question_answer

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

A) \[\frac{1}{2}M{{R}^{2}}\]

B) \[\frac{3}{2}M{{R}^{2}}\]

C) \[M{{R}^{2}}\]

D) \[M{{R}^{2}}/9\]

Correct Answer: B

Solution :

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 (in figure) is R, the radius of the ring.

Using the parallel axes theorem.

\[{{i}_{\tan gent}}={{I}_{dia}}+M{{R}^{2}}=\frac{M{{R}^{2}}}{2}+M{{R}^{2}}\]

\[=\frac{3}{2}M{{R}^{2}}\]