AP Physics C Mechanics Rotational Inertia MCQ

Rotational Inertia AP  Physics C Mechanics MCQ – Exam Style Questions etc.

Rotational Inertia AP  Physics C Mechanics MCQ

Unit 5: Torque and Rotational Dynamics

Weightage : 10-15%

AP Physics C Mechanics Exam Style Questions – All Topics

Question

A girl of mass m and a boy of mass 2m are sitting on opposite sides of a see-saw with its fulcrum in the center. Right now, the boy and girl are equally far from the fulcrum, and it tilts in favor of the boy. Which of the following would NOT be a possible method to balance the see-saw?
(A) Move the boy to half his original distance from the fulcrum.
(B) Move the girl to double her original distance from the fulcrum.
(C) Allow a second girl of mass m to join the first.
(D) Move the fulcrum to half its original distance from the boy.

Answer/Explanation

Ans:D
To balance the see-saw you need to balance the torques. Since T = Fr sin θ, the boy currently provides double the torque. Choices (B) and (C) would double the torque on the girl’s side, and (A) would cut the boy’s torque in half. Choice (D) would cut the boy’s torque in half, but it would also increase the girl’s torque, creating a new imbalance.

Question

A mechanical wheel initially at rest on the floor begins rolling forward with an angular acceleration of rad/\(s^{2}\). If the radius of the wheel is 0.5 m, what is the linear velocity of the wheel after 5 s ?
(A) 0.5 m/s
(B) 1 m/s
(C) 5 m/s
(D) 10 m/s

Answer/Explanation

Ans:C
Apply Big Five #2 for rotational motion:
\(\omega=\omega_{0}+at=0+2rad/s^{2}(5s)=10rad/s\)

The angular velocity can be related to linear velocity with the equation:

\(v=r\omega=(0.5m)(10rad/s)=5m/s\)

Question

 A 5 kg box is connected to a pulley with rope in the diagram shown below. If the radius of the pulley is 0.5 m, what is the torque generated by the box on the pulley?

(A) 10 N∙m
(B) 25 N∙m
(C) 50 N∙m
(D) 75 N∙m

Answer/Explanation

Ans:B
The weight of the box pulls down on the rope, producing tension that creates a torque on the pulley. This torque is equal to

\(\tau =rF=rF_{g}=rmg=0.5m(5kg)(10m/s^{2})=25N\cdot m\)

Question

Three equal-mass objects (A, B, and C) are each initially at rest horizontally on a pivot, as shown in the figure. Object A is a 40 cm long, uniform rod, pivoted 10 cm from its left edge. Object B consists of two heavy blocks connected by a very light rod. It is also 40 cm long and pivoted 10 cm from its left edge. Object C consists of two heavy blocks connected by a very light rod that is 50 cm long and pivoted 20 cm from its left edge. Which of the following correctly ranks the objects’ angular acceleration about the pivot point when they are released?
(A) A = B > C
(B) A > B = C
(C) A < B < C
(D) A > B > C

Answer/Explanation

Ans:

D—Angular acceleration is net torque divided by rotational inertia, \(\alpha =\frac{\tau _{net}}{I}\). Imagine each object has total mass of 2 kg. Begin by comparing objects A and B: To find the net torque on object A, assume the entire 20 N weight is concentrated at the dot representing the rod’s center of mass. That’s located 10 cm from the pivot, giving a net torque of 200 N∙cm. For object B, consider the torques provided by each block separately. The right block provides a torque of (10 N)(30 cm) = 300 N∙cm clockwise; the left block provides a torque of (10 N)(10 cm) = 100 N∙cm counterclockwise. That makes the net torque 200 N∙cm, the same as for object A. But object B has more rotational inertia, since its 2 kg of mass are concentrated farther away from the pivot than object A’s mass. So the denominator of the angular acceleration equation is bigger for object B with the same numerator, which means A > B. Now consider object C. It experiences less net torque than A and B. Calculate (10 N)(30 cm) = 300 N∙cm clockwise, and (10 N)(20 cm) = 200 N∙cm counterclockwise for a net torque of 100 N∙cm. And object C has the same mass distributed even farther from the pivot point than either of the other two objects, giving C an even bigger rotational inertia. In the angular acceleration equation, object C gives a smaller numerator and a bigger denominator than object B, meaning C < B. Put it all together to get A > B > C.

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