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(C) The total energy $E$ of a rolling sphere is the sum of its translational kinetic energy and rotational kinetic energy.$E = K_{trans} + K_{rot} = \

Vedclass · Accepted Answer

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

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(C) The total energy $E$ of a rolling sphere is the sum of its translational kinetic energy and rotational kinetic energy.$E = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since the sphere rolls without slipping,$\omega = \frac{v}{R}$ and the moment of inertia $I = mK^2$.Substituting these into the equation:$E = \frac{1}{2}mv^2 + \frac{1}{2}(mK^2)\left(\frac{v}{R}ight)^2 = \frac{1}{2}mv^2 \left(1 + \frac{K^2}{R^2}ight) = \frac{1}{2}mv^2 \left(\frac{R^2 + K^2}{R^2}ight)$Rotational kinetic energy $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(mK^2)\left(\frac{v^2}{R^2}ight) = \frac{1}{2}mv^2 \left(\frac{K^2}{R^2}ight)$The fraction of total energy in the form of rotational kinetic energy is:$	ext{Fraction} = \frac{K_{rot}}{E} = \frac{\frac{1}{2}mv^2 (K^2/R^2)}{\frac{1}{2}mv^2 (R^2 + K^2)/R^2} = \frac{K^2}{K^2 + R^2}$

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

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