Physics - Rotation Question with Solution | TestHub

The system consists of three identical rings, each of mass $M$ and radius $R$, lying in the same plane. The axis $Y{Y}^{\prime }$ is in the plane of the rings.

For ring A, the axis $Y{Y}^{\prime }$ passes through its center and acts as a diameter. The moment of inertia of a ring about its diameter is:
$I_A=\frac{1}{2}M{R}^2$

For ring B, the axis $Y{Y}^{\prime }$ is tangent to the ring. The distance from the center of ring B to the axis $Y{Y}^{\prime }$ is $R$. Using the parallel axis theorem, the moment of inertia of ring B about the axis $Y{Y}^{\prime }$ is the sum of its moment of inertia about a parallel diameter and the product of its mass and the square of the distance between the axes:
$I_B=I_{\text{diameter}}+M{R}^2$
$I_B=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

Similarly, for ring C, the axis $Y{Y}^{\prime }$ is also a tangent, and its center is at a distance $R$ from the axis. Its moment of inertia about the axis $Y{Y}^{\prime }$ is:
$I_C=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

The total moment of inertia of the system about the axis $Y{Y}^{\prime }$ is the sum of the individual moments of inertia:
$I_{\text{total}}=I_A+I_B+I_C$
$I_{\text{total}}=\frac{1}{2}M{R}^2+\frac{3}{2}M{R}^2+\frac{3}{2}M{R}^2$
$I_{\text{total}}=\frac{7}{2}M{R}^2$

Answer: $\frac{7M{R}^2}{2}$