Physics - Rotation Question with Solution | TestHub

Moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is given by:

$I=\frac{2}{5}M{R}^2$

When the sphere is melted and recast into a disc, the mass $M$ remains constant.

Moment of inertia of a uniform solid disc of radius $r$ about an axis passing through its center and perpendicular to its plane is $\frac{1}{2}M{r}^2$.

Using the parallel axis theorem, the moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is:

$I_{disc}=I_{cm}+M{r}^2=\frac{1}{2}M{r}^2+M{r}^2=\frac{3}{2}M{r}^2$

It is given that the moment of inertia of the disc is equal to the moment of inertia of the sphere:

$\frac{3}{2}M{r}^2=\frac{2}{5}M{R}^2$

$\frac{r^2}{R^2}=\frac{4}{15}$

$\frac{r}{R}=\frac{2}{\sqrt{15}}$

Answer: $\frac{2}{\sqrt{15}}$