A child is standing with his two arms outstretched at the centre of a turntable that is rotating about its central axis with an angular speed `omega_0`. Now, the child folds his hands back so that moment of inertia becomes `3` times the initial value. The new angular speed is.

Options

A. `3omega_o`
B. `omega_o`/3
C. `6omega_o`
D. `omega_o`/6

Answer

B

Explanation

Here, Initial angular speed, `omega_(i) = omega_(0)`
Initial moment of inertia =`I_(i)`
Final moment of inertia `I_(f)=3I_(i)`
According to the law of conservation of angualr momentum, we get
`L_(i)=L_(f)`or `I_(i)omega_(i)=I_(f)omega_(f)`
`omega_(f) =(I_(i)omega_(i))/(I_(f)) = (I_(i))/(3I_(i))omega_(0)=omega_(0)/3`