Solve the following :

Question:

A flywheel of moment of inertia $5.0 \mathrm{~kg}-_{-} \mathrm{m}^{2}$ is rotated at a speed of $60 \mathrm{rad} / \mathrm{s}$. 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 its stops rotating.

Solution:

$\omega_{0}=60 \frac{\mathrm{rad}}{\mathrm{s}} ; \omega=0 ; t=5 \mathrm{~min}=300 \mathrm{sec}$

$\omega=\omega_{0}+\alpha t$

$\alpha=-\frac{1}{2} \mathrm{rad} / \mathrm{sec}^{2}$

(a) $\tau=I \alpha$

$=5\left(\frac{-1}{5}\right)$

$=-1 \mathrm{~N}-\mathrm{m}$

(b) $W_{a l l}=\Delta \mathrm{KE}$

$W_{f f}=\frac{1}{2} W^{2}$

$\frac{1}{2}(5)(60)^{2}$

$=9 \mathrm{KJ}$

$=9$

$=5(\overline{5})$

$=-1 \mathrm{~N}-\mathrm{m}$

(b) $W_{a \|}=\Delta \mathrm{KE}$

$W_{f f} \stackrel{-1}{2} w^{2}$

\begin{aligned} & \frac{1}{2}(5)(60)^{2} \\=& 9 \mathrm{KJ} \end{aligned}

(c) $\omega_{0}=60 \frac{\mathrm{rad}}{\mathrm{sec}} ; t=4 \mathrm{~min}=240 \mathrm{sec} ; \alpha=\frac{-1}{5} \frac{\mathrm{rad}}{\mathrm{s}^{2}}$

$\omega=\omega_{0}+\alpha t$

$\omega=12 \mathrm{rad} / \mathrm{sec}$

Angular momentum

$\mathrm{L}=\left.\right|^{\omega}$

$=5(12)$

$=60 \frac{K g-m^{2}}{s e c}$