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A child is standing with folded hands at the centre of the platform rotating about its central axis. The kinetic energy of the system is ' $\mathrm{K}$ '. The child now stretches his arms so that the moment of inertia of the system becomes double. The kinetic energy of system now is
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The correct answer is:
$\frac{\mathrm{K}}{2}$
$$
\begin{aligned}
\mathrm{K} & =\frac{1}{2} \mathrm{I} \omega^2 \\
\mathrm{I}^{\prime} & =2 \mathrm{I}
\end{aligned}
$$
By law of conservation of angular momentum
$$
\begin{aligned}
& \mathrm{I}^{\prime} \omega^{\prime}=\mathrm{I} \omega \\
& \therefore 2 \mathrm{I} \omega^{\prime}=\mathrm{I} \omega \\
& \therefore \omega^{\prime}=\frac{\omega}{2} \\
& \mathrm{~K}^{\prime}=\frac{1}{2} \mathrm{I}^{\prime} \omega^{\prime 2}=\frac{1}{2}(2 \mathrm{I})\left(\frac{\omega}{2}\right)^2=\frac{1}{2}\left(\frac{1}{2} \mathrm{I} \omega^2\right)=\frac{\mathrm{K}}{2}
\end{aligned}
$$
\begin{aligned}
\mathrm{K} & =\frac{1}{2} \mathrm{I} \omega^2 \\
\mathrm{I}^{\prime} & =2 \mathrm{I}
\end{aligned}
$$
By law of conservation of angular momentum
$$
\begin{aligned}
& \mathrm{I}^{\prime} \omega^{\prime}=\mathrm{I} \omega \\
& \therefore 2 \mathrm{I} \omega^{\prime}=\mathrm{I} \omega \\
& \therefore \omega^{\prime}=\frac{\omega}{2} \\
& \mathrm{~K}^{\prime}=\frac{1}{2} \mathrm{I}^{\prime} \omega^{\prime 2}=\frac{1}{2}(2 \mathrm{I})\left(\frac{\omega}{2}\right)^2=\frac{1}{2}\left(\frac{1}{2} \mathrm{I} \omega^2\right)=\frac{\mathrm{K}}{2}
\end{aligned}
$$
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