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CHAPTER 10: Rotational Motion
Solutions to Assigned Problems
5.
 2500 rev  2 rad  1min 


  261.8 rad sec  260 rad sec
 1min  1rev  60s 
(a)   
(b) v   r   261.8 rad sec  0.175 m   46 m s
aR   2 r   261.8 rad sec   0.175 m   1.2  104 m s 2
2
15. (a) The direction of 1 is along the axle of the wheel, to the left. That is the iˆ direction. The
direction of 2 is also along its axis of rotation, so it is straight up. That is the kˆ direction.
That is also the angular velocity of the axis of the wheel.
(b) At the instant shown in the textbook, we have the vector relationship
z
as shown in the diagram.
 44.0 rad s 2   35.0 rad s 2
  12  22 
  tan 1
 56.2 rad s
2
35.0
 tan 1
 38.5
2
44.0
(c) Angular acceleration is given by α 
35.0kˆ rad s , α 
dω1
dt
dω
dt

2

x
1
. Since ω  ω1  ω2 , and ω 2 is a constant
. ω1 is rotating counterclockwise about the z axis with the angular
velocity of 2 , and so if the figure is at t = 0, then ω1  1   cos 2tˆi  sin 2tˆj .
α
dω
dt

d  ω1  ω2 
dt

dω1
dt



d 1  cos 2tˆi  sin 2tˆj 
dt

 12 sin 2tˆi  cos 2tˆj

 
α  t  0   12  ˆj    44.0 rad s  35.0 rad s  ˆj  1540 rad s 2 ˆj
  0   t  t 
  0
6.807 rad s

 9.9 s

0.6857 rad s 2
25. Each force is oriented so that it is perpendicular to its lever arm. Call counterclockwise torques
positive. The torque due to the three applied forces is given by the following.
 applied   28 N  0.24 m   18 N  0.24 m    35 N  0.12 m   1.8 m N
forces
Since this torque is clockwise, we assume the wheel is rotating clockwise, and so the frictional
torque is counterclockwise. Thus the net torque is as follows.
 net   28 N  0.24 m   18 N  0.24 m    35 N  0.12 m   0.40 m N  1.4 m N
 1.4 m N , clockwise
30. For each torque, use Eq. 10-10c. Take counterclockwise torques to be positive.
(a) Each force has a lever arm of 1.0 m.
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304
Physics for Scientists & Engineers, 4th Edition
Giancoli
 about   1.0 m  56 N  sin 30  1.0 m  52 N  sin 60  17m N
C
(b) The force at C has a lever arm of 1.0 m, and the force at the top has a lever arm of 2.0 m.
 about    2.0 m  56 N  sin 30  1.0 m  65 N  sin 45  10 m N (2 sig fig)
P
The negative sign indicates a clockwise torque.
40. (a) To calculate the moment of inertia about the y axis (vertical), use the following.
2
2
2
2
I   M i Rix2  m  0.50 m   M  0.50 m   m 1.00 m   M 1.00 m 
  m  M   0.50 m   1.00 m     5.3kg   0.50 m   1.00 m    6.6 kg m 2
2
2
2
2
(b) To calculate the moment of inertia about the x-axis (horizontal), use the following.
I   M i Riy2   2m  2 M  0.25m   0.66 kg m2
2
(c) Because of the larger I value, it is ten times harder to accelerate the array about
the vertical axis .
46. (a) The free body diagrams are shown. Note that only the forces producing
torque are shown on the pulley. There would also be a gravity force on
the pulley (since it has mass) and a normal force from the pulley’s
suspension, but they are not shown.
(b) Write Newton’s second law for the two blocks, taking the positive x
direction as shown in the free body diagrams.
mA :  Fx  FTA  mA g sin  A  mA a 
FTA  mA  g sin  A  a 

FTA
FNA
y
x
A
A
mAg
FTA
FTB

  8.0kg   9.80 m s 2 sin 32  1.00 m s 2   49.55 N
 50 N
mB :
F
x
 2 sig fig 
FNB
FTB
 mB g sin  B  FTB  mBa 
FTB  mB  g sin  B  a 

y
x

 10.0kg   9.80 m s sin 61  1.00 m s   75.71N
2
2
B
B
m Bg
 76 N
(c) The net torque on the pulley is caused by the two tensions. We take clockwise torques as
positive.
   F
TB
 FTB  R   75.71N  49.55 N  0.15m   3.924m N  3.9m N
Use Newton’s second law to find the rotational inertia of the pulley. The tangential acceleration
of the pulley’s rim is the same as the linear acceleration of the blocks, assuming that the string
doesn’t slip.
a
  I  I R   FTB  FTB  R 
I
 FTB  FTB  R 2
a

 75.71N  49.55 N  0.15m 2
1.00 m s
2
 0.59 kg m 2
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305
Chapter 10
Rotational Motion
I  Mk 2  MR02  k  R0
49. (a) Thin hoop, radius R0
(b) Thin hoop, radius R0 , width w
I  Mk 2  12 MR02  121 Mw2  k 
(c) Solid cylinder
I  Mk 2  12 MR02  k 
(d) Hollow cylinder
I  Mk 2  12 M R 2  R 2
1
2
(e) Uniform sphere
I  Mk 2  25 Mr02  k 
(f)
I  Mk 2  121 M l

Long rod, through center
(g) Long rod, through end
I  Mk 2  13 M l
(h) Rectangular thin plate
I  Mk 2 
1
12

2
 w2

1
2
R12  R22 
2
5 r0
1
12
 k
2
R02  121 w2
R0
 k
 k
2
M l

1
2
1
2
1
3
l
l
 k
1
12
l
2
 w2

57. (a) The parallel axis theorem (Eq. 10-17) is to be applied to each sphere. The distance from the
center of mass of each sphere to the axis of rotation is h  1.5r0 .
I for one  I CM  Mh 2  25 Mr02  M 1.5r0   2.65Mr02  I total  5.3Mr02
2
sphere
(b) Treating each mass as a point mass, the point mass would be a distance of 1.5r0 from the axis of
rotation.
I approx  2  M 1.5r0    4.5Mr02
2
 I approx  I exact 
 4.5Mr02  5.3Mr02 
4.5  5.3 
100



100   
100 

2


I exact
5.3Mr0
 5.3 




% error  
 15%
The negative sign means that the approximation is smaller than the exact value, by about 15%.
67. The only force doing work in this system is gravity, so mechanical energy is conserved. The initial
state of the system is the configuration with mA on the ground and all objects at rest. The final state
of the system has mB just reaching the ground, and all objects in motion. Call the zero level of
gravitational potential energy to be the ground level. Both masses will have the same speed since
they are connected by the rope. Assuming that the rope does not slip on the
pulley, the angular speed of the pulley is related to the speed of the masses
by   v R . All objects have an initial speed of 0.
Ei  E f
1
2
M

R
mA vi2  12 mBvi2  12 I i2  mA gy1i  mB gy2 i  12 mA v 2f  12 mBv 2f  12 I  2f
 mA gy1 f  mB gy2 f
mB gh  12 mA v 2f  12 mBv 2f  12

1
2
 v 2f 
 R2   mA gh
 
MR 2 
mB
mA
h
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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Physics for Scientists & Engineers, 4th Edition
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vf 
2  mB  mA  gh
 mA  mB 
1
2

2  38.0 kg  35.0 kg  9.80 m s 2

M
  2.5 m   1.4 m s
 38.0 kg  35.0 kg    3.1 kg 
1
2
75. Use conservation of mechanical energy to equate the energy at points
A and B . Call the zero level for gravitational potential energy to be
the lowest point on which the ball rolls. Since the ball rolls without
slipping,   v r0 .
EA  EB  U A  U B  K B final  U B  K B CM  K B rot 
mgR0  mgr0  12 mvB2  12 I B2
 mgr0  mv 
1
2
2
B
1
2

2
5
v 
mr   B 
 r0 
F
Mg R 2   R  h 
 vB 
10
7
R2   R  h  0
2
2

Rh
R0
C
y=0
B
2
2
0
g  R0  r0 
93. The wheel is rolling about the point of contact with the step, and so
all torques are to be taken about that point. As soon as the wheel is
off the floor, there will be only two forces that can exert torques on
the wheel – the pulling force and the force of gravity. There will
not be a normal force of contact between the wheel and the floor
once the wheel is off the floor, and any force on the wheel from the
point of the step cannot exert a torque about that very point.
Calculate the net torque on the wheel, with clockwise torques
positive. The minimum force occurs when the net torque is 0.
  F  R  h   mg
A
F
Rh
R
h
mg
R2   R  h 
2
Mg 2 Rh  h 2
Rh
100. (a) The disk starts from rest, and so the velocity of the center of mass is in the direction of the net
t
force: v  v 0  at  v  Fnet . Thus the center of mass moves to the right.
m
(b) For the linear motion of the center of mass, we may apply constant acceleration equations,
F
where the acceleration is .
m
F
 35 N 
5.5m   4.282 m s  4.3m s
m
 21.0 kg 
(c) The only torque is a constant torque caused by the constant string tension. That can be used to
find the angular velocity.
   0   I  Fr    Frt  Frt  2 Ft
  I  I 

1
I
mr 2
mr
 t  t
2
The time can be found from the center of mass motion under constant acceleration.
v 2  v02  2ax  v  2
x  v0t  12 at 2 
1
2
F
m
x  2
t2  t 
2mx
F
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307
Chapter 10
Rotational Motion

2 Ft
mr

2F
2mx
mr
F

2
2 F x
r
m

2
 0.850 m 
2  35.0 N  5.5 m 
 21.0 kg 
 2 sig fig 
 10.07 rad s  10 rad s
Note that v  r since the disk is NOT rolling without slipping.
(d) The amount of string that has unwrapped is related to the angle through which the disk has
turned, by the definition of radian measure, s  r  . The angular displacement is found from
constant acceleration relationships.
2mx
F
2
2
Ft
Ft

t 
F  2 x
  12 0    t  12 t  12 


mr
mr
r
 mr 
s  r  r
2 x
r
 2 x  11m
103. Since there is no friction at the table, there are no horizontal forces on the rod, and so the center of
mass will fall straight down. The moment of inertia of the rod about its center of mass is 121 M l 2 .
Since there are no dissipative forces, energy will be conserved during the fall. Take the zero level of
gravitational potential energy to be at the tabletop. The angular velocity and the center of mass
v
velocity are related by CM  1CM .
2 l 
Einitial  Efinal  U release  K final  Mg  12 l
o
Mg  12 l
2
 12  121 M l
  12 MvCM
2


  v l  

CM
1
2
2
2
 12 I CM
  12 MvCM

2

2
 gl  43 vCM
 vCM 
3
4
gl
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
308