Find the out the moment of inertia of a ring having uniform mass distribution of mass `M` and radius `R` about an axis which is tangent ot the ring and `a` in the plane of the ring `b`. perpendicular to the plane of the ring.

Explanation

The moment of inertia of a ring about an axis passing through centre and perpendicular to plain of ring is `I_(x)=I_(y)` and using perpendicular axis theorem.

`I_(z)=I_(x)+I_(y)impliesMR^(2)=2I_(x)`
`impliesI_(x)=(MR^(2))/2=I_(0)`
For case a using parallel axis theorem
`I_(1)=I_(0)+(MR^(2))/2+MR^(2)=3/2MR^(2)`
For case `b` using parallel axis theorem,
` I_(2)=I_(z)+MR^(2)=MR^(2)+MR^(2)=2MR^(2)`