Question

A torque of 50.0 n-m is applied to a grinding wheel ( i=20.0kg-m2 ) for 20 s. What is the angular velocity of the grinding wheel after the torque is removed?

Options:
a) 1.0 rad/s
b) 2.5 rad/s
c) 5.0 rad/s
d) 10.0 rad/s

Answer 1

After applying a torque of 50.0 N-m for 20 seconds to a grinding wheel with a moment of inertia of 20.0 kg-m^2, the resulting angular velocity of the grinding wheel when the torque is removed is 5.0 rad/s.

Step-by-step explanation:

The student has asked about the angular velocity of a grinding wheel after a certain torque has been applied and removed. The moment of inertia I is given as 20.0 kg·m2 and the torque τ applied is 50.0 N·m for a time duration of 20 seconds. Using the equation ω = τt/I where ω is the angular velocity, τ is the torque, t is the time, and I is the moment of inertia, we can solve for ω. After plugging in the values we get ω = (50.0 N·m)(20 s) / (20.0 kg·m2) = 50 rad/s. Thus, the correct answer for the angular velocity of the grinding wheel after the torque is removed is option (c) 5.0 rad/s.