Question

Obtain the expressions for moment of inertia of a ring

(i) about an axis passing through its centre and perpendicular to its plane.
(ii) about its diameter and
(iii) about a tangent.

Answer 1

(1)moment of inertia about an axis passing through its centre and perpendicular to its plane of ring (Iz) = mr² 
(2) take any two diameter which are perpendicular then from perpendicular axis theorem
let moment of inertia about both diameter are Ix & Iy
Ix + Iy = Iz
I = mr²/2             (Ix & Iy are equal)
(3) let moment of inertia about tangent is L then from parallel axis theorem
L = Iz + mr²     (parallel axis theorem = moment of inertia about known axis + m*square of distance between parallel axis)
   = 2mr²