Talk:Centrifugal force/Archive 17

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Now what?

I have changed the article so that it reflects current thinking in the subject but what now? Should we expand some sections, if so which, or should we merge? Martin Hogbin (talk) 16:47, 25 November 2014 (UTC)

Accelerating?

Ther has been some discussion over the words, 'Centrifugal force is an outward force apparent in a reference frame which is accelerating and also rotating'.

Thers is certainly no need for the words 'which is accelerating', in fact it is not cleart what it intended to mean. I assume that 'accelerating' is intended to mean 'linear acceleration'. All that is required for CF is a rotating reference frame, nothing else. It is true, that every point fixed in a rotating reference frame, (except one - the centre of rotation) is accelerating but that is not what those words suggest to me. Martin Hogbin (talk) 08:28, 25 May 2015 (UTC)

I agree. The general formula relating the coordinates of a point in two different coordinate systems with orthogonal axes and the same unit of distance is r_x = r₀ + Ω x, where r₀ is a row column vector whose components are the coordinates of the origin of one system in the other, r_x and x are row column vectors whose components are coordinates of the point in the two systems, and Ω is an orthogonal matrix (also called a "rotation matrix"—i.e. one which satisfies the equation Ω^-1 = Ω^T). In the terminology I am familiar with, the coordinate systems are said to be rotating with respect to each other if Ω is not constant (i.e. its first derivative, Ω̇ , is not zero), or accelerating with respect to each other if the second derivative of r₀ is non-zero. If one of the coordinate systems is inertial then the first condition (i.e. Ω̇ ≠ 0) is necessary and sufficient for centrifugal forces to appear in the other. While they will also still appear if ${\ddot {\mathbf {r} }}_{0}\neq 0$ (i.e. the non-inertial frame, as well as rotating, also has an accelerating origin), this is simply not necessary.

David Wilson (talk · cont) 09:49, 25 May 2015 (UTC)

On pondering Montyv's edit summaries, I think I can see what he is driving at. He will no doubt provide a correction if I have misunderstood. From his comment about a frame being "pinned to a centripetally accelerating object" in "the most common scenario", he would appear to be here thinking of a frame whose origin is fixed with respect to this centripetally accelerating object. In this scenario it is true that the frame (i.e. its origin) is accelerating, and that this will give rise to a pseudoforce in the frame's coordinate system. I think it's probably also true that in some circumstances this pseudoforce might sometimes be loosely referred to as a "centrifugal force". However, it is most definitely not what this article or the article Centrifugal force (rotating reference frame) is referring to as a "centrifugal force" for the frame in question.

David Wilson (talk · cont) 14:52, 25 May 2015 (UTC)

As you say above in mathematical language, the correct situation is very simple, if a frame is rotating there will always be centrifugal forces acting (on any massive particles not on the axis of rotation), if the frame is not rotating there will never be. Martin Hogbin (talk) 18:44, 25 May 2015 (UTC)

Hi guys. Yeah. I know it's normal to to say things like "rotating frame", etc. here. But, I think such terminology often gets used carelessly in the WP articles on this subject. I hope I can convince you that we all need to be more careful about it. Here I go....

Case 1). Imagine a linearly accelerating rocket in free space. Then imagine the smallest (almost zero) rotation rate of the rocket around an axis perpendicular to the thrust. This is a rotation rate of the direction of acceleration. The previously linear acceleration is now a centripetal acceleration no matter how small the omega as long as omega is not actually zero. The rocket is now always pointing at the center of a circle (the center of curvature of the path). That center will indeed in all probability have a non-zero velocity compared to one's preferred "non-moving" frame, but it's still the center of the circular motion of which we speak. All the "rotation" does here (of the direction of acceleration, not of a frame) is to keep a linearly accelerating object (the rocket) inside a comparatively confined space whereas it would otherwise run rapidly away to infinity. Remember again that it's the direction of acceleration that's rotating, and not necessarily the frame. Whether the frame is also rotating has no bearing on the "production" of centrifugal force (CF). Since CF is by definition aligned away from the center and along the line between the center and the rocket, that line between those two points fixes the direction of the CF and there's no need to use a frame that's rotating along with it. Rotating the frame merely changes how you specify that direction. The frame need only be attached to the rocket for centrifugal pseudoforce to appear, the frame doesn't need to be rotating along with the rocket.
CF is so closely related to linear acceleration that they're essentially the same situation. The example with the near-zero omega of the rocket illustrates this. In fact, like I've said in the edit summaries, it's the acceleration that makes the pseudoforce (or the reaction force depending on your world view). It's the acceleration of the object/frame that is the heart of the phenomena. The only thing rotation has to do with it is to add the "centripetal" (of "centrifugal") adjective to what is in essence a linear acceleration. Rotation of the frame is necessary for Coriolis force (I think), but not for centrifugal force.

Case 2). Here's another good one. Consider objects that are not moving and a frame that is rotating who's center is also not moving. Here we have a rotating frame and no centrifugal force. In fact, we now must invent a centripetal pseudoforce to get the objects to stay in their now-circular motion around the frame's origin. How wacky is that?!  :-) Surprising huh? It's my example to show that "rotating frame" doesn't automatically make for centrifugal force. If centrifugal force were to be a thing in this example, it would have to be a pseudo-pseudoforce, and that makes my brain hurt!

Yours, Montyv (talk) 05:57, 26 May 2015 (UTC)

I thought it was only the Euler force that needed an acceleration separate from the rotation. Dbfirs 07:02, 26 May 2015 (UTC)

Yes, as the article makes clear, the Euler force, 'is an acceleration that appears when a non-uniformly rotating reference frame is used for analysis of motion and there is variation in the angular velocity of the reference frame's axes'. In other words it applies specifically to angular acceleration. Martin Hogbin (talk) 07:13, 26 May 2015 (UTC)

Montyv, it really is as simple as I say above, '...if a frame is rotating there will always be centrifugal forces acting (on any massive particles not on the axis of rotation), if the frame is not rotating there will never be'. Centrifugal force is a direct result of doing mechanics in a rotating reference frame.

Problems like the ones you give above are routinely given to undergraduates to test their understanding of the subject. In all cases the problem can be analysed in both rotating and non-rotating reference frames without conflict. What you are trying to add to the article is essentially your own original research. Martin Hogbin (talk) 07:08, 26 May 2015 (UTC)

There are so many statements in your first example which I don't understand that I can't see how I can provide any response that would be helpful. For your second example, however, your statement that there are no centrifugal forces is simply wrong (at least for the concept of centrifugal force which is the subject of this article). If the rate of rotation is ω , then any object of mass m at distance r from the axis of rotation will be subject in the rotating frame to a centrifugal force of magnitude m ω² r directed radially outward from the axis of rotation. Since the object will be moving with a velocity v = ω r in the rotating frame around the circumference of a circle perpendicular to, and centred on, the axis of rotation, it is also subject in that rotating frame to a coriolis force of magnitude 2 m ω v = 2 m ω² r directed radially inward towards the axis of rotation. The resultant of these two pseudoforces is one with magnitude  m ω² r directed radially inward towards the axis of rotation, which is exactly what is needed to produce the acceleration of magnitude ω² r in that direction which the object is subject to in the rotating system. No invention of any other centripetal pseudoforces is at all necessary.

David Wilson (talk · cont) 15:20, 26 May 2015 (UTC)

Regarding Case-2, DW said "If the rate of rotation is ω , then any object of mass m at distance r from the axis of rotation will be subject in the rotating frame to a centrifugal force of magnitude m ω2 r", but that can't be. Maybe I didn't describe the scenario well. The objects are motionless in an inertial frame, the rotating frame's origin is also motionless in that same inertial frame. The pseudoforce in that rotating frame needed to describe the object's now-circular motions in that frame must be centripetal, not centrifugal.
The "kind" of centrifugal force which is the subject of this article is both the pseudoforce perspective and the reaction perspective (they're actually the same thing, but that's a debate for another time). In both perspectives for this case-2 scenario, there is no centrifugal force, just a centripetal pseudoforce.

Regarding Case-1 "original research", yes of course it's an example invented by me to illustrate a point on a talk page. It's something people frequently do on talk pages. It would only be the forbidden "OR" if it was included it in the text, which it isn't. The example is given just to show how it's the instantaneous acceleration, which is momentarily linear, of a frame attached to the object (or of the object itself) that is the genesis of centrifugal force. Acceleration is the property of the frame (or state of the object) that matters.

Regarding rotation of a frame vs. Rotation of the direction of the velocity vector. MH said "...if a frame is rotating there will always be centrifugal forces acting (on any massive particles not on the axis of rotation), if the frame is not rotating there will never be". Rotation of the frame does not matter. The only rotation that matters is the rotation of the direction of the centripetal component of acceleration, which is the same as the rotation of the direction of the velocity vector. And, all that rotation does is allow us to call the forces "centripetal" or "centrifugal" and whatnot, and fix the direction of those forces.
There is simply no need to rotate a frame at the same rate to produce those central-oriented forces and pseudoforces. Consider a frame pinned to an object in circular motion (such as a ball-on-string). Make that frame rotate at a rate faster than the rate of change of the direction of the velocity vector, say by a factor of 1.5. What would be the centrifugal force in that frame? Answer: the centrifugal force would be the same magnitude and direction no matter the rate of rotation of the frame. The choice of frame rotation rate only effects the numbers used to specify the direction. If the rotation rate of the frame is chosen as zero, the centrifugal forces are not zero because they depend not on the rotation of the frame but on the rotation of the direction of the velocity vector.

Regarding Euler force. Euler force is a needless digression off topic for this article and should not be mentioned in this article IMHO. My arguments above are generally for a constant omega.

The omega illusion. Omega (of the direction of the velocity vector) does indeed enter into some equations, such as mω2r and mVω, but only because ω=V/r. Appearance of omega (of the direction of the velocity vector) in those equations gives the illusion that omega has a hand in the genesis of centrifugal force, but it does not. Omega can always be removed and replaced with V/r. Omega can also be made very small (as discussed above in Case-1) so the situation is essentially a linear acceleration. Focusing on equations-only can be misleading as to what's really going on.

Montyv (talk) 18:33, 26 May 2015 (UTC)

Montyv: "The objects are motionless in an inertial frame, the rotating frame's origin is also motionless in that same inertial frame."

Yes, I understood that's what you meant.

Montyv: "The pseudoforce in that rotating frame needed to describe the object's now-circular motions in that frame must be centripetal, not centrifugal.

But as I tried to explain in my previous comment, in the scenario you describe, the centrifugal force is not the only pseudoforce operating on any inertially stationary objects which do not lie on the axis of rotation. Because such objects are moving in the rotating frame in a direction at right angles to the axis of rotation they must also be subject to a coriolis force. In the scenario you describe, this coriolis force (as I already tried to explain above) is twice the size of the centrifugal, and acts in exactly the opposite direction. The total pseudoforce—viz. the vector sum of the centrifugal and coriolis forces—is therefore centripetal. As I already tried to explain above, it has exactly the right size and direction to produce the acceleration which any inertially stationary object is observed to have in the rotating frame.

David Wilson (talk · cont) 23:42, 26 May 2015 (UTC)

I don't understand where the Centrifugal PF and Coriolis PF (with no R-dot) came from that you to add up to a Centripetal PF, but yes, the net pseudoforce is centripetal. Anyone can invent any arbitrary set of pseudoforces that add up to the net result, but that doesn't mean they're "present". Here, the net pseudoforce is centripetal, not centrifugal. But most importantly, the whole point of the example is to give an "existence proof" of a situation with a "rotating" frame which does not have centrifugal forces, pseudoforces, or reactions. And, given that, we can't say "rotating frames always have centrifugal pseudoforces". If a ref says that, we should not cut and paste it in by rote, and the reliability of that aspect of the ref should be reviewed. We can say that about rotating velocity directions, but not about rotating frames. Montyv (talk) 01:06, 27 May 2015 (UTC)

Montyv, you say, 'I don't understand where the Centrifugal PF and Coriolis PF (with no R-dot) came from...'. The subject has been well understood by mathematicians and physicists for many years now. I, and probably some others, would be happy to try to explain it to you but this is not the right place to do that. I suggest your talk page or mine would be a better place. Martin Hogbin (talk) 09:08, 27 May 2015 (UTC)

Hi guys. The subject is indeed well understood by all of us. I'm a 30-year mechanical engineer fully versed and experienced in the subject (and much more so now!). But, the matter at hand is how to communicate the subject in this article, which, if you will, is the main purpose of a talk page, so I will continue here.
I think I see where you're coming from now. In the example of case 2, there is no R-dot, so there is no actual Coriolis force. But, what you are saying is that if there was an R-dot, there would be a Coriolis force. I can agree with that. Indeed, the equation for magnitude of Coriolis pseudoforce is m*(R-dot)*omega and it would be perpendicular to R. So you are right, there is an equation for Coriolis pseudoforce in that situation. But, in that situation there is no R-dot so there is no actual Coriolis pseudoforce.
I'm glad to see that the argument is only about phraseology used in this article and that we do understand Case-2 in the same way. I hope so at least! Almost all readers are not as discerning as we are. When we say things like "in a rotating reference frame, there is a (pseudoforce of some kind)", most readers (including me) take that as meaning the only thing required is the rotating frame. Whereas, for Case-2 Coriolis, a non-zero R-dot would additionally be required for Coriolis (pseudo)force to actually be present -- and since R-dot is zero by definition, there is no actual Coriolis (pseudo)force present. Similarly, in Case-2, while there may be an equation that can be used to compute centrifugal pseudoforce, there is no actual centrifugal pseudoforce because it is zeroed-out or made negative via the given parameters of the case (i.e. the objects are not accelerating in inertial space nor in relation to the frame's origin/rotationpoint). So, in Case-2, while a centripetal pseudoforce is present, a centrifugal pseudoforce is not. By existence of Case-2 as a counterexample, it cannot be said or implied that "in (all) rotating frames centrifugal (pseudo)force will exist".
My bottom-line point is: saying "in a rotating reference frame, there is a centrifugal (pseudo)force" is misleading phraseology. It should be refined so it does not imply that it's the mere rotation of the frame that always induces a centrifugal (pseudo)force to be present. There is certainly always an equation that one can use to compute the centrifugal (pseudo)force, but the mere use of a rotating frame doesn't mean a centrifugal (pseudo)force will actually be present and positive (a negative centrifugal force would be centripetal). We shouldn't imply that there's something magic in the rotation of a frame that by itself induces (positive) centrifugal (pseudo)force. There are other parameters in those equations besides omega.
I only care here about communicating better to our lay audience. And like I've said, the wording needs some improvement so it doesn't equate the mere rotation of a frame automatically and always with the presence of centrifugal forces. And, I've given a proof-by-counterexample showing that the two do not go together.

Montyv (talk) 05:16, 31 May 2015 (UTC)

Monty, I am going to respond here even though this converstation should really take place in user space. I am happy to continue here but, as you say, this page is for discussing how we communicate ideas, via the article, not for discussion of the subject itself. I personally am happy to talk here but others may not like it; we shall see.

You say above, 'When we say things like "in a rotating reference frame, there is a (pseudoforce of some kind)", most readers (including me) take that as meaning the only thing required is the rotating frame'. As two people have pointed out, the pseudoforce known as centrifigal force is solely the result of working in an rotating reference frame so it is true (for any particle with mass not on the axis of rotation) that the only thing required is the rotating frame' Martin Hogbin (talk) 10:48, 31 May 2015 (UTC)

Inertial frames

Monty, I have started this section to discuss something that I hope we will agree on and that is this:

In Newtonian mechanics, when working in an inertial frame of reference there is no such thing as 'centrifugal force'. Newton's laws can be used, in their standard form, to describe every kind of motion, there is never any need for pseudoforces such a 'centrifugal force'. Martin Hogbin (talk) 10:53, 31 May 2015 (UTC)

Replaced "rotating" with "centripetally accelerating"

I went ahead and replaced "rotating" frame with "centripetally accelerating" frame where it was needed. If you don't have centripetally acceleration of a frame (or object), you do not have (net) centrifugal force of any kind. The idea of rotation of a frame being immaterial to centrifugal force is left unsaid. It's a complex idea to sell and is probably beyond the scope of the article. Leaving it out should satisfy you. The idea that a "centripetally accelerating" (frame or object) results in centrifugal force is true, although you-all might not believe it to be complete. Certainly, stating that all rotating frames must always result in the presence of positive net centrifugal forces is erroneous. If anybody wants to reinstate that, you better have some damn-reliable references for it. In one case (Coriolis) "rotating frame" was apropos. The paragraph talking about it and Euler force, etc. is off topic in my opinion. I'll leave that to you-all to decide whether to keep it or not.

Montyv (talk) 06:57, 31 May 2015 (UTC)

I do not know exactly what you mean by, 'positive net centrifugal forces'; it is not standard terminology. There is always a centrifugal force acting in a rotating frame. There may not always be a net force acting away from the axis or rotation because other forces (real an inertial) need to be taken into consideration. Martin Hogbin (talk) 11:01, 31 May 2015 (UTC)

Just to be clear here, there is always a centrifugal force acting in a rotating frame but the net force is the sum of all the real and inertial forces. In your case 2 above, the net force (which is the sum of just the centrifigal and Coriolis forces) is centripetal, exactly as David Wilson describes. Martin Hogbin (talk) 12:15, 1 June 2015 (UTC)

Yes. Good. More evidence that we understand the situation the same way. It is just a matter of how it should be said.  :-)
Regarding "positive net centrifugal". It looked like DW (and/or you) might have been trying to assert that there was a net centrifugal force according to his equations, but that that net centrifugal force was negative. Just to be clear, I was "heading off at the pass" any idea that we could get away with calling a centripetal force "centrifugal but negative". That's why I was careful to call it positive. I was also careful to call it "net" because it seemed he was also trying to say that a component of the net pseudoforce was centrifugal therefore centrifugal pseudoforce is produced. In response had I made the point that it's the net pseudoforce that matters. So, to be sure we're all talking about the same thing, I clarified that the centrifugal pseudoforce we're referring to in the text must be "net" (not a component) and "positive" to rule out trying to call a centripetal force "centrifugal".
Regarding "There is always a centrifugal force acting in a rotating frame." I've given Case-2 as a counterexample of a situation where there are no (positive net) centrifugal forces in a rotating frame. That's the crux here. We can't just say "There may not always be a net force acting away from the axis or rotation because other forces (real an inertial) need to be taken into consideration". It's the net force that matters. When we are referring to the net pseudoforce, we are taking those other forces into consideration. One might be able to invent an equation that computes a (positive) centrifugal component of pseudoforce which is simultaneously in the presence of a larger centripetal component, but we can't say that a centrifugal pseudoforce is present because of that. Any of an infinite number of valid equations can also be developed for that situation which create any number of components of the net centripetal result, but we can't say that those components are actually present. It's the net pseudoforce that matters. It's the net pseudoforce that is used in the frame for the purpose that pseudoforces are used for -- to compute the trajectories of objects using that frame.

Montyv (talk) 15:01, 31 May 2015 (UTC)

Incidentally, it was me who made the edit by the IP. I had forgotten to log on. Sorry about that. Montyv (talk) 15:06, 31 May 2015 (UTC)

We all do that from time to time. Martin Hogbin (talk) 15:16, 31 May 2015 (UTC)

You seem to be using the term 'net centrifugal force' to mean net force directed away from the axis of rotation. I agree that any force acting away from an axis of rotation can be described as 'centrifugal' but, in this context that is very confusing. In current physics, mathematics, and engineering the term 'centrifugal force' is used only to refer to the pseudoforce acting away from the axis of rotation caused by doing your calculation in a rotating frame. To use 'centrifugal force' or even 'net centrifugal force' to mean the sum of all (real and inertial) forces acting away from the axis is not standard terminology and very confusing. Martin Hogbin (talk) 15:16, 31 May 2015 (UTC)

I noticed that you added a dubious tag (which I removed) to the lead. I am puzzled as to how you can say you agree and add such a tag.

The pseudoforce normally referred to as the 'centrifugal force' is caused only by doing a calculation in a rotating reference frame. Using 'centrifugal force' to refer to anything else is nonstandard. Martin Hogbin (talk) 15:42, 31 May 2015 (UTC)

We are in agreement, yet not. You need to find some damn-good refs, or agree to removal or rewording of the assertion that "rotating reference frame always makes centrifugal". Please don't harp on so called "non-standard terminology" used on a talk page. It's a distraction. It's disruptive and wasteful to nitpick like that on a talk page. I've made my point and the point is valid and no amount of additional quibbling and WP:DONTGETIT matters at this point. If you really want to stick in whole to your stance that all the changes I made should be reverted, I suggest you begin by finding some strongly reliable refs for "rotating reference frame always makes centrifugal" and adding them inline (with page numbers). In the mean time, I can tag that terminology as dubious and/or unsourced. If good refs aren't added in a reasonable amount of time, anyone will be justified in deleting "rotating RF = centrifugal force" altogether. Then, according to WP:burden it will be required to bring in good refs before reinstating that assertion. Does that work for you?
What do you suggest should be my course of action? I can either 1) add "dubious" and "citation needed" tags with (possible) subsequent removal, or 2) revert the wholesale summary reversion you made of the set of changes I made yesterday. We can then review them more carefully, civilly, and one at a time. Pick 1) or 2). I will oblige either way. (I suggest option 2 because it's more cooperative and civil). Without good refs, "rotating RF = centrifugal force" cannot be allowed to stay. It's a matter of how we go about it at this point.
Montyv (talk) 16:26, 31 May 2015 (UTC)

No, we are not in agreement. When I say you are using a non-standard term I mean that you are using the words 'centrigfugal force' to mean something different from all other scientists and mathematicians. It is not a quibble it is what the article is about.

Any good classical mechanics book will tell you what centrifugal force is. We already have in the body of the article (my bold), 'In a rotating reference frame, all objects appear to be under the influence of a radially (from the axis of rotation) outward force that is proportional to their mass, the distance from the axis of rotation of the frame, and to the square of the (angular velocity) of the frame'. The original reference was Encyclopedia Britanica, I have just added Feynman. Would you like some more? Martin Hogbin (talk) 21:17, 31 May 2015 (UTC)

Merge?

It would now seem logical to merge this article with Centrifugal force (rotating reference frame). Martin Hogbin (talk) 12:22, 2 June 2015 (UTC)

(Sorry, didn't see this until today.) I agree with a merge of the two articles, per WP:PRIMARYTOPIC. If we consider both the usage and long-term significance aspects, all other uses of the term "centrifugal force" are minor and esoteric, or a perpetuation of bad physics. --FyzixFighter (talk) 01:10, 8 July 2015 (UTC)

I agree as well. This article is one aspect of the major muddle surrounding all the various centrifugal force articles in wikipedia. Every school of thought, or rather every "variant of muddlement" has it's own article managed by it's own priesthood. — Preceding unsigned comment added by 24.235.64.187 (talk) 07:01, 11 July 2015 (UTC)

So, I suggest that we merge this article first with Centrifugal force (rotating reference frame) then we include any appropriate material from Reactive centrifugal force in the History of conceptions of centrifugal and centripetal forces section. The final article should be called 'centrifugal force'. Agreed? Martin Hogbin (talk) 12:05, 11 July 2015 (UTC)

I also suggest that we leave the lead until the merge is completed. The lead should be an overview and a summary of the body of the article. Martin Hogbin (talk) 12:08, 11 July 2015 (UTC)

Looking at Centrifugal force (rotating reference frame), what are others' thoughts on the derivation, examples, and applications sections? The size of the examples section makes that article seem far too textbook-y and the application section imo only serves to further confuse between the real physics concept with the incorrect concept (a la David Tombe). I think a short Other uses of the term type section would be more appropriate for Lagrangian and reactive CFs - at least in the case of the Lagrangian use of the term, it appears to be more contemporary than historical. --FyzixFighter (talk) 15:56, 11 July 2015 (UTC)

I agree with you about the examples; they are not clear and are overcomplicated. We should start with a stone on a string. The derivation and application sections seem OK to me. What is missing is a simple description of what CF is. Several readers have said that this article that it is too technical and that they cannot understand it. Martin Hogbin (talk) 17:55, 11 July 2015 (UTC)

I agree with all Martin's proposals. Dolphin (t) 06:33, 12 July 2015 (UTC)

There seems to be a consensus to merge. Should we just do a simple merge of Centrifugal force (rotating reference frame) into this article and then work on the result, or should we prepare the new article outside main space and then move it here when ready? I think the former might be best. Martin Hogbin (talk) 18:10, 12 July 2015 (UTC)

I propose the strategy that he who does the work can choose how he does it. Dolphin (t) 06:40, 13 July 2015 (UTC)

That is a good idea. I may do a simple merge and we can take it from there. Martin Hogbin (talk) 11:57, 13 July 2015 (UTC)

Please go ahead. I'm happy to peruse your simple merge and offer constructive comment when you are ready for it. Dolphin (t) 14:00, 13 July 2015 (UTC)

I have now added all the content of the 'Centrifugal force (rotating reference frame)' except the examples, which nobody liked, and obvious duplication. I have redirected the 'Centrifugal force (rotating reference frame)' page to here but left all the original content in place for the moment. Martin Hogbin (talk) 16:16, 13 July 2015 (UTC)

So-called "reactive centrifugal force" is simply unnotable all around, barely worth mention.

"Reactive" centrifugal force is a centrifugally-pointing force, but it's not the "centrifugal force" we seek, grasshopper. It's a "real" force and it's centrifugally-directed, but calling it "centrifugal force" is nothing more than word play.

It's almost completely immaterial except for the need to caution readers to not think it's the centrifugal force the whole world means when they say "centrifugal force".

If the tone of the "reactive" CF article is to be believed, so-called "reactive" CF is a thing, a kind of special thing. It's a thing so akin to the pseudoforce that it warrants it's own article and mention in this article in a section just as long as the main article. (Crazy)

If I understand what they're actually saying, since "centrifugal" means "acting in a direction away from a center", then any (real) force that happens to point away from the center is "centrifugal force". Well, yes and no of course. In the case of a ball on a string, of course there are the two directions of the string's tension, one centripetal and one centrifugal. But that's trivial and of zero relevance to the actual subject at hand. It's a completely different thing than the pseudoforce which is the source of the centrifugally-directed half of the string's tension.

I would agree that such a thing should be mentioned, but only in a "not to be confused by" framework. And preferably too, it shouldn't even be given a name. The framework that exists now (or rather "frame-up job") implies that that "reactive" CF is some kind of important alternative and valid way of understanding "centrifugal force". It's a real force and it's centrifugally-directed, but calling the outward pointing part of the string tension "centrifugal force" is merely muddled word play, and it's irresponsible to hand-wave around that like both articles do. 72.74.19.224 (talk) 02:26, 1 April 2015 (UTC)

I agree with you, in fact, I would go further and ban the use of "centrifugal" to describe any force considered from outside the rotating frame. Unfortunately, some authors still use the term in differing ways, and once a concept has entered the language it is almost impossible to get rid of it. Wikipedia just reports what is written elsewhere. (Also, just occasionally, provided that one is aware of an artificial frame of reference, centrifugal force can be a useful concept for certain limited problems that never go outside the rotating frame.) Dbfirs 06:10, 1 April 2015 (UTC)

72.74.19.224, I agree with you too. There are some sources that use the term 'centrifugal force' to refer to the reaction force but they are few and far between and are not particularly authoritative. Use of the other meaning is unhelpful and confusing.

If you think this page is bad you should see it before I changed it. Martin Hogbin (talk) 23:15, 1 April 2015 (UTC)

I took care of some of the worst of it with two large removals. Somebody was working himself into a tizzy trying to justify his analysis, but all of it was unsupported and original. I expect some pushback however, and might need some backup. Montyv (talk) 06:06, 24 May 2015 (UTC)

Wow. I'm looking over what's left. Usually, large removals leave voids and awkward flow, but now the "reactive" section reads so much better. Now it's short and sweet and supported. It doesn't go hither and yon into someone's confused distraction. Nice! (If I do say so.)  :-) Montyv (talk) 06:19, 24 May 2015 (UTC)

Well done. Martin Hogbin (talk) 10:20, 24 May 2015 (UTC)

Yes, definitely an improvement. Dbfirs 11:02, 24 May 2015 (UTC)

Hello All

I have been following this discussion and I have a question about the math

In the velocity section of the centrifugal force article, the math usually begins with the equation,

${\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}=\left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\boldsymbol {\omega }}\times {\boldsymbol {r}}\ ,$

The first term on the right hand side is supposed to represent the velocity of the particle relative to the rotating frame. The second term on the right hand side represents the velocity of the fixed point in the rotating frame, relative to the stationary frame. Hence this approach is using the vector r for two different points, yet pretending that the two r vectors are the same vector. For this equation to be correct, we should be using an r' in the first term on the right hand side, as its a different vector.

If the two r vectors were the same vector, then the first term on the right hand side must be purely in the radial direction. This math however is allowing it to be in any direction. It is using one vector to represent two different points each with its own velocity. The lead to the article becomes a nonsense when it implies that observations from a rotating frame cause an actual outward effect.

All The Best — Preceding unsigned comment added by 2601:449:4001:1F70:5991:9730:51C3:B49A (talk) 16:53, 17 July 2015 (UTC)

Frames of reference

Should there be an elementary discussion of what a frame of reference is? Martin Hogbin (talk) 18:39, 14 July 2015 (UTC)

Does centrifugal force exist?

How best do we give an answer to non-technical readers to this very common question? Martin Hogbin (talk) 08:13, 15 July 2015 (UTC)

Here are some questions from various forums:

Currently in my last year of high school, and I have always been told that centrifugal force does not exist by my physics teachers. Today my girlfriend in the year below asked me what centrifugal force was, I told her it didn't exist, and then she told me her textbook said it did, and defined it as "The apparent force experienced towards the outside of a circle is the centrifugal force and is due to the mass of the object resisting the inward centripetal acceleration that the object is experiencing". I was pretty shocked to hear this after a few years of being told that it does not exist.

And here is one of the better answers:

As you know, Newton's laws work in so-called "inertial frames of reference". However, a point on the surface of the Earth is not really an inertial frame of reference because it is spinning around the center of the Earth. (So you can think of it as a rotating coordinate system.) So Newton's mechanics don't apply if you want to describe motion and use a reference point on the Earth. This is quite inconvenient, because we mostly want to engineer things that work on the Earth.

Fortunately, there is a trick: you can use a point on the surface of the Earth as your reference and pretend that it's an inertial frame of reference, if you also pretend that some external "imaginary" (fictious) forces exist in addition to the real ones.

Here is another question:

Back in my first physics class in college, a few years ago, my prof said that centrifugal force does not exist, and something about there is something called the centrifugal force effect that we seem to think we feel, but it really is centripetal force.

I took this as physics dogma, and today I spouted my mouth off while watching a TV show, and everyone thought I was lying so I looked it up in my physics book, and centrifugal force was no where to be found, which was good for me. Then I looked on the net, and found tons of mentions of it.

And finally I saw this:

We all know that centrifugal force exists don’t we? It’s the outward force on an object traveling in a circle. Pretty hard to deny it exists when we’ve all felt it every time we go around a tight turn in a car.

Except it really doesn’t exist, and you can prove it to yourself. Take a small object and tie a string to it. Then swing it in a circle (hey if you’re actually going to do this don’t go berserk and hurt someone or yourself). Now that you have the object swinging around in a circle, ask yourself this question; are you pushing on the string or pulling on it? You are pulling on it of course, you can’t push something around through a string. What happens when you suddenly let go (safety precautions again of course, and if you break a window or hit someone I am not responsible). As soon as you stop pulling on the string (applying an inward, not outward force) the object will fly off in a straight line. No more force inward, no more going in a circle.

But the fact is it feels like there is an outward force when the car goes around a tight corner. What is actually happening is that your body has momentum and is trying to go in a straight line. What you are feeling is your momentum as the car starts to move sideways underneath you. You’re not being thrown up against the side of the car. The side of the car is moving over and pushing on you to make you go in a circle too. The force it applies is called centripetal force.

With this comment!!

'Why don’t you send this to Wikipedia!'

We really do need a clear description for the layman of what CF is and is not. Martin Hogbin (talk) 10:03, 15 July 2015 (UTC)

My view on the subject is there are centripetal forces directed towards the center of rotation; and there are other forces equal in magnitude but opposite in direction as predicted by Newton's third law of motion. These forces directed away from the center of rotation don't have a formal name. Many people, and even some authors, call these forces centrifugal. However, the current consensus of opinion among the scientific community is that these forces shouldn't be called centrifugal; we should acknowledge their existence but accept that they don't have a convenient name; they are just the forces equal and opposite to centripetal forces. The concept of centrifugal force is a very old one that was given to certain fictitious forces that arise from certain choices of reference frame. So there is always a force equal in magnitude and opposite in direction to a centripetal force but we don't give it a special name. Dolphin (t) 12:30, 15 July 2015 (UTC)

What forces exactly are yo referring to? Martin Hogbin (talk) 12:41, 15 July 2015 (UTC)

See my first sentence. That says it all. Dolphin (t) 13:53, 15 July 2015 (UTC)

You are referring to what is sometimes called the reactive centrifugal force. As you say, this term is not widely used and its use is now deprecated in education. We do mention it later in the article.

So, leaving the reactive CF aside, how do we deal with the statement, 'Centrifugal force does not exist'? Martin Hogbin (talk) 17:31, 15 July 2015 (UTC)

Perhaps something along the lines of: "In dynamics, the centrifugal force is one of the corrections that is added to the sum of forces in Newton's second law of motion so that the law can be valid within rotating coordinate systems. In stationary (inertial) frames, no corrections are needed for Newton's laws of motion to be true and so the centrifugal force only exists in rotating frames. Motion attributed to the CF in the rotating frame is seen as a result of an objects inertia in the stationary frame." Thoughts? --FyzixFighter (talk) 02:00, 16 July 2015 (UTC)

I think there is no difficulty is describing exactly what CF is in technical language. The problem is how to explain what CF is to the layman. Most people who want to know what CF is do not realise that physics is normally done only in inertial frames, so that Newton's laws for example are, in their standard form, only applicable to inertial frames unless inertial forces are included. That fact is intuitively obvious, but only when you stop to think about it. The idea of playing a game like snooker or pool, for example, on a table on the back of a vehicle driving round town is quite absurd to everyone.

We know that the answer to the question, 'Does CF exist?' is simple, "In an inertial frame, 'no', in an rotating frame, 'yes' ", but to many people that answer seems like just weasel words. In teaching engineering, maths, and physics, the best approach is not no mention CF at all until you reach rotating frames of reference, having already been told that physics is normally done in inertial frames. In other words it is best to tell students at the start that 'CF does not exist'. The problem is that, in rotating frames, it does exist.

So what do we do? Do we say, 'Not our problem, we just state the facts, as given in reliable sources, and if our readers cannot understand it that is too bad'? Some would support that view but I would like to think that we can do better. Martin Hogbin (talk) 08:01, 16 July 2015 (UTC)

I have rewritten the introduction, hopefully to make the meaning of centrifugal force clearer to the general reader. Martin Hogbin (talk) 17:13, 27 July 2015 (UTC)

Assessment comment

The comment(s) below were originally left at Talk:Centrifugal force/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

The psychological aspects of this article relate to (i) the health related aspects of space stations and artificial gravity and (ii) the pedagogical aspects of teaching a topic with an abstract interpretation that sometimes conflicts with instinctive reactions. Brews ohare (talk) 15:58, 29 May 2008 (UTC)

Last edited at 17:47, 30 September 2009 (UTC). Substituted at 20:18, 2 May 2016 (UTC)

Inertia and real forces

Here is how we now start:

In Newtonian mechanics, the term centrifugal force is used to refer to an inertial force (also called a "fictitious" force) that appears to act on all objects when viewed in a rotating reference frame, drawing them away from the axis.

First of all, the fictitious forces do not follow Newton's third law, and appear or disappear as we change frames (either in linear or rotating systems). Since objects never feel fictitious forces (no mechanical stresses) they don't care if some frames see them and others do not. In this sense, they aren't real. Forces that follow Newton's third law and have an action/reaction pair, ARE real. There's not a thing you can do to change them by changing frames: they are invariant. And rightly so, since the g-force or weight on an astronaut or person in a rotating space station, is the same no matter what frame you wish to view the astronaut.

So the forces that follow all Newton's laws, including the 3rd, are real. They are not fictitious. This worry about which is which, can be cleared up by fixing the lede so it isn't wrong. Fictitious centrifugal forces of the same magnitude as the real centrifugal forces (the reactive force) only appear in the special case of rotating frames where an object in a circular trajectory (as in a centrifuge) is viewed in the rotating frame where it is motionless. Not otherwise. If you view it in any other frame, it goes one way or the other in a circle, and in that case, the force on it may be centrifugal of various magnitudes, OR (finally) even centripetal when you view it in frames rotating even faster than the centrifuge.

As you are standing in the middle of a centrifuge that is running at some fixed angular velocity w, and begin to spin yourself into a rotating frame, you at first see the object at the rim slow down. During this phase, the mechanical force inward on the object is too large to explain its motion in your new frame, so a fictitious centrifugal force must be added to counter it-- but at first, it's small. It reaches a maximum when you (the observer) spin so fast that the rim is motionless for you. Fine-- now the fictitious force is as large as the real inward force (and also the real outward reactive force), and in the same direction as the real outward force. Of course this causes confusion all around, but it shouldn't, as the object only feels the real force-pair inward/outward (between itself and the rim), and is unaware of the fictional one-- so it doesn't care if both act in the same direction at the same time, as it feels only the real. And you are happy, as the half-arrow force pair add to zero in the free-body view of this motionless object. Real inward force equals unreal outward centrifugal force, and net is zero. That's why we invented fictitious forces.

But now, observe what happens as you (the accelerated observer at the center) spin even faster than the centrifuge. The objects at the rim now begin to go slowly the other way in a circle, and now the fictitious centrifugal force must drop off again, as you go faster. Finally, as you (your rotating reference frame) reach twice the centrifuge rotational angular velocity w, the fictitious centrifugal force has gone to zero again. Really! To you, it now looks as though the centrifuge is spinning exactly as fast as it was at first when you viewed it from a nonaccelerated frame, but THE OTHER WAY. You don't need any fictitious forces to explain that motion. None. It's as though your reference frame is not spinning at all. However, you're spinning pretty fast, yet no fictitious forces appear. Explain that, lede.

Now we start to strain your brain and go even faster than twice the centrifuge angular velocity. Now, the objects go faster and faster for you the observer, yet the inward real centripital force (what centrifuged objects feel) has never changed throughout this entire exercise. So as they go faster in each new faster frame than at the first baseline speed, the (constant) inward real force on them IS NOT ENOUGH to explain their new angular velocity, that is now faster than baseline "w." So now, we need to posit an (extra) fictitious CENTRIPETAL force, acting inward to explain the new motion. And the faster we go from here, the larger that fictitious centripetal force must get to explain the faster motion. Explain that, lede.

The easy way to see that fictitious forces need not be centrifugal is to simply do this when the centrifuge isn't running. You sit in the middle and spin. Now you see the rim circling with no real forces on it at all. You can only explain this motion by a single fictitious CENTRIPETAL force, directed inward. It disappears when you leave your rotating frame, and otherwise, there are no forces on the rim (we pretend our centrifuge is out in space at zero-g). Explain that, lede.

I'll stop here. We need to continue to explain that forces that are needed to change an object's velocity because of inertia, are always real. It's completely wrong to associate fictitious forces with inertia. Inertia is inertia, and doesn't care about your choice of frame. The forces that arise due to the inertia of a body are real forces, and always come in pairs. They are real. SBHarris 03:51, 31 July 2015 (UTC)

The current article, including the start of the lead, is based on standard physics as may be found in any current text book on the subject. I am not sure what exactly you are saying and what it is based on. Martin Hogbin (talk) 16:39, 31 July 2015 (UTC)

Perhaps you could read it again? In the most simple rotating frame and setup you can think of, one where you have extended set of objects at rest and nothing moves in the non-rotating frame, when you go to a rotating frame, all objects not at the origin move in circles. The fictitious force needed to explain this motion is clearly directed inward. It is thus a centripetal, not a centrifugal fictitious force. That alone invalidates your lede. I don't really care if you have a text that says otherwise; that only means the authors are clueless (the equation for the direction of a fictitious force from rotation can produce a force in either direction-- in or out). Here also is a standard text that refers to fictitious forces as NON-inertial forces: [1]. That also contradicts your present lede (but is correct, as ficititous forces are indeed noninertial). Here is another text [2] that claims that you can have a centrifugal force which is fictitious, but that this has disappeared from education, and the only "true" centrifugal force is the familiar outward contact-reaction force in a centrifuge, which of course is not fictitious at all. So everybody disagrees. SBHarris 18:48, 31 July 2015 (UTC)

I don't understand why you think that either of your linked articles contradicts the lead paragraph in Wikipedia. John Roche gives a very clear explanation which we could possibly use as a guide for our article. (He mentions Tom Duncan who taught me to teach Physics, by the way.) I assume that you wish to give more emphasis to the engineers' approach? The whole area is liable to misunderstandings. I recall upsetting another teacher of Physics, many years ago, by telling him that if his explanation of a centrifuge were correct then cream would go to the outside in a milk separator. I agree that everyone disagrees with everyone else on the best way to explain this topic, and that use of the adjective inertial is ambiguous. Your examples have observer and object in different frames so fictitious centripetal forces can certainly appear to be observed there, just as, when I accelerate in my car, roadside objects appear to have a backwards force, but that situation is not the one considered in this article. Dbfirs 21:06, 31 July 2015 (UTC)

The only clear contradiction in the articles is the inertial vs non-inertial distinction. This is worse in the separate article on reactive centrifugal force, which actually is the engineer's force, and the only one that Roche says is "real." As in a real force. He is right in that-- engineers do not care about fictitious forces, as they do not induce stresses (g-force induced weights and so on), and so they don't give a fig about fictitious centrifugal forces.

So Wikipedia has a whole article on a fictitious force and calls it centrifugal force. But it is not a force so much as a mathematical device to make Newton's second law valid in accelerated frames. If you must do that, you need to invent "forces" out of thin air that nobody can feel. There should be one article that points this all out, and most of the work there is done in the reactive centrifugal force article.

The major problem with the lede is it insists that fictitious centrifugal forces arise in all rotating frames, which is wrong by simple inspection. I don't need a textbook to say this, as it's so obvious that nobody bothers. If you sit on your office chair and spin, the room moves in a circle around you. Since it moves in a circle it accelerates TOWARD you and thus (if you believe Newton) needs to have a force doing that. That force is fictitious, and since the acceleration is inward, this force must points inward (not outward as your lede says). It is not centrifugal, but centripetal. The average text doesn't treat such things because they typically examine only the case where you are already in a spinning room (or space station or carousel) and you spin yourself so that you are exactly in frame where objects have no rotation. In that case (and that one only) the fictitious force created is equal in direction and magnitude to the engineer's real (reactive) centrifugal force. But that is a coincidence, and happens only in that situation and that particular frame. It's not a general truth, as your lede pretends. The article body says flatly that all objects in rotating references frames appear to be under the influence of a radially OUTWARD force proportional to the square of the angular velocity of the frame. WRONG. This is true only of objects FIXED in rotating frames so that they do not move, like a ball skewered on a rotating reference frame coordinate rod! All other objects in motion in rotating frames (moving in circles faster or slower, one way or the other) experience quite different fictitious forces, sometimes centrifugal, and (if you rotate fast enough) entirely centripetal.

The article states as a general truth what is true only in a very, very special case. So anyway, it's all screwed up. If you don't consider the general case in the article, SAY that. As it stands, the article treats a special case as if it were generally true, and then (to top it off) ignores the most important actual centrifugal reactive forces, shoving them off into another article. SBHarris 21:53, 31 July 2015 (UTC)

In your counterexample, the sum of the fictitious forces is centripetal, but you are forgetting about the fact that there are other fictitious forces involved. In this case the Coriolis force must be considered since now the object is in motion in the rotating reference frame. In your counterexample it is twice the magnitude of the centrifugal force and points inward so that the net fictitious force is centripetal but the same magnitude as the centrifugal force contribution. (I remember finding a short physics article during the DTombe debates that addresses your counterexample and makes this observation, but I'll need to do a deep dive in the archive to find it again.) More importantly what we call the Coriolis, centrifugal, and Euler (fictitious) forces comes from transforming the acceleration vector from a stationary frame to a general rotating frame. In this expansion, you end up with three terms - one that depends on the object's velocity in the rotating frame, one that depends on its position (and is always a centrifugal contribution), and one that depends on the time derivative of the rotating frame's angular velocity, which are respectively the Coriolis, centrifugal, and Euler forces. This is how all classical mechanics textbooks break down the correction terms (the fictitious forces) that we add to the net force to bootstrap Newton's laws of motion to rotating frames. Yes, in some instances the net fictitious force is centripetal but the net fictitious force can always be decomposed into Coriolis, centrifugal, and Euler contributions which always obey the same equations. Because of that I don't see an issue with current text, except maybe to say that it is one of the inertial forces and the only one that is always present in all rotating frames, nor do I see the current text as being at odds with any reliable source. --FyzixFighter (talk) 23:33, 31 July 2015 (UTC)

You are quite right that you can break down the fictitious forces into this triad, and that some of the outward and inward pseudoforces in my example (I neglect angular acceleration and so neglect Euler) are technically Coriolis forces acting in the same (or opposite) direction as the "centrifugal force" which remains constant and outward, so long as the frame rotation rate and radial distance are constant. But now we're using a definition of "centrifugal force" as being the physics mRw^2 centrifugal force, and not the total of outward forces on a rotating object. And it's certainly not inertial and it's certainly does not have to be the same as what you call the reactive centrifugal force, which may be anything. Also, this is not the object's weight unless it happens to be motionless in the frame. This means that it is not applicable to most objects in a rotating frame, because most of them will be moving unless the frame rotation is adjusted to "fix" one of them.

Again, suppose we have an astronaut in a centrifugal spinning space station rim, which is spinning fast enough (velocity +V and radius R) to give him a cruise weight (which we will suppose is small enough that it can be increased by a factor of 4). Call this weight +W (we use a positive sign for outward or centrifugal). If we look at him in the rotating station rim frame, where he is sitting in a chair fixed to the inside of the outer rim wall, motionless in our frame, the force pushing him inward is the reaction to his outward weight -W, and the centrifugal force then must be +W to explain why he feels force but no net force, and is (thus) without acceleration in our frame. This is the only time these weight forces approximately agree with the reactive centrifugal force article, and they require that the object of our study be pinned down and immobile in our coordinate system, so that he is the subject of a real contact force, and HAS a weight. Otherwise, nothing works. There are no "inertial forces" on an object in a rotating system unless you pin the object DOWN in the system so it cannot move with regard to the coordinates. In that case, the inertial forces are supplied by the mechanical contact force of the "pin" (in this case, the chair), not by any fictitious force. Centrifugal force is a fictitious force and (again) the only force that pushes a subject outward in a rotating system is the contact/reaction to the force you use to pin him in the system-- not the same thing and not even the same magnitude or direction except in this case. It is not centrifugal force except as defined in direction.

This needs an example, using Coriolis language. Suppose our intrepid astronaut manages to run with speed -V in the opposite direction to the space station wheel, so he no longer shares its motion, and is motionless in space with regard to the center of the wheel. Now, he floats inertially and simply watches the chairs and furniture fly past, under him. There is no real force on him at all, and he is weightless, experiencing zero-g, and also not moving radially. No centrifugal force pushes him out toward the rim, however close to it he comes. However, in our rotating frame, which we haven't changed, he now travels in a circle, and the force that keeps him doing so is entirely fictitious, and is composed of 2 parts -W inward from (of course fictitious) Coriolis, and one part W outward from (of course fictitious) centrifugal force. These add up to centripetal force -W inward, as they must to explain his motion. However, he feels none of it

Next, our intrepid astronaut catches the chairs to catch up with the ring (we ignore these Euler and Coriolis transients), feels weight W again as he lands in his chair, and then gets up and proceeds to run around the ring in the opposite direction-- this time the SAME direction it turns (like that great Frank Poole jogging scene in 2001, a Space Odyssey). He manages to do this at a final fixed relative speed of +V with regard to the floor. His total speed is now +2V (you can't fool the universe and fixed stars about your rotation) and his weight goes up to +4W (the running is hard-- stomp, stomp). Our force diagram now shows a reactive real component of +4W (his weight on his shoes), and this is because the ring is pushing him inward with a real contact force of -4W when he inertially wants to go in a straight line. That's a really real force-- the only inertial force in this problem. He cannot get away from real inertial forces, which he always feels as weight and g-force-- but THESE are the inertial forces, not the fictitious ones we look at next. Okay, now look what the fictitious forces do: we haven't left our initial rotating frame, in which Frank Poole is not moving when he is sitting and is only going V when jogging, and so his (fictitious) centrifugal force is +W (always outward) and his (fictitious) Coriolis force is +2W, this time also outward, since he's changed direction (cross product). These add to +3W outward. Combined with the -4W inward reaction from the floor, we have a total force of -W on Poole. This centripetally-directed net force is just what we need to have him go in a circle at V, which he is doing again (albeit in the opposite direction) in our usual rotating frame. So the math all works, for sure.

Except: please note that none of the fictitious forces here are real or inertial, and they change to add up to the force you need to get the motion you see, NOT to the weight the astronaut feels, or to any odd influence that you see in the frame. So where is the reactive centrifugal force? And how big is it? I have no idea. The physics centrifugal force in these three examples never changes (it is always +W outward), but it's not very much help in calculating anything real. It's the same whether Pool floats, sits in the chair, or works hard running at 4x normal weight. See the problem? This force our intro says "appears to act on all objects when viewed in a rotating frame, drawing them away from the axis." Well, I don't see any "action on all objects, drawing them away from the axis." Clearly sometimes the action is invisible, removed by other fictitious forces that go the other way. Our astronaut floated for awhile and centrifugal force couldn't didn't draw him away from the axis. Indeed, we had to muggle it up with other fictitious forces just to get it pointed inward toward the axis to explain why our floating astronaut was circling the axis in our frame, while floating weightless. And this lede says the name "centrifugal force" has been used to refer to the "reaction force to the centripital" force. Well, what is that force, in these examples? I get reaction forces to net centripetal of zero, +W, and +4W. But centrifugal force stays at +W. SBHarris 03:53, 1 August 2015 (UTC)

You said But now we're using a definition of "centrifugal force" as being the physics mRw^2 centrifugal force, and not the total of outward forces on a rotating object. - Yes, this is correct. It is conceivable that you could use the term "centrifugal force" to refer to any radially outward force. But 95% of the time in science, when someone says "centrifugal force" they are talking about the physics mrw^2 centrifugal force, and so, per WP:PRIMARY, that's what this article is about. Is there a part of this article where that isn't clear or where the "total of outward forces on a rotating object" seems to be implied?

You said ...it's certainly not inertial... - That depends on how you define "inertial". We could probably go round and round on this, but the fact of the matter is that in multiple reliable sources fictitious forces (eg Coriolis, centrifugal, Euler) are also called "inertial forces". We can check to see if any of them explain the rationale behind that name (I'd check now but it's getting late - perhaps tomorrow), but I seem to recall (but I could be wrong) one saying that the effects attributed to the fictitious forces in the rotating frame are attributed to the object's inertia in the stationary frame. Before we really dig into that debate, let's see if any reliable sources actually say this.

You said ...it's certainly does not have to be the same as what you call the reactive centrifugal force, which may be anything... - Very true. There is no guarantee that the two will be the same magnitude. Additionally, the physics centrifugal force appears to act on the object while the reactive centrifugal force does not act on the object but is the force the object exerts on the source of the centripetal force (forces that are action-reaction/3rd Law pairs do not act on the same object).

You said Also, this is not the object's weight unless it happens to be motionless in the frame. This means that it is not applicable to most objects in a rotating frame, because most of them will be moving unless the frame rotation is adjusted to "fix" one of them. - Where is the requirement that the centrifugal force be the object's "weight"? Is this implied somewhere in the article? I disagree that it is not applicable to objects moving in the frame - you just also have to consider the Coriolis force when analyzing the object motion relative to the rotating frame. As Taylor notes, "In a few simple cases, it is actually easier (as well as instructive) to analyze the motion in an inertial frame and then transform the results to the rotating frame...Nevertheless, the transformation between the two frames is usually so complicated that it is easier to work all the time in the rotating frame and to live with the "fictitious" Coriolis and centrifugal forces." I also disagree with your later sentiment that the whole business of the fictitious forces are ad hoc, changing to whatever they need to be to match the motion you see. Certainly they do that, but the forms of the equations were derived from the properties of the rotating frame without any a priori knowledge of the particles motion or velocity.

In your examples, you seem to have a misunderstanding Newton's third law. For example, the "cruise weight" +W, as a force exerted on the astronaut, does not exist in the non-rotating frame. In other words, if we did a free-body diagram of the forces acting on the astronaut in the non-rotating frame, we would only draw the -W exerted on him by the chair. The reaction to the -W is the force the astronaut exerts on the chair which is +W, but we don't include this force in the free-body diagram since we are only concerned with forces acting on the astronaut (if we were doing a free-body diagram of the chair then we would care about that force). Only in the rotating frame does a "cruise weight", as a force acting on the astronaut, of +W appear, which we would attribute to fictitious forces.

However, in the end we need to follow reliable sources. Multiple reliable sources use the term "inertial force" as synonymous with "fictitious force". Reliable sources also describe the centrifugal force as one of the fictitious force that acts on all objects in a rotating frame and having the form mrw^2. Trying to use our own examples and counter-examples to prove points just brings back the same scenario we had with Tombe where editors tried to teach one another "correct" physics. We all saw how well that worked. If you've got reliable sources that contradict or better elucidate the subject, then please share them. --FyzixFighter (talk) 06:30, 1 August 2015 (UTC)

First of all, in the examples above, I never treated the case of the inertial frame. I merely said the astronaut's weight was +W, which it is (his weight is a constant and the same in all frames you choose to view it, accelerated or not). My math above is all correct, I believe. I'm well aware of how to use free body diagrams, and yes, certainly the only force on the astronaut in an inertial frame (which again I did not discuss!) is the reactive force to his weight, which is the inward -W force from the chair. Only the chair sees weight +W, as you say-- but it's still the guy's weight in any frame. In inertial frames, objects moving in circles always have only one force on them, acting centripitally; I know that.

You ask: "Where is the requirement that the centrifugal force be the object's "weight"?". Answer: That is implied if you use the "historical" engineer's definition of centrifugal force as a reaction to the centripetal force, which you mention at the end of the lede. In that case (where the two definitions are the same, and the real centrifugal force is the same as the fictitous one), you must be in the rotating frame where your object of interest is at rest, and of course there, (at least in absence of gravity) the object's (artificial?) "weight" will be supplied by what you call the reactive centrifugal force, as in our astronaut in his space station chair. If the object is moving in the rotating frame, all bets are off, as in my examples where the astronaut is in notion WRT to the station rim (chair). This special case thing (i.e., rotating space station astronaut in chair, reference frame where he is motionless) is explained fairly well in the reactive centrifugal force article, but you have taken all that out of THIS article, which is the place where people will come to look for it! For my other complaints, see my reply to Martin Hogbin, below. SBHarris 02:47, 2 August 2015 (UTC)

Sbharris, I agree with FyzixFighter here. If you look at modern physics, mathematics, or engineering text books on classical mechanics (a few more practical books still use the deprecated 'reactive centrifugal force') you will see only the definition of CF given in this article.

It is an inertial force that act on all objects and in an outwards direction only when physics is done in a rotating reference frame. It is called inertial because it is a consequence of the fact that objects with mass have inertia. If there were no inertia there would be no CF. That term is widely used in reliable sources.

Your claimed counterexample, 'In the most simple rotating frame and setup you can think of, one where you have extended set of objects at rest and nothing moves in the non-rotating frame, when you go to a rotating frame, all objects not at the origin move in circles. The fictitious force needed to explain this motion is clearly directed inward', has been clearly explained by FyzixFighter. Because, in the rotating frame, the object is moving, we need to apply both the centrifugal and the Coriolis forces. When this is done there is a net inward force of the correct size and direction to explain the circular motion of the object in the rotating frame. This is a standard undergraduate physics question. There are no ad hoc forces; both the centrifugal and Coriolis forces are determined by well known and unchanging equations.

We really should not be discussing this here it as is all absolutely standard stuff. If you disagree in any way you are quite simple wrong. Martin Hogbin (talk) 11:17, 1 August 2015 (UTC)

Look, you must have half a dozen references in this very article on the use of "centrifugal force" to refer to the Newtonian reaction-force outward on an object in circular motion. If this meaning is so "deprecated" (as you repeat from the article) how come the only reference for your dread deprecation is one lonely reference from 1960? Something that deprecated should be deprecated again and again and again. So where's your references? I gave you one which you actually use in the article, and which you ignored. Here's the pdf: [3]. Roche says that the only "true" centrifugal force is the reactionary (outward) contact force. That, of course, is what you discuss in reactive centrifugal force. Roche, here writing in 2001, also claims that the definition you use in this article "has nearly disappeared from texts" and primary education because it is so complicated. Well, that would stick it to WP:PRIMARY if he's right. So again, let's see something published after 1960 which clearly deprecates and gives the raspberry to the reactive-real meaning, and says that if you use it, you will go to Physics Hell.

Roche and I like the "reactive force" meaning because it's a real Newtonian force that you can feel. It shows up in rotation as an invariant g-force that causes physical stresses, and not some fictitious mathematical force that appears or disappears according to who is looking from which frame. There are, of course, two kinds of forces, but if I were titling these articles and had to have two articles on centrifugal force, I'd call one of them Fictitious centrifugal force and the other Real centrifugal force. Evidently, so would physics teacher Roche (like the name?).

While I'm on the subject, I don't really like the language this article starts with: "[Centrifugal force appears to act on all objects] drawing them away from the axis." Okay, the fictitious centripetal force is in a direction away from the axis (what you mean to say), but it doesn't "draw objects" anywhere. That implies what isn't so. They stay radially right where they were in the old frame, or else move radially just as they always did. If you increase the fictitious centrifugal force by increasing frame rotation, an equally fictitious centripetal Coriolis force always appears to oppose the increase, and more. This article should mention centripetally-acting and centrifugally-acting Coriolis forces early and often, so this is clear. Coriolis himself wanted to call the fictitious force he discovered, the "compound centrifugal force", but alas they named it for him instead, so the role of Coriolis forces in opposing fictitious centrifugal forces as you change rotational frames, was made a lot less clear. Here, you have a chance to fix this. The example with a distant fixed object when you move to a rotating frame, is a good one. A fictitious centrifugal force appears, and a double dose of fictitious centripetal force also appears, to counter it and explain its new motion. The object feels nothing, and radially goes nowhere (certainly not "away from the axis.") If all this is "absolutely standard stuff," why isn't it in the article?? You've spent years arguing over this point, I see, and you might have saved yourself all of it. For every troublesome Tombe, you probably have a thousand students who read this, just scratched their heads, and went away. SBHarris 04:16, 2 August 2015 (UTC)

Let me start with a point of agreement. You say you do not like, "... drawing them away from the axis.", I agree. This is not my wording and it is rather mystical sounding. That the force 'acts in a direction way from the axis', or similar, would be much better.

The point about Coriolis force is a difficult one. The problem is that this article is not about rotating reference frames in general but specifically about the centrifugal force. To talk about a different force before the centrifugal force has been fully explained is very confusing, that is why the two examples that I give do not require the use of the Coriolis force.

Another two points on which I hope that we can agree are these. When working in an inertial frame, we only need to consider real forces and use Newton's laws. For the large majority of our readers, who are not physics, maths, or engineering graduates or undergraduates, it is by far the best option to always work in an inertial frame. As I try to make clear in this article, in elementary mechanics, an inertial reference frame is usually tacitly assumed. Do you agree with these two points? Martin Hogbin (talk) 09:04, 2 August 2015 (UTC)

Let me also say that it has never been my objective to completely remove any mention of the reactive CF from this article. There are sources which use this term. The main thing though is to describe this term in a proper context and not let it confuse our readers. Martin Hogbin (talk) 21:38, 2 August 2015 (UTC)

I basically agree. I also would not want to bring up the odd Coriolis inward-and-outward components that add and subtract from centrifugal forces when objects have an in-frame tangential velocity. However, after centrifugal forces are fully explained, there is a place for this. In fact, it’s even done NOW to some extent in the discussion of the fictitious forces on fixed stars in a rotating frame. So, just before that, we can do an example of a simple inertial object that is first viewed in the inertial frame, then seen from a rotating frame. This is very informative.

All of this leads naturally to a discussion of the two kinds of centrifugal “forces”—the real Newtonian one you can feel and measure (as one half of a real force-pair), and the unpaired fictitious mathematical one you don’t feel and can’t measure, but must infer. We just barely mention the first one in the lede, but it’s the one most people are familiar with, and the one they will (unfortunately) come here (be directed HERE) to learn about. And it will (at best!) take them a while to realize it’s way at the end of this article, and isn’t this article’s subject.

All this is an argument to put the two articles centrifugal force and reactive centrifugal force back together again, Or put a lot of summary of the other back into this one, and closer to the front. Then we can introduce the math meaning of the term as the 1) fictitious mathematical theoretical force you never feel, and at the same time take great pains to mention that the real force that squashes you in a centrifuge is something else, and is 2) the OTHER type of centrifugal force you of course CAN feel.

Only in the frame where objects have no tangential velocity are these two kinds of centrifugal force of the same direction and magnitude, but still one is invisible and not palpable, and the other one is busy causing tension in strings and squashing astronaut candidates with “g-force” accelerations in centrifuges. And (again) is the one most people have heard of, and want to know about. We humans tend to be interested in what squashes us (real forces), not particularly in the terms that make the math come out right but otherwise can be ignored. The article on reactive centrifugal force actually does a fine job of all this, but unfortunately isn’t where “centrifugal force” directs to!

Because of the special properties of the properly sped-up rotating-frame view of objects that ARE rotating (like a planet or satellite seen in a frame where they have no tangential velocity), we can point out that problems simplify a lot in that frame if you can find it, so there are no velocities that cause Coriolis forces. The inward-and-outward components of the Coriolis fictitious force, resulting from tangential velocity, transform neatly to a single fictitious centrifugal force in the rotating frame where that velocity is reduced to zero, and since this is also the frame where the real centrifugal (reaction) force is equal to the fictitious force in magnitude and direction (except that the reaction force is REAL), it’s the frame where you can say how much force things FEEL (what the g-force acceleration is, what the accelerometer reads, and how much people weigh and how much they are squashed-- all things only real forces do). All this needs to be said, and isn’t here. Some is in the reactive centrifugal force article, some in the Coriolis force article, and there is no unified explanation of these fictitious forces and how to deal with them, get rid of them, or find out what the residual g-force acceleration that makes you black out, will be.

Yes, An inertial frame is indeed usually tacitly assumed, except in the few cases where it is not, and hence this article. Often because we’ve enclosed things and put all the furniture and handholds and everything else into a rotating frame like our spinning space station. Or a student will imagine himself riding the centrifuge along with the bucket. And occasionally a student will visualize a ball on a tether from a rotating frame as though it was the Olympic hammer throw, and have to be deprogrammed. Or somebody will wonder why their satellite dish points out at a point in the sky where the geosynch comsat satellites just "hang there" in one spot in the sky. What keeps them up? You’re in a rotating frame, so it’s hard to explain. In the satellite dish frame, a fictitious centrifugal force keeps them up, so they don’t fall down due to gravity. Let’s write an article that covers all that, at least in summary. It’s not a text with problems, but we can make these distinctions between real and fictitious forces at every point where the reader is likely to become confused. It will take a few more examples, and material now offloaded to reactive centrifugal force, which at minimum needs a better summary here even if kept there, per WP:SS.

And thanks for being reasonable about all this. After all this work on these articles you’ve done, you might be a little too “close” to the subject. SBHarris 03:23, 3 August 2015 (UTC)

I think there is still at least one point of disagreement (or perhaps a misunderstanding or misinterpretation of your words on my part). In your astronaut example above, when you say "weight" are talking about the force that the astronaut exerts on the chair as he sits in it or the force that he would say he feels keeping him in the chair. From my reading of the sources, the former is the reactive centrifugal force while the latter is the fictitious centrifugal force. Why I think this is that if we have the astronaut stand up and jump, he will say that he feels a force that pulls him back down to the "ground", the same force that holds him in the chair when he sits back down. However, since he is no longer in contact with the chair or outer wall of the space station when he jumps, there can be no reactive centrifugal force. Similarly, in the classic example of a passenger in a car going around a curve but with the passenger not starting out sitting right next to the door. The passenger will say that he feels a force pushing him outwards before he comes into contact with the wall. Again, that force that he feels before coming into contact with the wall really cannot be the reactive centrifugal force but is the fictitious centrifugal force and is a misinterpretation of his own inertia based on an intuition developed in a (roughly) non-inertial frame. Do we disagree on whether this "force" that our rotating observers feel in these instances is the reactive or fictitious centrifugal force? --FyzixFighter (talk) 04:42, 3 August 2015 (UTC)

If we have a really long radius of rotation so we don’t have to worry about Coriolis effects, it’s a lot easier to analyze the problem you posit from a linearly accelerated frame. We start with an astronaut floating in air, weightless in the middle section of big rocket far out in space with everything inertial and weightless. There is a chair at the rocket rear end. Now we fire the rocket motor to accelerate it into the astronaut. The astronaut sees the chair accelerate toward him, but he feels nothing as he sees it approach (he might imagine himself "falling," but that's a mere point of view). Until he hits the chair, he remains weightless. After he hits the chair, there are real contact forces, and these give him weight.

Now, before he hits, he can analyze the situation from his frame or the (accelerated) rocket frame. We’ve seen what it’s like in his frame already. In the (accelerated) rocket frame, a fictitious gravity-like force appears that makes the astronaut “fall” down toward the chair. We can’t really call this force “weight” because before he hits the chair, the astronaut is in free fall, zero-g, and is weightless. He does not FEEL this fictitious force during his “fall.” And that’s basically it. The situation is the same if the motor is always on, and the astronaut simply takes a leap upward: he’s in free fall, like a trampoline artist, until he hits the chair again. If he were blindfolded and couldn’t see the chair, he would conclude that there were no forces on him throughout his upward and downward arc (his is the happy weightlessness of the man in Einstein’s falling elevator).

So, to take your questions one-by-one, the man in the chair in a centrifuge feels the chair pushing up against him. That’s a real honest-to-god force, acting centripetally. There is also a Newtonian reaction force to it (as usual, which real force-pair you choose as “action” and which “re-action” are arbitrary) which the astronaut exerts against the chair. That’s a centripetally-directed reaction force, and is again real, not fictitious. It’s the subject of our reactive centrifugal force article.

Finally, in the chair frame, there is a fictitious centrifugal outward force on the man which explains why the chair pushes upward/inward on him, yet he does not accelerate and is motionless (in the non-rotating frame he isn’t motionless but travels in a circle, so doesn’t need this force, and it doesn’t exist). As usual, the astronaut cannot feel the fictitious centrifugal force; the only force he feels is the real centripetal force upward from the chair. (In somewhat the same way, we never feel gravity; we only feel our chairs. Hence it’s not too surprising that if you simply remove the chair, we are completely weightless, as the chair was the only force we could feel, and we interpreted that contact force as “weight”).

In the case of a passenger in a car, if he were properly weightless (the car/rocket is inertial before it decides to take a curve in outer space) he would simply continue to feel weightless while the door accelerated towards him, but there would be no impression of some force on him, pushing him toward the door. If he were blindfolded, he’d never care about the turn, or have any way of knowing it was happening, until the door hit/contacted him. We do not feel fictitious forces (the pure force that appears in the car frame, that “pushes” the passenger into the wall). The impression of such a thing in real life is due to friction on the seats of our pants as the car is accelerated sidewise toward us, and a friction normal-force informs our rear ends that we are being accelerated across the seat surface by a real force pushing the seat sideways, not a fictitious force. But this friction force from the seat (which is quite real, not fictitious, and slows us before we hit the door) is a complication of gravitational weight and standard sliding friction, and it only serves to fool us into thinking we can “feel” a fictitious force. We can’t. We feel friction-force if it is supplied by the seat and sliding weight, but it’s easy to construct this problem so that no such thing ever appears until we contact the door.

So I’m not sure if we disagree or not. I deny that anybody not touching a seat feels ANY force at all, and indeed feels weightless. And indeed is subject to no forces at all, in his own frame. In the frame of the seat, he is subject to a single fictitious force pushing him toward the seat, but of course he doesn’t feel that. Before he contacts the seat, he feels nothing. SBHarris 05:44, 3 August 2015 (UTC)

... so if I fall from my roof, I'm weightless with no forces (other than negligible air resistance) acting on me until the ground accelerates up to hit me? I'm not denying the truth of this in my own reference frame, it's just not how an outside observer sees the situation. If I happen to live on the equator instead of near a pole, the ground doesn't accelerate upwards quite as fast because it's turning away from me. What do you call this effect? Dbfirs 07:21, 3 August 2015 (UTC)

@SBHarris - by that argument then gravity and any other interaction-at-a-distance force are not real forces; the only real forces would be contact forces. An astronaut in orbit would say that they feel weightless but gravity is acting on him. Just because someone feels weightless does not mean that there are no real forces are acting on her nor that the net force is zero. Getting back to the astronaut on the space station - would not he say that there is a force holding him in the chair? Assuming he says yes, is that force the reactive centrifugal or the fictitious centrifugal force? Or more terrestrial, if you sit in an aircraft accelerating rapidly toward takeoff, then would you not say there is a force that pushes you back into your seat? Is that force real or fictitious? --FyzixFighter (talk) 12:18, 3 August 2015 (UTC)

Interesting though this discussion is, could we please try to get on with improving the article. In Newtonian physics gravity is regarded as a real force. How we feel or perceive a force is a complex issue involving the compression of nerve endings and our mental interpretation of the signals that this produces. Exactly how we interpret those signals is a matter for psychologist and is not relevant to physics.

We all agree that where there is a centripetal force there is an equal and opposite reaction force (which is sometimes called the reactive centrifugal force). No one is trying to say that this force does not exist. Giving it a special name, however, does cause a lot of confusion for reasons that I have stated before. To give this force a special name suggests that it is in some way special or unexpected but this is not the case. All forces have a reaction force and mostly they do not have special names. You have all managed to have a discussion about how we feel forces without using the term 'reactive centrifugal force'. This the way that physics and engineering has been taught for the last half century or so. All you ever need is Newton's laws and the real forces.

The other reason for not giving the centripetal reaction force a special name is that we do need a special name for the fictitious force that we have to invent when we work in a rotating reference frame. Nothing like this exists in an inertial frame; a new force has to be invented to make the physics work.

This is all well established physics that I hope that we can all agree on so that we can devote our energy to deciding how we can best explain all this as clearly as possible to our readers. Martin Hogbin (talk) 07:39, 4 August 2015 (UTC)

"All forces have a reaction force" you say? How do you define "reaction force"? It's easy for contact forces where there is no delay in force-action, but not so easy for G fields. If I'm being "pulled" by a gravitational force from someplace far, far away, where is the reaction force if I decide to resist (or cease resisting)? On the local gravitational field itself? The source of the field may be some ways away, too far at the speed of light to feel whatever I do immediately. It will need to propagate back. So to avoid instantaneous action at a distance, I must exert a force on the G field? But I do not do work on the field (in non-curved field approximation). The work that happens as a result of my inertial resistance to the field acceleration, goes into my increase in kinetic energy, 1/2mv^2 = force x distance. The problem is kinetic energy is a system property-- it has no location. It's a property of extended space-time. I don't see any simple application for Newton's third law here. The field doesn't store my momentum change. Locally, before my motion is felt by the source, I not only store the work done in space time by means of kinetic energy, but also I store my momentum for Newton's third law, by means of mass x velocity. That's in space-time in the system, too, as v is relative. I can easily choose a frame where I have no momentum or kinetic energy. SBHarris 02:55, 14 August 2015 (UTC)

Improving the article

If you Google 'centrifugal force' you will see all sorts of answers ranging from the currently accepted meaning to the plain wrong. This makes the subject very confusing to the beginner. It would be good for this article to be a beacon of clarity, so that readers of all types can find out exactly what CF is? That is not easy though. Here is my take on the subject. Martin Hogbin (talk) 09:26, 3 August 2015 (UTC)

The current meaning

As the subject is taught today, CF refers exclusively to the fictitious or inertial force that is necessary to invent of you want to use Newton's laws in their standard form in a rotating frame of reference.

Centrifugal force does not exist

If you have not done an engineering science or maths degree or are not studying for one it is unlikely that you will want to (or be able to) work in a rotating reference frame. All elementary work is done (often without explicitly stating so) in an inertial frame. This means that for the general reader, including students of elementary classical mechanics, centrifugal force does not exist. When working in an inertial frame, all mechanics problems can be dealt with using Newton's laws and real forces. There is never any need to have a centrifugal force. For this reason, many teachers of elementary mechanics tell their students that there is no such thing as CF. This lie-to-children it is told for good reason; to show how powerful Newton's laws and (real) forces are in mechanics problems and to avoid confusion and error when students attempt to apply various misunderstandings of CF to the problem.

Our problem is that saying that CF does not exist is wrong. On the other hand, it is the best place to start if you are a beginner in mechanics. Martin Hogbin (talk) 09:26, 3 August 2015 (UTC)

Other meanings

Reactive centrifugal force

This is the real force which is the reaction to the (real) centripetal force that needs to be applied to a body to cause it to move in a circle. It acts, not on the body moving in a circle, but in the body that is causing it to move in a circle (even though that body itself may be moving in a circle).

The current opinion on this is that to give a special name to a force that is just the reaction to another force in unnecessary and potentially misleading. All forces, according to Newton's laws, have reactions and there is no reason to give the reaction to the centripetal force a special name, in fact there is good reason not to.

My personal theory (which is OR and will not go in the article) is that some people, often engineers did not want to be bothered with rotating frames, as such but knew that centrifugal force, as an outward force acting on the object itself, does not exist but knew the term CF was in common use. Their answer is to say that the term refers to an outward reaction force. This provides a convenient explanation of the existence/non-existence of CF.

I do not subscribe to the theory that there is a difference in terminology between physicists/mathematicians and engineers. I am sure that if you study engineering today you will be taught the current meaning of CF. The problem is that reference to the reactive is found in some, mainly practically oriented, engineering text books although this is less than might appear at first sight. Martin Hogbin (talk) 09:26, 3 August 2015 (UTC)

Wrong meanings

There are plenty of these to be found in all sorts of places. Martin Hogbin (talk) 09:26, 3 August 2015 (UTC)

What to do?

Any suggestions. Martin Hogbin (talk) 09:26, 3 August 2015 (UTC)

Well, XKCD has a comment: [4] SBHarris 02:50, 4 August 2015 (UTC)

Excellent! That explains the situation very well. Centrifugal force does exist, but only when you do physics in a rotating reference frame. Martin Hogbin (talk) 08:31, 5 August 2015 (UTC)

Yes, I liked it too! I suggest that we continue to reflect the mainstream terminology found in basic texts, and ignore self-published criticism and idiosyncratic views. It would be better if we could agree on a form of words that doesn't annoy anyone, but is this possible? Dbfirs 16:10, 5 August 2015 (UTC)

Probably not, unfortunately. Many people have misconceptions as to what 'centrifugal force' is. Just Google the term and see what you get; all sorts of bizarre interpretations and misunderstandings. Martin Hogbin (talk) 14:20, 7 August 2015 (UTC)

How about a section on 'Misconceptions of centrifugal force'?

Misconceptions about CF seem widespread in the net. Should we have a section discussing them? Martin Hogbin (talk) 12:32, 10 August 2015 (UTC)

I have listed below the first few hit from Googling 'centrifugal force' and tried to categorise them with my comments. This shows exactly what this article is up against. Martin Hogbin (talk) 16:49, 10 August 2015 (UTC)

An outward force acting on objects that are in circular motion

These are just plain wrong.

[5] 'The apparent force, equal and opposite to the centripetal force, drawing a rotating body away from the center of rotation, caused by the inertia of the body'.

It least it calls it an apparent force and mentions inertia. Martin Hogbin (talk) 16:50, 10 August 2015 (UTC)

The reactive centrifugal force

These are all perfectly logical and correct. but pointless, and with all this confusion, who needs another correct answer. It is clear that we do need mention this usage in the article in some detail though, but in the right context. Martin Hogbin (talk) 16:53, 10 August 2015 (UTC)

[6] 'centripetal force and centrifugal force, action-reaction force pair associated with circular motion.'

[7] 'Many students call centripetal force centrifugal force, but the two are not the same. Centrifugal force is the reaction to centripetal force as explained by Newton's Third Law of Motion '

Plain crazy

[8] 'CENTRIFUGAL FORCE: Centrifugal force is generally not believed to be a source of energy. It is simply a FORCE that acts internally on all rotating objects, and therefore cannot be harnessed for the production of energy. Or is it?...In the 1960’s, Bruce DePalma ran some extraordinary experiments on the nature of rotation itself... '

Shut up and calculate

Uniform circular motion (centrifugal force) Calculator Who cares what it is so long as it acts outwards and we can calculate it. Martin Hogbin (talk) 16:54, 10 August 2015 (UTC)

Close, but no cigar

These, in my opinion, are the worst sources because they may, at first sight, appear to be correct, however they are quite misleading. I have added individual comments. Martin Hogbin

[9] 'Centrifugal force is a fictitious force associated with a rotating system, such as a merry-go-round on a playground. When the system stops rotating, the force seems to disappear.'

So what do they mean be 'rotating system'? Do they mean the merry-go-round? It does not matter whether the merry-go-round is rotating or not, the force is still ther if you use a rotating reference frame. Martin Hogbin (talk) 17:03, 10 August 2015 (UTC)

[10] 'An object traveling in a circle behaves as if it is experiencing an outward force. This force, known as the centrifugal force, depends on the mass of the object, the speed of rotation, and the distance from the center. It is important to note that the centrifugal force does not actually exist. We feel it, because we are in a non-inertial coordinate system.'

The first sentence is just wrong. Why are we 'in' a non-inertial coordinate system? We can chose whatever coordinate system we like. If it is non-inertial than we will get some extra (fictitious) forces.

[11] 'Centrifugal force is sometimes referred to as a 'fictitious' force, because it is present only for an accelerated object and does not exist in an inertial frame.'

If only it had said 'accelerated frame', that would have been closer. The last 7 words are good though.

[12] 'a fictitious force, peculiar to a particle moving on a circular path, that has the same magnitude and dimensions as the force that keeps the particle on its circular path (the centripetal force) but points in the opposite direction.'

The path of the object does not matter! It is the frame of reference that we use to measure the path that counts. Martin Hogbin (talk) 17:03, 10 August 2015 (UTC)

[13] 'An outward-directed "fictitious force" exerted on a body when it moves azimuthally in a noninertial rotating reference frame.' But then we have the correct, 'Centrifugal force is a fictitious force because it is a by-product of measuring coordinates with respect to a rotating coordinate system as opposed to an actual "push or pull." '.

The body does not need to move azimuthally. If the frame is rotating we get the CF. Martin Hogbin (talk) 17:03, 10 August 2015 (UTC)

Yes!

It is good to see that the Googler has some chance of hitting on the right answer.

[14]'Whereas the centripetal force is seen as a force which must be applied by an external agent to force an object to move in a curved path, the centrifugal and coriolis forces are "effective forces" which are invoked to explain the behavior of objects from a frame of reference which is rotating.'

[15] 'an outward force on a body rotating about an axis, assumed equal and opposite to the centripetal force and postulated to account for the phenomena seen by an observer in the rotating body. '

I could quibble about 'seen by an observer' rather than 'measured relative to' but let us give this one the benefit of the doubt. Martin Hogbin (talk) 17:07, 10 August 2015 (UTC)

[16] Speaks for itself. Martin Hogbin (talk) 17:07, 10 August 2015 (UTC)

[17] 'I don't have time to write a full answer right now, but I will mention that the term "centrifugal force" does refer to a real and precisely defined physical concept. We tell intro-level physics students that it's an illusion just because that's a simpler way to keep them from getting confused than discussing noninertial reference frames. – David Z '

There were many answers here but this was one of the best ones. If you do not know what a 'rotating reference frame' is you do not need centrifugal force. Forget about it! Martin Hogbin (talk) 17:07, 10 August 2015 (UTC)

Discussion

Except that all the time, very many people screw up and imagine themselves in rotating frames. So you can't just ignore this problem, until these people become physics students calculating things in rotating frames. Our Good Readers and Intelligent Laypersons imagine themselves in a rotating space station, a centrifuge, or throwing the hammer in the Olympics. But these are all rotating frames, with new laws, that bite. Or people may forget that the reference frame they inhabit here on Earth is already slowly rotating, so if we believe in Newton and gravity, we need some reason why geostationary satellites just "hang up there," in the same place that our satellite dishes point, unmoving, and don't fall straight down on us. A centrifugal force (CF) is the answer to all this, but we need to differentiate it from "real" forces you can feel, and that will squash you.

CF and other similar "fictitious forces" (Euler, Coriolis) are theoretical forces introduced to "explain" why (as viewed from the ground) those geo-synch satellites don't simply fall straight down from gravity. But any force that appears in a rotational motion that you actually FEEL (as in a centrifuge or fun-wheel or whatever) is not the physics special definition of (fictitious) centrifugal force, but rather is a real force that happens to be centrifugally (outward) directed. It is technically not a CF but a Newton's 3rd law reaction force to a real centripetal force, which is also a real force, being applied to you by a real object (if it is applied by a long range field, like gravity, no "reactive" force appears, so there is no real force directed outwardly on the field).

There also exist real forces where (as a point-particle) you can't even feel the primary force, like the mentioned forces from long-distance force fields (gravity). But these aren't CF either. Since CF completely goes away for a non-rotating person, we know it's only a function of point-of-view, and in that sense is not real. All other types of forces (the ones that squash you) don't depend on point of view, and still exist for every observer. Even gravity leaves some vestiges (tidal forces) for accelerated observers between any two points. These cannot be gotten entirely rid of by changing frames. This is a tell-tale sign that gravity is a real force, and not (wholey) fictitious and observer-dependent, like CF. SBHarris 01:23, 14 August 2015 (UTC)

I do not quite understand the point that you are making. People who are not (undergrad or above) physics, maths or engineering students are very strongly encouraged to work only in inertial frames. So what keeps the geostationary satellite in orbit is the fact that it the Earth's gravity provides the necessary centripetal force to keep it in orbit and prevent it continuing in a straight line and flying out into space. That is by far the best and simplest explanation as it requires only Newton's laws. No special forces of any kind are required.

What forces you feel depends on how your brain interprets the signals it receives from nerves and sense organs. This is a question of psychology rather than engineering or physics. For example, if you are in a vehicle which accelerates, many people describe feeling a strange force acting upon their body and pushing them back in their seats. In fact many sources show this force. In the other hand, if someone sneaks up behind you and pushes you you generally feel a force pushing you from behind.

I do agree that there are some very important psychological issues that we maybe should discuss in the article, if we can find suitable sources.

Finally, I do not understand what you mean when you say, 'any force that appears in a rotational motion that you actually FEEL (as in a centrifuge or fun-wheel or whatever) is not the physics special definition of (fictitious) centrifugal force, but rather is a real force that happens to be centrifugally (outward) directed'. What real 'centrifugally (outward) directed' force are you referring to? Martin Hogbin (talk) 14:40, 20 August 2015 (UTC)

Why Not

In another article https://en.wikipedia.org/wiki/Polar_coordinate_system#cite_ref-angular_19-0 it says, The term $r{\dot {\varphi }}^{2}$ is sometimes referred to as the centrifugal term, and the term $2{\dot {r}}{\dot {\varphi }}$ as the Coriolis term. For example, see Shankar. Although these equations bear some resemblance in form to the centrifugal and Coriolis effects found in rotating reference frames, nonetheless these are not the same things.

Why not?

2601:449:4001:1F70:C834:76B7:76C:AF06 (talk) 15:57, 22 July 2015 (UTC)

The line after, 'For example, the physical centrifugal and Coriolis forces appear only in non-inertial frames of reference. In contrast, these terms, that appear when acceleration is expressed in polar coordinates, are a mathematical consequence of differentiation; these terms appear wherever polar coordinates are used. In particular, these terms appear even when polar coordinates are used in inertial frames of reference, where the physical centrifugal and Coriolis forces never appear', explains why. Martin Hogbin (talk) 18:52, 22 July 2015 (UTC)

Please note the comment at the top of this page, 'This is not a forum for general discussion about centrifugal force'. Martin Hogbin (talk) 18:57, 22 July 2015 (UTC)

But in both cases isn't it about differentiation of a position vector? I can't see the difference, unless you meant to use r' instead of r in one of the terms on the right hand side, since you are referring to two different positions. One position r is for the stationary point on the rotating frame while the other position is moving in the rotating frame and so you should surely use r' to distinguish them.

the argument about the Coriolis force is wrong. The mistake is at equation (11.28) in the Atam P. Arya reference mentioned above. The dash should be on the vector and not on the differential operator. I raised this matter above but nobody responded 2601:449:4001:1F70:8040:8E94:FEA8:A072 (talk) 16:47, 3 August 2015 (UTC)

I'm certain that Atam Arya's equation (11.28) is correct as written. The vector r in the equation is the displacement vector from the common origin of the two coordinate systems to the location of the object whose motion it describes. I have no idea why you think its appearance in the term ω × r is supposed to be describing the location of a different point. It represents exactly the same point there that it also does in the rest of the equation.

David Wilson (talk · cont) 01:03, 6 August 2015 (UTC)

Hello

First , thank you Martin for moving my last post to the right place. I am not expert at navigating here on Wiki

Second , below , is an answer to Davids question/comments There are two positions involved. One of them is the position of the moving particle and the other is the position of the fixed point in the rotating frame. The particle moves relative to that fixed point. ω × r is the velocity of the fixed point in the rotating frame, but the particle is in motion relative to this point. You can't use the r vector in both cases. — Preceding unsigned comment added by 70.193.210.97 (talk) 11:18, 11 August 2015 (UTC)

To the extent there was any question implied by my comment, you have not actually answered it. You have simply asserted that "[t]here are two positions involved", one of which is "the position of the fixed point in the rotating frame", without either identifying which particular '"fixed point" you have in mind, or why you think any such fixed point is "involved".

The only "fixed point" Arya mentions anywhere in his exposition is the point Q, which he introduces on page 432 (of his second edition) for the sole purpose of deriving the equations of motion of the unit vectors along the axes of the rotating system. It has no other connection whatever to the point P with position vector r, which he introduces on page 430, and its position vector B has no other connection with r. Arya places no explicit restrictions on the motion of P—which is completely independent of that of Q—but implicitly assumes that its motion is "smooth"—i.e.that r is a differentiable vector function of time. The key to his exposition is his derivation, on pp.431–2, of his equation (11.26):

${\frac {d{\boldsymbol {\operatorname {A} }}}{dt}}={\frac {d'{\boldsymbol {\operatorname {A} }}}{dt}}+{\boldsymbol {\omega }}\times {\boldsymbol {\operatorname {A} }}\ .$

In this equation, the symbol "A" represents an arbitrary differentiable vector function of time. It has exactly the same meaning everywhere it occurs in the derivation, including its occurrence in the term "ω × A" of the final equation, so it is not even correct, let alone necessary, to distinguish this latter occurrence by appending a dash to it. Also A is not necessarily a position vector at all—though that possibility is certainly not excluded—so there is no question of any "fixed point in the rotating frame" being associated with it. Since r is a differentiable vector function of time, the above equation holds just as much for it as it does for any other differentiable vector function, and Arya obtains his equation (11.28) simply by substituting r for A in It.

David Wilson (talk · cont) 04:31, 18 August 2015 (UTC)

I would have to disagree with David. It seems to me that the maths is right but only in a situation like the weather where everything is joining in the rotation of the Earth. The maths only concerns one position. I seem to remember the term Hamiltonian frame, but perhaps that is something different again. 94.173.45.184 (talk) 13:02, 22 August 2015 (UTC)

Derivation of Acceleration

We define: ${\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}=\left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\boldsymbol {\omega }}\times {\boldsymbol {r}}$

And calculate the acceleration:

${\begin{aligned}{\boldsymbol {a}}&={\frac {\operatorname {d} ^{2}{\boldsymbol {r}}}{\operatorname {d} t^{2}}}={\frac {\operatorname {d} }{\operatorname {d} t}}{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}={\frac {\operatorname {d} }{\operatorname {d} t}}\left(\left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\boldsymbol {\omega }}\times {\boldsymbol {r}}\ \right)\\&=\left[{\frac {\operatorname {d} ^{2}{\boldsymbol {r}}}{\operatorname {d} t^{2}}}\right]+{\frac {\operatorname {d} {\boldsymbol {\omega }}}{\operatorname {d} t}}\times {\boldsymbol {r}}+2{\boldsymbol {\omega }}\times \left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times {\boldsymbol {r}})\ .\end{aligned}}$

By just inserting the definition I cannot explain myself where the 2 before the ${\boldsymbol {\omega }}$ comes from in the very last line. I think there must be something wrong or it's not clearly explained. — Preceding unsigned comment added by 92.204.72.219 (talk) 14:23, August 19, 2015‎ (UTC)

Yes, the derivation given in the article isn't very clear. The first equation isn't a definition, it's a theorem, and it applies not just to the vector function r, but to any vector function A of time:

${\frac {d{\boldsymbol {\operatorname {A} }}}{dt}}=\left[{\frac {d{\boldsymbol {\operatorname {A} }}}{dt}}\right]+{\boldsymbol {\omega }}\times {\boldsymbol {\operatorname {A} }}$

.

When you compute the acceleration you have to apply this identity with A = $\left[{\frac {d{\boldsymbol {r}}}{dt}}\right]$:

${\begin{aligned}{\frac {d}{dt}}\left(\left[{\frac {d{\boldsymbol {r}}}{dt}}\right]+{\boldsymbol {\omega }}\times {\boldsymbol {r}}\ \right)&=\left[{\frac {d^{2}{\boldsymbol {r}}}{dt^{2}}}\right]+{\boldsymbol {\omega }}\times \left[{\frac {d{\boldsymbol {r}}}{dt}}\right]+{\frac {d{\boldsymbol {\omega }}}{dt}}\times {\boldsymbol {r}}+{\boldsymbol {\omega }}\times {\frac {d{\boldsymbol {r}}}{dt}}\ .\end{aligned}}$

If the identity in its original form (i.e. with A = r ) is now applied to the last term on the right-hand side of the above equation, it becomes :

$\left[{\frac {d^{2}{\boldsymbol {r}}}{dt^{2}}}\right]+{\boldsymbol {\omega }}\times \left[{\frac {d{\boldsymbol {r}}}{dt}}\right]+{\frac {d{\boldsymbol {\omega }}}{dt}}\times {\boldsymbol {r}}+{\boldsymbol {\omega }}\times \left(\left[{\frac {d{\boldsymbol {r}}}{dt}}\right]+{\boldsymbol {\omega }}\times {\boldsymbol {r}}\right)\ .$

And now, on multiplying out the last term in this expression and collecting like terms together, you obtain the last line of your series of equations above.

David Wilson (talk · cont) 15:43, 19 August 2015 (UTC)

It only works for a point that has the angular speed in question. In other words it is the straight line path and the effects of inertia which draw it away from the centre. 94.173.45.184 (talk) 14:53, 25 August 2015 (UTC)

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Looks OK to me. - dcljr (talk) 14:34, 24 June 2016 (UTC)

Poles vs. equator

In the Weighing an object at the Earth's poles and on the equator section, the text used to say, "for this reason the object will weigh about 0.5% less at the equator" before an IP editor changed the number to "0.35%", explaining, "Two factors for weight reduction: 0.35 % due to rotation, 0.15 % due to ellipsoidity." Problem is, the 0.35% value is not to be found anywhere in the cited source (although it may be supported by additional calculations). Before I looked into the page history, I changed it to say "0.53%", since that's the number given in the source (on pages 2, 6, and 26). The source does note (on p. 2) that: "This is in part due to the centrifugal force resulting from the rotation of the earth… and in part due to the bulge of the earth at the equator". So, we need to do one of the following:

1. explain that the 0.53% value is actually mixing two effects;
2. cite two separate numbers for the two effects;
3. cite the correct number for the "latitude effect" alone.

Presumably, options 2 and 3 could be based on the current source, being calculated from equations on p. 7 and/or 8 (or the combined version on p. 13), but if so, it should be explained in the 'ref' that a calculation was used (as opposed to simply citing a number given in the text of the source). Ideally, though, a different source should be found that does explicitly state the two values. I would argue against option 3, since it would be misleading (giving the impression that the number given for the "latitude effect" is the true percent change in the object's weight, whereas it's only a portion of that change). And Option 1 seems a bit unsatisfying, since it wouldn't indicate how much of the change in weight was due to one effect versus the other. - dcljr (talk) 17:51, 23 June 2016 (UTC)

Now that the second source (previously a dead link) has been "rescued" by Cyberbot II, I have changed the section again to describe both effects, citing numbers from both sources. - dcljr (talk) 14:31, 24 June 2016 (UTC)

This source is not very clear and must have been be incorrectly inperpreted. From nasa I get primary data polar radius 6356.752 km and equatorial radius 6378.137 km. So the relative gravity difference is the square of the ratio between them minus one, giving -0.67% at the equator.

The effect of rotation only is -6378137 * (2π / (24 * 3600))² m/s², divided by gravity at the pole of 9.83 m/s² is indeed -0.34% as stated.

Together -1.01%

Furthermore it is incorrectly stated that the only force on the object at the equator is gravity. There is also the horizontal friction that actually forces it to rotate with Earth instead of staying put in space and sliding off the scale.

−Woodstone (talk) 05:50, 26 June 2016 (UTC)

If your interpretation is correct, then surely you can find your number in a reliable source… - dcljr (talk) 06:24, 26 June 2016 (UTC)

No friction is needed because the earth's angular velocity is not changing. The only acceleration is towards the centre of the earth, and gravity is in that direction. Dbfirs 07:06, 26 June 2016 (UTC)

Using the ratio of the squared radiuses to estimate the ratio of the accelerations attributable to the Earth's gravitation at the poles and equator is quite inaccurate. That calculation implicitly takes the Earth's gravitational attraction at the poles to be that of a sphere of uniform density (or, at least, one with a spherically symmetric distribution) and with radius equal to the Earth's polar semiaxis, and its gravitational attraction at the equator to be likewise that of a sphere with radius equal to the Earth's equatorial radius. But the Earth is an oblate spheroid, and using a spherical approximation to estimate the accelerations due to gravitational attraction on the Earth's surface is sufficiently inaccurate to account for the observed discrepancy.

Somewhat better estimates can be obtained by taking the Earth to be an oblate ellipsoid of revolution, with two semiaxes equal to the Earth's equatorial radius and the third equal to its polar semiaxis. According to my calculations, this gives an estimate of 9.825 m/s² for the Earth's (effective) polar gravitational acceleration and 9.784 m/s² for its effective equatorial gravitational acceleration. Compared to the actual values of 9.832m/s² and 9.780 m/s² given by the cited reference's formula, g = 9.80613 ( 1 – 0.0026235 cos(2 x latitude) ), these estimates are only 0.07% too low and 0.04% too high, respectively. The spherical approximations, on the other hand, produce values of 9.864 m/s² and 9.765 m/s² which are 0.3% too high and 0.15% too low, respectively—i.e. percentage errors some 3 to 4 times larger than those of the ellipsoidal approximation, and certainly sufficient to account for the discrepancy you're concerned about.

David Wilson (talk · cont) 23:29, 27 June 2016 (UTC)

Thanks for this clarification. −Woodstone (talk) 06:02, 28 June 2016 (UTC)

Should some of this clarification be added to the article — say, in the ref(s)? - dcljr (talk) 06:13, 28 June 2016 (UTC)

Velocity proof

In the velocity derivation, the wording and formulæ are contradictory in terms of what applies to rotating versus stationary reference frames. —DIV (120.19.177.139 (talk) 02:45, 6 September 2016 (UTC))

Centrifugal force mv²⁄r

Neither v_c=mv²⁄r nor a_c=v²⁄r is anywhere in this article. I feel that at least one of them should be somewhere. — Preceding unsigned comment added by 72.172.59.119 (talk) 22:05, 24 July 2016‎

There not in this article because F_c and a_c are the centripetal force and acceleration. --FyzixFighter (talk) 13:05, 29 July 2016 (UTC)

You have no basis for that knock-down. The magnitude of the centrifugal fictitious force is the same as the magnitude of the centripetal force. Any equation giving the magnitude of the centripetal force also gives the magnitude of centrifugal fictitious force. The direction of each vector is exactly opposite the other. 72.172.59.119 didn't specify which direction (petal or fugal) was positive, so how would you know that positive was "petal"?

Both v_c=mv²⁄r and a_c=v²⁄r are accurate and notable and what people need to see.

98.216.249.147 (talk) 08:08, 11 March 2018 (UTC)

The fact that speed (v), presumably the speed in the inertial, non-rotating frame appears makes those equations only always true for the "petal" case. Additionally, what the OP listed as v_c should more properly be listed as F_c since its units are that of a force, and not a speed. Only when the rotating frame is attached to the rotating object, so that the object appears stationary in the rotating frame (which is only true for circular motion in the inertial frame) is the magnitude of the centrifugal force equal to the magnitude of the centripetal force. Only in that case could the OP's equation be valid for both "petal" and "fugal" cases, looking at just magnitude and ignoring direction. The general concept of the centrifugal force, the one taught in all physics textbooks, applies to all objects whose motion is described in a rotating frame, regardless of how they are moving in the rotating frame. The extreme example of this is an object that is stationary (net force is zero) in the inertial, non-rotating frame appears to be moving in a circle in the rotating frame. To describe the motion in the rotating frame, an observer would have to add both the centrifugal force and the Coriolis force to the left hand side of F=ma to make that law still hold true within the rotating frame. If you've got a source that says otherwise, please share. --FyzixFighter (talk) 08:38, 11 March 2018 (UTC)

Okay, so the OP made a typo and I didn't catch it. You know what he was talking about. Certainly such equations would appear in proper explanatory context (that they only apply to uniform circular motion). As you say in "The general concept of the centrifugal force" there are complications, but the fact remains that in a very common special case (as it would be clarified in the text), those equations apply and are apt.

By the way, I think the example of a non-accelerating rotating-only frame is a cool one. If one stands on the North pole and watches the stars going around, the centripetal force holding the stars on the circular path around you is fictitious! Cool huh? 98.216.249.147 (talk) 09:26, 11 March 2018 (UTC)

The problem is that pedagogically those equations muddy the water. This article (yes, I know it's behind a paywall - sorry) argues that when you use the object's velocity to define the centrifugal force, the origin of the fictitious force is not clear. When we write it with omega and clearly state that omega is the rotation rate of the frame, it becomes clear that the centrifugal force appears because the frame has a rotational acceleration relative to the inertial frame. Special cases can be mentioned and noted, but they should not be used to define or explain the overall concept.

Regarding the stars, we did at one time have a section that had that situation as an example. The observed centripetal force is indeed fictitious, and is the sum of both the centrifugal force and Coriolis force - in this case the Coriolis force is inward and twice the magnitude of the centrifugal force. --FyzixFighter (talk) 13:17, 11 March 2018 (UTC)

Nah, those equations don't muddy the water at all. Arcane monolithic "explanations" muddy the water. In the ball-on-string-uniform-circular-motion, the magnitude of the centrifugal (and petal) accel is V2/r, is it not? I.e "shut up and calculate". That's a practical equation that adds a lot of clarity and it's one of the first things everyone learns. And, the ω is Vr, it can't be anything else. Are you trying to generalize to an ω that isn't necessarily Vr? Because if you are, I would argue strongly that that's way out of scope for this article. We're not writing an article to be perfectly perfect and righteously right and completely complete. Generalizing to an arbitrary ω is way off topic, or would need to be relegated to a small section at the bottom.

About the stars: Huh? Coriolis force/accel is a result of rdot. In the star example there is no rdot. And, Coriolis is perpendicular to both r and ω. Coriolis can't add to a vector aligned with r to reverse its direction (fugal to petal). What are you smoking?  :-)    98.216.249.147 (talk) 00:19, 12 March 2018 (UTC)

You got sources for any of those statements? I got plenty of sources such as university physics textbooks and journal articles that support mine. If don't have any then, based on my previous experience with other editors pushing similar statements, this conversation about improving the article is going to go nowhere. --FyzixFighter (talk) 00:51, 12 March 2018 (UTC)

You got a source for the ridiculous "centripetal force (is) the sum of both the centrifugal force and Coriolis force"? You can't have it both ways, buddy. You can't go on and on (and on and on) bullying your way around a talk page with your arcane mumbojumbo and then fall back on "You got a source for that?" when someone calls you on a part of it. Stop playing the disruptive "find a source or go away" game. You spout your collection of sources like an absolute authority that you pretend to be. I think you really are smoking something -- pride. 98.216.249.147 (talk) 02:10, 12 March 2018 (UTC)

Ask, you shall receive - The Mechanical Universe, pp. 398-399. The example considered beginning at the bottom of page 398 is for a particle which remains at rest in the inertial frame, and describing that motion in a rotating frame. The centrifugal force is mω^2r outward, the Coriolis force is 2mω^2r inward, so in the rotating frame the net force (real plus fictitious) is mω^r inward - exactly what the rotating observer would require to make Newton's 2nd law work in the rotating frame.

Again, you got sources for your statements? --FyzixFighter (talk) 03:55, 12 March 2018 (UTC)

This isn't the game you think it is, friend. You think we're arguing about the nature of reality. You're pulling out all your stops to show "everybody watching" how very smart and perfectly perfect and righteously right and completely complete you are.

You've lost sight of the purpose of the talk page which is discuss making the article better. Making it better means discussing civilly how to better communicate the matter to lay readers. Lay readers. It doesn't mean slamming down everyone who comes here and doesn't agree with you how to do that, implying they're yet another fringe-purveying idiot and constantly demanding sources on the talk page. It's wonderful how much you know and how very very many sources you have at your disposal, but you're being a real jerk about it. You've got a huge chip on your shoulder, buddy.      98.216.249.147 (talk) 04:29, 12 March 2018 (UTC)

So, what your saying is you have no sources to back up the improvements you are suggesting for the article. Got it. I'm all for making the article more accessible, but not at the cost of being incorrect. I don't think the inclusion of the equations for centripetal force and acceleration are helpful because it is limited to the special case of objects that appear stationary in a rotating frame. The definition of the centrifugal force applies to all objects, both stationary and non-stationary, in the rotating frame. For non-stationary objects, the centripetal equations are not applicable. I also agree with the Kobayashi reference on the pedagogy of how we define the centrifugal force.

Your common sense says I'm wrong, my common sense says you're wrong - one of the ways to resolve this is via sources, there are other dispute resolutions avenues if you'd like to pursue those. I don't think I've been uncivil in anyway - feel free to point out where I may have been uncivil. If you feel that I have been uncivil in any way then report me or request a third opinion. --FyzixFighter (talk) 12:17, 12 March 2018 (UTC)

Ferris Wheel, Omega Clarity

A Ferris-wheel chair is an example of a practical non-rotating frame in circular motion. In that case, the centripetal force and centrifugal fictitious forces exist because the frame is accelerating toward the wheel's axle, not because the frame is rotating. Allow the chair (and its frame) to rock or even to rotate all the way around at a rate faster or slower than wheel's rate, and the axle of the chair experiences the same centripetal/centrifugal forces regardless. In that arbitrarily rotating chair-frame, the direction of those centripetal/centrifugal forces is constantly changing, but they surely exist -- and without respect of the rotation of the frame.

Fear not! There's a solution to this!

In the article, we just need to be more clear when we're assuming a few things: 1) when the "rotating frames" we speak of are attached to and moving with objects that are moving in a circle or curved trajectory, and 2) when the "rotating frames" we speak of are rotating at the same rate as the velocity vector, and 3) when we're assuming that the point in question isn't actually on the axis.

It's often a pretty big leap for readers to make those assumptions, and we really should be more clear on it.

For example: The most fundamental statement in the article is "Centrifugal force is an outward force apparent in a rotating reference frame". That's incomplete unless the frame is also 1) moving along a curved path in inertial space, 2) rotating at the same rate as the angle of the velocity vector. Just saying "rotating" while leaving out the other criteria is misleading.

For another example, in the rω^2 equation, ω is really the rotation rate of the velocity vector, not the frame. If viewed that way, the meaning is clear. If not, since any frame can be rotating arbitrarily, then the frame's ω in this case needs to be specified as matching the velocity vector's ω.

Most lay-readers might assume that we are always talking about such conditions, or even know when we are and when we aren't. But, "most might understand" is a very low standard for clarity. Too low. The article really needs improvement there.

98.216.249.147 (talk) 07:43, 11 March 2018 (UTC)

The current text of the article follows closely textbook and other scientific sources. None of the ones I've encountered mention these additional assumptions. If you've got a source that says as much, please feel free to share. I'm open to the idea that I've missed one. I think there might be confusion as to what is meant by "rotating frame". Most of the similar examples in the article would say this frame is the one with an origin on the axle of the Ferris wheel and rotating around so that the chairs appear stationary. That's the usual and probably the most intuitive frame for analyzing the motion of chairs and riders. Let's expand the example - if riders started throwing balls from chair to chair, they would have to account for both a Coriolis force and a centrifugal force to bootstrap Newton's 2nd law to work in that frame. The centrifugal force would be the one using the omega of the Ferris wheel, and none of your assumptions necessarily apply to the balls being thrown around. That is, they aren't stationary (what I think you mean be "attached to") in the frame, their velocity vector isn't really rotating at the same rate (at least not in the inertial frame) as the frame, and even if they did pass through the origin of the frame the formula for centrifugal force still works (it goes to zero at the origin). And yet, with the centrifugal force (and Coriolis force) for the ferris wheel frame, they could accurately predict the motion of the balls.

Another "rotating frame" could be the local frame of the chair, which may not only rotating about its own axis but also whose origin/axis is also moving and accelerating relative to the inertial frame. I've seen at least one source use the term "centrifugal force" for the fictitious force in that case, but I think the more general term "d'Alembert force" applies since it's more akin to the fictitious force felt in linearly accelerating frames, like an elevator.

However, if you really want to make changes and include these assumptions and you want them to stick, then you really need reliable sources to back the edits up. This article has been the focus of fringe science pushing editors in the past, which required a fairly strict adherence to WP:RS and most editors still hold any new additions strongly to that content guideline. --FyzixFighter (talk) 09:03, 11 March 2018 (UTC)

Good points. What I'm really getting at is that refs are being misrepresented, paraphrased not carefully enough. Knowledgeable editors need to interpret sources (sometimes very carefully) in order to separate out the badly-stated and fringy stuff, right? It takes science knowledge and language interpretation to be able to that well.

I looked up the ref that DickLyon just added. It's hidden, but I found the 1965 version online. In it, the authors say "as we shall see in the next section the centrifugal force is a fictitious force introduced to describe motion in a rotating, and therefore non-inertial reference frame" and then give two examples and then say "a centrifugal force arises only in a rotating reference frame". That's pretty clear they say that. BUT, they also only give examples of frames that are attached to a "stone on a string" and are rotating at the same rate. The context of those broad statements about arising only in a rotating frame suggests they're really only considering frames under those special conditions. We can't use that ref to support anything more general than the special cases conveyed in the full context of it.

That ref is (currently) being used to justify "Centrifugal force is an outward force apparent in a rotating reference frame." That wording isn't supported by the source (unless the wording changed in the 2nd edition). The source says "only in a rotating reference frame" not "always in a rotating reference frame" as is essentially said in the text. So, that's an example of less-then-careful-enough interpretation. DickLyon was clearly annoyed, calling it "silly", then he hastily sought a source to justify the words, not noticing the difference between "only in a RRF" and "(always) in a RRF".

98.216.249.147 (talk) 10:47, 11 March 2018 (UTC)

(Friendly hint - please see WP:THREAD) I think it's more a case of references assuming that when you look at the derivation they give, they are giving it for the universal case which implies "always". It is possible to find sources that are more explicit. For example, Taylor in his "Classical Mechanics" textbook says that "in order to use Newton's second law in a rotating frame ... we must introduce two inertial forces, the centrifugal and the Coriolis forces." I think Taylor's "must" qualifies as a "always". Kobayashi in this article explicitly states "Noninertial frame S' turned relative to inertial frame S will always cause a centrifugal force no matter what the point mass is rotating in S', whereas the Coriolis force requires a point mass to be in motion relative to the noninertial frame rotating with a velocity that is not parallel to the axis of rotation." I'll go ahead and add both Taylor and Kobayashi to that statement.

Do you have any sources that contradict the statement that the centrifugal force is always present in a rotating frame? Or do you have sources that say those three assumptions you mentioned above are needed for a correct understanding of the centrifugal force? --FyzixFighter (talk) 13:31, 11 March 2018 (UTC)

The most striking example of the omnipresence of the centrifugal and coriolis forces in a rotating frame is their effect on the distant stars as observed in a frame attached to the rotating Earth. Somehow its explanation was removed long ago from the article on coriolis force. I have restored it there. −Woodstone (talk) 18:09, 11 March 2018 (UTC)

Your restoration didn't last very long. Those who argued that the explanation is bogus appear to cling to their erroneous opinions with some stubbornness, so I suspect you'll have a fight on your hands keeping it in. Good luck.

David Wilson (talk · cont) 02:31, 12 March 2018 (UTC)

We're not talking about the "find an impossible source or go away" game. We're talking about more careful paraphrasing of (existing) sources. Like I said, knowledgeable editors like you (and maybe me) need to interpret sources sometimes very carefully to separate out the imperfectly-stated and to get at full context of a source. I offer two areas where the article has been careless that way...

1) Fictitious forces are generally necessary to be added to EOM to describe motion in a rotating frame, of course. And, fictitious petal and fugal forces (force components aligned with r) are generally a part of that. But, fugal and petal fictitious forces aren't needed everywhere in a rotating frame. Particularly, they're never needed when the object is on the axis. And so,

wording should be careful to avoid the implication that fugal fictitious forces exist everywhere in a frame just because the frame is rotating -- regardless of how uncareful a source is in that regard.

2) Fugal fictitious forces are always pointed away from the axis, never toward (or in any other direction). That's not physics, it's just definition. But, it means that if at one moment the net r-aligned fictitious force is toward the axis, then there's no actual "fugal" fictitious force. For example, from the center of a merry-go-round, a tree needs a petal f-force, not a fugal f-force. A rotating frame just isn't the one and only condition for a net centrifugal f-force. Admittedly, that's much a matter of definition (taking r-aligned inward f-forces out of consideration). But again,

wording should be careful to avoid the implication that fugal fictitious forces exists always in a frame just because the frame is rotating -- regardless of how uncareful a source is in that regard.

98.216.249.147 (talk) 01:50, 12 March 2018 (UTC)

The more correct statement would be that the fictitious centrifugal force goes to zero at the axis, which is what the equation predicts. The centrifugal force equation is valid everywhere in a rotating frame. Not all fictitious forces are either -fugal or -petal. Only the centrifugal force is strictly fugal. The coriolis force will point in a direction perpendicular both to the object's velocity in the frame and the angular velocity of the frame, so there are situations where it could be -fugal or -petal or tangential or a combination of both. The tree and merry-go-round example is handle in this source which states that the net -petal fictitious force is a combination of the outward centrifugal force and an inward coriolis force. --FyzixFighter (talk) 04:04, 12 March 2018 (UTC)

I agree with most of what you say (above) with regard to the nature of reality. We need to be careful how we word things though. If we mean to say "the equations are valid everywhere, but the the fugal/petal forces go to zero at the axis, then say that. I actually like that. That's different than saying "that appears to act on all objects when viewed in a rotating frame" because a force that's zero-valued doesn't actually act on anything. Saying such a force (appears to) actually act on all objects in a RF implies the force is always non-zero, which makes no sense to a reader who knows that there is no fugal/petal f-force at the axis.

I'm really confused about this "so there are situations where (Coriolis) could be -fugal or -petal or tangential or a combination of both" business. Coriolis is always perpendicular to r and proportional to rdot. How it can rotate 90 degrees to be parallel to r is mind boggling. I'm curious though, so I'll look it up. 98.216.249.147 (talk) 05:00, 12 March 2018 (UTC)

No, the Coriolis force is perpendicular to the particle's velocity $\mathbf {v}$, not necessarily to $\mathbf {r}$. It is only perpendicular to the latter when the vector $\mathbf {r}$ is parallel to the plane spanned by the vector $\mathbf {v}$ and the angular velocity of the rotating coordinate system.

David Wilson (talk · cont) 06:03, 12 March 2018 (UTC)

I'm using "r" as the vector perpendicular to the axis from the axis to the object (or pertinent point in space), not as from some presumed origin somewhere on the axis. That might make a difference in the discussion. 98.216.249.147 (talk) 05:20, 12 March 2018 (UTC)

It makes no difference. The Coriolis force is not necessarily perpendicular to that vector, either.

David Wilson (talk · cont) 06:03, 12 March 2018 (UTC)

Let me try and recalibrate the conversation at least for myself. What exactly are the current statements in the article that you believe are incorrect? What exactly is the wording you are proposing for improving the article? --FyzixFighter (talk) 12:24, 12 March 2018 (UTC)

Equatorial railway and relativity

What happens in this thought experiment with the train around the equator when in the train is a pendulum clock? When the train drives faster, the pendulum clock slows down. Surprise, the slow down of the clock is calculated with exactly the same formula as the time dilation in the theory of relativity. Only the speed of light is replaced by the orbit speed. Pege.founder (talk) 07:00, 24 March 2018 (UTC)

Are you asking if this is just a coincidence or something more fundamental? At a certain speed, the clock will slow to a stop, of course, and as the speed increases further, you will need to turn the clock upside down to make it work again. I know of no equivalent in relativity. Dbfirs 08:25, 24 March 2018 (UTC)

Inapplicable to planetary orbits?

Since the article mention centrifugal force only applies to rotating reference frames it seems inapplicable to a (revolving) planetary orbit. I would suggest removing planetary orbits from this article. — Preceding unsigned comment added by 137.75.200.138 (talk) 20:39, 15 March 2018 (UTC)

One coordinate sytem which can be used to analyse planetary motion has its origin at the planetary system's centre of mass and one of its axes oriented so that it always passes through an orbiting planet. Under normal circumstances such a coordinate system will be rotating, and therefore give rise to a centrifugal force operating on the planet.

David Wilson (talk · cont) 22:19, 15 March 2018 (UTC)

Planetary orbits are also routinely used as a simple example when first introducing the centrifugal force, as in Goldstein's text. It is also used for identifying and analyzing Lagrangian points. @anon137 - What do you mean by "rotating" and "revolving", and why do you draw a distinction between the two? --FyzixFighter (talk) 01:56, 16 March 2018 (UTC)

A rotating example is the Earth (a cohesive object) spinning on its axis. A revolving example is the Earth moving around the sun under the influence of gravity. There is some Wikipedia Talk about this overall question here: https://en.wikipedia.org/?title=Talk:Centrifugal_force_(rotating_reference_frame)/Archive_14 In the same archive it is mentioned that the planetary orbits section was in the process of being reduced (and it seems subsequently removed). We might thus complete this process and remove the mention of centrifugal force from the introduction? — Preceding unsigned comment added by 174.16.225.249 (talk) 21:29, 17 March 2018 (UTC)

Without getting too far afield, one motivation of this is to help clarify whether the mentioning of centrifugal forces in the Earth's tides (as we see in some oceanographic and meteorological references) is a clearer explanation rather than one involving the gradients of gravity and the equipotential surface of the Earth's gravity simply being distorted by the moon (more prevalent in astronomy textbooks and in the Wikipedia Tides article). It's also worth noting that centrifugal force isn't mentioned in the Wikipedia article on Planetary Orbits. Looking at this another way, if the Earth orbiting around the sun represents traveling in a straight line in curved space-time, perhaps this is another reason why the fictitious centrifugal force is inapplicable to motion in a gravitational field.— Preceding unsigned comment added by 174.16.225.249 (talk) 21:42, 17 March 2018 (UTC)

Whether or not the system is cohesive or not does not really have any direct bearing on if the centrifugal force appears. If we analyze a system in a rotating frame, we have to include the centrifugal force in the equations of motion. That rotating frame can be the earth's rotating frame or a frame centered on the sun that follows a planet. You can use a rotating frame (and therefore the centrifugal force) for planetary orbits, but usually for simple situations you don't need to - the non-rotating frame is sufficient. However, when you start talking about the stability and location of Lagrangian points, it's a lot easier to pick a frame where the two large bodies are stationary. Perhaps in cohesive systems it is more natural to use a rotating frame than a revolving system, but the centrifugal force still only appears because of the use of a rotating frame. If we analyze the rotating, cohesive system from a non-rotating frame, we wouldn't have to include the centrifugal force. Additionally, any effects attributed to the centrifugal force in the rotating frame are explained as an effect of inertia and F=ma in the non-rotating frame.

In terms of the article on tides, I agree it would be wrong to attribute tides to the centrifugal force. The only reason to include the centrifugal force is if we assume a rotating frame centered on the earth and that co-rotates with the moon. It would make the math easier but would not explain the tides (if we ignored the moon, this analysis we would expect an oblate spheroid of water per Clairaut's theorem). But including an explanation of why the centrifugal explanation for tides is wrong starts to border on WP:TEXTBOOK imo and the explanation really should be backed up by a source like a journal on physics teaching or pedagogy that addresses the misconception. --FyzixFighter (talk) 00:45, 18 March 2018 (UTC)

Oh my - I just found an example of what you are talking about, from the NOAA website no less [18]. It's sad how bad it is. The centrifugal forces shown in its figure 1 are completely wrong and inconsistent with the text - for example, the text says that the centrifugal force "is always directed away from the center of revolution", which should be G per their description, but then shows the Fc at A pointing towards G, not away. The text also talks about the centrifugal force is real, but is only "real" in a rotating frame. In the moon/earth example, the logical rotating frame would be one where the moon and earth are stationary, which is satisfied as long as the axis lies along the line connecting the earth's and the moon's centers of mass and matches their angular velocity. The frame that is consistent with how the direction of the Fc arrows in figures 1 and 2 are consistent with a frame actually centered on or near the moon's center of mass. In figure 2, Fc at B is actually a bit larger than Fc at A. Figure 2 actually works as an explanation if we are clear that we are in a rotating frame centered on the moon. To sum up, tides can be explained using rotating frames and the centrifugal force and gradient of the gravitational field, but you have to be very careful about defining the rotating frame. If you use a non-rotating (-ish, ignoring orbital motion around the sun) centered on the system center of mass, then gravity and Fnet=ma is all that is needed to explain tides. this NOAA explanation is a better, though I would quibble a bit with some of the comments about inertia that make inertia sound like a force. --FyzixFighter (talk) 13:56, 18 March 2018 (UTC)

Very nice summary of some of the key NOAA links. I recently found a better NOAA-based link here There are some web pages where opinions about using the centrifugal force to explain tides are discussed. We might consider how authoritative they are. Here is one. — Preceding unsigned comment added by 174.16.225.249 (talk) 23:14, 18 March 2018 (UTC)

That link is better but, relative to improving this article, 1) I don't think it would qualify as a reliable source, and 2) it gets some things about centrifugal wrong. I think the author hits on some important points that the typical oceanography textbooks seem to miss, in particular the fact that the centrifugal force is a fictitious force only relevant in a rotating frame and does not appear in an inertial frame. However, in the section "A closer look at centrifugal forces", he makes the comment that because "every point on or within the earth to move in an arc of the same radius" therefore "every point on or within the earth experiences the same size centrifugal force at any given time", which logic I strongly disagree with. It's very weird that he spends so much time emphasizing that the centrifugal force only appears in rotating frames and then invokes the centrifugal force without defining the rotating frame for this argument the centrifugal force is equal across every point on or within the earth. The "r" in the equation for the centrifugal force is not the radius of the arc of the objects motion, but the distance from the object to the axis of rotation of the frame. Since the points on and within earth are different distances from the axis, then the centrifugal force will be different across those points. I do think his list of references and further reading (assuming they satisfy WP:RS) may be a good start for potential improvements to the tide article.

I don't mind the tide discussion, but I worry we are getting far afield of the original intent of improving this article - I don't think tides deserve mention (as we agree it is not necessary to the explanation of tides, and even misleading in explanations when it is invoked) except maybe as an example of a misapplication/misunderstanding of the concept (and would need good RS's). The concept of the centrifugal force also does not rely on any distinction between "rotating" and "revolving" systems, as Simanek gets right. If the frame for analyzing the system is rotating, the centrifugal force appears regardless of the nature of the system. I wouldn't mind seeing a return of a short planetary orbit example as that can easily be referenced to some textbooks and journal articles to show a large scale, gravitational, "revolving" system example. --FyzixFighter (talk) 12:46, 21 March 2018 (UTC)

It has occurred to me that the distinction the original poster intended to draw might be that between a rotating coordinate sytem and one whose origin is revolving about some other point. A system of the latter type might be rotating as well, but does not necessarily have to be. In either case, bodies in such a system will be subject to a pseudoforce whose magnitude per unit mass is equal to that of the system origin's acceleration, and acts in exactly the opposite direction. Unfortunately this pseudoforce is sometimes misleadingly referred to as being "centrifugal", even though—unlike a true centrifugal force—it has the same direction and magnitude per unit mass everywhere in the system where it acts. It is a pseudoforce of this form which is generally resorted to in perfectly reasonable explanations of the tides, and the conflation of it with some other vaguely specified—and supposedly centrifugal—pseudoforce which is usually responsible for the many badly muddled attempts to explain them (amongst which I would certainly include Simanek's).
David Wilson (talk · cont) 12:32, 25 March 2018 (UTC)

Yes! That fits the pseudoforce to which all these sources are referring - it's the same magnitude and direction throughout the frame. In an earth-centered frame revolving around the barycenter and rotating such that the moon is stationary, there would be this pseudoforce plus the centrifugal force on all objects. It's unfortunate that so many sources go through the derivation of the centrifugal force and then jump over to this other pseudoforce while keeping the term "centrifugal". Well done sir. --FyzixFighter (talk) 21:59, 25 March 2018 (UTC)

At least some of the confusion arises, I believe, because the standard definition of "centrifugal force" is not the most natural in the circumstance where it makes any difference—namely, when a coordinate system is rotating about a fixed axis which does not pass through its origin. In this case, the standardly defined centrifugal force, $-m\,{\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \mathbf {r} )$, is directed not away from the axis of rotation, but away from a line through the origin parallel to it. The acceleration of the coordinate system's origin, however, in these circumstances, is ${\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times \mathbf {r} _{a})$, where $\mathbf {r} _{a}$ is a displacement vector from the origin to any point on the axis of rotation. The vector sum of the pseudoforce resulting from this acceleration and the standardly defined centrifugal force is therefore $-m\,{\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times (\mathbf {r} -\mathbf {r} _{a}))$. This latter force is directed away from the axis of rotation; it is precisely what the centrifugal force would be if the origin of the coordinate system were to be shifted so as to lie on that axis. To avoid future circumlocution, let me coin the term "axiofugal" to refer to this pseudoforce. It has the following properties:

• For a coordinate system rotating about a fixed axis, it coincides everywhere with the standardly defined centrifugal force for any other system at rest with respect to the first, but with its origin on the axis of rotation. The axiofugal force is thus everywhere and always directed away from the axis of rotation.
• In particular, for a coordinate system rotating about an axis through its origin, the axiofugal force coincides everywhere with the centrifugal.
• At the origin of a coordinate system rotating about a fixed axis, the axiofugal force coincides in magnitude and direction with the pseudoforce arising from the origin's acceleration (if any). This is also true at any point on a line through the origin parallel to the axis of rotation, but not elsewhere.

When writers appear to be using the term "centrifugal force" mistakenly in a coordinate sytem rotating about an axis not passing through its origin, I suspect what they are very often doing is using it to refer to the axiofugal force.

It's possible that this has occurred in at least one place in the article. In the section Vehicle driving round a curve we have:

"This is the fictitious centrifugal force. It is needed within the passenger's local frame of reference to explain his sudden tendency to start accelerating to the right relative to the car … ."

Now, we can certainly give the author of this text the benefit of the doubt by interpreting "the passenger's local frame" to mean one whose origin lies at the centre of the curve being traversed by the vehicle, but that doesn't seem to me to be the most natural way to interpret it. If it weren't for the fact that the fictitious force being referred to would not be the standardly defined centrifugal force if the writer of this passage intended the origin of his "local frame of reference" to be located at the observer, I would tend to presume that is exactly what he did intend.

David Wilson (talk · cont) 08:46, 30 March 2018 (UTC)

acceleration

1. The equation for the absolute acceleration is wrong. You forgot the acceleration of the origin of the reference frame.

2. the passenger's local frame of reference is not a reference frame with the origin in the turning center. The origin of such a lokal frame is perhaps the center of gravity force.--Wruedt (talk) 19:09, 29 April 2019 (UTC)

Could you tell us which part of the article you think is wrong? Dbfirs 19:36, 29 April 2019 (UTC)

I expect user Wruedt is referring to section 2.1 of the article, which is the only place in the article where the term "passenger" appears. He or she is probably making the same point that I have earlier—namely, that if the origin of the passenger's local frame is taken to be right where the he is located, then the pdeudo-force he feels is not the centrifugal force (as standardly defined) in his local frame—which is always zero at the origin. It is instead a different pseudo-force resulting from the fact that the origin of the passenger's reference frame is accelerating towards the centre of the curve around which the car is moving. While it's true that this force is often loosely referred to as a "centrifugal force", that abuse of terminology is probably responsible for much of the confusion surrounding the concept.

David Wilson (talk · cont) 00:28, 30 April 2019 (UTC)

No, it's not wrong. The passenger's local frame is moving with the car, rotating about a point at the center of the curve, not about the car's position. I'm not sure what you mean by "origin", but if you mean center of rotation, it's not in the car; if you want to put an origin in the car, you still need to compute the CF with respect to the center of rotation, which is not at the origin in that case. In this rotating frame, the CF is the fictional force as described. Dicklyon (talk) 04:04, 30 April 2019 (UTC)

The most common use of centrifugal force is: F_centrifugal=-F_Centripetal which is not present in this article. And the origin of a rotating referenc frame fixed to the car is the CG. In this point the centrifugal force is zero.--Wruedt (talk) 05:37, 30 April 2019 (UTC)

By "origin", I mean the origin of the passenger's coordinate system. If this is taken to be the same as that of the inertial system, then the pseudo-force apparently acting on him is the centrifugal force as that is standardly defined. However, if the origin of his coordinate system does not coincide with that of the inertial system, or lie on a line through the inertial system's origin parallel to the axis of rotation, then it is absolutely not the centrifugal force, as that is standardly defined. It is a sum of two components, one of which is the centrifugal force (as standardy defined), and the other of which is a result of the fact that the origin of his coordinate system is accelerating. It is the force for which I coined the term "axiofugal" above. When the passenger is located right at the origin of his coordinate sytem, the centrifugal force (as standardly defined) vanishes entirely, and the only pseudo-force apparently acting on him is that resulting from the acceleration of his coordinate system's origin. As I have already indicated, I realise that many people do (unfortunately) refer to this as a "centrifugal force", and I did not say that this was "wrong". It is, however, in my opinion, a very sloppy use of terminology, and one that should never be used when explaining the concepts to anyone not already familiar with them.

David Wilson (talk · cont) 06:23, 30 April 2019 (UTC)

What people call "centrifugal force" is the opposit of the centripetal force. You call it "reactive centrifugal force". But this is not a reactive force, it is the "force of inertia" (see Lanczos or german article or this).--Wruedt (talk) 07:02, 30 April 2019 (UTC)

The vehicle example is wrong. The origin of the passengers lokal frame is never in the center of rotation. You may call such a system "radius frame" or what so ever but not lokal passenger frame. This is only an assumption to be able to descripe the feelings of the passenger as centrifugal force. True would be, that the centrifugal force in the lokal frame is zero.--Wruedt (talk) 10:03, 30 April 2019 (UTC)

It would be better to call it the rotating frame, the rotating frame with fixed center in which the car is not moving and the passenger feels a centrifugal force. Dicklyon (talk) 12:57, 21 May 2019 (UTC)

Semi-protected edit request on 20 January 2020

Please change "parallel to the axis of rotation" to "perpendicular to the axis of rotation" because the former is wrong and the latter is correct. See the example under 'Applications' regarding the creation of pressure gradients. 2603:9000:7180:139:7561:5AAC:88AE:1380 (talk) 22:18, 20 January 2020 (UTC)

 Not done: please provide reliable sources that support the change you want to be made. I think you may be misinterpreting the text. It says the force ...is directed away... from the axis of rotation, ie. perpendicular. Eggishorn (talk) (contrib) 14:06, 21 January 2020 (UTC)

Yes, the adjective "parallel" is intended to qualify the noun "axis", and the sentence is correct if read in that sense. However, the sentence is also grammatically ambiguous, since "parallel to the axis of rotation" could be read as a predicate of "It is", and the ambiguity can only be resolved by someone who already knows which of the two contradictory readings is correct. I'd therefore suggest modifying the sentence to read

"It is directed away from an axis which is parallel to the axis of rotation but which passes through the coordinate system's origin."

which expresses the correct sense unambiguously.

2001:8003:1847:3A00:F005:8134:1A8:C530 (talk) 03:22, 19 March 2020 (UTC)

Equatorial railway paradox

I have some doubts regarding this example. The force from the track is a constraint force, therefore it's equal in value and direction but inverted to the applied force. In this case it will be |Ft| = |Fg| - |Fcf| (Ft = track, Fg = gravity, Fcf = centrifugal force). Moreover in a movement parallel to rotation the Coriolis force is zero. Therefore there is no Coriolis force and no need of it to explain the paradox. Am I correct? zimbricchio — Preceding unsigned comment added by Zimbricchio (talk • contribs) 11:45, 29 November 2018 (UTC)

No, the article is correct. In the reference frame rotating with the Earth, the train is travelling perpendicular to, not parallel to, the axis of rotation. The Coriolis force is indeed twice as large in magnitude as the centrifugal force, and in the opposite direction to it.

David Wilson (talk · cont) 13:13, 29 November 2018 (UTC)

I'm sorry, I don't get this! Doesn't the train need to go up or down to experience the Coriolis force? I thought that in the thought experiment the railway has no gradients. The Coriolis force is apparent for bodies constrained in movement perpendicular to the axis of rotation, that is, getting closer to it or moving away from it. My understanding is that, on the train traveling in the same direction as the Earth's rotation, a bag of grapes would weigh less than it would when traveling with the surface of the earth at one of the stations. On the on the train that is stationary due to traveling in the opposite direction to the earths rotation the bag of grapes will be heavier again. There would also be marginally more wear on the track carrying the train opposing the earths rotation for the same reason. Coriolis forces wouldn't have any effect. </ref>https://www.youtube.com/watch?v=49JwbrXcPjc</ref> Shoddie (talk) 17:20, 24 June 2020 (UTC)

This is all academic - without any inline citations/Reliable Source, this goes! 50.111.59.129 (talk) 12:09, 1 July 2020 (UTC)

Semi-protected edit request on 19 February 2021

Jonbradleymiller (talk) 16:55, 19 February 2021 (UTC)

This page is totally incorrect. It gives the formula F=m(omega)^2r as the formula for Centrifugal force. This is the formula for Centripetal Force. It cant be the same formula for two terms. There is NO formula for Centrifugal Force because there is no such thing as Centrifugal Force

 Not done: As this article explains, a centrifugal force arises in a rotating (and so non-inertial) frame of reference, this is what we mean by "fictitious". Also, the centrifugal force has the same magnitude as the centripetal force, they're just in the opposite direction. The article is correct. Volteer1 (talk) 17:37, 19 February 2021 (UTC)

Echoing the response above, the article is clear that this is defined in a rotating reference frame. Small point of clarification, the centrifugal force and the centripetal force equal in magnitude and opposite in direction objects that are stationary in the rotating frame. For objects that are moving in the rotating frame, this will not be the case. --FyzixFighter (talk) 22:16, 19 February 2021 (UTC)

Oh, sure. Volteer1 (talk) 02:11, 20 February 2021 (UTC)

Alternative formula and proof

There is also another formula for centrifugal acceleration which is v²/r when v is the tangential velocity and r is the radius of the orbit. There is also a visual proof of this formula but I don't think it's necessary. But it is important to write this formula as well. 122.181.34.116 (talk) 07:08, 18 July 2021 (UTC)

Snowboarder image

I have some concerns about the snowboarding images added here and here. For one thing, it seems a bit odd to have a discussion in the text regard a passenger in a vehicle and have an image of a snowboarder with a caption that mentions a motorcyclist. Doesn't that seem too disjointed? Second, we should be clear that the centrifugal force only exists within the snowboarder's frame of reference. To an external stationary observer, the snowboarder is leaning in so that the normal force from the ground on the snowboarder is at the correct angle to provide sufficient centripetal force for the turn. I also find the attempt at a free body diagram in the second edit to be incomplete as it excludes the normal force from the ground on the snowboarder and is not clear that again this is in the snowboarder's non-inertial frame of reference in which the resultant force should be zero (assuming constant speed around the turn). Anyways, that's why I've removed the recent addition. Thoughts? --FyzixFighter (talk) 17:46, 11 February 2022 (UTC)