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Question

Physics Question on System of Particles & Rotational Motion

A solid sphere rolls without slipping on the roof. The ratio of its rotational kinetic energy and its total kinetic energy is

A

44597

B

44656

C

44599

D

44627

Answer

44599

Explanation

Solution

Kinetic energy of sphere Kr0=12Iω2K_{r_{0}}=\frac{1}{2} I \omega^{2} \therefore Moment of inertia of sphere, I=25MR2I=\frac{2}{5} M R^{2} \therefore Rotational kinetic energy of sphere Kr0=12MR2ω2K_{r_{0}}=\frac{1}{2} M R^{2} \omega^{2} Total energy of sphere Kr0=12Iω2+12Mv2K_{r_{0}} =\frac{1}{2} I \omega^{2}+\frac{1}{2} M v^{2} =12×25MR2ω2+12MR2ω2=\frac{1}{2} \times \frac{2}{5} M R^{2} \omega^{2}+\frac{1}{2} M R^{2} \omega^{2} =MR2ω2(15+12)=M R^{2} \omega^{2}\left(\frac{1}{5}+\frac{1}{2}\right) =710MR2ω2=\frac{7}{10} M R^{2} \omega^{2} Total energy of sphere Kt0=710MR2ω2K_{t_0}=\frac{7}{10} M R^{2} \omega^{2} Kr0Kt0=15MR2ω2710MR2ω2=27\frac{K_{r_{0}}}{K_{t_{0}}}=\frac{\frac{1}{5} M R^{2} \omega^{2}}{\frac{7}{10} M R^{2} \omega^{2}}=\frac{2}{7}