Prove that moment of inertia of uniform ring of mass M and radius R about its geometric axis is `MR^(2)`

Explanation

Assume the dx is the element of length dx on the ring as shown in fig.
The mass per unit length of the ring.

`lambda=(M)/(2piR)`
`therefore` Mass of length dx is dx, `m=lambdadx`
`therefore m=(M)/(2piR).dx....(1)`
`therefore` The moment of inertia about the axis ZZ.
`dI=mR^(2)`
`=(M)/(2piR)R^(2)dx`
`therefore dI=(MR)/(2pi)dx`
`therefore` For the moment of inertia along the whole ring take the integral from `x=0` to `x=2piR`
`therefore I=intdI`
`=underset(0)overset(2piR)int(M)/(2pi)Rdx=(MR)/(2pi)underset(0)overset(2piR)intdx`
`=(MR)/(2pi)[x]_(0)^(2piR)=(MR)/(2pi)[2piR-0]`
`therefore I=MR^(2)`