Physics - Rotation Question with Solution | TestHub

Let the mass of the complete uniform circular disc be $M$ and its radius be $R$.

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

The removed sector AOC subtends an angle of $120^{\circ }$ at the centre. The fraction of the disc that is removed is:
$\text{Fraction removed}=\frac{120^{\circ }}{360^{\circ }}=\frac{1}{3}$

Since the disc is uniform, the moment of inertia is directly proportional to the mass (and thus the angle) for a given radius. The moment of inertia of the removed portion is:
$I_{removed}=\frac{1}{3}{I}_{total}=\frac{1}{3}\left(\frac{1}{2}M{R}^2\right)=\frac{1}{6}M{R}^2$

The moment of inertia of the remaining portion is the difference between the total moment of inertia and the moment of inertia of the removed portion:
$I_{remaining}=I_{total}-I_{removed}$

$I_{remaining}=\frac{1}{2}M{R}^2-\frac{1}{6}M{R}^2$

$I_{remaining}=\left(\frac{3-1}{6}\right)M{R}^2=\frac{2}{6}M{R}^2=\frac{1}{3}M{R}^2$

Alternatively, the mass of the remaining portion is $M^{\prime }=\frac{2}{3}M$. The moment of inertia of a sector of a disc is $\frac{1}{2}\times \text{mass}\times {\text{radius}}^2$.
$I_{remaining}=\frac{1}{2}{M}^{\prime }{R}^2=\frac{1}{2}\left(\frac{2}{3}M\right)R^2=\frac{1}{3}M{R}^2$

Answer: $\frac{M{R}^2}{3}$