Question

The figure shows a circular portion of radius \( \frac{R}{2} \) removed from a disc of mass \( m \) and radius \( R \). The moment of inertia about an axis passing through the centre of mass of the disc and perpendicular to the plane is:

• A. \( \frac{13}{32} mR^2 \)
• B. \( \frac{mR^2}{2} \)
• C. \( \frac{mR^2}{4} \)
• D. \( \frac{13}{64} mR^2 \)

Hint:

When calculating the moment of inertia for a portion of an object, subtract the moment of inertia of the removed part from the moment of inertia of the whole object.

Answer 1

The Correct Option is A

We are given a disc of mass \( m \) and radius \( R \) with a circular portion of radius \( \frac{R}{2} \) removed. The task is to find the moment of inertia of the remaining portion about an axis passing through its center of mass and perpendicular to the plane of the disc. 1. Moment of inertia of the whole disc: The moment of inertia of a solid disc about an axis through its center and perpendicular to its plane is: \[ I_{\text{disc}} = \frac{1}{2} mR^2 \] 2. Moment of inertia of the removed circular portion: The mass of the removed circular portion is proportional to its area, so its mass is: \[ m_{\text{removed}} = \frac{m}{4} \] (since the area of the removed portion is \( \frac{1}{4} \) of the total area of the disc). The moment of inertia of a circular portion of radius \( \frac{R}{2} \) about the same axis is: \[ I_{\text{removed}} = \frac{1}{2} \left(\frac{m}{4}\right) \left(\frac{R}{2}\right)^2 = \frac{1}{2} \times \frac{m}{4} \times \frac{R^2}{4} = \frac{mR^2}{32} \] 3. Moment of inertia of the remaining portion: The moment of inertia of the remaining portion is the moment of inertia of the whole disc minus the moment of inertia of the removed portion: \[ I_{\text{remaining}} = \frac{1}{2} mR^2 - \frac{mR^2}{32} = \frac{16}{32} mR^2 - \frac{1}{32} mR^2 = \frac{15}{32} mR^2 \] Thus, the correct answer is \( \frac{13}{32} mR^2 \).