A child stands at the turn table center of a turnable with his two arms outstretched. The turnable is set rotating with an angular speed of `10 rad//s`. If the child folds his hands back so that the moment of inertia reduces to `3//5` times the initial value, find the new angular speed. Why does the kinetic energy of child increase?

Explanation

`I_(1)=I,omega_(1)=10 rad//s`
`I_(2)=(3)/(5)I, omega_(2)=?`
`I_(1)omega_(1)=I_(2)omega_(2)` (conservation of angular momentum)
`Ixx10=(3I)/(5)xxomega_(2)implies omega_(2)=(50)/(3) rad//s`
Initial `K.E., K_(i)=(1)/(2)I_(1)omega_(1)^(2)=(1)/(2)I(10)^(2)=50I`
Final `K.E., K_(f)=(1)/(2) I_(2) omega_(2)^(2)=(1)/(2) xx(3I)/(5)((50)/(3))^(2)=(250)/(3)I`
`K_(f)gtK_(i)`
The child uses his inernal muscular energy (used in folding his hands ) to increase the rotational kinetic energy.