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Question

(C) In pure rolling motion,the total kinetic energy is the sum of translational and rotational kinetic energy.
Translational $KE = \frac{1}{2}mv^2$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) In pure rolling motion,the total kinetic energy is the sum of translational and rotational kinetic energy.
Translational $KE = \frac{1}{2}mv^2$.
Rotational $KE = \frac{1}{2}I\omega^2 = \frac{1}{2}(mK^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(\frac{K^2}{R^2})$.
Total $KE = \frac{1}{2}mv^2 + \frac{1}{2}mv^2(\frac{K^2}{R^2}) = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2}) = \frac{1}{2}mv^2(\frac{R^2+K^2}{R^2})$.
The fraction of total energy associated with rotation is $\frac{Rotational \ KE}{Total \ KE} = \frac{\frac{1}{2}mv^2(\frac{K^2}{R^2})}{\frac{1}{2}mv^2(\frac{R^2+K^2}{R^2})} = \frac{K^2}{K^2+R^2}$.

$A$ uniform solid sphere of radius $R$ and radius of gyration $K$ about an axis passing through the centre of mass,is rolling without slipping. Then the fraction of total energy associated with its rotation will be

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