A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is $\frac{x}{5}$. The value of $x$ is _____ .

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{\frac{1}{2} \mathrm{I} \omega^2}{\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} \mathrm{mv}^2}=\frac{\left(\frac{1}{2}ight)\left(\frac{2}{3} \mathrm{mR}^2ight) \omega^2}{\left(\frac{1}{2}ight)\left(\frac{2}{3} \mathrm{mR}^2ight) \omega^2+\frac{1}{2} \mathrm{~m}(\mathrm{R} \omega)^2} \ & =\frac{\frac{2}{3}}{\frac{2}{3}+1}=\frac{2}{5} \ & \mathrm{x}=2\end{aligned}$

A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic…

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