8 Rotational Motion of Solid Objects
Answers to Questions
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Rotational velocity is rotational displacement divided by time, but m/s is linear velocity and rev/s2 is rotational
acceleration.
Rad/s and rev/m2 are inappropriate. Rad/s is angular velocity and rev/m2 is meaningless.
Yes. Any body whose rotational velocity is changing is exhibiting rotational acceleration.
Yes. Any change in rotational velocity is a rotational acceleration, even if it is caused by a torque in the
direction opposing the rotation.
Yes. The merry-go-round is a rigidly rotating body. All parts along a given line from center to edge are
rotating at exactly the same rotational velocity.
No. Since the relation for the distance along the arc of a circle is equal to the radius times the rotational
displacement in radians, the inner arc is smaller than the outer arc. Therefore the outer child’s linear
velocity is greater than that of the inner child.
No. Constant acceleration results in constantly changing velocity.
The ball experiences both linear and angular acceleration. The ball travels one circumference along the
incline during one revolution, as measured along the incline. The magnitude of the linear velocity is equal
to the radius of the ball multiplied by the magnitude of the angular velocity. The direction of the linear
velocity is perpendicular to the direction of the angular velocity.
The force applied at the end of the wrench will provide the greater torque because the length of the lever
arm associated with that force is greater.
F2. F1 has no lever arm with respect to the given axis.
F1 provides the larger torque. F2 has a smaller component perpendicular to the radius.
Yes, if the respective distances (lever arms) from the fulcrum are chosen properly. The ratio of the distances
will be in the inverse ratio of the weights.
Yes, if they do not act along a common line. A simply visualized example would be a cube with some force,
F, directed along the lower front edge towards the left in combination with an equal force, F, directed along
the upper rear edge towards the right. There are other arrangements also possible.
To get a magnification of force, the fulcrum should be placed closer to the rock, so that the lever arm for the
applied force is greater than the lever arm for the opposing weight.
No. A net torque of zero (i.e., balance) at this point indicates that the center of gravity for this pencil is
located at the same place the fulcrum is positioned.
The plank can be pushed to its center point, for that is where the center of gravity is located. Beyond that
point less of the plank’s mass will be supported by the platform than will be hanging over the edge. This
condition will result in a net torque downward.
No. The shape of the wire and distribution of its mass are such that the point about which its own weight
exerts no net torque is not on the wire. Therefore the center of gravity will not lie on the wire itself.
No. A net torque produces an angular acceleration.
The tall crate will be the easiest to tip over because the moment arm is the greatest when the forward edge
is the axis of rotation.
Body A will have a smaller rotational inertia, so that for a given torque it will acquire a larger angular
acceleration than if the torque were applied to B.
Yes, if the same mass is distributed at different distances from the axis of rotation in the two cases.
Yes. Extending one's arms will increase the rotational inertia compared to the case of the arms at one's side.
The hollow sphere has the greater rotational inertia.
No. Angular momentum is only conserved when there is no net torque.
No. The distribution of mass gives a different rotational inertia. Therefore the angular momentum will also be
different.
It will decrease. After the child is near the edge, the rotational inertia of the system will be greater; so to
conserve angular momentum, the final rotational velocity must be lower.
The increased mass will slow it down.
Yes. He can change his rotational inertia by changing the extension of his arms. His rotational velocity
changes according to conservation of angular momentum.
The rotational velocity will increase due to the decrease in rotational inertia.
The direction of the angular momentum is perpendicular to the plane of the wheel at any instant, so in turning
a corner the direction will change by 90o.
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Q31
Q32
Q33
Q34
By definition the clockwise direction as viewed from above results in the rotational velocity vector pointing
upward. Therefore the angular momentum vector is pointing upward because it is parallel to the velocity
vector, and since the skater is rotating counterclockwise the velocity vector is pointing upward.
a. Away from the observer.
b. No. During the fall there was an angular torque.
Due to the conservation of angular momentum, a spinning top maintains its orientation until frictional forces
slow the top significantly. As the spin slows, the top begins to precess until gravitational torque is dominant
enough to topple the toy over.
A sleeping yo-yo requires the string to slip on the axis of the yo-yo. If it is tied tightly, it will wind back up after
it is dropped.
Answers to Exercises
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a. 0.1667 rev/s
b. 1.05 rad/s
a. 0.75 rev/s
b. 3.75 revolutions
a. 18.8 rad
b. 4.71 rad/s
a. 0.3 rad/s2
a. 4.8 rev/s
b. 9.6 rev
-0.25 rev/s2
a. 1 rev/s
b. 2.5 rev
a. 12 N•m
b. 6 N•m
15 cm
6N
a. 96 N•m
b. -60 N•m
c. 36 N•m
5 rad/s2
13.5 N•m
a. 50 N•m
b. 25 kg•m2
0.10 kg•m2
a. 0.20 kg•m2
b. 0.60 kg•m2/s
a. 0.08 kg•m2
b. 1.6 kg•m2/s
6.28 rad/s = 60 rpm
Answers to Synthesis Problems
SP1
SP2
SP3
a.
b.
c.
d.
a.
b.
c.
a.
b.
c.
d.
132 N•m; Directed along the merry-go-round axis.
0.147 rad/s2
2.2 rad/s
-1.3x10-2 rad/s2; 165 s after pushing is stopped.
80 N•m
0.53 m
He could have somebody else be prepared to hold down the other end if the plank started to tip.
960 kg•m2; 2460 kg•m2
1560 kg•m2
1.89 rad/s
Yes. During the time the children are moving, friction between their feet and the merry-go-round produces
the accelerating torque.
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SP4
a. 0 kg•m2/s upward from plane of wheel.
b. .33 rad/s in the direction of the original angular velocity.
c. the student exerts forces on the handles when he flips the wheel, producing the torque.
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