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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.


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.


Grade:upto college level

1 Answers

Dushyanth N Raj
21 Points
6 years ago
Given, w1 = 40 rev/min Let initial moment of inertia of child is I . Then, I1 = I I2 = (2/5)I Use, law of conservation of angular momentum, I1w1 = I1w2 w2 = (I1/I2)w1 = 5/2 × 40 = 100w2 = 100 rev/min (B) final kinetic energy of rotation/initial kinetic energy of rotation = {(1/2)I1w1²}/{(1/2)I2w2²} = (2/5I)(100)²/(I)×(40)² = 2/5 × 10000/1600 = 5/2 Hence, final kinetic energy of rotation = 5/2( initial kinetic energy of rotation) it means final KE of rotation of the child is greater then initial . it is obtained from muscular energy of the child when he fold back his hands.

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