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Three rings each of mass ' $M$ ' and radius ' $R$ ' are arranged as shown in the figure. The moment of inertia of system about axis YY' will be
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Verified Answer
The correct answer is:
$\frac{7}{2} \mathrm{MR}^2$
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 $\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3$
$$
=\frac{\mathrm{MR}^2}{2}+\frac{3}{2} \mathrm{MR}^2+\frac{3}{2} \mathrm{MR}^2=\frac{7}{2} \mathrm{MR}^2
$$
$$
\mathrm{I}_1=\frac{\mathrm{MR}^2}{2}
$$
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 $\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3$
$$
=\frac{\mathrm{MR}^2}{2}+\frac{3}{2} \mathrm{MR}^2+\frac{3}{2} \mathrm{MR}^2=\frac{7}{2} \mathrm{MR}^2
$$
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