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. $3 \,MR^2$
• B. $3/2\, MR^2$
• C. $5 \,MR^2$
• D. $7/2 \,MR^2$

Answer 1

Answer: (D)

Clearly $I_{Y Y'}=I_{1}\left(I_{2}+I_{3}\right)$
where $I_{1}=M I$
about diameter $=\frac{1}{2} M R^{2}$
and $I_{2}=I_{3}=M I$
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,
$I_{Y Y'}=\frac{1}{2} M R^{2}+\int\left(\frac{3}{2} M R^{2}+\frac{3}{2} M R^{2}\right)$
$=\frac{7}{2} M R^{2}$