Understanding Foucault Pendulums

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A Foucault pendulum swings in a fixed plane that does not rotate with the Earth due to the absence of forces acting on it to cause rotation. When released from a displaced position, it exhibits elliptical motion rather than a straight line because of its initial sideways velocity relative to a non-rotating frame. The oscillation consists of two simple harmonic motions that are 90 degrees out of phase, resulting in a stable ellipse. However, the axes of this ellipse remain constant over time, leading to the pendulum appearing to rotate once per day relative to the Earth. Understanding the concept of reference frames is crucial to grasping this phenomenon.
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Simple question - I'm having trouble conceptually understanding why the plane that a Foucault pendulum swings in does not rotate with the earth. I understand how the Earth turns under the pendulum, but isn't the pendulum rotating with the Earth when it's released?
 
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As a simple case, consider the pendulum is at the north or south pole.

If you pull the pendulum away from the vertical, hold it steady relative to the earth, and then release it, you are right that in relative to the non-rotating reference frame, initially it has a "sideways" component of velocity equal to omega.x, where omega is the Earth's rotation speed and x is the displacment.

Because of that, it will oscillate by moving round a very thin ellipse, not back and forth in a straight line. Ignoring the finite length of the pendulum, the motion is just two simple harmonic oscillations 90 degrees out of phase, i.e. x = A cos pt and y = B sin pt where A is very much bigger than B.

But the axes of the ellipse won't rotate with time (relative to the non-rotating reference frame) because there are no forces to make them rotate.

So relative to the relative to the earth, the pendulum oscillates in almost a straight line, and that line rotates once per day.

Hope that helps.
 
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