Question

derive an expression for moment of inertia of circular ring​

Answer 1

Consider a circular ring of mass M and radius R. Let the ring be divided into several parts of small and equal masses 'dm'.

The moment of inertia of one such particle of mass 'dm' and radius R is,

$dI = dm\ R^2$

Then the total moment of inertia of the ring for the total mass M will be,

$\displaystyle I=\int\limits_0^M dm\ R^2\\\\\\\boxed {I=MR^2}$

Answer 2

$\textbf{Answer is in attachment!}$