Moment of inertia of a ring of mass M and radius R about an axis passing through the centre and perpendicular to the plane is

Explanation

Remember that in case of continuous mass distribution, we use the formula `I = int (dm)r^(2)` to find out the moment of inertia of the body. AA. is the axis about which rotation of the ring is being considered.
Mass of the ring = M, circumference of the ring `= 2 pi R`
Mass per unit length of the ring `= ((M)/(2 pi R)) = lambda` (say)
Consider a small element of the ring at an angle `theta` from a particular reference radius. The element subtends an angle `d theta` at the centre.
Length of the element `= R d theta`
Mass of the element `= (lambda R d theta)`
Moment of inertia of the element `= (lambda R d theta) R^(2)`
Moment of inertia of the ring `= int_(0)^(2pi) (lambda R d theta) R^(2) = MR^(2)`