2
$\begingroup$

From the wiki page:

If the plane of swing was north-south at the outset, it is east-west one sidereal day later. This implies that there has been exchange of momentum; the Earth and the pendulum bob have exchanged momentum. The Earth is so much more massive than the pendulum bob that the Earth's change of momentum is unnoticeable. Nonetheless, since the pendulum bob's plane of swing has shifted, the conservation laws imply that there must have been exchange.

What exactly does this mean? Let's say we had a pendulum so massive that the momentum change of Earth becomes noticeable. What do we notice? The planet rotates slower? It starts rotating along a different axis?

$\endgroup$
1
  • $\begingroup$ If the Earth wouldn't be massive, there would be a beating between the pendulum and the planet's axis of rotation. One can probably find a similar phenomenon in tidal locked bodies with roughly equal masses, although I don't know if there is a known example in the solar system. The moon may exert tiny variations on Earth's rotation at the month timescale that way. $\endgroup$
    – CuriousOne
    Feb 4, 2016 at 19:16

2 Answers 2

1
$\begingroup$

This answer is very late; I'm only adding it for the sake of completeness.

To bring the important features of the case to the foreground imagine the following case: a Foucault pendulum is constructed on a very small celestial body, say, an asteroid. For simplicity, let's make that asteroid spherical.

First consider the motion when the asteroid is not rotating, and the pendulum is not swinging. Then the tension of the wire points in the direction of gravitational pull.

Next consider the pendulum swinging. When the pendulum bob is not at the plum position the tension of the wire is not in the same direction as the direction of gravitational pull, so there is a resultant sideways pull on the pendulum wire's upper attachment point. So: if the mass of the pendulum bob is a non-neglible percentage of the mass of the asteroid the asteroid will have a measurable wobble as the pendulum swings. That's exchange of momentum right there.

Next consider the case of the pendulum swinging with the asteroid rotating, and the pendulum is not located on either of the poles, and not on the equator. In that case, when the pendulum is swinging north-south the distance of the pendulum bob to the asteroid's axis of rotation changes over the course of each swing, and that has an effect on how the pendulum bob and the asteroid are exchanging momentum.

(From here on I will refer to a pendulum located in a zone not on either of the poles or on the equator as a zonal pendulum.)

Now back to the Foucault pendulum on Earth. All the above reasoning applies, the only difference is that the change of momentum of the Earth is immeasurably small because of the Earth's far larger mass. Still, the pendulum bob and the Earth do exchange momentum.

The effect of that exchange of momentum happens to show up vividly on the vary latitude of Paris. At that latitude a full precesssion cycle takes about 32 hours. That means that after 24 hours a Foucault pendulum will have completed 3/4 of its precession cycle. Hence, if the pendulum was swinging north-south at the start, after 24 hours, with the Earth back to the same orientation relative to the stars, the pendulum is now swinging east-west. The dynamical interaction between the pendulum bob and the Earth has shifted the pendulum's plane of swing.

It would take an astronomical amount of stubbornness to keep insisting that for a zonal pendulum the plane of swing does not change direction relative to the stars. It does. On the latitude of Paris, after 24 hours the Earth is back to the same orientation as 24 hours before, and during that time the pendulum has progressed along 3/4 of its precession cycle.

$\endgroup$
0
$\begingroup$

You need to read the rest of that page you quoted. In fact, there is only a perceived momentum transfer in the Earth's frame of reference, but since the Earth is rotating, that's a lousy (so to speak) frame to work with.

Further, as the page states, a day later the pendulum's back where it started. The only way to effect an actual momentum transfer would be to somehow "lock" the pendulum into its direction of swing at the one-day point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.