Physics - Rotation Question with Solution | TestHub

The moment of inertia of a solid sphere of mass $m$ and radius $R$ about an axis passing through its center of mass is given by:

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

When the sphere is recast into a disc of radius $r$ and thickness $t$, its mass $m$ remains the same. The moment of inertia of the disc 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 rim and perpendicular to its plane is:

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

According to the given condition, the moment of inertia remains $I$. Therefore, we equate the two moments of inertia:

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

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

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

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