The gyroscopic stabilization

  • A
  • Thread starter wrobel
  • Start date
  • #1
wrobel
Science Advisor
Insights Author
994
850
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.
 

Answers and Replies

  • #4
A.T.
Science Advisor
11,529
2,897
This is not a very famous effect but it is really amazing.
Intuitively, when B is very strong it will force m on tiny circles anytime it tries to fall from the the North Pole.
 
  • #5
35,919
12,746
Oh, didn't notice the username.

Yes, magnetic fields tend to stabilize charged things, that should not be surprising.
 
  • #6
wrobel
Science Advisor
Insights Author
994
850
Intuitively, when B is very strong it will force m on tiny circles anytime it tries to fall from the the North Pole.
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 :)

es, magnetic fields tend to stabilize charged things, that should not be surprising.
so it makes wonder only me
 
  • #7
35,919
12,746
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.
 
  • #8
35,919
12,746
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.
 
  • #9
wrobel
Science Advisor
Insights Author
994
850
O, I am sorry, I deleted my last post containing error.
If you add a damping term, no matter how small, the stable point should become unstable.
this is true

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
in accordance with definition it is not obliged to come back https://en.wikipedia.org/wiki/Stability_theory
 
  • #10
A.T.
Science Advisor
11,529
2,897
It is not a fully stable point in the classical sense
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.
 
  • #11
wrobel
Science Advisor
Insights Author
994
850
the proof
 

Attachments

  • gyr_eng.pdf
    49.6 KB · Views: 203

Related Threads on The gyroscopic stabilization

  • Last Post
Replies
4
Views
736
  • Last Post
Replies
0
Views
6K
  • Last Post
Replies
6
Views
7K
  • Last Post
Replies
7
Views
5K
  • Last Post
  • Optics
Replies
4
Views
2K
  • Last Post
Replies
6
Views
10K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
890
Top