kg)(2 kg) (0.05 m) 2

3 kg 0.05 m 2 kg 0.05 m 6 kg


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*10-41. On a distant planet, the acceleration due to gravity is 5.00 m/s2. and the radius of the planet is roughly 4560 m. Use the law of gravitation to estimate the mass of this planet. mg

=

Gmm p R

2

;

=

m p

gR 2

=

G

(5.00 m/s 2 )(4560 m)2 -11

6.67 x 10

2

N m / kg

2

;

m p = 1.56 x 1024 kg

*10-42. The mass of the earth is about 81 times the mass of the moon. If the radius of the earth is 4 times that of the moon, what is the acceleration due to gravity on the moon? me = 81mm; Re = 4 Rm ; Consider test mass m on moon and then on earth: mg m

gm

=

ge

=

Gmmm

and g m =

2 m

R

mm Re2 2 m

me R

=

mm (4 Rm ) 2 2 m

81mm R

Gmm 2 m

R

; mg e

g m

;

9.8 m/s

2

=

Gmme

and ge =

2 e

R

= 0.1975 ;

Gme Re2

g m = 1.94 m/s2

*10-43. At 60-kg mass and a 20-kg mass are separated by 10 m. At what point on a line joining these charges will another mass experience zero resultant force? [ F2 = F6 ] Gm2 m ' (10 − x)

2

=

Gm6 m ' x

2

;

x 2 (10 − x)

2

=

m6

60 kg

m2

x F 6

x (10 − x)

=

m6 m2

=

60 kg 20 kg

= 1.732 ;

x = 1.732(10 – x);

x = 6.34 m from the 60-kg mass.

Kepler’s Laws and Satellites

m’

10 - x F 2

x = 17.32 – 1.732 x

20 kg


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10-44. (Cont.)

F c = F g ;

mv 2 R

=

Gmme R

2

;

Gme

v=

(6.67 x 10-11 N m2 / kg 2 )(5.98 x 1024 kg)

v=


7.18 x 106 m

R

;

;

v = 7450 m/s

10-45. The mass of the Jupiter is 1.90 x 1027 kg and its radius is 7.15 x 107 m. What speed must a spacecraft have to circle Jupiter at a height of 6.00 x 107 m above the surface of Jupiter? 7

7

8

R = R j + h = 7.15 x 10 m + 6 x 10 m; R = 1.315 x 10 m ;

v=

Gme R + h

=

(6.67 x 10-11 )(1.9 x 1027 kg)

mv 2 R

=

Gmme R2

;

v =31,400 m/s

1.315 x 108 m

This represents a speed of approximately 69,800 mi/h.

10-46. What is the orbital speed of a satellite that moves in an orbit 1200 km above the earth’s surface?

Note that : R = Re + h = (6.38 x 106 m) + (1.2 x 106 m) = 7.58 x 106 m F c = F g ;

v=

mv 2 R

=

Gmme R

2

;

(6.67 x 10-11 N m2 / kg2 )(5.98 x 1024 kg) 7.58 x 106 m

Gme

v=

R

;

;

v = 7254 m/s

10-47. The radius of the moon is 1.74 x 106 m and the acceleration due to its gravity is 1.63 m/s2. Apply the law of universal gravitation to find the mass of the moon. Gmmm

gRm2

(1.63 m/s 2 )(1.74 x 106 m)2

7.40 x 1022 kg


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*10-48.A satellite is located at a distance of 900 km above the earth’s surface. What is the period of the satellites motion? [ R = 6.38 x 106 m + 0.9 x 106 m = 7.28 x 106 m ] T

2

 4π 2  3 4π 2 (7.28 x 106 )3 =  R = (6.67 x 10-11 N m2 /kg 2 )(5.98 x 1024 kg) ; Gm  e  T = 3.82 x 107 s2 ;

T = 6180 s

2

7

T = 3.82 x 10 s

2

(about an hour and 43 minutes)

*10-49. How far above the earth’s surface must a satellite be located if it is to circle the earth in a time of 28 h? T

2

T = 28 h (3600 s/h) = 1.01 x 105 s;

 4π 2  3 =  R ;  Gme 

3

R

=

GmeT 2 4π 2

R = 3 1.03 x 1023 m3 ;

=

T 2 = 1.02 x 1010 s2

(6.67 x 10-11 N m2 /kg2 )(5.98 x 1024 kg)(1.02 x 1010 s2 ) 4π 2

R = 4.69 x 107 m;

h = R – R e = 4.05 x 107 m

Challenge Problems 10-50. At what frequency should a 6-lb ball be revolved in a radius of 3 ft to produce a centripetal acceleration of 12 ft/s2? What is the tension in the cord? ac

= 4π 2 f 2 R;

f2

T

=

ac 2

4π R

=

(12 ft/s 2 ) 2

4π (3 ft)

; f

=

6 lb  (12 ft/s 2 ) ; = mac =  2   32 ft/s 

0.1013 s-2 ;

f = 0.318 rev/s

T = 2.25 lb

10-51. What centripetal acceleration is required to move a 2.6 kg mass in a horizontal circle of radius 300 mm if its linear speed is 15 m/s? What is the centripetal force?


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10-52. What must be the speed of a satellite located 1000 mi above the earth’s surface if it is to travel in a circular path? [ R = 4000 mi + 1000 mi = 5000 mi ; 5000 mi = 2.64 x 107 ft ] mv 2 R

=

v

Gmme R2

Gme

v=

;

R

;

 0.3048 m  = 8.047 x 106 m   1 ft 

R = 2.64 x 107 ft 

(6.67 x 10 -11 N m 2 / kg 2)(5.98 x 10 24 kg)

=

8.047 x 10 6 m

;

v = 7041 m/s

10-53. A 2-kg ball at swings in a vertical circle at the end of a cord 2 m in length. What is the critical velocity at the top if the orbit is to remain circular? T

+ mg =

vc

=

mv 2

gR

R

=

T mg

; Critical speed vc is when T = 0

(9.8 m/s2 )(2.0 m) ;

R

vc = 4.42 m/s

*10-54. A 4-kg rock swings at a constant speed of 10 m/s in a vertical circle at the end of a 1.4 m cord. What are the tensions in the cord at the top and bottom of the circular path? At Top: T

+ mg =

mv 2 R

and T

=

mv 2

− mg

R

v T

T =

(4 kg)(10 m/s)

2

− (4 kg)(9.8 m/s2 ) ;

(1.4 m)

At Bottom: T

− mg =

mv 2 R

and T

=

mv 2 R

T = 247 N

+ mg

mg

T

v mg

T =

(4 kg)(10 m/s) (1.4 m)

2

+ (4 kg)(9.8 m/s2 ) ;

T = 325 N

R


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*10-55. What frequency of revolution is required to raise the flyweights in Fig. 10-16 a vertical distance of 25 mm above their lowest position. Assume that L = 150 mm. L

h = 150 mm – 25 mm = 124 mm; h = 0.124 m f =

1

g

2π

h

=

1

9.8 m/s2

2π

0.125 m

;

θ

h

R

f = 1.41 rev/s = 84.6 rpm

*10-56. The combined mass of a motorcycle and driver is 210 kg. If the driver is to negotiate a loop-the-loop of radius 6 m, what is the critical speed at the top? vc

=

gR

=

(9.8 m/s2 )(6.0 m) ;

vc = 7.67 m/s

*10-57. If the speed at the top of the loop in Prob. 10-54 is 12 m/s, what is the normal force at the top of the loop? T

At Top: N

N

=

+ mg =

(210 kg)(12 m/s) 2 (6 m)

mv 2 R

and N

mv 2 =

R

− (210 kg)(9.8 m/s2 ) ;

−

mg

mg

N

R

= 2980 N

10-58. The speed limit at a certain turn of radius 200 ft is 45 mi/h. What is the optimum banking angle for this situation. Are roads actually constructed at the optimum angles? v = 45 mi/h = 66.0 ft/s;

tan θ =

v2 gR

=

(66 ft/s) 2 2

(9.8 m/s )(200 ft)

;

θ = 34.20;

*10-59. For the figure shown in Fig. 10-17, assume that a = 2 m and L = 4 m.

NO

a

Find


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Critical Thinking Questions *10-60. A coin rests on a rotating platform distance of 12 cm from the center of rotation. If the coefficient of static friction is 0.6, what is the maximum frequency of rotation such that he coin does not slip? Suppose the frequency is cut in half. Now how far from the center can F c = F s

the coin be placed? The maximum frequency occurs when F c = Fs Fc

= 4π 2 f 2 mR; f =

f 2

=

1.114 rev/s 2

R2

=

4π 2 f 2 mR = µs mg ;

F s =µ s mg ;

µ s g

4π 2 R

=

(0.6)(9.8 m/s 2 ) 4π 2 (0.12 m)

= 0.557 rev/s;

(0.6)(9.8 m/s 2 ) 4π 2 (0.557 rev/s)2

;

;

4π 2 f 22 mR2

f

=

µ s g

4π 2 R

f = 1.11 rev/s

= µ s mg ;

R = 0.480 m;

R2

=

µ s g

4π 2 f 22

R = 48.0 cm

*10-61. The laboratory apparatus shown in Fig. 10-19 allows a rotating mass to stretch a spring so that the supporting cord is vertical at a particular frequency of rotation. Assume the mass of the bob is 400 g and the radius of revolution is 14 cm. With a stop watch the time for 50 revolutions is found to be 35 s. What is the magnitude and direction of the force acting on the bob? First find f in (rev/s). f =

50 rev 35 s

= 1.429 rev/s .

Fc

= 4π 2 f 2 mR

14 cm


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*10-62. In Problem 10-2 above, assume that a 100-g mass is added to the 400-g bob? The force required to stretch the spring should be the same as before, but the rotating mass has increased. What changes when the experiment is performed again so that the centripetal force is the same as before? On what does the centripetal force act in this experiment? Since the centripetal force must be the same, it is

14 cm

necessary that the velocity be reduced so that:

100 g

F c =

2 1 1

mv

mv 2 R

; The product mv2 must be the same.

=mv

2 2 2

v2

=

2 2

; v

=

m1v12 m2

(400 g)v12 (500 g)

=

=

(400 g)v12 (500 g)

0.894 v1 ;

400 g

v1 = 2π f 1 R = 2π (1.429 rev/s)(0.14 m) = 1.26 m/s

v2 = (0.894)(1.26 m/s);

v2 = 1.13 m/s

Thus, the object moves slower and the frequency of revolution decreases so that the 2

centripetal force acting ON the bob does not change: m1v1

= m2 v22

*10-63. A 10-in. diameter platform turns at 78 rpm. A bug rests on the platform 1 in. from the outside edge. If the bug weighs 0.02 lb, what force acts on it? What exerts this force? Where should the bug crawl in order to reduce this force by one-half? f = 78 rev/min = 1.30 rev/s;

m = W/g = 0.02 lb/32 ft/s2; m = 0.000625 slugs

R = 5 in. – 1 in. = 4 in.; R = (4/12) ft = 0.333 ft;

F c = 4π 2 f 2 mR

F = 4π 2 (1.3 rev/s)2 (0.000625 slug)(0.333 ft) ;

F = 0.0139 lb


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R2

=

(4 in.)

=

2


2 in. ; The bug should crawl to a point 2 cm from the center.

*10-64. The diameter of Jupiter is 11 times that of the earth, and its mass is about 320 times that of earth. What is the acceleration due to gravity near the surface of Jupiter? m j = 11me = 320(5.98 x 1024 kg); m j = 1.914 x 1027 6

mg j

7

R j = 11 Re = 11(6.38 x 10 m); R j = 7.018 x 10 m ;

g j

=

Gm j 2 j

R

=

(6.67 x 10-11 N m 2 /kg2 )(1.914 x 1027 kg) 7

2

(7.018 x 10 m)

;

=

Gmm j R j2

g j = 25.9 m/s2

*10-65. Assume that L = 50 cm and m = 2 kg in Fig. 10-16. How many revolutions per second are needed to make the angle θ = 300? What is the tension in the supporting rod at that point? [ h = (50 cm) cos 300 = 0.433 m; m = 2.0 kg; tan θ

=

v2 gR

; v=

v = 1.19 m/s;

T cos θ = mg;

L

θ = 300 ]

gR tan θ = (9.8 m/s2 )(0.25 m) tan(300 )

v = 2π fR; f =

T =

mg cosθ

=

v 2π R

=

(1.19 m/s) 2π (0.25 m)

(2 kg)(9.8 m/s2 ) cos 300

;

;

θ

h

R

f = 0.757 rev/s

T = 22.6 N

*10-66. A 9-kg block rests on the bed of a truck as it turns a curve of radius 86 m. Assume that µk = 0.3 and that µs = 0.4. Does the friction force on the block act toward the center of the


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=

µ s gR

=

2

(0.4)(9.8 m/s )(86 m) ;




vc

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vc = 18.4 m/s

Fc Fs

Fk =

µk mg = (0.3)(9 kg)(9.8 m/s 2);

Fk =

FR = 26.5 N


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