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(A) The moment of inertia of a solid sphere is $I_{sphere} = \frac{2}{5} M_1 R^2$,where $M_1 = \frac{4}{3} \pi R^3 \rho$.So,$I_{sphere} = \frac{2}{5}

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Solid sphere

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(A) The moment of inertia of a solid sphere is $I_{sphere} = \frac{2}{5} M_1 R^2$,where $M_1 = \frac{4}{3} \pi R^3 \rho$.So,$I_{sphere} = \frac{2}{5} (\frac{4}{3} \pi R^3 \rho) R^2 = \frac{8}{15} \pi \rho R^5$.The moment of inertia of a solid cylinder is $I_{cylinder} = \frac{1}{2} M_2 R^2$,where $M_2 = \pi R^2 L \rho$. Given $L = R$,$M_2 = \pi R^3 \rho$.So,$I_{cylinder} = \frac{1}{2} (\pi R^3 \rho) R^2 = \frac{1}{2} \pi \rho R^5$.Comparing the two: $\frac{I_{sphere}}{I_{cylinder}} = \frac{\frac{8}{15} \pi \rho R^5}{\frac{1}{2} \pi \rho R^5} = \frac{16}{15}$.Since $\frac{16}{15} > 1$,it follows that $I_{sphere} > I_{cylinder}$.

If a solid sphere and a solid cylinder of the same radius and density rotate about their own axes,which one will have a greater moment of inertia? (Assume $L=R$ for the cylinder)

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