Where does the K.E go? [RotMech]

  • #1
412
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?
 
Last edited:

Answers and Replies

  • #2
67
0
erm the kinetic energy of a rotation is 1/2I*w^2
 
  • #3
Doc Al
Mentor
44,993
1,266
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]
OK. Let's at least imagine that he's holding weights in his hands or something to give him a bit of rotational inertia to play with.

Also,

[tex]
K = \frac{1}{2}{\omega_1}^2
[/tex]
That should be:
[tex]K_1 = \frac{1}{2}I{\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]
That should be:
[tex]K_2 = \frac{1}{2}I_2{\omega_2}^2 = \frac{1}{2}( \frac{1}{2}I{\omega_1}^2)[/tex]


So, the energy difference is:

[tex]
K - K_2 = \frac{3}{8}{\omega_1}^2
[/tex]
[tex]K_2 = \frac{1}{2} K_1[/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?
Mechanical energy is not conserved. (Consider the work he must do as he extends his hands.)
 
  • #4
412
2
OK. Let's at least imagine that he's holding weights in his hands or something to give him a bit of rotational inertia to play with.

That should be:
[tex]K_2 = \frac{1}{2}I_2{\omega_2}^2 = \frac{1}{2}( \frac{1}{2}I{\omega_1}^2)[/tex]

[tex]K_2 = \frac{1}{2} K_1[/tex]

Mechanical energy is not conserved. (Consider the work he must do as he extends his hands.)
umm.. right.. i made that slip there.. it's not 1/4th but half of the K.E. but that apart, you mean to say that the work done by the boy to move his hands apart equals the difference [taking the general assumptions of no air friction and such] of the 2 kinetic energies? So, if he brings his hands back together, the work done will translate into kinetic energy again and bring it back to the previous kinetic energy 'K'.

But here, the boy has done work.. but still the system remains unchanged. Where does the energy consumed by the boy go?

By Work energy theorem, the change in Kinetic energy is given by the total work done. Which means, the work done in this case is zero??

Could somebody please clarify this a little bit..
 
  • #5
99
0
well, in that case, i guess the boy applied two non-conservative forces, making the system to apparently(literally) conserve itself...

Whoever, energy was lost to the boy make that movements. In general, Men system are non conservative, as it was a great process to make him move(or being alive), heat is only one of the K.E. lost...
 
  • #6
1,914
46
umm.. right.. i made that slip there.. it's not 1/4th but half of the K.E. but that apart, you mean to say that the work done by the boy to move his hands apart equals the difference [taking the general assumptions of no air friction and such] of the 2 kinetic energies? So, if he brings his hands back together, the work done will translate into kinetic energy again and bring it back to the previous kinetic energy 'K'.

But here, the boy has done work.. but still the system remains unchanged. Where does the energy consumed by the boy go?

By Work energy theorem, the change in Kinetic energy is given by the total work done. Which means, the work done in this case is zero??

Could somebody please clarify this a little bit..
When he brings his hands back, he makes mechanical work to the system, which then increases its kinetic energy; when he lets his hands moving apart, the work he makes on the system is negative, that is, the system makes work on him.

Think of a stretched spring connected with the two masses: when the spring is released and can contract making the massess approach, its previous elastic potential energy is converted into system's kinetic energy; if you instead had a non-stretched spring, then centrifugal force makes it stretch and system's kinetic energy is converted into spring potential energy.
(There are no tricks! Energy is neither created nor destroyed!)
 
Last edited:

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