Question

The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis touching the disc at its diameter and normal to the disc is :

• A. $\frac{1}{2}MR^2$
• B. $MR^2$
• C. $\frac{2}{5}MR^2$
• D. $\frac{3}{2}MR^2$

Answer 1

The Correct Option is D

To determine the moment of inertia of a uniform circular disc about an axis touching the disc at its diameter and normal to the disc, we need to use the concept of the parallel axis theorem.

1. The moment of inertia of a disc about its center (axis perpendicular to the plane) is given by the formula: I_{\text{center}} = \frac{1}{2}MR^2.
2. According to the parallel axis theorem, the moment of inertia about an axis parallel to an axis through the center of mass (CM), at a distance d away, is given by: I = I_{\text{CM}} + Md^2.
3. Here, the distance d from the center of the disc to the point on its diameter, as described, is the radius R itself.
4. Substituting the values, we have: I = \frac{1}{2}MR^2 + MR^2.
5. Simplifying this expression gives: I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2.

Therefore, the moment of inertia of the disc about the specified axis is \frac{3}{2}MR^2.

Thus, the correct answer is the option \frac{3}{2}MR^2.