A Merry -go-round, made of a ring-like plarfrom of radius `R and mass M`, is revolving with angular speed `omega`. A person of mass `M` is standing on it. At one instant, the person jumps off the round, radially awaay from the centre of the round (as see from the round). The speed of the round after wards is

Options

A. `2 omega`
B. `omega`
C. `(omega)/(2)`
D. `0`

Answer

A

Explanation

As no external torque acts on the system, amgular momentum should be conserved .
Hence `I omega` = constant
When, I is moment of inertia of the system and `omega` is angular velocity of the system
From Eq. (i) `I_(1)omega_(1)=I_(2)omega_(2)`
( where `omega_(1)` and `omega_(2)` are angular velocities before and after jumping )
`impliesIomega=(I)/(2)xxomega_(2)`
( as mass reduced to half, hence, moment of inertia also reduced to half )
`implies `omega_(2)=2omega`