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CHAPTER
9 ROTATIONAL DYNAMICS
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION The magnitude of the torque produced by a force F is
given by the magnitude of the force times the lever arm. When a long pipe is slipped over a
wrench handle, the length of the wrench handle is effectively increased. The same force can
then be applied with a larger lever arm. Therefore, a greater torque can be used to remove
the "frozen" nut.
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2.
REASONING AND SOLUTION
where is the lever arm.
The torque produced by a force F is given by τ = F
a. A large force can be used to produce a small torque by choosing a small lever arm. Then
the product F is small even though the magnitude of the force, F, is large. If the line of
action passes through the axis of rotation, = 0 m, and the torque is zero regardless of how
large F is.
b. A small force may be used to produce a large torque by choosing a large lever arm .
Then, the product F can be large even though the magnitude of the force F is small.
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3.
REASONING AND SOLUTION In a cassette deck, the magnetic tape applies a torque to
the supply reel. The magnitude of the torque is equal to the product of the tension FT in the
tape, assumed to be constant, and the lever arm . As shown below, the lever arm is the
radius of the tape remaining on the supply reel. Figure A shows the reel and tape at an
earlier time; figure B shows the tape and the reel at a later time.
F
T
F
T
Figure A
Figure B
As the reel becomes empty, the lever arm gets smaller, as suggested in the figures above.
Therefore, the torque exerted on the reel by the tape becomes smaller.
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Chapter 9 Conceptual Questions
4.
425
REASONING AND SOLUTION A flat rectangular sheet of plywood is fixed so that it can
rotate about an axis perpendicular to the sheet through one corner, as shown in the figure. In
order to produce the largest possible torque with the force F (acting in the plane of the
paper), the force F must be applied with the largest possible lever arm. This can be attained
by orienting the force F so that it is perpendicular to the diagonal that passes through the
rotation axis as shown below. The lever arm is equal to the length of the diagonal of the
plywood.
Axis of
rotation
F
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5.
REASONING AND SOLUTION To unscrew either cap, you apply a force tangentially to it
and rotate it about an axis perpendicular to the end of the cap at its center. It is the torque
created by this force that turns the cap. The greater the torque, the more easily is the cap
unscrewed. For a given force, a greater torque is created when the force has a greater lever
arm than when it has a smaller lever arm. In the case of the two caps, the lever arms are the
radii of the caps, and the “easy-off” cap has a greater radius than the normal cap. Thus, it is
the greater radius of the “easy-off” cap that makes it easier to open.
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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7.
REASONING AND SOLUTION If we assume that the branch is fairly uniform, then in
the spring, just when the fruit begins to grow, the center of gravity of the branch is near the
center of the branch. As the fruit grows, the mass at the outer end of the branch increases.
Therefore, as the fruit grows, the center of gravity of the fruit-growing branch moves along
the branch toward the fruit.
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8.
SSM REASONING AND SOLUTION For a rigid body in equilibrium, there can be no
net torque acting on the object with respect to any axis. Consider an axis perpendicular to
426 ROTATIONAL DYNAMICS
the plane of the paper at the right end of the rod. The lines of action of both F1 and F3 pass
through this axis. Therefore, no torque is created by either of these forces. However, the
force F2 does create a torque with respect to this axis. As a result, there is a net torque
acting on the object with respect to at least one axis, and the three forces shown in the
drawing cannot keep the object in equilibrium.
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9.
REASONING AND SOLUTION The
forces that keep the right section of the
ladder in equilibrium are shown in the
free-body diagram at the right.
They are: the weight of the section, mg,
the normal force exerted on the ladder by
the floor, FN, the tension in the crossbar,
FT, and the reaction force due to the left
side of the ladder, FR.
FR
FT
mg
FN
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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
FN
Note that the total weight of the wine rack
and the bottle acts at the center of gravity
(CG), which must be located exactly above
the point where the normal force acts.
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11. REASONING AND SOLUTION The torque required to give the triangular sheet an
angular acceleration α is given by the rotational analog of Newton's second law for a rigid
body: ∑ τ = Iα . The greatest net torque will be required about the axis for which the
moment of inertia I is the greatest. The greatest value for I occurs when the greatest portion
of the mass of the sheet is far from the axis. This will be for axis B. Therefore, for axis B,
the greatest net external torque is required to bring the rectangular plate up to 10.0 rad/s in
10.0 s, starting from rest.
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12. REASONING AND SOLUTION The object has an angular velocity and an angular
acceleration due to torques that are present.
Chapter 9 Conceptual Questions
427
a. If additional torques are applied so as to make the net torque suddenly equal to zero, the
object will continue to rotate with constant angular velocity. The angular velocity will
remain constant at the value that it had at the instant that the net torque became zero.
b. If all the torques are suddenly removed, the effect is the same as when the net torque is
zero as in part (a) above. That is, the object will continue to rotate with constant angular
velocity. The angular velocity will remain constant at the value that it had at the instant that
the torques are removed.
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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 TR, each in the same
direction, so that the net torque on the probe is 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. SSM 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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15. REASONING AND SOLUTION The translational kinetic energy of a rigid body depends
only on the mass and the speed of the body. It does not depend on how the mass is
distributed. Therefore, for purposes of computing the body's translational kinetic energy,
the mass of a rigid body can be considered as concentrated at its center of mass.
Unlike the translational kinetic energy, the moment of inertia of a body does depend on the
location and orientation of the axis relative to the particles that make up the rigid body.
Therefore, for purposes of computing the body's moment of inertia, the mass of a rigid body
cannot be considered as concentrated at its center of mass. The distance of any individual
mass particle from the axis is important.
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428 ROTATIONAL DYNAMICS
16. REASONING AND SOLUTION A thin sheet of
plastic is uniform and has the shape of an equilateral
triangle, as shown in the figure at the right. The axes
A and B are two rotation axes that are perpendicular to
the plane of the triangle. If the triangle rotates about
each axis with the same angular speed ω, the
rotational kinetic energy of the sheet is given by
1
2
KE = Iω . The rotational kinetic energy will be
B
A
2
greater if the rotation occurs about the axis for which
the triangle has the larger moment of inertia I. The value for I will be greater when more of
the mass is located farther away from the axis, as is the case for axis B. Therefore, the
triangle has the greater rotational kinetic energy when it rotates about axis B.
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17. REASONING AND SOLUTION With feet flat on the floor and body straight, you can
lean over until your center of gravity is above your toes. If you lean any farther, it is no
longer possible to balance the torque due to your weight with a torque created by the upward
reaction force that pushing downward with your toes produces. Bob’s center of gravity
must be closer to the ground. The reason is that, with the center of gravity closer to the
ground, a greater amount of leaning can occur before it moves beyond his toes.
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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 ) Iω 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 ω is the angular speed
about the center of mass at the bottom. From conservation of mechanical energy
Mgh = 21 Iω 2 + 21 Mv 2
Since the objects roll down the incline, ω = 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)
Chapter 9 Conceptual Questions
429
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.
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19. REASONING AND SOLUTION A woman sits on the spinning seat of a piano stool with
her arms folded. Let the woman and the piano stool comprise the system. When the woman
extends her arms, the torque that she applies is an internal torque. Since there is no net
external torque acting on the system, the angular momentum of the woman and the piano
stool, Iω, must remain constant.
a. With her arms extended, the woman's moment of inertia I about the rotation axis is
greater than it was when her arms were folded. Since the product of Iω must remain constant
and I increases, ω must decrease. Therefore, the woman's angular velocity decreases.
b. As discussed above, since there is no net external torque acting on the system, the
angular momentum of the system remains the same.
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20. REASONING AND SOLUTION Consider the earth to be an isolated system. Note that
the earth rotates about an axis that passes through the North and South poles and is
perpendicular to the plane of the equator. If the ice cap at the South Pole melted and the
water were uniformly distributed over the earth's oceans, the mass at the South Pole would
be uniformly distributed and, on average, be farther from the earth's rotational axis. The
moment of inertia of the earth would, therefore, increase. Since the earth is an isolated
system, any torques involved in the redistribution of the water would be internal torques;
therefore, the angular momentum of the earth must remain the same. If the moment of
inertia increases, and the angular momentum is to remain constant, the angular velocity of
the earth must decrease.
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21. REASONING AND SOLUTION Note that the earth rotates about an axis that passes
through the North and South poles and is perpendicular to the plane of the equator. When
rivers like the Mississippi carry sediment toward the equator, they redistribute the mass
from a more uniform distribution to a distribution with more mass concentrated around the
equator. This increases the moment of inertia of the earth. If the earth is considered to be an
isolated system, then any torques involved in the redistribution of mass are internal torques.
Therefore, the angular momentum of the earth must remain constant. The moment of inertia
increases, and the angular momentum must remain the same; therefore, the angular velocity
must decrease.
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430 ROTATIONAL DYNAMICS
22. REASONING AND SOLUTION Let the system be the rotating cloud of interstellar gas.
The gravitational force that pulls the particles together is internal to the system; hence, the
torque resulting from the gravitational force is an internal torque. Since there is no net
external torque acting on the collapsing cloud, the angular momentum, Iω , of the cloud
must remain constant. Since the cloud is shrinking, its moment of inertia I decreases.
Since the product Iω remains constant, the angular velocity ω must increase as I decreases.
Therefore, the angular velocity of the formed star would be greater than the angular velocity
of the original gas cloud.
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23. SSM 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.
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24. REASONING AND SOLUTION The
box that creates the greatest torque is the
one that appears shaded in the drawing at
the right.
All boxes give rise to the same force mg.
Each box produces a torque about the
axle. The box that produces the greatest
torque is the one for which the lever arm
is the greatest. By inspection of the
drawing, the lever arm relative to the axle
is greatest for the shaded box.
mg
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Chapter 9 Problems
431
CHAPTER 9 ROTATIONAL DYNAMICS
PROBLEMS
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1.
SSM REASONING According to Equation 9.1, we have
Magnitude of torque = F
where F is the magnitude of the applied force and is the lever arm. From the figure in the text,
the lever arm is given by = (0.28 m) sin 50.0° . Since both the magnitude of the torque and
are known, Equation 9.1 can be solved for F.
SOLUTION Solving Equation 9.1 for F, we have
45 N ⋅ m
= 2.1×102 N
(0.28 m) sin 50.0°
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F=
2.
Magnitude of torque
REASONING AND SOLUTION
=
According to Equation 9.1, we have
Magnitude of torque = F
where F is the magnitude of the applied force and is the lever arm. For an axis passing
through the cable car at its center and perpendicular to the ground, the lever arm is one-half
the length of the car or 4.60 m. Both people rotate the car in the same direction, so the
torques that they create reinforce one another. Therefore, using Equation 9.1 to express the
magnitude of each torque, we find that the magnitude of the net torque is
Magnitude of net torque = (185 N )( 4.60 m ) + (185 N )( 4.60 m ) = 1.70 × 103 N ⋅ m
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3.
REASONING The drawing shows the wheel as it rolls to the
right, so the torque applied by the engine is assumed to be
clockwise about the axis of rotation. The force of static friction
that the ground applies to the wheel is labeled as fs. This force
produces a counterclockwise torque τ about the axis of
rotation, which is given by Equation 9.1 as τ = fs , where is
the lever arm. Using this relation we can find the magnitude fs
of the static frictional force.
Axis of
rotation
r
fs
SOLUTION The countertorque is given as τ = fs , where fs is the magnitude of the static
frictional force and
is the lever arm. The lever arm is the distance between the line of
432 ROTATIONAL DYNAMICS
action of the force and the axis of rotation; in this case the lever arm is just the radius r of
the tire. Solving for fs gives
τ 295 N ⋅ m
fs = =
= 843 N
0.350 m
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4.
REASONING In both parts of the problem, the magnitude
of the torque is given by Equation 9.1 as the magnitude F
of the force times the lever arm . In part (a), the lever
arm is just the distance of 0.55 m given in the drawing.
However, in part (b), the lever arm is less than the given
distance and must be expressed using trigonometry as
= ( 0.55 m ) sin θ . See the drawing at the right.
axis
θ
0.55 m
lever arm
49 N
SOLUTION
a. Using Equation 9.1, we find that
Magnitude of torque = F = ( 49 N )( 0.55 m ) = 27 N ⋅ m
b. Again using Equation 9.1, this time with a lever arm of
= ( 0.55 m ) sin θ , we obtain
Magnitude of torque = 15 N ⋅ m = F = ( 49 N )( 0.55 m ) sin θ
sin θ =
5.
15 N ⋅ m
( 49 N )( 0.55 m )
or
⎡
⎤
15 N ⋅ m
⎥ = 34°
⎣⎢ ( 49 N )( 0.55 m ) ⎦⎥
θ = sin −1 ⎢
SSM REASONING The torque on either wheel is given by Equation 9.1, τ = 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
τ truck F truck Frtruck rtruck 0.25 m
=
=
=
=
= 1.3
τ car
F car
Frcar
rcar
0.19 m
Chapter 9 Problems
6.
433
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
T = (2.50 m) cos 32.0°
W
T
W = (2.50 m) sin 32.0°
SOLUTION
a. Using Equation 9.1, we find that the magnitude of the torque due to the weight W is
Magnitude of torque = W
W
= (10 200 N )( 2.5 m ) sin 32° = 13 500 N ⋅ m
b. Using Equation 9.1, we find that the magnitude of the torque due to the thrust T is
Magnitude of torque = T
T
= ( 62 300 N )( 2.5 m ) cos 32° = 132 000 N ⋅ m
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7.
SSM 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 add to zero.
F
2
pinned
end
30.0°
2
1
F
1
TOP VIEW
SOLUTION If we take counterclockwise torques as positive, then the torque of the first
force about the pin is
τ 1 = F1 1 = (2.00 N)(1.00 m) = 2.00 N ⋅ m
The torque due to the second force is
τ 2 = – F2 (sin 30.0 ° )
2
= – (6.00 N)(sin 30.0° )
2
= –( 3.00 N)
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
434 ROTATIONAL DYNAMICS
2 .00 N ⋅ m +[– ( 3.00 N)
Thus,
2
=
2
]=0
2.00 N ⋅ m
= 0.667 m
3.00 N
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8.
REASONING Each of the two forces produces a torque about the axis of rotation, one
clockwise and the other counterclockwise. By setting the sum of the torques equal to
zero ( Στ = 0 ) , we will be able to determine the distance x in the drawing
90°
2
F2 = 6.00 N
Meter Stick
60.0°
x
F1 = 4.00 N
Axis of
Rotation
Table (Top view)
The torque τ1 produced by the force F1 is given by Equation 9.1 as
SOLUTION
τ1 = +F1
1,
where the lever arm is
1
= 1.00 m. It is a positive torque, since it tends to
produce a counterclockwise rotation. The torque τ2 produced by F2 is τ2 = −F2
2,
where
= x sin 60.0°. It is a negative torque, since it tends to produce a clockwise rotation.
Setting the net torque equal to zero, we have
2
+ F1
1
+ ( − F2
2
) =0
or
∑τ
+F1 (1.00 m ) − F2 ( x sin 60.0° ) = 0
1
2
Solving for x gives
x=
F1 (1.00 m )
=
( 4.00 N )(1.00 m ) =
( 6.00 N ) sin 60.0°
0.770 m
F2 sin 60.0°
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9.
REASONING AND SOLUTION The torque produced by each force of magnitude F is
given by Equation 9.1, τ = F , where is the lever arm and the torque is positive since
each force causes a counterclockwise rotation. 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.
Chapter 9 Problems
435
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 + τ C = 0 + FL = 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 + τ C = 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 + τ C = FL + 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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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
11. REASONING The drawing shows
the forces acting on the person. It
also shows the lever arms for a
rotational axis 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.
and b = 0.300 m
FHANDS
FFEET
Axis
W
W = 0.840 m
HANDS = 1.250 m
436 ROTATIONAL DYNAMICS
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
W
FHANDS =
W
W
W
− FHANDS
=
HANDS
=0
b584 N gb 0.840 mg = 392 N
HANDS
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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12. 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 y 1 + W2 y 2 + W3 y 3
W1 + W2 + W3
b 438 N gb1.28 mg + b144 N gb 0.760 m g + b87 N gb 0.250 mg =
1.03 m
438 N + 144 N + 87 N
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13. SSM REASONING The drawing shows the bridge and the four forces that act on it: the
upward force F1 exerted on the left end by the support, the force due to the weight Wh of the
hiker, the weight Wb of the bridge, and the upward force F2 exerted on the right side by the
support. Since the bridge is in equilibrium, the sum of the torques about any axis of rotation
must be zero ( Στ = 0 ) , and the sum of the forces in the vertical direction must be
(
)
zero ΣFy = 0 . These two conditions will allow us to determine the magnitudes of F1 and
F2.
Chapter 9 Problems
F1
+y
F2
Axis
Wh
1
5
437
+τ
+x
L
1
2
L
Wb
SOLUTION
a. We will begin by taking the axis of rotation about the right end of the bridge. The torque
produced by F2 is zero, since its lever arm is zero. When we set the sum of the torques equal
to zero, the resulting equation will have only one unknown, F1, in it. Setting the sum of the
torques produced by the three forces equal to zero gives
Στ = − F1L + Wh
( 54 L ) + Wb ( 12 L ) = 0
Algebraically eliminating the length L of the bridge from this equation and solving for F1
gives
F1 = 54 Wh + 12 Wb = 54 ( 985 N ) + 12 ( 3610 N ) = 2590 N
b. Since the bridge is in equilibrium, the sum of the forces in the vertical direction must be
zero:
Σ Fy = F1 − Wh − Wb + F2 = 0
Solving for F2 gives
F2 = − F1 + Wh + Wb = −2590 N + 985 N + 3610 N = 2010 N
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14. REASONING Multiple-Concept Example 8 discusses the static stability factor (SSF) and
rollover. In that example, it is determined that the maximum speed v at which a vehicle can
negotiate a curve of radius r is related to the SSF according to v = rg ( SSF ) . No value is
given for the radius of the turn. However, by applying this result separately to the sport
utility vehicle (SUV) and to the sports car (SC), we will be able to eliminate r algebraically
and determine the maximum speed at which the sports car can negotiate the curve without
rolling over.
SOLUTION Applying v = rg ( SSF ) to each vehicle, we obtain
vSUV = rg ( SSF )SUV
and
vSC = rg ( SSF )SC
438 ROTATIONAL DYNAMICS
Dividing these two expressions gives
vSC
vSUV
=
rg ( SSF )SC
rg ( SSF )SUV
vSC = vSUV
or
(SSF)SC
= (18 m/s )
(SSF )SUV
1.4
= 24 m/s
0.80
15. REASONING Since the forearm is in equilibrium, the sum of the torques about any axis of
rotation must be zero ( Στ = 0 ) . For convenience, we will take the elbow joint to be the axis
of rotation.
SOLUTION Let M and F be the magnitudes of the forces that the flexor muscle and test
apparatus, respectively, exert on the forearm, and let M and F be the respective lever
arms about the elbow joint. Setting the sum of the torques about the elbow joint equal to
zero (with counterclockwise torques being taken as positive), we have
Στ = M
Solving for M yields
M=
F
F
M
=
M
−F
F
=0
(190 N )( 0.34 m ) = 1200 N
0.054 m
The direction of the force is to the left .
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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
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
______________________________________________________________________________
Chapter 9 Problems
17. SSM 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.
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 = Lx − U x = 0 or Lx = U x
∑ Fy = Ly − W = 0
or
Ly = W = 140 N
439
Ux
h = 2.1 m
W
Ly
(1)
Lx
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
⎝2⎠
(3)
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 (140 N)(0.81 m)
=
= 27 N
2h
2(2.1 m)
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 on the upper hinge. This
force points horizontally to the right , according to Newton's third law.
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.
440 ROTATIONAL DYNAMICS
From the Pythagorean theorem,
Dx
D=
D + D = ( −27 N) + ( −140 N) =
2
x
2
y
2
2
143 N
θ
The angle θ is given by
θ = tan −1
FG –140 N IJ = 79 °
H –27 N K
D
y
The force is directed 79° below the horizontal .
18. 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 M = 1.20 × 103 N
______________________________________________________________________________
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
where W is the weight of the plane, and
w
w
+ Ff
and
f
f
=0
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 the data gives
Chapter 9 Problems
441
ΣFy = 1.60 × 105 N + 2Fr − 1.00 × 106 N = 0
Fr = 4.20 ×105 N
______________________________________________________________________________
20. REASONING The drawing shows the beam and the five forces that act on it: the horizontal
and vertical components Sx and Sy that the wall exerts on the left end of the beam, the
weight Wb of the beam, the force due to the weight Wc of the crate, and the tension T in the
cable. The beam is uniform, so its center of gravity is at the center of the beam, which is
where its weight can be assumed to act. Since the beam is in equilibrium, the sum of the
torques about any axis of rotation must be zero ( Στ = 0 ) , and the sum of the forces in the
(
)
horizontal and vertical directions must be zero ΣFx = 0, ΣFy = 0 . These three conditions
will allow us to determine the magnitudes of Sx, Sy, and T.
T
+y
50.0°
+τ
30.0°
+x
Wc
Sy
Wb
Axis
30.0°
Sx
SOLUTION
a. We will begin by taking the axis of rotation to be at the left end of the beam. Then the
torques produced by Sx and Sy are zero, since their lever arms are zero. When we set the
sum of the torques equal to zero, the resulting equation will have only one unknown, T, in it.
Setting the sum of the torques produced by the three forces equal to zero gives (with L equal
to the length of the beam)
Στ = −Wb
( 12 L cos 30.0°) − Wc ( L cos 30.0° ) + T ( L sin 80.0°) = 0
Algebraically eliminating L from this equation and solving for T gives
442 ROTATIONAL DYNAMICS
T=
=
Wb
( 12 cos 30.0°) + Wc ( cos 30.0°)
sin 80.0°
(1220 N ) ( 12 cos 30.0° ) + (1960 N )( cos 30.0° )
sin 80.0°
= 2260 N
b. Since the beam is in equilibrium, the sum of the forces in the vertical direction must be
zero:
Σ Fy = + S y − Wb − Wc + T sin 50.0° = 0
Solving for Sy gives
S y = Wb + Wc − T sin 50.0° = 1220 N + 1960 N − ( 2260 N ) sin 50.0° = 1450 N
The sum of the forces in the horizontal direction must also be zero:
Σ Fx = + S x − T cos 50.0° = 0
so that
S x = T cos 50.0° = ( 2260 N ) cos 50.0° = 1450 N
______________________________________________________________________________
21. SSM WWW 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.
2.00 m
person's feet
person's head
x
425 N
center of gravity
of person
315 N
W
a. Applying Newton's second law to the vertical direction gives
315 N + 425 N – W = 0
or
W = 7.40 ×102 N, downward
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
or
x = 0.851 m
______________________________________________________________________________
Chapter 9 Problems
443
0.61 m
22. REASONING The arm, being stationary, is
in equilibrium, since it has no translational or
angular acceleration. Therefore, the net
0.28 m
external force and the net external torque
98 N
acting on the arm are zero. Using the fact
0.069 m
that the net external torque is zero will allow
us to determine the magnitude of the force
cg
Axis
M. The drawing at the right shows three
forces: M, the 47-N weight of the arm acting
29°
47 N
at the arm’s center of gravity (cg), and the
98-N force that acts upward on the right end
M
of the arm. The 98-N force is applied to the
arm by the ring. It is the reaction force that
(0.070 m) sin 29°
arises in response to the arm pulling downward on the ring. Its magnitude is 98 N, because
it supports the 98-N weight hanging from the pulley system. Other forces also act on the
arm at the shoulder joint, but we can ignore them. The reason is that their lines of action
pass directly through the axis at the shoulder joint, which is the axis that we will use to
determine torques. Thus, these forces have zero lever arms and contribute no torque.
SOLUTION The magnitude of each individual torque is the magnitude of the force times
the corresponding lever arm. The forces and their lever arms are as follows:
Force
Lever Arm
98 N
0.61 m
47 N
0.28
M
(0.069 m) sin 29°
Each torque is positive if it causes a counterclockwise rotation and negative if it causes a
clockwise rotation about the axis. Thus, since the net torque must be zero, we see that
( 98 N )( 0.61 m ) − ( 47 N )( 0.28 m ) − M ⎡⎣( 0.069 m ) sin 29°⎤⎦ = 0
Solving for M gives
M=
( 98 N )( 0.61 m ) − ( 47 N )( 0.28 m ) = 1400 N
( 0.069 m ) sin 29°
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 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× 103 N .
444 ROTATIONAL DYNAMICS
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× 103 N . Since the forces must add to give a
net force of zero, we know that the direction of F is downward .
______________________________________________________________________________
24. 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 × 102 N .
______________________________________________________________________________
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 .
______________________________________________________________________________
26. 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 following figure.
In the vertical direction −W + FN = 0, so that
Chapter 9 Problems
FN = W = mg = (10.0 kg)(9.80 m/s2) = 98.0 N
445
R
In the horizontal direction it is clear that R = T. The net
torque about the base of the ladder is
o
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
FN
Substituting for W and using R = T, we obtain
T
( 9.80 N )( 2.00 m ) cos 75.0°
= 17.5 N
( 3.00 m ) sin 75.0°
______________________________________________________________________________
T=
27.
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 µsP2 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.
Wall 1
P1
Wall 2
1
µ s P2
W
θ
P2
SOLUTION Then, since the horizontal forces balance to zero, we have
P1 – P2 = 0
(1)
µsP2 – 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⎞
W ⎜ ⎟ ( cos θ ) − P1L ( sin θ ) = 0
⎝2⎠
(3)
Rearranging Equation (3) gives
tan θ =
W
2 P1
(4)
446 ROTATIONAL DYNAMICS
But W = µsP2 according to Equation (2), and P2 = P1 according to Equation (1). Therefore,
W = µsP1, which can be substituted in Equation (4) to show that
tan θ =
or
µs P1
2 P1
=
µs
2
=
0.98
2
θ = tan–1(0.49) = 26°
From the drawing at the right,
cos θ =
d
L
L
Therefore, the longest board that can be propped between
the two walls is
θ
d
1.5 m
d
=
= 1.7 m
cos θ cos 26°
______________________________________________________________________________
L=
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 I disk = 12 M disk R 2 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
I stool = I disk + 3 I rod = 21 M disk R 2 + 3 M rod R 2
=
1
2
b1.2 kg gb 0.16 mg + 3b 0.15 kg gb 0.16 m g
2
2
= 0.027 kg ⋅ m 2
______________________________________________________________________________
29. REASONING According to Newton’s second law for rotational motion, Στ = Iα, the
angular acceleration α of the blades is equal to the net torque Στ applied to the blades
divided by their total moment of inertia I, both of which are known.
SOLUTION The angular acceleration of the fan blades is
Στ
1.8 N ⋅ m
=
= 8.2 rad/s 2
(9.7)
2
I
0.22 kg ⋅ m
______________________________________________________________________________
α=
Chapter 9 Problems
447
30. REASONING According to Table 9.1 in the text, the moment of inertia I of the disk is
1
2
I = MR 2 , where M is the disk’s mass and R is its radius. Thus, we can determine the mass
from this expression, provided we can obtain a value for I. To obtain the value for I, we will
use Newton’s second law for rotational motion, Στ = Iα (Equation 9.7), where Στ is the net
torque and α is the angular acceleration.
SOLUTION From the moment of inertia of the disk, we have
1
2
I = MR 2
or
M=
2I
R2
(1)
Using Newton’s second law for rotational motion, we find for I that
Στ = I α
or
I=
Στ
α
Substituting this expression for I into Equation (1) gives
M=
2 I 2Στ
=
R 2 R 2α
(2)
The net torque Στ is due to the 45-N force alone. According to Equation 9.1, the magnitude
of the torque that a force of magnitude F produces is F , where is the lever arm of the
force. In this case, we have = R , since the force is applied tangentially to the disk and
perpendicular to the radius. Substituting Στ = FR into Equation (2) gives
M=
2 ( 45 N )
2Στ 2 FR 2 F
=
=
=
= 5.0 kg
R 2α R 2α Rα ( 0.15 m ) 120 rad/s 2
(
)
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
______________________________________________________________________________
448 ROTATIONAL DYNAMICS
32. REASONING The rotational analog of Newton's second law is given by Equation 9.7,
∑ τ = Iα . Since the person pushes on the outer edge of one section of the door with a force
F that is directed perpendicular to the section, the torque exerted on the door has a
magnitude of FL, where the lever arm L is equal to the width of one section. Once the
moment of inertia is known, Equation 9.7 can be solved for the angular acceleration α.
The moment of inertia of the door relative to the rotation axis is I = 4IP, where IP is the
moment of inertia for one section. According to Table 9.1, we find I P = 13 ML2 , so that the
rotational inertia of the door is I = 43 ML2 .
SOLUTION Solving Equation 9.7 for α, and using the expression for I determined above,
we have
α=
4
3
FL
=
ML2
4
3
F
68 N
= 4
= 0.50 rad/s 2
ML 3 (85 kg)(1.2 m)
______________________________________________________________________________
33. SSM 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
I hoop = MR 2 = (1.20 kg)(0.330 m)2 = 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
I rod = 13 ML2 = 13 (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 = I hoop + 50 I rod = 0.131 kg ⋅ m 2 + 50(3.6 × 10 –4 kg ⋅ m 2 ) = 0.149 kg ⋅ m 2
______________________________________________________________________________
34. REASONING The angular acceleration α that results from the application of a net external
torque Στ to a rigid object with a moment of inertia I is given by Newton’s second law for
Στ
(Equation 9.7). By applying this equation to each of the two
rotational motion: α =
I
situations described in the problem statement, we will be able to obtain the unknown
angular acceleration.
Chapter 9 Problems
SOLUTION The drawings at the right illustrate the
two types of rotational motion of the object. In
applying Newton’s second law for rotational motion,
we need to keep in mind that the moment of inertia
depends on where the axis is. For axis 1 (see top
drawing), piece B is rotating about one of its ends,
so according to Table 9.1, the moment of inertia is
449
B
Axis 1
1
3
I for axis 1 = M B L2B , where MB and LB are the mass
Axis 2
and length of piece B, respectively. For axis 2 (see
bottom drawing), piece A is rotating about an axis
through its midpoint. According to Table 9.1 the
moment of inertia is I for axis 2 =
1
12
M A L2A . Applying
A
the second law for each of the axes, we obtain
α for axis 1 =
Στ
I for axis 1
and
α for axis 2 =
Στ
I for axis 2
As given, the net torque Στ is the same in both expressions. Dividing the expression for
axis 2 by the expression for axis 1 gives
Στ
α for axis 2 I for axis 2 I for axis 1
=
=
Στ
α for axis 1
I for axis 2
I for axis 1
Using the expressions from Table 9.1 for the moments of inertia, this result becomes
α for axis 2 I for axis 1
=
=
α for axis 1 I for axis 2
1
M B L2B
3
1
M A L2A
12
=
4M B L2B
M A L2A
Noting that M A = 2M B and LA = 2 LB , we find that
α for axis 2 4 M B L2B
4M B L2B
1
=
=
=
2
2
2
α for axis 1 M A LA 2M ( 2 L )
B
B
1
2
α for axis 2 = α for axis 1 =
4.6 rad/s 2
= 2.3 rad/s 2
2
450 ROTATIONAL DYNAMICS
35. REASONING
a. The angular acceleration α is defined as the change, ω − ω0, in the angular velocity
divided by the elapsed time t (see Equation 8.4). Since all these variables are known, we can
determine the angular acceleration directly from this definition.
b. The magnitude τ of the torque is defined by Equation 9.1 as the product of the magnitude
F of the force and the lever arm . The lever arm is the radius of the cylinder, which is
known. Since there is only one torque acting on the cylinder, the magnitude of the force can
be obtained by using Newton’s second law for rotational motion, Στ = F = Iα .
SOLUTION
a. From Equation 8.4 we have that α = (ω – ω0)/t. We are given that ω0 = 76.0 rad/s,
ω = 12 ω0 = 38.0 rad/s , and t = 6.40 s, so
α=
ω − ω0
t
=
38.0 rad/s − 76.0 rad/s
= −5.94 rad/s 2
6.40 s
The magnitude of the angular acceleration is 5.94 rad/s 2 .
b. Using Newton’s second law for rotational motion, we have that Στ = F = Iα . Thus, the
magnitude of the force is
F=
Iα
0.615 kg ⋅ m 2 )( 5.94 rad/s 2 )
(
=
= 44.0 N
0.0830 m
______________________________________________________________________________
36. REASONING The drawing shows the two identical sheets and the axis of rotation for each.
Axis of Rotation
L1 = 0.40 m
L2 = 0.20 m
The time t it takes for each sheet to reach its final angular velocity depends on the angular
acceleration α of the sheet. This relation is given by Equation 8.4 as t = (ω − ω0 ) / α , where
ω and ω0 are the final and initial angular velocities, respectively. We know that ω0 = 0 rad/s
in each case and that the final angular velocities are the same. The angular acceleration can
be determined by using Newton’s second law for rotational motion, Equation 9.7, as
α = τ / I , where τ is the torque applied to a sheet and I is its moment of inertia.
Chapter 9 Problems
451
SOLUTION Substituting the relation α = τ / I into t = (ω − ω0 ) / α gives
t=
(ω − ω0 ) = (ω − ω0 ) = I (ω − ω0 )
⎛τ ⎞
⎜ ⎟
⎝I⎠
α
τ
The time it takes for each sheet to reach its final angular velocity is:
tLeft =
I Left (ω − ω0 )
and
τ
tRight =
I Right (ω − ω0 )
τ
The moments of inertia I of the left and right sheets about the axes of rotation are given by
the following relations, where M is the mass of each sheet (see Table 9.1 and the drawings
above): I Left = 13 ML12 and I Right = 13 ML22 . Note that the variables M, ω, ω0, and τ are the
same for both sheets. Dividing the time-expression for the right sheet by that for the left
sheet gives
I Right (ω − ω0 )
tRight
I Right 13 ML22 L2
τ
=
=
=
= 22
1 ML2
tLeft
I Left
I Left (ω − ω0 )
L1
1
3
τ
Solving this expression for tRight yields
⎛ L22 ⎞
( 0.20 m )2
tRight = tLeft ⎜ 2 ⎟ = ( 8.0 s )
= 2.0 s
2
⎜L ⎟
0.40
m
(
)
⎝ 1⎠
______________________________________________________________________________
37. SSM 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
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
2.67 kg ⋅ m 2
I
R=
=
= 1.16 m
m
2.00 kg
______________________________________________________________________________
452 ROTATIONAL DYNAMICS
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.
38. REASONING AND SOLUTION
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
Setting ω0 = 0 rad/s and ω = vT/r, the force becomes
⎛v ⎞
I⎜ T ⎟
0.065 kg ⋅ m 2 ( 5.0 m/s )
r t ⎠ I vT
⎝
=
=
= 460 N
M=
L
Lrt ( 0.025 m )( 0.28 m )( 0.10 s )
______________________________________________________________________________
(
)
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,
α=
ω − ω0
t
=
3.7 rad/s − 13.1 rad/s
= −3.1 rad/s 2
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 = mr 2 . The net
torque that acts on the wheel due to the brake pads is, therefore,
∑ τ = Iα = (mr 2 )α
(1)
From Equation 9.1, the net torque that acts on the wheel due to the action of the two brake
pads is
(2)
∑ τ = –2 f k
where fk 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)
Chapter 9 Problems
f k = µk FN
453
(3)
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( µk FN ) = (mr 2 )α
– mr 2α –(1.3 kg)(0.33 m) 2 (–3.1 rad/s 2 )
=
= 0.78 N
2µk
2(0.85)(0.33 m)
______________________________________________________________________________
FN =
40. REASONING
To determine the angular
acceleration of the pulley and the tension in the
cord attached to the block, we will apply
Newton’s second law to the pulley and the
block separately. Only one external forces acts
on the pulley, as its free-body diagram at the
right shows. This force T is due to the tension
in the cord. The torque that results from this
force is the net torque acting on the pulley and
obeys Newton’s second law for rotational
motion (Equation 9.7). Two external forces act
on the block, as its free-body diagram at the
right indicates. These are (1) the force T′ due to
the tension in the cord and (2) the weight mg of
the block. The net force that results from these
forces obeys Newton’s second law for
translational motion (Equation 4.2b).
R
+y
+τ
T
+x
Free-body diagram for pulley
T′
mg
Free-body diagram for block
SOLUTION Applying Newton’s second law for rotational motion to the pulley gives
Στ = +TR = Iα
(1)
where we have written the torque as the magnitude of the tension force times the lever arm
(the radius) as specified by Equation 9.1. In addition, we have assigned this torque a
positive value, since it causes a counterclockwise rotation of the pulley. To obtain a value
for T, we note that the tension has the same magnitude everywhere in the massless cord, so
that T = T ′ . Thus, by applying Newton’s second law for translation (Equation 4.2b), we
obtain
ΣF = T ′ − mg = −ma
or
T = T ′ = mg − ma
454 ROTATIONAL DYNAMICS
where we have used a to denote the magnitude of the vertical acceleration of the block and
included the minus sign to account for the fact that the block is accelerating downward.
Substituting this result for T into Equation (1) gives
( mg − ma ) R = Iα
(2)
To proceed further, we must deal with a. Note that the pulley rolls without slipping against
the cord, so a and α are related according to a = Rα (Equation 8.13). With this
substitution, Equation (2) becomes
⎡⎣ mg − m ( Rα ) ⎤⎦ R = Iα
mgR = mR 2α + Iα
or
Solving for α, we find that
(
)
( 2.0 kg ) 9.80 m/s2 ( 0.040 m )
mgR
α=
=
= 180 rad/s 2
2
2
3
2
−
mR + I ( 2.0 kg )( 0.040 m ) + 1.1× 10 kg ⋅ m
The value for the tension can be now obtained by substituting this value for α into
Equation (1):
TR = Iα
or
(
)(
)
1.1× 10−3 kg ⋅ m 2 180 rad/s 2
Iα
T=
=
= 5.0 N
R
0.040 m
41. 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
______________________________________________________________________________
Chapter 9 Problems
455
42. REASONING The drawing shows the drum, pulley, and the crate, as well as the tensions
in the cord
r2
T1
Pulley
m2 = 130 kg
T2
T2
T1
r1
Crate
m3 = 180 kg
Drum
m1 = 150 kg
r1 = 0.76 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 = I1α1,
where I1 is the moment of inertia of the drum, and α1 is its angular acceleration. If we
assume that the cable does not slip, then Equation 9.7 can be written as
c
− T1 r1 + τ = m1 r12
∑τ
I1
h FGH ra IJK
(1)
1
α1
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:
+ T1 r2 − T2 r2 =
∑τ
c
1
2
m2 r22
I2
h FGH ra IJK
(2)
2
α2
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
456 ROTATIONAL DYNAMICS
+ T2 − m3 g = m3 a
(3)
∑ Fy
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
τ = r1 ⎡⎣ a ( m1 + 12 m2 + m3 ) + m3 g ⎤⎦
(
)
= (0.76 m) ⎡ (1.2 m/s 2 ) 150 kg + 12 130 kg + 180 kg + (180 kg)(9.80 m/s 2 ) ⎤ = 1700 N ⋅ m
⎣
⎦
______________________________________________________________________________
43. SSM 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 ω in rad/s. This quantity can then
be converted to rev/min.
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
h FGH 21πrevrad IJK FGH 160mins IJK =
6.1 × 10 5 rev / min
______________________________________________________________________________
44. REASONING
a. The kinetic energy is given by Equation 9.9 as KE R = 21 Iω 2 . Assuming the earth to be a
uniform solid sphere, we find from Table 9.1 that the moment of inertia is I = 25 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 = 21 Iω 2 . 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
Chapter 9 Problems
457
2
of inertia in this case is just I = MRorbit
. 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 Iω 2 =
=
1
2
2
5
1
2
c
2
5
h
MR 2 ω 2
c5.98 × 10
24
hc
kg 6.38 × 10
6
L F 2π rad IJ FG 1 day IJ FG 1 h IJ OP
mh MG
N H 1 day K H 24 h K H 3600 s K Q
2
2
= 2 .57 × 10 29 J
b. According to Equation 9.9, we have
KE R = 21 Iω 2 =
=
1
2
1
2
c MR hω
c5.98 × 10
2
orbit
24
hc
2
kg 1.50 × 10
11
L F 2π rad IJ FG 1 yr IJ FG 1 day IJ FG 1 h IJ OP
mh MG
N H 1 yr K H 365 day K H 24 h K H 3600 s K Q
2
2
= 2 .67 × 10 33 J
______________________________________________________________________________
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:
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 v 12 + 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
458 ROTATIONAL DYNAMICS
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 Iω 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).
______________________________________________________________________________
46. REASONING Each blade can be approximated as a thin rod rotating about an axis
perpendicular to the rod and passing through one end. The moment of inertia of a blade is
given in Table 9.1 as 13 ML2 , where M is the mass of the blade and L is its length. The total
moment of inertia I of the two blades is just twice that of a single blade. The rotational
kinetic energy KER of the blades is given by Equation 9.9 as KE R = 12 I ω 2 , where ω is the
angular speed of the blades.
SOLUTION
a. The total moment of inertia of the two blades is
I = 13 ML2 + 13 ML2 = 23 ML2 =
2
3
( 240 kg )( 6.7 m )2 =
7200 kg ⋅ m 2
b. The rotational kinetic energy is
KE R = 12 I ω 2 =
1
2
( 7200 kg ⋅ m2 ) ( 44 rad/s )2 = 7.0 × 106 J
______________________________________________________________________________
47. REASONING The rotational kinetic energy of a solid sphere is given by Equation 9.9
as KE R = 12 I ω 2 , where I is its moment of inertia and ω its angular speed. The sphere has
translational motion in addition to rotational motion, and its translational kinetic energy is
KE T = 12 mv 2 (Equation 6.2), where m is the mass of the sphere and v is the speed of its
center of mass. The fraction of the sphere’s total kinetic energy that is in the form of
rotational kinetic energy is KER/(KER + KET).
SOLUTION The moment of inertia of a solid sphere about its center of mass is I = 52 mR 2 ,
where R is the radius of the sphere (see Table 9.1). The fraction of the sphere’s total kinetic
energy that is in the form of rotational kinetic energy is
Chapter 9 Problems
1 Iω 2
KE R
2
=
=
1
KE R + KE T 2 I ω 2 + 12 mv 2
(
1 2 mR 2
2 5
1 2 mR 2 ω 2
2 5
(
)
)ω 2
+ 12 mv 2
=
459
2 R 2ω 2
5
2 R 2ω 2 + v 2
5
Since the sphere is rolling without slipping on the surface, the translational speed v of the
center of mass is related to the angular speed ω about the center of mass by v = Rω (see
Equation 8.12). Substituting v = Rω into the equation above gives
KE R
KE R + KE T
=
2 R 2ω 2
5
2 R 2ω 2 + v 2
5
=
2
5
2
5
R 2ω 2
R ω + ( Rω )
2
2
2
=
2
7
______________________________________________________________________________
48. REASONING The drawing shows the rod in its initial
(dashed lines) and final (solid lines) orientations. Since
both friction and air resistance are absent, the total
mechanical energy is conserved. In this case, the total
mechanical energy is
E =
1
2
Iω 2 +
Rotational
kinetic energy
Center of mass
mgh
Pivot
Gravitational
potential energy
In this expression, m is the rod’s mass, I is its moment of
inertia, ω is its angular speed, and h is the height of its
center of mass above a reference level that we take to be
the ground. Since the rod is uniform, its center of mass is
located at the center of the rod. The energy-conservation
principle will guide our solution.
hf
Center of mass
h0
v0
SOLUTION Conservation of the total mechanical energy dictates that
1
2
I ωf2 + mghf =
Final total
mechanical energy
1
2
I ω02 + mgh0
(1)
Initial total
mechanical energy
Since the rod is momentarily at rest in its final orientation, we know that ωf = 0 rad/s, so
that Equation (1) becomes
mghf = I ω02 + mgh0
1
2
or
1
2
I ω02 = mg ( hf − h0 )
(2)
We can relate the initial angular speed ω0 to the initial linear speed v0 by using
v
Equation 8.9: ω0 = 0 . With this substitution, the fact that hf − h0 = L (see the drawing),
L
1
3
and the fact that I = mL2 (see Table 9.1 in the text), Equation (2) indicates that
460 ROTATIONAL DYNAMICS
1
2
I ω02
= mg ( hf − h0 )
Solving for v0 gives
(
1 1
mL2
2 3
or
)
(
2
⎛ v0 ⎞
⎜ ⎟ = mgL
⎝L⎠
or
1 2
v
6 0
= gL
)
v0 = 6 gL = 6 9.80 m/s 2 ( 0.80 m ) = 6.9 m/s
49. SSM 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
mghs = 12 mv02 + 12 I sω02
(1)
where, from Table 9.1, I s = 12 mr 2 . In this expression, v0 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 v0 are
related according to Equation 8.12, ω0 = v0 / r . Then, solving Equation (1) for hs , we obtain
hs =
3v02
4g
Repeating the above for the hollow cylinder and using I h = mr 2 we have
hh =
v02
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
h
h
and
ss = s
sh = h
sin θ
sin θ
Dividing these gives
ss
= 3/4
sh
______________________________________________________________________________
The conservation of energy gives for the cube
50. REASONING AND SOLUTION
2
(1/2) mvc = 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) mR2. Combining the last three equations gives
Chapter 9 Problems
vm =
461
10 gh
7
Thus,
vc
= 7 = 1.18
5
vm
______________________________________________________________________________
51. REASONING
We first find the speed v0 of the ball when it becomes airborne using the
conservation of mechanical energy. Once v0 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 m 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 this instant its total energy is 12 mv02 + 12 I ω 2 . 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
mv02
2
+ Iω =
1
2
2
1
mv02
2
⎛v
1 2
( mr 2 ) ⎜ 0
2 3
r
+
⎝
2
⎞
5
2
⎟ = 6 mv0
⎠
Therefore, from conservation of mechanical energy, he have
5
6
mgh = mv02
or
v0 =
6 gh
5
The range of the tennis ball is given by x = v0 xt = v0 (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
t=
v − v0 y
ay
=
(−v0 y ) − v0 y
ay
=–
2v0 sin θ
ay
Therefore, the range of the tennis ball is
⎛ 2v sin θ
x = v0 xt = v0 (cos θ ) ⎜ – 0
⎜
ay
⎝
⎞
⎟
⎟
⎠
462 ROTATIONAL DYNAMICS
If we take upward as the positive direction, then using the fact that a y = – g and the
expression for v0 given above, we find
2
⎛ 2cos θ sin θ ⎞ 2 ⎛ 2cos θ sin θ ⎞ ⎛ 6 gh ⎞ 12
x=⎜
⎟⎟ = h cos θ sin θ
⎟ v0 = ⎜
⎟ ⎜⎜
5
5
g
g
⎝
⎠
⎝
⎠ ⎝
⎠
12
(1.8 m) (cos 35°)(sin 35°)= 2.0 m
5
______________________________________________________________________________
=
52. REASONING Before any sand strikes the disk, only the disk is rotating. After the sand has
landed on the disk, both the sand and the disk are rotating. If the sand and disk are taken to
be the system of objects under consideration, we note that there are no external torques
acting on the system. As the sand strikes the disk, each exerts a torque on the other.
However, these torques are exerted by members of the system, and, as such, are internal
torques. The conservation of angular momentum states that the total angular momentum of a
system remains constant (is conserved) if the net average external torque acting on the
system is zero. We will use this principle to find the final angular velocity of the system.
SOLUTION The angular momentum of the system (sand plus disk) is given by
Equation 9.10 as the product of the system’s moment of inertia I and angular velocity ω, or
L = Iω. The conservation of angular momentum can be written as
Iω
Final angular
momentum
=
I 0ω0
Initial angular
momentum
where ω and ω0 are the final and initial angular velocities, respectively, and I and I0 are final
and initial moments of inertia. The initial moment of inertia is given, while the final moment
of inertia is the sum of the values for the rotating sand and disk, I = Isand + I0. We note that
the sand forms a thin ring, so its moment of inertia is given by (see Table
2
, where Msand is the mass of the sand and Rsand is the radius of the
9.1) I sand = M sand Rsand
ring. Thus, the final angular velocity of the system is, then,
⎛
⎞
⎛
⎞
I0
I0
⎛ I0 ⎞
⎟
⎟⎟ = ω0 ⎜⎜
⎟ = ω0 ⎜⎜
2
⎟
+
I
I
+
M
R
I
⎝ I ⎠
⎝ sand 0 ⎠
⎝ sand sand 0 ⎠
ω = ω0 ⎜
⎡
⎤
0.10 kg ⋅ m 2
⎥ = 0.037 rad/s
= ( 0.067 rad/s ) ⎢
⎢⎣ ( 0.50 kg )( 0.40 m )2 + 0.10 kg ⋅ m 2 ⎥⎦
______________________________________________________________________________
Chapter 9 Problems
463
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 = L final , or
I A ω A + I B ω B = ( I A + I B ) ω 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
Fω
GH ω
final
B
–ω A
– ω final
I = ( 3.4 kg ⋅ m ) L –2.4 rad / s – 7.2 rad / s O =
JK
MN –9.8 rad / s – (–2.4 rad / s) PQ
2
4.4 kg ⋅ m 2
______________________________________________________________________________
54. 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
FG 40.0 kg IJ LM 1.25 m
H 1.00 × 10 kg K MN b 2.00 mg
2
2
OP 2.00 m / s = −0.500 rad/s
g
PQ b
The magnitude of ωd is 0.500 rad/s . The negative sign indicates that the disk rotates in a
direction opposite to the motion of the person.
______________________________________________________________________________
55. REASONING The rod and bug are taken to be the system of objects under consideration,
and we note that there are no external torques acting on the system. As the bug crawls out to
the end of the rod, each exerts a torque on the other. However, these torques are internal
torques. The conservation of angular momentum states that the total angular momentum of a
system remains constant (is conserved) if the net average external torque acting on the
system is zero. We will use this principle to find the final angular velocity of the system.
SOLUTION The angular momentum L of the system (rod plus bug) is given by
Equation 9.10 as the product of the system’s moment of inertia I and angular velocity ω, or
L = Iω. The conservation of angular momentum can be written as
Iω
Final angular
momentum
=
I 0ω0
Initial angular
momentum
464 ROTATIONAL DYNAMICS
where ω and ω0 are the final and initial angular velocities, respectively, and I and I0 are the
final and initial moments of inertia. The initial moment of inertia is given. The initial
moment of inertia of the bug is zero, because it is located at the axis of rotation. The final
moment of inertia is the sum of the moment of inertia of the bug and that of the rod;
I = Ibug + I0. When the bug has reached the end of the rod, its moment of inertia is
Ibug = mL2, where m is its mass and L is the length of the rod. The final angular velocity of
the system is, then,
⎛
⎞
⎛
⎞
I0
I
⎛ I0 ⎞
⎟ = ω0 ⎜ 2 0
⎟
⎟ = ω0 ⎜⎜
⎜ mL + I ⎟
I bug + I 0 ⎟
⎝ I ⎠
0
⎝
⎠
⎝
⎠
ω = ω0 ⎜
⎡
⎤
1.1×10−3 kg ⋅ m 2
⎥ = 0.26 rad/s
= ( 0.32 rad/s ) ⎢
⎢ 4.2 × 10−3 kg ( 0.25 m )2 + 1.1× 10−3 kg ⋅ m 2 ⎥
⎢⎣
⎦⎥
______________________________________________________________________________
(
)
(
)
56. REASONING Since friction and air resistance play no role in this problem, no net external
torque acts on the system comprised of the carousel and the person on it. Therefore, the
total angular momentum of the system is conserved, and it is via this conservation principle
that we will determine the person’s mass. In applying this principle, we note that the
moment of inertia of the carousel itself remains constant as the person moves toward the
center, but the moment of inertia of the person changes.
SOLUTION Since the total angular momentum of the system is conserved, we have
I Cωf + I Pf ωf = I Cω0 + I P0ω0
Final total angular
momentum
(1)
Initial total angular
momentum
where IC is the moment of inertia of the carousel, IPf and IP0 are, respectively, the final and
initial moments of inertia of the person, and ωf and ω0 are, respectively, the final and initial
angular velocities. As the person (mass = m) moves, his distance from the center changes to
a final value of rf from an initial value of r0. Correspondingly, his moment of inertia
changes to a final value of mrf2 from an initial value of mr02 , according to Equation 9.6.
Substituting these values for IPf and IP0 into Equation (1) gives
( )
( )
I Cωf + mrf2 ωf = I Cω0 + mr02 ω0
( )
( )
(
or I Cωf − I Cω0 = mr02 ω0 − mrf2 ωf = m r02ω0 − rf2ωf
Solving for m gives
m=
I C (ωf − ω0 )
r02ω0 − rf2ωf
=
(125 kg ⋅ m2 ) ( 0.800 rad/s − 0.600 rad/s )
(1.50 m )2 ( 0.600 rad/s ) − ( 0.750 m )2 ( 0.800 rad/s )
= 28 kg
)
Chapter 9 Problems
465
57. SSM 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 v0 and vf.
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 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 ωf, and the momentum of
inertia is If = 2M(L/4)2. The conservation of angular momentum indicates that
Ifω f
= I 0ω 0
Final angular
momentum
Initial angular
momentum
LM 2 M FG L IJ OPω = LM 2 M FG L IJ OPω
MN H 4 K PQ MN H 2 K QP
2
2
f
0
ω f = 4ω 0
According to Equation 8.9, ωf = vf/(L/4) and ω0 = v0/(L/2). With these substitutions, the
result that ωf = 4ω0 becomes
vf
=4
FG v IJ
H L /2K
b
g
or v f = 2 v 0 = 2 17 m / s = 34 m / s
L/4
______________________________________________________________________________
0
58. 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
⎛ I0 ⎞
⎟⎟
⎝ If ⎠
ωf = ω0 ⎜⎜
where for a rod of mass M and length L, I 0 =
1
12
ML2 . 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
I f = ⎢ ⎜⎜
⎢⎣ 12 ⎝ 2
⎞⎛
⎟⎜
⎟⎜
⎠⎝
L⎞
⎟
2 ⎟⎠
2⎤
⎡ ⎛ M ⎞ ⎛ L ⎞2 ⎤ 1
ML2
⎥ + ⎢2 ⎜ ⎟ ⎜ ⎟ ⎥ =
4
4
24
⎥⎦ ⎢⎣ ⎝ ⎠ ⎝ ⎠ ⎥⎦
466 ROTATIONAL DYNAMICS
and
⎡ 1 ML2 ⎤
⎥ = (7.0 rad/s)(2) = 14 rad/s
ωf = ω0 ⎢ 12
⎢ 1 ML2 ⎥
⎣ 24
⎦
______________________________________________________________________________
59. 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. Thus,
µs g = rf ωf2
(1)
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:
µs g = rf ω02
r04
rf4
Solving for rf gives
1/3
1/3
⎛ ω02 r04 ⎞
⎡ (2.2 rad/s) 2 (0.30 m) 4 ⎤
rf = ⎜
⎟ =⎢
⎥ = 0.17 m
2
⎜ µ g ⎟
⎣ (0.75)(9.80 m/s ) ⎦
⎝ s ⎠
______________________________________________________________________________
60. REASONING AND SOLUTION After the mass has moved inward to its final path the
centripetal force acting on it is T = 105 N.
r
105 N
Chapter 9 Problems
467
Its centripetal acceleration is
ac = v2/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
______________________________________________________________________________
61. SSM WWW REASONING The maximum torque will occur
when the force is applied perpendicular to the diagonal of the
square as shown. The lever arm
is half the length of the
diagonal. From the Pythagorean theorem, the lever arm is,
therefore,
1
2
(0.40 m)2 + (0.40 m)2 = 0.28 m
0.40 m
=
0.40 m
axis
Since the lever arm is now known, we can use Equation 9.1 to
obtain the desired result directly.
F
SOLUTION Equation 9.1 gives
τ = F = (15 N)(0.28 m) = 4.2 N ⋅ m
______________________________________________________________________________
62. REASONING The net torque Στ acting on the CD is given by Newton’s second law for
rotational motion (Equation 9.7) as Στ = Ι α, where I is the moment of inertia of the CD and
α is its angular acceleration. The moment of inertia can be obtained directly from Table 9.1,
and the angular acceleration can be found from its definition (Equation 8.4) as the change in
the CD’s angular velocity divided by the elapsed time.
SOLUTION The net torque is Στ = Ι α. Assuming that the CD is a solid disk, its moment of
inertia can be found from Table 9.1 as I = 12 MR 2 , where M and R are the mass and radius
of the CD. Thus, the net torque is
Στ = I α =
( 12 MR2 )α
468 ROTATIONAL DYNAMICS
The angular acceleration is given by Equation 8.4 as α = (ω − ω0 ) / t , where ω and ω0 are
the final and initial angular velocities, respectively, and t is the elapsed time. Substituting
this expression for α into Newton’s second law yields
Στ =
( 12 MR2 )α = ( 12 MR2 ) ⎛⎜⎝
(
ω − ω0 ⎞
t
⎟
⎠
)(
)
2 ⎤ ⎛ 21 rad/s − 0 rad/s ⎞
⎡
−4
= ⎢ 12 17 × 10−3 kg 6.0 × 10−2 m ⎥ ⎜
⎟ = 8.0 × 10 N ⋅ m
0.80 s
⎣
⎦⎝
⎠
______________________________________________________________________________
63. 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.
______________________________________________________________________________
64. REASONING When the board just begins to tip, three forces act on the board. They are
the weight W of the board, the weight WP of the person, and the force F exerted by the right
support. At the moment tipping begins, the left support exerts no force on the board and,
therefore, is omitted from the following drawing.
F
1.4 m
x
W = 225
WP
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
Chapter 9 Problems
469
−WP x + W (1.4 m) = 0
This expression can be solved for x.
SOLUTION Solving the expression above for x, we obtain
W (1.4 m) (225 N)(1.4 m)
=
= 0.70 m
WP
450 N
______________________________________________________________________________
x=
65. SSM 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 = , where 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
2
the faces. From the Pythagorean theorem, d =
+ 2 =
2.
Axis
d
d
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 2 ) + 2(md 2 ) = 4(m 2 ) + 2(2m 2 ) = 8m
2
or
I = 8(0.12 kg)(0.25 m) 2 = 0.060 kg ⋅ m 2
______________________________________________________________________________
66. REASONING When the wheel is resting on the ground it is in equilibrium, so the sum of
the torques about any axis of rotation is zero ( Στ = 0 ) . This equilibrium condition will
provide us with a relation between the magnitude of F and the normal force that the ground
exerts on the wheel. When F is large enough, the wheel will rise up off the ground, and the
normal force will become zero. From our relation, we can determine the magnitude of F
when this happens.
470 ROTATIONAL DYNAMICS
SOLUTION The free body diagram
shows the forces acting on the wheel: its
weight W, the normal force FN, the
horizontal force F, and the force FE that
the edge of the step exerts on the wheel.
We select the axis of rotation to be at
the edge of the step, so that the torque
produced by FE is zero. Letting
r
FN
F
Axis of Rotation
FE
h
W
N , W , and F represent the lever
arms for the forces FN, W, and F, the sum of the torques is
r2 − (r − h)
Στ = − FN
2
r2 − (r − h)
+W
N
2
− F (r − h) = 0
F
W
Solving this equation for F gives
(W − FN )
F=
r 2 − (r − h)
2
r−h
When the bicycle wheel just begins to lift off the ground the normal force becomes zero
(FN = 0 N). When this happens, the magnitude of F is
(W − FN )
F=
r2 − (r − h)
2
( 25.0 N − 0 N ) ( 0.340 m )2 − ( 0.340 m − 0.120 m )2
=
= 29 N
r−h
0.340 m − 0.120 m
__________________________________________________________________________
67. SSM 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 µsV 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 V
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
µsV – P = 0
P
θ
W
µ sV
(1)
The vertical forces also balance to zero giving
V – W = 0
(2)
Chapter 9 Problems
471
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
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 = µsV according to Equation (1), and W = V according to Equation (2). Substituting
these results into Equation (4) gives
1
V
=
tan θ =
2 µsV 2 µs
Therefore,
⎛ 1 ⎞
⎤
⎡
1
⎟ = tan −1
= 37.6°
θ = tan −1 ⎜
⎢
⎝ 2 µs ⎠
⎣ 2(0.650) ⎥⎦
______________________________________________________________________________
68. 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 WB 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 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 WL 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.
472 ROTATIONAL DYNAMICS
SOLUTION The directions upward and to the right will be taken as the positive directions.
In the x direction we have
(1)
∑ Fx = H − T cos θ = 0
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 WL
is L cos φ , while the lever arm for WB 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 ⎜
⎟ − WL L cos φ + TL sin (φ − θ ) = 0
⎝ 2 ⎠
Solving for T, we have
1 W +W
L
T= 2 B
cos φ
(3)
sin(φ –θ )
a. From Equation (3) the tension in the support cable is
T=
1 (3600
2
N) + 4800 N
sin(48° – 32°)
cos 48° = 1.6 × 104 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 × 104 N cos 32° = 1.4 × 104 N
and, from Equation (2)
(
)
V = WL + WB + T sin θ = 4800 N + 3600 N + 1.6 × 104 N sin 32° = 1.7 × 104 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 ×104 N) 2 + (1.7 × 104 N) 2 = 2.2 × 104 N
Chapter 9 Problems
473
______________________________________________________________________________
69. SSM 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.
SOLUTION Since angular momentum is conserved,
I final ω final = I 0 ω 0
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,
ω final
I
3.24 × 10 9 kg ⋅ m 2
= 1.08
= 0 =
I final
ω0
3.00 × 10 9 kg ⋅ m 2
This fraction represents a percentage increase of 8 percent .
______________________________________________________________________________
70. 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) mR2.
Substitution of the last two equations into the first and rearrangement gives
v = v02 − 10
gh =
7
( 3.50 m/s )2 − 107 ( 9.80 m/s 2 ) ( 0.760 m ) =
1.3 m/s
______________________________________________________________________________
71. SSM WWW 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
474 ROTATIONAL DYNAMICS
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
⎛L⎞
+ (178 N)(sin 30.0°) ⎜ ⎟ + fs(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
(178 N)(sin 30.0°) – 44.5 N
= 51.4 N
cos 30.0°
______________________________________________________________________________
fs =
72. REASONING AND SOLUTION Newton's law applied to the 11.0-kg object gives
T2 − (11.0 kg)(9.80 m/s2) = (11.0 kg)(4.90 m/s2)
or
T2 = 162 N
or
T1 = 216 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)
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.
Chapter 9 Problems
475
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
+ FB
A
FA L + (12 N )
B
− FD
D
=0
( 12 L ) − (12 N ) L = 0
The length L can be eliminated algebraically from this result, which can then be solved for
FA:
FA = − (12 N )
( 12 ) + (12 N ) =
6.0 N (pointing toward corner B)
______________________________________________________________________________
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
Right design
b
gb
g b gb
g b
g
b525 N gb 0.400 mg + b 60.0 N gb 0.600 mg = 189 N
F=
Στ = − 525 N 0.400 m − 60.0 N 0.600 m + F 1.300 m = 0
1.300 m
b
gb
g b
g
b 60.0 N gb 0.600 m g = 27.7 N
F=
Στ = − 60.0 N 0.600 m + F 1.300 m = 0
1.300 m
476 ROTATIONAL DYNAMICS
______________________________________________________________________________
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.
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 sin θ + 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.
W
Torques
c L h − VL = 0
1
2
Since L can be eliminated algebraically, Equation (3) may be solved immediately for V:
V = 21 W =
1
2
b 340 N g =
170 N
Substituting this result into Equation (2) gives
P sin θ + 21 W − W = 0
P=
W
340 N
=
= 270 N
2 sin θ 2 sin 39 °
Substituting this result into Equation (1) gives
( 3)
Chapter 9 Problems
477
FG W IJ cosθ − H = 0
H 2 sinθ K
W
340 N
=
= 210 N
H=
2 tan θ 2 tan 39 °
______________________________________________________________________________
76. CONCEPT QUESTIONS a. Equation 8.7 from the equations of rotational kinematics
indicates that the angular displacement θ is θ = ω 0 t + 21 αt 2 , where ω0 is the initial angular
velocity, t is the 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 = 21 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 = ω 0 t + 21 αt 2
or
α=
c
2 θ − ω 0t
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
⎡ 2 (θ − ω0t ) ⎤
Hoop Στ = I hoopα = MR 2 ⎢
⎥
t2
⎣⎢
⎦⎥
⎡
⎤
2 2 (13 rad ) − 2 ( 0 rad/s )( 8.0 s )
⎥ = 0.20 N ⋅ m
= ( 4.0 kg )( 0.35 m ) ⎢
2
⎢⎣
⎥⎦
8.0
s
(
)
⎡ 2 (θ − ω0t ) ⎤
Disk Στ = I diskα = 12 MR 2 ⎢
⎥
t2
⎣⎢
⎦⎥
=
1
2
⎡
⎤
2 2 (13 rad ) − 2 ( 0 rad/s )( 8.0 s )
⎢
⎥ = 0.10 N ⋅ m
4.0
kg
0.35
m
(
)(
)
2
⎢⎣
⎥⎦
(8.0 s )
478 ROTATIONAL DYNAMICS
______________________________________________________________________________
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.
b. Since the moment of inertia for rod A is greater, its kinetic energy is also greater,
according to Equation 9.9 KE R = 21 Iω 2 .
SOLUTION Using Equation 9.9 to calculate the kinetic energy, we find
Rod A
Rod B
c ML hω
b 0.66 kg gb 0.75 mg b 4.2 rad / sg
I ω = c ML hω
b 0.66 kg gb 0.75 mg b 4.2 rad / sg
KE R = 21 I A ω 2 =
=
1
2
KE R =
1
2
2
1
2
2
2
2
B
1
2
1
3
2
2
= 3.3 J
2
= 16
= 1.1 J
______________________________________________________________________________
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.
Chapter 9 Problems
I carousel ω f + I person ω f
= I carousel ω 0
Initial total
angular momentum
Final total
angular momentum
ωf =
479
I carousel ω 0
I carousel + I person
=
I carousel ω 0
I carousel + MR 2
c125 kg ⋅ m hb 3.14 rad / sg
=
125 kg ⋅ m + b 40.0 kg gb1.50 m g
2
2
2
= 1.83 rad / s
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79. CONCEPT QUESTIONS a. Two identical systems of objects may have different moments
of inertia, depending on where the axis of rotation is located. The moment of inertia depends
on the mass of each object and its distance from the axis of rotation. According to
Equation 9.6, the moment of inertia for the three-ball system is I = m1r12 + m2 r22 + m3r32 ,
where m1, m2, and m3 are the masses of the balls and r1, r2, and r3 are the distances from the
axis. In system A, the ball whose mass is m1 does not contribute to the moment of inertia,
because the ball is located on the axis and r1 = 0 m. In B, the ball whose mass is m3 does not
contribute to the moment of inertia, because it is located on the axis and r3 = 0 m.
b. Even though the force acting on both systems is the same, the torque is not the same. The
magnitude of the torque is equal to the magnitude F of the force times the lever arm
(Equation 9.1). In system A the lever arm is = 3.00 m. In B the lever arm is = 0 m,
since the line of action of the force passes through the axis of rotation.
c. The systems have different angular accelerations. According to Newton’s second law for
rotational motion, Equation 9.7, the angular acceleration α is given by α = ( Στ ) / I , where
Στ is the net torque and I is the moment of inertia. Since the net torque and moment of
inertia are different for the two systems, the angular accelerations are different. The angular
velocity ω is given by Equation 8.4 as ω = ω0 + α t, where ω0 is the initial angular velocity
and t is the time. Since the angular accelerations are different for the two systems, the
angular velocities will also be different at the same later time.
SOLUTION
a. The moment of inertia for each system is
System A
I = m1r12 + m2 r22 + m3r32
= ( 9.00 kg )( 0 m ) + ( 6.00 kg )( 3.00 m ) + ( 7.00 kg )( 5.00 m ) = 229 kg ⋅ m 2
2
2
2
480 ROTATIONAL DYNAMICS
I = m1r12 + m2 r22 + m3r32
System B
= ( 9.00 kg )( 5.00 m ) + ( 6.00 kg )( 4.00 m ) + ( 7.00 kg )( 0 m ) = 321 kg ⋅ m 2
2
2
2
As anticipated, the two systems have different moments of inertia.
b. The torque produced by the force has a magnitude that is equal to the product of the force
magnitude and the lever arm:
τ = − F = − ( 424 N )( 3.00 m ) = −1270 N ⋅ m
System A
The torque is negative because it produces a clockwise rotation about the axis.
τ = F = ( 424 N )( 0 m ) = 0 N ⋅ m
System B
As anticipated, the torques in the two cases are not the same.
c. The final angular velocity ω is related to the initial angular velocity ω0, the angular
acceleration α, and the time t by ω = ω0 + α t (Equation 8.4). The angular acceleration is
given by Newton’s second law for rotational motion as α = ( Στ ) / I , where Στ is the net
torque and I is the moment of inertia. Since there is only one torque acting on each system,
it is the net torque, so Στ = τ . Substituting this expression for α into Equation 8.4 yields
⎛τ ⎞
ω = ω0 + α t = ω0 + ⎜ ⎟ t
I
⎝ ⎠
In both cases the initial angular velocity is ω0 = 0 rad/s, since the systems start from rest.
The final angular velocities after 5.00 s are:
⎛ −1270 N ⋅ m ⎞
⎛τ ⎞
ω = ω0 + ⎜ ⎟ t = ( 0 rad/s ) + ⎜
( 5.00 s ) = −27.7 rad/s
2 ⎟
I
System A
⎝ 229 kg ⋅ m ⎠
⎝ ⎠
⎛
⎛τ ⎞
ω = ω0 + ⎜ ⎟ t = ( 0 rad/s ) + ⎜
I
0 N⋅m ⎞
( 5.00 s ) = 0 rad/s
2⎟
321
kg
m
⋅
⎝ ⎠
⎝
⎠
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System B
80. CONCEPT QUESTIONS
a. The rolling wheel has the greater total kinetic energy. Its kinetic energy is the sum of its
translational 12 mv 2 and rotational 12 I ω 2 kinetic energies. The sliding wheel only has
(
)
(
)
translational kinetic energy, since it does not rotate.
Chapter 9 Problems
481
b. The rolling wheel has the greater potential energy when it comes to a halt. As the wheels
move up the incline plane, the total mechanical energy is conserved, since only the
conservative force of gravity does work on each wheel. Thus, the initial kinetic energy at
the bottom of the incline is converted entirely into potential energy when the wheels come
to a momentary halt. Since the rolling wheel has the greater kinetic energy to begin with, it
also has the greater potential energy.
c. The rolling wheel rises to the greater height. The potential energy PE is given by
Equation 6.5 as PE = mgh, where h is the height of the wheel above an arbitrary zero level.
Since the rolling wheel has the greater potential energy when it comes to a halt, it also rises
to a greater height.
SOLUTION
a. The total kinetic energy KE of the rolling wheel is the sum of its translational and
rotational kinetic energies
KE = 12 mv 2 + 12 I ω 2
where ω is the angular speed of the wheel. Since the wheel is a disk, its moment of inertia is
I = 12 mR 2 (see Table 9.1), where R is the radius of the disk. Furthermore, the angular speed
ω of the rolling wheel is related to the linear speed v of its center of mass by Equation 8.12
as ω = v/R. Thus, the total kinetic energy of the rolling wheel is
Rolling
Wheel
KE =
1 mv 2
2
1
2
+
= 34 mv 2 =
3
4
Iω =
2
1 mv 2
2
+
(
1 1 mR 2
2 2
( 2.0 kg )( 6.0 m/s )2 =
)
⎛v⎞
⎜ ⎟
⎝R⎠
2
54 J
The kinetic energy of the sliding wheel is
Sliding
Wheel
KE =
1 mv 2
2
=
1
2
( 2.0 kg )( 6.0 m/s )2 =
36 J
As expected, the rolling wheel has the greater total kinetic energy.
b. As each wheel rolls up the incline, its total mechanical energy is conserved. The initial
kinetic energy KE at the bottom of the incline is converted entirely into potential energy PE
when the wheels come to a momentary halt. Thus, the potential energies of the wheels are:
Rolling
Wheel
PE = 54 J
Sliding
Wheel
PE = 36 J
482 ROTATIONAL DYNAMICS
As anticipated, the rolling wheel has the greater potential energy.
The potential energy of a wheel is given by Equation 6.5 as PE = mgh, where g is the
acceleration due to gravity and h is the height relative to an arbitrary zero level. Therefore,
the height reached by each wheel is as follows:
Rolling
Wheel
h=
PE
54 J
=
= 2.8 m
mg
( 2.0 kg ) 9.80 m/s2
Sliding
Wheel
h=
PE
36 J
=
= 1.8 m
mg
( 2.0 kg ) 9.80 m/s2
(
(
)
)
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