Question

For mass 20 kg, torque \( \tau = 5 \, \text{Nm} \), find angular acceleration \( \alpha \).

Hint:

Angular acceleration is found using \( \alpha = \frac{\tau}{I} \), where \( I \) is the moment of inertia. For a point mass, \( I = mr^2 \).

Answer 1

Angular acceleration follows from \( \alpha = \tau/I \), where the moment of inertia of a point mass is \( I = mr^2 \). Here the torque is 5 N·m and the mass is 20 kg, but the question doesn't state the radius \( r \) about which the mass rotates, so the answer stays in terms of \( r \): \( \alpha = 5/(20r^2) = 0.25/r^2 \).