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 . The value of is

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Solution

Q.
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 _______ . [2024]

Ans.
(2) $\frac{K E_{Rot}}{K E_{Trans}} = \frac{\frac{1}{2} I ω^{2}}{\frac{1}{2} I ω^{2} + \frac{1}{2} m v^{2}} = \frac{(\frac{1}{2}) (\frac{2}{3} m R^{2}) ω^{2}}{(\frac{1}{2}) (\frac{2}{3} m R^{2}) ω^{2} + \frac{1}{2} m {(R ω)}^{2}}$

$\Rightarrow \frac{K E_{Rot}}{K E_{Trans}} = \frac{\frac{2}{3}}{\frac{2}{3} + 1} = \frac{2}{5}$

Hence, $x = 2$

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