A solid disk of mass 3 kg and radius 0.20 m has a constant torque of 0.60 N·m applied. What is its angular acceleration?

A solid disk of mass 3 kg and radius 0.20 m has a constant torque of 0.60 N·m applied. What is its angular acceleration?

• A. 5 rad/s^2
• B. 15 rad/s^2
• C. 10 rad/s^2
• D. 20 rad/s^2

Answer: (C)

Torque sets how quickly the disk spins up, via tau = I alpha. For a solid disk, the moment of inertia about its center is I = (1/2) M R^2. With M = 3 kg and R = 0.20 m, I = 0.5 × 3 × (0.20)^2 = 0.06 kg·m^2. The applied torque is 0.60 N·m, so the angular acceleration is alpha = tau / I = 0.60 / 0.06 = 10 rad/s^2. Since the torque is constant, the angular acceleration is constant as well. Answer: 10 rad/s^2.

Torque sets how quickly the disk spins up, via tau = I alpha. For a solid disk, the moment of inertia about its center is I = (1/2) M R^2. With M = 3 kg and R = 0.20 m, I = 0.5 × 3 × (0.20)^2 = 0.06 kg·m^2. The applied torque is 0.60 N·m, so the angular acceleration is alpha = tau / I = 0.60 / 0.06 = 10 rad/s^2. Since the torque is constant, the angular acceleration is constant as well. Answer: 10 rad/s^2.