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)` time the initial value ? Assume that the turn table 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.` ?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the Initial Angular Speed
The initial angular speed of the turntable is given as \( \omega_i = 40 \, \text{rpm} \). We need to convert this into radians per second.

\[ \omega_i = 40 \, \text{rpm} \times \frac{2\pi \, \text{radians}}{1 \, \text{revolution}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} = \frac{40 \times 2\pi}{60} \, \text{radians/second} \]
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