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zanick

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- Thread starter zanick
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zanick

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Of course, seen from an observer in the rest frame of the center of the Earth (which can be taken as an inertial frame for this purpose), the pendulum has a tangential velocity since it has none in the rest frame of the observer on Earth. This you have to take into account whenever you want to solve the Foucault-pendulum equation of motion in the center-of-Earth inertial frame. Interestingly enough, I've never seen a solution of the problem in this frame. Maybe it's worth while studying it.

Note that in general due to the centrifugal force in the surface-of-the-Earth frame the stable equilibrium position of the pendulum is not along the direction of the Earth's radius through the point where the pendulum is fixed either, though in your special case where the fixed point is above the North Pole that's the case!

Note that in general due to the centrifugal force in the surface-of-the-Earth frame the stable equilibrium position of the pendulum is not along the direction of the Earth's radius through the point where the pendulum is fixed either, though in your special case where the fixed point is above the North Pole that's the case!

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- #3

zanick

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- #4

mitochan

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On North or South pole, set the ceiling rotating at rate of 1 cycle/24H with reverse direction against the Earth rotation around the pole axis ,i,e, still in a inertial reference frame. Set the end of separation string on the ceiling and burn it. I hope it would help you to solve your case.

- #5

zanick

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- #6

mitochan

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Say a pendulum in a reference frame of reference. It moves to and fro in a plane. Let it have initial velocity tangent to the plane when it is at the extreme point. How this initial tangential velocity changes the motion? This motion in the frame of reference easily would be interpreted in a rotating frame of reference. I set the question to investigate how tangential speed effects but have not investigated it by myself.

- #7

jbriggs444

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The situation is most easily analyzed from the inertial frame. We have an object moving under the influence of a field that goes as ##F=-k\vec{r}##. This force law can equally well be broken down into components: ##F_x=-kx## and ##F_y=-ky##. The forces in the two component directions are independent and give rise to a simple differential equation. The solution to the equation is simple harmonic motion.Say a pendulum in a reference frame of reference. It moves to and fro in a plane. Let it have initial velocity tangent to the plane when it is at the extreme point. How this initial tangential velocity changes the motion? This motion in the frame of reference easily would be interpreted in a rotating frame of reference. I set the question to investigate how tangential speed effects but have not investigated it by myself.

Since the solution for x and the solution for y are independent, periodic and share the same period. It follows that the orbit is a closed path. It does not precess -- assuming zero drag and the ideal ##F=-k\vec{r}## force law.

- #8

mitochan

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Thanks for advice on the case of my previous post.

I think Lagrangean of the case is

[tex]L=\frac{ml^2}{2}(\dot{\theta}^2+\dot{\phi}^2sin^2\theta)+mgl \ cos \theta[/tex]

where parameters are those of polar coordinates and gravity accerelation works +z direction. It seems tiresome to get solution.

I think Lagrangean of the case is

[tex]L=\frac{ml^2}{2}(\dot{\theta}^2+\dot{\phi}^2sin^2\theta)+mgl \ cos \theta[/tex]

where parameters are those of polar coordinates and gravity accerelation works +z direction. It seems tiresome to get solution.

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- #9

mitochan

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Correction: spherical coordinates

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