Download CHAPTER 9 ROTATIONAL DYNAMICS

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 τ = F l, where l is 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 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 The magnitude of the torque produced by a force F is given
by τ = F l where l is the lever arm.
a. A large force can be used to produce a small torque by choosing a small lever arm. Then the
product Fl is small even though the magnitude of the force, F, is large. If the line of action passes
through the axis of rotation, l = 0, 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 l. Then,
the product Fl 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 l . 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
l
T
F
T
l
F ig ur e A
F ig ur e B
As the reel becomes empty, the lever arm l gets smaller, as suggested in the figures above.
Therefore, the torque exerted on the reel by the tape becomes smaller.
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ROTATIONAL DYNAMICS
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4.
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.
A x is o f
ro t a t io n
l
F
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5.
REASONING AND SOLUTION Work and torque are both the product of force and
distance. Work and torque are distinctly different physical quantities, as is evident by considering
the distances in the definitions. Work is defined by W = ( F cos θ )s , according to Equation 6.1,
where F is the magnitude of the force, θ is the angle between the force and the displacement, and s
is the magnitude of the displacement. The magnitude of the torque is defined as the magnitude of
the force times the lever arm, according to Equation 9.1. In the definition of work, the "distance" is
the magnitude of the displacement over which the force acts. In the definition of torque, the
distance is the lever arm, a "static" distance. The lever arm is not the same physical quantity as the
displacement. Therefore, work and torque are different quantities.
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6.
REASONING AND SOLUTION A person stands on a train, both feet together, facing a
window. The front of the train is to the person's left. When the train starts to move forward, a
force of static friction is applied to the person's feet. This force tends to produce a torque about
the center of gravity of the person, causing the person to fall backward. If the right foot is slid out
toward the rear of the train, the normal force of the floor on the right foot produces a counter
torque about the center of gravity of the person. The resultant of these two torques is zero, and
the person can maintain his balance.
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Chapter 9 Conceptual Questions
7.
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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.
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 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 freebody 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
N
FN
Note that the weight acts at the center of
gravity (CG), which must be located exactly
above the point where the normal force acts,
as is the case in Figure 9.10c.
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ROTATIONAL DYNAMICS
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 = I α. 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.
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 magnitude TR, each in the same
direction, so that the net torque on the probe has magnitude 2TR. Since there is a net torque on
the probe, there will be an angular acceleration. The angular velocity of the probe will increase.
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14. REASONING AND SOLUTION When you are lying down with your hands behind your
head, your moment of inertia I about your stomach is greater than it is when your hands are on your
stomach. Sit-ups are then more difficult because, with a greater I, you must exert a greater torque
about your stomach to give your upper torso the same angular acceleration, according to Equation
9.7.
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Chapter 9 Conceptual Questions
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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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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
1
is given by KE = Iω2 . The rotational kinetic energy
B
A
2
will be 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
2
2
( 1 / 2 ) I ω + (1 / 2 ) Mv , 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 Iω2 + 12 Mv 2
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ROTATIONAL DYNAMICS
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)
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 about the rotation axis, I, 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
Chapter 9 Conceptual Questions
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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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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. 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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ROTATIONAL DYNAMICS
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 l is the
greatest. By inspection of the drawing, the
lever arm relative to the axle is greatest for
the shaded box.
mg
l
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