Linear momentum conservation puzzle

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

  1. Q-reeus Valued Senior Member

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    The expression looks to be one appropriate for a rotating shaft subject to presumably the first transverse flexural mode instability. Such that the shaft bows in the middle. Which is why engineers are careful to operate either well below resonant frequency or have the machinery driven up fast to well above it. Above/below critical speed is iirc how it's usually termed. Being close to critical speed would lead to small, 2nd-order axial displacements. But sorry to say such has no connection to the COM issue of #1.
     
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  3. Q-reeus Valued Senior Member

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    Nov 13 to Dec. 31, 2016 - Over 6 weeks. As promised, here's my answer (first part anyway) to the problem posed in #1:

    It will simplify to have the common transmission shaft linking the LHS motor and RHS generator to be a thin-walled tube of mean radius r, instantaneous angular velocity ω, and subject to an axially uniform torque τ generating a spatially uniform shear stress of magnitude σ everywhere within the tube wall.

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    As shown in the figure, consider a thin annular section of the shaft wall - a ring - of width b, and in particular a small square shaped segment of it - wall element s of edge lengths a, two such edges oriented parallel to the shaft axis.

    With the LHS sourced driving torque τ directed as shown, antiparallel edge traction forces f1, f2, acting on s to form a CW couple, will be as shown. Assuming negligible shaft elastic energy storage capacity, power notionally flowing in to element s via f1 is exactly compensated by an equal withdrawal via f2. Being in shear equilibrium, an opposing ACW couple owing to axially oriented antiparallel edge forces f3, f4 must also be present. Which pair being normal to the peripheral velocity v = ω × r of s play no role in the notional power flows. In the lab frame S where system COM is stationary, static equilibrium requires at all times |f1|=|f2|=|f3|=|f4|, whether τ or ω are time varying or not (restricting evaluation to where wave propagation i.e axial phase delay is at all times negligible.)

    Initial clue in #1 was to ponder F = dp/dt. It implies any shaft axial force will necessarily be directly proportional to time-rate-of-change of power flow dP/dt = d(τ.ω)/dt from source to load. Hence zero whenever P = τ.ω is constant not just zero. Suggesting examining just two complimentary cases:

    1: Fixed ω, ramp torque dτ/dt,
    2: Fixed τ, ramp angular velocity dω/dt

    Case 1 first. In lab frame S, time varying τ thus σ thus edge forces f1, f2, f3, f4 offer no evident avenue for an axial force as they all vary uniformly. But now move into the instantaneous proper frame S' of s, in S moving vertically up at v = ω × r as seen in the figure. In S' the relevant quantity is the temporal term of Lorentz boost:
    t' = γ(t-v.x/c²) - (1)
    As only the pair f3, f4, have a separation distance b along v, only they are affected by (1) such that a differential in magnitude exists between them in S'. Their equal purely time rate of change df/dt in S now having a spatial gradient added in S'. Setting t = 0, x = 0, for f3 in S', gives
    f3' = f3'(t'=0) - (2)
    f4' = -(f3'(t'=0) + γdf/dt(-v.b/c²)) - (3)
    δf' = f4'+f3' = γdf/dt(v.b/c²) - (4)

    The γ factor in (4) is unimportant and cancels out on transforming back into frame S. However the crucial residual force δf, owing to nonsimultaneity does not and is responsible for 'rescuing' conservation of momentum. it also nicely acts against the nominal rate of change of power flow, as required. The force density is just (4) divided by the separation distance b, (and γ factored back out ) i.e.
    |ρ| = |δf/b| = |df/dt(v/c²)| - (5)
    Which by simple inspection matches and cancels the rate of change of momentum transfer through element s heading from motor to load. Obviously since their is a balance across any given element of shaft, integrating over the whole will maintain that situation.
    How about that. No recourse to heavy higher maths. I did leave out the connection to shear waves where body forces explicitly are proportional to the shear stress gradient, but the reader can pick that sort of thing up in e.g. eq'n. (1) here: http://petrowiki.org/Compressional_and_shear_velocities
    The interesting and subtle difference to this case is that the shear gradient is rotated orthogonal relative to the much more familiar shear wave situation - as needed.

    Sorry but lots of trouble to deal with last few days threw my plans out, so no time to get case 2: dealt with before 2017 forum time arrives. While I'm recovering, maybe at last someone here can now figure out how to do that one. Back later....
     
    Last edited: Dec 31, 2016
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  5. Q-reeus Valued Senior Member

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    Continuing on then from last post, let's look at Case 2: Fixed τ, ramp angular velocity dω/dt.
    Which for tube wall element s shown in #182, translates to an azimuthal acceleration a = dv/dt = d(ω × r)/dt. For +ve sign of dω/dt, a is as for v previously, directed upward in the figure. In the accelerated proper frame S'' of s, neglecting any inconsequential centripetal component, there is an equivalent essentially uniform gravitational field g = -a. Edge traction force f4 is in a higher gravitational potential than f3 by the amount -|gb| = |ab|. Not just time but mass-energy and transverse force is gravitationally dilated or 'redshifted' to the same degree. Something easily proven by comparing the relative energy change owing to work done by an azimuthal force over a set azimuthal distance, at two different elevations in a Schwarzschild exterior metric. The relative change in forces is equal to the corresponding frequency redshift between the two elevations. Which for weak gravity case is just fractionally -gh/c².

    In the torsion shaft case, it means in S'', f4 fractionally exceeds f3 by the factor 1+|ab|/c². Hence the difference is
    δf = f4-f3 = f4(|ab|/c²) - (1)
    Again, directed back towards the motor. But the rate of change of input power to s is just d(f1.v)/dt = f1.a. Which 'flows' in a notional sense across the axial distance b before being extracted at the RHS edge via f2. Hence magnitude of the rate of change of transfer of power across the axial distance b is f1.ab. Dividing by c² to give the mass equivalent also gives numerical equivalence to (1), but as required has the opposite sense such that no overall change in COM has occurred.

    Which about wraps it up, unless someone has some basic objections. Which I doubt since my analysis though brief and unpolished gets the right results in both cases.
    As a footnote, I will add that the original inspiration for this thread was after viewing an essentially identical case posed over at PhysicsForums by a Jonathan Scott here:
    https://www.physicsforums.com/threads/transfer-of-momentum-along-axis-by-torque.891359/
    Sadly not one of the many qualified physicists attempted an answer, and the 'hints' mentioned by JS in post #11 there were actually mine, but owing to the 3rd party way they had to be delivered, he could not have known that. I know of at least 8 or so PF members who are also SF members. Please, at least one of you do the right thing - alert JS to this thread and in particular to the last few posts here. Such a poor record he had no effective help over there.
     
    Last edited: Jan 1, 2017
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  7. The God Valued Senior Member

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    The OP took everyone, including some not so bad in physics guys, in tangential (no pun) direction. Never came out of battery mass loss. Moreover you had thwarted any attempt which thought of shear in shaft. Not exactly, that also put off some creative thinking. Still this transformation stuff looks overkill for this problem.

    Nice that you kept your multi extended deadline.
     
  8. Q-reeus Valued Senior Member

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    No. You took yourselves in tangential directions. I gave all the necessary info. Wrong thinking made no good use of it. Simple as that.
    What do you mean? Transfer of mass-energy from one battery to another was integral to the scenario.
    How do you figure that? When one or two started heading in the right direction, I said so. But no-one kept going the right way.
    He he he. Actually I took the very simplest possible route. An experienced physicist or mathematician would want it all in advance tensor formulations. Be thankful.
    The best of plans sometimes go astray. They did for me.
     
    Last edited: Jan 1, 2017
  9. danshawen Valued Senior Member

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    Idiot. Your so-called 'answer' does not even mention COM, as promised. Who gives a fart about the forces or torque involved, or even gravitational red shift? I don't. Or were you simply not finished? If you had invoked tensors, it would have been even less relevant. You seem to be saying, no change in the center of mass, and the batteries aren't even a consideration. What is it exactly that makes you think this problem is a distinguished one? Not a differential element of a shaft experiencing torque, surely. Your classical physics bindings are laid bare here, and it is as unsightly as your 19th century Newtonian math. Relativity happened. It's not just a theory like any other math construct; it relates to the actual, physical world. Your spinning shaft is infinitely rigid, and this is as implausible as it is impractical. It won't work for a galaxy sized motor or shaft. If the math works at all here, it won't work at relativistic speeds. You said you were sharing a relativity problem, didn't you?

    It reminds me of a former colleague well published in the Royal Society who still believes the Cavendish experiment needs to be done in an orbiting spacecraft, to determine if gravitational potential and time dilation are maximal or minimal at the centers of large gravitating masses. Now THAT is what a distinguished physics problem looks like. Yours is an engineering question at best. If you wanted to spin up an atomic clock's rubidium or cesium atoms with your motor shaft to test the principle of equivalence or something, that would be a more interesting PHYSICS experiment, but not something like this. Are you an automotive engineer or a member of SAE?

    My colleague was right, I figured out later. The time dilation is the same at the center as it is for free space outside of the influence of the gravity well. But how he got there was unbelievably contorted in terms of supporting math, just like your answer. If something no longer experiences a force, then it is at the center of mass and there is no acceleration, and zero time dilation relative to outside of the isolated system. It could still have red shift. Depends on the state of motion of the whole mass.

    Happy New Year all the same. I still greatly enjoyed your thread, even if the answer you have at length provided is something of a disappointment.
     
    Last edited: Jan 1, 2017
  10. Q-reeus Valued Senior Member

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    'Happy' New Year? After that nasty outburst? Completely missing that what I gave answered the original problem quite adequately? YOU as usual are the disappointment Dan. If you can't recognize that the solution for an arbitrary element of the shaft is also a general one, it's your limitations being exposed.
     
  11. danshawen Valued Senior Member

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    I recognized that feature (a differential element of torque for a portion of the drive shaft) just fine. Not completely general though, is it really?

    "...assuming the entire system is isolated, conservation of momentum requires <the state of motion of> the centre of mass is not changed by the transfer of mass-energy from battery to battery." ... "The problem is how, or even if, that is possible given power transmission is via pure torsion in a rotating drive shaft."

    I guess you showed that. If that's all you wanted to show, you succeeded. Was it worth it?

    If the shaft was designed to eject small masses or pellets as it turned, would linear momentum still be conserved? The differential force diagram for the shaft then becomes unbalanced each time a pellet is ejected, transferring impulses to the pellets and the COM moves faster as more pellets are ejected.

    Conservation of linear momentum still works, and it even works for systems that are not isolated, like this one, or guns or rockets. Conservation of momentum does not depend on isolation to work.

    So, why do you suppose the batteries threw us all a curve like that? They make no difference at all. You could come to exactly the same conclusion with an impossible perpetual motion machine in place of the motor and generator and it would make no difference at all to the solution you proposed. Something still bothers me about that.
     
    Last edited: Jan 2, 2017
  12. Q-reeus Valued Senior Member

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    You asking? I stipulated two complementary cases, analyzed and solved for both. Which by linear combination covered the general case. What do YOU claim was missing?
    Well let's see. After more than six weeks, no-one here was up to offering an answer anywhere near right. You recall that footnote piece in #183? No-one over at the vastly more talent filled PF was either up to or willing to give an answer to the same problem. From that you might conclude the problem was not an easy or obvious one to solve.
    It took me iirc at least an hour to figure out (but only for the first case covered) back in the 90's sometime when I first thought of it as an issue.
    There is a certain intuitive feel required for connecting basic physics to a mundane yet in a way novel physical situation evidently not covered in a textbook or uni course.
    Sounds like a trivial variation to me and unless you can suggest something truly special, why even ask the question?
    No but if you wish to solve such a problem, treating it as an isolated system is THE rational way of doing things. I do recall one higher up poster here trying to muddy the waters way back in another thread, by deliberately introducing extraneous matters designed to needlessly complicate and confuse. Not smart. And btw, while everything works out fine in the case considered, it's fairly easy to show that situations exist where standard theory actually predicts a breakdown in energy-momentum conservation - or at least shows itself to have internal inconsistencies. But I won't wake that sleeping dog again here - given the all-round inane hostility it evoked when I first raised the matter way back.
    Nonsense. A source and sink for mass-energy would always be needed, and I later explicitly showed that it need not be electrical/electrochemical in nature - e.g. flywheel/disk-brake. Which situation should have been obvious from the start.
    The essence of the solution was as shown owing to SR applied to the shaft under dynamical load conditions where dP/dt was nonzero. Failure to home in on that was not my problem or doing. It was always there in front of you. But none here or over at PhysicsForums saw it. It will be 'explained in-principle' by some elegant use of stress-energy-momentum tensor, but the actual application has to be recognized!
     
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  13. danshawen Valued Senior Member

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    Can you really compare dP/dt of the charge carriers in the wires with the dP/dt of the shaft (or make it a spring if you need to)? One is relativistic. One usually considered not to be, but both actually are. Both the electrons and the shaft are hard limited by c, but neither actually approach that, for any practical load. But the electrons can't slow down very much, or be at rest, like the shaft can, until both batteries are discharged, which is the real end state. The electrons revert to their respective bound chemical state.

    OK, you convinced me it was worth it. It's not a perpetual motion machine, and there isn't anything magic about shaft rotation in that regard either. When the shaft comes to a stop due to friction, it is entirely equivalent to a pair of batteries being discharged, or a flywheel grinding to a stop, and regenerative braking makes no difference one way or the other.

    A flywheel all by itself (no friction) can rotate forever under no load conditions, and a superconducting wire ring can likewise carry an electric current indefinitely outside of the influence of stray magnetic fields to change it.

    Likewise, atomic structure was once considered to be self-sustaining and immune to outside influences, or it was until the Higgs boson was discovered. Now we have more force pairs to consider, and the way they interact (forces, inertia) are up for grabs. Energy still needs to be conserved, and gravity based perpetual motion machines don't work either. You aren't going to solve this one as easily. Is it isolated or not? How much energy is in the vacuum, or how much do you actually need to maintain atomic structure fo an entire universe? How does this compare to dark energy?
     
  14. The God Valued Senior Member

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    Q-reeus,

    You have to realize that all this battery stuff put many in tangent.

    I very specifically told you that as battery discharges, voltage reduces and that gives you variable power, your dP/dt. You bulldozed it or so I inferred. Basically I was expecting something which would have made me look short of basics of physics. But no it did not happen.
     
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  15. danshawen Valued Senior Member

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    Great differential force diagram, incidentally, Q-reeus. Very, very clear.
     
  16. danshawen Valued Senior Member

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    A design for a space trebuchet that spins up spacecraft to boost them to escape velocity is solar powered. A counterbalancing pair of rotating tanks of a heavy liquid is pumped by means of solar powered pumps to maintain constant angular velocity for launches, and the spacecraft is extended to release position outside of these by means of retractable tethers. Counterbalancing weights on superconducting bearings are spun up (also solar powered) in a reaction ring for both attitude control and flywheel storage of flung spacecraft propulsion energy. The angular velocity of the flywheel ring doesn't really matter, since it is uninhabited.

    Since weightless settling of spinal fluid is now recognized as a major medical problem on weightless interplanetary space voyages, this technology, based heavily on conservation of linear / angular momentum, is a natural fit to continued and sustained human space exploration.

    Cool update and use of the space station technology / architecture and also space elevators designs suggested by Arthur Clarke, eh? When I first saw it, I was glad to see an update to Fountains of Paradise on the drawing boards. The tethers needn't be Earthbound to be useful.

    This is just one practical real world application of Q-reeus's problem scenario, only a little less isolated, and using solar panels in lieu of batteries.
     
    Last edited: Jan 2, 2017
  17. Q-reeus Valued Senior Member

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    Nice of you to say so but the illustrated forces were merely the usual tractive forces from mechanics. The residual force wasn't shown but derived from relativistic treatment of those shown.
     
  18. Q-reeus Valued Senior Member

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    Amazing! I never would have made such a connection!!

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  19. sweetpea Registered Senior Member

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