Physics
Chapter 8: Torque and Angular Momentum
Chapter 8
TORQUE AND ANGULAR MOMENTUM
Problems
1. Strategy and Solution I has units kg  m2.  2 has units (rad s)2. So, 12 I  2 has units
kg  m2  rad 2 s2  kg  m2 s2  J, which is a unit of energy.
9. (a) Strategy and Solution Since a significant fraction of the wheel’s kinetic energy is rotational, to model it as
if it were sliding without friction would be unjustified. So, the answer is no.
(b) Strategy Use Eq. (8-1) and form a proportion.
Solution Find the fraction of the total kinetic energy that is rotational.
K rot

K total
 12 I 2   1  1  1
1 Mv 2  4 1 I  2
 2  4MvI  1 1  4I (Mvv R ) 1  MR4I
2
4
2
2
2
2
2
2

1
1
(1300 kg)(0.35 m)2
4(0.705 kgm2 )
 0.017
25. (a) Strategy Use the work-kinetic energy theorem.
Solution Find the work done spinning up the wheel.
1
1
W  K  I f2  (MR 2 )f2
2
2
1
 (182 kg)(0.62 m) 2[(120 rev min)(1 60 min s)(2 rad rev)]2  5.5 kJ
2
(b) Strategy Use the equations for rotational motion with constant acceleration and the relationship between
work, torque, and angular displacement.
Solution Find the torque.
W     (avt ), so  
W
5.5 103 J

 29 N  m .
avt (120 rev min)(1 60 min s)(2 rad rev)(30.0 s) 2
35. Strategy Use Eqs. (8-8).
Solution Choose the axis of rotation at the hinge.
1
Chapter 8: Torque and Angular Momentum
Physics
  0  T (2.38 m) sin 35  (80.0 N)(1.50 m)  (120.0 N)(3.00 m), so
(80.0 N)(1.50 m)  (120.0 N)(3.00 m)
T
 350 N .
(2.38 m) sin 35
Find Fx and Fy .
Fx  0  T cos 35  Fx and Fy  0  Fy  T sin 35  80.0 N  120.0 N, so
Fx  T cos 35  (350 N) cos 35  290 N and Fy  (351.6 N) sin 35  80.0 N  120.0 N  2 N .
The magnitude of Fy is small compared to that of Fx and T.
45. Strategy Refer to Figure 8.32. First find the magnitude of the force exerted by the back Fb by analyzing the
torques about an axis at the sacrum; then, find the horizontal component of the extreme force on the sacrum Fs .
Use Eqs. (8-8).
Solution Sum the torques to find Fb .
  0  Fb (44 cm)sin12  (10 kg)(9.80 m s2 )(76 cm)  (55 kg)(9.80 m s2 )(38 cm), so
(10 kg)(9.80 m s2 )(76 cm)  (55 kg)(9.80 m s2 )(38 cm)
 3053 N.
(44 cm)sin12
The only forces with components in the horizontal direction are those due to the back and the sacrum. Find the
horizontal component of the extreme force, Fsx .
Fb 
Fx  0  Fsx  Fb cos12, so Fsx  Fb cos12  (3053 N)cos12  3.0 kN .
(3053 N) cos12
 5.5, so the force is about 5.5 times larger than that from his torso alone!
540 N
57. Strategy Follow the steps to derive the rotational from of Newton’s second law.
Solution
(a) According to Newton’s second law, Fi  mi ai , so ai  Fi mi .
(b) The torque is the product of the perpendicular component of the force and the shortest distance between the
rotation axis and the point of application of the force, so  i  Fi ri  mi ai ri .
(c) The tangential acceleration is related to the angular acceleration by ai  ri , so  i  mi (ri )ri  mi ri 2 .
(d) Summing the torques and using the definition of rotational inertia, we have
N
N
N

  i   mi ri 2    mi ri 2    I  .
i 1
i 1
 i 1

69. Strategy The rotational inertia of a uniform disk is I  12 MR2 . Use Eq. (8-14).
Solution Find the magnitude of the angular momentum of the turntable.
1
1
L  I   MR2  (5.00 kg)(0.100 m)2 (0.550 rev s)(2 rad rev)  0.0864 kg  m2 s
2
2
2
Physics
Chapter 8: Torque and Angular Momentum
81. Strategy The average torque is equal to the magnitude of the change in angular momentum divided by the time
interval.
Solution Let Li  L in the +y-direction. Then L has components Lx  L sin  and
Ly  L cos  L  L(cos 1). So,
L  (L sin  )2  [L(cos  1)]2  L sin 2 60.0  (cos 60.0 )2  1.00L.
Compute the magnitude of the required torque.


2
1
L 1.00L 1.00I  2 mr 



t
t
t
t
(1.00 105 kg)(2.00 m) 2 (300.0 rpm)  2 rad  1 min 
6
 rev  60 s   2.10 10 N  m
2(3.00 s)



93. (a) Strategy The rotational inertia of a uniform solid disk is I  12 MR2 .
Solution Compute the rotational inertia.
1
1
I  MR2  (200.0 kg)(0.40 m)2  16 kg  m2
2
2
(b) Strategy Use Eq. (8-1).
Solution Compute the initial rotational kinetic energy.
1
1
Krot  I  2  (16 kg  m2 )(3160 rad s)2  8.0 107 J
2
2
(c) Strategy and Solution The ratio of the rotational to the translational kinetic energies is
Krot
K
2(8.0 107 J)
 rot2 
 320 .
2
1 mv
K tr
(1000.0
kg)(22.4
m
s)
2
(d) Strategy Set the work done by air resistance equal to the stored energy in the flywheel.
Solution Find the distance d the car can travel.
Fd  K rot , so d 
K rot 8.0 107 J

 120 km .
F
670.0 N
109. Strategy Since the bike travels with constant velocity, the acceleration is zero and   0.
Solution Find the magnitude of the force with which the chain pulls.
r
  0  fr2  FCr1, so FC  2 f  6.0(3.8 N)  23 N .
r1
3