Question

An arc making an angle \(15^\circ\) at the center is removed from a ring of mass \( M \) and radius \( R \). The moment of inertia of the remaining ring about an axis passing through its center and perpendicular to its plane is:

• A. \( \frac{23}{24} M R^2 \)
• B. \( \frac{M R^2}{24} \)
• C. \( \frac{M R^2}{23} \)
• D. \( \frac{24}{23} M R^2 \)

Hint:

The moment of inertia is additive. When a part is removed, its inertia is subtracted from the total inertia.

Answer 1

The Correct Option is A

Step 1: Moment of Inertia of a Complete Ring

- The moment of inertia of a complete ring about an axis passing through its center and perpendicular to its plane is: \[ I_{\text{full}} = M R^2 \]
Step 2: Mass of Removed Arc

- The full ring corresponds to an angular span of \( 360^\circ \). - The removed arc subtends an angle of \( 15^\circ \), so its mass is: \[ M_{\text{arc}} = M \times \frac{15}{360} = \frac{M}{24} \]
Step 3: Moment of Inertia of Removed Arc

- Since mass is uniformly distributed, the moment of inertia of the removed arc is: \[ I_{\text{arc}} = \frac{M}{24} R^2 = \frac{1}{24} M R^2 \]
Step 4: Moment of Inertia of Remaining Ring

- The moment of inertia of the remaining part is: \[ I_{\text{remaining}} = I_{\text{full}} - I_{\text{arc}} \] \[ = M R^2 - \frac{1}{24} M R^2 \] \[ = \frac{23}{24} M R^2 \]
Step 5: Conclusion

Since the remaining moment of inertia is \( \frac{23}{24} M R^2 \), Option (1) is correct.