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. $3/2M{R}^2$
• C. $5M{R}^2$
• D. $7/2M{R}^2$

Answer 1

Answer: (D)

Clearly $I_{Y{Y}^{\prime }}=I_1\left(I_2+I_3\right)$
where $I_1=MI$
about diameter $=\frac{1}{2}M{R}^2$
and $I_2=I_3=MI$
about tangent $I_2=I_3=\frac{1}{2}M{R}^2+M{R}^2$
(By parallel axis theorem)
$\therefore I_2=I_3=\frac{3}{2}M{R}^2$ So,
$I_{Y{Y}^{\prime }}=\frac{1}{2}M{R}^2+\int \left(\frac{3}{2}M{R}^2+\frac{3}{2}M{R}^2\right)$
$=\frac{7}{2}M{R}^2$