### Q.13:- (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?

**Answer:-**(a) 100 rev/min

Initial angular velocity, ω_{1}= 40 rev/min

Final angular velocity = ω_{2}

The moment of inertia of the boy with stretched hands = *I*_{1}

The moment of inertia of the boy with folded hands = *I*_{2}

The two moments of inertia are related as:

*I*_{2} = (2/5) *I*_{1}

Since no external force acts on the boy, the angular momentum *L* is a constant.

Hence, for the two situations, we can write:

*I*_{2}ω_{2} = *I*_{1} ω_{1}

ω_{2} = (*I*_{1}/*I*_{2}) ω_{1}

= [ *I*_{1} / (2/5)*I*_{1} ] × 40 = (5/2) × 40 = 100 rev/min

(b)** **Final K.E. = 2.5 Initial K.E.

Final kinetic rotation, *E*_{F} = (1/2) *I*_{2} ω_{2}^{2}

Initial kinetic rotation, *E*_{I} = (1/2) *I*_{1} ω_{1}^{2}

E_{F} / E_{I} = (1/2) *I*_{2} ω_{2}^{2 }/ (1/2) *I*_{1} ω_{1}^{2}

= (2/5) *I*_{1} (100)^{2} / *I*_{1} (40)^{2}

= 2.5

∴ *E*_{F} = 2.5 *E*_{1}

The increase in the rotational kinetic energy is attributed to the internal energy of the boy.