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

A disc with mass \( m \) and radius \( R \) has a circular section of radius \( \frac{R}{2} \) removed. Calculate the moment of inertia of the remaining portion about an axis through its center of mass, perpendicular to the disc's plane. 1. Moment of Inertia of the Full 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 Portion: The mass of the removed circular portion is proportional to its area, resulting in a mass of: \[ m_{\text{removed}} = \frac{m}{4} \] This is because the removed area is \( \frac{1}{4} \) of the total disc area. The moment of inertia of this removed circular portion (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 calculated by subtracting the moment of inertia of the removed portion from that of the whole disc: \[ I_{\text{remaining}} = \frac{1}{2} mR^2 - \frac{mR^2}{32} = \frac{16}{32} mR^2 - \frac{1}{32} mR^2 = \frac{15}{32} mR^2 \] The final result is \( \frac{15}{32} mR^2 \).