A child stands at the centre of a turn table with his two arms outstretched. The turn table is set rotating with an angular speed of `40` rpm. How much is the angular speed of the child, if he folds his hands back reducing the moment of inertia to `(2//5)` time the initial value ? Assume that the turn table rotates without friction.
(b) Show that the child's new `K.E.` of rotation is more than the initial `K.E.` of rotation. How do you account for this increase in `K.E.` ?

Explanation

(a) Suppose, initial moment of inertia of the child is l1 Then final moment of inertia,
`I_(2)=(2)/(5)I_(l)`
Also, `v_(1)=40 rev min^(-1)`
By using the principle of conservation of angular momentum, we get
`I_(1)omega_(1)=(I_(2)omega_(2) " or " I_(1)(2piv_(1))=I_(2)(2piv_(2))`
or `v_(2)=(I_(1)v_(1))/(I_(2))=(I_(1)xx40)/((2)/(5)xxI_(1))=100 "rev min"^(-1)`
(b) ("Final K.E. of rotation")/("Initial K.E. of rotation")=((1)/(2)I_(2)omega_(2)^(2))/((1)/(2)I_(1)omega_(1)^(2))=((1)/(2)I_(2)(2piv_(2))^(2))/((1)/(2)I_(1)(2pi v_(1))^(2))=(I_(2)v_(2)^(2))/(I_(1)v_(1)^(2))=((2)/(5)I_(1)xx(100)^(2))/((2)/(5)I_(1)xx(40)^(2))=2.5`
Clearly, final `(K.E.)_("not")` becomes more because the child uses his internal energy when he folds his hands to increase the kinetic energy.