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Chapter 10 Solutions
10.
Picture the Problem: The floppy disk rotates about its axis at a constant angular speed.
Strategy: Use equation 10-5 to relate the period of rotation to the angular speed. Then use equation 10-12 to find the linear
speed of a point on the disk’s rim.
2
2

 31.4 rad/s
T
0.200s
Solution: 1. (a) Solve equation 10-5 for  :

2. (b) Apply equation 10-12 directly:
vT  r   12  3.5 in   31.4 rad/s   55 in/s  1 m 39.4 in  1.4 m/s
3. (c) A point near the center will have the same angular speed as a point on the rim because the rotation periods are the same.
18.
Picture the Problem: The bicycle wheel rotates about its axis, slowing down with constant angular acceleration before coming
to rest.
Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to find the angular acceleration and the time
elapsed.
02   6.35 rad/s 
 2  0 2

  0.226 rad/s 2
2   0  2 14.2 rev  2 rad rev 
2
25.
Solution: 1. (a) Solve equation 10-11 for  :

2. (b) Solve equation 10-8 for t:
t
  0 0  6.35 rad/s

 28.1 s

 0.226 rad/s 2
Picture the Problem: The saw blade rotates about its axis, slowing its angular speed at a constant rate until it comes to rest.
Strategy: Use the kinematic equations for rotation (equations 10-8 through 10-11) to find the angular acceleration of the saw
blade and the angle through which the blade spins during this interval. Then use equation 10-2 to convert the angular distance to
a linear distance.
  0
0   4440 rev/min 1 min 60 s 
Solution: 1. (a) Solve equation 10-8 for  :

2. (b) Use equation 10-9 to find  :
  12   0  t 
3. Convert  to s:
s  r    12 10.0 in 12 in/ft   92.5 rev  2 rad/rev   242 ft
t

2.50 s
1
2
  29.6 rev/s 2
 0  4440 rev/min  2.50 s 1 min
60 s   92.5 rev
4. (c) The blade completes exactly 92.5 revolutions, so a point on the rim ends up exactly opposite of where it started. Its
displacement is therefore one blade diameter or 10.0 in.
34.
Picture the Problem: Jeff clings to a vine and swings along a vertical arc as depicted in the figure
at right.
Strategy: Use equation 10-12 to find the angular speed from the knowledge of the linear speed and
the radius. Use equation 6-15 to find the centripetal acceleration from the speed and the radius of
motion.
vt 8.50 m/s

 1.18 rad/s
r
7.20 m
Solution: 1. (a) Solve equation
10-12 for  :

2. (b) Apply equation 6-15 directly:
v 2 8.50 m/s 
acp  t 
 10.0 m/s 2
r
7.20 m
2
3. (c) The centripetal force required to keep Jeff moving in a circle is provided by the vine.
36. Picture the Problem: The compact disk rotates about its axis, increasing its angular speed at a constant rate.
Strategy: Find the angular acceleration of the disk using equation 10-8, and then find the tangential acceleration from
equation 10-14.
  0  4.00 rev/s  2 rad rev   0
Solution: 1. (a) Solve equation 10-8 for  :


 8.38 rad/s 2
t
3.00 s
2. Apply equation 10-14 directly:
at  r 
1
2
 0.12 m  8.38 rad/s2  
0.503 m/s 2
3. (b) The tangential acceleration depends only on the rate of change of angular speed, not the particular value of the
angular speed. Therefore, the tangential acceleration remains 0.503 m/s2 irregardless of the angular speed.
48.
Picture the Problem: The drive wheel of the tricycle rolls without slipping at constant speed.
Strategy: Because the wheel rolls without slipping, equation 10-15 describes the direct relationship between the center of mass
speed and the angular velocity of the driving wheel.
Solution: Apply equation 10-15 directly:
50.
vt  r   0.260 m  0.373 rev/s  2 rad rev   0.609 m/s
Picture the Problem: Your car’s tires roll without slipping, increasing their velocity at a constant rate.
Strategy: Use the fact that the tires roll without slipping to find the angular acceleration and angular velocity from their linear
counterparts. Then use the kinematic equations for rotation (equations 10-8 through 10-11) to determine the angle through
which the tire rotated during the specified interval.
v0 17 m/s

 52 rad/s
r 0.33 m
Solution: 1. Solve equation 10-12 for 0 :
0 
2. Solve equation 10-14 for  :

3. Apply equation 10-10 directly:
  0t  12  t 2   52 rad/s  0.65 s   12  3.4 rad/s 2   0.65 s 
a 1.12 m/s 2

 3.4 rad/s 2
r
0.33 m
 35 rad  5.5 rev
2
57.
Picture the Problem: The fan blade rotates about its axis with a constant angular speed.
Strategy: Use equation 10-17 for the kinetic energy of a rotating object to solve for the moment of inertia.
I
Solution: Solve equation 10-17 for I:
60.
2K

2

2  4.6 J 
13 rad/s 
2
 0.054 kg  m 2
Picture the Problem: The compact disk rotates about its central axis at a constant angular speed.
Strategy: First find the moment of inertia of the CD, modeling it as a uniform disk. Then find the kinetic energy using equation
10-17. Finally, use a ratio to find the new angular speed that would double the rotational kinetic energy.
Solution: 1. (a) Apply equation 10-17
directly, using I  12 MR 2 for the CD:
2. (b) Use a ratio to find the new  :
K  12 I  2 

1
4
1
2

1
2
MR 2   2
 0.012 kg  0.060 m  34 rad/s 
2
2
 0.012 J = 12 mJ
2 K new I
new
K new
2 K old



 2
old
K old
K old
2 K old I
new  2old  2  34 rad/s   48 rad/s
75.
Picture the Problem: The cylinder rolls down the ramp without
slipping, gaining both translational and rotational kinetic energy.
Strategy: Use conservation of energy to find total kinetic energy at
the bottom of the ramp. Then set that energy equal to the sum of
the rotational and translational energies. Because the cylinder rolls
without slipping, the equation   v r can be used to write the
expression in terms of linear velocity alone. Use the resulting
equation to find expressions for the fraction of the total energy that
is rotational and translational kinetic energy.
Solution: 1. (a) Set Ei  Ef and solve for K f :
U i  Ki  U f  K f
mgh  0  0  K f
K f  mgh   2.0 kg   9.81 m/s 2   0.75 m 
 14.7 J  15 J
2. (b) Set K f equal to K t  K r :
K f  12 mv 2  12 I  2  12 mv 2  12  12 mr 2   v r 
2
 12 mv 2  14 mv 2  34 mv 2
3. Determine K r from steps 1 and 2:
Kr  14 mv 2  13  34 mv 2   13 Kf  13 14.7 J   4.9 J
4. (c) Determine K t from steps 1 and 2:
K t  12 mv2 
2
3

3
4
mv 2   32 Kf 
2
3
14.7 J  
9.8 J