Question

Two circular rings each of mass M and radius R are attached to each other at their rims and their planes perpendicular to each other. The moment of inertia of system about a diameter of one of the rings and passing through the point of contact is:

• A. $\frac{3}{2}M{R}^2$
• B. $\frac{3}{4}M{R}^2$
• C. $\frac{5}{2}M{R}^2$
• D. $\frac{5}{4}M{R}^2$

Answer 1

The correct option is C $\frac{5}{2}M{R}^2$

Moment of Inertia of a ring passing through its diameter $=\frac{M{R}^2}{2}$
By parallel axis theorem ,Moment of Inertia about the rim and perpendicular to its plane is:
$M{R}^2+I=M{R}^2+M{R}^2=2M{R}^2$
$\therefore TotalMomentofInertia=2M{R}^2+\frac{M{R}^2}{2}=\frac{5M{R}^2}{2}$