Question

a. A child stands at the center of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of $40\dfrac{{rev}}{{\min }}$. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to $\dfrac{2}{5}$ 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 1

In this question, we will first determine the total initial moment of inertia before stretching the arms, followed by the final moment of inertia. We will then apply the Law of Conservation of Angular Momentum to obtain the new angular speed, and in the second part of the question, we will determine whether the initial and final kinetic energy of rotation and compare them.

Complete answer:
a.We have been given,
The boys' arms are stretched out at their initial angular velocity is ${\omega _1} = 40\dfrac{{rev}}{{\min }}$
let the Final angular velocity after folding his hands, ${\omega _2}$
The boy with stretched hands' initial moment of inertia is ${I_1}$
The boy's final moment of inertia occurs when his hands are folded ${I_2}$
so, The two moments of inertia are related as
${I_2} = \dfrac{2}{5}{I_1}$
The product of moment of inertia $I$ and angular velocity $\omega$ equals angular momentum $L$.
The angular momentum of the boy is constant because no external force acts on him.
Hence, we can write:
${I_2}{\omega _2}\;{\text{ }} = \;\;{I_1}\;{\omega _1}$
$\Rightarrow {\omega _2}\; = \left( {\dfrac{{{I_1}}}{{{I_2}}}} \right){\text{ }}{\omega _1}$
$\Rightarrow {\text{ }}{\omega _2} = \dfrac{{{I_1}}}{{\left( {2/5} \right){I_1}}} \times \;40\;{\text{ }}\;$
$\Rightarrow 100\dfrac{{rev}}{{\min }}$
Hence, the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to $\dfrac{2}{5}$ times the initial value is $100\dfrac{{rev}}{{\min }}$

b. Now finding the initial and final kinetic energy of the child.
kinetic rotation finally, ${E_F}\; = \dfrac{1}{2}{I_2}\;{\omega _2}^2$
kinetic rotation initially, ${E_I}\; = \;\dfrac{1}{2}{I_1}\;{\omega _1}^2$
Taking their ratio we get
$\dfrac{{{E_F}}}{{{E_I}}} = \dfrac{{\dfrac{1}{2}\;{I_2}\;{\omega _2}^{2\;}}}{{\dfrac{1}{2}{I_1}\;{\omega _1}^2}}$

$\Rightarrow \dfrac{{\dfrac{2}{5}\;{I_1}\;{{\left( {100} \right)}^2}\;}}{{\;{I_1}\;{{\left( {40} \right)}^2}}}$
$\Rightarrow 2.5$
$\therefore \;{E_F}\; = {\text{ }}2.5\;{E_1}$
Hence we can see that the child’s new(final) kinetic energy of rotation is more than the initial kinetic energy of rotation.
When things are farther away from the axis, moment of inertia increases, and when they are closer, it decreases. As a result, extending your hands increases your moment of inertia, causing you to slow down in order to maintain your angular momentum constant, which it must be because it is a conserved quantity. When you pull your hands in, the opposite happens.

Note:
When the net external torque operating on a system is zero, the total angular momentum of the system is conserved and so does not change, according to the law of conservation of angular momentum.