Constrained Pendulum and Earth Rotation

[JAParticipant @ja]

I have not really been following the forum for the last couple of days due to another commitment, a motorcycle show, so I am now catching up on interesting threads.

I think only Martin has mentioned the magic phrase "conservation on angular momentum". The Foucault pendulum demonstrates this. If one considers a clock with a free swinging pendulum at one of the poles the direction of swing will change considerably during the day relative to the clock's mechanism. This in itself will give the clock designer serious problems.

A balance wheel on a horizontal spindle will have similar problems which, I would suggest, is why Harrison used a pair of opposing torsional balance "weights" on his early chronometers. Later chronometers used a balance on a vertical spindle which gets over this problem. However they may still be effected by Coriolis effects as they move over the surface of the Earth. Again Coriolis forces are the result of the Earth's rotation and conservation of angular momentum.

Conservation of angular momentum can become horribly complex particularly with gyroscopes leading to most engineering students switching off when they reach the subject. Highly distinguished learned persons have been known to be total confused by the subject in public, one suggested that gyroscopes had magical powers.

I hope that I have got all this right!

JA (from his armchair)

[Russell EberhardtParticipant @russelleberhardt48058]

Posted by JA on 06/02/2017 13:20:32:If one considers a clock with a free swinging pendulum at one of the poles the direction of swing will change considerably during the day relative to the clock's mechanism. This in itself will give the clock designer serious problems.

Not really a problem. They just need to use a suspension which is free in one direction but stiff at right angles to it such as these.

The double compound pendulum used in Harrison's ships chronometers was intended to compensate for the rolling movement of the ship not the rotation of the earth.

Russell.

[SillyOldDufferModerator @sillyoldduffer]

Also, from an armchair.

I couldn't find anything on the web that directly compares the periods of the two types of pendulum.

But I did find an equation for the angular frequency of a Foucault Pendulum that is a modification of that of the standard pendulum. The equations are different, which implies that their periods are also different.

Not surprisingly in both equations, the frequency (and therefore period) depends on gravity and the length of the pendulum. The pendulums must be similar, but they are not identical.

I am notoriously bad at maths, but I read the Foucault equation as: "the period is determined by the square root of the acceleration due to gravity divided by the length of the pendulum but it is varied + and – by the effects of precession."

I think this means that the individual period of each swing of a Foucault pendulum is continually changing. It speeds up and slows down as the earth rotates and I assume the pluses and minuses average out over 24 hours. The 'tick' of a Foucault pendulum is modulated by the rotation of the earth.

Or, just as I remember being bemused by spherical geometry at school, am I hopelessly confused again? Help!

Dave

[Martin KyteParticipant @martinkyte99762]

I would be handy to know where you found this equation and what omega and lambda refer to.

regards Martin

[JAParticipant @ja]

Russell

I think you would need a suspension which is free in both directions for a pendulum to swing successfully at the poles.

Granted your comment about Harrison's double pendulum.

Dave

What are the definitions for the symbols in the equations other than gravity and length?

There is some sense in ignoring angular momentum, particularly for the effects it has on one's brain. Those students were right. Now to do something simple, unload the car and go food shopping.

JA

[DrDaveParticipant @drdave]

Further to SillyOldDuffer's formulae, I found an interesting derivation of the period of a Foucault pendulum on Warwick University's website. Their final equation is where w1 is pendulum frequency, w is the rotational frequency of the earth (1 rev per day) and w0 is the pendulum frequency at the equator. The final Greek character (whose name I am sure someone will tell me) is the latitude. So, a Foucault pendulum will slow down the further it is from the equator, but the difference is very small.

And we are no closer to finding out how a normal pendulum would change frequency with latitude. I would surmise that, because to pendulum is forced to swing in one plane and have no Coriolis acceleration, it should be independent of latitude. Lowers head and prepares to be shot at!

[SillyOldDufferModerator @sillyoldduffer]

Posted by Martin Kyte on 06/02/2017 14:36:01:

I would be handy to know where you found this equation and what omega and lambda refer to.

regards Martin

Yes, sorry about that Martin. The equation comes from this page at the University of Texas which refers back to this one on Coriolis Force

Ω is the 'constant angular velocity about an axis passing through the origin of the inertial frame', and λ defines the angle to the reference frame.

As to JA's most excellent suggestion "Now to do something simple, unload the car and go food shopping", I tried that yesterday lunchtime. At about 7pm a neighbour rang my doorbell to ask if knew my car boot had been open all afternoon. Proof that unloading a car is now beyond me too!

Dave

[Martin ConnellyParticipant @martinconnelly55370]

Could it be that the unknown variables in Dave's equation are drag from the air which would depend on the configuration of the pendulum and the maximum angle from vertical. As I recall these equations are generally given for ideal pendulums in a vacuum and in the real world the further you go away from very small angles the bigger the difference between theory and practice.

The precession of a true Foucault pendulum at a pole is one revolution per sidereal day of 23 hr, 56 min, 4.1 sec. since the earth revolves 366.25 times on its axis each year as viewed from a distant point outside the solar system.

It should be noted that the Science museum Foucault pendulum is reset every day and is given an impulse to keep it going that results in a bob motion that is a shallow figure of 8 in the vertical plane. The mechanism was devised by Professor Sir Brian Pippard in 1983.

Also details from the Science Museum state that because the Science Museum is not at a pole their pendulum turns slower than once per day with a rotation period of 30 hours and 40 minutes and they also state that at the equator the pendulum would not turn at all.

Martin

[SillyOldDufferModerator @sillyoldduffer]

Posted by Martin Connelly on 06/02/2017 15:28:26:

Could it be that the unknown variables in Dave's equation are drag from the air

…

Martin

No, but that's yet another valid complication to worry about. One of the page links I provided for Martin K mentions that the equations don't account for air resistance.

Which leads me into an example of faulty logic. As the earth exists in a very large vacuum, I suggest that drag on the pendulum can be neglected. After all, the average air pressure in our universe must be very close to zero!

Interesting that the Science Museum data shows that their pendulum takes over 30 hours to rotate. That's not what I imagined at all. Back to the drawing board – I don't understand it!

Dave

[DrDaveParticipant @drdave]

Posted by SillyOldDuffer on 06/02/2017 15:47:26:

Interesting that the Science Museum data shows that their pendulum takes over 30 hours to rotate. That's not what I imagined at all. Back to the drawing board – I don't understand it!

Dave

Dave,

If you look at the bottom of the page that I linked to above, they give the period of rotation as T = 24/sin(latitude). For the Science Museum, this gives T = 24/sin(51.4978) = 30.67 hours.

Dave

[JAParticipant @ja]

The important thing to know here is what "w" is in Dave's Foucault equation. I think DrDave is right. "w" has two components: that for a simple pendulum which is independent of latitude and that due to the period of rotation. To put it another way "w" is the time take for the pendulum, simple or Foucault, to return to the same position.

Now someone is going to say "but a Foucault pendulum is a simple pendulum". I just hope we don't meet on a moonless night in a dark ally otherwise the pendulum from my regulator will have a different period.

JA

[John HaineParticipant @johnhaine32865]

Sorry to be pedantic but if by "w" you mean Greek "omega", then it is by convention angular frequency, which equals 2 x pi/period measured in seconds.

[Neil WyattModerator @neilwyatt]

Posted by DrDave on 06/02/2017 14:57:52:

Further to SillyOldDuffer's formulae, I found an interesting derivation of the period of a Foucault pendulum on Warwick University's website. Their final equation is where w1 is pendulum frequency, w is the rotational frequency of the earth (1 rev per day) and w0 is the pendulum frequency at the equator. The final Greek character (whose name I am sure someone will tell me) is the latitude. So, a Foucault pendulum will slow down the further it is from the equator, but the difference is very small.

And we are no closer to finding out how a normal pendulum would change frequency with latitude. I would surmise that, because to pendulum is forced to swing in one plane and have no Coriolis acceleration, it should be independent of latitude. Lowers head and prepares to be shot at!

Wikipedia gives a formula:

w = 360 degrees * sin latitude/day

the term after sin squared is latitude in your equation.

It seems that the rate of the pendulum slows by one swing per rotation, which is one swing per day at the poles, and half a swing per day at 30 degrees latitude (sin 30 = 0.5)

Neil

[Neil WyattModerator @neilwyatt]

Just to make things worse, the 'day' is a sidereal day not a solar day as the rotation is relative to the 'frame' of the universe

[Tim StevensParticipant @timstevens64731]

There may be some confusion here over the different Greek letters. Omega has two forms, normally called (by typographers at least) 'upper and lower case' The upper case is the familiar ohms symbol – a circle broken at the base with short extensions sideways. Lower case omega looks for all the world like a seated generous lady's bottom, from, er, behind. It also resembles a handwriting or italic type small w.

The other symbol which looks like a badly written lower-case italic p, is phi – and the upper case version looks like a tall capital I with a fat capital O printed over it.

Not that this helps much with Foucaults pendulosities, but at least it gives a clue to how to pronounce them.

I think that all pendulums act as Foucaults did, but you can only notice the effect (of change in the line of swing) if the pendulum can swing as it likes, and act for a long time (several minutes at least) without any outside influence to keep it going. This can be achieved easily if you have a very tall building and so can have a very slow moving but heavy bob.

Never mind pendulums for a moment, just think about the same effect on the balance wheel of a wrist watch as you play tennis. Other games are available …

Cheers, Tim

Edited By Tim Stevens on 06/02/2017 18:19:19

[duncan webster 1Participant @duncanwebster1]

I'm probably being a bit thick here, but if we go back to the pendulum at the pole, the point of suspension has no linear movement relative to the earth's axis, and so if the suspension has zero rotational stiffness in the axis of the earth's rotation how does the pendulum know that the earth is moving? If the suspension has rotational stiffness it isn't a Foucault pendulum. If the pendulum doesn't know the earth is moving, then that movement can't affect it's period.

[JAParticipant @ja]

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

[Martin KyteParticipant @martinkyte99762]

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.

[John HaineParticipant @johnhaine32865]

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.

[Martin KyteParticipant @martinkyte99762]

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

[John HaineParticipant @johnhaine32865]

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.

[Martin KyteParticipant @martinkyte99762]

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

[John HaineParticipant @johnhaine32865]

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.

[Martin KyteParticipant @martinkyte99762]

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

Martin

[Geoff TheasbyParticipant @geofftheasby]

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

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