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 m s-1 relative to the ground. Time taken by the man to complete one revolution is
- s
- 3 2 s
- 2 s
- 2 s
Answer
(3) 2 s
Solution
There is no external torque about the axle, so angular momentum is conserved (initially zero):
mvR = I p
50 1 2 = 200 p
p = 0.5 rad s-1 (platform turns opposite to the man)
The angular speed of the man (w.r.t. ground) is m = vR = 12 = 0.5 rad s-1, so relative to the platform it is rel = m + p = 1 rad s-1.
Time for one revolution on the platform T = 2 rel = 2 s
Answer: (3)