Physics - Rotation Question with Solution | TestHub

Let the mass of the disc be $M$.

The moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane is given by:
$I=\frac{1}{2}M{R}^2$

The volume of the disc is:
$V_{\text{disc}}=\pi R^{2}t=\pi R^2\left(\frac{R}{6}\right)=\frac{\pi R^3}{6}$

Let the radius of the solid sphere formed by melting the disc be $r$. The volume of the solid sphere is:
$V_{\text{sphere}}=\frac{4}{3}\pi r^3$

Since the volume remains constant during recasting:
$\frac{4}{3}\pi r^3=\frac{\pi R^3}{6}$

$r^3=\frac{R^3}{8}\Rightarrow r=\frac{R}{2}$

The moment of inertia of the solid sphere about its diameter is:
$I_{\text{sphere}}=\frac{2}{5}M{r}^2$

Substituting $r=\frac{R}{2}$:
$I_{\text{sphere}}=\frac{2}{5}M{\left(\frac{R}{2}\right)}^2=\frac{2}{5}M\left(\frac{R^2}{4}\right)=\frac{M{R}^2}{10}$

Since $I=\frac{1}{2}M{R}^2$, we can write:
$I_{\text{sphere}}=\frac{1}{5}\left(\frac{1}{2}M{R}^2\right)=\frac{I}{5}$

Answer: $\frac{I}{5}$