8 Rotational Motion of Solid Objects
Answers to Questions
Q1 Rotational velocity is rotational displacement divided by time, but m/s is linear velocity and rev/s2 is rotational acceleration.
Q2Rad/s and rev/m2 are inappropriate. Rad/s is angular velocity and rev/m2 is meaningless.
Q3Yes. Any body whose rotational velocity is changing is exhibiting rotational acceleration.
Q4Yes. Any change in rotational velocity is a rotational acceleration, even if it is caused by a torque in the direction opposing the rotation.
Q5Yes. 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.
Q6No. 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.
Q7No. Constant acceleration results in constantly changing velocity.
Q8The 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.
Q9The 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.
Q10F2. F1 has no lever arm with respect to the given axis.
Q11F1 provides the larger torque. F2 has a smaller component perpendicular to the radius.
Q12Yes, 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.
Q13Yes, 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.
Q14To 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.
Q15No. 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.
Q16The 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.
Q17No. 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.
Q18No. A net torque produces an angular acceleration.
Q19The 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.
Q20Body 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.
Q21Yes, if the same mass is distributed at different distances from the axis of rotation in the two cases.
Q22Yes. Extending one's arms will increase the rotational inertia compared to the case of the arms at one's side.
Q23The hollow sphere has the greater rotational inertia.
Q24No. Angular momentum is only conserved when there is no net torque.
Q25No. The distribution of mass gives a different rotational inertia. Therefore the angular momentum will also be different.
Q26It 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.
Q27The increased mass will slow it down.
Q28Yes. He can change his rotational inertia by changing the extension of his arms. His rotational velocity changes according to conservation of angular momentum.
Q29The rotational velocity will increase due to the decrease in rotational inertia.
Q30 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.
Q31By 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.
Q32a. Away from the observer.
b. No. During the fall there was an angular torque.
Q33Due 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.
Q34A 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
E1a.0.1667 rev/s
b.1.05 rad/s
E2a.0.75 rev/s
b.3.75 revolutions
E3a.18.8 rad
b.4.71 rad/s
E4a.0.3 rad/s2
E5a.4.8 rev/s
b.9.6 rev
E6-0.25 rev/s2
E7a.1 rev/s
b.2.5 rev
E8a.12 N•m
b.6 N•m
E915 cm
E106 N
E11a.96 N•m
b.-60 N•m
c.36 N•m
E125 rad/s2
E1313.5 N•m
E14a.50 N•m
b.25 kg•m2
E150.10 kg•m2
E16a.0.20 kg•m2
b.0.60 kg•m2/s
E17a.0.08 kg•m2
b.1.6 kg•m2/s
E186.28 rad/s = 60 rpm
Answers to Synthesis Problems
SP1a.132 N•m; Directed along the merry-go-round axis.
b.0.147 rad/s2
c.2.2 rad/s
d.-1.3x10-2 rad/s2; 165 s after pushing is stopped.
SP2a.80 N•m
b.0.53 m
c.He could have somebody else be prepared to hold down the other end if the plank started to tip.
SP3a.960 kg•m2; 2460 kg•m2
b.1560 kg•m2
c.1.89 rad/s
d.Yes. During the time the children are moving, friction between their feet and the merry-go-round produces the accelerating torque.
SP4a.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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