Question

Three rings each of mass M and radius R are arranged as shown in the figure. The moment of inertia of the system about YY' will be

• A. $3M{R}^2$
• B. $\frac{3}{2}M{R}^2$
• C. $5M{R}^2$
• D. $\frac{7}{2}M{R}^2$

Answer 1

The correct option is D $\frac{7}{2}M{R}^2$

M.I of system about YY' $I=I_1+I_2+I_3$
here $I_1$ = moment of inertia of ring about diameter, $I_2=I_3=M.I.$ of inertia of ring about a tangent in a plane
$\therefore$ $I=\frac{1}{2}m{R}^2+\frac{3}{2}m{R}^2+\frac{3}{2}m{R}^2=\frac{7}{2}m{R}^2$