Question

A man stands on a rotating platform, with his arms stretched horizontally holding 5kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from \[90\;cm\] to \[20\;cm\]. The moment of inertia of the man together with the platform may be taken to be constant and equal to \[7.6\;kg{m^2}\]
(a) What is the new angular speed? (Neglect friction)
(b) Is kinetic energy conserved in the process if not, from where does the change come about?

Answer 1

Hint: n this question first of all we will find the total initial moment of inertia before stretching the arms and then the final moment of inertia then we will apply the Law of conservation of angular momentum to get the new angular velocity and the second part of the question we will find initial and final kinetic energy to see whether these are equal or not.

Formula used:
1. \[I = m{r^2}\]
Where \[I\] is the moment of inertia, \[m\] is mass of the body and \[r\] is the perpendicular distance of a body from the axis of rotation.
2. \[{I_1}{\omega _1} = {I_2}{\omega _2}\]
Where \[{I_1}\], \[{I_2}\], \[{\omega _1}\]and \[{\omega _2}\] are initial moment of inertia, final moment of inertia, initial angular speed and final angular speed respectively.
3. \[k = \dfrac{1}{2}I{\omega ^2}\]
Where \[k\], \[I\] and \[\omega \] are kinetic energy, moment of inertia and angular speed of a body respectively.

Complete step by step answer:
Given, Weight hold by man \[m = 5\;kg\]
Angular speed of platform \[{\omega _1} = 30\;rev/\min = \dfrac{{30 \times 2\pi }}{{60}} = \pi \;radian/\sec \]
Initial distance of each weight from axis of rotation \[{r_i} = 90\;cm = 0.9\;m\]
Final distance of each weight from axis of rotation \[{r_f} = 20\,cm = 0.2\;m\]
Moment of inertia of a man together with platform \[I = 7.6\;kg{m^2}\]
Initial moment of inertia of each weight
Part (A):
 \[\Rightarrow {I_i} = m{r_i}^2 = 5\;kg \times {(0.9\;m)^2} = 4.05\;kg{m^2}\]
Total Initial moment of inertia
\[\Rightarrow {I_1} = I + {I_i} + {I_i} = (7.6 + 4.05 + 4.05)\;kg{m^2} = 15.70\;kg{m^2}\]
After when the man stretches his arms close to axis of rotation
Final momentum of inertia of each weight
\[\Rightarrow {I_f} = m{r_f}^2 = 5\,kg \times {(0.2\,m)^2} = 0.2\;kg{m^2}\]
Total final momentum of inertia
\[{I_2} = I + {I_f} + {I_f} = (7.6 + 0.2 + 0.2)\,kg{m^2} = 8.0\,kg{m^2}\]
Now by using Law of conservation of angular momentum
\[\Rightarrow {I_1}{\omega _1} = {I_2}{\omega _2}\]
$\Rightarrow 15.70 \times \pi = 8.0 \times \omega_2 $
$\Rightarrow \omega_2= \dfrac{150 \times \pi}{8.0}$
$\Rightarrow \omega_2=1.9625 \pi \,rad/\sec$

Hence the New angular speed is \[1.9625\,\,\pi \,rad/\sec \] or \[58.875\,\,rev/\min \].

Part (B):
Now initial kinetic energy
\[{k_i} = \dfrac{1}{2}{I_1}{\omega _1}^2 = \dfrac{1}{2} \times 15.70\,kg{m^2} \times {(\pi \,rad/\sec )^2} = 7.85\,{\pi ^2}joule\]
And final kinetic energy
\[{k_f} = \dfrac{1}{2}{I_2}{\omega _2}^2 = \dfrac{1}{2} \times 8.0\,kg{m^2} \times {(1.9625\,\pi \,rad/\sec )^2} = 15.405625\,{\pi ^2}joule\]

Here kinetic energy is not conserved. The change in kinetic energy comes due to the internal energy which is provided by the man in stretching his arms closer increases the final energy.

Note:
In this type of question, we should have the knowledge about moment of inertia, Law of conservation of angular momentum and the way how this law applies to the real time problems and the idea of kinetic energy in terms of the moment of inertia and angular speed of rotation.