A child stands at the centre of turntable with his two arms out stretched. The turntable is set rotating with an angular speed of 40 rpm. How much is the angular speed of the child if he folds his back and thereby reduces his moment of inertia to `(2)/(3)` times the initial value ? Assume that the turntable rotates without friction

Explanation

Let `I_(1) and I_(2)` be initial and final moment of inertia of child `omega_(1)and omega_(2)` be initial and final angular velocities
`omega_(1)=(40)/(60) rps, I_(2)=2//5I_(1)`
As there is no external torque acting in the process therefore, L is constant.
Thus, by lew of conservation of angular momentum
`:. I_(1)omega_(1)=I_(2) omega_(2)`
`omega_(2)=(I_(1))/(I_(2)) omega_(1)=(5)/(2)xx40`
`omega_(2)=100 rpm`
(b) Initial `K.E.E_(1)=(1)/(2)I_(1)omega_(1)^(2)=(1)/(2)I_(1)((40)/(60))^(2)`
Final `K.E. E_(2)=(1)/(2)I_(1)omega_(2)^(2)`
`=(1)/(2)*(2)/(5)I_(1) omega_(2)^(2)=(1)/(5)I_(1)xx((100)/(60))^(2) " " [ :. I_(2)=(2)/(5)I_(1)]`
`(E_(2))/(E_(1))=(1)/(5)(I_(1)xx100xx100xx2xx60xx60)/(60xx60xx1xxI_(1)xx40xx40)`
`(E_(2))/(E_(1))=2.5`
`E_(2)=2.5E_(1)`
The final kinetic energy of rotation is 2.5 times the initial kinetic energy of rotation. The child uses his internal energy to fold back his arms,