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CHAPTER
9 ROTATIONAL DYNAMICS
CONCEPTUAL QUESTIONS
5.
REASONING AND SOLUTION Work and torque are both the product of force and
distance. Work and torque are distinctly different physical quantities, as is evident by
considering the distances in the definitions. Work is defined by W  (F cos  )s , according
to Equation 6.1, where F is the magnitude of the force,  is the angle between the force and
the displacement, and s is the magnitude of the displacement. The magnitude of the torque
is defined as the magnitude of the force times the lever arm, according to Equation 9.1. In
the definition of work, the "distance" is the magnitude of the displacement over which the
force acts. In the definition of torque, the distance is the lever arm, a "static" distance. The
lever arm is not the same physical quantity as the displacement. Therefore, work and torque
are different quantities.
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6.
REASONING AND SOLUTION A person stands on a train, both feet together, facing a
window. The front of the train is to the person's left. When the train starts to move forward,
a force of static friction is applied to the person's feet. This force tends to produce a torque
about the center of gravity of the person, causing the person to fall backward. If the right
foot is slid out toward the rear of the train, the normal force of the floor on the right foot
produces a counter torque about the center of gravity of the person. The resultant of these
two torques is zero, and the person can maintain his balance.
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10. REASONING AND SOLUTION
Treating the wine rack and the bottle as a
rigid body, the two external forces that
keep it in equilibrium are its weight, mg,
located at the center of gravity, and the
normal force, FN, exerted on the base by
the table.
CG
mg
N
FN
Note that the weight acts at the center of
gravity (CG), which must be located
exactly above the point where the normal
force acts, as is the case in Figure 9.10c.
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13. REASONING AND SOLUTION The space probe is initially moving with a constant
translational velocity and zero angular velocity through outer space.
a. When the two engines are fired, each generates a thrust T in opposite directions; hence,
the net force on the space probe is zero. Since the net force on the probe is zero, there is no
translational acceleration and the translational velocity of the probe remains the same.
b. The thrust of each engine produces a torque about the center of the space probe. If R is
the radius of the probe, the thrust of each engine produces a torque of magnitude TR, each in
the same direction, so that the net torque on the probe has magnitude 2TR. Since there is a
net torque on the probe, there will be an angular acceleration. The angular velocity of the
probe will increase.
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14. REASONING AND SOLUTION When you are lying down with your hands behind your
head, your moment of inertia I about your stomach is greater than it is when your hands are
on your stomach. Sit-ups are then more difficult because, with a greater I, you must exert a
greater torque about your stomach to give your upper torso the same angular acceleration,
according to Equation 9.7.
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18. REASONING AND SOLUTION Let each object have mass M and radius R. The objects
are initially at rest. In rolling down the incline, the objects lose gravitational potential
energy Mgh , where h is the height of the incline. The objects gain kinetic energy equal to
(1 / 2 ) I 2  (1 / 2 ) Mv 2 , where I is the moment of inertia about the center of the object, v is
the linear speed of the center of mass at the bottom of the incline, and  is the angular speed
about the center of mass at the bottom. From conservation of mechanical energy
Mgh  21 I 2  21 Mv 2
Since the objects roll down the incline,  = v/R. For the hoop, I  MR 2 , hence,
Mgh  21 ( MR 2 )
v2 1
 2 Mv 2
2
R
Solving for v gives v  gh for the hoop. Repeating the calculation for the solid cylinder,
the spherical shell, and the solid sphere gives
v
gh
(hoop)
v  1.15 gh
(solid cylinder)
v  1.10 gh
(spherical shell)
v  1.20 gh
(solid sphere)
The solid sphere is faster at the bottom than any of the other objects. The solid cylinder is
faster at the bottom than the spherical shell or the hoop. The spherical shell is faster at the
bottom than the hoop. Thus, the solid sphere reaches the bottom first, followed by the solid
cylinder, the spherical shell, and the hoop, in that order.
23. REASONING AND SOLUTION A person is hanging motionless from a vertical rope over
a swimming pool. She lets go of the rope and drops straight down. Consider the woman to
be the system. When she lets go of the rope her angular momentum is zero. It is possible
for the woman to curl into a ball; however, since her angular momentum is zero, she cannot
change her angular velocity by exerting any internal torques. In order to begin spinning, an
external torque must be applied to the woman. Therefore, it is not possible for the woman to
curl into a ball and start spinning.
PROBLEMS
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5.
REASONING The torque on either wheel is given by Equation 9.1,   F , where
F is the magnitude of the force and is the lever arm. Regardless of how the force is
applied, the lever arm will be proportional to the radius of the wheel.
SSM
SOLUTION The ratio of the torque produced by the force in the truck to the torque
produced in the car is
 truck
 car

F
F
truck
car

Frtruck
Frcar

rtruck
rcar

0.25 m
 1.3
0.19 m
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7.
REASONING AND SOLUTION The torque due to F1 is
1 = F1L1 = (20.0 N)(0.500 m) = 10.0 Nm (CW)
The torque due to F2 is
2 = (F2 cos 30.0°)L2 = (35.0 N)(1.10 m) cos 30.0° = 33.3 Nm (CCW)
The net torque is therefore,
 = 1 + 2= 10.0 Nm + 33.3 Nm = 23.3 N  m, counterclockwise
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16. REASONING AND SOLUTION The net torque about an axis through the contact point
between the tray and the thumb is
 = F(0.0400 m)  (0.250 kg)(9.80 m/s2)(0.320 m)  (1.00 kg)(9.80 m/s2)(0.180 m)
 (0.200 kg)(9.80 m/s2)(0.140 m) = 0
F  70.6 N, up
Similarly, the net torque about an axis through the point of contact between the tray and the
finger is
 = T(0.0400 m)  (0.250 kg)(9.80 m/s2)(0.280 m)  (1.00 kg)(9.80 m/s2)(0.140 m)
 (0.200 kg)(9.80 m/s2)(0.100 m) = 0
T  56.4 N, down
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23.
REASONING Since the man holds the ball motionless, the ball and the arm are in
equilibrium. Therefore, the net force, as well as the net torque about any axis, must be zero.
SSM
SOLUTION
a. Using Equation 9.1, the net torque about an axis through the elbow joint is
 = M(0.0510 m) – (22.0 N)(0.140 m) – (178 N)(0.330 m) = 0
Solving this expression for M gives M  1.21 103 N .
b. The net torque about an axis through the center of gravity is
 = – (1210 N)(0.0890 m) + F(0.140 m) – (178 N)(0.190 m) = 0
Solving this expression for F gives F  1.01 103 N . Since the forces must add to give
a net force of zero, we know that the direction of F is downward .
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31. REASONING AND SOLUTION
a. The net torque on the disk about the axle is
 = F1R  F2R = (0.314 m)(90.0 N  125 N) = 11 N m
b. The angular acceleration is given by  = /I. From Table 9.1, the moment of inertia of
the disk is
I = (1/2) MR2 = (1/2)(24.3 kg)(0.314 m)2 = 1.20 kgm2
 = ( Nm)/(1.20 kgm2) =
–9.2 rad / s 2
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34. REASONING AND SOLUTION
a. We know that  = ( – o)/t. But  = (1/2)o, so
 = (1/2)o/t = (1/2)(88.0 rad/s)/(5.00 s) = 8.80 rad/s2
The magnitude of  is 8.80 rad/s2 .
b.  = I and FR = I. The magnitude of the frictional force is
F = I/R = (0.850 kgm2)(8.80 rad/s2)/(0.0750 m) =
99.7 N
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39.
REASONING The angular acceleration of the bicycle wheel can be calculated
from Equation 8.4. Once the angular acceleration is known, Equation 9.7 can be used to
find the net torque caused by the brake pads. The normal force can be calculated from the
torque using Equation 9.1.
SSM
SOLUTION The angular acceleration of the wheel is, according to Equation 8.4,

 0
t

3.7 rad/s 13.1 rad/s
2
 3.1 rad/s
3.0 s
If we assume that all the mass of the wheel is concentrated in the rim, we may treat the
wheel as a hollow cylinder. From Table 9.1, we know that the moment of inertia of a
hollow cylinder of mass m and radius r about an axis through its center is I  mr2 . The net
torque that acts on the wheel due to the brake pads is, therefore,
2
   I  (mr )
(1)
From Equation 9.1, the net torque that acts on the wheel due to the action of the two brake
pads is
   –2 fk
(2)
where f k is the kinetic frictional force applied to the wheel by each brake pad, and
 0.33 m is the lever arm between the axle of the wheel and the brake pad (see the
drawing in the text). The factor of 2 accounts for the fact that there are two brake pads. The
minus sign arises because the net torque must have the same sign as the angular acceleration.
The kinetic frictional force can be written as (see Equation 4.8)
f k  k FN
(3)
where k is the coefficient of kinetic friction and FN is the magnitude of the normal force
applied to the wheel by each brake pad.
Combining Equations (1), (2), and (3) gives
–2( k FN )  (mr2 )
– mr 2  –(1.3 kg)(0.33 m) 2 (–3.1 rad/s 2 )
FN 

 0.78 N
2k
2(0.85)(0.33 m)
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43.
REASONING The kinetic energy of the flywheel is given by Equation 9.9. The
moment of inertia of the flywheel is the same as that of a solid disk, and, according to Table
9.1 in the text, is given by I  21 MR 2 . Once the moment of inertia of the flywheel is known,
Equation 9.9 can be solved for the angular speed  in rad/s. This quantity can then be
converted to rev/min.
SSM
SOLUTION Solving Equation 9.9 for  , we obtain,

2 ( KE R )
I

2 ( KE R )
1
2
MR 2

4(1.2  10 9 J)
 6.4  10 4 rad / s
(13kg)(0.30m) 2
Converting this answer into rev/min, we find that
c
  6.4  10 4 rad / s
1 rev I F60 s I
G
hF
H2 rad J
KG
H1 min J
K
6.1  10 5 rev / min
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45.
SSM REASONING AND SOLUTION
a. The tangential speed of each object is given by Equation 8.9, v T  r . Therefore,
For object 1:
vT1 = (2.00 m)(6.00 rad/s) =
12.0 m / s
For object 2:
vT2 = (1.50 m)(6.00 rad/s) =
9.00 m / s
For object 3:
vT3 = (3.00 m)(6.00 rad/s) =
18.0 m / s
b. The total kinetic energy of this system can be calculated by computing the sum of the
kinetic energies of each object in the system. Therefore,
KE  21 m1 v12  21 m2 v 22  21 m3 v 32
KE 
1
2
( 6.00 kg)(12.0 m / s) 2  ( 4 .00 kg)(9.00 m / s) 2  ( 3.00 kg)(18.0 m / s) 2  1.08  10 3 J
c. The total moment of inertia of this system can be calculated by computing the sum of the
moments of inertia of each object in the system. Therefore,
I   mr 2  m1 r12  m2 r22  m3 r32
I  ( 6.00 kg)(2.00 m) 2  ( 4 .00 kg)(1.50 m) 2  ( 3.00 kg)(3.00 m) 2  60.0 kg  m 2
d. The rotational kinetic energy of the system is, according to Equation 9.9,
KE R  21 I 2  21 (60.0 kg  m 2 )(6.00 rad / s) 2  1.08  10 3 J
This agrees, as it should, with the result for part (b).
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53.
Let the two disks constitute the system. Since there
SSM WWW REASONING
are no external torques acting on the system, the principle of conservation of angular
momentum applies. Therefore we have Linitial  Lfinal , or
I A  A  I B B  ( I A  I B ) final
This expression can be solved for the moment of inertia of disk B.
SOLUTION Solving the above expression for I B , we obtain
IB  IA
F
G
H
final
B
– A
–  final
I  ( 3.4 kg  m ) L–2.4 rad / s – 7.2 rad / s O 4.4 kg  m
M
J
K
N–9.8 rad / s – (–2.4 rad / s) P
Q
2
2
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58. REASONING When the modules pull together, they do so by means of forces that are
internal. These pulling forces, therefore, do not create a net external torque, and the angular
momentum of the system is conserved. In other words, it remains constant. We will use the
conservation of angular momentum to obtain a relationship between the initial and final
angular speeds. Then, we will use Equation 8.9 (v = r) to relate the angular speeds 0 and
f to the tangential speeds v0 and vf.
SOLUTION Let L be the initial length of the cable between the modules and 0 be the
initial angular speed. Relative to the center-of-mass axis, the initial momentum of inertia of
the two-module system is I0 = 2M(L/2)2, according to Equation 9.6. After the modules pull
together, the length of the cable is L/2, the final angular speed is f, and the momentum of
inertia is If = 2M(L/4)2. The conservation of angular momentum indicates that
I f  f  I 0 0


Final angular
momentum
Initial angular
momentum
L
L2 M F
LI O
FL I O
2 M GJ P
 M

G
JKP
M
H
K
H
4
2
M
P
M
P
N
Q N
Q
2
2
f
0
 f  4 0
According to Equation 8.9, f = vf/(L/4) and 0 = v0/(L/2). With these substitutions, the
result that f = 40 becomes
vf
L/4
4
v I
F
G
HL / 2 J
K or
0
b
g
v f  2 v 0  2 17 m / s  34 m / s
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