Question

A rod of length L and mass M is bent to form a circular ring. The moment of inertia of the ring about its diameter will be​

Answer 1

Answer:

I=ML^2/4pi^2

Explanation:

L=2piR (L=circumference of circle)

R=L/2pi

I=MR^2=ML^2/4pi^2

Answer 2

Answer:

The moment of inertia of the circular ring about its diameter will be $\frac{ML^2}{8\pi^2}$.

Explanation:

The rod of length "L" turns into a circular ring of radius "r". So,

$L=2\pi r$        (1)

Where,

L=length of the rod

r=radius of the circular ring

And for the moment of inertia of the circular ring about the diameter is given as,

$I=\frac{1}{2} Mr^2$        (2)

Where,

I=moment of inertia of the ring

M=mass of the ring

r=radius of the ring

Equation (1) can also be written as,

$r=\frac{L}{2\pi}$        (3)

By substituting equation (3) in equation (2) we get;

$I=\frac{1}{2} M\times (\frac{L}{2\pi} )^2$

$I=\frac{ML^2}{8\pi^2}$

Hence, the moment of inertia of the circular ring about its diameter will be $\frac{ML^2}{8\pi^2}$.