Solucionario Raymond Serway - Capitulo 10

Chapter 10 Solu tions

10.1

( a ) α =

(b )

10.2

*10.3

ω – ω i

=

t

θ = ω it +

12.0 12.0 rad/ s = 4.00 4.00 rad/ s 2 3.00 s

1 1 α t t 2 = (4.0 (4.00 0 rad/ s 2)(3.00 s) 2 = 18.0 ra d 2 2

( a ) ω =

2π r a d 1 day 1 h = 1.99 365 da ys 24 h 3600 s

(b )

2 π r a d 1 day 1 h = 2.65 × 10 –6 r a d / s 27.3 27.3 day s 24 h 3600 s

ω =

× 10–7 r a d / s

ω i = 200 2000 rad/ s α = – 80.0 80.0 rad/ s 2 t = [2000 ( a ) ω = ω i + α t [2000 – (80.0)(10. (80.0)(10.0)] 0)] = 1200 ra d / s

(b )

t 0 = ω i + α t t =

10.4

ω i

–α

=

2000 = 25.0 s 80.0

( a ) Let ω h an d ω m be the angular angular speeds of the hour hand and m inute hand, so that that ω h =

2π r a d π = r ad ad / h 12 h 6

an an d

ω m = 2π r ad / h

Then if θ h an d θ m are the angular positions positions of the hour han d an d m inute hand with respect to the 12 o'clock position, we have θ h = ω ht

an d

θ m = ω mt

For the two han ds to coincide, coincide, we need θ m = θ h + 2π n , where n is a positive integer. Therefore, Therefore, we may write ω m t – ω ht = 2 π n, or

t n =

2π n 2π n 12n = = h π ω m – ω h 2π – 11 6

© 200 2000 0 by H arcourt College Publishers. All rights reserved.

2

Chapter 10 Solutions

Construct the following following table: t n (h) 0 .0 0 1.09 2.18 3.27 4.36 5.45 6.55 7.64 8.73 9.82 1 0 .9 1

n

0 1 2 3 4 5 6 7 8 9 10 (b )

tim e (h:m in:s) 12:00:00 1:05:27 2:10:55 3:16:22 4:21:49 5:27:16 6:32:44 7:38:11 8:43:38 9:49:05 10:54:33

Le t θ s an d ω s be the angular position position and angu lar speed of the sec second ond hand , then ω s = 2π rad/ min = 12 120π r ad ad / h

an an d

θ s = ω st

k (k is any positive integer) at any of For all three hands to coincide, we need θ s = θ m + 2 π k the times given above. That is, we need

ω st n – ω m t n = 2 π k k , o r k =

ω s – ω m

2π

t n =

118 12 (3)(4)(59)n n= 2π 11 11

to be an integer. This is is possible only for for n = 0 or 11. 11. Therefore, all three hand s coincide coincide only when straight up at 12 o'c o'cloc lock k .

10.5

ω i =

  100 rev    1.00   2π r a d   10π 1.00 m in   ad / s, ω f = 0  1.00 min   60.0 s   1.00 rev  = 3 r ad ω f – ω i

( a ) t =

(b )

*10.6

α

=

  

0 – 10π / 3 s = 5.24 s –2.00

ω f + ω i – θ = ω t = 2

   10π    10π   t = r ad a d / s   6   6 s =

2 7. 7. 4 r a d

ω i = 3600 3600 rev/ min = 3.77 3.77 × 102 r a d / s θ = 50.0 rev = 3.14 2

× 102 rad

and

ω f = 0

2

ω f = ω i + 2 αθ

0 = (3.77

(3.14 × 102 rad) × 102 r aadd / s )2 + 2 α (3.14

α = –2.26

× 102 r a dd// s 2


Chapter

10.7

( a ) θ t = 0 = 5.00 ra d ω t = 0 =

(b )

θ d θ d t

d ω d t

α t = 0 =

= 10.0 + 4.00 t t = 0 = 10.0 10.0 rad/ s t = 0

= 4.00 4.00 rad/ s 2 t = 0

θ t = 3.00 s = 5.00 5.00 + 30.0 + 18.0 18.0 = 53.0 r ad θ d θ d t

ω t = 3.00 s =

= 10.0 + 4.00 t t = 3.00 s = 22.0 22.0 rad/ s t = 3.00s

d ω d t

α t = 3.00 s = *10.8

10 Solutions

= 4.00 4.00 rad/ s 2 t = 3.00s

ω = 5.00 5.00 rev/ rev/ s = 10.0 10.0 π r a d / s

We will break the m otion into two stages: (1) (1) an acceleration acceleration p eriod an d (2) a deceleration period. While speeding up, 0 + 10.0 π r a d / s – θ 1 = ω t = (8.00 s) = 40.0π ra d 2 While slowing down, 10.0π r ad ad / s + 0 – θ 2 = ω t = (12.0 s) = 60.0π ra d 2 So , θ τotal = θ 1 + θ 2 = 100π rad = 50.0 rev

10.9

θ – θ i = ω it +

1 α t 2 2

and an d

ω = ω i + α t

are two equations equations in two unknow ns ω i an d α . ω i = ω – α t θ – θ i = ( ω – α t )t +

37.0 rev

 2π r a d    1 rev 

1 1 α t 2 = ω t t – α t 2 2 2 = (98.0 (98.0 rad/ s)(3.00 s)(3.00 s) –

1 α (3.00 s)2 2

232 rad = 294 rad – (4.50 (4.50 s2)α


3

4


α =

61.5 rad = 13.7 13.7 rad/ s 2 4.50 s 2


Chapter

*10.10

(b )

*10.11

∆θ = 1 rev ∆ t 1 d a y

( a ) ω =

2 π r a d = 7.27 × 10 –5 r a d / s 86400 s

 2π r a d  107° ∆t = ∆ω θ = 7.27 × 10–5 r a d / s  360°   =

2.57

× 10 4 s

(428 min)

Estimate the tire’s radius at 0.250 m and miles driven as 10,000 per year. θ =

  

s 1.00 × 104 m i 1609 m = r 0.250 m 1 mi

θ = 6.44 × 107

or

*10.12

=

~ 1 07

  

ra d 1 rev y r 2π r a d

  ra d  = 6.44 × 107 y r

  rev  = 1.02 × 107 y r

r ev yr

Main Rotor:

  

ω = (3.80 m) 450 v = r ω

  

v = 179

m s

r ev min

    v sound     343 m/ s =

   2π r a d    1 min     1 rev   60 s  =

179 179 m/ s

0.522v sound

Tail Rotor:

  

ω = (0.510 m) 4138 v = r ω

  

v = 221

10.13

    v sound     343 m/ s =

ω ; ω = ( a ) v = r ω

(b )

10.14

m s

ar =

v = 36.0

ω =

r ev min

   2π r a d    1 min     1 rev   60 s  =

221 221 m/ s

0.644v sound

v 45.0 5.0 m/ s = = 0.18 0.180 0 rad / s r 250 m

v2 (45. (45.0 0 m/ s)2 = = 8.10 8.10 m/ s 2 toward the center of track track r 250 m

  

km 1 h h 3600 s

   103 m   10.0 0 m/ s   1 km  = 10.

v 10.0 0.0 m/ s = = 40.0 40.0 rad/ s r 0.250 m


10 Solutions

5

6

10.15


Given r = 1.00 m, α = 4. 4.00 rad/ s 2, ω i = 0, and θ i = 57.3° = 1.00 rad ( a ) ω = ω i + α t = 0 + α t At t = 2.00 s, ω = (4. (4.00 00 rad / s 2)(2. )(2.00 00 s) = 8.00 8.00 rad / s (b )

ω = (1. v = r ω (1.00 00 m)(8.0 m)(8.00 0 rad / s) = 8.00 8.00 m/ s ω 2 = (1.00 ar = r ω (1.00 m)(8. m)(8.00 00 rad/ s) 2 = 64.0 64.0 m/ s 2 α = (1.00 at = r α (1.00 m)(4. m)(4.00 00 rad/ s 2) = 4.00 4.00 m/ s 2

The mag nitud e of the total acceleration acceleration is: 2

a=

2

a r + a t =

(64. (64.0 0 m/ s 2)2 + (4.0 (4.00 0 m/ s 2) 2 = 64. 64.1 1 m/ s 2

The direction of the total acceleration vector makes an angle φ with respect to the radius to point P: φ = tan –1

(c) (c )

10.16

θ = θ i + ω it +

( a ) ω = (b )

2

2

ω f = ω i + 2α (∆ θ )

α =

10.17

1 1 α t 2 = (1.00 (1.00 rad ) + (4.0 (4.00 0 rad/ s 2)(2.00 s) 2 = 9.00 ra d 2 2

v 25. 25.0 m/ s = = 25.0 25.0 rad/ s r 1.00 m

2

(c) (c )

 at    4.00  –1  = tan  a r   64.0 = 3.58°

2

ω f – ω i

2( ∆ θ )

=

(25. (25.0 0 rad/ s)2 – 0 = 39.8 39.8 rad/ s 2 2[(1.25 rev)(2 π r a d / r ev ev )] )]

∆ t = ∆α ω = 39.8 39.8 rad/ s 2 25.0 25.0 rad/ s

= 0.628 s

– ( a ) s = v t = (11.0 (11.0 m/ s)(9.00 s)(9.00 s) = 99.0 99.0 m θ =

(b )

ω =

s 99.0 m = = 341 341 rad = 54.3 rev r 0.290 m v 22.0 2.0 m/ s = = 75.9 75.9 rad/ s = 12.1 12.1 rev/ s r 0.290 m


Chapter

10.18

10 Solutions

7

K A + U A = K p + U p P

1 2 6.00 × 9.80 × 5.00 = (6.00) v p + 6.00 × 9.80 × 2.00 2

ar

2

v p = 58.8 m 2 / s 2 g

Radial acceleration at P , 2

ar =

v p R

= 29.4 29.4 m/ s 2

Tangential acceleration acceleration at P, at = g = 9.80 9.80 m/ s 2

10.19

f = ( a ) ω = 2 π

2π r ad 1200 rev/ rev/ s = 126 126 rad/ s r ev 60.0

(b )

v = ω r r = (126 (126 rad/ s)(3. s)(3.00 00 × 10–2 m) = 3.77 3.77 m/ s

(c) (c )

ar = ω 2r = (126)2(8.00

× 10–2) =

1260 1260 m/ s 2

= 1.26 1.26 km/ s 2

r = ω t r = (126 rad / s)(2.00 (d ) s = θ r s)(2.00 s)(8.00 s)(8.00 × 10–2 m ) = 20.1 m

10.20

Just before it starts to skid ,

∑F r = m ar f =

mv2 = µ sn = µ smg r 2

ω 2 r (ω 2 – ω i ) r 2αθ r v2 r 2at θ µ s = = = = = rg g g g g µ s =

*10.21

2(1. 2(1.70 70 m/ s 2)(π / 2) = 0.545 9.8 9.80 m/ s 2

( a ) x = r cos θ = (3.00 m) cos (9.00 rad) = (3.00 m) cos 516 ° = –2.73 m (3.00 m) sin (9.00 (9.00 rad ) = (3.00 (3.00 m) sin 516 ° = 1.24 m y = r sin θ = (3.00 r = xi + y j = (–2.73 i + 1.24 j) m

(b)

516˚ 516˚ – 360 360˚˚ = 156˚ 156˚.. This This is between 90.0 90.0˚˚ and 180˚ 180˚,, so so the objec objectt is in the second second qu adrant . The vector r makes an angle of 156 156°

with the positive x -axis or 24.3 ° with the negative

x -axis.


8

Chap ter 10 Solutions Solutions

(d ) The direct direction ion of motion motion (i.e (i.e., ., the directi direction on of the velocity vector is at 156 ° + 90.0° = 246° from the

m

positive x axis. The direction of the acceleration acceleration vector is at 156 ° + 180° = 336° from the positive positive x axis. (c) (c )

a

156° x

v

v = [(4.5 j [(4.50 0 m/ s) cos cos 246°]i + [(4. [(4.50 50 m/ s) sin sin 246°] j = (–1.85i 1.85i – 4.10 j) j) m / s

(e )

a=

v2 (4.5 (4.50 0 m/ s)2 = = 6.7 6.75 5 m/ s 2 directed tow ard th e center center or at 336 336 ° 3.00 m r

a = (6. j) m / s 2 (6.75 75 m/ s 2)(i )(i cos 336° + j sin 336°) = (6.15i (6.15i – 2.78 j) (f) *10.22

kg)[(6.15i – 2.78 j) j) m / s 2] = ∑F = m a = (4.00 kg)[(6.15i

When comp letely letely rewoun d, the tape is a hollow cylinder cylinder w ith a difference difference between between th e inner and ou ter radii of ~1 cm. cm. Let N represent the num ber of revolutions revolutions through w hich hich the d riving riving spind le turns in 30 minu tes (and (and hence the nu mber of layers of tape on th e spool). spool). We can determine N from: N =

∆θ = ω (∆t )

2π

=

2π

Then, thickness ~

10.23

(24.6i (24.6 i – 11.1 j) j) N

(1 rad/ s)(3 s)(30 0 min)(60 min)(60 s/ min) = 286 rev π r ad 2π r ad / r ev ev

1 cm ~10 –2 cm N

m 1 = 4.00 kg, r 1 = y 1 = 3.00 m; m 3 = 3.00 kg, r 3 = y 3 2

m 2 = 2.00 kg, r 2 = y 2

ω = ω = 2.0 2.00 0 rad/ rad/ s about the x -axis

= 4.00 m; 2

= 2.00 m;

2

( a ) I x = m 1r 1 + m 2r 2 + m 1r 2 = (4.00)(3.00)2 + (2.00)(2.00)2 + (3.00)(4.00)2 y

I x = 92.0 kg · m 2

1 1 K R = I x ω 2 = (92.0)(2.00) 2 2 (b )

4.00 kg 2

y = 3.00 m

= 184 J

ω = (3.00 v 1 = r 1ω = (3.00)(2 )(2.00) .00) = 6.00 6.00 m/ s

x

O

2.00 kg

y =

-2.00 m

y

-4.00 m

ω = (2.00 v 2 = r 2ω = (2.00)(2 )(2.00) .00) = 4.00 4.00 m/ s 3.00 kg

ω = (4.00 v 3 = r 3ω = (4.00)(2 )(2.00) .00) = 8.00 8.00 m/ s

© 2000 2000 by Har court College Publishers. All rights reser ved.

=


K 1 =

1 1 2 m 1v 1 = (4.00)(6.00) 2 2

2

= 72.0 J

K 2 =

1 1 2 m 2v 2 = (2.00)(4.00) 2 2

2

= 16.0 J

K 3 =

1 1 2 m 3v 3 = (3.00)(8.00) 2 2

2

= 96.0 J

K = K 1 + K 2 + K 3 = 72.0 72.0 + 16.0 + 96.0 96.0 = 184 J

10.24

RATIO =

10.25

( a ) I =

1 2

1 ω 2 I ω 2 1 mv2 2 2 5

1 I ω 2 2 x

ω = ω = 12 125 rad/ rad/ s

v = 38. 38.0 0 m/ s

RATIO =

=

9

=

(3.80

  25 m r 2  ω 2 1 2

mv2

× 10 –2 )2 (125)2 (38.0) 2

=

1 160

∑ j m j r j2

y (m )

4

In this case, 3.00 kg

r 1 = r 2 = r 3 = r 4

(3.00 m )2 + (2.00 m)

r =

2.00 kg 2

2

=

13.0 m

1 x (m)

I = [ 13.0 m ]2 [3.00 + 2.00 + 2.00 + 4.00]kg

0

1

3

= 143 kg · m 2 (b )

K R =

1 1 ω2 = (143 kg · m 2)(6. I ω )(6.00 00 rad/ s) 2 2 2

= 2.57 10.26

2.00 kg

× 10 3 J

The moment of inertia inertia of a thin rod abou t an axis through one end is I = rotational kinetic energy is given as K R =

1 1 I hω 2h + 2 I mω 2m , w i t h 2 2

I h =

m h Lh

3

=

(60.0 kg)(2.70 kg)(2.70 m )2 = 146 kg m 2, and 3

2

I m =

m m Lm

3

4.00 kg

=

(100 kg)(4.50 kg)(4.50 m )2 = 675 kg m 2 3


1 M L2. The total 3

10


In addition,

10.27

ω h =

(2π (2 π rr a d ) 1 h = 1.45 × 10–4 rad/ s, (12 h) 3600 s

whil while

ω m =

(2π (2 π rr a d ) 1 h = 1.75 × 10–3 rad/ s. Therefore, Therefore, ( 1 h ) 3600 s

K R =

1 (146)(1.45 2

1

× 10–4) 2 + 2 (675)(1.75 × 10–3) 2 =

1.04

× 10 –3 J

I = M x 2 + m ( L L – x )2

x

d I = 2 M x – 2m ( L (for an extrem um ) L – x ) = 0 (for dx mL

L

∴ x = M + m

m

M

d 2 I = 2m + 2 M ; therefore I is minimum w hen dx 2 mL the axis of rotation rotation passes throu gh x = M + m

L − x

x

wh ich ich is also also the center of mass of the system. The mom ent of inertia inertia abou t an axis pa ssing through x is

I CM = M

2  m L  2  m   + 1 m –  M + m   M + m  L2

Mm M m = L2 = µL 2 M + m

where µ = 10.28

Mm M + m

We assume the rods are thin, with radius m uch less than L. Call the jun jun ction of the rod s the origin origin of coordinates, and the axis of rotation the z -axis. For the rod along the y -axis, I =

1 m L2 from the table. 3

For the rod parallel to the z -axis, the parallel-axis theorem gives

I =

    2 1   ≅ 4

1 L mr 2 + m 2 2

ax

is

ro t o f

mL 2


at i

on


11

In the rod along the x -axis, the bit of material between x an d x + dx has mass ( m / L)dx and is at distance r = x 2 + ( L / 2 )2 from the axis of rotation . The total rotation al inertia is: 1 1 mL2 + mL2 + 3 4

I total =

=

=

*10.29

⌠ ⌠ L / 2 ( x 2 + L2/ 4) (m / L)dx ⌡ – L / 2

    x 3 L / 2   3 –

7 m mL 2 + 12 L

L / 2

+

mL x 4

L / 2

– L / 2

7 11 m L2 mL 2 mL 2 mL 2 + + = 12 12 4 12

Treat the tire tire as consisting consisting of three parts. The two sidew alls alls are each treated as a hollow cylind cylind er of inn inn er rad ius 16.5 16.5 cm, cm, ou ter rad ius 30.5 30.5 cm, cm, an d h eight 0.635 0.635 cm. The tread region is treated as a hollow cylind er of inn inn er rad ius 30.5 30.5 cm, cm, ou ter rad ius 33.0 33.0 cm, cm, and height 20.0 cm. Use I =

1 2 2 m ( R1 + R2 ) for the moment of inertia of a hollow cylinder. 2

Sidewall: π [(0.305 m = π [(0.3 05 m )2 – (0.165 (0.165 m )2] (6.35 × 10–3 m)(1.10 × 103 k g / m 3) = 1.44 kg I side =

1 (1.44 kg ) [(0.165 [(0.165 m )2 + (0.305 m )2] = 8.68 × 10–2 kg ⋅ m 2 2

Tread: π [(0.330 m = π [(0.3 30 m )2 – (0.305 (0.305 m )2] (0.200 m)(1.10 I tread =

× 103 k g / m 3) = 11.0 kg

1 (11.0 kg) [(0.330 [(0.330 m )2 + (0.305 m)2] = 1.11 kg ⋅ m 2 2

Entire Tire: I total = 2 I side + I tread = 2(8.68 × 10–2 kg ⋅ m 2) + 1.11 kg ⋅ m 2 = 1.28 kg

10.30

( a ) I = I CM + M D 2 =

1 3 MR M R2 + MR 2 = M R 2 2 2

(b ) I = I CM + M D 2 =

2 7 MR 2 + MR 2 = M R 2 5 5


⋅ m2

12

*10.31


Model you r bod y as a cylinder cylinder of mass 60.0 60.0 kg and circumference circumference 75.0 75.0 cm. cm. Then its radius is 0.750 m = 0.120 m 2π and its moment of inertia is is 1 1 MR 2 = (60.0 kg)(0.120 m) 2 = 0.432 kg m 2 ~ 10 0 kg · m 2 = 1 kg · m 2 2 2

10.32

τ = 0 = mg (3r ) – Tr ∑τ = M g sin 45.0 2T – Mg 45.0° = 0 T =

45.0° 1500 kg (g) sin 45.0° M g sin 45.0 = = (530)(9.80) N 2 2

m=

T 530g = = 177 kg 3g 3g 3r

r

m

1500 kg

θ

θ =

45 °



10.33

∑τ = (0.100 m)(12.0 N) – (0.250 m)(9.00 N) – (0.250 m)(10.0

13

10.0 N

N) 30.0 °

= –3.55 N · m

a O

The thirty-degree angle is un necessary necessary information.

12.0 N b

9.00 N

Goal Solution simp ly examining examining the magnitud es of the forces forces and their respecti respective ve lever lever arms, it it appears G: By simp that the w heel will rotate cloc clockwise, kwise, and the net torqu e app ears to be about 5 Nm. find the net torque, we simply add the individu individu al torques, torques, remembering remembering to apply the O : To find convention that a torque producing clockwise rotation is negative and a counterclockwise torque is positive. positive. A:

τ = ∑Fd ∑τ = τ = (12.0 N)(0.100 m) – (10.0 N)(0.250 m) – (9.00 N)(0.250 m) ∑τ = τ = –3.55 N ⋅ m ∑τ = The minus sign m eans perp endicularly endicularly into the p lane of the paper, or it means clockwise. clockwise.

L: The resulti resulting ng torque has a reasonable reasonable magnitud e and prod uces clockwis clockwisee rotation rotation as expected. expected. Note that th e 30° angle was not requ ired for the solution since each each force acted acted p erpend icular icular to its lever lever arm . The 10-N 10-N force is to the right, bu t its torqu e is negat ive – that is, clockwise, clockwise, just just like the torque of the d own ward 9-N 9-N force. force.

10.34

Resolve Resolve the 100 100 N force force into comp onents perpendicular to and parallel to the rod, as F pa r = (100 N) cos 57.0° = 54.5 N

2.00 m

20.0 ° 37.0 °

20.0 °

an d F perp = (100 N) sin 57.0° = 83.9 N

Torque of F pa r = 0 since its line line of action p asses through the pivot point. Torque of F perp is τ = τ = (83.9 (83.9 N )(2.00 m ) = 168 N · m (cloc clockwise kwise))


100 N

14

*10.35


The normal force force exerted exerted by the groun d on each w heel is is mg (1500 (1500 kg)(9.80 kg)(9.80 m/ s 2) = = 3680 N 4 4

n=

The torqu e of fricti friction on can be as large as τ m ax = f m ax r = ( µ sn )r = (0.800)(36 (0.800)(3680 80 N )(0.300 )(0.300 m) = 882 N

⋅m

The torque of the axle on the w heel can be equally as large as the light wh eel starts starts to turn without slipping. *10.36

We calculate calculated d th e maximum torque that can be ap plied withou t skidding in Problem 35 to be 882 N · m. This same torque is to be ap plied by the frictional frictional force, force, f , between the brake pad and the rotor for this this wheel. Since the wh eel is sli slipp pp ing against the brake pad , we use the coefficient of kinetic friction to calculate the normal force. τ = τ = fr = ( µ k n )r , so n =

10.37

m = 0.750 kg

882 N ⋅ m τ = = 8.02 × 103 N = 8.02 kN µ k r (0.500)(0.220 m)

F = 0.800 N

τ = rF = (30.0 ( a ) τ = (30.0 m )(0.800 )(0.800 N ) = 24.0 N · m

*10.38

rF 24.0 τ = = = 0.03 0.0356 56 rad / s 2 2 I m r (0.750)(30.0) 2

(b )

α = α =

(c) (c )

(0.0356)(30.0) 0.0) = 1.07 m / s 2 aT = α r = (0.0356)(3

τ = τ = 36.0 N · m = I α

ω f = ω i + α t

10.0 10.0 rad/ s = 0 + α (6.00 (6.00 s) α = α =

10.00 r a d / s 2 = 1.6 1.67 7 rad/ s2 6.00

( a ) I = (b )

36.0 N · m τ = = 21.6 kg · m 2 α 1.67 1.67 rad/ s 2

ω f = ω i + α t 0 = 10.0 + α (60.0) (60.0) α = α = – 0.16 0.167 7 rad/ s 2 m)(0.16 167 7 rad/ s 2) = 3.60 N · m τ = τ = I α = α = (21.6 kg · m)(0.



( c) c)

15

N u m b er er of of r ev ev ol olu t io io ns ns θ = θ = θ i + ω it +

1 α t 2 2

During first 6.00 s θ = θ =

1 (1.67)(6.00) 2

2

= 30.1 rad

Du ring n ext 60.0 60.0 s 1 (0.167)(60.0) 2

θ = θ = 10.0(60.0) –

θ total = (329 (329 rad ) 10.39

For m 1:

∑F y = ma y

2

= 299 rad

r ev = 52.4 rev π r a d 2π r

+ n – m 1g = 0

T1

I, R

m1

n = m 1g = 19.6 N

T2

f k = µ k n = 7.06 N

∑F x = ma x

–7.06 N + T 1 = (2.00 kg)a (1)

m2

For the pulley t

τ = I α α ∑τ =

–T 1 R + T 2 R =

–T 1 + T 2 =

     

1 a M R 2 R 2

1 (10.0 (10.0 kg ) a 2

–T 1 + T 2 = (5 .00 kg)a

(2)

n2

T2

n

f k T1 f k

T1 n T2

m 1g

M g M g


m 2g

16


θ = 0 For m2: + n – m 2g cos θ = n = 6.00 6.00 kg(9.80 kg(9.80 m/ s2) cos 30.0° = 50.9 N f k = µ k n = 18.3 N

θ = m 2a –18.3 N – T 2 + m 2 g sin θ = (6.00 kg)a –18.3 N – T 2 + 29.4 N = (6.00

(3)

(a ) Add equations equations (1 (1) (2 (2) and and (3) (3):

–7.06 N – 18.3 N + 29.4 N = (13.0 kg)a a=

4.01 N = 0.30 0.309 9 m/ s 2 13.0 kg

2.00 kg (0.30 (0.309 9 m/ s2) + 7.06 7.06 N = 7.67 N T 1 = 2.00

(b )

T 2 = 7.67 7.67 N + 5.00 kg(0.30 kg(0.309 9 m/ s2) = 9.22 N

10.40

I =

1 1 mR 2 = (100 kg)(0.500 kg)(0.500 m ) 2 = 12.5 kg ⋅ m 2 2 2 n

ω i = 50. 50.0 0 rev/ rev/ min = 5.2 5.24 4 rad/ s α = α =

ω f – ω i t

=

= 70.0 N

0.500 m

0 – 5.2 5.24 rad/ rad/ s = –0.87 0.873 3 rad/ s 2 6.00 s

f = µn

τ = τ = I α = α = (12.5 kg ⋅ m 2)(–0.87 0.873 3 rad/ s 2) = –10.9 N ⋅ m The magnitud e of the torque is given given by fR = 10.9 N ⋅ m, where f is the force of friction. Therefore, f =

f = µ k n

10.41

10.9 N ⋅ m , and 0.500 m yields

µ k =

f 21.8 N = = 0.312 n 70.0 N

I = M R 2 = 1.80 kg(0.320 m)2 = 0.184 kg · m 2

τ = I α α ∑τ = ( a ) F a(4.50 × 10–2 m ) – 120 N(0.320 m) = 0.184 kg · m 2(4.5 (4.50 0 rad/ s 2) F a =

(b )

(0.829 N · m + 38.4 N · m ) = 872 N 4.50 × 10 –2 m

F b(2.80 × 10–2 m ) – 38.4 N · m = 0.829 N · m F b = 1.40 kN



10.42

We assume the rod is thin. thin. For the comp comp oun d object object

 

 

I =

1 2 M rod L2 + M ball R 2 + M ball D 2 3 5

I =

1 2 1.20 kg (0.240 m)2 + 20.0 kg(4.00 × 10–2 m )2 + 20.0 kg(0.280 m) 2 3 5

I = 1.60 kg ⋅ m 2

( a ) K f + U f = K i + U i + ∆ E 1 ω2 + 0 = 0 + M rod g( L/ 2) + M ballg( L + R ) + 0 I ω 2 1 (1.60 kg ⋅ m 2) ω 2 = 1.20 1.20 kg(9.80 kg(9.80 m/ s2)(0.12 )(0.120 0 m) + 20.0 kg(9.80 m/ s 2)(0.280 )(0.280 m ) 2 1 (1.60 kg ⋅ m 2) ω 2 = 56.3 J 2 (b )

ω = ω = 8.38 8.38 rad/ s

(c) (c )

ω = (0. v = r ω = (0.28 280 0 m)8.38 m)8.38 rad / s = 2.35 2.35 m/ s

(d )

v 2 = v i + 2a( y – y i)

2

v=

0 + 2(9. 2(9.80 80 m/ s 2)(0.28 )(0.280 0 m) = 2.34 2.34 m/ s

The speed it attains in swinging is greater by 2.35 2.35// 2.34 2.34 = 1.00140 1.00140 times 10.43

Choose the zero gravitational potential energy at the level where the masses pass. K f + U gf = K i + U gi + ∆ E

1 1 1 ω 2 = 0 + m 1gh 1i + m 2gh 2i + 0 m 1v 2 + m 2v 2 + I ω 2 2 2

 

1 1 1 (15.0 + 10.0) v 2 + (3.00) R 2 2 2 2

  v   2   R  = 15.0(9.80)(1.50) + 10.0(9.80)(–1.50)

1 (26.5 kg) v 2 = 73.5 J ⇒ v = 2.36 .36 m/ s 2


17

18

10.44


Choose the zero gravitational potential energy at the level where the masses pass. K f + U gf = K i + U gi + ∆ E

1 1 1 ω 2 = 0 + m 1gh1i + m 2gh 2i + 0 m 1v 2 + m 2v 2 + I ω 2 2 2

 

  v   2  d     d   m g m g = +   R  1  2  2  – 2 

1 1 1 M R 2 ( m 1 + m 2) v 2 + 2 2 2

  

  

     

d 1 1 m 1 + m 2 + M v 2 = ( m 1 – m 2)g 2 2 2

v=

10.45

( m 1 – m 2 ) g d 1 2

m 1 + m 2 + M

( a ) 50.0 – T =

50.0 a 9.80

Fg

a α = I TR = I α = R I =

ω

1 MR M R 2 = 0.0938 kg ⋅ m 2 2

50.0 – T = 5.10

 T R 2   I 

3 kg

n

T

0.250 m

T = 11.4 N

a=

50.0 – 11.4 = 7.57 7.57 m/ s 2 5.10

v=

2a( y i – 0) = 9.53 9.53 m/ s



(b)

Use Use cons conser ervat vatiion of energy energy:: (K + U )i = ( K + U ) f m gh =

1 1 ω 2 mv 2 + I ω 2 2

 v 2    R 2

2m gh = m v 2 + I

  

= v2 m +

v=

2m g h m +

I

I R 2

   2(50.0 N)(6.00 m)

=

5.10 kg +

R 2

0.0938

= 9.53 9.53 m/ s

(0.250)2

Goal Solution G: Since Since the rotational inertia inertia of the reel will slow the fall fall of the weight, w e should expect the dow nw ard a ccele cceleration ration to be less less than g. If the reel did n ot rotate, the tension in the string wou ld be equ al to the weight of the objec object; t; and if the reel disapp disapp eared, the tension wou ld be zero. Therefore, T < mg for the given given pr oblem. With similar similar reasoning, reasoning, the final final speed m ust be less than if the weight were to fall freely: v f <

2gy

≈ 11 m/ s

can find find th e acce accelerati leration on and tension tension using the rotational rotational form form of Newton ’s second second law. O : We can The final final speed can be found from the kinematics equation equation stated above and from conservation conservation of energ y. Free-body diagr am s w ill greatly assist in analyzing the forces. forces. A : ( a ) U se

τ = I α α to find T an d a. ∑τ =

First find I for the reel, reel, which we assum e to be a un iform iform d isk: I =

1 1 MR 2 = 3.00 kg (0.250 m)2 = 0.0938 kg ⋅ m 2 2 2

The forces forces on the reel are show show n, includ includ ing a norm al force force exerted exerted by its axle. From the diagram, w e can see that the tension is the only u nbalanced force force causing the reel to rotate.

T

6.00 m

50.0 N


19

20


τ = I α becomes α becomes ∑τ = (0.250 m) = (0.0938 kg ⋅ m 2)(a/ 0. 0 .250 m ) n (0) + F g(0) + T (0.250

(1)

α to the p oint of contact where w e have app lied lied at = r α to contact between string and reel. reel. The falling weight has mass F g

=

50.0 N = 5.10 kg 9.80 .80 m/ s 2

For this mass,

∑F y = ma y becomes

m=

g

+ T – 50.0 50.0 N = (5.10 kg)(–a)

(2)

Note that since we have defined up ward s to be positive, positive, the minus sign shows that its accel accelerati eration on is dow nw ard. We now have our two equations in in the unknow ns T an d a for the two linked linked objects. objects. Substituting T from equ ation (2) into equ ation (1), (1), we ha ve: [50.0 N – (5.10 kg)a](0.250 ](0.250 m) = 0.0938 0.0938 kg ⋅ m 2 12.5 N ⋅ m – (1.28 kg ⋅ m )a = (0.375 kg ⋅ m )a 12.5 N ⋅ m = a(1.65 kg ⋅ m)

or

a 0.250 m

7.57 7 m/ s2 a = 7.5

an d T = 50.0 N – 5.10 5.10 kg(7. kg(7.57 57 m/ s 2) = 11.4 N For the motion of the weight, 2

2

v f = v i + 2 a( y y f – y i) = 0 2 + 2(7. 2(7.57 57 m/ s 2)(6.00 m) v f = 9.5 9.53 3 m/ s

(b)

The work-energy theorem can can take account account of mu ltiple ltiple objec objects ts more easi easily ly than than N ewton's second second law . Like Like your bratty cousins, cousins, the wor k-energy theorem keeps growing betw een visits. visits. Now it reads: (K 1 + K 2,rot + U g1 + U g2)i = ( K 1 + K 2,rot + U g1 + U g2) f 0 + 0 + m 1gy 1i + 0 = ω = Now note that that ω =

v as the string string u nw inds from the reel. reel. Making substitutions: substitutions: r

50.0 N(6.00 m) = 300 N ⋅ m =

v f =

1 1 2 2 m 1v 1 f + I 2 ω 2 f + 0 + 0 2 2

  

v f 1 1 2 (5.10 kg) v f + (0.0938 kg ⋅ m 2) 2 2 0.250 m

  2 

1 1 2 2 (5.10 kg) v f + (1.50 kg) v f 2 2

2(300 N ⋅ m ) = 9.5 9.53 3 m/ s 6.60 kg

L: As we shou ld expect, expect, both both method s give the same final final speed for the fall falling ing objec object. t. The The acceleration acceleration is less than g, and the tension is less than the object ’s weight as we predicted. Now that w e und erstand the effec effectt of the reel reel ’s moment of inertia, this problem solution could be app lied lied to solve other real-world real-world pu lley lley systems with masses that should not be ignored.



10.46

τ · θ = θ =

1 ω 2 I ω 2

(25.0 N · m)(15.0 · 2π ) =

1 (0.130 kg · m 2) ω 2 2

ω = ω = 190 190 rad/ s = 30.3 30.3 rev/ s 10.47

From conservation of energy,

    2 1   + 2

1 v I 2 r

I

v2 mg h – m v 2 = 2 mgh r 2

I = m r 2

10.48

E =

P

=

 2 g h    v 2 – 1   

   3000 × 2π   2   60.0  = 6.17 × 106 J

1 1 1 ω 2 = I ω M R 2 2 2 2

∆ E 4 ∆ t = 1.00 × 10 J/ s

∆t = 10.49

m v 2 = m gh

∆ E P

=

6.17 × 10 6 J = 617 s = 10.3 min mi n 1.00 × 10 4 J/ s

( a ) Fin d t h e v el elo ci cit y of of th th e CM CM (K + U )i = ( K + U ) f 0 + mgR =

ω = ω =

1 ω 2 I ω 2

2m gR = I

v CM = R

(b )

v L = 2 v CM = 4

(c) (c )

v CM =

4g = 2 3 R

Pivot

2m gR 3 mR2 2

R g 3

Rg R g 3

2mg R = 2m

R g


R

g

21

22

*10.50


The moment of inertia of the cylinder is I =

1 1 kg)(1.50 m ) 2 = 91.8 kg · m 2 mr 2 = (81.6 kg)(1.50 2 2

and the angu lar accel accelerati eration on of the m erry-go-round erry-go-round is found found as α = α =

τ I

=

(Fr ) (50.0 N)(1.50 m) = = 0.8 0.817 17 rad/ s2 I (91.8 kg · m 2)

At t = 3.00 s, we find the angular velocity ω = ω = ω i + α t ω = ω = 0 + (0.8 (0.817 17 rad / s 2)(3.00 )(3.00 s) = 2.45 2.45 rad / s an d K =

10.51

mg

l

2

α = α =

1 1 ω 2 = (91.8 kg · m 2)(2. )(2.45 45 rad/ s) 2 = 276 J I ω 2 2

θ = sin θ =

1 m l2 α 3

3g sin θ 2 l

at =

 3 g   θ sin  2 l  r

Then

θ

at

g sin θ

g

 3 g    2 l r > g sin θ

for r >

θ

2 l 3

About ∴ About

1/ 3 the length length of the chimn chimn ey

will have a tangential tangential acce accele leratio ration n greater than than

g sin θ .

*10.52

ω , where r is the rad ius of each ball’s A v 2. Here v = r ω The resistive force on each ball is R = D ρ τ = rR , so the total resistive pa th. The resistive resistive torqu e on each ball is τ = resistive torqu e on the thr ee ball system is τ total = 3 rR . The power required to m aintain aintain a constant rotation rotation rate is ω = 3 rR ω . This required pow er may be written as P = τ totalω = P

ω = 3 r [ D ρ ω )2]ω = ω = (3 r 3 DA A (r ω D A ω 3)ρ = τ totalω =

With ω =

P

or

P

 2π r   3 r ev    1 min    1000π π r a d  1000 π   10 = ad / s,  1 rev   m in   60.0 s  30.0   r ad

= 3(0.100 m) 3(0.600)(4.00

)(1000 π / 30.0 30.0 s)3ρ × 10–4 m 2)(1000π

ρ where ρ is ρ is the density of the resisting = (0.827 m 5 / s 3)ρ where resisting med ium.



ρ = 1.2 ( a ) In a ir , ρ = 1.20 0 kg/ m 3, and P

(b )

10.53

= (0.827 m 5 / s 3)(1. )(1.20 20 kg/ m 3) = 0.992 N ⋅ m/ s = 0.992 0.992 W

ρ = 100 In w at ater , ρ = 1000 0 kg/ m 3 an d

α = α =

0.600 τ = = 122 122 rad/ s2 I 4.90 × 10 – 3

α = α =

∆ω ∆ t

∆ t = ∆α ω =

10.54

θ = θ =

1200

  2π   60

122

= 1.03 s

1 1 α t t 2 = (122 (122 rad/ s)(1. s)(1.03 03 s) 2 = 64.7 64.7 rad = 10.3 rev 2 2

For a sp herical shell

I ∫ dI =

I =

= 827 W

1 1 MR 2 = (2.00 kg )(7.00 )(7.00 × 10–2 m ) 2 = 4.90 × 10–3 kg ⋅ m 2 2 2

( a ) I =

(b )

P

2 2 [(4π r dm r 2 = [(4π r 2dr )ρ ]r 2 3 3

2

r 2 ) r 2ρ (r )dr (4π r ∫ 3 (4π

   3 k g   r  ⌠ ⌠ R 2 (4π  ⌡ (4π r 14.2 11.6 10 r 4 )  dr –  3 m 3 R   0

=

I =

 2  2 R 5  R 5 3) 3)   π π – 4 (14.2 ( 14.2 10 4 (11.6 ( 11.6 10 × ×  3  3 5 6   

8π 14.2 11.6 (103) R 5 – 3 5 6

M = ∫ dm=

⌠ ⌠ R 4π r r 2  14.2 – 11.6 r   103 dr ⌡  R 

= 4π 4 π × 103

I = M R 2

  

0

 14.2 11.6  3  3 – 4  R

11.6 –   14.2 5 6  2  .907  = 0.330 =   14.2 11.6 3 1.83 3 3 2 – 4 π × 10 R R   8π (10 (1 0 3 ) R 5 3

3

4


23

24


∴ I =

0.330 M R 2



10.55

( a ) W =

1

1

2 ω 2 – I ω ω i ∆K = 2 I ω 2

=

1 2 I (ω 2 – ω i ) 2

=

 1   1   r a d   2  2   8.00 (1.00 kg)(0.500 m) – 0 =  2  2  s 

I =

1 mR 2 2

4. 00 00 J

ω – 0 ω r r (8.00 (8.00 rad / s)(0.5 s)(0.500 00 m) = = = 1.60 s 2 α 2.50 .50 m/ s a

(b)

t =

(c) (c )

θ = θ = θ i + ω it +

θ = θ =

where

1 α t t 2; θ i = 0; ω i = 0 2

  

  

1 1 2.50 .50 m/ s 2 α t (1.60 s) 2 = 6.40 rad t 2 = 2 2 0.500 m

θ = (0.50 s = r θ = (0.500 0 m)(6.40 m)(6.40 rad ) = 3.20 m < 4.00 m Yes

10.56

( a ) I =

1 1 mr 2 = (200 kg)(0.300 kg)(0.300 m ) 2 = 9.00 kg 2 2

ω = ω = (10 π rad)// 1 rev = 105 (1000 00 rev/ min)(1 min/ 60 s)(2 s)(2 π rad) 105 rad/ rad/ s

(b )

W = K =

(c) (c )

*10.57

⋅ m2

1 1 ω 2 = (9.00)(104.7) I ω 2 2

500 rev/ min ω f = 500

2

= 49.3 kJ

s(2π rad/ rev) = 52 52.4 rad/ s π rad/ 1 rev) × 1 min/ 60 s(2

K f =

1 ω 2 = 12.3 kJ I ω 2

W =

∆K = 12.3 kJ – 49.3 kJ =

–37.0 kJ

α = α = –10.0 ω / dt 10.0 rad/ s 2 – (5.0 (5.00 0 rad/ s 3)t = d ω

⌠ ⌠ω ⌡

ω = d ω

65.0

ω = ω =

⌠ ⌠t [–10.0 – 5.00t ]dt = –10.0t – 2.50t 2 = ω – 65.0 ⌡ 5.0 rad/ rad/ s 0

θ d θ = 65. 65.0 0 rad/ rad/ s – (10. (10.0 0 rad/ s 2)t – (2.5 (2.50 0 rad/ s 3)t 2 d t

( a ) A t t = 3.00 s, ω = ω = 65. 65.0 0 rad/ s – (10. (10.0 0 rad/ s 2)(3.00 s) – (2.5 (2.50 0 rad/ s 3)(9.00 s 2) = 12.5 12.5 rad/ s


25

26


(b )

⌠ ⌠θ d θ θ = ⌠ ⌠t ω dt = ⌡⌠ ⌠t [65. ⌡ ⌡ [65.0 0 rad/ s – (10. (10.0 0 rad/ s 2)t – (2.5 (2.50 0 rad/ s 3)t 2]dt 0

0

o

θ = θ = (65 (65.0 .0 rad/ s)t – (5.0 (5.00 0 rad/ s 2)t 2 – (0.8 (0.833 33 rad/ s 3)t 3 At t = 3.00 s, (65.0 rad/ s)(3. s)(3.00 00 s) – (5.0 (5.00 0 rad/ s 2)9.00 s 2 – (0.8 (0.833 33 rad/ s 3)27.0 s 3 θ = θ = (65.0 θ = θ = 128 ra d

10.58

10.59

M R 2 , 2

K =

(b ) M K 2 =

M L2 , 12

K =

(c) (c ) M K 2 =

2 M R 2, 5

K = R

(a )

M K 2=

R

2 L 3 6

2 5

y

(a ) Since ince only only conserva conservati tive ve forc forces es act, act, ∆ E = 0, so K f + U f = K i + U i

    1 I , where =   3

L 1 ω2 + 0 = 0 + mg I ω 2 2

L

mL 2 x

pivot

ω = ω =

L 3g / L

  L  mL2  2  = 3

τ = τ = I α α so that in the h orizontal, m g

(c) (c )

ω 2 = a x = ar = r ω

(d)

3 U si sin g N ew ew t on on ' s s ec eco n d la la w , we have R x = ma x = – m g 2

  L  2  2  ω =

R y – mg = –ma y

or

–

3g 2

R y =

α α = α =

3g 2 L

(b )

  L   2  =

α = –α a y = –at = –r α =

–

3g 4

1 mg 4



10.60

The first first dr op h as a velocity velocity leaving the w heel given by

v1 =

1 2 m v i = mgh 1, so 2

2(9. 2(9.80 80 m/ s 2)(0.5 )(0.540 40 m) = 3.25 3.25 m/ s

2gh1 =

The second second drop has a velocity velocity given given by v2 =

v , we find r

ω = From ω =

ω 1 =

v1 r

=

2

or 10.61

α = α =

2(9. 2(9.80 80 m/ s 2)(0.5 )(0.510 10 m) = 3.16 3.16 m/ s

2gh2 =

3.2 3.25 m/ s = 8.53 8.53 rad/ s 0.381 m 2

ω 2 – ω 1 2θ

=

and

ω 2 =

v2 r

=

3.16 3.16 m/ s = 8.2 8.29 9 rad/ s 0.381 m

(8.2 (8.29 9 rad/ s)2 – (8.5 (8.53 3 rad/ s)2 = – 0.32 0.322 2 rad/ rad/ s 2 4π

At the instant it comes comes off the wh eel, eel, the first first dr op h as a velocity velocity v 1, directed directed up ward . The magnitude of this velocity is found from K i + U gi = K f + U gf

1 2 m v 1 + 0 = 0 + mgh1 2

or

v1 =

2 gh 1

and the angular velocity of the wheel at the instant the first drop leaves is

ω 1 =

v1 R

2 gh 1

=

R 2

Similarly for the second drop: v 2 =

2gh 2 an d ω 2 =

v2 R

=

2 gh 2 R 2

The angu lar accelerati acceleration on of the wh eel is is then 2

α = α =

10.62

2

ω 2 – ω 1 2θ

=

2gh 2 / R 2 – 2gh 1 / R 2 2(2 π )

=

g (h 2 – h 1)

2π R 2 A′

Work done = Fs = (5.57 N)(0.800 m) = 4.46 J and Work = ∆K =

1 1 2 2 ω f – I ω ω i I ω 2 2

F

(The last term is zero because the top starts from rest.) Thus, 4.46 J =

1 2 (4.00 × 10–4 kg ⋅ m 2) ω f 2


A

27

28


and from this, this, ω f = 149 149 rad/ s



10.63

K f =

1 1 1 1 2 2 2 2 ω f : U f = Mgh f = 0; K i = M v i + I ω ω i = 0 M v f + I ω 2 2 2 2

ω = U i = ( Mgh)i: f = µN = µMg cos θ ; ω = (a )

∆ E = E f – E i – fd =

1 v θ an d I = mr 2 ; h = d sin θ an 2 r

– fd = K f + U f –K i – U i

or

1 1 2 2 ω f – Mgh M v f + I ω 2 2

µMg cos θ )d = –( µMg

 

1 m M + 2 2

 

1 M v 2 + ( mr 2/ 2)(v 2 / r 2)/ 2 – Mgd sin θ 2

v 2 = Mgd sin θ – ( µMg µMg cos θ )d or

(sin θ – µ cos θ ) v 2 = 2 Mgd (m/ 2) + M

 

v d = 4 gd

(b )

M (sin θ – µ cos co s θ ) (m + 2 M )

2

2

v 2 = v i – 2as, v d = 2ad 2

a=

10.64

( a ) E =

E =

(b )

 1/ 2 

v d

2 d

= 2g

  

  M    m + 2 M  (sin θ – µ cos θ )   

1 2 M R 2 (ω 2 ) 2 5 1 2 ⋅ (5.98 2 5

 2π  2

× 1024)(6.37 × 106) 2  86400 

    

d E d 1 2 = M R 2 d t d t 2 5

× 10 29 J

   2π   2    T  

=

dT 1 (2 π )2(–2T –3) M R 2(2π 5 d t

=

1 2π MR M R 2 T 5

    2  –2  dT    T  d t

= (2.57 × 1029 J)

d E = –1.63 d t

= 2.57

  –2     10 × 10–6 s   (86400 00 s/ s/ day)  86400 s  3.16 × 107 s (864

× 1017 J/ d a y


29

30

10.65


θ = ω t t ∆θ = t =

v=

∆θ = (31.0°/ 360°) rev ω

900 900 rev/ 60 s

= 0.00574 s

0.800 m = 139 139 m/ s 0.00574 s

∆θ = 31° v

ω

d

10.66

(a ) Each ach spoke counts as a thi thin n rod pivote pivoted d at one end. end.

I = M R 2 + n

(b)

By t h e p a r a ll ll el el -a -a xi xi s th th e or or e m , I = M R 2 +

nm R 2 R 2 + ( M + nm ) R 3

= 2 M R 2 +

*10.67

mR2 3

4 nm R 2 3

Every particle particle in in the d oor could be slid straight dow n into a h igh-density igh-density rod across across its bottom, bottom, withou t changing the par ticle ticle ’s distance from from th e rotation axis of the doo r. Thu s, a rod 0.87 0.870 0 m long w ith mass 23.0 23.0 kg, pivoted about one end , has the same rotational inertia as the door: I =

1 1 ML 2 = (23.0 kg)(0.870 m ) 2 = 5.80 kg · m 2 3 3

The height of the door is unn ecess ecessary ary

data.



10.68

31

τ f will oppose the torque causing causing the motion: M

τ = I α = α = TR – τ f ⇒ τ f = TR – I α α ∑τ = Now find find T , I an d equation (1)

(1 )

α in given or known terms and substitute into m R / 2

∑F y = T – mg = –ma then T = m (g – a) also

∆ y = v it +

at 2 2

2 y

⇒ a = t 2

2 y a α = = 2 an d α = R R t

I =

(2)

R / 2

(3 )

(4 )

 

    2   

R 1 M R 2 + 2 2

=

5 MR M R 2 8

(5 )

Substituting (2), (3), (4) and (5) into (1) we find

  

τ f = m g –

10.69

(a)

  

    

  

2 y 5 M R 22 y 2 y 5 M y = R m g – 2 – R – 2 2 8 ( R t ) 4 t 2 t t

 

W h il ile d e ce ce llee ra r a ttii n g, g, τ f = I α α ' = (20 000 kg ⋅ m 2)

 2.00 2.00 rev/ min (2π (2π rad/ rad/ rev)( rev)(1 min/ min/ 60 s)    10.0 s

τ f = 419 N ⋅ m While accelerating, τ = I α α ∑τ =

or

τ – τ f = I (∆ω / ∆t )

τ = τ = 419 N ⋅ m + (20 000 000 kg ⋅ m 2)

τ = τ = 2.16

(b )

P

 10.0 10.00 0 rev/ min (2π (2π rad/ rad/ rev)( rev)(1 min/ min/ 60 s)    12.0 s

× 10 3 N ⋅ m

 

ω = (419 N ⋅ m ) 10.0 = τ f ⋅ ω =

r ev min

 2π r  1 min    π r a d    1 rev   60 s   =

4 39 39 W


(≈ 0.6 0.6 hp )

y

32


10.70

( a ) W =

∆K + ∆U R

W = K f – K i + U f – U i m

0=

1 1 1 ω 2 – mgd sin θ – mv 2 + I ω kd 2 2 2 2

k

θ

1 2 1 ω ( I + mR 2) = mgd sin θ + θ + kd 2 2 2

ω =

(b )

10.71

θ + kd 2 2 m gd si n θ + I + m R 2

ω = ω =

2(0.500 2(0.500 kg)(9.80 m/ s 2)(0.200 m)(sin 37.0 °) + (50.0 (50.0 N / m)(0.200 m)(0.200 m) 2 1.00 kg · m 2 + (0.500 kg)(0.300 m) 2

ω = ω =

1.18 + 2.00 = 1.05

3.04 3.04 = 1.74 1.74 rad / s

( a ) m 2g – T 2 = m 2a

2.00 2.00 m/ s 2

T 2 = m 2(g – a) = (20.0 kg )(9.80 – 2.00 2.00)m/ )m/ s 2 = 156 N

37.0° + m 1a T 1 = m 1g sin 37.0 2.00)m/ )m/ s 2 = 118 N T 1 = (15.0 kg)(9.80 sin 37.0° + 2.00

(b )

T 1

T 2

15.0 kg m1

m2

20.0 kg

37.0 °

 a    R 

α = I ( T 2 – T 1) R = I α =

I =

(T 2 – T 1 ) R 2 a

=

(156 – 118)N 118)N (0.250 (0.250 m )2 = 1.17 kg · m 2 2.0 2.00 m/ s2

Goal Solution G: In earli earlier er problems, problems, we assumed that the tensio tension n in a string string was the same on either either side side of a pu lley. lley. Here w e see see that the moment of inertia inertia changes changes that assump tion, but w e should still still expect expect the tensions tensions to be similar similar in magn itude (about the w eight of each each m ass ~150 ~150 N), and T 2 > T 1 for the p ulley to rotate clockwise clockwise as shown . If we kn ew th e mass of the pu lley, we could calculate calculate its mom ent of inertia, but since we only know the acceleration, it is difficult to estimate I . We at least least know that I must have un its its of kgm 2, and a 50-cm 50-cm d isk probab ly has a m ass less than 10 kg, so I is probably less than 0.3 kgm 2. O:

For each each block, block, we know its mass and accel accelerati eration, on, so so we can can use Newton ’s second second law to find the net force, and from it the tension. The difference difference in the two tensions causes the p ulley to rotate, so so this net torque an d the resulting angular accelerati acceleration on can be u sed to find the pulley ’s mom ent of inertia. inertia.



A:

( a ) A p p ly

∑F = ma

33

to each block to find each string tension.

The forces acting on the 15-kg block are its weight, the normal support force from the incline, and T 1. Taking the po sitive x axis as directed directed u p the incline, incline, ∑F x = m a x yields:

–(m 1g) x + T 1 = m 1(+ a) –(15.0 (15.0 kg)(9.80 kg)(9.80 m/ s 2) sin 37° + T 1 = (15.0 (15.0 kg)(2.00 kg)(2.00 m/ s 2) T 1 = 118 N

Similarly for the counterweight, we have

o r T 2 – m 2g = m 2(–a) ∑F y = m a y , or

T 2 – (20.0 (20.0 kg)(9.80 kg)(9.80 m/ s 2) = (20.0 kg)( –2.00 2.00 m/ s 2)

So, T 2 = 156 N (b )

τ = r (T 2 – T 1) = I α α . We may choose to call clockw N o w fo fo r t he he p u ll lle y, y, ∑τ = clockw ise positive. The angu lar acceleration acceleration is α = α =

a 2.00 .00 m/ s2 = = 8.0 8.00 0 rad/ rad/ s 2 r 0.250 m

τ = I α α ∑τ = I =

L:

10.72

(0.250 (0.250 m )(156 )(156 N – 118 N) = I (8.0 ( 8.00 0 rad/ s 2)

or

9.38 N ⋅ m = 1.17 kg ⋅ m 2 8.00 8.00 rad/ s 2

The tensions are close to the weight of each mass and T 2 > T 1 as expected. expected. How ever, the mom ent of inertia inertia for for the pu lley lley is about 4 times times greater than expected. expected. Unless we mad e a mistake in solving solving th is problem, our resu lt means that the p ulley has a m ass of 37. 37.4 4 kg (about 80 lb), lb), which means that the p ulley is is probably mad e of a dense m aterial, aterial, like like steel steel.. This is certainly certainly not a pr oblem oblem w here the m ass of the pulley can can be ignored since since the pu lley lley has more m ass than th e combination combination of the tw o blocks! blocks!

For the board just starting to move, τ = I α α ∑τ = l

mg

α = α =

2

θ = cos θ =

 1 2   3 m  α l

R mg

     

3 g cos θ 2 l


34


The tangential acceleration of the end is α = at = lα =

3 g cos θ 2

θ = Its vertical vertical compon ent is a y = at cos θ =

3 g cos2 θ 2

If this is greater than g , the board will pull ahead of the ball in falling: (a )

3 2 g cos2 θ ≥ g ⇒ cos2 θ ≥ 2 3 cos θ ≥

so (b )

2 3

and

θ ≤ 35.3°

θ = 35.3° (⇒ cos 2 θ = θ = 2/ 3), W h e n θ = 3), the cup cup will land land und erneath the release-poi release-point nt of the ball if cos2 θ 2l θ = = r c = l cos θ = cos θ 3 cos θ l

(c)

W h en

l

θ = 35.3° = 1.00 1.00 m, an d θ =

r c =

2(1.00 2(1.00 m )

= 0.816 m

3 2/ 3

w hich is (1.00 m – 0.816 0.816 m) = 0.184 0.184 m from th e moving en d 10.73

At t = 0, ω = 3.50 0 rad/ rad/ s = ω 0e0. Thus, ω 0 = 3.5 3.50 0 rad/ rad/ s ω = 3.5 (9.30 s), yieldin ω = 2.0 σ = 6.02 × 10–2 s –1 At t = 9.30 s, ω = 2.00 0 rad/ s = ω 0e–σ (9.30 yieldin g σ =

α = ( a ) α =

ω d (ω 0e–σ t t ) d ω = = ω 0(–σ )e–σ t t d t d t

At t = 3.00 s, α = (3.5 (3.50 0 rad/ s)(–6.02

(b )

θ = θ =

× 10–2 s–1)e–3.00(6.02 × 10–2) =

0.176 6 rad/ s 2 –0.17

ω ω ⌠ ⌠ ⌡t ω 0e–σ t t dt = –σ 0 [e–σ t t – 1] = σ 0 [1 – e–σ t t ] 0

At t = 2.50 s, θ = θ =

3.5 3.50 rad/ s –2 [1 – e–(6.02 × 10 )(2.50)] = 8.12 8.12 rad = 1.29 re v 2 – (6.02 × 10 )1/ s



(c)

A s t → ∞, θ →

ω 0 σ

(1 – e–∞) =

3.5 3.50 rad/ s = 58.2 58.2 rad = 9.26 rev 6.02 × 10 – 2 s – 1


35

36

10.74


Consider the total weight of each hand to act at the center of gravity (mid-point) of that hand . Then the total torque (taking (taking CCW as p ositive) ositive) of these hand s about the center of the clock is is given b y

τ = τ = –m hg

  L h    Lm  –g θ θ m g – sin sin = (m h Lh sin θ h + m m Lm sin θ m) h m m  2   2  2

If we take t = 0 at 12 o'clock, then the angular positions of the hands at time t ar e θ h = ω ht , where ω h =

π r ad ad / h an a n d θ m = ω mt , where ω m = 2π 2 π rr a d / h 6

Therefore,

  

τ = τ = – 4.90

or

m s2

  ( π t (100 kg)(4.50 kg)(4.50 m) sin 2π 2 π t t / 6) + (100 t ]  [(60.0 kg)(2.70 m) sin (π

m[sin(π t 2.78 sin 2π 2 π t ] , where t is in in h ours. τ = τ = –794 N ⋅ m[sin(π t / 6) + 2.78 t ],

(a ) (i)

A t 3:00, t = 3.00 h, so τ = τ = –794 N ⋅ m [sin( π [sin(π / 2) + 2.78 2.78 sin 6π 6π ] = –794 N

( i i) i)

A t 5:15, t = 5 h +

15 h = 5.25 5.25 h, and sub stitution gives: gives: 60

τ = τ = –2510 N

(b)

⋅m ⋅m

( i i i ) A t 6: 6:00,

τ = τ = 0 N

( iv i v ) A t 8:20,

τ = τ = –1160 N

⋅m

(v )

τ = τ = –2940 N

⋅m

A t 9:45,

⋅m

The tota totall torque torque is is zero zero at those those ti times when t / 6) + 2.78 t = 0 sin(π sin(π t 2.78 sin sin 2π 2 π t

We p roceed n um erically, erically, to find 0, 0.515 0.51529 2955 55,, ..., ..., corresp corresp ond ing to th e times 12:00:00 2:33:25 4:58:14 7:27:36 10:02:59

12:30:55 2:56:29 5:30:52 8:03:05 10:27:29

12:58:19 3:33:22 6:00:00 8:26:38 11:01:41

1:32:31 3:56:55 6:29:08 9:03:31 11:29:05

1:57:01 4:32:24 7:01:46 9:26:35


Solucionario Raymond Serway - Capitulo 10

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