Physics - Rotation Question with Solution | TestHub

The moment of inertia of the upper ring about its diameter is given by $${\mathrm{I}}_1=\frac{{\mathrm{MR}}^2}{2}$$ The moment of inertia of the two lower rings about a tangent in their plane is given by $${\mathrm{I}}_2={\mathrm{I}}_3=\frac{3}{2}{\mathrm{MR}}^2$$ Total moment of inertia $\mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3$ $$=\frac{{\mathrm{MR}}^2}{2}+\frac{3}{2}{\mathrm{MR}}^2+\frac{3}{2}{\mathrm{MR}}^2=\frac{7}{2}{\mathrm{MR}}^2$$