Torque and Work AP Physics 1 MCQ – Exam Style Questions etc.
Torque and Work AP Physics 1 MCQ
Unit 6: Energy and Momentum of Rotating Systems
Weightage : 10-15%
Exam Style Practice Questions, Torque and Work AP Physics 1 MCQ
Question
A 50-kilogram person is sitting on a seesaw 1.2 meters from the balance point. On the other side, a 70-kilogram person is balanced. How far from the balance point is the second person sitting?
(A) 0.57 m
(B) 0.75 m
(C) 0.63 m
(D) 0.86 m
▶️Answer/Explanation
Ans:(D) Balanced means the torques are equal and opposite:
50 kg · g · 1.2 = 70 kg · g · x
x = 6/7 = 0.86
Question
A solid sphere \((I=0.06kg\cdot m^{2})\) spins freely around an axis through its center at an angular speed of 20 rad/s. It is desired to bring the sphere to rest by applying a friction force of magnitude 2.0 N to the sphere’s outer surface, a distance of 0.30 m from the sphere’s center. How much time will it take the sphere to come to rest?
(A) 4 s
(B) 2 s
(C) 0.06 s
(D) 0.03 s
▶️Answer/Explanation
Ans:
B—This is a calculation using \(\tau _{net}=Ia\). The net torque on the sphere is force times distance from the center, or (2.0 N)(0.30 m) = 0.60 m∙N. Now the angular acceleration can be calculated: (0.60 m∙N) = (0.06 \(kg\cdot m^{2}\))(a), so a = 10 rad/s per second. Use the definition of angular acceleration: the sphere loses 10 rad/s of speed each second. It started with 20 rad/s of speed, so after 1 s it has 10 rad/s of speed; after 2 s it will have lost all its speed.
Questions (a) and (b) refer to the following information:
A bicycle wheel of known rotational inertia is mounted so that it rotates clockwise around a vertical axis, as shown in the first figure. Attached to the wheel’s edge is a rocket engine, which applies a clockwise torque τ on the wheel for a duration of 0.10 s as it burns. A plot of the angular position θ of the wheel as a function of time t is shown in the graph.
Question(a)
In addition to the wheel’s rotational inertia and the duration of time the engine burns, which of the following information from the graph would allow determination of the net torque the rocket exerts on the wheel?
(A) The area under the graph between t = 0 s and t = 3 s
(B) The change in the graph’s slope before and after t = 2 s
(C) The vertical axis reading of the graph at t = 3 s
(D) The vertical axis reading of the graph at t = 2 s
▶️Answer/Explanation
Ans:B
Newton’s second law for rotation says \(\tau _{net}=I\alpha\), where α is the angular acceleration, or change in the wheel’s angular velocity per time. To find the wheel’s angular velocity, look at the slope of the angular position versus time graph. The slope changes after the torque is applied; so the change in the slope is the change in the angular velocity, which (when divided by the 0.10 s duration of the rocket firing) gives the angular acceleration.