Answer
$-14.8 \frac{rad}{s^2}$. 1.52 s.
Work Step by Step
Apply Newton’s second law for rotations.
First find the moment of inertia.
$$I=\frac{2}{5}(0.225kg)(1.50\times10^{-2}m)^2=2.025\times10^{-5}kg\cdot m^2$$
Calculate the angular acceleration.
$$\alpha=\frac{\tau}{I}$$
$$\alpha=\frac{-(0.0200N) (1.50\times10^{-2}m)}{I}=-14.81\frac{rad}{s^2}$$
This angular acceleration tends to slow down the rotation because it is a frictional torque.
Find the time.
$$t=\frac{\Delta \omega}{\alpha}=\frac{-22.5 rad/s}{-14.81\frac{rad}{s^2}}=1.52 s$$
