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This is the setup of my Foucault pendulum:
Foucault pendulum

Assume we have a Foucault pendulum suspended in the north pole. the pivot is just above the the axis of rotation (allowing the pendulum's plane to be in any direction relative to the apparatus). Before it starts swinging, the pendulum is attached to one of the supporting bars (the black lines) by a small string. To begin the motion of the pendulum you burn that string (this is the mechanism Foucault used).

While the pendulum is connected to the supporting bars it is moving with the apparatus, thus possessing both a velocity and an acceleration, the moment it is released, it has an initial velocity in the direction of rotation. what happens to this velocity?? All explanations and illustrations I see, assume from the beginning that the plane of the pendulum in the north pole is fixed relative to the "fixed stars" (I am aware of the philosophical debates), somehow neglecting that initial velocity.

I guess that I lack the understanding of the pivot mechanism itself, so if any one can shed light on my question and on the mechanism of the pivot, I would be very Thankful!

Edit:

I think I'm not so clear about my question and explanation of the situation. I found an incredible wiki article: Foucault pendulum Under this section: Relative motion of the plane of the pendulum swing to the surface of the earth, at the north pole:

Once the bob is displaced from the central axis of the pendulum and then released there no longer is a force acting on the bob that causes it to revolve about the central axis of the pendulum and rotate (turn) with the Earth. As observed from an end-view of the swinging bob, the swing of the bob will always line up or swing towards one star (just like the axis of the Earth points at one star for the time periods considered) as the bob swings through the central axis of the pendulum. There can be a slight ellipsoid swing if the initial conditions of angular motion are not cancelled but there is no longer a force acting on the bob causing it to have an angular velocity after the bob is released. The plane of the swing of the pendulum bob is now independent of the surface of the earth which was imparting a force to the bob before it was released (through the holding point). As noted previously, the bob is still spinning with the Earth (a spot of the bob will spin with the Earth), even though the bob is no longer turning with the Earth. Thus the Earth continues to turn underneath the swing of the pendulum while the swing of the pendulum remains in a fixed plane that doesn't rotate (turn). The point of significance is that the force imparting an angular velocity to the pre-released bob is no longer acting on the swinging bob. At the North Pole, this force takes one day for the direction of the force to complete a full circle since it takes one day for the Earth to rotate.

The initial angular momentum is clearly mentioned here, and it should be cancelled, how did Foucault cancel it?

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Suppose the Earth wasn't rotating, but instead you impart a small sideways velocity to the pendulum bob as you release it. What you would have is a conical pendulum that traces out an ellipse instead of a straight line.

Now start the Earth rotating. The point of the Foucault's pendulum is that the rotation of the Earth doesn't affect the motion of the pendulum, so the pendulum behaves exactly as it did in the thought experiment we started with. The bob traces out a (very narrow) ellipse, and the axes of the ellipse rotate relative to the Earth.

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  • $\begingroup$ That thought experiment is a great way to look at it, thank You, I think it is clear now! $\endgroup$ – Daniel s Aug 27 '14 at 16:32
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Nothing happens to the initial angular momentum. It is simply irrelevant.

Let's imagine that the pendulum is placed on a non-rotating body (or at the earth's equator). When the string is cut, the pendulum falls straight to the center. The path it traces is a straight line.

In the non-equatorial case, the pendulum structure has some rotation. At the time of release, there is some tangential speed. This causes the path that that the pendulum traces to be an ellipse instead of a line. The pendulum does not go through the center.

But even if there is a small ellipse, it is minimal and does not change the experiment. You can still easily determine the long axis of this ellipse and observe how the axis changes over time. So the initial velocity changes the shape of the path, but cannot cause the path to change later (which is what is being observed).

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  • $\begingroup$ I would mark this as the answer too if I could, thank You! do you know whether the first foucault's pendulum had this "minimal" elipse? $\endgroup$ – Daniel s Aug 27 '14 at 16:34
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You do have to define the velocity with respect to something. The idea (I think) behind defining wrt fixed stars is that the Earth is in motion wrt the stars, too. Barring friction, the moment the rope is burned, the pendulum is only under the influence of gravity and the restraining force of the suspension rope -- analogous to swinging a weight around yourself and letting go of the string. The pendulum mass (undoubtedly held up with a massless, inextensible string connected to a frictionless u-joint :-) ) will follow the standard Newtonian force laws, so just add its initial velocity vector to whatever gravity causes.

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    $\begingroup$ Thanks for the comment, I'll try to refine my question. Lets forget about the stars, assume there is absolute space and the axis of rotation of the earth is fixed in space. we analyse it in the axis f.o.r: The whole apparatus is revolving around the axis, the pendulum is not just under the influence of gravity, it is also accelerating and moving around the axis, at the momemnt the rope is burned it posseses a velocity w.r.t to the axis, i.e w.r.t to "absolute space"(or the stars that we forgot), unlike any illustration I've seen on the internet.. Didn't I understand you Correctly? $\endgroup$ – Daniel s Aug 27 '14 at 11:54
  • $\begingroup$ I think so :-) . The stars are just considered a fixed reference to define a static set of axes. $\endgroup$ – Carl Witthoft Aug 27 '14 at 13:17
  • $\begingroup$ I know that... What I don't undertsand is why in any illustration and explanation of the pendulum in the north pole, the initial velocity(w.r.t to the static axes) of the pendulum is not mentioned.. just like you've said - swinging a weight around you and letting it go, it has an initial velocity w.r.t to you... while the pendulum is connected to the supporting bar it spins with the earth and has a velocity w.r.t to the axis, when the rope is burned it has velocity w.r.t to the axis(never seen an illustration where it is mentioned).. $\endgroup$ – Daniel s Aug 27 '14 at 13:33

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