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(a) A child stands at the centre of a tu...

(a) 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.
(b) 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?

Text Solution

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(a) 100 rev/min (use angular momentum conservation).
(b) The new kinetic energy is 2.5 times the initial kinetic energy of rotation. The child uses his internal energy to increase his rotational kinetic energy.
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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) times the initial value? Assume that turntable 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.?

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) times the initial value? Assume that turntable 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.?

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