Question

Consider a uniform thin circular disk of radius \( R \) and mass \( M \). A concentric square of side \( R/2 \) is cut out from the disk (see figure). What is the moment of inertia of the resultant disk about an axis passing through the centre of the disk and perpendicular to it? 

• A. \( I = \frac{MR^2}{4} \left( 1 - \frac{1}{48\pi} \right) \)
• B. \( I = \frac{MR^2}{2} \left( 1 - \frac{1}{48\pi} \right) \)
• C. \( I = \frac{MR^2}{2} \left( 1 - \frac{1}{24\pi} \right) \)
• D. \( I = \frac{MR^2}{2} \left( 1 - \frac{1}{24\pi} \right) \)

Hint:

When calculating the moment of inertia of a composite object, subtract the moment of inertia of removed parts from the total.

Answer 1

The Correct Option is A

Step 1: Moment of inertia of the original disk.
The moment of inertia of a full disk of mass \( M \) and radius \( R \) about an axis perpendicular to the disk is given by \( I_{\text{disk}} = \frac{1}{2} M R^2 \).
Step 2: Moment of inertia of the cut square.
The moment of inertia of the square is \( I_{\text{square}} = \frac{M_{\text{square}} R^2}{4} \).
Step 3: Resultant moment of inertia.
The moment of inertia of the resultant disk is obtained by subtracting the moment of inertia of the square from the moment of inertia of the full disk. This gives the final answer as \( I = \frac{MR^2}{4} \left( 1 - \frac{1}{48\pi} \right) \).
Step 4: Conclusion.
Therefore, the correct answer is option (A).