Question

The wheel on the grinder is a uniform 0.9kg disk of 8cm radius it costs uniformly to rest from 1400rpm in a time of 35 sec. How large a friction torque slows its motion?

Answer 1

The Newton's second law says

$$\tau=I\alpha$$

The moment of inertia of a disk

$$I=\frac{1}{2}mR^2=\frac{1}{2}\times 0.9\times 0.08^2=2.88\times 10^{-3}\:\rm kg\cdot m^2$$

The angular acceleration

$$\alpha=\frac{\Delta\omega}{\Delta t}=\frac{1400/60\times 2\pi}{35}=4.19\:\rm rad/s^2$$

Hence, the torque

$$\tau=2.88\times 10^{-3}\times 4.19=0.012\:\rm N\cdot m$$