Q.A uniform disc of mass $4\ \text{kg}$ and radius $0.25\ \text{m}$ spins at $10\ \text{rad/s}$ about a frictionless axle through its centre, perpendicular to its plane. A constant frictional torque of $0.5\ \text{N}\,\text{m}$ then brings it to rest. Find

The disc's moment of inertia about its central axis:

(a) Angular deceleration, from $\tau=I\alpha$:

(the friction torque opposes the spin, so this is a deceleration, i.e. $\omega$ decreases at $4\ \text{rad/s}^2$). …

Find $I=\tfrac12MR^2$, then $\alpha=\tau/I$ from the rotational form of Newton's second law, then $t=\omega_0 …

• Forgetting the friction torque opposes the existing spin, and treating $\alpha$ as a speeding-up acceleration instead of a deceleration. …