MHT CET · Physics · Rotational Motion
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
- A \(5 \mathrm{MR}^2\)
- B \(\frac{7}{2} \mathrm{MR}^2\)
- C \(\frac{3}{2} \mathrm{MR}^2\)
- D \(3 \mathrm{MR}^2\)
Answer & Solution
Correct Answer
(B) \(\frac{7}{2} \mathrm{MR}^2\)
Step-by-step Solution
Detailed explanation
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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