Foucault Pendulum Release Velocity in Spherical Coordinates

[zanick]

If the pendulum on the North pole is released to start it via burning the separation string, (to avoid adding an additional tangental force), wouldn't the pendulum at the time of release, already have a tangential velocity? does that effect the outcome of the experiment ? since the plane of movement is supposed to be fixed in space (an inertial frame of refence), if it has a some velocity at release as it would, should be considered , correct?

 

[vanhees71]

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!

 

[zanick]

The point here is, how to we give credibility to this experiment when those that might not understand it, will object to the error possibly being in one of the control variables. now, my knee jerk answer might be that the tangential velocity is so small and would only add a path shape variant and would not effect the fixed plane motion by adding precession to it. With both being very slight, it is expected to object to this as being a possible reason for the apparent precession around the 360 degree of position points.

 

[mitochan]

Hi.
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.

 

[zanick]

that's what I was thinking, but what if you don't... can it add the amount of precession to match the Earth rotation? (or some factor.) doesn't seem like other tests completed have seen any issues with extreme errors.

 

[mitochan]

Hi.
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.

 

[jbriggs444]

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.

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.

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.

 

[mitochan]

Thanks for advice on the case of my previous post.

I think Lagrangean of the case is
L=\frac{ml^2}{2}(\dot{\theta}^2+\dot{\phi}^2sin^2\theta)+mgl \ cos \theta
where parameters are those of polar coordinates and gravity accerelation works +z direction. It seems tiresome to get solution.

 

[mitochan]

Correction: spherical coordinates