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.
____________________________________________________________________________________________
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.
____________________________________________________________________________________________
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, F
N
, exerted on the base by the table.
CG
mg
N
F
N
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 9.10
c.
___________________________________________________________________________________________
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 2 TR . Since there is a net torque on the probe, there will be an angular acceleration. The angular velocity of the probe will increase.
____________________________________________________________________________________________
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.
____________________________________________________________________________________________
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
I

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

1 I
2

2 
1
2
Mv
2
Since the objects roll down the incline,

= v / R . For the hoop, I

MR
2
, hence,
Mgh

1
2
( MR
2
) v
2
R 2

1
2
Mv
2
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 v

1 15
(hoop) gh (solid cylinder) v

1 10 v

1 20 gh gh
(spherical shell)
(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.
____________________________________________________________________________________________
5.
SSM
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.
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

Fr truck
Fr car
 r truck r car

0.25 m
0.19 m
 1.3
_____________________________________________________________________________________________
7. REASONING AND SOLUTION The torque due to F
1
is

1
=

F
1
L
1
=

(20.0 N)(0.500 m) =

10.0 N
 m (CW)
The torque due to F
2
is

2
= (F
2
cos 30.0°)L
2
= (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
 
____________________________________________________________________________________________
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/s
2
)(0.320 m)

(1.00 kg)(9.80 m/s
2
)(0.180 m)

(0.200 kg)(9.80 m/s
2
)(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/s
2
)(0.280 m)

(1.00 kg)(9.80 m/s
2
)(0.140 m)

(0.200 kg)(9.80 m/s
2
)(0.100 m) = 0
T  56.4 N, down
_____________________________________________________________________________________________
23. SSM 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.
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
 10
3
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
 10
3
N . Since the forces must add to give a net force of zero, we know that the direction of F is downward
.
_____________________________________________________________________________________________
31. REASONING AND SOLUTION a. The net torque on the disk about the axle is

= F
1
R

F
2
R = (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) MR
2
= (1/2)(24.3 kg)(0.314 m)
2
= 1.20 kg
 m
2

= (

N
 m)/(1.20 kg
 m
2
) = –9.2 rad / s 2
_____________________________________________________________________________________________

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/s
2
. 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
_____________________________________________________________________________________________
39.
SSM
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.
SOLUTION The angular acceleration of the wheel is, according to Equation 8.4,
 
   t
0 
3.7 rad/s  13.1 rad/s
3.0 s
  3.1 rad/s
2
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  mr
2
. The net torque that acts on the wheel due to the brake pads is, therefore,
 I   ( mr
2
) 
(1)
From Equation 9.1, the net torque that acts on the wheel due to the action of the two brake pads is
 –2 f k (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
F
N (3)

where
 k
is the coefficient of kinetic friction and F
N
is the magnitude of the normal force applied to the wheel by each brake pad.
Combining Equations (1), (2), and (3) gives
–2(  k
F
N
)  ( mr
2
) 
F
N

– mr 2 
2  k

–(1.3 kg)(0.33 m) 2 (–3.1 rad/s 2
2(0.85)(0.33 m)
)
 0.78 N
_____________________________________________________________________________________________
43.
SSM
REASONING The kinetic energy of the flywheel is given by Equation 9.9. The converted to rev/min.
SOLUTION Solving Equation 9.9 for

, we obtain, 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

1
2
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
 
2 ( KE
R
)
I

2 ( KE
R
1
2
MR 2
)

9
2

6.4

10 rad / s
Converting this answer into rev/min, we find that
  c
6.4

10 rad / s h
F
H
1 rev
I
K
G IK  
_____________________________________________________________________________________________
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: v
T1
= (2.00 m)(6.00 rad/s) = 12.0 m / s
For object 2: v
T2
= (1.50 m)(6.00 rad/s) = 9.00 m / s
For object 3: v
T3
= (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

1 m v
2 1 1
2 
1 m v
2 2 2
2 
1
2 m v
2
3 3

KE

1
2
( .
kg)(12.0 m / s)
2 
kg)(9.00 m / s)
2 
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  m r
 m r
 m r
1 1
2
2 2
2
3 3
2
I

kg)(2.00 m) 2 
kg)(1.50 m) 2 
kg)(3.00 m) 2  2 d. The rotational kinetic energy of the system is, according to Equation 9.9,
KE
R

1
2
I

2 
1
2
This agrees, as it should, with the result for part (b).
2 
1.08

10
3
J
_____________________________________________________________________________________________
53. SSM WWW REASONING Let the two disks constitute the system. Since there are no external torques acting on the system, the principle of conservation of angular momentum applies. Therefore we have L initial

L final
, 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
I
B

I
A
F

H final
B
–
–


A final
I
J   2
)
M
–2.4 rad / s – 7.2 rad / s
–9.8 rad / s – (–2.4 rad / s)
O
Q
 2
_____________________________________________________________________________________________
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 v
0
and v f
.
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 I
0
= 2 M ( 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 I f
= 2 M ( L /4)
2
. The conservation of angular momentum indicates that
I f
 f

I


Final angular momentum
Initial angular momentum
L
2 M M
F
H
L
4
J
2
O

Q f

L
2 M M
F
H
L
2
J
2
O

Q 0
 f

4

0
According to Equation 8.9,
 f
= v f
/( L /4) and

0 result that
 f
= 4

0
becomes v
F v
L / f
4

4
/
0
2
J
or v f

2 v
0

= v
0
/( L /2). With these substitutions, the b
m / s g

34 m / s
_____________________________________________________________________________________________