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No, you don’t. You just need to include the potential of the centrifugal force in the total potential.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.Agreed, however if we want to find the equipotential surface we have to use a different definition or water will flow uphill.
I don't think we are disagreeing about what you have to do, it's just a matter of consistent definitions.No, you don’t. You just need to include the potential of the centrifugal force in the total potential.
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.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.
Ah yes - so by definition, the puck wouldn't move on a 'level surface'?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.
Yes, absolutely.The point is that the apparent gravitational field in the rotating system includes the centrifugal force.
Yes, I am sorry if I have caused or added to any confusion.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.
You are right, it probably isn't any different in theory.Why is this any different from a Foucault Pendulum?
that is NOT due to earth rotationIt works a lot better if you use rocks in the desert...
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https://www.nationalparks.org/connect/blog/sailing-stones-death-valley
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 )Why is this any different from a Foucault Pendulum? Build a frctionless circular arena with perfect bumpers and perfect ice. Check for precession.
Anyone else here been there? It's a long bumpy jeep road.It works a lot better if you use rocks in the desert...
View attachment 303745
https://www.nationalparks.org/connect/blog/sailing-stones-death-valley
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 $$\vec {v'} \neq 0$$the puck wont suffer Coriolis effects, because it is not moving as noted by the OP
Times change.The word meteorology itself is such a tease. Turns out it has very little to do with actual meteors …