question_answer

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

Answer:

                M.I. of a ring about a tangent in its plane. Refer to Fig. Let \[{{I}_{T}}\] be the moment of inertia of the ring about the tangent EBF. Applying the theorem of parallel axes, we get \[{{I}_{T}}=M.I.\,about\,diameter\,CD+M{{R}^{2}}\] \[=\frac{1}{2}M{{R}^{2}}+M{{R}^{2}}\] or\[{{I}_{T}}=\frac{3}{2}M{{R}^{2}}.\]