Our Foucault pendulum rotated the wrong direction - why?

[Oystein]

You can, if you like, entirely forget about the earth's movement and work in a frame where it is not rotating. In that frame there is a Coriolis force, a force proportional to mass and the cross product of the velocity with a fixed vector (the rotation axis of the earth). That's the force responsible for the precession of the plane of the pendulum.

When you released the pendulum, if you gave it a push that wasn't towards the center, as 69dodge said the pendulum would then move in an ellipse (at least approximately). With no friction, I think Coriolis force would simply cause the major and minor axes of that ellipse to precess - although I haven't checked. But I think that's the only thing giving it a push would do - it could not cause the pendulum to swing in a plane and that plane to precess the wrong way.

Ah - thanks, yes, that sounds like it makes sense.

I suppose if you didn't notice that the motion was ellipsoidal you might make a measurement error that could make you think the precession was opposite to what it really was (you could measure the direction of motion along two different parts of the ellipse). Could that have been it? 10 degrees isn't all that much.

Don't nail me on the 10 degrees. Like I said, I don't recall the exact duration of our experiment, and we didn't precisely measure the precession, 10 degrees is ballpark. If it was only 8° or 7°, that wouldn't change my story.

 

[Oystein]

casebro has that backwards - east/west motion maximizes the force.

Really?? I was surprised that there is a difference, but if there is one, intuition tells me that n-s should maximize the force, as there is a greater difference between the radius of earth between the north and south ends of my path than between east and west.

Explanation?

 

[sol invictus]

Really?? I was surprised that there is a difference, but if there is one, intuition tells me that n-s should maximize the force, as there is a greater difference between the radius of earth between the north and south ends of my path than between east and west.

Explanation?

Imagine a pendulum swinging in a N-S plane at the equator. As the earth rotates, the plane will remain N-S - there's zero Coriolis force. Technically speaking the force is mass times the cross product of the velocity with the rotation vector of the earth, and that cross product is zero for N-S motion on the equator.

Away from the equator it's not zero, but it's suppressed by the sine of the latitude relative to the force that would result from the same speed E-W (the N-S force is equal in magnitude to the E-W force only at the poles).

Come to think of it though, for a Foucault pendulum we don't care about the overall magnitude of the force, we only care about the component of it that is responsible for the precession (i.e. not vertical). A pendulum swinging E-W at the equator would have Coriolis force, but it would be vertical, so it also wouldn't cause precession. So now I'm unsure whether I'm right in the general case all things considered - and don't have the time right this moment to work it out.

My guess - but it's just a guess for the moment - is that these effects cancel each other, and the relevant part of the Coriolis force is independent of direction (i.e. a Foucault pendulum precesses through 360 degrees at a constant rate).

 

[Beanbag]

The best way to "Start" a foucault pendulum is to tie a thread to the pendulum bob, pull the bob to one side, and tie the thread to a stationary anchor. Wait until the system is free from any vibration or oscillation, then use a small flame (like a match or candle) to burn through the thread without touching anything. This usually guarantees a smooth start right through the center of oscillation.

Scientific American magazine's The Amateur Scientist column had a bunch of information and how-to's for foucault pendulums from the 50's and 60's. It might be worth a visit to the library's bound periodicals section to do some reading.

Beanbag

 

[69dodge]

Really?? I was surprised that there is a difference, but if there is one, intuition tells me that n-s should maximize the force, as there is a greater difference between the radius of earth between the north and south ends of my path than between east and west.

Explanation?

Yeah, something moving east or west also experiences Coriolis force. There doesn't have to be a difference in the distance to the Earth's axis. You can think of the Coriolis force in that case as an increase or decrease in the centrifugal force. For example, if you're moving east, you're rotating faster than the Earth, so there's a greater force than usual pushing you outwards, i.e., partly up and partly south.

 

[GodMark2]

Imagine a pendulum swinging in a N-S plane at the equator. As the earth rotates, the plane will remain N-S - there's zero Coriolis force. Technically speaking the force is mass times the cross product of the velocity with the rotation vector of the earth, and that cross product is zero for N-S motion on the equator.

Actually, any pendulum at the equator experiences no precession. N-S, E-W: doesn't matter. The equator is the one special case that makes the math go bonkers (if you'll excuse the technical language).

The other extreme (at the N/S pole) is the easiest, the earth turns, and the pendulum stays in the same inertial plane. In-between, you get a combination of those two outcomes (much like any oscillator).

Look at it in the inertial frame, and you'll see that the strength of the effect is independent of the direction of the initial plane of swing. It is dependent only on the mass involved (like all fictitious forces), and it's latitude in the spinning reference.

 

[sol invictus]

Actually, any pendulum at the equator experiences no precession. N-S, E-W: doesn't matter.

Yes, I pointed that out in the post you quoted. To first approximation there will be no Coriolis force on a N-S swinging pendulum at the equator. There is Coriolis force on an E-W pendulum at the equator, but it is directed vertically and therefore will not cause precession.

Look at it in the inertial frame, and you'll see that the strength of the effect is independent of the direction of the initial plane of swing. It is dependent only on the mass involved (like all fictitious forces), and it's latitude in the spinning reference.

That's not particularly obvious in either frame, at least not to me. However it does turn out to be true at least approximately, because the decrease in the magnitude of the force on a N-S swinging pendulum is compensated for by the increase in the component perpendicular to the plane of the swing.

 

[Oystein]

Guys, my head is spinning, I think I'll let something oszillate inside while I sleep over it

Thanks for the replies so far. I will dig out this thread in 11-12 days, after my return from a skiing vacation.

 

[Beerina]

ETA:
I think it did not run long enough. If it had enough angular momentum to go the wrong way, it would take a long time to disipate.

That's the first thing that pops into my mind. Also, if it's lightly powered, it could be pushing ever so slightly in the wrong way. Just a little off-center should vastly dominate the natural swing wobble.

 

[Meridian]

That's not particularly obvious in either frame, at least not to me. However it does turn out to be true at least approximately, because the decrease in the magnitude of the force on a N-S swinging pendulum is compensated for by the increase in the component perpendicular to the plane of the swing.

Isn't it clearer to express the `compensation' as follows: the Coriolis force is Omega cross v. In the first order approximation where v is always horizontal, the horizontal component of Omega contributes a vertical force which is (to first order) irrelevant. What remains is (vertical component of Omega) cross v, which has magnitude a constant times v and is always horizontal and orthogonal to v, so the rate of rotation is constant. In other words, write it as the vector sum of two effects, one of which is irrelevant to first order and the other is obviously rotationally symmetric (within the `lab' frame).

 

[Dancing David]

How so? How signifficant would that be?
We constructed the pendulum in the main building of our school, which features a 3-floors-high atrium, with classrooms and halls all around on the 3 floors. We attached a little construction featuring a horizontal steel beam, maybe 1.5 meters long, to the handrail on the upper floor, with a steel wire hooked to its end, and an iron ball of some kilograms hanging on the wire 5-6 meters below. When the pendulum swung, the beam wasn't totally steady, it moved back and forth in the direction of the pendulum by some millimeters, I'd guess.



I don't remember anymore how exactly we released the pendulum, but let's assume we were careful, but not entirely precise. What is the significance of "me" (or whoever released the pendulum) standing outside the circle? I'd say if you stand inside, you'll get hit in short order!


Basically what I am looking for is the several ways you could tamper with the experiment, and what their respective influence and probability of reversing the rotation is. In particular, can someone explain to me this thing about starting with the ball at rest relative to the floor on the edge of the circle and thus with an angular velocity equal to (?) that of the floor.

Sounds very steady so not an issue, if you stand outside the circle and make a slanted release the off momentum will be higher.

Did you read the quote about vector velocity?

 

[NobbyNobbs]

It means you're having a girl. Mazel tov.

 

[Cuddles]

Guys, my head is spinning

Which direction?

 

[sol invictus]

Isn't it clearer to express the `compensation' as follows: the Coriolis force is Omega cross v. In the first order approximation where v is always horizontal, the horizontal component of Omega contributes a vertical force which is (to first order) irrelevant. What remains is (vertical component of Omega) cross v, which has magnitude a constant times v and is always horizontal and orthogonal to v, so the rate of rotation is constant. In other words, write it as the vector sum of two effects, one of which is irrelevant to first order and the other is obviously rotationally symmetric (within the `lab' frame).

Yes, that's simpler.