A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given.

Mass of Removed Part ($$m$$):

$$The\ mass\ of\ a\ uniform\ disc\ is\ proportional\ to\ its\ area\ (\text{Area}=\pi R^2):$$

$$m = \frac{\pi (R/2)^2}{\pi R^2} \times M = \frac{M}{4}$$

Moment of Inertia of the Removed Part ($$I_{rem}$$):

The center of the removed part is at a distance $$d = R/2$$ from the center of the original disc. Using the Parallel Axis Theorem:

$$I_{rem} = I_{cm} + md^2$$

$$I_{rem} = \frac{1}{2} m \left(\frac{R}{2}\right)^2 + m \left(\frac{R}{2}\right)^2 = \frac{3}{2} m \left(\frac{R}{2}\right)^2$$

$$I_{rem} = \frac{3}{2} \left(\frac{M}{4}\right) \frac{R^2}{4} = \frac{3}{32} MR^2$$

Moment of Inertia of the Remaining Part ($$I_{final}$$):

$$I_{final} = I_{original} - I_{rem}$$

$$I_{final} = \frac{1}{2} MR^2 - \frac{3}{32} MR^2$$

$$I_{final} = \frac{16}{32} MR^2 - \frac{3}{32} MR^2 = \frac{13}{32} MR^2$$