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(C) For a body rolling without slipping,the total kinetic energy $E_{	ext{total}}$ is the sum of translational kinetic energy $E_t$ and rotational ki

Vedclass · Accepted Answer

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

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(C) For a body rolling without slipping,the total kinetic energy $E_{	ext{total}}$ is the sum of translational kinetic energy $E_t$ and rotational kinetic energy $E_r$.$E_t = \frac{1}{2}mv^2$$E_r = \frac{1}{2}I\omega^2$Since $I = mK^2$ and $\omega = \frac{v}{R}$,we have $E_r = \frac{1}{2}(mK^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(\frac{K^2}{R^2})$.$E_{	ext{total}} = E_t + E_r = \frac{1}{2}mv^2 + \frac{1}{2}mv^2(\frac{K^2}{R^2}) = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2})$.The fraction of total energy associated with rotational kinetic energy is $\frac{E_r}{E_{	ext{total}}}$.$\frac{E_r}{E_{	ext{total}}} = \frac{\frac{1}{2}mv^2(\frac{K^2}{R^2})}{\frac{1}{2}mv^2(1 + \frac{K^2}{R^2})} = \frac{\frac{K^2}{R^2}}{\frac{R^2 + K^2}{R^2}} = \frac{K^2}{K^2 + R^2}$.

$A$ ball rolls without slipping. The radius of gyration of the ball about an axis passing through its center of mass is $K$. If the radius of the ball is $R$,what fraction of its total energy is associated with its rotational kinetic energy?

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