Question

A thin ring has mass 0.25 kg and radius 0.5 m. Calculate the moment of inertia about i) an axis passing through its centre and perpendicular to the plane and ii) an axis passing through a point on its circumference, perpendicular to its plane iii) diameter.

Answer 1

moment of inertia of ring about axis passing through its centre and perpendicular to its plane is given by 

                  I = mr²

where m is the mass of ring and r is the minimum seperation between axis of rotation and point of observation { e.g., radius of ring }

   here, m = 0.25 Kg 

            r = 0.5 m 

now, I = 0.25 × (0.5)² = 0.25 × 0.25 = 0.0625 Kg.m²