Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass ?M and radius ?R? about an axis perpendicular to the hoop’s plane at an edge.

Solution 54E Step 1: The moment of inertia of the hoop about the axis passing through it’s centre which is parallel to the plane of the hoop is, 2 Iz= mr . By perpendicular axis theorem, Iz= I x I .y Where I , IxandyI are tze respective moment of inertias along the three perpendicular axes. But by symmetry, I = I x y So, I = 2I z x 1 I =xI /2z= mr .2 2