Constrained Pendulum and Earth Rotation

When I used "w" I meant the time period. I have yet to find the correct Greek letter on my keyboard or as a symbol for use in this script.

I wish I had not entered this thread. Re-reading my first posting I realise that a simple pendulum clock will happily work at the poles. However there will be a very very small twisting force on the suspension.

I think I will leave this thread to others and joining those students who ignored the conservation of angular moment. You cannot ignore angular momentum.

JA

Edited By JA on 06/02/2017 19:27:01

I did not understand the Foucault pendulum initially. Now I think I am clearer.
The equation of precession for a Foucault pendulum (Warwick site) is for a rotation frame of reference and does accurately describe what you see. What is doesn't do is clarify real forces.

My contention now is that the apparent motion at the pole is just the rotation of the earth, the plane of oscillation does not rotate and there is no coupling between the pendulum and the bob at this position. At the equator the pendulum and the earth are fully coupled and there is no differential rotation between the earth and the plane of oscillation (irrespective of its initial orientation) The rotating suspension couples into the bob.The pendulum rotates with the earth. The change of sign of apparent rotation moving the pendulum from the North to the South hemisphere is just the inversion of the pendulum with respect to North.
I clearly made the error of not sticking to inertial frames. The Warwick University page gives the calculations for a rotating frame with the associated apparent forces caused by the curved motion.
So in essence all the action takes place on the equator and none at the poles which answers the question "if the pendulum is suspended from a point on the rotation axis how does it know that the earth is spinning.

I would be grateful for comments on the following statements.
The coupling must vary as the cosine(squared)? of the Latitude to be consistent with the precession equation.
The nature of the coupling is gyroscopic.
The best way to think about a pendulum swinging in a plane given the gyroscopic comment is as the special case of a conical pendulum with amplitude zero at right angles to the plane of oscillation.
Pendulum clocks generate no torque in the suspension spring at the equator and maximum at the poles varying with the sine(squared)? of Latitude sufficient to create one extra beat per day.
There is no variation in observed period (standing on the earth) with Latitude for constrained pendulums as the suspension spring compensates for this. I am assuming that the two effects vary as Sin^2 +Cos^2 = 1

Best regards Martin

Do say if you are fed up yet.

Imagine that the clock is fixed at the centre of a spinning wheel, turning at 1 rev per day, so that the pendulum vertical plane is perpendicular to the wheel. Further assume that the pendulum is replaced by a rotor that is spinning at some rate with the same rotational axis. This forms a gyroscope which is being rotated around an axis orthogonal to its spin axis, which generates a torque around the other mutually orthogonal axis. So that torque can have no effect on the spin speed of the gyro. Now assume that the flywheel is spinning against a spiral spring, so this becomes a balance wheel. Though its angular momentum keeps changing as energy is exchanged with the spring, the same argument applies at any instant. Now remove most of the rim of the wheel and replace the spring with gravity. It still looks the same. So there is no effect on the timekeeping of a pendulum constrained to swing in a plane, at least if located at a pole.

I don't think that models it John. A pendulum swinging back a forth is a special case of a conical pendulum where the bob moves in a circle. The rotational axis is vertical. The bob motion can be represented as a sinusoidal and cosinusoidal. Reduce the amplitude of one of the components to zero and you have a pendulum swinging in a plane.

I think you are absolutely right in the balance wheel statement that rotation in the same plane as the balance wheel will not affect the rate as the wheel will track with the escapement. I think this gyroscopic precession is what made Harrison move from vertical opposed oscillating bars to a horizontal balance wheel. The turning and corkscrewing of ships causing errors.

regards Martin

Your original posting said that it was constrained to swing in one plane, not like a conical pendulum. That's the example I'm considering. If it can swing in 2 planes it's a Foucault pendulum.

Hi John

You are correct regarding constraint. I am just querying your assumed rotational axis. I think it is perpendicular to your spinning wheel ie is coaxial with the wheel's rotational axis. This I understand to be at right angles to your assumed rotational axis. Correct me if I am wrong. I quoted the conical pendulum to identify the rotational axis which does not change when only swinging in a plane. I think you model could be a good one I'm just not sure you are right about the axis which would lead you erroneous conclusions. However I am not clear I have really understood what you mean. Word diagrams are often difficult to interpret correctly.

regards Martin

Not at all. It's just like the pendulum, the wheel's axis is horizontal, parallel to and fixed w.r.t. the plane. The plane is rotating aroud a vertical axis through the centre of the wheel, but perpendicular to the rotation axis. I'll try to produce a picture.

Yes, but the the pendulums rotational axis is vertical through the suspension point.

Martin

I discovered the Coriolis effect as a child, playing on a roundabout. Trying to kick the central pillar whilst facing inwards, my foot was deflected to the left.

In John Haine's posting just above, consider a small scaffold on a record turntable. A ball bearing suspended above the centre is set swinging. Now start the turntable and watch.

Geoff

This is what I mean.

The plane is XY (which could be the earth's surface near to the pole). The vertical axis is Z, and the plane rotates around Z. The wheel axis in this case is horizontal and parallel to the Y axis. The wheel rotates around this. The vertical axis through the wheel's CG is actually the Z axis, but the wheel is not rotating around it.

Saved from a life of delinquency by the coriolis effect!

Reminds me a of an avatar cartoon I saw recently - Mother to errant child: "Look here young, Copernicus, you'll have to learn the world does not revolve around you."

Neil

You mean I'm not a senile delinquent, even after trying so hard for so long? Oh drat! And double drat!!

Geoff