Question 617: Centre of Mass and Rotational Motion

Question

A sphere of radius $R$ is cut from a larger solid sphere of radius $2 R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the $Y$-axis is:

Accepted Answer

For larger solid sphere about diameter $Y$-axis,
$$I_{	ext {whole }}=\frac{2}{5} M(2 R)^2=\frac{8}{5} M R^2$$
Density of sphere is uniform
$$\begin{aligned}
& \Rightarrow \frac{M}{V_{	ext {whole }}}=\frac{M_{	ext {smaller }}}{V_{	ext {smaller }}} \Rightarrow \frac{M}{\frac{4}{3} \pi(2 R)^3}=\frac{M^{\prime}}{\frac{4}{3} \pi R^3} \
& \Rightarrow M^{\prime}=\frac{M}{8}
\end{aligned}$$
Using parallel axis theorem for smaller sphere,
$$\begin{aligned}
& I^{\prime}=I_{\mathrm{cm}}+M^{\prime} R^2=\frac{2}{5} \frac{M R^2}{8}+\frac{M R^2}{8}=\frac{7}{40} M R^2 \
& 	herefore 	ext { Ratio }=\frac{I_{	ext {smaller }}}{I_{	ext {remaining }}}=\frac{I^{\prime}}{I_{	ext {whole }}-I^{\prime}}=\frac{\frac{7}{40} M R^2}{\left(\frac{8}{5}-\frac{7}{40}ight) M R^2}=\frac{7}{64-7}=\frac{7}{57}
\end{aligned}$$

Answer

$\frac{7}{57}$

Answer

$\frac{7}{64}$

Answer

$\frac{7}{8}$

Answer

$\frac{7}{40}$

Centre of Mass and Rotational Motion

A sphere of radius R is cut from a larger solid sphere of radius 2 R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:

Choose the correct answer: