1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A The gyroscopic stabilization

  1. May 15, 2016 #1
    948bd5e598cc.png

    This is not a very famous effect but it is really amazing. Consider a pendulum which consists of a massless rod of length ##r## and a point of mass ##m##; the system is in the standard gravitational field ##\boldsymbol g##. So the point ##m## moves on the sphere of radius ##r##.
    It is clear, the equilibrium when the point rests in the North Pole of the sphere is unstable.

    Introduce a Cartesian inertial frame ##OXYZ## with origin in the point of suspension and the axis ##OZ## is vertical such that ##\boldsymbol g=-g\boldsymbol e_z##.
    Now let us switch on a Lorentz force ##\boldsymbol F=\boldsymbol B\times\boldsymbol v## which acts on ##m##. The vector ##\boldsymbol B=B\boldsymbol e_z## is constant.

    Theorem. Assume that ##B## is sufficiently big:
    ##\frac{B^2}{8m}>\frac{mg}{2r}##
    then the North Pole equilibrium is stable.
     
  2. jcsd
  3. May 15, 2016 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Is this a homework problem?
     
  4. May 15, 2016 #3
    No I just thought that it would be interesting for PF. I planned to write the proof.
     
  5. May 15, 2016 #4

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    Intuitively, when B is very strong it will force m on tiny circles anytime it tries to fall from the the North Pole.
     
  6. May 15, 2016 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Oh, didn't notice the username.

    Yes, magnetic fields tend to stabilize charged things, that should not be surprising.
     
  7. May 15, 2016 #6
    Yes, something like that is going on. But a pretty thing is: the Lorentz force does not do the work but it turns unstable equilibrium into the stable one :)

    so it makes wonder only me
     
  8. May 15, 2016 #7

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    It is not a fully stable point in the classical sense: if you displace the pendulum a bit, it won't come back to the center, it will rotate around it in a sequence of small "u"-patterns.

    If you add a damping term, no matter how small, the stable point should become unstable.
     
  9. May 15, 2016 #8

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    That surprises me with the damping. Gravity and the magnetic field do not change the total energy, while damping can only reduce it. And the north pole is the state of maximal energy. I would expect some downwards spiral if the initial position is slightly off.

    m=1g, r=1m, B=1T leads to 8.8 mC charge (B should be Bq I guess). Hmm, not practical on a large scale.
     
  10. May 15, 2016 #9
    O, I am sorry, I deleted my last post containing error.
    this is true

    in accordance with definition it is not obliged to come back https://en.wikipedia.org/wiki/Stability_theory
     
  11. May 15, 2016 #10

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    Seems similar to the stability of Lagrange points, which aren't minima of the graviational+centrifugal potential, but stuff stays there due to the Coriolis force, which is analogous to Lorentz force here.
     
  12. May 17, 2016 #11
    the proof
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: The gyroscopic stabilization
  1. Gyroscopic precession (Replies: 7)

  2. Principle of Gyroscope (Replies: 3)

  3. Gyroscope precession (Replies: 3)

  4. Gyroscopeic precession (Replies: 9)

Loading...