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

Explanation

Consider a uniform ring of mass M and radius R . Let us find the moment of inertia of thering about an axis passing through its centre 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, that is the radius of the ring from the axis as shown in figure.

The moment of inertia (dI) of this small mass (dm) is
`dI=(dm)R^(2)`
The length of the ring is its circumference (2R). As the mass is uniformly distributed, the mass per unit length `(lamda)` is
`lamda=("mass")/("length")=M/(2pir)`
The mass (dm) of the infinitesimally small length is dm `=lamda dx=M/(2pir)dx`
Now the moment of inertia (I) of the entire ring is
`I=intdI=int(dm)R^(2)`
`=int(M/(2piR)dx)R^(2)`
`I=(MR)/(2pi)int dx`
In order to cover the entire length of the ring, the limits of integration are taken from 0 to `2piR`.
`I=(MR)/(2pi)int_(0)^(2pir)dx`
`I=(MR)/(2pi)[x]_(0)^(2piR)=(MR)/(2pi)[2piR-0]`
`I=MR^(2)`