Derive the expression for moment of inertia of a uniform ring about an axis passing thorugh the centre and perpendicular to the plane.

Explanation

Let us consider a uniform ring of mass M and radius R. To find the moment of inertia of the ring about an axis passing through its center and perpendicular to the plane, let us take an infinitesimally small mass (dm) of length (dx) of the ring. This (dm) is located at a distance R, which is the radius of the ring from the axis as shown in figure.
The moment of inertia (dl) of this small mass (dm) is.
`dI=(dm)R^(2)`

The length of the ring is its circumference `(2piR)`. As the mass is uniformly distributed, the mass per unit length `(lambda)` is,
`lambda=("mass")/("length")=(M)/(2piR)`
The mass (dm) of the infinitesimally small length is,
`I= int dI= int (dm)R^(2)= int ((M)/(2piR)dx)R^(2)`
`II=(MR)/(2pi)int dx`
To cover the entire length of the ring, the limits of integration are taken from 0 to `2piR`.
`I=(MR)/(2pi) overset(2piR) underset(0) int dx`
`I=(MR)/(2pi)[x]_(0)^(2piR)=(MR)/(2pi)[2piR-0]`
`I=MR^(2)`