A uniform circular board of mass `M` and radius `R` is fixed on a horizontal plane and free to rotate along a vertical axis through its center. A man of mass `m` walks round the edge of the board without slipping, when he has walked completely round the board to his starting point, find the angle turned by the board.

Explanation

Let the man walk with speed `v` relative to board, in anticlockwise direction, the board will rotate in clockwise direction to keep the angular momentum of system zero.
By the conservation of angular momentum
`0=m(v-Romega)R-(1)/(2)MR^(2)omega`
`omega=(2mv)/((M+2m)R)`
Time taken by the man to complete one revolution relative to platform
`t=(2piR)/(v)`
Angle turned by platform in this duration
`theta=omega t=(2mv)/(M+2m)xx(2piR)/(v)`
`=((4m)/((M+2m)))pi`