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Is this contraption theoretically possible?


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#21 Bender

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Posted 19 January 2017 - 11:19 PM

On what basis do you make that claim? I see no violation of conservation of angular momentum or conservation of energy.

 

It would really help if you attempted to do some math. Perhaps that way it would be easier to point out your misconceptions, at least if you are here to learn.


Edited by Bender, 19 January 2017 - 11:22 PM.

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#22 Delburt Phend

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Posted 20 January 2017 - 01:02 AM

This is of course a copy of a yo-yo despin device; I commonly cite the Dawn Mission yo-yo de-spin. For the math I use F = ma: when a small object gives its motion to a larger object only linear Newtonian momentum is conserved. The sphere's mass is 1 / 4.5 the total mass. The total motion is conserved and only linear Newtonian momentum can do this back and forth restoration of motion.

 

Energy conservation would predict large losses; but there are no losses of motion.


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#23 Bender

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Posted 20 January 2017 - 09:46 AM

Let's calculate the angular momentum in the two extremes (assuming a thin cylinder and small balls with respect to cylinder radius R):

L_1=4.5 R^2 \omega_1

L_2=2.5 R^2 \omega_{2,cylinder}+2 (R+l_{rope})^2 \omega_{2,balls}

 

The kinetic energy:

E_1=\frac{4.5 R^2 \omega_1^2}{2}

E_2=\frac{2.5 R^2 \omega_{2,cylinder}^2}{2}+\frac{2 (R+l_{rope})^2 \omega_{2,balls}^2}{2}

 

Since angular momentum and energy have to be conserved, we get two equations from which \omega_{2,cylinder} and \omega_{2,balls} can easily be calculated:

4.5 R^2 \omega_1=2.5 R^2 \omega_{2,cylinder}+2 (R+l_{rope})^2 \omega_{2,balls}

\frac{4.5 R^2 \omega_1^2}{2}=\frac{2.5 R^2 \omega_{2,cylinder}^2}{2}+\frac{2 (R+l_{rope})^2 \omega_{2,balls}^2}{2}

 

Now let's look at the linear momentum:

p_{cylinder}=mv=m \cdot 0=0

p_{balls}=m_{ball}v_1+m_{ball}v_2=m_{ball}v_1-m_{ball}v_1=0

As you see, there is no linear momentum. Could you clarify how you would make the calculations with linear momentum if there is no linear momentum?

 

 

note: earlier, I assumed a quasi-static system where the balls are released slowly and remain at the same orientation with respect to the central mass, in that case there is indeed a loss in kinetic energy. There are no significant losses in the system in the video.


Edited by Bender, 20 January 2017 - 09:47 AM.

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#24 imatfaal

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Posted 20 January 2017 - 10:45 PM

!

Moderator Note

I have removed Speculative post which seemed to challenge the conservation of angular momentum to the correct forum.  Only mainstream physics in the main physic forum please.

 

http://www.sciencefo...gular-momentum/


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#25 imatfaal

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Posted Yesterday, 07:52 AM

!

Moderator Note

Additional whinge by hijacker also split off - please take it as read that complaints about moderation should not be in the thread.


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A little learning is a dangerous thing; drink deep, or taste not the Pierian spring:
there shallow draughts intoxicate the brain, and drinking largely sobers us again.

- Alexander Pope

 

feel free to click the green arrow  ---->

 





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