Question

Two identical uniform rings each of mass $m$ with their planes mutually perpendicular, radius $R$ are welded at their point of contact $O$ . If the system is free to rotate about an axis passing through the point $P$ perpendicular to the plane of the paper, the moment of inertia of the system about this axis is equal to:

• A. $6.5m{R}^2$
• B. $12m{R}^2$
• C. $6m{R}^2$
• D. $11.5m{R}^2$

Answer 1

The correct option is B $12m{R}^2$

$\begin{array}{c}\text{We know}\\ \text{moment of Inertia of a Ring about its axis}\\ I_0=M{R}^2\\ \text{By parallel axis theorem}\\ \text{moment of Inertia of a Ring parallel to its}\\ \text{axis at a distance (d) is -}\end{array}$

$\begin{array}{c}I=I_0+M{d}^2\\ \Rightarrow \text{For Ring}A(d=R)\\ \Rightarrow I_A=M{R}^2+M{R}^2=2M{R}^2\\ \ast \text{FoR}\mathrm{Ring}B(d=3R)\end{array}$

$\begin{array}{c}\Rightarrow I_B=M{R}^2+M(3R{)}^2=10M{R}^2\\ \Rightarrow \text{Moment of Inertia of the System}\\ I=I_A+I_B=12M{R}^2\end{array}$