Question

A wedge of $60^\circ$ representing one sixth portion of a circular disc of mass $M$ and radius $R$ is cut. It is rotated about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is:

• A. $\frac{1}{2}MR^2$
• B. $\frac{1}{8}MR^2$
• C. $\sqrt{2}\,MR^2$
• D. $\frac{1}{12}MR^2$

Hint:

For uniform bodies: - Moment of inertia scales linearly with mass, - Fraction of body $\Rightarrow$ same fraction of total moment of inertia (about same axis).

Answer 1

The Correct Option is D

Concept: The moment of inertia of a full circular disc about an axis perpendicular to its plane and passing through its center is: \[ I_{\text{disc}} = \frac{1}{2}MR^2 \] For a uniform disc, mass is uniformly distributed, so the moment of inertia is proportional to its mass.

Step 1: Identify the portion of the disc.
A $60^\circ$ wedge is: \[ \frac{60}{360} = \frac{1}{6} \text{ of the full disc} \]

Step 2: Find mass of the wedge.
If total mass of disc is $M$, then mass of wedge: \[ M_{\text{wedge}} = \frac{M}{6} \]

Step 3: Use proportionality of moment of inertia.
Moment of inertia is directly proportional to mass for the same geometry and axis: \[ I_{\text{wedge}} = \frac{1}{6} \times \frac{1}{2}MR^2 \]

Step 4: Simplify.
\[ I_{\text{wedge}} = \frac{1}{12}MR^2 \]

Step 5: Final conclusion.
Thus, the required moment of inertia is: \[ \frac{1}{12}MR^2 \]