A child stands at the centre of turn table with his two arms out stretched. 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 and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turn table rotates without friction (ii) Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Explanation

Here `omega_(1)=40 rpm, I_(2)=(2)/(5)I_(1)`
By the principle of conservation of angular momentum
`I_(1) omega_(1)=I_(2) omega_(2) or I_(1)xx40=(2)/(5)I_(1) omega_(1) or omega_(2)=100` rpm.
(ii) Initial kinetic energy of rotation
`(2)/(5)I_(1) omega_(1)^(2)=(2)/(5)I_(1)(40)^(2)=800I_(1)`
New kinetic energy of rotation
`(2)/(5)I_(2) omega_(2)^(2)=(1)/(5)xx(2)/(3)I_(1)(100)^(2)=2000I_(1)`
`("New K.E.")/("Initial K.E.")=(2000I_(1))/(800I_(1))=2.5`
hus the child.s new kinetic energy of rotation is 2.5 times its initial kinetic energy of rotation. This increase in kinetic energy is due to the internal energy of the child which he uses in folding his hands back from the outstretched position.