Question

How to solve this?

Answer 1

From law of conservation of angular momentum in vertical direction (axis of rotation), we have for a closed system where there is no external torque applied,

`L_"initial"=L_"final"` .......(1)

Both platform and the man standing on it are initially at rest. As such, initial angular momentum of the system is zero.
From equation (1) it follows that the final angular momentum of the system is also zero.

However, final angular momentum has two constituents

1. angular momentum of the platform and
2. angular momentum of man.
   `=>` Sum of these two angular momenta should also be zero.

Let `P` and `M` subscripts denote platform and man respectively. We have the equation

`L_"final"=L_P+L_M=0`
`=>L_P=-L_M` ....(2)
Now

`L_P=Iomega` .....(3)
where `I` is moment of inertia of platform and `omega` its angular velocity.

Also

`L_M=mRv` ......(4)
where `m` is mass of man, `v` his velocity and `R` radius of circle of his motion.

In the given question `R=` radius of platform as man is walking on the edge of platform.
Inserting given values we get

`L_M=50xx2xx1=100" kg m"^2"s"^-1`

Using (2) and (3) we get
`omega=-100/200=-0.5 " radian s"^-1`
Velocity of edge of platform relative to ground `v_P=romega=-2xx0.5=-1" m s"^-1`

Relative velocity of man with respect to platform`=1-(-1)=2" m s"^-1`

Time taken by man to complete one revolution`="Circumference"/"velocity"`
`=(2pir)/2=(2pixx2)/2=2pi" s"`