CHAPTER 9 ROTATIONAL DYNAMICS
PROBLEMS
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1.
SSM
REASONING AND SOLUTION Solving Equation 9.1 for the lever arm l , we obtain
τ 3.0 N ⋅ m
= 0.25 m
l= =
F
12 N
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2.
REASONING The torque is given by Equation 9.1, τ = F l , where F is the magnitude of the
applied force and l is the lever arm. From the figure in the text, the lever arm is given by
l = ( 0.28 m) sin 50.0 ° . Since both τ and l are known, Equation 9.1 can be solved for F.
SOLUTION Solving Equation 9.1 for F, we have
F=
τ
45 N ⋅ m
=
= 2.1 ×10
l (0.28 m) sin 50.0 °
2
N
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3.
REASONING AND SOLUTION
where L is the lever arm.
The torque about the center of the cable car is τ = FL,
τ = FL = 2(185 N)(4.60 m) = 1.70 × 10 3 N ⋅ m
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4.
REASONING To calculate the torques, we need to determine the lever arms for each of the
forces. These lever arms are shown in the following drawings:
Axis
2.50 m
Axis
32.0°
32.0°
2.50 m
lT = (2.50 m) cos 32.0°
W
lW = (2.50 m) sin 32.0°
T
SOLUTION
a. Using Equation 9.1, we find that the magnitude of the torque due to the weight W is
320
ROTATIONAL DYNAMICS
b
gb
g
τ W = Wl W = 10 200 N 2 .5 m sin32 ° = 13 500 N ⋅ m
b. Using Equation 9.1, we find that the magnitude of the torque due to the thrust T is
b
gb
g
τ T = Tl T = 62 300 N 2 .5 m cos 32 ° = 132 000 N ⋅ m
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5.
REASONING The torque on either wheel is given by Equation 9.1, τ = F l, where F is
the magnitude of the force and l 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
τtruck
τ car
=
F l truck
F l car
=
Fr truck
Fr car
=
rtruck
r car
=
0.25 m
0.19 m
= 1.3
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6.
REASONING To calculate the torque,
we will need the lever arm of the 110 N
force. The drawing identifies the lever
arm l.
15°
3.0 m
W =110 N
SOLUTION According to Equation
9.1, the magnitude of the torque is
b
gb
l = (3.0 m)cos 15°
g
τ = Wl = 110 N 3.0 m cos 15°= 320 N ⋅ m
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7.
REASONING AND SOLUTION The torque due to F1 is
τ1 = −F1L1 = −(20.0 N)(0.500 m) = −10.0 N⋅m (CW)
The torque due to F2 is
τ2 = (F2 cos 30.0°)L2 = (35.0 N)(1.10 m) cos 30.0° = 33.3 N⋅m (CCW)
The net torque is therefore,
Chapter 9 Problems
321
Στ = τ1 + τ2 = −10.0 N⋅m + 33.3 N⋅m = 23.3 N ⋅ m, counterclockwise
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8.
REASONING AND SOLUTION The torque produced by a force of magnitude F is given by
Equation 9.1, τ = F l, where l is the lever arm. In each case, the torque produced by the couple
is equal to the sum of the individual torques produced by each member of the couple.
a. When the axis passes through point A, the torque due to the force at A is zero. The lever arm
for the force at C is L. Therefore, taking counterclockwise as the positive direction, we have
τ = τ A + τC = 0 + τC = FL
b. Each force produces a counterclockwise rotation. The magnitude of each force is F and each
force has a lever arm of L / 2 . Taking counterclockwise as the positive direction, we have
L
L
τ = τ A + τC = F   + F   = FL
2
2
c. When the axis passes through point C, the torque due to the force at C is zero. The lever arm
for the force at A is L. Therefore, taking counterclockwise as the positive direction, we have
τ = τ A + τC = τA + 0 = FL
Note that the value of the torque produced by the couple is the same in all three cases; in other
words, when the couple acts on the tire wrench, the couple produces a torque that does not
depend on the location of the axis.
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9.
REASONING Since the meter stick does not move, it is in equilibrium. The forces and
the torques, therefore, each add to zero. We can determine the location of the 6.00 N force, by
using the condition that the sum of the torques must vectorially add to zero.
SSM
F
pinned
end
2
30.0°
l2
l1
F
TOP VIEW
1
SOLUTION If we take counterclockwise torques as positive, then the torque of the first force
about the pin is
τ 1 = F1 l 1 = (2.00 N)(1.00 m) = 2.00 N ⋅ m
322
ROTATIONAL DYNAMICS
The torque due to the second force is
τ 2 = – F2 (sin 30.0 ° ) l 2 = – (6.00 N)(sin 30.0 ° ) l 2 = –( 3.00 N) l 2
The torque τ 2 is negative because the force F2 tends to produce a clockwise rotation about the
pinned end. Since the net torque is zero, we have
2 .00 N ⋅ m +[– ( 3.00 N) l 2 ] = 0
Thus,
l2 =
2.00 N ⋅ m
= 0.667 m
3.00 N
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10. REASONING AND SOLUTION The net torque about the axis in text drawing (a) is
Στ = τ1 + τ2 = F1b − F2a = 0.
Considering that F2 = 3F1, we have b – 3a = 0. The net torque in drawing (b) is then
Στ = F1(1.00 m − a) − F2b = 0
or
1.00 m − a − 3b = 0.
Solving the first equation for b, substituting into the second equation and rearranging, gives
a=
0.100 m
0.300 m
and b =
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11. REASONING AND SOLUTION
a. Taking torques about an axis through the rear axle of the car gives
2Ff(2.54 m) − (1160 kg)(9.80 m/s2)(1.52 m) = 0
or
Ff =
3.40 × 10 3 N
Fr =
2.28 × 10 N
b. Taking torques about an axis through the front axle gives
(1160 kg)(9.80 m/s2)(1.02 m) − 2Fr(2.54 m) = 0
or
3
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12. REASONING The drawing
shows the forces acting on the
person. It also shows the lever
arms for a rotational axis
FHANDS
FFEET
Axis
W
Chapter 9 Problems
323
perpendicular to the plane of the
paper at the place where the
person’s toes touch the floor.
Since the person is in equilibrium,
the sum of the forces must be
zero. Likewise, we know that the
sum of the torques must be zero.
SOLUTION Taking upward to be the positive direction, we have
FFEET + FHANDS − W = 0
Remembering that counterclockwise torques are positive and using the axis and the lever arms
shown in the drawing, we find
Wl W − FHANDS l HANDS = 0
FHANDS =
Wl W
l HANDS
=
b584 N gb0 .840 mg= 392 N
1.250 m
Substituting this value into the balance-of-forces equation, we find
FFEET = W − FHANDS = 584 N − 392 N = 192 N
The force on each hand is half the value calculated above, or 196 N . Likewise, the force on
each foot is half the value calculated above, or 96 N .
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13.
REASONING The minimum value for the coefficient of static friction between the ladder
and the ground, so that the ladder does not slip, is given by Equation 4.7:
SSM
f s MAX = µs FN
SOLUTION From Example 4, the magnitude of the force of static friction is Gx = 727 N. The
magnitude of the normal force applied to the ladder by the ground is Gy = 1230 N. The minimum
value for the coefficient of static friction between the ladder and the ground is
µs =
f sMAX
FN
=
Gx
Gy
=
727 N
= 0 . 591
1230 N
324
ROTATIONAL DYNAMICS
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14. REASONING When the board just begins to tip, three forces act on the board. They are the
weight W of the board, the weight W P of the person, and the force F exerted by the right support.
F
1.4 m
x
W = 225 N
W P = 450 N
Since the board will rotate around the right support, the lever arm for this force is zero, and the
torque exerted by the right support is zero. The lever arm for the weight of the board is equal to
one-half the length of the board minus the overhang length: 2.5 m – 1.1 m = 1.4 m .
The lever arm for the weight of the person is x. Therefore, taking counterclockwise torques as
positive, we have
− W P x + W (1.4 m) = 0
This expression can be solved for x.
SOLUTION Solving the expression above for x, we obtain
x=
W (1.4 m)
WP
=
(225 N)(1.4 m)
450 N
= 0.70 m
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Chapter 9 Problems
15.
REASONING The figure at the right shows the door
and the forces that act upon it. Since the door is uniform, the
center of gravity, and, thus, the location of the weight W, is at
the geometric center of the door.
325
SSM
Let U represent the force applied to the door by the upper
hinge, and L the force applied to the door by the lower hinge.
Taking forces that point to the right and forces that point up as
positive, we have
∑ Fx = L x − U x = 0
∑ Fy = Ly − W = 0
or L x = U x
or L y = W = 140 N
U
x
h = 2.1 m
W
L
y
Lx
(1)
d = 0.81 m
(2)
Taking torques about an axis perpendicular to the plane of the door and through the lower hinge,
with counterclockwise torques being positive, gives
d
∑ τ = −W   + U x h = 0
(3)
2
Equations (1), (2), and (3) can be used to find the force applied to the door by the hinges;
Newton's third law can then be used to find the force applied by the door to the hinges.
SOLUTION
a. Solving Equation (3) for Ux gives
Ux =
Wd
2h
=
(140 N)(0.81 m)
2(2.1 m)
= 27 N
From Equation (1), Lx = 27 N.
Therefore, the upper hinge exerts on the door a horizontal force of 27 N to the left
.
b. From Equation (1), Lx = 27 N, we conclude that the bottom hinge exerts on the door a
horizontal force of 27 N to the right .
c. From Newton's third law, the door applies a force of 27 N
force points horizontally to the right
on the upper hinge. This
, according to Newton's third law.
326
ROTATIONAL DYNAMICS
d. Let D represent the force applied by the door to the lower hinge. Then, from Newton's third
law, the force exerted on the lower hinge by the door has components Dx = – 27 N and
Dy = – 140 N.
From the Pythagorean theorem,
D
D=
D +D =
2
x
2
y
( −27 N) + ( −140 N) =
2
2
x
143 N
θ
The angle θ is given by
θ = tan −1
D
F–140 N I = 79 °
G
H–27 N JK
The force is directed 79 ° below the horizontal
y
.
16. REASONING AND SOLUTION The net torque about an axis through the contact point
between the tray and the thumb is
Στ = 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
Στ = 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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17. REASONING AND SOLUTION When the arm is in equilibrium, the sum of the torques is
zero. Taking torques about the elbow joint with counterclockwise torques taken as positive, we
have
− F lF + M lM = 0
Solving for M,
 lF 
 0.30 m 
M=F 
= 1.0 × 10 3 N
 = (170 N) 
0.051 m 
 lM 
The direction of the force is to the left .
Chapter 9 Problems
327
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18. REASONING Although this arrangement of body parts is vertical, we can apply Equation 9.3 to
locate the overall center of gravity by simply replacing the horizontal position x by the vertical
position y, as measured relative to the floor.
SOLUTION Using Equation 9.3, we have
y cg =
=
W1 y1 + W2 y 2 + W3 y 3
W1 + W2 + W3
b438 N gb1. 28 mg+ b144 N gb0.760 m g+ b87 N gb0.250 m g=
438 N + 144 N + 87 N
1.03 m
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19. REASONING
The jet is in equilibrium, so the sum of the external forces is zero, and the sum of the external
torques is zero. We can use these two conditions to evaluate the forces exerted on the wheels.
SOLUTION
a. Let Ff be the magnitude of the normal force that the ground exerts on the front wheel. Since the
net torque acting on the plane is zero, we have (using an axis through the points of contact between
the rear wheels and the ground)
Στ = −W lw + Ff lf = 0
where W is the weight of the plane, and lw and lf are the lever arms for the forces W and Ff ,
respectively. Thus,
Στ = −(1.00 × 106 N)(15.0 m − 12.6 m) + Ff (15.0 m) = 0
Solving for Ff gives Ff = 1.60 × 10 5 N .
b. Setting the sum of the vertical forces equal to zero yields
ΣFy = Ff + 2Fr − W = 0
where the factor of 2 arises because there are two rear wheels. Substituting in the data,
ΣFy = 1.60 × 105 N + 2Fr − 1.00 × 106 N = 0
328
ROTATIONAL DYNAMICS
Fr =
4.20 × 10
5
N
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20. REASONING AND SOLUTION
a. Taking torques about an axis through the bolt we have
−(4.5 × 103 N)(2.50 m) − (12.0 × 103 N)(3.50 m) + T(5.00 m) sin 25.0° = 0
Solving for T yields
T=
4
2.52 × 10
N
b. Applying Newton's second law in the horizontal direction gives −T cos 25.0° + Fh = 0, so
Fh = 2.28 × 104 N
In the vertical direction Fv − WL − WB + T sin 25.0° = 0, and
Fv = 5.85 × 103 N
Now the force exerted on the beam by the bolt is
F =
Fh2 + F v2 = 2.35 × 10
4
N
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21.
REASONING AND SOLUTION The figure below shows the massless board
and the forces that act on the board due to the person and the scales.
SSM
WWW
2.00 m
person's feet
person's head
x
center of gravity
of person
425 N
315 N
W
a. Applying Newton's second law to the vertical direction gives
315 N + 425 N – W = 0
or
W = 7.40
× 10 2 N, downward
Chapter 9 Problems
329
b. Let x be the position of the center of gravity relative to the scale at the person's head. Taking
torques about an axis through the contact point between the scale at the person's head and the
board gives
(315 N)(2.00 m) – (7.40 × 102 N)x = 0
x = 0.851 m
or
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22. REASONING AND SOLUTION The net torque about an axis through the knee joint is
Στ = M(sin 25.0°)(0.100 m) − (44.5 N)(cos 30.0°)(0.250 m) = 0
M = 228 N
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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
Στ = 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
Στ = – (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
.
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24. REASONING If we assume that the system is in equilibrium, we know that the vector sum of all
the forces, as well as the vector sum of all the torques, that act on the system must be zero.
The figure below shows a free body diagram for the boom. Since the boom is assumed to be
uniform, its weight W B is located at its center of gravity, which coincides with its geometrical
center. There is a tension T in the cable that acts at an angle θ to the horizontal, as shown. At the
hinge pin P, there are two forces acting. The vertical force V that acts on the end of the boom
330
ROTATIONAL DYNAMICS
prevents the boom from falling down. The horizontal force H that also acts at the hinge pin
prevents the boom from sliding to the left. The weight W L of the wrecking ball (the "load") acts at
the end of the boom.
θ=
32°
T
φ– θ
WB
V
θ=
32°
WL
P
φ= 48°
H
By applying the equilibrium conditions to the boom, we can determine the desired forces.
SOLUTION The directions upward and to the right will be taken as the positive directions. In
the x direction we have
∑ Fx = H − T cos θ = 0
(1)
while in the y direction we have
∑ Fy = V − T sin θ − WL − WB = 0
(2)
Equations (1) and (2) give us two equations in three unknown. We must, therefore, find a third
equation that can be used to determine one of the unknowns. We can get the third equation from
the torque equation.
In order to write the torque equation, we must first pick an axis of rotation and determine the lever
arms for the forces involved. Since both V and H are unknown, we can eliminate them from the
torque equation by picking the rotation axis through the point P (then both V and H have zero lever
arms). If we let the boom have a length L, then the lever arm for W L is L cos φ , while the lever
arm for W B is ( L / 2) cos φ . From the figure, we see that the lever arm for T is L sin( φ – θ) . If
we take counterclockwise torques as positive, then the torque equation is
 L cos φ 
∑ τ = −WB 
− W L L cos φ+ TL sin (φ −θ ) = 0
 2 
Solving for T, we have
T =
1
2
WB + WL
sin( φ – θ )
cos φ
(3)
Chapter 9 Problems
331
a. From Equation (3) the tension in the support cable is
T =
1
2
(3600 N) + 4800 N
sin(48 ° – 32 °)
cos 48 ° = 1.6 × 10
4
N
b. The force exerted on the lower end of the hinge at the point P is the vector sum of the forces H
and V. According to Equation (1),
(
H = T cos θ = 1.6 ×10
4
)
N cos 32 ° = 1.4 × 10 4 N
and, from Equation (2)
(
)
V = W L + W B + T sin θ = 4800 N + 3600 N + 1.6 × 10 4 N sin 32 ° = 1.7 × 10 4 N
Since the forces H and V are at right angles to each other, the magnitude of their vector sum can
be found from the Pythagorean theorem:
FP =
H
2
+ V 2 = (1.4 × 10 4 N) 2 + (1.7 × 10 4 N) 2 = 2.2 × 10 4 N
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25. REASONING AND SOLUTION Taking torques on the right leg about the apex of the “A”
gives
Στ = −(120 N) L cos 60.0° − FL cos 30.0° + T(2L) cos 60.0° = 0
where F is the horizontal component of the force exerted by the crossbar and T is the tension in the
right string. Due to symmetry, the tension in the left string is also T. Newton's second law applied to
the vertical gives
2T − 2W = 0
so T = 120 N
Substituting T = 120 N into the first equation yields F = 69 N .
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ROTATIONAL DYNAMICS
332
26. REASONING The drawing shows the forces acting on
the board, which has a length L. Wall 2 exerts a normal
force P2 on the lower end of the board. The maximum
force of static friction that wall 2 can apply to the lower
end of the board is µs P2 and is directed upward in the
drawing. The weight W acts downward at the board's
center. Wall 1 applies a normal force P1 to the upper end
of the board. We take upward and to the right as our
positive directions.
P
Wall 1
1
Wall 2
µ s P2
W
θ
P2
SOLUTION Then, since the horizontal forces balance to zero, we have
P 1 – P2 = 0
(1)
µs P2 – W = 0
(2)
The vertical forces also balance to zero:
Using an axis through the lower end of the board, we now balance the torques to zero:
 L 
(cos θ ) − P1 L (sin θ ) = 0
2
W
Rearranging Equation (3) gives
W
tan θ =
(3)
(4)
2P1
But W = µs P2 according to Equation (2), and P2 = P1 according to Equation (1). Therefore, W =
µs P1, which can be substituted in Equation (4) to show that
tan θ =
µ s P1
2P1
=
µs
2
=
0.98
2
or
θ = tan–1(0.49) = 26°
Chapter 9 Problems
333
From the drawing at the right,
cos θ =
d
L
L
Therefore, the longest board that can be propped between the
two walls is
L=
d
1.5 m
=
=
cos θ cos 26 °
θ
d
1.7 m
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27. REASONING AND SOLUTION The weight W of the left side of the ladder, the normal force
FN of the floor on the left leg of the ladder, the tension T in the crossbar, and the reaction force R
due to the right-hand side of the ladder, are shown in the figure below. In the vertical direction −W
+ FN = 0, so that
FN = W = mg = (10.0 kg)(9.80 m/s2) = 98.0 N
R
In the horizontal direction it is clear that R = T.
The net torque about the base of the ladder is
75.0
Στ = – T [(1.00 m) sin 75.0o] – W [(2.00 m) cos 75.0o]
+ R [(4.00 m) sin 75.0o] = 0
W
F
Substituting for W and using R = T, we obtain
T =
(98.0 N )(2.00 m) cos 75.0 °
(3.00 m) s in 75.0 °
o
N
T
= 17.5 N
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28. REASONING The moment of inertia of the stool is the sum of the individual moments of inertia
of its parts. According to Table 9.1, a circular disk of radius R has a moment of inertia of
Idisk = 21 Mdisk R2 with respect to an axis perpendicular to the disk center. Each thin rod is
attached perpendicular to the disk at its outer edge. Therefore, each particle in a rod is located at
a perpendicular distance from the axis that is equal to the radius of the disk. This means that each
of the rods has a moment of inertia of Irod = Mrod R2.
SOLUTION Remembering that the stool has three legs, we find that the its moment of inertia is
334
ROTATIONAL DYNAMICS
I stool = I disk + 3 I rod = 12 M disk R 2 + 3 M rod R 2
=
1
2
b1.2 kg gb0.16 mg + 3b0.15 kg gb0 .16 m g =
2
2
0. 027 kg ⋅ m 2
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29.
REASONING AND SOLUTION For a rigid body rotating about a fixed axis,
Newton's second law for rotational motion is given by Equation 9.7, ∑ τ = I α, where I is the
moment of inertia of the body and α is the angular acceleration expressed in rad/s2. Equation 9.7
gives
SSM
I=
∑τ
α
=
10 .0 N ⋅ m
8.00 rad/s
2
= 1.25 kg ⋅ m 2
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30. REASONING AND SOLUTION
We know
Στ = Iα = (0.16 kg⋅m2)(7.0 rad/s2) = 1.1 kg⋅m2/s2 = 1.1 N ⋅ m
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31. REASONING AND SOLUTION
a. The net torque on the disk about the axle is
Στ = F1R − F2R = (0.314 m)(90.0 N − 125 N) = − 11 N ⋅ m
b. The angular acceleration is given by α = Στ/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
α = (−11 N⋅m)/(1.20 kg⋅m2) =
–9.2 rad / s 2
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32. REASONING
The rotational analog of Newton's second law is given by Equation 9.7,
∑ τ = I α. Since the person pushes on the outer edge of one pane of the door with a force F that
is directed perpendicular to the pane, the torque exerted on the door has a magnitude of FL, where
the lever arm L is equal to the width of one pane of the door. Once the moment of inertia is
known, Equation 9.7 can be solved for the angular acceleration α.
Chapter 9 Problems
335
The moment of inertia of the door relative to the rotation axis is I = 4 IP , where IP is the moment
1
of inertia for one pane. According to Table 9.1, we find IP = 3 ML 2 , so that the rotational inertia
4
of the door is I = 3 ML 2 .
SOLUTION Solving Equation 9.7 for α, and using the expression for I determined above, we
have
α=
FL
4
3
ML
=
2
F
4
3
ML
=
68 N
4
3
(85 kg)(1.2 m)
= 0.50 rad/s
2
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33.
REASONING The figure below shows eight particles, each one located at a different
corner of an imaginary cube. As shown, if we consider an axis that lies along one edge of the cube,
two of the particles lie on the axis, and for these particles r = 0. The next four particles closest to
the axis have r = l , where l is the length of one edge of the cube. The remaining two particles
have r = d , where d is the length of the diagonal along any one of the faces. From the
Pythagorean theorem, d = l 2 + l 2 = l 2 .
SSM
l
Axis
d
d
l
l
l
According to Equation 9.6, the moment of inertia of a system of particles is given by I = ∑ mr 2 .
SOLUTION Direct application of Equation 9.6 gives
I = ∑ mr 2 = 4(m l2 ) + 2(md 2 ) = 4(m l 2 ) + 2(2m l 2 ) = 8m l2
or
I = 8( 0.12 kg)(0.25 m)
2
= 0.060 kg ⋅ m 2
_________________________________________________________________________________________
____
34. REASONING AND SOLUTION
336
ROTATIONAL DYNAMICS
a. We know that α = (ω – ωo)/t. But ω = (1/2)ωo, so
α = −(1/2)ωo/t = −(1/2)(88.0 rad/s)/(5.00 s) = −8.80 rad/s2
The magnitude of α is 8.80 rad/s
2
.
b. Στ = Iα and FR = Iα. The magnitude of the frictional force is
F = Iα/R = (0.850 kg⋅m2)(8.80 rad/s2)/(0.0750 m) =
99.7 N
_________________________________________________________________________________________
____
35. REASONING AND SOLUTION We know Στ = Iα, where
I = (1/2) mr 2 = (1/2)(0.400 kg)(0.130 m) 2 = 3.38 × 10–3 kg⋅m2
Also, ω = ωo + αt, so that α = (85.0 rad/s − 262 rad/s)/(18.0 s) = −9.83 rad/s2. Thus, the net
torque is
Στ = Iα = (3.38 × 10–3 kg⋅m2)(−9.83 rad/s2) = − 3.32 × 10 –2 N ⋅ m
_________________________________________________________________________________________
____
36. REASONING The force applied to the baton creates a net torque that gives the rod an angular
acceleration, according to Newton’s second law as expressed for rotational motion (Στ = Iα).
From the data given, we will calculate the moment of inertia I and the angular acceleration α.
Newton’s second law, then, will allow us to obtain the net torque Στ. From the net torque, we will
be able to determine the force, since the magnitude of the torque in this case is just the magnitude of
the force times the lever arm of 2.0 cm.
SOLUTION The momentum of inertia of the baton is the sum of the individual moments of inertia
of its parts. For a thin rod of length L rotating about an axis perpendicular to the rod at its center,
the moment of inertia is I rod = 121 M rod L2 . For each ball (assumed to be very small compared to
the length of the rod), the moment of inertia is Iball = Mball (L/2)2. The total moment of inertia, then,
is
I baton = I rod + 2 I ball = 121 M rod L2 + 2 M ball
c Lh =
1
2
2
1
12
M rod L2 + 21 M ball L2
Using Equation 8.4 from the equations of kinematics for rotational motion, we can determine the
angular acceleration as
Chapter 9 Problems
α=
337
ω − ω0
t
where ω and ω0 are, respectively, the final and initial angular velocities. Since only one force of
magnitude F creates the torque and since Equation 9.1 gives the magnitude of the torque as the
magnitude of the force times the lever arm l, Newton’s second law becomes Στ = Fl= Iα.
Substituting our expressions for the moment of inertia and angular acceleration of the baton into the
second law gives
Fl = I baton α =
c
1
12
M rod L2 + 12 M ball L2
ω− ω I
hF
G
H t JK
0
In Newton’s second law the angular acceleration α must be expressed in rad/s2. Therefore, the
values for the angular speeds ω and ω0 must be expressed in rad/s. Recognizing that 2.0 rev/s is
4.0 π rad/s and solving for F, we find that
F=
=
c
hFω − ω I
G
l
H t JK
b0 .13 kg gb0.81 mg + b0.13 kg gb0 .81 mg F
4 π rad / s − 0 rad / s I
G
JK=
H 0.40 s
0.020 m
1
12
1
12
M rod L2 + 12 M ball L2
2
0
2
1
2
78 N
_________________________________________________________________________________________
____
37.
REASONING AND SOLUTION
a. From Table 9.1 in the text, the moment of inertia of the rod relative to an axis that is
perpendicular to the rod at one end is given by
SSM
I rod = 13 ML2 = 13 ( 2 .00 kg)(2.00 m) 2 = 2 .67 kg ⋅ m 2
b. The moment of inertia of a point particle of mass m relative to a rotation axis located a
perpendicular distance r from the particle is I = mr 2 . Suppose that all the mass of the rod were
located at a single point located a perpendicular distance R from the axis in part (a). If this point
particle has the same moment of inertia as the rod in part (a), then the distance R, the radius of
gyration, is given by
R=
I
=
m
2.67 kg ⋅ m 2
= 1.16 m
2.00 kg
_________________________________________________________________________________________
____
338
ROTATIONAL DYNAMICS
38. REASONING AND SOLUTION The final angular speed of the arm is ω = vT/r, where
r = 0.28 m. The angular acceleration needed to produce this angular speed is α = (ω − ω0)/t. The
net torque required is Στ = Iα. This torque is due solely to the force M, so that Στ = ML. Thus,
ω − ω0
I
t
∑τ
M=
=
L
L
F
G
H
I
JK
Setting ω0 = 0 and ω = vT/r, the force becomes
I
M=
Fv I
G
Hr t JK= Iv
T
L
c0. 065 kg ⋅ m hb5.0 m / sg=
=
Lr t b
0. 025 m g
b0.28 mgb0.10 sg
2
T
460 N
_________________________________________________________________________________________
____
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,
α=
ω − ω0
t
=
3.7 rad/s − 13 .1 rad/s
3.0 s
2
= −3.1 rad/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 = mr 2 . The net torque that acts on the wheel
due to the brake pads is, therefore,
2
∑ τ = I α= (mr ) α
(1)
From Equation 9.1, the net torque that acts on the wheel due to the action of the two brake pads is
∑ τ = –2 fk l
(2)
where f k is the kinetic frictional force applied to the wheel by each brake pad, and l = 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 = µk FN
(3)
Chapter 9 Problems
339
where µk 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
–2(µk FN ) l = (mr )α
FN =
– mr 2 α
2 µk l
=
–(1.3 kg)(0.33 m)
2
(–3.1 rad/s
2(0.85) ( 0.33 m)
2
)
= 0.78 N
_________________________________________________________________________________________
____
40. REASONING AND SOLUTION The moment of inertia of a solid cylinder about an axis
coinciding with the cylinder axis which contains the center of mass is Icm = (1/2) MR2. The parallel
axis theorem applied to an axis on the surface (h = R) gives I = Icm + MR2, so that
I=
3
2
MR
2
_________________________________________________________________________________________
____
41. REASONING AND SOLUTION For door A, θ = (1/2) α A tA 2. For door B, θ = (1/2) α BtB2.
Equating gives
αA
t B = tA
αB
We need the angular accelerations. Applying Newton's second law for rotation to door A
FL = (1/3)ML2α A
and to door B
F(1/2)L = (1/12)ML2α B
Division of the previous two equations yields α A /α B = 1/2. Now
tB = tA / 2 = (3.00 s)/(1.41) =
2.12 s
_________________________________________________________________________________________
____
42. REASONING The drawing shows the drum, pulley, and the crate, as well as the tensions in the
cord
340
ROTATIONAL DYNAMICS
r2
T1
Pulley
m
2
T2
= 1 3 0 kg
T2
T1
r1
Cr a t e
Dr u m
m
3
= 1 8 0 kg
m1 = 150 kg
r
1
= 0 .7 6 m
Let T1 represent the magnitude of the tension in the cord between the drum and the pulley. Then,
the net torque exerted on the drum must be, according to Equation 9.7, Στ = I1α 1, where I1 is the
moment of inertia of the drum, and α 1 is its angular acceleration. If we assume that the cable
does not slip, then Equation 9.7 can be written as
FI
hG
H JK
a
− T1 r1 + τ = m1 r12
14243 123 r1
23
∑τ
I 1 1α
1
c
(1)
where τ is the counterclockwise torque provided by the motor, and a is the acceleration of the
cord (a = 1.2 m/s2). This equation cannot be solved for τ directly, because the tension T1 is not
known.
We next apply Newton’s second law for rotational motion to the pulley in the drawing:
FI
hG
H JK
a
+ T1 r2 − T2 r2 = 12 m2 r22
144244
3 1
424
3 r2
123
I2
∑τ
α2
c
(2)
where T2 is the magnitude of the tension in the cord between the pulley and the crate, and I2 is the
moment of inertial of the pulley.
Finally, Newton’s second law for translational motion (ΣFy = m a) is applied to the crate, yielding
+ T2 − m3 g = m3 a
14
4244
3
∑ Fy
(3)
Chapter 9 Problems
341
SOLUTION Solving Equation (1) for T1 and substituting the result into Equation (2), then solving
Equation (2) for T2 and substituting the result into Equation (3), results in the following value for the
torque
c
h
) c150 kg +
τ = r1 a m1 + 21 m2 + m3 + m3 g
= ( 0. 76 m) (1.2 m / s 2
h
130 kg + 180 kg + ( 180 kg)(9.80 m / s 2 ) = 1700 N ⋅ m
1
2
_________________________________________________________________________________________
____
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 = 12 MR 2 . Once the moment of inertia of the flywheel is known, Equation 9.9 can be
solved for the angular speed ω in rad/s. This quantity can then be converted to rev/min.
SSM
SOLUTION Solving Equation 9.9 for ω, we obtain,
2 ( KE R )
ω=
I
=
2 ( KE R )
1
2
MR 2
=
4(1.2 × 10 9 J)
= 6.4 × 10 4 rad / s
2
(13 kg)(0.30 m)
Converting this answer into rev/min, we find that
c
ω = 6.4 × 10 4 rad / s
1 rev I F60 s I
hF
G
H2 π rad JKG
H1 min JK=
6.1 × 10 5 rev / min
_________________________________________________________________________________________
____
44. REASONING The rotational kinetic energy of each object is given by Equation 9.9,
KE R = 12 Iω2 . To solve this problem, we need only set the two kinetic energies equal, being sure
to use the proper moment of inertia for each object, as given in Table 9.1.
SOLUTION
According to Table 9.1 the moment of inertia of the solid sphere is
I solid = MR , while the moment of inertia of the shell is I shell = 23 MR 2 . Using these
expressions in Equation 9.9 for the rotational kinetic energy, we obtain
2
5
2
ROTATIONAL DYNAMICS
342
I solid ω2solid = 12 I shell ω2shell
14243 14243
1
2
Rotational KE
of solid sphere
1
2
c MR hω
2
5
2
2
5
ωsolid =
5
3
Rotational KE
of spherical shell
2
solid
=
1
2
c MR hω
2
3
2
2
shell
ω2solid = 23 ω2shell
ωshell =
5
3
b7. 0 rad / sg=
9. 0 rad / s
_________________________________________________________________________________________
____
45.
SSM REASONING AND SOLUTION
a. The tangential speed of each object is given by Equation 8.9, v T = rω. 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 = 12 m1 v 12 + 12 m2 v 22 + 12 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 Iω2 = 12 (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).
Chapter 9 Problems
343
_________________________________________________________________________________________
____
46. REASONING
a. The kinetic energy is given by Equation 9.9 as KE R = 12 Iω2 . Assuming the earth to be a
uniform solid sphere, we find from Table 9.1 that the moment of inertia is I =
2
5
MR 2 . The mass
and radius of the earth is M = 5.98 × 1024 kg and R = 6.38 × 106 m (see the inside of the text’s
front cover). The angular speed ω must be expressed in rad/s, and we note that the earth turns
once around its axis each day, which corresponds to 2π rad/day.
b. The kinetic energy for the earth’s motion around the sun can be obtained from Equation 9.9 as
KE R = 12 Iω2 . Since the earth’s radius is small compared to the radius of the earth’s orbit
(Rorbit = 1.50 × 1011 m, see the inside of the text’s front cover), the moment of inertia in this case
2
is just I = MR orbit
. The angular speed ω of the earth as it goes around the sun can be obtained
from the fact that it makes one revolution each year, which corresponds to 2π rad/year.
SOLUTION
a. According to Equation 9.9, we have
KE R = 21 Iω2 =
1
2
c MR hω
2
2
5
2
LF2 π rad I F1 day IF
1 h IO
mh M
G
JP
G
J
G
J
NH1 day KH24 h KH3600 s KQ
2
=
1
2
2
5
c5.98 × 10
24
hc
kg 6. 38 × 10
6
2
= 2 .57 × 10 29 J
b. According to Equation 9.9, we have
KE R = 21 Iω2 =
1
2
cMR hω
2
orbit
2
LF2 π rad I F 1 yr I F1 day I F 1 h I O
mh M
H24 h JKG
H3600 s JKP
H1 yr JKG
H365 day JKG
NG
Q
2
=
1
2
c5. 98 × 10
24
hc
kg 1.50 × 10
11
2
= 2 .67 × 10 33 J
_________________________________________________________________________________________
____
47. REASONING AND SOLUTION The conservation of energy gives for the cube (1/2) mvc2 =
mgh, so vc = 2gh . The conservation of energy gives for the marble (1/2) mvm2 + (1/2) Iω2 =
mgh. Since the marble rolls without slipping: ω = v/R. For a solid sphere we know I = (2/5) mR 2.
Combining the last three equations gives
344
ROTATIONAL DYNAMICS
vm =
10gh
7
=
7 = 1.18
5
Thus,
vc
vm
_________________________________________________________________________________________
____
The rotational kinetic energy of the shell is KEr = (1/2) Iω2
and the translational kinetic energy is KEt = (1/2) Mv2. Now, v = Rω since the sphere rolls
without slipping, so
KEt = (1/2) MR2ω2
The desired fraction is
KEr/KE = I/(I + MR2) = 2/5
48. REASONING AND SOLUTION
_________________________________________________________________________________________
____
49.
REASONING AND SOLUTION The only force that does work on the cylinders as
they move up the incline is the conservative force of gravity; hence, the total mechanical energy is
conserved as the cylinders ascend the incline. We will let h = 0 on the horizontal plane at the
bottom of the incline. Applying the principle of conservation of mechanical energy to the solid
cylinder, we have
1
1
2
2
mgh s = 2 mv 0 + 2 Is ω0
(1)
SSM
1
2
where, from Table 9.1, Is = 2 mr . In this expression, v 0 and ω0 are the initial translational and
rotational speeds, and hs is the final height attained by the solid cylinder. Since the cylinder rolls
without slipping, the rotational speed ω0 and the translational speed v 0 are related according to
Equation 8.12, ω0 = v 0 / r . Then, solving Equation (1) for hs , we obtain
hs =
3 v 02
4g
Repeating the above for the hollow cylinder and using Ih = mr 2 we have
hh =
v 20
g
The height h attained by each cylinder is related to the distance s traveled along the incline and the
angle θ of the incline by
Chapter 9 Problems
ss =
hs
sin θ
Dividing these gives
sh =
and
ss
345
hh
sin θ
= 3/4
sh
_________________________________________________________________________________________
____
50. REASONING AND SOLUTION The conservation of energy gives
mgh + (1/2) mv2 + (1/2) Iω2 = (1/2) mvo2 + (1/2) Iωo2
If the ball rolls without slipping, ω = v/R and ωo = vo/R. We also know I = (2/5) mR 2.
Substitution of the last two equations into the first and rearrangement gives
v=
v 20 −
10
7
gh =
b3.50 m / sg − c9 .80 m / s hb0 .760 m g=
2
2
10
7
1.3 m / s
_________________________________________________________________________________________
____
51. REASONING
We first find the speed v 0 of the ball when it becomes airborne using the
conservation of mechanical energy. Once v 0 is known, we can use the equations of kinematics to
find its range x.
SOLUTION When the tennis ball starts from rest, its total mechanical energy is in the form of
gravitational potential energy. The gravitational potential energy is equal to mgh if we take h = 0
at the height where the ball becomes airborne. Just before the ball becomes airborne, its
mechanical energy is in the form of rotational kinetic energy and translational kinetic energy. At
1
1
this instant its total energy is 2 mv 20 + 2 Iω2 . If we treat the tennis ball as a thin-walled spherical
shell of mass m and radius r, and take into account that the ball rolls down the hill without slipping,
its total kinetic energy can be written as
1
2
2
mv 0 +
1
2
2
Iω =
1
2
2
mv 0 +
1
2
2
2
( mr )
3
2
5
 v0 
2
=
mv 0
 r 
6
Therefore, from conservation of mechanical energy, he have
mgh =
5
6
2
mv 0
or
v0 =
6 gh
5
The range of the tennis ball is given by x = v0 x t = v 0 (cos θ ) t , where t is the flight time of the ball.
From Equation 3.3b, we find that the flight time t is given by
346
ROTATIONAL DYNAMICS
t =
v − v0 y
ay
=
( −v 0 y ) − v 0 y
ay
=–
2v 0 sin θ
ay
Therefore, the range of the tennis ball is
 2v0 sin θ 

ay


x = v0 x t = v0 (cos θ)  –
If we take upward as the positive direction, then using the fact that a y = – g and the expression for
v 0 given above, we find
 2 cos θ sin θ  2  2 cos θ sin θ   6 gh  2 12
x =
 v0 = 
 
 = h cos θ sin θ
g
g
5 
5



 
= 12 (1.8 m) (cos 35 °)(sin 35 ° ) = 2.0 m
5
_________________________________________________________________________________________
____
52. REASONING AND SOLUTION The angular momentum of the satellite is conserved if only the
gravitational force of the earth is acting on it. We have LA = LP or mv A rA = mv PrP. Now
rP = (v A /v P)rA = (1/1.20)(1.30 × 107 m) = 1.08 × 10 7 m
_________________________________________________________________________________________
____
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 Linitial = Lfinal , or
I A ω A + I Bω B = ( I A + I B ) ωfinal
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 = IA
Fω
G
Hω
final
B
– ωA
– ωfinal
I = ( 3.4 kg ⋅ m ) L –2.4 rad / s – 7.2 rad / s O=
JK
M
N–9.8 rad / s – (–2.4 rad / s) P
Q
2
4.4 kg ⋅ m 2
_________________________________________________________________________________________
____
54. REASONING AND SOLUTION
Angular momentum is conserved, Iω = I0ω0, so
Chapter 9 Problems
FI I F5.40 kg ⋅ m Ib5.00 rad / sg=
ω = G Jω = G
HI K H3.80 kg ⋅ m JK
347
2
0
0
2
7 .11 rad / s
_________________________________________________________________________________________
____
55. REASONING AND SOLUTION
a. Angular momentum is conserved, so that Iω = Ioωo, where
I = Io + 10mbRb2 = 2100 kg⋅m2
ω = (Io/I)ωo = [(1500 kg⋅m2)/(2100 kg⋅m2)](0.20 rad/s) = 0.14 rad/s
b. A net external torque must be applied in a direction that is opposite to the angular deceleration
caused by the baggage dropping onto the carousel.
_________________________________________________________________________________________
____
56. REASONING AND SOLUTION
The conservation of angular momentum applies about the indicated axis
Idωd + MpRp2ωp = 0
where ωp = vp/Rp
Now Id = (1/2) MdRd2. Combining and solving yields
ωd = −2(Mp/Md)(Rp/Rd2)vp = −2
F 40.0 kg I L
1.25 m O
M
Pb2 .00 m / sg= −0.500 rad/s
G
J
H1.00 × 10 kg KM
Nb2.00 m g P
Q
The magnitude of ωd is 0.500 rad/s
2
2
. The negative sign indicates that the disk rotates in a
direction opposite to the motion of the person.
_________________________________________________________________________________________
____
57.
REASONING Let the space station and the people within it constitute the system. Then
as the people move radially from the outer surface of the cylinder toward the axis, any torques that
occur are internal torques. Since there are no external torques acting on the system, the principle
of conservation of angular momentum can be employed.
SSM
SOLUTION Since angular momentum is conserved,
I final ω final = I 0ω0
ROTATIONAL DYNAMICS
348
Before the people move from the outer rim, the moment of inertia is
2
I 0 = I station + 500 mperson rperson
or
I 0 = 3.00 × 10 9 kg ⋅ m 2 + ( 500)(70.0 kg)(82.5 m) 2 = 3.24 × 10 9 kg ⋅ m 2
If the people all move to the center of the space station, the total moment of inertia is
I final = I station = 3.00 × 10 9 kg ⋅ m 2
Therefore,
ωfinal
ω0
=
I0
I final
=
3.24 × 10 9 kg ⋅ m 2
= 1. 08
3.00 × 10 9 kg ⋅ m 2
This fraction represents a percentage increase of 8 percent .
_________________________________________________________________________________________
____
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 = rω) to relate the angular speeds ω0 and ωf to the tangential speeds v 0
and v f.
SOLUTION Let L be the initial length of the cable between the modules and ω0 be the initial
angular speed. Relative to the center-of-mass axis, the initial momentum of inertia of the twomodule 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 ωf, and the momentum of inertia is
If = 2M(L/4)2. The conservation of angular momentum indicates that
I f ω f = I 0ω0
12
3
123
Final angular
momentum
L2 M FL I O
JKPω
M G
H
4
M
P
N
Q
2
f
Initial angular
momentum
L FL I O
JKPω
M G
H
2
M
P
N
Q
2
= 2M
0
ωf = 4 ω0
According to Equation 8.9, ωf = v f/(L/4) and ω0 = v 0/(L/2). With these substitutions, the result
that ωf = 4ω0 becomes
Chapter 9 Problems
vf
L/4
=4
Fv I
G
HL / 2 JK
0
or
b
349
g
v f = 2 v 0 = 2 17 m / s = 34 m / s
_________________________________________________________________________________________
____
59. REASONING AND SOLUTION
After the mass has moved inward to its final path the
centripetal force acting on it is T = 105 N.
r
10 5 N
Its centripetal acceleration is
ac = v 2/R = T/m
Now
v = ωR
so
R = T/(ω2m)
The centripetal force is parallel to the line of action (the string), so the force produces no torque on
the object. Hence, angular momentum is conserved.
Iω = Ioωo
so that
ω = (Io/I)ωo = (Ro2/R2)ωo
Substituting and simplifying
R3 = (mRo4 ωo2)/T, so that R =
0.573 m
_________________________________________________________________________________________
____
60. REASONING AND SOLUTION The block will just start to move when the centripetal force on
the block just exceeds f smax . Thus, if rf is the smallest distance from the axis at which the block
stays at rest when the angular speed of the block is ωf, then µs FN = mrf ωf2, or
µs mg = mrf ωf2
µs g = rf ωf2
(1)
350
ROTATIONAL DYNAMICS
Since there are no external torques acting on the system, angular momentum will be conserved.
I0 ω0 = If ωf
where I0 = mr02, and If = mrf2. Making these substitutions yields
r02 ω0 = rf2 ωf
(2)
Solving Equation (2) for ωf and substituting into Equation (1) yields:
2
µ s g = rf ω 0
r04
r f4
Solving for rf gives
 ω 2 r 4  1/3  (2.2 rad/s) 2 (0.30 m) 4  1/3
0
0
 =
rf = 
 = 0.17 m
µ
g
 (0.75)(9.80 m/s 2 ) 
 s 
_________________________________________________________________________________________
____
REASONING
The maximum torque will occur
when the force is applied perpendicular to the diagonal of the square as
shown. The lever arm l is half the length of the diagonal. From the
Pythagorean theorem, the lever arm is, therefore,
SSM
WWW
l=
1
2
0.40 m
( 0.40 m ) 2 + ( 0.40 m ) 2 = 0.28 m
Since the lever arm is now known, we can use Equation 9.1 to obtain
the desired result directly.
0.40 m
61.
axis
SOLUTION Equation 9.1 gives
F
τ = Fl = (15 N)(0.28 m) = 4.2 N ⋅ m
_________________________________________________________________________________________
___
62. REASONING AND SOLUTION For a solid disk we know I = (1/2)MR2 so M = 2I/R2.
Newton's second law for rotation yields I = (Στ)/α, so M = 2(Στ)/(αR2).
α = (ω − ωo)/t = (−3.49 rad/s)/(15.0 s) = −0.233 rad/s2, so that
Chapter 9 Problems
M =
(
−
2 −6.20 × 10 3 N ⋅ m
)
( −0.233 rad/s 2 )( 0.150 m ) 2
351
= 2.37 kg
_________________________________________________________________________________________
____
163. REASONING AND SOLUTION
the forces must be zero:
The sum of
F
ΣFy = FW + FC – mg = 0
F
W
Therefore,
1.40 m
FW + FC = mg
C
0.60 m
mg
= (71.0 kg)(9.80 m/s2) = 696 N
The sum of the torques must also be zero. Taking the axis for torques at the point just below the
boulder, we have
Στ = FC (0.60 m) – FW (1.40 m) = 0.
a. Substituting for FW yields
FC(0.60 m) – (696 – FC)(1.40 m) = 0
Solving for FC gives FC =
487 N
b. Also, FW = 696 N – FC =
.
209 N
_________________________________________________________________________________________
____
64. REASONING AND SOLUTION Use conservation of angular momentum, Ioωo = (Io + I) ω,
where
Io = 1.2 × 10–3 kg⋅m2 and ωo = 0.40 rad/s
The moment of inertia of the bug is
I = mL2 = (5.0 × 10–3 kg)(0.20 m) 2 = 2.0 × 10–4 kg⋅m2
The final angular velocity is
ω = Ioωo/(Io + I) = 0.34 rad/s
352
ROTATIONAL DYNAMICS
_________________________________________________________________________________________
____
65.
REASONING AND SOLUTION
a. The rim of the bicycle wheel can be treated as a hoop. Using the expression given in Table 9.1
in the text, we have
SSM
Ihoop = MR
2
2
= (1.20 kg)(0.330 m)
= 0.131 kg ⋅ m 2
b. Any one of the spokes may be treated as a long, thin rod that can rotate about one end. The
expression in Table 9.1 gives
Irod =
1
3
2
1
ML = 3 (0.010 kg)(0.330 m)
2
= 3.6 × 10 –4 kg ⋅ m 2
c. The total moment of inertia of the bicycle wheel is the sum of the moments of inertia of each
constituent part. Therefore, we have
I = Ihoop + 50 I rod = 0.131 kg ⋅ m
2
+ 50(3.6 × 10 – 4 kg ⋅ m 2 ) = 0.149 kg ⋅ m 2
_________________________________________________________________________________________
____
66. REASONING AND SOLUTION The net torque about an axis through the elbow joint is
Στ = (111 N)(0.300 m) − (22.0 N)(0.150 m) − (0.0250 m)M = 0 or
3
M = 1.20 × 10 N
_________________________________________________________________________________________
____
67.
REASONING The drawing shows the forces acting
on the board, which has a length L. The ground exerts the
vertical normal force V on the lower end of the board. The
maximum force of static friction has a magnitude of µs V and
acts horizontally on the lower end of the board. The weight W
acts downward at the board's center. The vertical wall applies
a force P to the upper end of the board, this force being
perpendicular to the wall since the wall is smooth (i.e., there is
no friction along the wall). We take upward and to the right as
our positive directions. Then, since the horizontal forces
balance to zero, we have
µs V – P = 0
SSM
P
V
θ
W
µs V
(1)
The vertical forces also balance to zero giving
V – W = 0
(2)
Chapter 9 Problems
353
Using an axis through the lower end of the board, we express the fact that the torques balance to
zero as
L
PL sin θ − W
cos θ = 0
(3)
2
F
I
G
H JK
Equations (1), (2), and (3) may then be combined to yield an expression for θ.
SOLUTION Rearranging Equation (3) gives
tan θ =
W
2P
(4)
But, P = µs V according to Equation (1), and W = V according to Equation (2). Substituting these
results into Equation (4) gives
V
1
tan θ =
=
2 µsV 2 µs
Therefore,
1
1
θ = tan −1
= tan − 1
= 37.6°
2 µs
2 ( 0.650)
F I
G
H JK
L
M
N
O
P
Q
_________________________________________________________________________________________
____
68. REASONING AND SOLUTION
a. The net torque about an axis through the point of contact between the floor and her shoes is
Στ = − (5.00 × 102 N)(1.10 m)sin 30.0° + FN(cos 30.0°)(1.50 m) = 0
FN = 212 N
b. Newton's second law applied in the horizontal direction gives Fh − FN = 0, so Fh = 212 N
c. Newton's second law in the vertical direction gives Fv − W = 0, so Fv = 5.00 × 10 2 N
_________________________________________________________________________________________
____
69. REASONING AND SOLUTION Since the change occurs without the aid of external torques,
the angular momentum of the system is conserved: Ifωf = I0ω0. Solving for ωf gives
I 
ωf = ω0  0 
 If 
ROTATIONAL DYNAMICS
354
where for a rod of mass M and length L, I 0 =
1
12
ML 2 . To determine If, we will treat the arms of
the "u" as point masses with mass M/4 a distance L/4 from the rotation axis. Thus,
 1  M  L  2    M   L  2  1
2
I f =       + 2      =
ML
12  2   2 





  4 4  24
and
 1

2
 12 ML 
ωf = ω0 
 = (7.0 rad/s)(2) = 14 rad/s
1
2
 ML 
24
_________________________________________________________________________________________
____
70. REASONING AND SOLUTION The conservation of energy gives (1/2) Iω2 = mg[(1/2) L].
Now the tip of the rod has, as it hits the ground, a speed of v = Lω, so ω = v/L. The moment of
inertia of the rod is given by I = (1/3) mL2. Substituting the last two equations into the first equation
and simplifying yields
v=
c
3gL = 3 9.8 m / s 2
hb2.00 mg=
7.67 m / s
_________________________________________________________________________________________
____
71.
REASONING AND SOLUTION Consider the left board, which has a length L
and a weight of (356 N)/2 = 178 N. Let Fv be the upward normal force exerted by the ground on
the board. This force balances the weight, so Fv = 178 N. Let fs be the force of static friction,
which acts horizontally on the end of the board in contact with the ground. fs points to the right.
Since the board is in equilibrium, the net torque acting on the board through any axis must be zero.
Measuring the torques with respect to an axis through the apex of the triangle formed by the
boards, we have
SSM
WWW
L
+ (178 N)(sin 30.0°)   + f s(L cos 30.0°) – Fv (L sin 30.0°) = 0
2
or
44.5 N + fs cos 30 .0 ° – FV sin 30.0 ° = 0
so that
fs =
(178 N)(sin 30.0 °) – 44.5 N
cos 30.0 °
= 51.4 N
_________________________________________________________________________________________
____
72. REASONING AND SOLUTION Newton's law applied to the 11.0-kg object gives
Chapter 9 Problems
T2 − (11.0 kg)(9.80 m/s2) = (11.0 kg)(4.90 m/s2)
or
355
T2 = 162 N
A similar treatment for the 44.0-kg object yields
T1 − (44.0 kg)(9.80 m/s2) = (44.0 kg)(−4.90 m/s2)
or
T1 = 216 N
For an axis about the center of the pulley
T2r − T1r = I(−α) = (1/2) Mr2 (−a/r)
Solving for the mass M we obtain
M = (−2/a)(T2 − T1) = [−2/(4.90 m/s2)](162 N − 216 N) = 22.0 kg
__________________________________________________________________________________________
___
73. CONCEPT QUESTIONS a. The magnitude of a torque is the magnitude of the force times the
lever arm of the force, according to Equation 9.1. The lever arm is the perpendicular distance
between the line of action of the force and the axis. For the force at corner B, the lever arm is half
the length of the short side of the rectangle. For the force at corner D, the lever arm is half the
length of the long side of the rectangle. Therefore, the lever arm for the force at corner D is greater.
Since the magnitudes of the two forces are the same, this means that the force at corner D creates
the greater torque.
b. As we have seen, the torque at corner D (clockwise) has a greater magnitude than the torque at
corner B (counterclockwise). Therefore, the net torque from these two contributions is clockwise.
This means that the force at corner A must produce a counterclockwise torque, if the net torque
produced by the three forces is to be zero. The force at corner A, then, must point toward corner
B.
SOLUTION Let L be the length of the short side of the rectangle, so that the length of the long
side is 2L. We can use Equation 9.1 for the magnitude of each torque. Remembering that
counterclockwise torques are positive and that the torque at corner A is counterclockwise, we can
write the zero net torque as follows:
Στ = FA l A + FB l B − FD l D = 0
b gc L h− b12 N gL = 0
FA L + 12 N
1
2
The length L can be eliminated algebraically from this result, which can then be solved for FA :
b gch+ b12 N g=
FA = − 12 N
1
2
6.0 N (pointing toward corner B)
356
ROTATIONAL DYNAMICS
__________________________________________________________________________________________
___
74. CONCEPT QUESTIONS a. In the left design, both the 60.0 and the 525-N forces create
clockwise torques with respect to the rotational axis at the point where the tire contacts the ground.
In the right design, only the 60.0-N force creates a torque, because the line of action of the 525-N
force passes through the axis. Thus, the total torque from the two forces is greater for the left
design.
b. The man is supporting the wheelbarrow in equilibrium. Therefore, his force must create a torque
that balances the total torque from the other two forces. To do this, his torque must be
counterclockwise and have the same magnitude as the total torque from the other two forces. Since
the two forces create more torque in the left design, the magnitude of the man’s torque must be
greater for that design.
c. The magnitude of a torque is the magnitude of the force times the lever arm of the force,
according to Equation 9.1. The drawing shows that the lever arms of the man’s forces in the two
designs is the same (1.300 m). Thus, to create the greater torque necessitated by the left design, the
man must apply a greater force for that design.
SOLUTION The lever arms for the forces can be obtained from the distances shown in the
drawing for each design. We can use Equation 9.1 for the magnitude of each torque.
Remembering that counterclockwise torques are positive, we can write an expression for the zero
net torque for each design. These expressions can be solved for the man’s force F in each case:
Left design
b gb
g b gb
g b
b525 N gb0.400 mg+ b60.0 N gb0.600 mg= 189 N
F=
g
Σ τ = − 525 N 0. 400 m − 60.0 N 0. 600 m + F 1.300 m = 0
1. 300 m
Right design
b gb
g b
b60.0 N gb0 .600 m g= 27.7 N
F=
g
Σ τ = − 60.0 N 0.600 m + F 1. 300 m = 0
1.300 m
__________________________________________________________________________________________
___
75. CONCEPT QUESTIONS a. The forces in part b of the drawing cannot keep the beam in
equilibrium (no acceleration), because neither V nor W, both being vertical, can balance the
horizontal component of P. This unbalanced component would accelerate the beam to the right.
b. The forces in part c of the drawing cannot keep the beam in equilibrium (no acceleration). To
see why, consider a rotational axis perpendicular to the plane of the paper and passing through the
pin. Neither P nor H create torques relative to this axis, because their lines of action pass directly
through it. However, W does create a torque, which is unbalanced. This torque would cause the
beam to have a counterclockwise angular acceleration.
Chapter 9 Problems
357
c. The forces in part d are sufficient to eliminate the accelerations discussed above, so they can
keep the beam in equilibrium.
SOLUTION Assuming that upward and to the right are the positive directions, we obtain the
following expressions by setting the sum of the vertical and the sum of the horizontal forces equal to
zero:
Horizontal forces
P cosθ − H = 0
( 1)
Vertical forces
P sinθ + V − W = 0
( 2)
Using a rotational axis perpendicular to the plane of the paper and passing through the pin and
remembering that counterclockwise torques are positive, we also set the sum of the torques equal to
zero. In doing so, we use L to denote the length of the beam and note that the lever arms for W
and V are L/2 and L, respectively. The forces P and H create no torques relative to this axis,
because their lines of action pass directly through it.
Torques
W
c L h− VL = 0
1
2
(3 )
Since L can be eliminated algebraically, Equation (3) may be solved immediately for V:
V = 12 W =
1
2
b340 N g=
170 N
Substituting this result into Equation (2) gives
P sinθ + 12 W − W = 0
P=
W
340 N
=
= 270 N
2 sinθ 2 sin39 °
Substituting this result into Equation (1) gives
W I
F
G
H2 sinθ JKcos θ − H = 0
H=
W
340 N
=
= 210 N
2 tanθ 2 tan 39 °
__________________________________________________________________________________________
___
76. CONCEPT QUESTIONS a. Equation 8.7 from the equations of rotational kinematics indicates
that the angular displacement θ is θ = ω0 t + 12 αt 2 , where ω0 is the initial angular velocity, t is the
358
ROTATIONAL DYNAMICS
time, and α is the angular acceleration. Since both wheels start from rest, ω0 = 0 rad/s for each.
Furthermore, each wheel makes the same number of revolutions in the same time, so θ and t are
also the same for each. Therefore, the angular acceleration α must be the same for each.
b. For the hoop, all of the mass is located at a distance from the axis equal to the radius. For the
disk, some of the mass is located closer to the axis. According to Equation 9.6, the moment of
inertia is greater when the mass is located farther from the axis. Therefore, the moment of inertia of
the hoop is greater. Table 9.1 indicates that the moment of inertia of a hoop is I hoop = MR 2 , while
the moment of inertia of a disk is I disk =
1
2
MR 2 .
c. Newton’s second law for rotational motion indicates that the net external torque is equal to the
moment of inertia times the angular acceleration. Both disks have the same angular acceleration, but
the moment of inertia of the hoop is greater. Thus, the net external torque must be greater for the
hoop.
SOLUTION Using Equation 8.7, we can express the angular acceleration as follows:
θ = ω0 t + 12 αt 2
or
α=
c
2 θ − ω0 t
t
h
2
This expression for the acceleration can now be used in Newton’s second law for rotational motion.
Accordingly, the net external torque is
L2 cθ − ω t hO
M t
P
M
P
N
Q
L2 b13 rad g− 2 b0 rad / sgb8 .0 sgO
=b
4 .0 kg g
P= 0.20 N ⋅ m
b0.35 mg M
M
8
.
0
s
P
b
g
N
Q
L2 cθ − ω t hO
Στ = I α = MR M
P
t
M
P
N
Q
L2 b13 rad g− 2 b0 rad / sgb8 .0 sgO
= b
4 .0 kg g
P= 0.10 N ⋅ m
b0.35 mg M
M
8
.
0
s
P
b
g
N
Q
Hoop Στ = I hoop α = MR 2
0
2
2
2
Disk
disk
1
2
1
2
0
2
2
2
2
__________________________________________________________________________________________
___
77. CONCEPT QUESTIONS a. According to Equation 9.6, the moment of inertia for rod A is just
that of the attached particle, since the rod itself is massless. For rod A with its attached particle,
then, the moment of inertia is I A = ML2 . According to Table 9.1, the moment of inertia for rod B
is I B = 13 ML2 . The moment of inertia for rod A with its attached particle is greater.
Chapter 9 Problems
359
b. Since the moment of inertia for rod A is greater, its kinetic energy is also greater, according to
Equation 9.9 KE R = 12 Iω2 .
SOLUTION Using Equation 9.9 to calculate the kinetic energy, we find
Rod A
=
Rod B
cML hω
b0.66 kg gb0 .75 mgb4 .2 rad / sg =
3.3 J
c ML hω
b0 .66 kg gb0. 75 mgb4 .2 rad / sg =
1.1 J
KE R = 12 I A ω2 =
1
2
1
6
2
2
2
KE R = 12 I Bω2 =
=
1
2
1
2
1
3
2
2
2
2
2
__________________________________________________________________________________________
___
78. CONCEPT QUESTIONS a. The force is applied to the person in the counterclockwise
direction by the bar that he grabs when climbing aboard.
b. According to Newton’s action-reaction law, the person must apply a force of equal magnitude
and opposite direction to the bar and the carousel. This force acts on the carousel in the clockwise
direction and, therefore, creates a clockwise torque.
c. Since the carousel is moving counterclockwise, the clockwise torque applied to it when the
person climbs aboard decreases its angular speed.
SOLUTION We consider a system consisting of the person and the carousel. Since the carousel
rotates on frictionless bearings, no net external torque acts on this system. Therefore, angular
momentum is conserved, and we can set the total angular momentum of the system before the
person hops on equal to the total angular momentum afterwards. Afterwards, the angular
momentum includes a contribution from the person. According to Equation 9.6, his moment of
inertia is I person = MR 2 , since he is at the outer edge of the carousel.
I carousel ωf + I person ωf
144424443
= I carousel ω0
1424
3
Initial total
angular momentum
Final total
angular mo mentum
ωf =
I carousel ω0
I carousel + I person
=
I carousel ω0
I carousel + MR 2
c125 kg ⋅ m hb3.14 rad / sg =
=
125 kg ⋅ m + b
40. 0 kg g
b1.50 m g
2
2
2
1.83 rad / s
360
ROTATIONAL DYNAMICS
__________________________________________________________________________________________
___