Constrained Pendulum and Earth Rotation

[Martin KyteParticipant @martinkyte99762]

OK chaps how about this for something to muse on.

Consider a pendulum constrained to swing with the rotation of the Earth (as normal clock arrangement). How does the period vary from that of a Foucault pendulum of the same length (which is not so constrained). Also what is the effect or not of latitude. I would be interested on anyones take on energy/momentum considerations.

Intuitively it must slow down as a result of earth rotation and I would say by a factor of one complete period per year.

Before anyone tells me I know this is totally irrelevant regarding clock making as it has to be a constant effect (subject to the rotation staying practically constant). It's just something I happened to be musing on.

Have a good weekend

Martin

[Martin KyteParticipant @martinkyte99762]

[Martin KyteParticipant @martinkyte99762]

I obviously meant one period per day.

Martin

[John HaineParticipant @johnhaine32865]

When you say "with the rotation of the Earth", do you mean, say, east-west at the equator?

[David JuppParticipant @davidjupp51506]

What exactly do you mean by 'slow down' ? The period of a pendulum is essentially constant (as long as the angle of swing is modest), as the energy in the system decays, the height of swing and speed at mid swing both reduce but period remains constant.

I can imagine additional energy losses at the pivot for the fixed (not Foucault) pendulum – but the period should not change. That's if I'm remembering the derivation of the equation for period of a pendulum correctly from school Physics lessons many years ago.

[SillyOldDufferModerator @sillyoldduffer]

This sort of thing hurts my head. The Earth is spinning on it's axis and orbiting the sun. The sun is in a Galaxy, and the whole Universe is expanding. Everything is moving.

The forces acting on a unconstrained pendulum (ie one only partly effected by the earth's movement) must be different from those acting on a constrained pendulum where the whole assembly is physically linked to the earth and there is only one degree of freedom.

For many years I was happy with the pendulum formula taught at school until I discovered by accident that it's only an approximation. A very good one for small displacements, but nonetheless slightly inaccurate. The same book said that plumb lines are only approximately vertical, another illusion shattered.

Latitude might make a difference too. Gravity weakens as you move towards the poles because the earth is an oblate spheroid.

I've no idea how big a difference any of this would make to the periods of pendulums in practice. Any clever people out there?

Dave

[john carruthersParticipant @johncarruthers46255]

I've often wondered about buying a precious material (say, gold) at the equator, and selling it at the poles?

[duncan webster 1Participant @duncanwebster1]

Posted by john carruthers on 03/02/2017 17:56:42:

I've often wondered about buying a precious material (say, gold) at the equator, and selling it at the poles?

If the bullion dealer has any sense he will measure the MASS using standard weights, and so different gravity won't matter

[pgk pgkParticipant @pgkpgk17461]

It makes my head hurt to think about such things but presumably there's also a need to account for altitude and any physical effects acting on the pendulum material since persumably even non-magnetic materials can have eddy currents induced in them…and I suppose the shape of the pendulum might effect how ions flow.

[Gary WoodingParticipant @garywooding25363]

Does the swing of a Foucault pendulum at the Equator appear to rotate?

[martin perman 1Participant @martinperman1]

Martin,

Are you describing the very large pendulum that hangs in the Science Museum in London, that swings with relation to the earth rotation.

Martin P

[Geoff TheasbyParticipant @geofftheasby]

Yes. But rather the earth rotates underneath it… Anyway, surely, a clock pendulum is constrained in that it can only swing in one plane, unlike the Foucault. Any tendency to twist from that plane is resisted by its construction and suspension. Yes?

Geoff

[John HaineParticipant @johnhaine32865]

Posted by Martin Kyte on 03/02/2017 15:27:36:

Intuitively it must slow down as a result of earth rotation and I would say by a factor of one complete period per year day.

Martin

Martin, please say why it is intuitive? A Foucault pendulum swinging in the equatorial plane at the equator would be completely "unaware" that it isn't constrained in the north-south direction. And why would it be intuitive that any slowing effect is just one period irrespective of the pendulum period?

[martin perman 1Participant @martinperman1]

Gentlemen,

There is a good explanation of how a Foucault Pendulum works here https://en.wikipedia.org/wiki/Foucault_pendulum

Martin P

[Martin KyteParticipant @martinkyte99762]

Hi Chaps. What I was really interested in was a way of thinking about the problem.

OK. so I've thought about it some more and my conclusion is that there is no variation between the two systems as far as pendulum period is concerned. Here is my thought process.Assume that the period of the pendulum is much smaller than the period of earth rotation. Put the pendulum at the pole for simplicity. Rotation axis vertical through the mount period one day. Take a plane of reference the initial plane of the pendulum swing. An unconstrained pendulum swings for P periods per day and remains in the plane of reference. The constrained pendulum also swings for P periods but effectively completes one extra period in the day due to the rotation forced on it by the mount. (That's why I initially thought the pendulum must be slowed as it has to travel further) The moves in a curve close to an ellipse but precessing through 360 degrees in 24hours, think spirograph patterns. Each period the bob has travelled further than one complete swing of an unconstrained pendulum but has been accelerated a small amount by the torque in the suspension mount (usually a flat spring strip for non clock makers) so the period has remained the same. The suspension torque in this thought model compensates for the additional distance travelled by the bob. So there is no difference intrinsically in the period between the two systems (constrained and unconstrained).

For the clock, however the escapement rotates with the mount so it must count one less swing per day.

I am happy to listen to better explanations to my musings and as I said I'm mainly interested in a way of thinking about it.

regards Martin

[Geoff TheasbyParticipant @geofftheasby]

I think that the lack of accuracy in a clockwork assembly would be much greater than the difference of one extra swing of the pendulum. If it swings once a second, that is 60 x 60 x 24 times per day, or 86,400, but he clock mechanism is accurate only to within several seconds a day, even in the best clocks. One extra swing of the pendulum is 0.001% of the day.

Geoff

[Martin KyteParticipant @martinkyte99762]

I did say at the start that this has no bearing on practical clockmaking Geoff. It's just an exercise in thinking about the dynamics of pendulums. I still haven't sorted the mechanism by which the precession of Foucalt systems change with latitude so I've got a long way to go yet.

:0)

Martin

[Russell EberhardtParticipant @russelleberhardt48058]

Posted by Geoff Theasby on 06/02/2017 09:58:02:

I think that the lack of accuracy in a clockwork assembly would be much greater than the difference of one extra swing of the pendulum. If it swings once a second, that is 60 x 60 x 24 times per day, or 86,400, but he clock mechanism is accurate only to within several seconds a day, even in the best clocks. One extra swing of the pendulum is 0.001% of the day.

Really? A relatively recent reconstruction of Harrison's 250 year old design of pendulum clock lost less than one second in a 100 day trial at the Royal Observatory. Conventional English Regulator clocks can easily achieve 1/2 s a week.

Russell.

[Martin KyteParticipant @martinkyte99762]

It's irrelevant to clockmaking because any change in period would be constant.

Martin

[roy entwistleParticipant @royentwistle24699]

Removed because covered in original post

Edited By roy entwistle on 06/02/2017 11:08:14

[Russell EberhardtParticipant @russelleberhardt48058]

Posted by Martin Kyte on 06/02/2017 09:24:17:

For the clock, however the escapement rotates with the mount so it must count one less swing per day.

Yourbargument seems right to me. However that effect is much less than that due to the change of gravity with altitude and with latitude. The latter being a result of the rotation of the earth flattening the earth to an oblate spheroid.

Russell.

P.S. Your effect will vary with the sine of the latitude.

Edited By Russell Eberhardt on 06/02/2017 11:55:16

[roy entwistleParticipant @royentwistle24699]

Don't forget also that a clock pendulum is considerably shorter than a Foucault

[Geoff TheasbyParticipant @geofftheasby]

OK, Russell, I didn't look up exactly how accurate the very best clocks are.

Sorry.

Geoff

[Martin KyteParticipant @martinkyte99762]

Yes Russell. I'm only interested in the effect of constraining. Lets just say the distance from the center of mass is a constant.

Roy, I did say that. If the periods were the same the pendulum would bob would describe a circle and would not move relative to the escapement (counting device) and the clock would not tick.

It has been pointed out to me that the Foucault pendulum precession varies with latitude, something I did not initially recall, although it is clear it must change direction at the equator and vary in a continuous relationship to latitude. This implies that the motion of the mount of a Foucault pendulum affects the bob and causes it to move in an ellipse the minor axis of which varies as the sine of the latitude and is maximum at the pole and zero at the equator.

Martin

[duncan webster 1Participant @duncanwebster1]

It is much easier to think about Foucault pendulum at the poles, where the suspension point just rotates once per day. It is rotating about the axis of symmetry of the pendulum swing, I think it is easy to see that it doesn't affect the period of the pendulum. At the equator there is no Foucault effect. This being the case I don't think there is any effect anywhere else. As others have said, the period will be affected by the apparent variaton in g (the accelearation due to gravity) caused by altitude, the oblate shape of the earth, and to a very small degree by centripetal force due to the earth's rotation

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