A sphere of radius $R$ is cut from a larger solid sphere of radius $2 R$ as shown in the figure. The

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

Professor PC Thomas Classes · Answer

(a) : Moment of inertia of a solid sphere of mass $M$ and radius $R$ is $ I=\frac{2}{5} M R^2 $ \ Mass of smaller sphere, $M_S=\frac{M}{8}$ \ Mass of remaining part, $M_r=M-\frac{M}{8}=\frac{7 M}{8}$ \ Using parallel-axes theorem, moment of inertia of smaller sphere about $y$-axis, 
$I_S=\frac{2}{5} M_S R^2+M_S R^2=\left(\frac{2}{5}+1ight) M_S R^2=\frac{7}{5} M_S R^2$
or $I_S=\frac{7}{5} 	imes \frac{M}{8} 	imes R^2=\frac{7}{40} M R^2$
\ Moment of inertia of the larger part of the sphere about $y$-axis, $ I_L=\frac{2}{5} M(2 R)^2=\frac{8}{5} M R^2 $ Now, moment of inertia of the rest part of the sphere is 
$I_r=I_L-I_S=\frac{8}{5} M R^2-\frac{7}{40} M R^2=\frac{57}{40} M R^2$
$	herefore \frac{I_S}{I_r}=\frac{\frac{7}{40} M R^2}{\frac{57}{40} M R^2}=\frac{7}{57}$