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(B) For a solid sphere,the moment of inertia $I = \frac{2}{5} MR^2$.Since the sphere rolls without slipping,the condition $V = \omega R$ holds,which i

Vedclass · Accepted Answer

$28.6$

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(B) For a solid sphere,the moment of inertia $I = \frac{2}{5} MR^2$.Since the sphere rolls without slipping,the condition $V = \omega R$ holds,which implies $\omega = \frac{V}{R}$.The rotational kinetic energy is $E_r = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{2}{5} MR^2) (\frac{V}{R})^2 = \frac{1}{5} MV^2$.The translational kinetic energy is $E_t = \frac{1}{2} MV^2$.The total kinetic energy is $E_{total} = E_r + E_t = \frac{1}{5} MV^2 + \frac{1}{2} MV^2 = \frac{7}{10} MV^2$.The ratio of rotational kinetic energy to total energy is $\frac{E_r}{E_{total}} = \frac{\frac{1}{5} MV^2}{\frac{7}{10} MV^2} = \frac{1}{5} 	imes \frac{10}{7} = \frac{2}{7}$.To find the percentage,we calculate $\frac{2}{7} 	imes 100 \approx 28.57\%$,which rounds to $28.6\%$.

$A$ solid sphere rolls down without slipping on an inclined plane,then the percentage of rotational kinetic energy of the total energy will be ........ $\%.$

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