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A child stands at the centre of a turnta...

A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. 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 turntable rotates without friction.

Text Solution

Verified by Experts

Here, initial angular speed `omega_(1)=40` rev/min, `omega_(2)=?`
Final moment of inertia, `l_(2)=(2)/(5)I`, final moment of inertia
As no external torque acts in the process, therefore
I = constant
i.e. `I_(2)omega_(2)=I_(1)omega_(1)`
`omega_(2)=(I_(1))/(I_(2))omega_(1)=(5)/(2)xx40`
= 100 rpm.
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