Question

Two rings of the same radius and mass are placed such that their centres are at common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass of ring = $m$, radius = $r$)

• A. $\frac{1}{2}m{r}^2$
• B. $m{r}^2$
• C. $\frac{3}{2}m{r}^2$
• D. $2m{r}^2$

Answer 1

Answer: (C)

$\frac{3}{2}m{r}^2$

Moment of Inertia of two rings system is sum of moment of inertia of individual rings.

Moment of inertia of the ring where axis is perpendicular to the plane of the ring = $M{R}^2$

Moment of inertia of the ring where axis is in the plane of the ring: $I_1=M{R}^2$

Moment of inertia of the 2nd ring about axis passing through the center of the 1st ring and perpendicular to the plane of the 1st ring $I_2=\frac{M{R}^2}{2}$ where the the factor of $\frac{1}{2}$ is due to perpendicular axis theorem.

Moment of inertia of the system$I=I_1+I_2=M{R}^2+\frac{M{R}^2}{2}=\frac{3}{2}M{R}^2$