# Solve the following :

Question:

A boy is standing on a platform which is free to rotate about its axis. The boy holds an open umbrella coincides with that of the platform. The moment of inertia of " the platform plus the boy system" is $3.0$

$\times 10^{-3} \mathrm{~kg}_{-} \mathrm{m}^{2}$ and that of the umbrella is $2.0^{\times} 10^{-3} \mathrm{~kg}^{-} \mathrm{m}^{2}$. The boy starts spinning the umbrella about the axis at an angular speed of $2.0 \mathrm{rev} / \mathrm{s}$ with respect to himself. Find the angular velocity imparted to the platform.

Solution:

Since, $\tau_{\text {ext }}=0$

$\therefore L_{i}=L_{f}$

$0+0=L_{\text {boy-final }}+L_{u m b-f \text { inal }}-(\mathrm{i})$

So, Angular speed of umbrella with respect to boy will be in the opposite direction. Let angular speed of boy be $\omega$

Now, $\omega_{u m b-d i s c}=\omega_{u m b}-\omega_{d i s c}$

$-2=\omega_{u m b}-\omega$

$\omega_{u m b}=\omega-2$

In equation (i),

$0=\left(3 \times 10^{-3}\right)(\omega)+\left(2 \times 10^{-3}\right)(\omega-2)$

$\omega=0.8 \mathrm{rev} / \mathrm{sec}$