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

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