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

Explanation

(i) Consider a uniform ring of mass M and radius R as shown in figure.

(ii) The moment of inertia of the ring is found about the axis, which passes through its center and perpendicular to the plane.
(iii) Now the moment of inertia of an infinitesimal small mass 'dm' of length dx of the ring, which is at a distance 'R' from the center can be expressed as,
`d I = (dm)R^(2)" "...(1)`
The moment of inertia (I) of the entire ring can be found by integrating the equation (1) as,
`I = int dI = underset(0)overset(2piR)(int)(dm)R^(2)" "...(2)`
(v) If `lambda` is linear mass density (i.e. `lamda = (M)/(2piR)`), the small mass dm can be written as,
`dm = lambda dx = (M)/(2piR)dx`
(vi) Substituting the 'dm' value in equation (2), we get,
`I = underset(0)overset(2piR)(int) ((M)/(2piR)dx)R^(2)=(MR)/(2pi) underset(0)overset(2piR)(int)dx`
`I = (MR)/(2pi) [x]_(0)^(2piR), I = (MR)/(2pi) [2pi R - 0]`
`I = MR^(2)`