Question

A flywheel of moment of inertia 5⋅0 kg-m² is rotated at a speed of 60 rad/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

Let the angular deceleration produced due to frictional force be α.
Initial angular acceleration, ${\omega }_0=60\mathrm{rad}/\mathrm{s}$
Final angular velocity, $\omega =0$
t = 5 min =300 s
We know that:
$\omega ={\omega }_0+\alpha t$
$$\begin{gathered} \Rightarrow \alpha =-\frac{{\omega }_0}{t} \\ \Rightarrow \alpha =-\left(\frac{60}{300}\right)=-\frac{1}{5}\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

(a) Torque produced by the frictional force (R),
$$\begin{gathered} \tau =I\alpha =5\times \left(\frac{-1}{5}\right) \\ =1\mathrm{N}-\mathrm{m}\mathrm{opposite}\mathrm{to}\mathrm{the}\mathrm{rotation}\mathrm{of}\mathrm{wheel} \end{gathered}$$

(b) By conservation of energy,
Total work done in stopping the wheel by frictional force = Change in energy
$$\begin{gathered} \mathrm{W}=\frac{1}{2}\mathrm{I}{\omega }^2 \\ =\frac{1}{2}\times 5\times \left(60\times 60\right) \\ =9000\mathrm{joule}=9\mathrm{kJ} \end{gathered}$$


(c) Angular velocity after 4 minutes,

$$\begin{gathered} \omega ={\omega }_0+\alpha t \\ =60-\frac{4\times 60}{5} \\ =\frac{60}{5}=12\mathrm{rad}/\mathrm{s} \end{gathered}$$
So, angular momentum about the centre,
$$\begin{gathered} L=I\omega \\ =5\times 12=60\mathrm{kg}-{\mathrm{m}}^2/\mathrm{s} \end{gathered}$$