Why doesn't the pendulum's swing plane rotate at the equator?

At the equator, the horizontal component of the Coriolis force—which is responsible for the sideways push on the pendulum bob—vanishes. The formula Ω_eff = Ω_E sin|λ| shows that when latitude λ is 0°, sin(0°) = 0, resulting in zero effective precession. The Coriolis force still acts vertically at the equator, affecting objects moving up/down, but it does not deflect the horizontal swing of a pendulum.

Is the rotation of the swing plane proof that Earth rotates, and not that the universe rotates around us?

Yes. The pendulum's behavior is most simply explained by assuming we are observing it from a rotating reference frame (Earth). In an inertial frame (like the distant stars), the pendulum's swing plane remains fixed while Earth rotates beneath it. The pendulum's motion is consistent with Newton's laws only if we include the fictitious Coriolis force, which is a direct consequence of our frame's acceleration (rotation).

What simplifications does this simulator make compared to a real Foucault pendulum?

This model ignores air resistance and damping, so the swing amplitude does not decay. It also treats the pendulum as a simple planar swing, neglecting the more complex 3D spherical pendulum motion that can cause elliptical orbits. Furthermore, it assumes a perfectly symmetric setup and a point-mass bob, whereas real pendulums require careful design and a driving mechanism to sustain oscillation.

How does the precession rate change if I move the pendulum from the North Pole to a mid-latitude?

The precession rate decreases as you move away from the pole. At the North Pole (λ=90°), the rate is maximum at 15°/hour. At a latitude of 45°, sin(45°) ≈ 0.707, so the rate is about 10.6°/hour. The sine function in Ω_eff = Ω_E sin|λ| captures this dependence, showing the precession is slower at lower latitudes until it reaches zero at the equator.

Foucault Pendulum (Sketch)

A Foucault pendulum provides a direct, visual demonstration of Earth's rotation. This simulator models the pendulum's motion from a top-down perspective, focusing on the apparent rotation of its swing plane. The core physics principle is the Coriolis effect, an inertial force that arises in rotating reference frames. For a pendulum swinging on Earth's surface, the effective precession rate of its swing line is given by Ω_eff = Ω_E sin|λ|, where Ω_E is Earth's angular rotation rate (approximately 15°/hour or 7.29 × 10⁻⁵ rad/s) and λ is the latitude. At the North or South Pole (λ = ±90°), sin|λ| = 1, and the swing plane rotates a full 360° in one sidereal day. At the equator (λ = 0°), sin|λ| = 0, and no rotation occurs. The simulator simplifies the real-world system by ignoring factors like air resistance, the spherical shape of the pendulum bob's path (spherical pendulum effects), and imperfections in the suspension. It assumes a perfectly symmetric, ideal pendulum on a uniformly rotating Earth. By interacting with the controls for latitude and time scale, students can observe how the Coriolis force magnitude changes with location and watch the swing line precess, solidifying their understanding of non-inertial reference frames and the tangible evidence for a rotating Earth.

Who it's for: High school and introductory undergraduate physics students studying rotational motion, inertial forces, and classical mechanics in a non-inertial frame.

Key terms

Foucault Pendulum
Coriolis Effect
Non-inertial Reference Frame
Precession
Angular Velocity
Latitude
Earth's Rotation
Classical Mechanics

How it works

Real Foucault motion is subtle on a rotating Earth; the textbook precession rate of the swing plane is Ω = Ω_E sin λ (latitude λ), Ω_E ≈ 7.29×10⁻⁵ rad/s. This sim is a didactic cartoon: the bob follows A cos(ω_p t) on a horizontal circle whose diameter line rotates at Ω_eff = Ω_E sin|λ| times a time multiplier so you can see a turn in minutes. Not a full Coriolis integration in 3D.

Key equations

Ω_F ≈ Ω_E sin λ · ω_p = √(g/L) (small-angle)

Frequently asked questions

Why doesn't the pendulum's swing plane rotate at the equator?: At the equator, the horizontal component of the Coriolis force—which is responsible for the sideways push on the pendulum bob—vanishes. The formula Ω_eff = Ω_E sin|λ| shows that when latitude λ is 0°, sin(0°) = 0, resulting in zero effective precession. The Coriolis force still acts vertically at the equator, affecting objects moving up/down, but it does not deflect the horizontal swing of a pendulum.
Is the rotation of the swing plane proof that Earth rotates, and not that the universe rotates around us?: Yes. The pendulum's behavior is most simply explained by assuming we are observing it from a rotating reference frame (Earth). In an inertial frame (like the distant stars), the pendulum's swing plane remains fixed while Earth rotates beneath it. The pendulum's motion is consistent with Newton's laws only if we include the fictitious Coriolis force, which is a direct consequence of our frame's acceleration (rotation).
What simplifications does this simulator make compared to a real Foucault pendulum?: This model ignores air resistance and damping, so the swing amplitude does not decay. It also treats the pendulum as a simple planar swing, neglecting the more complex 3D spherical pendulum motion that can cause elliptical orbits. Furthermore, it assumes a perfectly symmetric setup and a point-mass bob, whereas real pendulums require careful design and a driving mechanism to sustain oscillation.
How does the precession rate change if I move the pendulum from the North Pole to a mid-latitude?: The precession rate decreases as you move away from the pole. At the North Pole (λ=90°), the rate is maximum at 15°/hour. At a latitude of 45°, sin(45°) ≈ 0.707, so the rate is about 10.6°/hour. The sine function in Ω_eff = Ω_E sin|λ| captures this dependence, showing the precession is slower at lower latitudes until it reaches zero at the equator.

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