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(A) The moment of inertia of a solid sphere is $I = \frac{2}{5} MR^2$. Since the sphere rolls without slipping,the condition $v = R\omega$ holds. The

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$\frac{7}{2}$

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(A) The moment of inertia of a solid sphere is $I = \frac{2}{5} MR^2$. Since the sphere rolls without slipping,the condition $v = R\omega$ holds. The rotational kinetic energy $E_{rot}$ is given by $E_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{2}{5} MR^2) \omega^2 = \frac{1}{5} MR^2 \omega^2 = \frac{1}{5} Mv^2$. The total kinetic energy $E_{total}$ is the sum of translational and rotational kinetic energy: $E_{total} = \frac{1}{2} Mv^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} Mv^2 + \frac{1}{5} Mv^2 = (\frac{5+2}{10}) Mv^2 = \frac{7}{10} Mv^2$. The ratio of total kinetic energy to rotational kinetic energy is $\frac{E_{total}}{E_{rot}} = \frac{\frac{7}{10} Mv^2}{\frac{1}{5} Mv^2} = \frac{7}{10} 	imes 5 = \frac{7}{2}$.

$A$ solid sphere rolls without slipping on an inclined plane at an angle $\theta$. The ratio of total kinetic energy to its rotational kinetic energy is

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