Question 589: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

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
$$\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3$$
$$=\frac{\mathrm{MR}^2}{2}+\frac{3}{2} \mathrm{MR}^2+\frac{3}{2} \mathrm{MR}^2=\frac{7}{2} \mathrm{MR}^2$$

Answer

5 MR$$^2$$

Answer

$$\frac{7}{2}$$ MR$$^2$$

Answer

$$\frac{3}{2}$$ MR$$^2$$

Answer

3 MR$$^2$$

Centre of Mass and 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

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