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

[fahraynk]

TL;DR Summary

Dynamics question about seeing things move due to Earth's rotation

If you put a hockey puck on a flat Ice rink, will it move due to the Earths rotation? For example, if I make a mark and measure the distance moved with a caliper, would I notice a change?

 

[hutchphd]

Define "move" please. If the Earth is rotating we are all "moving". You need to ask a more specific question I believe.

Can you detect Earths Rotation Using Ice Field? Yes.

 

[fahraynk]

Define "move" please. If the Earth is rotating we are all "moving". You need to ask a more specific question I believe.

Can you detect Earths Rotation Using Ice Field? Yes.

I wrote in original question if you could measure distance moved by the puck due to rotation of the Earth with a caliper. So by that I mean the puck moving in the frame of the ice field by gaining acceleration due to rotation of Earth.

 

[russ_watters]

I don't see why the puck would move on the ice.

 

[hutchphd]

If the puck is moving, say, at the north pole (good place for ice sheets!) there will be "coriolis forces" causing the path on the ice to be a spiral. Actually the puck will travel "straight" and the Earth rotate under it. This assumes frictionless ice but I feel certain one could actualize this with a very heavy puck and very good ice.
At other latitudes there are similar effects although smaller and the ice more difficult to chill.

 

[Vanadium 50]

How fast do you think the puck will move with respect to the ice? (The Earth spins at around 1000 miles per hour)

 

[berkeman]

Summary: Dynamics question about seeing things move due to Earth's rotation

If you put a hockey puck on a flat Ice rink, will it move due to the Earths rotation?

It works a lot better if you use rocks in the desert...

https://www.nationalparks.org/connect/blog/sailing-stones-death-valley

 

[sophiecentaur]

I don't see why the puck would move on the ice.

There are two questions to answer. 1. Is there a force and 2. is there a situation where the friction forces are less than that force.

There will be a centrifugal force due to the Earth's rotation, and the horizontal component will depend on the observer's latitude. (Enough to make it an oblate spheroid - so it is significant.)

I think the coefficient of friction would need to be lower than ice on ice. You'd need to look at the angle of a suspended mass, perhaps. But local gravitational field variations could well be greater so could you even measure it?

 

[Orodruin]

at the north pole (good place for ice sheets!)

… wait a couple of years …

Still though, I have a hard time imagining the Coriolis force becoming the main contributor to deviation from straight motion over impurities, puck rotation and friction, etc, in a real experiment.

There will be a centrifugal force due to the Earth's rotation, and the horizontal component will depend on the observer's latitude. (Enough to make it an oblate spheroid - so it is significant.)

The horizontal component will be zero by definition of horizontal as orthogonal to the gravitational field in the local rest frame. In other words, if you freeze over a pond, the component of acceleration in the ice plane will be zero.

The effect you are considering here would be accounted and corrected for in the buildup of the ice sheet.

 

[Orodruin]

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

The linked article seems to suggests the motion is due to very particular weather effects, not due to Earth’s rotation.

 

[Vanadium 50]

With respect to a fixed point on the earth, the Coriolis force is zero. The Centrifugal force is an \Omega \times (\Omega \times R) acceleration. However, only a small component points towards the equator - it is mostly up. Further, the ice felt this force when it froze, so the puck needs to move 'uphill' with exactly the same resistance needed to cancel this force. This will be approximately the thickness of the ice sheet divided by the radius of the earth.

 

[sophiecentaur]

The effect you are considering here would be accounted and corrected for in the buildup of the ice sheet.

(Your) good thinking and my sloppy thinking . So the water / ice will have already experienced the same effect as the puck. But say the ice rink were tilted, after freezing, to be parallel with the equatorial plane. That would give the maximum effect. Whoops, the angle of g would also be changing. So go to near the North Pole and g would be normal to the centrifugal force and ignore the cos(small angle) factor at 1km from the pole.
a = ω² r
= 1.3e-10 X 1e3 ms^-2
and g = 10ms^-2

That would imply a coefficient of friction less than 1.3e-8
Pretty slippery.

 

[Orodruin]

With respect to a fixed point on the earth, the Coriolis force is zero.

While this is true for a stationary puck, it will not be true for a moving puck. Hence why I discussed deviation from a straight line and not motion in itself.

 

[Orodruin]

(Your) good thinking and my sloppy thinking . So the water / ice will have already experienced the same effect as the puck. But say the ice rink were tilted, after freezing, to be parallel with the equatorial plane. That would give the maximum effect. Whoops, the angle of g would also be changing. So go to near the North Pole and g would be normal to the centrifugal force and ignore the cos(small angle) factor at 1km from the pole.
a = ω² r
= 1.3e-10 X 1e3 ms^-2
and g = 10ms^-2

That would imply a coefficient of friction less than 1.3e-8
Pretty slippery.

If you know the exact angle to tilt you already know that the Earth is spinning. You cannot ignore the small deviation from the cosine. The effect is of magnitude 1/(2*6700^2) ~ 1e-8. It is the same as what you are trying to probe.

 

[Baluncore]

For example, if I make a mark and measure the distance moved with a caliper, would I notice a change?

Assuming that "on ice" means floating without friction.
At the equator, it would remain flat and so not move relative to the Earth.
At the poles, it would continue to rotate with the Earth.
I think that, in between, you could model the puck as the slow rotor in a gyrocompass, but with a vertical, not a horizontal axis. Maybe like a Foucault pendulum.

 

[berkeman]

The linked article seems to suggests the motion is due to very particular weather effects, not due to Earth’s rotation.

I debated whether to add a winkie emoji to my post...

You university professors are so serious and literal...

 

[Orodruin]

I debated whether to add a wink emoji...

You university professors are so serious and literal...

You just can’t tease us with what would be very cool stuff based on physics that upon examination is just cool stuff based on meteorology …

Edit: … except professors in meteorology I guess …

The word meteorology itself is such a tease. Turns out it has very little to do with actual meteors …

 

[Vanadium 50]

While this is true for a stationary puck, it will not be true for a moving puck.

That is true, but if you're going to have a moving puck, why not a cannon?

 

[Vanadium 50]

Oh, also the puck has a height - so it's center of mass is slightly too high relative to the velocity of the base.

 

[hutchphd]

Still though, I have a hard time imagining the Coriolis force becoming the main contributor to deviation from straight motion over impurities, puck rotation and friction, etc, in a real experiment.

Yeah for the specified hockey puck. Maybe some giant curling stones at high speed (with lots of broomsters). The ice would be perfect (by stipulation).

 

[jbriggs444]

Yeah for the specified hockey puck. Maybe some giant curling stones at high speed (with lots of broomsters). The ice would be perfect (by stipulation).

360 degrees per 23.9 hours. 0.1 degrees per 23.9 seconds. That sounds do-able. No need for high speed. The deflection angle get larger when the trajectory takes longer.

You control for any systematic error due to a side-ways tilt of the ice by running the puck both ways. Coriolis will cause a rightward deflection both ways. A tilt would cause a rightward deflection one way and a leftward deflection the other.

 

[N1206]

There are tides to consider.
Water slops along in tandem with the moon and sun.
It takes The mass of a whole ocean and the right underwater topography to get measurable tide.
The Great Lakes tidal height is about 5 cm.
Prevailing winds are generated by the atmosphere 'slipping' in regard to the rotation of the earth.
Again, massive amounts of mass.
I doubt that the mass of a hockey puck, even sitting on a perfectly frictionless surface in vacuum will move enough to be measurable, and if it did, you might not be able to sort it out from tidal effects.

Also, take your experiment up to geosynchronous orbit. Put the puck on the frictionless surface at rest. Both are being pulled by gravity. Both are rotating around the center of the Earth at the same speed. Both stay at the same point over the Earth's surface. The puck is an inch high, so if the frictionless surface is closer to the earth, then the puck's center of mass will be ~1.75 cm farther from the center of the Earth than the frictionless surface. You can do the Newton's classical force calculation for that. A relative force difference compared to the frictionless surface will exist, but it is going to very, very, very, very small. If you waited for long enough, you would eventually wind up a with measurable distance the puck moved relative to the surface.

If other effects like tidal forces don't swamp it.

 

[sophiecentaur]

You control for any systematic error due to a side-ways tilt of the ice by running the puck both ways.

How would you separate the effect of the 'sideways tilt' and the centrifugal force? What would be the reference level to determine tilt?

 

[pbuk]

There's some pretty dubious fictitious force analysis going on in this thread, and the OP doesn't seem interested any more.

Anyway, to pick up on a couple of recent points:

If other effects like tidal forces don't swamp it.

The force you have just described includes the tidal force.

How would you separate the effect of the 'sideways tilt' and the centrifugal force?

In exactly the way described you can't: centrifugal force forms part of the tilt (unless it is perfectly balanced, in which case there is no tilt).

What would be the reference level to determine tilt?

The geoid, by definition. Alternatively, the surface of a bucket/swimming pool/sea/ocean of water.

 

[jbriggs444]

How would you separate the effect of the 'sideways tilt' and the centrifugal force? What would be the reference level to determine tilt?

No need to distinguish. The intent is to measure Coriolis. Both tilt and centrifugal force result in a consistent sideways deflection in a single direction (e.g. southward). Coriolis delivers a sideways deflection that reverses when the puck traverses the track in the opposite direction (e.g. rightward).

The obvious way to obtain level ice is to start with water in a large flat rink. It'll naturally relax to become level. Then you freeze it. The tilt will naturally compensate for any centrifugal force.

[Then you scribe a fine line on the edge of the rink right at the surface of the ice. You label this line with the letter C. This will be referred to as "C level"]

 

[sophiecentaur]

The intent is to measure Coriolis. Both tilt and centrifugal force result in a consistent sideways deflection in a single direction (e.g. southward).

In this experiment we can't actually measure the tilt because the angle is small and the ice is flat over only a few metres max. The principle of equivalence tells us that acceleration and gravity are indistinguishable so how can you know if the force in one direction is due to the g component along the slope or the centrifugal force.

This thread is about the feasibility of a particular experiment. I still say that the quantities involved are way too small.

 

[jbriggs444]

In this experiment we can't actually measure the tilt because the angle is small and the ice is flat over only a few metres max. The principle of equivalence tells us that acceleration and gravity are indistinguishable so how can you know if the force in one direction is due to the g component along the slope or the centrifugal force.

This thread is about the feasibility of a particular experiment. I still say that the quantities involved are way too small.

Again, we are not attempting to measure either tilt or centrifugal force. We are attempting to measure Coriolis.

 

[pbuk]

The principle of equivalence tells us that acceleration and gravity are indistinguishable

No, I think the principle of equivalence tells us that 1: acceleration and 2: the net of gravity and centrifugal force are indistinguishable.

so how can you know if the force in one direction is due to the g component along the slope or the centrifugal force.

You calculate the constituents of $g = \dfrac{GMm}{r^2} - r \omega ^2$.

 

[sophiecentaur]

Again, we are not attempting to measure either tilt or centrifugal force. We are attempting to measure Coriolis.

The first post doesn't mention motion of the puck. But threads tend to change directions on the way.

You calculate the constituents of g=GMmr2−rω2.

Aren't we dealing with vectors here?

 

[pbuk]

Aren't we dealing with vectors here?

Well yes but we can analyse their components separately. By definition g is normal to the geoid. Centrifugal force is measured parallel to g, the component of intertial forces normal to g is Coriolis force.

 

[hutchphd]

Why is this any different from a Foucault Pendulum? Build a frctionless circular arena with perfect bumpers and perfect ice. Check for precession.

 

[sophiecentaur]

Well yes but we can analyse their components separately. By definition g is normal to the geoid. Centrifugal force is measured parallel to g, the component of intertial forces normal to g is Coriolis force.

And at latitude >0?

 

[pbuk]

And at latitude >0?

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.

 

[Orodruin]

By definition the net of gravity and centrifugal force only has a vertical component i.e. normal to the geoid.

The standard definition of the centrifugal force is $-m\vec\omega\times(\vec\omega\times\vec r)$, which is the fictitious force independent of velocity relative to a rotating reference frame. This is generally not parallel to the gravitational field $\vec g$.

The standard definition of the Coriolis force is the fictitious force dependent on the velocity relative to a rotating frame of reference and is given by $-2m \vec\omega\times \vec v$.

 

[pbuk]

The standard definition of the centrifugal force is $-m\vec\omega\times(\vec\omega\times\vec r)$, which is the fictitious force independent of velocity relative to a rotating reference frame. This is generally not parallel to the gravitational field $\vec g$.

Agreed, however if we want to find the equipotential surface we have to use a different definition or water will flow uphill.

 

[Orodruin]

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.

 

[pbuk]

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.

 

[Orodruin]

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.

 

[sophiecentaur]

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]

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]

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

that is NOT due to earth rotation

 

[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 )

 

[gmax137]

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

Anyone else here been there? It's a long bumpy jeep road.

 

[hutchphd]

the puck wont suffer Coriolis effects, because it is not moving as noted by the OP

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 $$\vec {v'} \neq 0$$

 

[DaveC426913]

The word meteorology itself is such a tease. Turns out it has very little to do with actual meteors …

Times change.

65 million years ago, meteorology was very much a descriptive term for weather.