Question

The moment of inertia of a ring of mass $M$ and radius $R$ about $PQ$ axis will be :

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

Answer 1

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

$\text{Explanation:-}$

We know that,

$I_{COM}=M{R}^2$

By perpendicular axis theorem ,

$I_X+I_Y=I_Z$

Here,

$I_D+I_D=I_{COM}=M{R}^2$

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

By parallel axis theorem ,

$I_{PQ}=I_{COM}+M{R}^2$

$I_{PQ}=\frac{M{R}^2}{2}+M{R}^2$

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

$\text{Hence the correct option is C}$