Would a Foucault Pendulum Work on a Rotating Flat Earth?

Title: Would a Foucault Pendulum Work on a Rotating Flat Earth? Category: physics

A Foucault pendulum reveals rotation because its swing direction remains approximately fixed in inertial space while the ground beneath it turns. That principle does not require a spherical surface. On an ideal flat surface rotating about an axis perpendicular to the plane, the pendulum would still show precession relative to the ground.

The predicted rate would differ from the familiar result on Earth. On the ideal flat surface, the swing plane would appear to turn at approximately the full rotation rate, in the direction opposite the surface’s rotation. On a spherical Earth, the rate depends on latitude.

Why the pendulum precesses

In a nonrotating frame, a small-angle pendulum swings back and forth in an approximately fixed plane. An observer standing on rotating ground sees that plane change direction relative to their surroundings. In the rotating observer’s frame, this effect is described by the Coriolis force.

For a pendulum whose local vertical is aligned with the rotation axis, after expanding about its equilibrium and neglecting centrifugal corrections of order Ω², the horizontal equations in the rotating frame can be written as

$$\begin{aligned} \ddot{x}+\omega_0^2x &= 2\Omega\dot{y},\\ \ddot{y}+\omega_0^2y &= -2\Omega\dot{x}, \end{aligned}$$

Here $x$ and $y$ are horizontal coordinates, $\omega_0$ is the pendulum’s small-oscillation angular frequency when rotational corrections are neglected, and $\Omega$ is the platform’s rotation rate. These equations describe an oscillation whose direction slowly turns relative to the platform at rate $\Omega$, opposite the platform’s rotation. The Coriolis acceleration contains a factor of two, but the resulting rotation rate of the swing plane is not twice the platform’s rate.

More generally, the Foucault precession depends on the component of the rotation vector along the local vertical. Its magnitude is

$$\omega_F = |\boldsymbol{\Omega}\cdot\hat{\mathbf{z}}|,$$

where $\hat{\mathbf{z}}$ points along the local vertical. The sign determines which way the plane appears to turn.

Flat surface versus spherical Earth

On the usual idealized rotating flat surface, the spin axis is perpendicular to the surface and the local vertical is parallel to it. The whole rotation vector then contributes, giving $\omega_F=\Omega$. The swing plane completes one apparent turn relative to the surface during one rotation of the surface, in the opposite direction.

For Earth, the local vertical is not generally parallel to Earth’s rotation axis. At latitude $\phi$, the vertical component of Earth’s angular velocity is $\Omega\sin\phi$, so the magnitude is

$$\omega_F = \Omega|\sin\phi|.$$

The rate is zero at the equator and greatest at the poles. It reverses direction between the Northern and Southern Hemispheres. This latitude dependence follows from the orientation of Earth’s rotation axis relative to the local vertical on a sphere.

Worked comparison

Suppose a hypothetical flat surface rotates once per sidereal day, about 23 hours 56 minutes. Its angular speed is approximately

$$\Omega = \frac{360^\circ}{23.93\ \text{hours}} \approx 15.04^\circ\text{ per hour}.$$

With the rotation axis aligned with the pendulum’s local vertical, the swing plane would appear to drift at about $15.04^\circ$ per hour in the opposite direction. It would take one sidereal day to complete a full turn relative to the surface.

For comparison, an Earth-based pendulum at latitude $45^\circ$ has a precession rate of about $15.04\sin(45^\circ)\approx10.64^\circ$ per hour. Its plane takes about 33.9 hours to complete a full turn relative to the ground.

What the result would—and would not—show

A pendulum on the hypothetical surface could detect that the surface rotates. But that observation would not, by itself, reveal whether the surface is flat or spherical: the pendulum measures rotation relative to inertial space. To infer the shape of a world, you would need other measurements as well.

There is also a physical caveat to the idealization. A real rotating disk’s centrifugal effect points outward and can change the direction of a freely hanging pendulum’s effective vertical, especially far from the center. The simple full-rate prediction assumes this tilt is negligible or that the restoring field is arranged to keep the pendulum’s vertical aligned with the spin axis. A real flat object’s gravity would also not generally provide the uniform, perpendicular restoring field assumed in the thought experiment.

Common mistakes

• Assuming the effect requires a sphere: It does not. The essential ingredients are a pendulum and a rotating support.
• Using Earth’s latitude formula unchanged: The factor $\sin\phi$ describes how Earth’s rotation axis projects onto the local vertical on a sphere. For an ideal flat surface with its normal aligned to the spin axis, the factor is 1.
• Saying the pendulum plane physically spins: It is the plane’s direction relative to the rotating ground that changes. In the ideal approximation, the swing direction remains approximately fixed in inertial space.
• Treating precession as proof of shape: Precession is evidence of rotation, not a standalone test of surface curvature.