Question

Calculate the moment of inertia of a ring of mass 500 g and radius 0.5 m about an axis of rotation passing through 

(i) its diameter 

(ii) a tangent perpendicular to its plane.

Answer 1

Data : M = 500 g 0.5 kg, R = 0.5 m

(i) The moment of inertia of the ring about its

(ii) The moment of inertia of the ring about a tangent perpendicular to its plane = 2MR² = 2 × 0.5 × (0.5)² = 0.25 kg.m²