Question

Find the moment of inertia of ring of mass $M$ and radius $R$ about the axis passing through the ring diametrically. If moment of inertia about the axis passing through its center of mass and perpendicular to the plane is $M{R}^2$.

• A. $M{R}^2$
• B. $2M{R}^2$
• C. $\frac{M{R}^2}{4}$
• D. $\frac{M{R}^2}{2}$

Answer 1

The correct option is D $\frac{M{R}^2}{2}$

Given, $\text{MOI}$ of ring about the axis passing through the center of masss and perpendicular to its plane is $I_z=M{R}^2$.

Now, $\text{MOI}$ of ring about the axis passing along its diameter is given by $I_x$ or $I_y$
Using perpendicular axis theorem we have
$I_z=I_x+I_y$
But, $I_x=I_y$, by symmetry
$\Rightarrow I_x=I_y=\frac{I_z}{2}$
$\Rightarrow I_x=I_y=\frac{M{R}^2}{2}$