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

Hint:

For an annular disc: \[ I=\frac12 M(R_1^2+R_2^2) \] where: \[ R_1=\text{inner radius} \] \[ R_2=\text{outer radius} \] This formula is extremely important for rotational motion problems.

Answer 1

The Correct Option is B

Concept: The given body is an annular disc (ring-shaped disc). Moment of inertia of a disc about its center: \[ I=\frac12 MR^2 \] For an annular disc with inner radius \(R_1\) and outer radius \(R_2\): \[ I=\frac12 M(R_1^2+R_2^2) \]

Step 1: Identifying inner and outer radii.
Outer radius: \[ R_2=R \] Inner radius: \[ R_1=\frac{R}{2} \]

Step 2: Substituting into the annular disc formula.
\[ I = \frac12 M \left[ \left(\frac{R}{2}\right)^2+R^2 \right] \] \[ = \frac12 M \left[ \frac{R^2}{4}+R^2 \right] \] \[ = \frac12 M \left[ \frac{5R^2}{4} \right] \] \[ I = \frac{5}{8}MR^2 \] Thus, the moment of inertia is: \[ \boxed{\frac{5}{8}MR^2} \]