Linear momentum conservation puzzle

Discussion in 'Physics & Math' started by Q-reeus, Nov 13, 2016.

  1. Confused2 Registered Senior Member

    Messages:
    443
    If it's length contraction in the shaft then I think the answer will contentious.
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. Q-reeus Valued Senior Member

    Messages:
    2,605
    No. See #54. No other kind of length contraction either.
     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. OceanBreeze Registered Member

    Messages:
    30
    Just some more thoughts, based on the clues so far:
    If the Lorentz transform is not applied to length, I don’t see it being applied to time, it must be applied to the mass being transferred along the shaft.
    If I solve that transform for v^2 and then differentiate that I have another expression for d(v^2) and I can plug those into the equation for Force = d/dt(mv) and get a resulting Force in terms of the mass, which I could not get via a purely mechanical method of solution.
    Getting warm?
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. Q-reeus Valued Senior Member

    Messages:
    2,605
    That suggests an axial energy/mass flow as determined in the centre-of-length frame of the shaft. I think not. Formally there is an energy flux, but for a spinning shaft under torsional load, nothing analogous to say a Poynting flux in EM can be locally identified. A read through section 3 here: http://web.mit.edu/2.151/www/Handouts/EnPwrFlow.pdf
    provides plenty of formal definitions for rotary power etc. but nowhere identifies an axial power flow density expression rigorously derivable from velocity and stress, that is easily found for say mechanical longitudinal wave propagation.
    Sorry but without identifying any vectorial relations and connections, that's just too vague to meaningfully comment on. So far I think you are headed in the wrong direction, but give me some fuller theory to judge that better.
     
    OceanBreeze likes this.
  8. hansda Valued Senior Member

    Messages:
    2,219
    Your link says axial force is due to torque. This was already told to you, ref post #85. Your link does not talk about twist of the shaft.
     
  9. Q-reeus Valued Senior Member

    Messages:
    2,605
    The latter claim first: see Example p16, finishing with eqn. (54). So you are wrong there.
    As to an axial force, can find no mention of such that section. You have probably confused reference there to the fornal power flow along the shaft axis - see e.g. Fig.11 p12.
    Finally please avoid such laziness and at the very least cite particular passages, eqn's etc. as I have done here.
     
    Last edited: Dec 10, 2016
  10. OceanBreeze Registered Member

    Messages:
    30
    That's a nice resource, thanks for the link.

    I will need some time to look it over and see if it relates to the approach I was working on.

    In the meantime, don't wait up for me, if you decide to post the answer before I solve it, I'll try not to be too upset.

    Please Register or Log in to view the hidden image!

     
  11. Q-reeus Valued Senior Member

    Messages:
    2,605
    No problem. I can wait!

    Please Register or Log in to view the hidden image!

     
  12. hansda Valued Senior Member

    Messages:
    2,219
    If the shaft is not twisted, the direction of force F(which is causing the torque) will be at normal to the shaft axis. When the shaft twists, the direction of force F will also change. It is not normal to the shaft axis. So this changed force F can be resolved into normal and axial components.
     
  13. Q-reeus Valued Senior Member

    Messages:
    2,605
    At first I thought you were talking nonsense, but in reading it again and trying to interpret your loose use of terms, you may mean the following:

    A small area element lying on the shaft surface, originally say square shaped when unstressed, deforms into a diamond under torsion. There are equal and opposite 'rotations' involved in that, and one could argue it leads to an initially normal-to-axis shear force on one such element edge reorienting slightly to give an axial component? That it?
    I seem to recall that being suggested earlier but anyway it's wrong. Shear strains in the shaft do not alter the symmetry of the induced stresses.

    And further, as brought up way back, even if a resolved axial force could be found somehow that route (would require at least the assumption of non-linearity to be introduced), it will be material dependent in magnitude. Contrary to the basic physics involved.
     
  14. The God Valued Senior Member

    Messages:
    3,546
    Q-reeus,

    Shoot it, how long will you keep this heavy bazooka in your hands.

    I am very confident that no meaningful physics will come out of it, still I am keen to know as I am curious, Q-reeus. In general this SF has tremendous lack of talent on the subject, but its not so bad either.
     
  15. Q-reeus Valued Senior Member

    Messages:
    2,605
    Well OceanBreeze has yet to get his theory together and give it another stab. And the more time goes on and a long record of failed attempts accumulates, the nicer the record becomes. I honestly thought someone would have an answer within a few days at most. Just goes to show.
    Are you indeed. Good thing I have no confidence in your confidence.
    No doubt. But have given up trying at any more mostly wild stabs at an answer? Guess so.
    The first part obviously true. The last bit means???? Never mind. Given recent happenings here at SF, I'm expecting to cop a life ban via one maneuver or another. So you may never get to know.....
     
  16. The God Valued Senior Member

    Messages:
    3,546
    To give you some benefit..

    Meaningful ~ Practical.

    And why you have to cop life ban? Already science is almost dead here, thanks to some self admitted unequipped mods, whose interventions are pathetic but they get gangahoo on some copy paster's posts. Quite comical!

    Rpenner has taken back seat, but he was kinda boring with his latex (!) in all posts. He should avoid heavy maths and come back.
     
  17. danshawen Valued Senior Member

    Messages:
    3,942
    I don't think there is a limit to how fast energy itself may spin, even if there is a limit to how fast it may propagate in linear mode or spin a physical shaft. But composite spins extend the idea to shafts replaced by quantum entanglement. Evidently there is nothing preventing the coupling of the spinning to energy stored in the rotation of the shaft to something as massive as a black hole flywheel.

    There is however no way a single battery will source all of its mc^2 energy enough to shift more than a tiny amount of its center of mass into the spin. The center of mass shifts more the faster or more massive the thing you are spinning, in this case, the generator, increases angular velocity. So, what of it? Are we including the flywheel arrangement as part of the generator? Makes sense to do that, like putting a hydroelectric generator close to the waterfall. Is the waterfall considered part of the generators?

    Maybe it would have helped to make the dynomotor arrangement frictionless, or even several generators connected to the same turning shaft. That way most of the energy isn't ultimately radiated away via friction as infrared energy to misdirect arriving at the solution you wanted. And I wouldn't have been as likely to think of the system as a dynomotor.

    A twisted spinning shaft is like a spring. You can store potential energy that way also, so the spin is actually composite in that situation as well. Not all of it is transferred to the flywheel or generator instantly.
     
    Last edited: Dec 16, 2016
  18. Q-reeus Valued Senior Member

    Messages:
    2,605
    See #30, #55, #83, #94, etc. Around in circles. The rest is pure danshawen so not worth my attempt to unscramble.
     
  19. danshawen Valued Senior Member

    Messages:
    3,942
    Ditto, #1. There is simply no reason for anyone to obsess over the center of mass shifting as a result of a spinning shaft by anyone familiar with the mechanical equivalent of heat (simple classical physics, not mc^2 or relativity). And furthermore, no such system is ever really "isolated" in any real sense. Partitioning such a system wrong is a classical thermodynamics mistake also. Sorry if I miss your point entirely. Apparently, this is mutual.
     
  20. danshawen Valued Senior Member

    Messages:
    3,942
    Here's a suggestion. Remove the batteries entirely. Use only superconducting wires and bearings to match for both the motor and the generator. Does your idea of a solution for an isolated system still work?

    The electric and magnetic fields remain isolated and do not radiate any photons. The bearings are all frictionless and do not radiate either. I have reduced your problem to a pair of spinning superconducting metal discs joined by a superconducting shaft. So, what's the big deal? Any change of the center of mass as a result of the spinning now?
     
    Last edited: Dec 17, 2016
  21. Q-reeus Valued Senior Member

    Messages:
    2,605
    Another danshawenism that could mean anything really. The problem is a real one, whether or not one 'obsesses' over it. It needs a real answer, not mumbo jumbo musings, Dan.
    Every mostly irrelevant criticism has been earlier raised and suitably answered time and again. Got something new on the table? Oh yes, your very next post....
     
    danshawen likes this.
  22. danshawen Valued Senior Member

    Messages:
    3,942
    . Nope. Finished.
     
    Q-reeus likes this.
  23. Q-reeus Valued Senior Member

    Messages:
    2,605
    Remove the source and sink for energy transfer? So we are left with a piece of dead machinery doing nothing? Why didn't I think of that!? Probably because I'm not Dan.
    That was a curved-ball late edit of your's Dan - after I prepared above. Why not just stick to #1 and use common sense? And try another peak at #72.
    Ah - just caught your last post. That sounds really good! Please stick to that one Dan. Please.
     
    danshawen likes this.

Share This Page