Question

A solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The first part has a mass of $\frac{7M}{8}$ and is converted into a uniform disc of radius $2R$. The second part is converted into a uniform solid sphere. Let $I_1$ be the moment of inertia of the new sphere about its axis. The ratio $I_1/I_2$ is $\frac{140}{x}$. Find the value of $x$.

Answer 1

Answer: 140

$I_1=(\frac{7M}{8})(2R{)}^2\frac{1}{2}=(\frac{7}{16}\times 4)M{R}^2=\frac{14}{8}m{R}^2$
$I_2=\frac{2}{5}(\frac{M}{8})r^2$
$\Rightarrow I_2=\frac{2}{5}(\frac{M}{8})(\frac{R^2}{4})=\frac{M{R}^2}{80}$
$[\frac{4}{3}\pi r^3\rho =\frac{1}{8}\frac{4}{3}\pi R^3\times \rho \Rightarrow r=R/2]$
$\frac{I_1}{I_2}=\frac{14\times 80}{8}=140$