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. $M{R}^2$
• B. $\frac{2}{5}M{R}^2$
• C. $\frac{3}{2}M{R}^2$
• D. $\frac{1}{2}M{R}^2$

Answer 1

The correct option is C $\frac{3}{2}M{R}^2$

Moment of inertia of the disc about the axis passing through its COM and perpendicular to the plane of disc $I_{CM}=\frac{1}{2}M{R}^2$

From parallel axis theorem, moment of inertia of the disc about the axis passing through its diameter and perpendicular to the plane of disc is given by $I=I_{CM}+M{d}^2$

So, $I=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$