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Question

Kinetic energy of rotation

$K_{r o t}=\frac{1}{2} I \omega^{2}=\frac{1}{2} M K^{2} \frac{v^{2}}{R^{2}}$

where $\mathrm{K}$ is radius of gyration

Vedclass · Accepted Answer

$\frac{{{K^2}}}{{{K^2} + {R^2}}}$

Vedclass · Answer

Kinetic energy of rotation

$K_{r o t}=\frac{1}{2} I \omega^{2}=\frac{1}{2} M K^{2} \frac{v^{2}}{R^{2}}$

where $\mathrm{K}$ is radius of gyration. Kinetic energy of translation,

$K_{t r a n s}=\frac{1}{2} M v^{2}$

Thus, total energy

$E=K_{r o t}+K_{t r a n s}$

$=\frac{1}{2} M K^{2} \frac{V^{2}}{R^{2}}+\frac{1}{2} M v^{2}$

$=\frac{1}{2} M v^{2}\left(\frac{K 62}{R^{2}}+1ight)$

$=\frac{1}{2} \frac{M_{v^{2}}}{R^{2}}\left(K^{2}+R^{2}ight)$

Hence $\frac{K_{r e t}}{K_{	ext {trans}}}=\frac{\frac{1}{2} M K^{2} \frac{v^{2}}{R^{2}}}{\frac{1}{2} \frac{M_{v}^{2}}{h^{2}}\left(K^{2}+R^{2}ight)}$

$\frac{K^{2}}{K^{2}+R^{2}}$

A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass $K$. If radius of the ball be $R$, then the fraction of total energy associated with its rotational energy will be

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