Question

A flywheel of moment of inertia $5.0kg-m^2$ is rotated at a speed of $60rad/s$ .Because of the friction at the axle, it comes to rest in $5.0$ minutes. Find (a) the average torque of the friction, (b) the total work done by the friction and (c) the angular momentum of the wheel $1$ minute before it stops rotating.

Answer 1

A flywheel of the moment of inertia $5kgm$ is rotated at a speed of $60rad/s$. The flywheel comes to rest due to the friction at the axle after $5$ minutes.
Therefore, the angular deceleration produced due to frictional force $=\omega ={\omega }_0+\alpha t$
$\Rightarrow {\omega }_0=-\alpha t(\omega =0+$
$\Rightarrow \alpha =-(60/5\times 60)=-1/5rad/s^2$.

a) Therefore total workdone in stopping the wheel by frictional force
$W=1/2I{\omega }^2=1/2\times 5\times (60\times 60)=9000Joule=9KJ$.

b) Therefore torque produced by the frictional force
$I_R=I\times \alpha =5\times (-1/5)=1N-m$ opposite to the rotation of wheel.

c) Angular velocity after $4$ minutes
$\Rightarrow \omega ={\omega }_0+\alpha t=60-240/5=12rad/s$
Therefore angular momentum about the centre $=1\times \omega =5\times 12=60kg-m^2/s$.