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Rotational Inertia AP  Physics 1 MCQ

Rotational Inertia AP  Physics 1 MCQ – Exam Style Questions etc.

Rotational Inertia AP  Physics 1 MCQ

Unit 5: Torque and Rotational Dynamics

Weightage : 10-15%

AP Physics 1 Exam Style Questions – All Topics

Exam Style Practice Questions, Rotational Inertia AP  Physics 1 MCQ

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 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

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

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