Question

What is the moment of inertia of a ring about a tangent to the circle of the ring?

Answer 1

The tangent to the ring in the plane of the ring is parallel to one of the diameters of the ring. The distance between these two parallel axes is $R$, the radius of the ring. Using the parallel axes theorem,
$I_{tangent}=I_{dia}+M{R}^2=\frac{M{R}^2}{2}+M{R}^2=\frac{3}{2}M{R}^2$.