AP Physics Practice Questions: Rotational Motion

AP Physics multiple choice questions on rotational motion, torque, and angular momentum. Each question includes detailed solutions at the bottom of the page.

1. A uniform disk of mass 2.0 kg and radius 0.5 m rotates about its central axis. What constant torque is needed to increase its angular speed from 0 to 10 rad/s in 5.0 s?

   A) 0.25 N·m
   B) 0.50 N·m
   C) 1.00 N·m
   D) 2.00 N·m
   E) 4.00 N·m

2. A solid sphere (I = 2/5 MR²) rolls without slipping down a 30° incline from rest. What is its acceleration down the incline?

   A) 2.45 m/s²
   B) 3.27 m/s²
   C) 3.50 m/s²
   D) 4.20 m/s²
   E) 4.90 m/s²

3. A figure skater with arms extended has moment of inertia I₁ = 4.0 kg·m² and rotates at 2.0 rad/s. When she pulls her arms in, her moment of inertia becomes I₂ = 1.0 kg·m². What is her new angular speed?

   A) 0.5 rad/s
   B) 2.0 rad/s
   C) 4.0 rad/s
   D) 8.0 rad/s
   E) 16.0 rad/s

4. A 0.5 m long wrench is used to apply a force of 40 N at an angle of 60° from the handle. What is the torque about the bolt?

   A) 8.7 N·m
   B) 10.0 N·m
   C) 17.3 N·m
   D) 20.0 N·m
   E) 34.6 N·m

5. A hollow cylinder and a solid cylinder with the same mass and radius roll down an incline from the same height. Which statement is correct?

   A) The solid cylinder reaches the bottom first
   B) The hollow cylinder reaches the bottom first
   C) They reach the bottom at the same time
   D) The hollow cylinder has greater linear acceleration
   E) The solid cylinder has greater rotational inertia

Solutions

1. B) 0.50 N·m
   Moment of inertia of disk: \( I = \dfrac{1}{2}MR^2 = \dfrac{1}{2} \times 2.0 \times (0.5)^2 = 0.25 \, \mathrm{kg·m}^2 \)
   Angular acceleration: \( \alpha = \dfrac{\Delta \omega}{\Delta t} = \dfrac{10}{5} = 2.0 \, \text{rad/s}^2 \)
   Torque: \( \tau = I\alpha = 0.25 \times 2.0 = 0.50 \, \mathrm{N·m} \)

2. C) 3.50 m/s²
   For rolling without slipping: \( a = \dfrac{g\sin\theta}{1 + \dfrac{I}{MR^2}} \)
   For solid sphere: \( \dfrac{I}{MR^2} = \dfrac{2}{5} \)
   \( a = \dfrac{g\sin30°}{1 + \dfrac{2}{5}} = \dfrac{0.5g}{1.4} = \dfrac{4.9}{1.4} = 3.5 \, \text{m/s}^2 \)

3. D) 8.0 rad/s
   Conservation of angular momentum: \( I_1 \omega_1 = I_2 \omega_2 \)
   \( 4.0 \times 2.0 = 1.0 \times \omega_2 \)
   \( \omega_2 = 8.0 \, \text{rad/s} \)

4. C) 17.3 N·m
   Torque = \( rF\sin\theta = 0.5 \times 40 \times \sin60° = 20 \times 0.866 = 17.32 \, \mathrm{N·m} \)

5. A) The solid cylinder reaches the bottom first
   When objects roll without slipping down an incline, their linear acceleration depends on how their mass is distributed, i.e. their rotational inertia (I).

   The acceleration of a rolling object is \( a = \dfrac{g \sin\theta}{1 + \dfrac{I}{mR^2}} \)

   Solid cylinder has \( I_{\text{solid}} = \tfrac{1}{2} mR^2 \)

   Hollow cylinder (thin-walled) has \( I_{\text{hollow}} = mR^2 \)

   Since \( I_{\text{hollow}} > I_{\text{solid}}, \)
   Solid cylinder has smaller moment of inertia than hollow cylinder , so it accelerates faster down the incline.