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If the Earth were to be flat, but still rotating with a constant angular frequency, would the Foucault pendulum still show it's precessing motion?

The angular velocity of the precession can be derived to be: $$ \omega = -(\Omega \cos\beta)\mathbf{k} $$ where $\Omega$ is the angular frequency of the earth, $\mathbf{k}$ is the apparent vertical and $\beta$ is the angle between the apparent vertical and the rotation axis.

On a rotating disk, with the same rotational frequency as the Earth, one would expect that the formula for $\omega$ would still be valid, albeit with a different apparent vertical $\mathbf{k}$.

EDIT: Gravity on the disk is assumed to be uniformly perpendicular to the disk, such as most flat-Earth theories suggest.

N.B. I do not believe the Earth to be flat, this question is purely out of interest.

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  • $\begingroup$ How does gravity work on your flat Earth, is it a homogeneous gravitational acceleration perpendicular to the disk? (Because that is not the field the disk would source itself.) $\endgroup$
    – Void
    Jan 25, 2018 at 9:58
  • $\begingroup$ I assumed gravity to be perpendicular to the disk. I made an edit to clarify. $\endgroup$
    – Simon
    Jan 25, 2018 at 10:01
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    $\begingroup$ Think of a horizontal disc rotating on a table underneath a simple pendulum whose point of suspension is fixed relative to the table. $\endgroup$
    – Farcher
    Jan 25, 2018 at 10:34
  • $\begingroup$ Do flat-earthers think that the Earth spins like this, or do they assume it tumbles in some way (which would, I suppose, explain the Sun's motion better)? A Foucault pendulum on a tumbling flat Earth would be interesting. $\endgroup$
    – user107153
    Jan 25, 2018 at 18:38
  • $\begingroup$ I'm not sure, since I'm not that well informed on flat-earth theory, but i found a couple of posts online where flat earthers 'debunked' the foucault pendulum experiment by saying that the flat earth rotated too. But I agree that a tumbling earth would be interesting.. $\endgroup$
    – Simon
    Jan 26, 2018 at 15:02

2 Answers 2

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Yes, it would work, but $\omega$ would be the same at all points on the disc, and equal to the rate of rotation of the disc itself. So although the pendulum would precess, you could very easily know if you were on a flat disc by moving it around and comparing precession rates,

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    $\begingroup$ For centrifugal forces absolutely negligible with respect to the gravitational force, the above is correct. However, it is not correct when the centrifugal acceleration becomes large enough to change the direction of the apparent vertical. Then there are corrections to the frequency of order $(a_c/a_g)^2 = (\Omega^2 r/g)^2$, so it is a question of how large the disk and the rotation are. $\endgroup$
    – Void
    Jan 25, 2018 at 18:23
  • $\begingroup$ @Void: thanks. I think that the normal foucault pendulum computation for a spherical Earth ignores centrifugal forces: is that right? If it is (or in either case) are you happy for me to add your comment to my answer, as it's not really right without it. $\endgroup$
    – user107153
    Jan 25, 2018 at 18:35
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    $\begingroup$ @Void +1. But, on earth, what we commonly call "gravity" is a mix of gravitational and centrifugal forces. That is the only definition under which gravity points "straight down". The same definition could be used on this hypothetical flat disc: the actual gravitational field could point more and more inward the further from the axis of rotation you go, such that "gravity" (which includes centrifugal force) always points down, just as it does on earth. $\endgroup$
    – Ben51
    Jan 25, 2018 at 18:38
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A rotating flat Earth can't actually exist. The bigger a planet is, the smaller the gravitational constant would have to be for it to have Earth's gravitational field strength and the slower it would have to be rotating for its centrifugal force to be balanced by gravity and be rotating barely slowly enough to form into a stable oblate shape. A flat Earth would be infinitely big and so would necessarily have a zero rotation rate.

Let's suppose it has a nonzero rotation rate and there's also some magic force it exactly balances the centrifugal force in its frame of reference so that the only fictitious force in its frame of reference that's not balanced by and equal and opposite force is the Coriolis force. According to this answer, precession can be caused by something other than the Coriolis force. Under certain conditions, if the Earth were rotating slowly enough, we still wouldn't be able to tell from watching the pendulum that Earth is rotating.

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