# A The gyroscopic stabilization

1. May 15, 2016

### wrobel

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. May 15, 2016

### Staff: Mentor

Is this a homework problem?

3. May 15, 2016

### wrobel

No I just thought that it would be interesting for PF. I planned to write the proof.

4. May 15, 2016

### A.T.

Intuitively, when B is very strong it will force m on tiny circles anytime it tries to fall from the the North Pole.

5. May 15, 2016

### Staff: Mentor

Yes, magnetic fields tend to stabilize charged things, that should not be surprising.

6. May 15, 2016

### wrobel

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

7. May 15, 2016

### 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.

8. May 15, 2016

### 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.

9. May 15, 2016

### wrobel

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

10. May 15, 2016

### A.T.

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. May 17, 2016

the proof

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