__Question 7.13__:(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)**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}] X 40 = (5/2) X 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.

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