Question

A thin rod of mass $M$ and length $L$ is bent into a circular ring. The expression for moment of inertia of ring about an axis passing through its diameter is:

• A. $\frac{M{L}^2}{2{\pi }^2}$
• B. $\frac{M{L}^2}{4{\pi }^2}$
• C. $\frac{M{L}^2}{8{\pi }^2}$
• D. $\frac{M{L}^2}{{\pi }^2}$

Answer 1

The correct option is C $\frac{M{L}^2}{8{\pi }^2}$

Length of a ring $L=2\pi R$

$\Rightarrow R=\frac{L}{2\pi }$

Let the moment of inertia about any diameter be $I_l$,

$\therefore$ by perpendicular axis theorem $2I_l=I$

$I_l=\frac{I}{2}$

where $I$ is the moment of Inertia perpendicular to the plane of the ring $=M{R}^2$

$\therefore I_l=\frac{M{R}^2}{2}=\frac{M{L}^2}{8{\pi }^2}$