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 m^(2) . 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 is

A circular platform is mounted on a frictionless vertical axle . Its r

1.	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 m^(2) . 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 is
Answer» `pi s ` `(3pi)/(2) s` `2pi s ` `(pi)/(2) s ` Solution :As the system is INITIALLY at REST , therefore , initial angular momentum `L_(i) =0` . According to the principle of conservation of angular momentum FINAL angular momentum `L_(f) = 0`. `therefore` Angular momentum of man on platform = Angular momentum of system i.e., `MVR = I omega` or `omega = (mv R)/(I) = (50 xx 1 xx 2)/(200) = (1)/(2)` rad `s^(-1)` Angular velocity of man relative to platform is `omega_(r) = omega + (v)/(R) = (1)/(2) + (1)/(2) = 1 rad s^(-1)` Time taken by the man to complete one revolution is `T = (2pi)/(omega_(r)) = (2pi)(1) = 2 pi s`