Question

A flywheel of moment of inertia $5.0kg-m^2$ is rotated at a speed of $10rad/s$. Because of the friction at the axis it comes to rest in $10s.$ The average torque of the friction in Nm is

Answer 1

$0={\omega }_0-\alpha t$
$\Rightarrow \alpha =\frac{{\omega }_0}{t}=\frac{10}{10}=1rad/s^2$
$\tau =I\alpha =5N.m$