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zenbudda
2010-Jul-08, 06:14 PM
I wondered to myself if things located near the equator experience more momentum than those things located further away from the equator. So I did some good searching and came across this link:

So, as I anticipated, the earth is indeed moving faster (relative to it's axis) at the equator than it's poles. This speed also transfers to any objects fixed to or on the surface of the earth. So, i thought to myself, if you were to measure a person's speed (relative to the earth's axis), they would technically be running faster at the equator than they would if they ran closer to the poles. but then i thought to myself, what if they ran east? west? north? south? bi-directional? what if they ran in opposing directions?

so let's take my thoughts a step further. say you build 2 perfectly flat racetracks 1km long. both tracks are at the same elevation and are in the same climate zone (really dry). you also build 2 identical vehicles that are perfectly tuned, and perfectly functioning and can precisely accelerate and maintain speeds. basically, whatever else is left that could un-level the playing field for this experiment you would accommodate for).

if both vehicles started at the same time, accelerated at the same rate, and traveled at precisely the same speed, and traveled in the same direction, they "should" finish at the same time. HOWEVER, relative to the earth's axis, the vehicle closest to the equator was traveling faster, right? if so, should the vehicle closer to the equator also experienced more momentum? i'm thinking the actual physical effects may not have been "felt" but i'm thinking it would be measurable, yes?

what if the 2 vehicles started at other ends of the tracks (one traveling east, the other traveling west). the one traveling east would be traveling in the same direction that the earth is turning on it's axis. the one traveling west would be going against the spin of the earth. here's another way i am looking at this. so if this were a merry-go-round spinning REALLY FAST, if you leaped off in the direction that the merry-go-round is spinning, you would automatically be traveling (at a minimum) at same speed the merry-go-round was spinning (and of course + whatever force you leaped forward at). but if you were to jump in the opposite direction, you would be working against the force of the merry-go-round, thereby negating some (most likely all) of the force exerted in the direction you leaped in (most likely causing you to still go in the direction of the merry-go-round). how does that apply to the vehicles on the tracks? shouldn't vehicles traveling in the direction of the earths rotation travel faster than those traveling against it?

i also understand that the earth's orbit around the sun, the solar system around the galaxy, and the galaxy around the universe (or some other unforeseen axis). does any of that affect our speed on the surface of our planet?

dgh64
2010-Jul-08, 07:38 PM
The cars will still finish their respective 1km tracks at the same time, regardless of where they are on Earth or which direction they're going. The Earth's rotation does have an effect on their speed relative to the Earth's axis, but remember that the tracks are also moving at the same speed as the Earth so it all cancels out.

Basically, there's no detectable difference between running (or driving, or flying) east and going west.

Your analogy with the merry-go-round breaks down, because you're actually jumping OFF onto the ground. Instead, think of simply walking around the perimeter of the spinning merry-go-round, either forward or backward. That's more like what the cars are doing. You'll find that it's equally easy to go against the rotation than it is to go in the same direction.

As for jumping off, you're right you will go farther jumping in the same direction as the rotation. This is like getting in a rocket and going to another planet. Earth's rotation DOES have an effect on a rocket's speed; it takes considerably less fuel to go into an orbit in the same direction as Earth's rotation than it does to go into a "backward" orbit. When going to, say, Mars, or sending probes to the outer solar system, Earth's motion around the sun is also used to reduce fuel requirements.

Ken G
2010-Jul-08, 10:18 PM
shouldn't vehicles traveling in the direction of the earths rotation travel faster than those traveling against it?Speed is relative, so it depends on what you are using as your reference. If you use the space in which the Earth spins, it's like the ground around the merry-go-round, and sure enough, the eastbound car is faster. If you use the surface of the Earth, there is no difference in relative speed. Which one "should" you use? You can use either, but which is more convenient depends on the context. As was mentioned, if you are launching a rocket into space, you do point it in the direction of the Earth's spin to pick up that extra speed. But if you want to know which car circumnavigates the Earth first, the answer is, it is a tie (there would be different coriolis forces, which will make the eastbound car seem a little lighter, but let's compensate for that or neglect it). Here's where it really gets interesting: the "explanation" for the tie sounds different in the two ways of looking at it-- for the surface of the Earth, you say it's a tie because both the distances, and the speeds, were the same, but for the space that the Earth is spinning in, you say it's a tie because the eastbound car was both faster and had to go farther to complete the trip. Such is relativity, the testable answers come out the same, but the whys sound totally different.

dgh64
2010-Jul-08, 10:34 PM
I've been thinking about this some more, and there is one instance where a car would notice the Earth's rotation. Imagine you built a track that went from the North pole to the equator. A car driving along that road (in either direction) will feel a sideways force, trying to push it off the side of the road, and a rotational force, trying to spin it around. This is barely noticeable, millions of times smaller than the amount of traction the wheels have, but in an ideal experiment where everything else is perfect, it would be measurable. This is the reason hurricanes spin opposite directions in the two hemispheres: air going north or south is "bent" by the Earth's rotation.

Geo Kaplan
2010-Jul-08, 11:27 PM
I've been thinking about this some more, and there is one instance where a car would notice the Earth's rotation. Imagine you built a track that went from the North pole to the equator. A car driving along that road (in either direction) will feel a sideways force, trying to push it off the side of the road, and a rotational force, trying to spin it around. This is barely noticeable, millions of times smaller than the amount of traction the wheels have, but in an ideal experiment where everything else is perfect, it would be measurable. This is the reason hurricanes spin opposite directions in the two hemispheres: air going north or south is "bent" by the Earth's rotation.

Ken G's answer already acknowledges this effect, in his reference to the Coriolis force.

dgh64
2010-Jul-09, 12:23 AM
Not really... all he said in relation to the Coriolis effect was that a car going east feels a little lighter than one going west. I was pointing out that a pole-to-equator or equator-to-pole path would feel sideways/rotational forces due to moving away/toward the Earth's axis, like a figure skater spinning faster as she pulls her ams in.

Ken G
2010-Jul-09, 12:27 AM
Not really... all he said in relation to the Coriolis effect was that a car going east feels a little lighter than one going west. I was pointing out that a pole-to-equator or equator-to-pole path would feel sideways/rotational forces due to moving away/toward the Earth's axis, like a figure skater spinning faster as she pulls her ams in.I think Geo Kaplan's point is that this is also due to the coriolis effect, but it is certainly true that the coriolis effect manifests differently for east-west motion (where it decreases or increases the object's apparent weight) as for north-south motion (where it deflects east or west), as for radial motion (where it increases the velocity going around, as for the skater pulling in her arms). So no one is really disagreeing.

mr obvious
2010-Jul-10, 02:58 PM
Please correct me if I'm wrong but my understanding of the physics suggests there's a mish-mash of concepts going on.

KenG's statement that an eastbound car is lighter than a westbound car does not have anything to do with the Coriolis force. The weight differences of the car is a result of centripetal acceleration remaining constant (gravity) but the angular speed increasing for the eastbound car relative to the westbound car. If the angular speed increases enough, then the car will be launched into space, which is presumably why rockets are launched eastward and part of the reason why the southeastern location for launches was chosen, at least in the US - closest to the equator and launches over ocean in case of disaster.

dgh64's statement that the pole-to-equator 'sideways force,' which is in fact what I understand the Coriolis force to be, has nothing to do with the figure skater spinning faster as her arms are pulled in. The latter is conservation of angular momentum, whereby if the radius of an effective mass decreases, the (rotational) velocity must increase since the total angular momentum remains constant, in the absence of external forces or torques (ignoring the contribution of ice friction and air resistance as being small).

In the case of an object moving from the pole (low angular velocity) to equator (high angular velocity), there must be a sideways acceleration on the object if the object does not slip on the surface. This is because, once the object is at the equator, it now has all the added momentum associated with the high angular velocity at the equator. This is thus not a conservation situation since there is clearly an external force acting on the object (friction). The net effect is that if I were to take a single step from the pole towards the equator, the foot that lands is landing on a surface that is moving eastward faster than the rear foot. I would then tend to be dragged towards one side (which side depends on whether you're going north or going south, and in which hemisphere). So, in the Northern hemisphere, an air packet moving south would be dragged eastward. An air packet moving north would be dragged westward - net result, hurricanes in the Northern hemisphere rotate counter-clockwise.

Ken G
2010-Jul-10, 03:02 PM
Please correct me if I'm wrong but my understanding of the physics suggests there's a mish-mash of concepts going on.

KenG's statement that an eastbound car is lighter than a westbound car does not have anything to do with the Coriolis force. The weight differences of the car is a result of centripetal acceleration remaining constant (gravity) but the angular speed increasing for the eastbound car relative to the westbound car.Believe it or not, what you are talking about is an example of the coriolis force! It's just said from a different reference frame.

dgh64's statement that the pole-to-equator 'sideways force,' which is in fact what I understand the Coriolis force to be, has nothing to do with the figure skater spinning faster as her arms are pulled in.Yet as I said above, all three are the coriolis effect. They are all forces that appear in the rotating frame, as a result of motion in that frame, which is just exactly what the coriolis force is. All you are doing is choosing to analyze two of them from the nonrotating frame, and reserve analysis in the rotating frame for just one (the north-south motion). This brings up a fascinating point about physics-- the answers sound a lot different when expressed from different points of view, and it can require some detective work to recognize they are in fact the same answer in different clothes.

dgh64
2010-Jul-10, 04:17 PM
dgh64's statement that the pole-to-equator 'sideways force,' which is in fact what I understand the Coriolis force to be, has nothing to do with the figure skater spinning faster as her arms are pulled in. The latter is conservation of angular momentum, whereby if the radius of an effective mass decreases, the (rotational) velocity must increase since the total angular momentum remains constant, in the absence of external forces or torques (ignoring the contribution of ice friction and air resistance as being small).

In the case of an object moving from the pole (low angular velocity) to equator (high angular velocity), there must be a sideways acceleration on the object if the object does not slip on the surface. This is because, once the object is at the equator, it now has all the added momentum associated with the high angular velocity at the equator. This is thus not a conservation situation since there is clearly an external force acting on the object (friction). The net effect is that if I were to take a single step from the pole towards the equator, the foot that lands is landing on a surface that is moving eastward faster than the rear foot. I would then tend to be dragged towards one side (which side depends on whether you're going north or going south, and in which hemisphere). So, in the Northern hemisphere, an air packet moving south would be dragged eastward. An air packet moving north would be dragged westward - net result, hurricanes in the Northern hemisphere rotate counter-clockwise.

Coriolis effect = air (or cars) moving north being dragged westward, air/cars moving south being dragged eastward. This is a form of conservation of angular momentum.

Think of it this way: Imagine everyone drove their cars to the equator, and then measured very carefully the Earth's rotational speed. Then if everyone drove to the north and south poles and measured the speed again, they'd find the Earth spinning slightly faster, because the mass of all those cars is now closer to the axis, like the skater pulling her arms in. During the drive, that extra velocity comes from an east-west force between the tires and the road. So, we're both right -- it's the Coriolis effect AND the ice-skater-pulling-in-arms effect.

mr obvious
2010-Jul-10, 04:38 PM
"They are all forces that appear in the rotating frame, as a result of motion in that frame, which is just exactly what the coriolis force is."

This is overly broad. For example, a car racing along the equator experiences friction. This is a force that appears in the rotating frame and is a result of motion in this frame but is not the Coriolis force.

The car going east is lighter because its 'centrifugal' force (inertial force) is higher, whereas the centripetal force (gravity) is the same. The same effect can be achieved by keeping the centrifugal force the same and decreasing Earth Mass (while keeping its rotation constant, so this is not a conserved situation). This is not a Coriolis force phenomenon. Actually the inertial (centrifugal) force is itself a force in the rotating reference frame that results from motion in the frame but is separate from the Coriolis force. I do not see how you can say they are the same. If I'm standing on the edge of a spinning disk, I'm experiencing the 'centrifugal' force but no Coriolis forces. This is because the Coriolis acceleration can be expressed in dynamics as 2wXv, where w is the rotation rate of the axis, v is the velocity relative to the rotating reference frame (zero in this case) and X is the cross product of these vector quantities. Since v is zero, so is the Coriolis acceleration.

So I'm not sure I understand your claim of equivalence.

mr obvious
2010-Jul-10, 04:55 PM
I realized after posting that you may claim foul that my calculation of Coriolis force on someone standing on a disk has no motion in the reference frame, and is thus not a valid rebuttal, but the point I was trying to make is that the car getting lighter is a result of amplification of the centrifugal force, which is not the Coriolis force, since I can zero one without zeroing the other.

Ken G
2010-Jul-10, 11:02 PM
For example, a car racing along the equator experiences friction. This is a force that appears in the rotating frame and is a result of motion in this frame but is not the Coriolis force.That would be there in the nonrotating frame too. I'm talking about forces that appear because the frame is rotating and because there is motion. That should clarify it.

mr obvious
2010-Jul-11, 01:55 AM
If I understand you correctly, then we agree on what the Coriolis force is. However, neither conservation of angular momentum, nor the change in "centrifugal" force appear because the frame is rotating. Both phenomenon are true in nonrotating reference frames. Thus, neither of those can be said to result from the Coriolis force, according to your definition.

Ken G
2010-Jul-11, 02:09 AM
If I understand you correctly, then we agree on what the Coriolis force is. However, neither conservation of angular momentum, nor the change in "centrifugal" force appear because the frame is rotating. Both phenomenon are true in nonrotating reference frames. Thus, neither of those can be said to result from the Coriolis force, according to your definition.What is confusing you is that in one frame, you have neither the centrifugal nor the coriolis. In another (a spinning frame), you have only centrifugal (you may have chosen a frame so there is no motion), and in another, you have both centrifugal and coriolis. So all the distinctions are frame-dependent. The frame I am choosing is the frame of the rotating Earth. In that frame, the lightening of east-west cars is indeed the coriolis force, not the centrifugal force. You can choose a different frame, one that makes the east-west car stationary, and then it will be centrifugal force.

mr obvious
2010-Jul-11, 03:56 AM
I see what you are saying (at least the main point). In the reference frame of the Earth, 2wxV is directed perpendicular to the Earth surface. Thus the Coriolis force may be contributing some or all of the weight-lightening effect. As I alter the rotation rate of the reference frame, this force can increase or decrease. I'm not sure how to reconcile that with the dynamics equation. Forgive me for cluttering with math terms, but it'll make my explanation easier (I hope) to understand.

In a reference frame rotating at rate w, a particle is moving in some path with velocity V and acceleration A relative to this rotating reference frame. The absolute acceleration of the particle in an inertial (nonrotating reference frame) is
a= wX(wXr) + 2wXV + A
where I ignore the rectilinear acceleration of the rotating reference frame and fix the rotation rate at a constant w (thus dw/dt = 0).

The wX(wXr) term is a major part of the centripetal acceleration we know and love. Centrifugal acceleration arises as the reaction to this acceleration. This stays constant at a given rotation rate of the rotating reference frame.

The term 2wXV is the Coriolis acceleration. As you correctly noted, if we shift the reference frame to a new w' such that the car is not moving in that reference frame, then V goes away and all we have is the centrifugal acceleration. In our Earth-stuck reference frame, however, the east-west car has some net velocity V, so the Coriolis acceleration is non-zero. Using vector math, one can derive that this acceleration is perpendicular to the Earth's surface and is presumably the reason you stipulate the car's weight is being affected.

We also have the last term, the relative acceleration. This term has magnitude V^2/r, where V is the net velocity of the car with respect to the Earth's surface (same V as in the Coriolis term). This term arises because of the centripetal acceleration of the car in the rotating reference frame. Because the car is not traveling in a straight line at constant velocity in the Earth-stuck reference frame, it must have some acceleration. This term is directed towards the Earth's center, which is the center of the radius of curvature of the car's path. This term is the second centripetal acceleration term.

If we assume the Earth is a perfect sphere and the car is moving along the equator, then all of the vectors are orthogonal and I think all vectors point to the Earth center. Thus, in the Earth stuck reference frame,
a = (w^2)r+2wV+(V^2)/r
This acceleration has two contributions from centripetal acceleration (and the corresponding centrifugal acceleration) and one term from the Coriolis contribution. Note that the Coriolis contribution is only part of the acceleration.

What if we were to stop the rotating reference frame so that the Earth spins in the non-rotating reference frame? Then the car is now zipping along the equator at velocity U=wr+V. The wr is from the Earth spinning and the V is from the car's motion relative to the Earth's surface. In this case, the acceleration of the car is (U^2)/r = [(wr)^2 + 2wrV+V^2 ]/r = (w^2)r+2wV+(V^2)/r
Hey, the accelerations match! In this case, there is no Coriolis component; it's all centripetal (you get the same thing in a rotating reference frame where the car is stationary). However, one can see the Coriolis contribution term lurking in the 2wV term.

Ok, my point is that centripetal acceleration (and it's inertial resulting centrifugal acceleration) is distinct from the Coriolis force. In neither the stationary frame, nor the Earth-attached frame do you ever have only Coriolis contribution. The lightening of the car is a result of all of these effects in combination. In some cases, you can explain it all with centripetal acceleration. In other cases, centripetal acceleration must be supplemented by the Coriolis effect (which is why it was made to begin with - to make sure that the answers match when using a non-inertial reference frame). I do not believe you can ever choose a purely rotating reference frame which yields only the Coriolis force and no centrifugal force. Ergo, the centripetal acceleration and Coriolis acceleration are NOT the same thing. The Coriolis acceleration is a correction term intended to 'fix' a gap created by using a non-inertial reference frame.

That's my understanding of it.

Ken G
2010-Jul-11, 07:39 AM
That's all true. Note I was not saying that the only thing that could lighten the car is the Coriolis effect, I said the Coriolis effect is the thing that is lightening the east-west car in comparison to the west-east car, in the frame rotating with the Earth. It's true one cannot only have a coriolis force and no centrifugal force, but if you stick to a single rotating frame throughout, the centrifugal force will never be different between the two cars, whereas the coriolis force will (as will the net acceleration, as you point out). That's what I was alluding to, differences that could affect the comparison of race times, as seen in a frame (rotating with the Earth) where the cars are traveling at the same speed.

But this is no doubt rather an aside to the needs of the original question-- we can probably say that all the rotational fictitious forces are insignificant to the OP, and just put the question to zenbudda: is your question now resolved by the recognition that in the nonrotating frame, the two cars travel different distances at different speeds and the race is a tie, and that in the rotating frame, they travel the same distance at the same speed?

mr obvious
2010-Jul-11, 10:32 AM
In the Earth-stuck reference frame, the relative acceleration (my A term) depends on V^2/r and is different between the two cars. Since this is a centripetal acceleration, your statement that the centrifugal force will never be different between the two cars is puzzling. Similarly in the inertial reference frame the centripetal acceleration is all there is, so that is definitely different between the two cars, and there is no Coriolis force component.

Basically I'm responding to the part where I distinguished conservation of angular momentum, centrifugal force and Coriolis forces as distinct entities and you claimed all three are the same. They are not the same. I have shown mathematically that it is not the case that the Coriolis force is the same thing as the centrifugal force, only in a different reference frame. We haven't discussed angular momentum but I hope it's clear that that isn't the same thing as the Coriolis force, either.

To address the relevant part of zenbudda's main question, these acceleration terms we discuss are all oriented perpendicular to the direction of travel by the cars, regardless of whether they are going east-west or west-east. Thus, these accelerations (not fictitious) do not contribute to the motion of the cars (other than perhaps changing the amount of work done against friction) and do not impart any boost to the car going in either direction.

Ken G
2010-Jul-11, 02:48 PM
In the Earth-stuck reference frame, the relative acceleration (my A term) depends on V^2/r and is different between the two cars.True.

Since this is a centripetal acceleration, your statement that the centrifugal force will never be different between the two cars is puzzling. Similarly in the inertial reference frame the centripetal acceleration is all there is, so that is definitely different between the two cars, and there is no Coriolis force component.This is getting a little technical, but the answer is that centripetal acceleration can be different between the cars, but centrifugal force will be the same unless you want to analyze the two cars in two different rotating reference frames. In effect the latter is just what you are doing, but I'm taking a single frame (the Earth-stuck frame). It's just plain true that in that frame, the centrifugal force is the same on both cars, and the coriolis force accounts for every difference between the two cars, regardless of whether they are driving north-south or east-west. Angular momentum is not conserved in the rotating frame, and in the non-rotating frame, where it is conserved, there are neither coriolis nor centrifugal forces.

mr obvious
2010-Jul-11, 03:20 PM
Centrifugal force is just the result of inertia acting on an object undergoing centripetal acceleration. In my experience, you cannot have one and not the other. Using only the Earth-stuck reference frame, the wXwXr term is the same for both cars, but the (V^2)/r term is not. But the latter term is centripetal acceleration, which contributes to centrifugal force. The Coriolis force is 2wXv and only accounts for PART of the additional force in this reference frame. This is unless you wish to claim that (V^2)/r is Coriolis as well, which I have never seen in any texts.

It is correct that angular momentum is conserved in a non-rotating reference frame, and that there are no Coriolis forces in that frame. That is why I'm puzzled by your claim that conservation of angular momentum is the 'same thing' as the Coriolis force. You said "Yet as I said above, all three are the coriolis effect. " But they are not. The Coriolis effect is a correction term, and is distinct from centrifugal "force" and angular momentum.

dgh64
2010-Jul-11, 03:36 PM
If Earth were a cylinder and not a sphere, there would be no Coriolis effect, correct?

This is why I thought the Coriolis effect was a form of angular momentum conservation -- you only feel the effect when going north or south on a sphere, where such motion also involves moving closer/farther from the axis. Am I wrong in this assumption?

mr obvious
2010-Jul-11, 03:53 PM
Incorrect. You can have a Coriolis effect in a rotating reference frame, even if the object you are tracking is moving in a straight line at constant velocity. The shape of the surface is not important. The Coriolis effect disappears when either the reference frame is non-rotating (w=0) or when the object being tracked has no velocity in a direction other than the direction of the rotation axis (wXv=0), within the rotating reference frame.

I also don't know what it means for a 'force' to be a "form of" a conservation law. This is like saying a length is a "form of" a pasta. A simple form of the conservation of angular momentum may be described as saying that in the absence of external moments, the angular momentum remains a constant value. It's a principle that constrains how systems behave, based on our models and assumptions.

Ken G
2010-Jul-11, 04:55 PM
Centrifugal force is just the result of inertia acting on an object undergoing centripetal acceleration.No, that is not the definition of centrifugal force. For one thing, that definition doesn't even mention the most important thing-- the reference frame from which the centrifugal force arises. The mathematical definition of centrifugal force can be looked up anywhere, and uses only the spin rate of the frame, and the location of the object within that frame (and its mass of course, but that's just because it's a fictitious force so must be proportional to mass).
Using only the Earth-stuck reference frame, the wXwXr term is the same for both cars, but the (V^2)/r term is not.And only the former is the centrifugal force, in that frame.
But the latter term is centripetal acceleration, which contributes to centrifugal force. No, the definition of centrifugal force is the former term and is not "contributed to" by any other term. Your confusion is that you are continually mixing reference frames. Stay in the spinning frame of the Earth, and all the language has a precise meaning. Mix frames, and the meanings get mixed up.

It is correct that angular momentum is conserved in a non-rotating reference frame, and that there are no Coriolis forces in that frame. That is why I'm puzzled by your claim that conservation of angular momentum is the 'same thing' as the Coriolis force.It is indeed the same thing, appearing in the different reference frames that one might possibly wish to choose (and stick with).

mr obvious
2010-Jul-11, 05:12 PM
I think I'm reached the limit of what can be communicated by message posting. We aren't communicating and I don't know why. I'm basing my comments on physics and dynamics. You say that centrifugal force uses only the spin rate of the frame (mass and location). You claim this can be looked up anywhere. I do not see that in my texts. I also used wikipedia, which I think counts as 'anywhere.' Here's what it says:

"In Newtonian mechanics, the term centrifugal force is used to refer to one of two distinct concepts: an inertial force (also called a "fictitious" force) observed in a non-inertial reference frame, and a reaction force corresponding to a centripetal force. "

Thus my statement that centrifugal force is a result of inertia acting on an object undergoing centripetal acceleration is correct. You have both a rotating reference frame plus an additional centripetal acceleration because the car is accelerating in the rotating reference frame. (V^2)/r is the latter contribution. It is not the Coriolis effect. If you say it is, then we are at an impasse since none of my texts say that. All references say that the Coriolis force is 2wXV that I'm aware of. *shrugs*

You say that the conservation of angular momentum is the same thing as the Coriolis effect, with just a reference frame shift. The two concepts are distinct and cannot be a reference frame shift. It's like saying length and velocity are the same thing, only in different reference frames. That makes no sense. One is a principle, the other is a mathematical term, not a formula. If you wish to say that angular momentum is the same thing as the Coriolis force, then again you have a units mismatch but I'd be more willing to consider that possibility.

In rectilinear coordinates, what you are saying is analogous to saying that conservation of linear momentum is the same thing as friction, only seen from a different reference frame. Friction is a force added to correct for the fact that objects in the physical world do not satisfy Newton's laws without it. The objects slow down 'by themselves' if you omit friction. Ergo, we need to add in a force that opposes the direction of motion, which is friction. Saying it's just the conservation of momentum, in a different reference frame, makes no sense to me. True, you need friction to make the conservation of momentum hold. But the conservation of momentum holds in the absence of friction - it is a principle that's far broader than friction. They are not the same thing. Similarly for Coriolis and conservation of angular momentum.

you state that the conservation of angular momentum is the same thing as Coriolis effect. The Coriolis acceleration, you agreed, is Ac=2wXV. You asked for a concrete question. Here it is:
How do you mathematically go from Fa=2wXV to the conservation of angular momentum (CAM) by changing only the reference frame? We start with your reference frame stuck to the rotating Earth. So, go ahead, show me the derivation of CAM by changing the reference frame, starting from Ac=2wXV. You may use Fc=2mwXV if you wish to use forces instead of accelerations; we ignore relativistic effects and assume mass is fixed.

Ken G
2010-Jul-11, 05:27 PM
I also used wikipedia, which I think counts as 'anywhere.' Here's what it says:

"In Newtonian mechanics, the term centrifugal force is used to refer to one of two distinct concepts: an inertial force (also called a "fictitious" force) observed in a non-inertial reference frame, and a reaction force corresponding to a centripetal force. "
Read on. That statement describes the centrifugal force from two different reference frames. The first version is described in the rotating frame, the second version is described in the nonrotating frame. You never get both at the same time, that is mixing frames, which is what I'm trying to get you to see you are doing. In the spinning frame of the Earth, the second part of that Wiki sentence simply does not appear, nor will that Wiki entry imply that it does.

You say that the conservation of angular momentum is the same thing as the Coriolis effect, with just a reference frame shift.And I clarified the meaning of that remark, which is that you can analyze the resulting motion in one frame via the coriolis force, and in another frame via the conservation of angular momentum. There is no problem with units mismatch one is never writing an equation where one is equated to the other-- indeed, we are not even considering reference frames where both exist at the same time.

If you wish to say that angular momentum is the same thing as the Coriolis force, then again you have a units mismatch but I'd be more willing to consider that possibility. I wish to say that the coriolis force is the force that causes east-west deflection of a north-south car in the spinning frame, which in the non-spinning frame can be analyzed as conservation of angular momentum, and it is the force that causes lightening of east-west cars in the spinning frames.

mr obvious
2010-Jul-11, 05:43 PM
Unfortunately, the wiki equation that follows did not separate the terms because they assumed the object was stationary in the rotating reference frame. Thus their end equation only had the centrifugal component and the Coriolis. Note that in my derivation, there's an additional A (relative acceleration term) which is in fact needed for this analysis. And yes, I know the frames aren't to be mixed. My mathematical analysis should've demonstrated that amply.

"I wish to say that the coriolis force is the force that causes east-west deflection of a north-south car in the spinning frame, which in the non-spinning frame can be analyzed as conservation of angular momentum, and it is the force that causes lightening of east-west cars in the spinning frames. "

Using different techniques to get the same answer does not mean equivalence of the techniques.
Bolding mine: conservation of angular momentum is not a force. That is why I say you have a units mismatch. You can analyze it using the conservation of momentum principle. The Coriolis force was added in to make sure the conservation principle (among other things) holds but is not the same thing as the conservation principle. See my friction argument. You are saying that an equation that obeys a principle is the same thing as the principle.

dgh64
2010-Jul-11, 05:45 PM
Incorrect. You can have a Coriolis effect in a rotating reference frame, even if the object you are tracking is moving in a straight line at constant velocity. The shape of the surface is not important. The Coriolis effect disappears when either the reference frame is non-rotating (w=0) or when the object being tracked has no velocity in a direction other than the direction of the rotation axis (wXv=0), within the rotating reference frame.

Are you saying that moving north or south on a spinning cylinder will cause east/west forces? I find that hard to believe.

I also don't know what it means for a 'force' to be a "form of" a conservation law. This is like saying a length is a "form of" a pasta. A simple form of the conservation of angular momentum may be described as saying that in the absence of external moments, the angular momentum remains a constant value. It's a principle that constrains how systems behave, based on our models and assumptions.

I may have mixed up my terminology, but I was saying the same thing that Ken G appears to be saying -- in a non-rotating reference frame, it looks like conservation of angular momentum, and in the frame rotating with Earth it's the Coriolis force. Both points of view are equally valid.

mr obvious
2010-Jul-11, 06:07 PM
I said you CAN have a Coriolis effect when moving in a straight line at constant velocity. I didn't say you always did. I state the exceptions explicitly: the Coriolis effect disappears, when the object being tracked has no velocity in a direction other than the direction of the rotation axis (this is just one condition, but it's the relevant condition here). Since moving north or south along an axially-spinning cylinder is in the direction of the rotation axis, the Coriolis effect disappears because wXV is zero.

I don't know what you mean by 'it looks like x' in reference to physical laws. You can apply the law or not; if the conditions under which the principle applies are true, then it the principle applies whether it looks like it to you or not. Saying that the two are different ways of solving the same thing does not make the two ways the "same thing." Perhaps you can tell me how you use the conservation of angular momentum in a non-rotating reference frame to show the eastward-moving car is lighter than the westward-moving car. Keep in mind the Conservation of angular momentum states that in the absence of external moments, that L=rXmv is constant. You may use the moment of inertia configuration if you wish.

Ken G
2010-Jul-11, 08:03 PM
Note that in my derivation, there's an additional A (relative acceleration term) which is in fact needed for this analysis. And yes, I know the frames aren't to be mixed. My mathematical analysis should've demonstrated that amply.Just choose a frame, say it's the Earth's spinning frame. Now write A=F/m in that frame. You will have 5 terms. One is the real forces per gram on the car (here the normal force upward from the ground). A second term is the gravitational acceleration (g) of the Earth. A third is the centripetal acceleration term (A). The last two are the accelerations from the coriolis force (2wXv) and the centrifugal force (wXwXr). The only two terms that are different between the east-west and west-east cars are the normal-force term (that's the lightening of the car I mentioned, weaker contact with the road) and the coriolis term (2wXr). Since only two terms are different, the one is the cause of the other. That's all there is to it.

mr obvious
2010-Jul-11, 08:52 PM
Yes, that is the analysis that I did before (I skipped gravity as understood), with one correction that if the east-bound and west-bound cars are moving at the same speed (but in opposite directions) then yes, the centripetal accelerations are identical - I was being more general than that, so I allowed the speeds to be different so long as the directions were opposite.

But in your case, with the assumption that the cars are moving at V eastward and V westward, it is true that all you have is the Coriolis force leading to the difference in lightening. This does not mean that the Coriolis force is the 'same' thing as 'centrifugal' force or conservation of angular momentum. It is that part I'm having the problem with since you say all are the same forces, but conservation of angular momentum is not a force, and your own mathematics shows that you have a centrifugal (centripetal) component distinct from the Coriolis. Further, you agreed previously, that you cannot have only Coriolis and no centrifugal forces.

Thus, I agree with your initial claim with a minor revision: that the lightening of the cars relative to each other can be explained using the Coriolis force, but I do not agree with your contention that these mean that all three concepts are the same thing; only that a small part of the overall lightening - the difference between the cars - can be swapped between Coriolis or centripetal forces inside the Earth or nonrotating frame, respectively. This is a highly particular situation to let it occur and should not be taken to mean that centrifugal forces = Coriolis forces = conservation of angular momentum. Actually, I fail to see how ang mom comes into this at all, so I'm waiting for dgh64's explanation for that.

Ken G
2010-Jul-11, 09:10 PM
But in your case, with the assumption that the cars are moving at V eastward and V westward, it is true that all you have is the Coriolis force leading to the difference in lightening.Excellent, I'm glad you have withdrawn your objection to that claim. That was my purpose to establish, along with the fact that it is probably not relevant to the OP anyway.
This does not mean that the Coriolis force is the 'same' thing as 'centrifugal' force or conservation of angular momentum.You merely misinterpreted my meaning, but I think that has now also been clarified, as the whole issue was that first point.

astromark
2010-Jul-11, 11:33 PM
Back at the OP... NO.

As Ken has covered this so well and has nailed the facts of it well.

Its true there is a effect. You can try to measure it... I will watch.:eh:

There is NO perceivable result or test. Ask a lecture hall filled with the knowledgeable to demon straight ... they can not.

You tell them of velocities of rotation and orbit and remove them from a view of the sky and they can try with much arm waving...

Show the mathematics and you might get a raised brow. How would you demon straight this ?

mr obvious
2010-Jul-11, 11:51 PM
Since the conservation of angular momentum has not been shown to relate to this problem I do not think it's safe to say that the whole issue was the first point. The claims made were far stronger than what the 'objection withdrawn' part was. It's as if you said that all x are y, and I said it's true that one x is y and you said, 'problem solved, that's all I was saying.' It's true you did say that, but it's also true you said things far stronger than that, which were and are not supported.

On the other hand, I don't particularly think it's terribly fun to keep this thread sidetracked..

Ken G
2010-Jul-12, 12:13 AM
Yeah, it's probably best to just see if the issue crops up again somewhere else, and if not, it probably doesn't matter so much. Everyone agrees on the answer to the OP, so hopefully the OPer is satisfied with these answers.

dgh64
2010-Jul-12, 01:49 PM
All I was really doing was trying to put it into some plain English that I could understand, because it's been a loooong time since I did any vector multiplication.

That's I meant by the "form of" and "looks like" comments that mr obvious took issue with, just trying to use non-mathy terms that more people can understand. It's not really necessary to figure out exact numbers/formulas for things, all that's required is to say "yes there's a force in X direction, it's because of Y law/principle". I guess what I meant was that the Coriolis effect is one of the ways in which angular momentum is conserved, in a different reference frame. So when someone says "X doesn't happen because of conservation of angular momentum, it happens because of the Coriolis effect." that's not quite true, they're just using a different reference frame, but the other one, the one where it's angular momentum, is equally valid.

And you two are right, the OP has been answered, and I'm tired of arguing about this, and having my grammar and word choice nitpicked all the time.

Ken G
2010-Jul-12, 03:07 PM
I guess what I meant was that the Coriolis effect is one of the ways in which angular momentum is conserved, in a different reference frame. So when someone says "X doesn't happen because of conservation of angular momentum, it happens because of the Coriolis effect." that's not quite true, they're just using a different reference frame, but the other one, the one where it's angular momentum, is equally valid.Yes, that was what I was trying to say also.

And you two are right, the OP has been answered, and I'm tired of arguing about this, and having my grammar and word choice nitpicked all the time.For my own part, I understood what you were saying. Sometimes communication is quite tricky.

zenbudda
2010-Jul-12, 09:45 PM
i am satisfied. i like details beyond a simple explanation. i do like simple explanations too, especially if they are accurate. :clap: ;)

after reading all of the responses and trying to absorb the major points of each, i have another question on this topic. do we currently measure the speed of the earths rotation on it's axis by the circumference of the earth and the length of our days? i have this visualization of some horrible experiment of measuring the earth's speed, resembling an airplane's wheels hitting the tarmac as the plane lands.

dgh64
2010-Jul-12, 11:37 PM
You mean, how do they calculate the speed of a point on the equator? They just divide the circumference by 24 hours (in a day). Distance/time=speed.

The airplane thing wouldn't work, because the air it flies through is going at (more or less, not counting wind) the same speed as the ground.

zenbudda
2010-Jul-13, 09:09 PM
my airplane wheels on the tarmac example was actually contrived to be much larger than an actual airplane. the airplane would be say the size of jupiter and defies the laws of gravity and hovers above the earth, not in orbit, that stuck a device out onto the earths surface and measured the speed. yeah, impossible. lol. i wonder what kind of drag that would create around the pole that's sticking out from this object into our atmosphere would be. :wall: lol.

Ken G
2010-Jul-13, 09:28 PM
Interestingly, even the ancient Greeks could have figured out how fast the equator was moving, if they had thought the equator was moving at all. They had the circumference of the Earth to pretty good accuracy, and they had the day of course. I can't help wondering if the result they got (almost 2000 kilometers an hour) was a speed so incomprehensibly fast to them, that it added to their conviction the Earth could not spin like that. Think about it-- if the fastest speed you yourself ever traveled in your life was like 20 kilometers per hour, or some such thing, how could you accept that you were already moving almost 100 times that speed all the time? And worse for an orbiting Earth-- then it's closer to 5000 times faster than your daily experience! It's no wonder they were skeptical.

grant hutchison
2010-Jul-13, 11:36 PM
Interestingly, even the ancient Greeks could have figured out how fast the equator was moving, if they had thought the equator was moving at all.Many of them, especially the Pythagoreans, thought that the Earth did rotate. Aristotle took a disapproving swipe at his contemporary, Heraclides, who advocated the rotation of the Earth as the cause of night and day. Eratosthenes worked out a good radius for the Earth just a century later, so I'm sure someone must have thought about rotation velocity, though I can't recall every encountering a record of a named individual. Aristotlean mechanics considered that a dropped object couldn't fall vertically in such a rotating frame, so there was a major impediment to acceptance even before the magnitude of the rotation velocity was considered.

There's a fine old Isis review of the history of the theory of diurnal rotation here (http://www.jstor.org/pss/224925), but you'll need an institution login (or \$14) to access it.

Grant Hutchison

Ken G
2010-Jul-14, 12:26 AM
That's interesting, and yes we do tend to treat the "ancient Greeks" as some monolithic body of thought, when I'm sure they had as many debates as we have today. I was just wondering if perhaps the huge speed required was fodder for the Aristotelian camp, just as was the huge distance required to the stars to not see stellar parallax. A lot of reasoning from incredulity might have been going on-- it's always a surprise how incredible things really are!

Jens
2010-Jul-14, 07:18 AM
after reading all of the responses and trying to absorb the major points of each, i have another question on this topic. do we currently measure the speed of the earths rotation on it's axis by the circumference of the earth and the length of our days?

I think the problem is that if you are really wanting to measure the speed of the earth's rotation, the proper way would be to use something like RPM. With the answer being something like 1/day. Because obviously, the speed you are moving will be different at the surface and as you go toward the core. By the time you get to the middle, you no longer have any speed.