Test to prove Earths spins:

Originally posted by sonhouse
But there would still be a force on the bob by the simple rotation of Earth I would think. There is a rotation in X direction around the equator and my bell jar is on the equator. I don't see how the bob on a 100 meter string could still point directly to the center of Earth, it seems there should be some deflection. I think the circular motion represents circular acceleration so why wouldn't the bob not deflect?

Consider the vectors in spherical coordinates. At the equator the vector along the radial direction has all the forces along it. The rate of change of azimuthal angle is just the angular velocity and more or less constant and because we've chosen coordinates so that the rotation is entirely around the azimuthal angle the other angle is unchanging. The only force at the equator (ignoring mountains) is directly down through the earth. Your set-up might work at 45 degrees of lattitude as the direction of the centrifugal force is perpendicular to the axis of rotation and in general not parallel to the direction of the gravitational force which is towards the centre of the Earth. For someone in the Northern hemisphere the effect should pull the bob southwards slightly. I'm not convinced by the experimental setup as it occurs to me that one has no really good way of establishing which direction is downwards with any kind of precision apart from by which way gravity pulls...

Originally posted by joe shmo
Its when the frame of reference is rotating ( specifically with a constant angular velocity I believe). I don't think it is when an object is rotating relative to another object. Even still...Is the Corialis force always zero at the equator for some reason that I'm missing? The mathematics of the ficticious force seem to indicate three possible scenarios ...[text shortened]... at the equator... So, since Deep Thought implied it, I must be wrong. I just still cant see it.

It's just a matter of getting your head around the coordinate change. This page on Wikipedia is probably a good place to start:

https://en.wikipedia.org/wiki/Fictitious_force#Rotating_coordinate_systems

Originally posted by joe shmo
Its when the frame of reference is rotating ( specifically with a constant angular velocity I believe). I don't think it is when an object is rotating relative to another object.

Yes, I didn't describe it too well.
I believe it involves an object moving relative to a rotating reference frame.

The answer is:
1) Ω=0 ...The Earth is not rotating (relative to an object at the equator).

The Coriolis force in the northern hemisphere is opposite to that in the southern hemisphere so it makes sense that there must be some point in between where it is zero.

Originally posted by DeepThought
The only force at the equator (ignoring mountains) is directly down through the earth.

And centrifugal force. But centrifugal force is a function of momentum along the axis of rotation so I am not certain that it works the same with a pendulum. The bob is trying to go straight off into space (at a tangent to the earths surface), but is being pulled back towards earth by gravity and back towards its anchor point by the pendulum wire. It seems to me it would tend to advance ahead of the earth.

Does anyone know whether or not a space elevator goes straight up or is angled?

Originally posted by DeepThought
I'm not convinced by the experimental setup as it occurs to me that one has no really good way of establishing which direction is downwards with any kind of precision apart from by which way gravity pulls...

That also concerned me, but if the forces on a liquid are not the same as those on a pendulum then we may be able to see a difference between a water surface and a pendulum.

Could this be tested experimentally with a horizontal pendulum and a rubber band attached to the centre point?

Originally posted by DeepThought
Consider the vectors in spherical coordinates. At the equator the vector along the radial direction has all the forces along it. The rate of change of azimuthal angle is just the angular velocity and more or less constant and because we've chosen coordinates so that the rotation is entirely around the azimuthal angle the other angle is unchanging. The ...[text shortened]... which direction is downwards with any kind of precision apart from by which way gravity pulls...

With GPS systems I imagine the bell jar could be made to be 100% vertical I would think anyway, then the bob would either be dead center as you all say or it would show some deflection away from dead center. I think it would deflect because the bob is on a long string and would not feel the same circular motion the same extent as ground motion.

I don't think this force would have anything to do with Coriolis forces.
I think, anyway🙂

If you did the same thing on a smaller wheel maybe with magenetic confinement of the bob in the jar, but now the whole thing is horizontal, wheel moving parallel to the ground and the bell jar going around with the wheel, if the magnetic confinement of the bob which would hold it in place was just strong enough to hold it, wouldn't it show a bias of force away from the direction of motion of the wheel?

so I imagine a wheel say 10 meters in diameter and a bell Jar sticking out 10 meters also and of course the system balanced to keep the turning steady with no wobble, now the magnet holds the bob in place just enough to keep it aiming at the center of the wheel.

Wouldn't the bob try to take off in the opposite direction of motion, or at least be able to be shown a force against the magnetic confinement field?

Originally posted by sonhouse
With GPS systems.....

Of course satellite orbits are the easiest proof that the earth is rotating. A satellite orbit is independent of rotation, so the earth as measured from satellites or vice versa demonstrates rotation.

Originally posted by twhitehead
Of course satellite orbits are the easiest proof that the earth is rotating. A satellite orbit is independent of rotation, so the earth as measured from satellites or vice versa demonstrates rotation.

Sure, you and I know it but I am talking about the idiot dyed in the wool flatasser🙂

Originally posted by sonhouse
Sure, you and I know it but I am talking about the idiot dyed in the wool flatasser🙂

And there is your problem. There are two types:
1. The person who makes the argument for the sake of trolling. ie they don't actually believe it, but think its funny to pretend that they do. Any argument you come up with will therefore be useless.
2. Much rarer, are people who actually believe it. This implies they are not scientific enough to understand any explanation you give.

Originally posted by twhitehead
And there is your problem. There are two types:
1. The person who makes the argument for the sake of trolling. ie they don't actually believe it, but think its funny to pretend that they do. Any argument you come up with will therefore be useless.
2. Much rarer, are people who actually believe it. This implies they are not scientific enough to understand any explanation you give.

And a third type, where it is their religion. Just like bible religions, the more you try to disprove their tales, the more they resist, they are programmed that way at an early age.

The sad part is the programming has worked splendidly.

Originally posted by sonhouse
And a third type, where it is their religion. Just like bible religions, the more you try to disprove their tales, the more they resist, they are programmed that way at an early age.

The sad part is the programming has worked splendidly.

Do you actually know anyone whose religion requires them to believe in a flat earth?

Originally posted by twhitehead
And centrifugal force. But centrifugal force is a function of momentum along the axis of rotation so I am not certain that it works the same with a pendulum. The bob is trying to go straight off into space (at a tangent to the earths surface), but is being pulled back towards earth by gravity and back towards its anchor point by the pendulum wire. It seem ...[text shortened]... of the earth.

Does anyone know whether or not a space elevator goes straight up or is angled?

I was assuming that the pendulum bob was just hanging there and not swaying. In which case its only component of centrifugal acceleration is outwards from the z-axis. The bob's instantaneous velocity is tangential to the Earth's surface. Gravity provides an inwards force normal to the instantaneous velocity. If gravity and friction were switched off and the wire cut then the bob would move as you described. But in the co-rotating frame it would seem to be losing horizontal and gaining vertical velocity.

Originally posted by DeepThought
I was assuming that the pendulum bob was just hanging there and not swaying. In which case its only component of centrifugal acceleration is outwards from the z-axis. The bob's instantaneous velocity is tangential to the Earth's surface. Gravity provides an inwards force normal to the instantaneous velocity. If gravity and friction were switched off ...[text shortened]... ut in the co-rotating frame it would seem to be losing horizontal and gaining vertical velocity.

I am still not entirely convinced. Centrifugal force relies on a force towards the centre. That is not the only force in this case.

Originally posted by twhitehead
Yes, I didn't describe it too well.
I believe it involves an object moving relative to a rotating reference frame.

The answer is:
1) Ω=0 ...The Earth is not rotating (relative to an object at the equator).

The Coriolis force in the northern hemisphere is opposite to that in the southern hemisphere so it makes sense that there must be some point in between where it is zero.

There is no way this answer you provided is correct.

"1) Ω=0 ...The Earth is not rotating (relative to an object at the equator)."

The Earth having a rotation is independent of where any body is placed on it.

The Coriolis Force doesn't disapear. It just becomes a radial force at the equator.

Lets align the standard coordinate system so that the k axis is colinear with the Earth axis of rotation, and conicident with the center of the Earth. Lets put a velocity vector tangent to equator aligned with the equatorial line in the ( i, j ) plane. It follows that:

Ω = < 0i + 0j+ Ωk>

and

v = < 0i + vj+ 0k>

Then,

Ω x v = -Ω*vi

That is a force with a magnitude Ω*v in the radial direction -i.

If you two are so sure I'm wrong, and the force goes to zero please elaborate on what I'm donig incorrectly...I honestly can't figure it out?