Question

(a) A child stands at the centre of a turntable with his arm 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 fold his hand back and thereby reduces his moment of inertia of 2/5 times the initial value? Assume that the 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 kinetic energy?

Answer 1

(a) Suppose, the initial moment of inertia of the child is I₁. Then the final moment of inertia,

I₂ = 2/5 I₁

Also, v₁ = 40 rev.min^-1

By using the principle of conservation of angular momentum, we get

I₁ω₁ = I₂ω₂

or, ω₂ = {I₁ω₁}/{I₂} = {I₁ x 40}/{2/5 x I₁} = 100 rev min^-1

(b) {Final K.E. of rotation}/{Initial K.E. of rotation}

or, (K.E.)_final = 2.5 (K.E.)_initial

Clearly, final (K.E,)_rot becomes more, because the child uses his internal energy when he folds his hands to increase the kinetic energy.