Question

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

• A. $\dfrac{15}{32}MR^2$
• B. $\dfrac{13}{32}MR^2$
• C. $\dfrac{5}{16}MR^2$
• D. $\dfrac{3}{8}MR^2$

Hint:

For an annular disc: \[ I = \frac{1}{2}M(R_1^2 + R_2^2) \] Always check whether the given mass belongs to the original disc or the remaining annular disc.

Answer 1

The Correct Option is A

Concept:
The given body is an annular disc (ring-shaped disc). Moment of inertia is calculated by subtracting the inner removed disc from the full disc. For a solid disc: \[ I = \frac{1}{2}MR^2 \] For an annular disc: \[ I = \frac{1}{2}M(R_1^2 + R_2^2) \] where:
• $R_1$ = outer radius
• $R_2$ = inner radius

Step 1: Identify radii. Outer radius: \[ R_1 = R \] Inner radius: \[ R_2 = \frac{R}{2} \]

Step 2: Substitute into annular disc formula. \[ I = \frac{1}{2}M \left( R^2 + \left(\frac{R}{2}\right)^2 \right) \] \[ I = \frac{1}{2}M \left( R^2 + \frac{R^2}{4} \right) \] \[ I = \frac{1}{2}M \left( \frac{5R^2}{4} \right) \] \[ I = \frac{5}{8}MR^2 \] This expression corresponds to mass of the annular disc itself. But in the question, $M$ denotes the mass of the original full disc before removal.

Step 3: Relate remaining mass with original mass. Area ratio: \[ \text{Remaining area} = \pi R^2 - \pi \left(\frac{R}{2}\right)^2 \] \[ = \pi R^2 - \frac{\pi R^2}{4} = \frac{3\pi R^2}{4} \] Thus remaining mass: \[ M' = \frac{3M}{4} \] Now substitute: \[ I = \frac{5}{8} \left( \frac{3M}{4} \right) R^2 \] \[ I = \frac{15}{32}MR^2 \]

Step 4: Conclusion. Hence, the correct answer is: \[ \boxed{\frac{15}{32}MR^2} \]