A boy standing at the centre of a turntable with his arms outstretched is set into rotation with angular speed $\omega \mathrm{rev} / \mathrm{min}$. When the boy folds his arms back, his moment of inertia reduces to $\frac{2}{5}$ th its initial value. Find the ratio of his final kinetic energy of rotation to his initial kinetic energy.
Q 110
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Data $: I_2=\frac{2}{5} l_1$
$L=l \omega$
Assuming the angular momentum $\vec{L}$ is conserved, in magnitude,
$
\begin{aligned}
& I_1 \omega_1=I_2 \omega_2 \\
\therefore & \frac{\omega_2}{\omega_1}=\frac{I_1}{I_2}=\frac{5}{2}
\end{aligned}
$
Rotational KE, $E=\frac{1}{2} I \omega^2$
$
\therefore \frac{E_2}{E_1}=\frac{I_2}{I_1}\left(\frac{\omega_2}{\omega_1}\right)^2=\frac{2}{5}\left(\frac{5}{2}\right)^2=\frac{5}{2}
$
This gives the required ratio.
$L=l \omega$
Assuming the angular momentum $\vec{L}$ is conserved, in magnitude,
$
\begin{aligned}
& I_1 \omega_1=I_2 \omega_2 \\
\therefore & \frac{\omega_2}{\omega_1}=\frac{I_1}{I_2}=\frac{5}{2}
\end{aligned}
$
Rotational KE, $E=\frac{1}{2} I \omega^2$
$
\therefore \frac{E_2}{E_1}=\frac{I_2}{I_1}\left(\frac{\omega_2}{\omega_1}\right)^2=\frac{2}{5}\left(\frac{5}{2}\right)^2=\frac{5}{2}
$
This gives the required ratio.
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