Question

The moment of inertia of a circular ring(radius R, mass M) about an axis which passes through tangentially and perpendicular to its plane will be?

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

Answer 1

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

Based on the above figure, Let's consider, $I_{AB}$= Moment of inertia of ring about tangent

$I_{AB}$= $I_Z$+$M{R}^2$

= $\frac{1}{2}M{R}^2+M{R}^2$

= $\frac{3}{2}M{R}^2$