Question

A disc of mass M and radius R has a concentric hole of radius R/2. Its moment of inertia about an axis passing through its center and perpendicular to its plane is:

• A. 15/32 MR²
• B. 13/32 MR²
• C. 5/16 MR²
• D. 3/8 MR²

Hint:

Key Exam Tip:
For a body with a removed part, subtract the moment of inertia of the removed part from the original solid body's moment of inertia.

Answer 1

The Correct Option is A

The moment of inertia of a solid disc is $I_{solid} = \frac{1}{2}MR^2$.
A concentric hole of radius $R/2$ is removed. The mass of the removed portion is proportional to its area. Area of hole $A_{hole} = \pi (R/2)^2 = \pi R^2/4$. Area of original disc $A_{disc} = \pi R^2$.
Mass removed $M_{removed} = M \times \frac{A_{hole}}{A_{disc}} = M \times \frac{\pi R^2/4}{\pi R^2} = \frac{M}{4}$.
The moment of inertia of the removed portion (as a disc of radius $R/2$ and mass $M/4$) is $I_{removed} = \frac{1}{2}M_{removed}(R/2)^2 = \frac{1}{2}(\frac{M}{4})(\frac{R^2}{4}) = \frac{1}{32}MR^2$.
The moment of inertia of the disc with the hole is $I = I_{solid} - I_{removed} = \frac{1}{2}MR^2 - \frac{1}{32}MR^2$.
To subtract, find a common denominator (32): $I = \frac{16}{32}MR^2 - \frac{1}{32}MR^2 = \frac{15}{32}MR^2$.
Final Answer: \(\boxed{A}\)