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How to calculate spinning disc torque which resists tilting of the disc's axis?

  1. May 11, 2012 #1
    Hello,

    From what I've read, I understand that a spinning disc has a torque which resists the tilting of the disc's axis. From what I understand, the higher the angular velocity of the disc is, the more the disc will resist tilting. Can anyone tell me how to calculate this torque? If there is not a simple explanation, could you give me a reference to a book or a paper that describes this?

    Thanks,
    Kevin
     
  2. jcsd
  3. May 11, 2012 #2

    haruspex

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    No, a spinning disc has angular momentum (ML[itex]^{2}[/itex]/T). A torque is a rate of change of angular momentum (ML[itex]^{2}[/itex]/T[itex]^{2}[/itex]). Like the difference between speed and acceleration.
    It's not that the disc resists tilting. This is a common misunderstanding of how gyroscopes work. Rather, the disc responds paradoxically to an attempt to tilt it. Imagine trying to push a trolley forwards, but it responds by moving sideways. (As indeed it might if the castors are stuck at an angle.) Your natural inclination would be to resist the sideways movement and persist in pushing forwards. Now you would feel resistance, but it's your own resistance to the sideways movement being relayed back to you.
    The same happens with a gyroscope. Suppose it's spinning on a vertical axis and you try to tip the top of it away from you. Instead, it will tend to tip to the left or right, depending on which way it's spinning. If mounted in a gimbal, so that it is free to tip left or right, you will not feel any resistance, but if holding it in your hand you will tend to resist the paradoxical response, so you do feel resistance.
    If the disc has moment of inertia M and spins at rate w then its angular momentum is Mw. If you rotate it at right angles to its spin axis at speed u (an angular velocity) then to inhibit the paradoxical response you will need to supply a transverse torque Mwu.
    (The M.I. of a uniform disc is its mass * radius squared / 2.)
     
  4. May 11, 2012 #3
    Hi haruspex,

    Thanks for your reply. I understand what you are saying, but I'm not quite sure how to apply this for my example. Maybe I should clarify my question.

    I am considering levitating one magnetic disk above another magnetic disk. The two disks repel each other, so that the downward gravitational force on the upper magnet cancels the upward levitation force. If the upper disk is not spinning, then a torque will act to tilt the upper magnet so that it tilts 180 degrees (and now the upper magnet will be attracted to the lower magnet, which obviously destroys the levitation). Electromagnetics is my specific area of study, so I can compute this torque easily. I have plotted the torque as a function of tilt angle and confirmed that the torque acts to flip the upper magnet upside down. (In order words the stable equilibrium occurs when the upper magnet tilts 180 degrees.)

    However, from what I've read, if the upper disk is spinning (as in Levitron devices), then this spin stabilizes the device from flipping upside down. However I don't have a mechanical engineering background, so I'm not sure how to model this mathematically. Do you have any thoughts for how I can combine the gyroscopic effects with the magnetic torque I know how to calculate and show that when spinning the upper disk won't flip upside down?

    Thanks again,
    Kevin
     
  5. May 11, 2012 #4

    haruspex

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    Can you define the magnetisation for me? Is each disc N on one face and S on the other (and you place them with two similar poles facing)?
     
  6. May 12, 2012 #5
    Yes, that's correct. The lower disc has magnetization in the positive vertical direction (North pole on the top surface and South pole on the bottom surface) while the upper disc has magnetization in the negative vertical direction (North pole on the bottom surface and South pole on the top surface).
     
  7. May 12, 2012 #6

    haruspex

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    I think the calculation would involve perturbation analysis. After all, that's why it's not stable when not spinning. I'll have a think about it, but it looks nasty.
    Meanwhile
     
    Last edited by a moderator: Sep 25, 2014
  8. May 12, 2012 #7

    haruspex

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    Well, I've thought about it a bit and more and realise I don't understand why it would flip over without spin. Why wouldn't it just slide off to one side, like a plate off a smooth dome?
    Maybe the lower disc is much wider, so the upper disc would have time to flip over before escaping to the side of the lower? That being the case, spinning it might inhibit the flip but still allow the slide.
     
  9. May 12, 2012 #8

    Cleonis

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    Hi Kevin,

    As Haruspex pointed out already, it's not so much a stabilisation.

    The property of a spinning gyroscope wheel of not yielding immediately to flip-over is subject to decay. As friction decreases the wheel's spin rate any wobble gets bigger and bigger.

    As Haruspex writes: the response to keeling of the spin axis in one direction is acceleration in a direction perpendicular to it. Because of that right angles effect the dynamics of gyroscopic precession is by nature cyclic.

    I've written before about gyroscopic effects, and for that posting I uploaded two images, so I refer to that post from november 2010 about gyroscope physcis
     
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