Question

Three rings each of mass $M$ and radius $R$ are arranged as shown. The moment of inertia of the system about $YY\text{'}$ 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 $\frac{7}{2}M{R}^2$


Total moment of inertia ($I_{total}$)$=I_1+I_2+I_3$

For ring-1:
Moment of inertia about $y{y}^{\prime }$, $I_{y{y}^{\prime }}=\frac{M{R}^2}{2}$

For ring-2:
$I_2=I_{A{A}^{\prime }}+I_{y{y}^{\prime }}=\frac{M{R}^2}{2}+M{R}^2=\frac{3}{2}M{R}^2$

For ring 3:
ring 3 is symmectrical as ring 2,
so,
$I_3=\frac{M{R}^2}{2}+M{R}^2=\frac{3}{2}M{R}^2$

Hence, $I_{total}=I_1+I_2+I_3$
$I_{total}=\frac{M{R}^2}{2}+\frac{3}{2}M{R}^2+\frac{3}{2}M{R}^2$
$I_{total}=\frac{7M{R}^2}{2}$