Physics - Rotation Question with Solution | TestHub

The mass of the original complete disc is $M$.

The radius of the original disc is $2R$.

The moment of inertia of the original disc about the axis passing through its center O and perpendicular to its plane is:

$I_{\text{original}}=\frac{1}{2}M(2R{)}^2=2M{R}^2$

The removed disc has a diameter of $2R$, which means its radius is $R$.

Since the disc is uniform, its mass is proportional to its area. The area of the original disc is $\pi (2R{)}^2=4\pi R^2$, and the area of the removed disc is $\pi R^2$.

Mass of the removed disc, $m=M\times \frac{\pi R^2}{4\pi R^2}=\frac{M}{4}$

The center of the removed disc is at a distance $R$ from O.

The moment of inertia of the removed disc about an axis passing through its own center of mass is:

$I_{\text{cm}}=\frac{1}{2}m{R}^2=\frac{1}{2}\left(\frac{M}{4}\right)R^2=\frac{M{R}^2}{8}$

Using the parallel axis theorem, the moment of inertia of the removed disc about the axis passing through O is:

$I_{\text{removed}}=I_{\text{cm}}+m{d}^2=\frac{M{R}^2}{8}+\left(\frac{M}{4}\right)R^2=\frac{3M{R}^2}{8}$

The moment of inertia of the remaining part about O is:

$I_{\text{remaining}}=I_{\text{original}}-I_{\text{removed}}=2M{R}^2-\frac{3M{R}^2}{8}=\frac{13M{R}^2}{8}$

Answer: $\frac{13M{R}^2}{8}$