Calculate the moment of inertia of
a. a ring of mass `M` and radius `R` about an axis coinciding with the diameter of the ring.
b. as thin disc about an axis coinciding with the diameter.

Explanation

Let `X` and `Y` axis be along two perpendicular diameters of the ring.
BY symmetry`I_(x)=I_(y)` and by perpendicular axis theorem `I_(z)=I_(x)+I_(y)`.
But we know that `I_(z)=MR^(2)`
`MR^(2)=I_(x)+I_(y)^(z)=2I_(x)`
`implies I_(x)=I_(y)=(MR^(2))/2`
Situation of a thin disc (i.e. a circular plate) the moment of inertia about a diameter is
`I=1/2(1/2MR^(2))=1/4MR^(2)`