- #1
rohanprabhu
- 414
- 2
Got this question somewhere: A boy is standing on a rotating platform. The system has a kinetic energy 'K'. Now, the boy stretches out is hand such that the Moment of Inertia of the system doubles. What is the Kinetic Energy of the system?
So.. i first applied conservation of angular momentum,
[tex]
I\omega_1 = 2I\omega_2
[/tex]
(here, [itex]\omega_1[/itex] and [itex]\omega_2[/itex] are the angular velocities before & after he stretches his hands).
So, we have:
[tex]
\omega_2 = \frac{\omega_1}{2}
[/tex]
Also,
[tex]
K = \frac{1}{2}{\omega_1}^2
[/tex]
and the new kinetic energy is:
[tex]
K_2= \frac{1}{2}{\omega_2}^2 = \frac{1}{8}{\omega_1}^2
[/tex]
So, the energy difference is:
[tex]
K - K_2 = \frac{3}{8}{\omega_1}^2
[/tex]
So.. where exactly does this energy go? By COE, this needs to be translated to the potential energy.. but does something called 'Rotational Potential Energy' even exist? How do i calculate it if it does?
So.. i first applied conservation of angular momentum,
[tex]
I\omega_1 = 2I\omega_2
[/tex]
(here, [itex]\omega_1[/itex] and [itex]\omega_2[/itex] are the angular velocities before & after he stretches his hands).
So, we have:
[tex]
\omega_2 = \frac{\omega_1}{2}
[/tex]
Also,
[tex]
K = \frac{1}{2}{\omega_1}^2
[/tex]
and the new kinetic energy is:
[tex]
K_2= \frac{1}{2}{\omega_2}^2 = \frac{1}{8}{\omega_1}^2
[/tex]
So, the energy difference is:
[tex]
K - K_2 = \frac{3}{8}{\omega_1}^2
[/tex]
So.. where exactly does this energy go? By COE, this needs to be translated to the potential energy.. but does something called 'Rotational Potential Energy' even exist? How do i calculate it if it does?
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