Question
A solid sphere of mas M and radius R is divided into two unequal parts. The smaller part having mass M/8 is converted into a sphere of radius r and the larger part is converted into a circular disc of thickness $t$ and radius $2R$. If $I_{1}$ is moment of inertia of a sphere having radius r about an axis through its centre and $I_{2}$ is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia $I_{2}/I_{1} =$ $\_\_\_\_$ .
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Solution
Sol. [Key Concept: $I_{1} = \frac{2}{5}(\frac{M}{8})r^{2}$ and $I_{2} = \frac{1}{4}(\frac{7M}{8})(2R)^{2}$; put $r = R/2$ and take the ratio.]
$$\begin{matrix} & I_{1} = \frac{2}{5}\left( \frac{M}{8} \right)r^{2} \\ & I_{2} = \frac{\frac{7M}{8} \times (2R)^{2}}{4} \end{matrix}$$
$$\begin{matrix} & M = \rho \times \frac{4\pi}{3} \times R^{3} \\ & \frac{M}{8} = \rho \times \frac{4\pi}{3} \times |r|^{3} \\ & r = \frac{R}{2} \end{matrix}$$
$${\frac{I_{1}}{I_{2}} = \frac{\frac{2}{5}\left( \frac{M}{8} \right)\left( \frac{R}{2} \right)^{2}}{\frac{7M}{8 \times 4} \times (2R)^{2}} = \frac{2}{5} \times \frac{1}{4 \times 7} = \frac{1}{70} }{\frac{I_{2}}{I_{1}} = \frac{70}{1} }$$
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