Rotational Motion — Medium JEE physics MCQ
The moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about an axis tangent to the disc and lying in its plane is:
- A. $\frac{1}{2}MR^2$
- B. $\frac{3}{4}MR^2$
- C. $\frac{5}{4}MR^2$
- D. $\frac{3}{2}MR^2$
Solution
The moment of inertia of a disc about a diameter is:
$$I_d = \frac{1}{4}MR^2$$
Using the parallel axis theorem, for an axis tangent to the disc and in its plane (distance $R$ from the diameter):
$$I = I_d + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$$