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

Vedclass · Accepted Answer

$K^2 / (K^2 + R^2)$

Vedclass · Answer

(B) For a body rolling without slipping,the total kinetic energy $E_{total}$ is the sum of translational kinetic energy $E_t$ and rotational kinetic energy $E_r$.$E_t = \frac{1}{2}mv^2$$E_r = \frac{1}{2}I\omega^2 = \frac{1}{2}(mK^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(\frac{K^2}{R^2})$$E_{total} = E_t + E_r = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2})$The fraction of total energy that is rotational is given by:$\frac{E_r}{E_{total}} = \frac{\frac{1}{2}mv^2(\frac{K^2}{R^2})}{\frac{1}{2}mv^2(1 + \frac{K^2}{R^2})}$$= \frac{K^2/R^2}{(R^2 + K^2)/R^2} = \frac{K^2}{K^2 + R^2}$

$A$ ball is rolling 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 the total energy is rotational kinetic energy?

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