A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre…

A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $\mathrm{K}$. If radius of the ball be $\mathrm{R}$, then the fraction of total energy associated with its rotational energy will be
  1. $\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}$
  2. $\frac{\mathrm{K}^{2}}{\mathrm{~K}^{2}+\mathrm{R}^{2}}$
  3. $\frac{R^{2}}{K^{2}+R^{2}}$
  4. $\frac{\mathrm{K}^{2}+\mathrm{R}^{2}}{\mathrm{R}^{2}}$

Solution

$\frac{\text { Rotational } \mathrm{KE}}{\text { Total } \mathrm{KE}}=\frac{\frac{1}{2} \mathrm{mv}^{2}\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)}{\frac{1}{2} \mathrm{mv}^{2}\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)}$
$=\frac{\mathrm{K}^{2}}{\mathrm{~K}^{2}+\mathrm{R}^{2}}$

Asked in: JEE Mains - Rotational Motion - Test 2

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