# Can you detect the Earth's Rotation Using a Hockey Puck on Ice?

• I
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Agreed, however if we want to find the equipotential surface we have to use a different definition or water will flow uphill.
No, you don’t. You just need to include the potential of the centrifugal force in the total potential.

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No, you don’t. You just need to include the potential of the centrifugal force in the total potential.
I don't think we are disagreeing about what you have to do, it's just a matter of consistent definitions.

My point is that you can't say "here I have a horizontal surface (i.e. a local portion of a geoid) and I can find an net force on a stationary object that is not normal to the surface". So either you define centrifugal force as parallel to gravitational force or you introduce another force to equalize things (and accept that gravity does not point straight down). Because this force is tiny it has never been given a name and so we ignore it, which is equivalent to saying centrifugal force is parallel to gravity.

Staff Emeritus
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I don't think we are disagreeing about what you have to do, it's just a matter of consistent definitions.

My point is that you can't say "here I have a horizontal surface (i.e. a local portion of a geoid) and I can find an net force on a stationary object that is not normal to the surface". So either you define centrifugal force as parallel to gravitational force or you introduce another force to equalize things (and accept that gravity does not point straight down). Because this force is tiny it has never been given a name and so we ignore it, which is equivalent to saying centrifugal force is parallel to gravity.
The point is that the apparent gravitational field in the rotating system includes the centrifugal force. The apparent gravitational force is obviously not parallel to neither the gravitational force nor to the centrifugal force except when those two are parallel. You cannot make the centrifugal force parallel to the gravitational force by wishing it to be so and if you want to change conventions around you are just going to confuse people. The centrifugal force has a commonly accepted definition, which is therefore what should be used.

hutchphd
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The same, please see the definition of the geoid e.g. https://en.wikipedia.org/wiki/Geoid

Think about a toy boat floating on a lake on a still day, initially stationary. If there was a force that would make it move to somewhere else on the lake then it would do so. Now take the boat out of the lake and consider the water that it was previously displacing. The same force would make that water move to somewhere else on the lake, changing the profile of the surface.

This is the definition of a geiod: it is an equipotential surface, taking into account all the forces at the surface. By definition the net of gravity and centrifugal force only has a vertical component i.e. normal to the geoid.
Ah yes - so by definition, the puck wouldn't move on a 'level surface'?

pbuk
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The point is that the apparent gravitational field in the rotating system includes the centrifugal force.
Yes, absolutely.

You cannot make the centrifugal force parallel to the gravitational force by wishing it to be so and if you want to change conventions around you are just going to confuse people.
Yes, I am sorry if I have caused or added to any confusion.

N1206
Why is this any different from a Foucault Pendulum?
You are right, it probably isn't any different in theory.
Instead of plumb bob hanging from a long wire to get as frictionless as possible, we have a puck bouncing back and forth between bumpers on a frictionless surface. Neat!

Foucault did his experiment in Paris. I wonder what the gravitational gradients are like there? When the Canadian west was surveyed, the survey team for the Fifth Meridian proceeded south from Stony Plain Alberta to the 49th parallel and their presumed intersection of the those two lines (the 5th and the border) was inaccurate a substantial distance (hundreds of feet) to the west from what the teams surveying east-to west from the First Meridian had chained measuring from the First Meridian along the 49th parallel.

Upon review, the mass of the Rocky Mountains had deflected the plumb bobs used enough to cause the discrepancy, and the Fifth Meridian was re-surveyed taking that into account and came out within inches of the east-west result. I guess when the pendulum is suspended in a fixed location, the gradient of what is 'down' would get baked in, and the precession would work out the same as at any other location on the same line of latitude.

zanick
Why is this any different from a Foucault Pendulum? Build a frctionless circular arena with perfect bumpers and perfect ice. Check for precession.
the movement for the pendulum is a precession due to a swinging plane deviating from surface alignment. the puck wont suffer Coriolis effects, because it is not moving as noted by the OP. coriolis is the perceived deviation from a path as viewed from the rotating reference frame, so the "puck" has to be moving to detect this deviation. it will not cause the puck to move on its own. (as mentioned by others and that centrifugal force component is too small )

Please indicate where this is stipulated. My recollection is that the OP was interested in "acceleration of the puck" but did not indicate the initial state of motion. Coriolus force $$-2m \vec {\omega} \times \vec {v'}$$ obviously requires $$\vec {v'} \neq 0$$