First slide
Law of conservation of angular momentum
Question

A circular platform is mounted on a frictionless vertical axle. Its radius R = 2 m and its moment of inertia about the axle is 200 kg m2. It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 ms-1 relative to the ground. Time taken by the man to complete one revolution relative to the platform is

Moderate
Solution

From conservation of angular momentum

                  I ω = mvr

               200 × ω = 50×1×2

                ω = 12rad/s

                 veloccity of edge of platform = rω = 1 m/s.So velocity of man relative to the platform = (1+1) m/s =2m/s

                  T = 2πr/2 =πr=2π s 

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