Question

Determine the moment of inertia of a thin ring about a tangent to the circle in the plane 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,

\(I_{tangent} = I_{dia} + MR^2\) = \(\frac{MR^2}{2} = MR^2 = \frac{3}{2}MR^2\)