Question

1. 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. 
2. 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 1

a. 100 rev/min
Initial angular velocity, \(\omega_1\)= \(40 \,rev/min\)
Final angular velocity = \(\omega_2\)
The moment of inertia of the boy with stretched hands = \(\text I_1\)
The moment of inertia of the boy with folded hands = \(\text I_2\)
The two moments of inertia are related as : 
\(\text I_2\) = \(\frac{2}{5}\text I_1\)
Since no external force acts on the boy, the angular momentum L is a constant. Hence, for the two situations, we can write : 
\(\text I_2\omega_2\)= \(\text I_1\omega_1\)
\(\omega_2\) = \(\frac{\text I_1}{\text I_2}\omega_1\) 

=\(\frac{\text I_1}{\frac{2}{5}I_1}\times40\) = \(\frac{5}{2}\times 40\) 

= \(100 \,rev/min\)

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(b) Final K.E. = 2.5 Initial K.E.
Final kinetic rotation, \(E_F\) = \(\frac{1}{2}\text I_2\omega^2_2\)

Initial kinetic rotation, \(E_I\)= \(\frac{1}{2}\text I_1\omega^2_1\)

\(\frac{E_F}{E_I}\) = \(\frac{\frac{1}{2}\text I_2\omega^2_2}{\frac{1}{2}\text I_1\omega^2_1}\)

= \(\frac{2}{5}\)\(\frac{\text I(100)^2}{\text I_1(40)^2}\) 

= \(\frac{2}{5}\times \frac{100\times 100}{40\times 40}\) 

= \(\frac{5}{2}\) = 2.5

∴ \(E_F\) = 2.5 \(E_1\) 
The increase in the rotational kinetic energy is attributed to the internal energy of the boy.