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How does the Levitron "break" Earnshaw Theorem?

  1. Apr 1, 2015 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 1, 2015 #2
    Hi everbody. This is my first post on the site.

    Recently i built a Levitron for my physics class in highschool. It levitates for like 1.20-1.30 minutes.

    Now i have to do the theoric part which I don´t understand very well.

    I know that Samuel Earnshaw is responsible for a theorem that goes like this:
    "A collection of electric charges or magnetic dipoles cannot be maintained in a stable stationary equilibrium condition by the application of static fields alone."

    So how does the levitron manage to disobey Earnshaw and levitate? I can tell by now it has something to do with the spinning, but I still dont understand it...

    Everytime i search about this, some strange mathematical MAMBOJAMBO i dont understand jumps in (Sorry for my ignorance) and draws my motivation away.

    So can anyone leave some words here in this thread mentioning without using many "advanced" physic terms, how does The Levitron constitue a loophole for Earnshaw theorem.

    (Sorry if my english sounds silly)
  4. Apr 26, 2015 #3

    I'm probably a little late for the purposes of your Physics class, nevertheless I'll respond and hopefully satisfy your curiosity...without being overly technical and detailed. To start, the Levitron does not defy Earnshaw's Theorem; it works around it. The spinning of the top adds a dynamic aspect, that is, the spinning top is not part of a static system; thus there is no violation of theorem. Going a bit further, while the top's spinning prevents its flipping over, it does something else: it prevents it from exiting the levitation field through sideward motion. This probably sounds a little strange, but there is something more to the top's rotation than meets the eye. There is a natural tendency for that rotation to be more-or-less wobbly. The technical name for this wobbling motion is "precession", but "wobble" will do. If the top were to spin perfectly smoothly such that the rotation of its ring magnet remained essentially planar, it would simply move outward from the center of the levitation field and crash to the side. The wobble, in effect, tilts the top's magnet to and fro (ultimately in every direction, through 360 degrees). When the top's magnet tilts, its field interacts with the field of the base magnet in such a way as to create resistance to movement in the direction of that tilting. Because this occurs over 360 degrees, the top remains centered within the field. If the top spins too fast, rotation is excessively smooth and the degree of wobble (and thus tilting and magnetic resistance to sideward movement) is inadequate; the top will exit the levitation field. If spinning is too slow, the rate at which tilting occurs will also be too slow and the top will either exit the field or simply flip over. While there are certain principles involved that have not been touched on, and the precision of the language could withstand a bit of refining, I think that what has been discussed should provide sound footing and a satisfying degree of understanding.


  5. Apr 26, 2015 #4
    THANKS for the answer. My presentation is only in the 12th of May.

    Now, (after reading your comment) im goin to the check the "wobble",

    Before you answered I made a little more research and got to the conclusion (probably wrong) that the key factor for the top stability was the torque.

    "If the top was spinning to slow, The torque would contribute for the top to get out of the center and fall afterwards, but if the top is spinning "fast", that torque would act gyroscopically and consequently give the top stability... "

    Is this completely wrong ? Probably... So, does the torque exist like i describe in the levitron, or the only key factor for stability is the wobble?

    Once again, thanks JC. (Sorry for my english, it is not my main language.)
  6. Apr 26, 2015 #5

    Torque is generally thought of as the twisting force that is applied to the spindle of the top to produce rotation. The greater the torque (or twisting force), the greater will be the resulting rate of rotation. Perhaps you are using the word "torque" where "angular momentum" is intended. In any event, the greater the speed of rotation (and angular momentum), the greater the resistance of the top to flipping. So, in that sense, a higher rate of rotation - that is, a faster top - means a more stable top. However, this is not the case when we are talking about lateral stability. The top will be stable within a certain range, but once the speed of rotation exceeds a certain limit - namely the point at which the noted wobble is as slight as it can be while still effectively resisting the top's tendency to move sideways - the top will move sideways and out of the levitation field. I apologize if this has gotten a bit more complicated than is helpful. By the way, your English is more than adequate.

  7. Apr 27, 2015 #6
    Once again, thanks for replying with this precious information.

    There´s just one thing left that still bothers me...

    In my presentation I would like to establish a relationship between the spinning top (Levitron) and a gyroscope. The thing I was looking for as to do with the way they spin.

    GYROSCOPE: When i watch videos on youtube or play with my gyroscope toy I notice that it doesnt spin with a vertical axis, so i guess it is more stable that way. Now, what I dont understand is...WHY?

    LEVITRON: Since the magnetic torque tends to turn the magnet in the spinning top (and consequently the top itself), i guess the top also doesnt spin completely vertical.
    Why is it more stable that way? Does that have something to do with the gyroscope with the fact of the gyroscope not spinning around a completely vertically axis?

    Thanks again for reading this.
  8. Apr 29, 2015 #7

    A gyroscope does not spin perfectly vertically because gravity is exerting torque on it. The result is the wobbling motion, that is, the precession that you observe. There is much that can be said to explain why the gyroscope remains upright, but frankly I don't think that I can do so while keeping the discussion "simple". As for the Levitron, you are correct in noting that the top's spin axis is not entirely vertical; gravitational and magnetic torque are at play. You ask why the Levitron's top is more stable as it spins with an axis that is not entirely vertical. The reason for this was touched on in my first response: When the top's magnet tilts as it does in precession, its field interacts with the field of the base magnet in such a way as to exert resistance - repelling force - in the direction of that tilting. This occurs throughout 360 degrees, keeping the top centered within the field. The rate of precession is critical to maintaining this centralization, and because this rate is dependent on rotational speed, the speed at which the top spins is also critical. For this reason the top is stable only within a certain range of rotational speed. To summarize, the laterally stabilizing effect of precession and the force that counters the torque exerted by gravity and magnetism so as to stabilize the top against flipping - the same force that keeps the gyroscope upright when spinning - coupled with the repellent force between confronting like poles of the base and top magnets, all work in coordination, allowing the Levitron's top to levitate.

    M, I apologize if I haven't made matters sufficiently clear, although I hope that I've helped to some extent. In any event, I'll finish by saying that I wish you good fortune in your Physics class and in your future.

  9. Apr 29, 2015 #8
    Thanks for answering.

    You definitely helped me.

    I think im able now to provide my classmates a satisfying explanation for why the levitron levitates.

    Im going to search for "simple" information that connects the levitron to the gyroscope and terminate my work.

    You were vital to this woks!
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