How does Foucault's pendulum show the earth's rotation?

[helpmeplz!]

As you may know, Foucault's pendulum is an easy way to verify that the Earth is rotating about its axis. Its a pendulum that is free to swing in any direction. Since Earth rotates under it, the position of the pendulum with respect to the ground changes after some time. So you could put pins on the ground to show that the pendulum knocks down pins at different locations as time goes by.

But my question is this, isn't the pendulum also rotating at the same rate and in the same direction as the earth? Because the pendulum is attached to a frame in some building, and so if Earth rotates, so does the building and so the pendulum does too? Wouldn't that mean that with respect to other objects in the room, the pendulum won't really move and so the pins shouldn't fall?

 

[russ_watters]

What force would make the pendulum rotate? It is held by a wire, not a rod.

 

[helpmeplz!]

What force would make the pendulum rotate? It is held by a wire, not a rod.

Well the wire is held on to the ceiling, and the ceiling is connected to the building which is on the earth. And so when Earth moves, the building moves, and so does the ceiling and so does the wire?

Oh I see, when the pendulum is first given a velocity, that velocity + velocity due to Earth's rotation - velocity due to Earth's rotation means its initial velocity is unchanged and so the effects can be seen. Is this true?

 

[russ_watters]

Well the wire is held on to the ceiling, and the ceiling is connected to the building which is on the earth. And so when Earth moves, the building moves, and so does the ceiling and so does the wire?

But held how? The wire is flexible, so the pendulum can swing in any direction or change direction without a force that would tend to keep it in one plane.

 

[helpmeplz!]

But held how? The wire is flexible, so the pendulum can swing in any direction or change direction without a force that would tend to keep it in one plane.

But if Earth and wire moved at different rates, wouldn't the wire in 24 hours be left way behind and eventually hit the sides of the building?

 

[russ_watters]

Behind what? And the building is built far enough away that the pendulum can't hit it. I'm not sure what you are getting at.

 

[helpmeplz!]

Behind what? And the building is built far enough away that the pendulum can't hit it. I'm not sure what you are getting at.

Do you agree that the pendulum and the Earth are moving initially at the same rate? Then when the pendulum is about to swing, it has a velocity = Earth's velocity. As it swings, it gets an additional velocity and now moves at a different rate than the ground (earth) does right? So that's the reason it hits the pins on the ground. Am I correct?

 

[A.T.]

I'm not sure what you are getting at.

I think he makes a valid point. It is often explained that a pendulum at the Pole would swing within a fixed plane, as seen from an inertial frame of reference. But if you release the pendulum at the highest point at rest relative to the ground, then in the inertial frame it has an initial tangential velocity. So it doesn't swing within one fixed plane in the inertial frame of reference, but in two fixed planes. In the primary plane it has a large amplitude. In the secondary plane, which is perpendicular to the primary one, it has just a tiny amplitude. But both planes are still fixed in the inertial frame of reference, so they do indicate the Earth's rotation.

 

[russ_watters]

Do you agree that the pendulum and the Earth are moving initially at the same rate?

No, I don't. The pendulum is swinging from side to side, the Earth is rotating. They aren't even moving in the same way, much less at the same rate.

Then when the pendulum is about to swing, it has a velocity = Earth's velocity. As it swings, it gets an additional velocity and now moves at a different rate than the ground (earth) does right? So that's the reason it hits the pins on the ground. Am I correct?

No, the pendulum moves from side to side, while the Earth rotates. Two different motions.

 

[russ_watters]

I think he makes a valid point. It is often explained that a pendulum at the Pole would swing within a constant plane, as seen from an inertial frame of reference. But if you release the pendulum at the highest point at rest relative to the ground, then in the inertial frame it has an initial tangential velocity. So it doesn't swing within a constant plane in the inertial frame of reference.

If the pendulum is released with a velocity component of revolution, it won't swing through any plane, but it will swing through a large ellipse; It won't quite swing through the center of its axis, but the plane that bisects the ellipse will still rotate. And if it is released without the revolution component being perfectly zero, it will be so close to zero that it will get damped-out quickly.

You can't release it in such a way that the tangential velocity component is big enough to matter; you can't make that tangential component cause circular rotation because it is way too slow.

 

[sophiecentaur]

I could point out that the mass on the end of the pendulum is rotating, initially, around a vertical axis with a period of 24 hours nothing about that will change over time. There is not even a net torque on the system to affect its net, 'independent' motion about the CM of the Earth..

 

[A.T.]

If the pendulum is released with a velocity component of revolution, it won't swing through any plane, but it will swing through a large ellipse;

This 3D motion can be decomposed in two 2D motions within two perpendicular planes. The key is that the pendulum reaches its maximal height at two points, which remain in a plane that is fixed in the inertial frame.

the plane that bisects the ellipse will still rotate.

In which reference frame will that plane rotate?

it will be so close to zero that it will get damped-out quickly.

There is no need to invoke damping, if the issue can be explained in principle, so it applies for an ideal pendulum too.

You can't release it in such a way that the tangential velocity component is big enough to matter; you can't make that tangential component cause circular rotation because it is way too slow.

There is no need to invoke the size of the effect in the case of the Earth, if the issue can be explained in principle, so it applies in general. As long as the path in the of the bob is not a perfect circle, but somewhat elliptical, you can use it to indicate the rotation of the ground.

 

[Sidhardha]

Assume that the pendulum oscillates in one fixed plane as it would on equator. Now when the pendulam is at the poles, the forces on the bob when it reaches one extreme position are (i) one due to gravity downwards and (ii) one perpendicular to the first force and in the horizontal direction which tries to push the bob out of the plane,due to the spin of the earth.
When the bob reaches the other extreme same two forces act in the same way. So now we have two pairs of forces- two downward forces due to gravity and two oppositely acting forces due to spin. The pair of two oppositely acting spin forces constitute a couple which gradually turn the plane of rotation of the pendulum. ie. The pendulum always oscillates in a plane, but the plane slowly spins about a vertical centre pole.

 

[AlephZero]

I'm not sure what you are getting at.

It's not quite so simple as text-books make out.

Suppose the pendulum is on the equator and swinging north-south. OK, I know that situation will NOT show the rotation of the earth, but ... after 12 hours, the support at the top of the pendulum has moved round 180 degrees relative to the swinging mass, because if not the pendulum would be swinging upside down!

Obviously the reason for this has to do with the fact that the weight of the pendulum always acts towards the center of the earth, but I don't criticize the OP for feeling that a textbook explanation might have skipped over some of the details.

If you just think about it without doing the math, it's not obvious (at least to me) why the pendulum would be still swinging in a "straight line" north-south after 6 hours, and not swinging around in a circle instead...