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Homework Help: Magnetic reductor

  1. Dec 17, 2014 #1
    1. The problem statement, all variables and given/known data

    It is a question I asked to myself.

    I studied the induction motor at school, and I asked to the teacher if it's possible to use magnets for build a reductor with magnets. She said it's possible but the energy lost in heating will be very high. I would like to know if my forces are correct and how to estimate the losses.

    http://imageshack.com/a/img537/1866/zqaJ3k.png [Broken]

    The Rotor1 rotates at 3w clockwise from an external device not drawn. The Rotor2 is push by the Rotor1. I guess the radius of the Rotor1 is 1 and the radius of the Rotor2 is 3. I would like to have w for the Rotor2, but I'm not sure if it is possible.

    I have some questions:

    1) I think it's possible to use it only if I start the device like I drawn, correct ? The magnet is pushing clockwise the Rotor2: the Rotor2 is acccelarating, the Rotor2 gives an energy to an external device (not drawn), magnets push counterclockwise the Rotor2: it is decelerating. The angular velocity of the Rotor2 is not constant but the mean can be at w, no ?

    2) Losses are very high, even w is low ? How to estimate them ? In theory is it possible to have an efficiency of 1 with this reductor ?

    3) Maybe this device exist, have you a link ?

    2. Relevant equations


    3. The attempt at a solution

    I drawn forces:

    http://imageshack.com/a/img901/3278/NOjZYC.png [Broken]

    My forces are correct ? The Rotor1 gives an energy F*r*3w*t and the Rotor2 can give F*3r*w*t, the efficiency can be at 1 only if magnets don't lost heating.

    For me the angular velocity is not constant but like:

    http://imageshack.com/a/img540/668/Pd5ALz.png [Broken]
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Dec 17, 2014 #2


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    Homework Helper

    If the inner rotor is turning steadily at constant angular velocity, I think the outer one will turn steadily at the same angular velocity. I picture an equilibrium where the N on the inner magnet is just a little closer to the outer N on the forward side than to the outer N on the reverse side. Just enough closer so the larger force forward is just enough to overcome whatever air and friction resistances are involved.

    I don't think there would be very much energy loss to friction with decent bearings so very little torque and rotational energy would be needed to keep it moving steadily. Anyone who has chased a magnet around with another magnet will have a good feel for this efficiency. Odd, I seem to recall doing this as a child with one magnet on a table top and the other under the table top, yet there wouldn't have been sufficiently strong magnets over 50 years ago. Maybe tabletops were thinner.

    Interesting application:
    http://www.magnomatics.com/technology/magnetic-gears.aspx [Broken]
    Last edited by a moderator: May 7, 2017
  4. Dec 17, 2014 #3
    For me the angular velocity of the Rotor2 is lower than the Rotor1 because the Rotor1 gives an energy F*3*w*t (the radius is R1=1 and the angular velocity is 3w), so the Rotor2 must give the same energy, like the radius is R2=3 and the force is the same, w must be lower no ?
  5. Dec 17, 2014 #4


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    Homework Helper

    I see your point. Like an automatic transmission torque converter. When a lot of output torque is needed, you have to run the engine faster to provide the necessary output power. However, with the magnetic coupling my mind sees catastrophic failure when the inner ring goes out of sync with the outer one. As the Inner N passes the outer N, there is no torque on the outer ring for a moment, then reverse torque for about as long as there will be forward torque.

    Too complicated for me, I guess.
  6. Dec 17, 2014 #5
    Yes like that. Thanks for your help.
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