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.