Calculate the moment of inertia of a ring about an axis passing through centre of the ring and perpendicular to the plane of the ring.

Explanation

Let M be the mass of the ring with radius r and centre O. Length of the ring = 2 `pir`
Mass per unit length of the ring = `(m)/(2pir)`
Now, mass of length element dx`= (m)/(2pir).dx`
Moment of inertia of this element about the axis
`=((m)/(2pir)dx)r^(2)=(mr)/(2pi).dx`
Hence moment of inertia of the entire ring about the axis,
`I=int_(0)^(2pir)(mr)/(2pi)dx=(mr)/(2pi)int_(0)^(2pir)dx`
`=(mr)/(2pi)(2pir-0) = mr^(2)`