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