Foucault pendulums (or “pendula” if you prefer, but I will pluralize them as “pendulums”) are without doubt most intriguing constructions, but I have a problem visualizing how they actually work in locations other than at the poles and at the Equator.
I totally get that at the North Pole, a Foucault pendulum will describe a clockwise path, with a complete revolution each sidereal day … at the South pole it will behave similarly, except the path will be anticlockwise . No problems there, clear as crystal.
I also have no problem understanding that at the equator , a Foucault pendulum will not describe any path other than a straight line. … or at least I understand that it will if its initial oscillation is east-west. I am not sure that I fully grasp the mechanics of what happens if the initial swing is set north-south.
What I don’t get is how the pendulum traces out a circle when it is located on a latitude, say, half way between the pole and the equator. For the sake of argument let us assume that the initial oscillation is set east-west. I am fully aware that it **does **trace out a circle, taking longer than a sidereal day to do so, I just don’t visualize how it does so.
What I would really like to know is whether there is an animation anywhere on the net which shows how this works, because although I am normally pretty good at mental visualization, I just cannot get my head round this one.
I am sure that there will be an “ahh moment” at some point on this thread when the penny drops. Sooner rather than later, hopefully.
interesting article and video … it doesn’t quite get me where I want to be.
I think one of the stumbling blocks I have is not knowing whether the pendulum swing at locations other than the poles and at the equator is actually a constant plane.
At the poles, you could take 3 suitable stars as co-ordinates and define the path along which the pendulum swings as being on the plane of these three stars. AFAIK the pendulum maintains that same plane throughout a complete rotation of the planet. Similarly at the equator (leastways as long as the path is east west).
What I don’t quite grasp is whether the same applies to the pendulum path when located at different latitudes. Does the path still lie in one fixed plane throughout the earth’s rotation, a plane which could be defined by three extra-terrestrial coordinates ?
It may help to think about what would happen if you start at the pole, then perform the same experiment fifty paces away from the pole (it’s going to be very similar, isn’t it?), then perform the same experiment half a mile from the pole, etc. There’s not going to be an abrupt point where the effect suddenly cuts off.
Working out the equations for the Foucalt pendulum at random latitudes is nontrivial. I recall that my freshman physics book, Kleppner and Kolenkow’s Introduction to Mechanics, did so. You can also look it up online, such as here:
Fascinating stuff, Cal …but she lost me as soon as she said “transformation matrix” … and I don’t understand her co-ordinate system (my fault, not hers, I know).
What I need is more an “Idiot’s Guide to the Foucault Pendulum”.
When the pendulum is first set in motion , the path described by the center of the bob and the center of the suspending cord is a plane which can be defined by three coordinates, these being:
1:the point from which the pendulum is suspended
2: the center of gravity of the earth
3: an arbitrarily chosen extraterrestrial coordinate … obviously a star would be a logical choice. The pendulum bob is “aimed” at the star when first set in motion, so to speak.
My question is: does the pendulum maintain its motion along this plane during the rotation of the earth, irrespective of where on earth it is located, and irrespective of the initial direction of motion, be it north/ south or east/ west.
I do get that the polar and equator locations are special cases, and (for me) much easier to visualize.
The math for the time it takes to describe the circle as a function of latitude is given here. An intuitive explanation for how this works is a little trickier but I think Mangetout makes a helpful observation when he says to think about it in terms of the behavior at the north pole and then what happens when you move south.
The key word here is “precession”. I don’t know if this is going to be helpful, but think about it intuitively in terms of the fact that when a pendulum is suspended at the pole, the point of suspension is rotating. Imagine a solid post planted at the north pole – it will rotate with the earth. Such a post – or such a point of suspension for a pendulum – planted at the equator will not rotate at all. The rotation of the earth carries the pendulum along but it doesn’t turn against its plane. And such a post planted elsewhere between a pole and the equator will rotate in lesser amounts as a function of latitude. This rotation is a way to visualize the precession. The plane of the pendulum swing prefers not to rotate, so the plane appears to move relative to the earth as a function of latitude.
A bit of a nitpick, but it’s really no more an “abrupt” cut-off than most other physical phenomena. The rate of precession approaches zero as you approach the equator, and then goes negative as you continue past. It will be exactly zero only when you are at the equator with infinite precision, which will never happen.
I was envisaging the suspension point as a highly lubricated ball inside a bearing … as near frictionless as humanly possible.
No answer yet to my question about the constancy of the plane?
I might as well throw in another supplementary question as well … as I said earlier, I do intuitively understand the mechanics of the pendulum behavior at the poles and also at the equator … as long as the equatorial pendulum is swinging in an east/west direction.
I am sure my intuition is in fact wrong, but it tells me that if the suspension cord were long enough, and the duration of oscillation were long enough, then, if the initial direction of the equatorial pendulum were north/south, that eventually the point of the bob would start to describe an elliptical path rather than a straight line.
Please confirm I am wrong (but don’t explain the math … it will go over my head).
Happy to confirm that you’re wrong! Where do you see the elliptical path coming from, and what difference would it make what direction the pendulum was swinging? If it was positioned with its center point exactly at the equator, there would be zero rotation around its plane, it would simply be carried round and round. The plane would move in space, but it wouldn’t rotate relative to the earth, hence it would keep tracing out the same straight line.
If I correctly understand your “constancy of the plane” question, the answer is: only at the poles (disregarding the orbit of the earth around the sun). At the poles the earth is rotating, the pendulum isn’t, and nor is it being moved laterally in a meridional direction as it would be everywhere else.
But are you aware that the straight line won’t be centered on the point below the fulcrum? (If my intuition serves me … please correct me if I’m wrong!)
I’m delighted to hear that, since I’ve wondered about this myself. While I can see how funny stuff starts happening as you move away from the poles and the equator, I’m far from able to figure out how to set up the equations, without a lot more work than I’m interested in doing for an armchair exercise!
BTW, the figure is never a perfect ellipse (except when it’s a line or a circle.) If you put an extensible pen tracing the path on a flat surface below the pendulum, it traces a Lissajous pattern. That is, it’s a rotating elipse. I’ve seen one of these, a huge pendulum tracing a furrow in sand IIRC, in some museum – might have been the Smithsonian. It was as a child in the 60’s and I can’t place it, but it did make an impression on me. At the time I really couldn’t figure it out intuitively, other than taking it for granted that it’s due to “the rotation of the Earth”.
I am certainly not going to be presumptous enough to correct you, but what my intuition tells me is that the the straight line would indeed be centered directly under the fulcrum …ie the point of suspension, the tip of the bob and the earth’s center of gravity would all lie in the same straight line (or plane).
Experience however, tells me that my intuition is often wrong …
I don’t have much to add except that the OP could have been written by me.
I have said the exact thing that I can easily imagine a Foucault pendulum on the poles or on the equator, but in-between is very hard to mentally visualize.
That and the Coriolis force. I have a hard time fully understanding that one too.
I am not talking about working out the math. I can work out the math and calculate forces, etc, but I don’t feel like I can really understand it unless I can see it in my head.
To get a really intuitive feel for the phenomenon, seek out a children’s playground roundabout of this type.
Start it turning, then jump on board and stand facing the central post (holding on to the rails). Try to kick the central post.
Your foot will tend to swerve alarmingly to one side of the post or the other - it’s actually really hard to kick it.
Your foot, the pendulum, is trying to move in a straight path (conservation of momentum), but within a rotating reference frame, so by the time it’s moved a bit (in a notionally straight line), all of the background has rotated out of place.
The same is true if you try to swing your foot when standing elsewhere on the disc - even though you are not trying to pass through the centre of rotation - the point you aim at isn’t there any more by the time you get to it.
However, at the equator, taking the orthogonal cases: the pendulum is either moving along the line of the equator (in which case the target may be higher up or lower down than it was at the start of the swing, but it’s not to the left or right) - or the pendulum is swinging parallel to the axis of rotation, in which case it’s all moving in what is essentially the same inertial reference frame, so although the target is moving to the right or left, so is the pendulum.
Non-orthogonal cases can be thought of as a blend of the two.
Interesting post, Mangetout. The equatorial pendulum could also swing at 45 degrees, or any other angle, to the axis of rotation and that is something which confuses my intuition.
Your children’s roundabout is interesting however.
Let’s assume that it is mechanized, so that it completes one revolution in, say, one hour.
Let us further imagine that we suspend a pendulum above the center axis, set it swinging and trace out its path. It will of course trace out a circle in one hour, analogous to the pendulum on the North or South pole.
Now, what happens if we take the same pendulum and mount it on the perimeter of the roundabout ? It will trace out a circle, but will it take longer to do so (analogous to the varying latitudes on the earth?)
If it does indeed take longer, then I am starting to see the light …if it doesn’t then I am back to square one.
It’s back to square one … i can figure out intuitively that the time taken for the perimeter pendulum to trace out a complete circle will be exactly the same as for the center pendulum.
That’s because the child’s playground roundabout thing is flat. The earth is a sphere.
Think about the “rotating at the pole”, “not rotating at all at the equator”, and “precessing in various amounts depending on latitude” argument I mentioned. It really did help me perfectly visualize it.
Where do you see the elliptical path coming from, and what difference would it make what direction the pendulum was swinging? If it was positioned with its center point exactly at the equator, there would be zero rotation around its plane, it would simply be carried round and round. The plane would move in space, but it wouldn’t rotate relative to the earth, hence it would keep tracing out the same straight line.