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428 CHAPTER 13 Dynamics of a Rigid Body

Answers to Checkups

Checkup 13.1 Checkup 13.3
1. You should place your hand at the end of the handle farthest 1. Since the angular momentum is L I and the angular
from the bolt; this will provide the largest R in Eq. (13.2) and speeds ( ) are equal, the hoop (with moment of inertia
2
maximize the torque. Similarly, your push should be perpendic- I MR ; see Table 12.3) has a larger angular momentum by a
ular to the wrench handle, in order to maximize sin to the factor of 2 compared with the uniform disk (which has
1
value sin 90 1 in Eq. (13.2). I MR 2 ).
2
2. The direction must be toward the axis (along a radius), so that 2. Since the angular velocities are equal and the angular momen-
sin sin 0 0; thus both the torque and the work done tum is L I , the car with the larger moment of inertia I
2
will be zero. MR has the greater angular momentum. Since the masses
are equal, this is the car on the outside, with the greater
3. Initially, when the stick is upright, the weight acts downward,
value of R.
along the radial direction, and so the torque is zero. As the
stick falls, the weight (mg) and the point at which it acts (R 3. Since there are no external torques on you, angular momen-
y l 2) remain constant. Only the angle between the force tum L I is conserved. Since you increase your moment
CM
2
and the radial line changes; the sine of this angle is maximum of inertia I by stretching your legs outward (increasing R ),
just as the meterstick hits the floor (when sin sin 90 1), your angular velocity must decrease.
so the torque is maximum then. 4. No. Since angular momentum L I is conserved and
1
4. (A) . The work done is W FR sin . In both she decreases her moment of inertia I, her angular
4
cases, pushing at right angles implies sin 1, and both velocity increases. But her rotational kinetic energy is
2
1
1
angular displacements are the same. But with half the K I I . Since I is constant and
2
2
force applied at half the radius for the second push, the work increases, the kinetic energy increases. Thus the skater
will be one-fourth of that for the first push. must do work to bring her arms close to her body.
5. (A) Frequency increases. The moment of inertia decreases
Checkup 13.2 when the children sit up, since more of their mass is closer to
the axis. Since the angular momentum L I is conserved,
1. The angular acceleration results from the torque exerted by a smaller moment of inertia requires a larger angular
gravity at the center of mass; this is maximum when the weight frequency.
is perpendicular to the radial direction (when sin 1). That
occurs when the meterstick is horizontal, just before it hits the Checkup 13.4
floor.
1. Yes; since the angular-momentum vector is L r p, it will
2. The translational kinetic energy is twice as large for a (uni-
be zero when r and p are parallel (or antiparallel).
form) rotating cylinder, because the rotational kinetic energy
2
2
2
2
1
1
1
1
1
is I MR (v>R) Mv . 2. Since the angular-momentum vector is L r p, L is
2
2
2
2
2
always perpendicular to p; the angle between them is 90 .
3. The rolling cylinder’s total kinetic energy is the same as for a
slipping cylinder; in each case, it is equal to the change in 3. The individual angular-momentum vectors will be inclined at
potential energy Mgh. For the rolling cylinder, one-third of an angle with respect to the z axis; each, however, will point
the total kinetic energy is rotational kinetic energy, and two- toward the z axis, like the angular-momentum vectors L and
1
thirds is translational kinetic energy; thus, the rolling cylin- L in Fig. 13.19. In this case, the horizontal components of
2
der’s translational speed is smaller when it reaches the bottom the two angular-momentum vectors will cancel, and the total
than that of a slipping cylinder (by a factor of 22>3 ). angular-momentum vector will point along the z axis.
4. The sphere and cylinder must have equal kinetic energies when 4. Yes; the total angular momentum is changing as the dumbbell
they reach the bottom; each kinetic energy is equal to the rotates about the z axis (because the direction of L is chang-
change in potential energy Mgh.The sphere’s moment of iner- ing), so a torque is required to produce that change in angular
1
2
tia is only MR 2 , compared with MR 2 for the cylinder, so the momentum.
5
2
sphere will achieve a higher speed and get to the bottom first. 5. (E) Perpendicularly into the face of the watch. By the right-
5. (A) Less than that of the cylinder. For the thin hoop (I hand rule, with r pointing along the minute hand and p in the
2
MR ), only one-half of its kinetic energy is translational; for direction of motion, the clockwise rotation implies that the
2
1
the cylinder (I Mr ) , two-thirds of its kinetic energy will angular-momentum vector L r p is perpendicularly into
2
be translational. Since the total kinetic energy in each case will the face of the watch.
equal the change in potential energy (Mgh), the speed of the
hoop will be smaller.