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 , angular momentum should be conserved .
Hence `I omega`= constant …(i)
where ,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 )
`implies I omega=(I)/(2)xxomega_(2)`
(as mass reduced to half hence moment of inertia also reduced to half )
`implies omega_(2)= 2omega`