Question
(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 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 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?
Solution
(a)
Let I1 be the moment of inertia of child and ω1 be the angular velocity of the table when the arms are outstretched.
Let, I2 be the moment of inertia of child and ω2 be the angular velocity of the table after the child folds his hands back.
According to the conservation of angular momentum,
(b)
Initial kinetic energy of child is,
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Kinetic energy of child after folding his hands is,
That is, the new kinetic energy of child becomes 2.5 times the initial kinetic energy and this increase in kinetic energy is at the cost of internal muscular energy of the child.
Let I1 be the moment of inertia of child and ω1 be the angular velocity of the table when the arms are outstretched.
Let, I2 be the moment of inertia of child and ω2 be the angular velocity of the table after the child folds his hands back.
According to the conservation of angular momentum,
(b)
Initial kinetic energy of child is,
Kinetic energy of child after folding his hands is,
That is, the new kinetic energy of child becomes 2.5 times the initial kinetic energy and this increase in kinetic energy is at the cost of internal muscular energy of the child.
