# Is the concept of reactive centrifugal force valid?

• Andrew Mason
In summary, the article is incorrect in its assertion that the reactive centrifugal force is always centrifugal. It depends on the context.
D H said:
So at best, if the term "reactionary centrifugal force" has any meaning ..., it only has meaning some times. Other times the third law reaction to a centripetal force is centripetal.
I agree with this.

DaleSpam said:
You still haven't read post 119, have you?
I have. I and have said I disagree with that analysis.

In the rotating uniform circle of molecules example, each molecule only has two forces acting on it. These are the molecular bonding forces from each laterally adjacent molecule. These forces are equal in magnitude and sum to the centripetal acceleration of the molecule. Now, you can say that each molecule is pulling on each other, so for each pull of molecule A on adjacent molecule B, there is an equal and opposite pull by molecule B on A. But the geometry shows that those forces cannot be equal and opposite. The fact that they are both accelerating in slightly different directions shows that they are not equal and opposite. So this "centrifugal force" that you postulate cannot be equal and opposite to the the centripetal force that the molecule experiences.

The equal and opposite force is the force acting on the molecule that is diametrically opposite. The centripetal force on that diametrically opposite molecule is the "reaction" force to the centripetal force on the first. Neither is centrifugal.

AM

Andrew Mason said:
Reduce the space station to a circular string of molecules.
Why so complicated? Reduce the space station just two molecules orbiting each other.

And what are trying to prove here? Nobody ever claimed that the reactive centrifugal force exists in every possible scenario. I told you that in post #18 already. I know this is about rotation, but you overdoing that going in circles a bit now.

rcgldr said:
In a 2 body system, where gravity is causing the objects to orbit in a circular path, the Newton third law pair of forces is the gravity force that each object exerts on the other. From each object's perspective, the gravity force that accelerates each object is towards the center of the orbit, so both gravity forces are centripetal.

So after 9 pages... what's wrong with this explanation again?

Andrew Mason said:
I have. I and have said I disagree with that analysis.
Sorry, I missed that. What was your disagreement with my analysis in 119?

Andrew Mason said:
Now, you can say that each molecule is pulling on each other, so for each pull of molecule A on adjacent molecule B, there is an equal and opposite pull by molecule B on A. But the geometry shows that those forces cannot be equal and opposite.
It is not me who says that, it is Newton's 3rd law. Are you disputing/disagreeing with Newton's 3rd law? If not then you should revise your comments because they communicate a disagreement with Newton that I hope is unintentional.

PS. I have not claimed that a centrifugal reaction force exists in all cases, so showing examples where it doesn't exist is just a red herring. The point is that in some cases they do exist, and A.T.'s example is one of those cases. The ring and the 2 orbiting point masses do not have centrifugal reaction forces, but A.T.'s astronaut example does have them. You need to address A.T.'s example on its own merits. It is a relatively simple scenario that Newton's laws can easily be applied to.

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D H said:
I truly despise the name "reactionary centrifugal force".
Why make it political?

D H said:
That said, I'm not a big fan of the term "centripetal force", either.
Now it's not only against Wikipedia put pretty much every physics book on the planet?

D H said:
So at best, if the term "reactionary centrifugal force" has any meaning (to me it doesn't; and no, your pictures don't sway me), it only has meaning some times.
The term has always a meaning according to the definition. But you will not always find something that fits that definition

D H said:
Other times the third law reaction to a centripetal force is centripetal.
Yes, that what I was saying all the time. It depends on the scenario.

Andrew Mason said:
...so for each pull of molecule A on adjacent molecule B, there is an equal and opposite pull by molecule B on A. But the geometry shows that those forces cannot be equal and opposite.
You lost me here. Are they equal and opposite or not?

Andrew Mason said:
The equal and opposite force is the force acting on the molecule that is diametrically opposite.
So the reaction to the force by an adjacent molecule, is the force on a diametrically opposite molecule, which doesn't even have a direct interaction with the considered molecule? Now, that is an interesting combo, to be sold as Newtons 3rd force pair.

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D H said:
I truly despise the name "reactionary centrifugal force". That said, I'm not a big fan of the term "centripetal force", either. "Centripetal acceleration" is fine; it is a kinematic description of motion. Forces are out of the picture in kinematics -- so why are we dragging forces into the mix?
I think you may have hit the nail on the head. [Others, including me, have been hitting a few thumbs]. This addresses the point that AT made, which I have not yet properly answered, questioning whether Newton's third law is only about accelerations. Newton spoke about "action" and "reaction", after all. He did not really deal with static forces. Static forces are, by definition, equal and opposite - otherwise they would not be static.

I think our discussion shows that we really run into problems if we try to apply Newton's third law to static forces. It becomes really difficult, if not arbitrary, to select third-law force pairs. Not so when there are accelerations. There can never be only one acceleration. There must always be a "reaction" acceleration. One mass accelerating all by itself is not possible under the laws of physics as we presently know them.

If we ask, where is the acceleration that is the third law reaction to the astronaut's centripetal acceleration, the answer is clear that it has to be a centripetal acceleration because that is the only kind of acceleration that occurs.

So at best, if the term "reactionary centrifugal force" has any meaning (to me it doesn't; and no, your pictures don't sway me), it only has meaning some times. Other times the third law reaction to a centripetal force is centripetal.
I think you are being too charitable. The reaction to a centripetal acceleration cannot be a static centrifugal force - ever.

AM

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Andrew Mason said:
I think you may have hit the nail on the head. [Others, including me, have been hitting a few thumbs]. This addresses the point that AT made, which I have not yet properly answered, questioning whether Newton's third law is only about accelerations. Newton spoke about "action" and "reaction", after all. He did not really deal with static forces. Static forces are, by definition, equal and opposite - otherwise they would not be static.
You are mixing up Newton's 3rd law, which is about forces not acceleration, with the 2nd law which deals with net force and the resulting acceleration. Things are 'static' only if the net force on them is zero--this has nothing to do with Newton's 3rd law.

Don't try to read into the rather archaic terminology of 'action' and 'reaction'--you cannot learn physics from a dictionary.

I think our discussion shows that we really run into problems if we try to apply Newton's third law to static forces. It becomes really difficult, if not arbitrary, to select third-law force pairs.
It's actually rather trivial.

Not so when there are accelerations. There can never be only one acceleration. There must always be a "reaction" acceleration. One mass accelerating all by itself is not possible under the laws of physics as we presently know them.
Again you confuse Newton's 2nd and 3rd laws.

If we ask, where is the acceleration that is the third law reaction to the astronaut's centripetal acceleration, the answer is clear that it has to be a centripetal acceleration because that is the only kind of acceleration that occurs.
Gibberish. Let's keep it simple: A exerts a force on B, thus B exerts an equal and opposite force on A. That's Newton's 3rd law. No mention of acceleration.

Andrew Mason said:
I think our discussion shows that we really run into problems if we try to apply Newton's third law to static forces.
It sounds like you do not understand basic Newtonian mechanics. Newtons third law has no problem with static forces. Newtons third law is used extensively in statics, and it is absolutely required for analyzing the stresses in a beam or structural member.

In fact, looking in my statics textbook the strong form of Newton's 3rd law is introduced on page 6. It is followed by an explanatory paragraph asserting "This principle holds for all forces, variable or constant, regardless of their source and holds at every instant of time. Lack of careful attention to this basic law is the cause of frequent error by the beginner."

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Andrew Mason said:
This addresses the point that AT made, which I have not yet properly answered, questioning whether Newton's third law is only about accelerations.
No it doesn't. A proper reference supporting your version of Newton's 3rd Law would address it.

Andrew Mason said:
Static forces are, by definition, equal and opposite - otherwise they would not be static.
Sounds like the common flawed reasoning, that the two equal and opposite forces are balancing each other.

Andrew Mason said:
I think our discussion shows that we really run into problems if we try to apply Newton's third law to static forces.
There are no problems. Static cases are classic examples of Newtons 3rd Law.

Andrew Mason said:
It becomes really difficult, if not arbitrary, to select third-law force pairs.
To people interested in problem solving this is not difficult. It is a feature, which makes the Law applicable on different levels of abstraction. It might be a problem to armchair philosophers though.

Andrew Mason said:
If we ask, where is the acceleration that is the third law reaction...
The third law is not about acceleration.

Andrew Mason said:
The reaction to a centripetal acceleration...
The third law is not about acceleration.

DaleSpam said:
I have not claimed that a centrifugal reaction force exists in all cases, so showing examples where it doesn't exist is just a red herring.

I found another Wikipedia article that should be deleted:
http://en.wikipedia.org/wiki/Cat

Consider two molecules orbiting each other. Where it the cat in that scenario? Cats cannot exist!

DaleSpam said:
It sounds like you do not understand basic Newtonian mechanics. Newtons third law has no problem with static forces. Newtons third law is used extensively in statics, and it is absolutely required for analyzing the stresses in a beam or structural member.
Are you suggesting that Newtonian mechanics is simple! I think discussion shows that it is far from simple.

It is not that Newton's third law has a problem with static forces. It is just that it does not say anything new about static forces. If there is no change in motion (ie. the forces are static) there is no net force, so all forces sum to zero. If all forces sum to zero, there is no change in motion. You get that from the first law. You do not need Newton's third law to analyse stresses and strains.

The essence of the third law, it seems to me, is that it establishes the principle that every change in motion gives rise, instantaneously, to offsetting change in motion (ie. of other bodies). The first two laws do not imply the law of conservation of momentum. The third does.

I think we all agree that rotation does not create a centrifugal acceleration. But it is suggested here that rotation - centripetal acceleration, gives rise (sometimes) to a static centrifugal force. I say that is incorrect. That can never occur. Centripetal acceleration cannot occur in just one body. The reaction to any acceleration is another acceleration ie of another body. That is what the third law says.

AM

Andrew Mason said:
It is not that Newton's third law has a problem with static forces. It is just that it does not say anything new about static forces. If there is no change in motion (ie. the forces are static) there is no net force, so all forces sum to zero.
This shows that you have the usual misunderstanding about Newton's 3rd law. If object A is static then all of the forces on A sum to zero, as required by Newton's 1st law. Newton's 3rd law describes the relationship of the force on A due to B and the force on B due to A. You can never sum a 3rd law pair of forces to get a net force because the forces act on different objects. The 1st and 3rd laws say completely different things.

Andrew Mason said:
The essence of the third law, it seems to me, is that it establishes the principle that every change in motion gives rise, instantaneously, to offsetting change in motion (ie. of other bodies). ... The reaction to any acceleration is another acceleration ie of another body. That is what the third law says.
This is also not correct, although at least it is an unusual misunderstanding. The third law says nothing about acceleration. It strictly deals with forces. There are many cases where the accelerations due to third-law pairs are unequal, and other cases where they are both zero but still interacting.

A bit off topic:

Andrew Mason said:
It is not that Newton's third law has a problem with static forces. It is just that it does not say anything new about static forces.
That static (and all) forces only exist in equal and opposing pairs?

Wiki article:

To every action there is always an equal and opposite reaction: or the forces of two bodies on each other are always equal and are directed in opposite directions

http://en.wikipedia.org/wiki/Newton's_laws_of_motion#Newton.27s_third_law

exceptional case for rotation and centrifugal acceleration ...

Andrew Mason said:
I think we all agree that rotation does not create a centrifugal acceleration.
What about the previously described case of an object sliding inside a frictionless cylinder that rotates end over end? This is an exceptional case though.

back on topic ...

Andrew Mason said:
suggested here that rotation - centripetal acceleration, gives rise (sometimes) to a static centrifugal force.
The point of the wiki articles is to define the term "reactive centrifugal force", and they include a couple of examples. What should be added to the article is that sometimes the equal and opposing force to a centripetal force is another centripetal force, such as a 2 body system orbiting due to charge, gravity, or magnetism. Looking at the discussion threads, the people involved at least agreed that the term "reactive centrifugal force" is valid, and with the qualifier "reactive" I doubt anyone will confuse it's usage with the term "fictitious centrifugal force".

DaleSpam said:
This shows that you have the usual misunderstanding about Newton's 3rd law. If object A is static then all of the forces on A sum to zero, as required by Newton's 1st law. Newton's 3rd law describes the relationship of the force on A due to B and the force on B due to A.
You have to read what I wrote. I think I made this point very clear in my first post.
You can never sum a 3rd law pair of forces to get a net force because the forces act on different objects. The 1st and 3rd laws say completely different things.
I don't think you meant to say that. I think you meant to say that you could not get a single third law pair forces NOT to produce a net force. You can never have a single pair of third law forces that do NOT produce acceleration.

This is also not correct, although at least it is an unusual misunderstanding. The third law says nothing about acceleration. It strictly deals with forces. There are many cases where the accelerations due to third-law pairs are unequal, and other cases where they are both zero but still interacting.
I didn't say that the accelerations are equal and opposite in third law pairs! I said that every change in motion gives rise "to offsetting change in motion (ie. of other bodies)".

AM

Andrew Mason said:
You can never have a single pair of third law forces that do NOT produce acceleration.
Acceleration is "produced" by net forces, not by third law force pairs. Most scenarios involve more than one single interaction, so in general you cannot say anything about acceleration, based solely on single pair of third law forces.

Andrew Mason said:
DaleSpam said:
You can never sum a 3rd law pair of forces to get a net force because the forces act on different objects. The 1st and 3rd laws say completely different things.
I don't think you meant to say that. I think you meant to say that you could not get a single third law pair forces NOT to produce a net force. You can never have a single pair of third law forces that do NOT produce acceleration.
I think Dale mean to say exactly what he said, Andrew: What Dale said is exactly right.

You seem to have a fundamental misunderstanding of Newton's laws. A lot of students do, too. That is perhaps because there are several implicit assumptions that underlie Newton's laws. Key amongst these underlying assumptions are
• The change in the motion of some object does not immediately depend on the forces that object exerts on other objects. That object A exerts some force on object B has no direct bearing on object A's motion. All that matters with respect to the immediate change in the motion of object A are the forces exerted by object B (and C and D and ...) on object A.

This is a source of confusion amongst many students. We repeatedly get students asking how anything can move given that forces are equal but opposite. The answer is that the force object A exerts on object B has nothing to do (directly) with object A's motion.
• "Force" is subject to the superposition principle. Newton's second law is a statement about the net force, or the superposition of all of the forces, acting on some body. It says nothing about the response of some body to an individual force acting on the body. Newton's third is a statement about an individual force between a pair of bodies. It too says nothing about the acceleration that results from this force.

Andrew Mason said:
You have to read what I wrote. I think I made this point very clear in my first post.
Yes, you have been inconsistent in the correct application of the 3rd law. The OP was correct, your more recent posts have not been.

Andrew Mason said:
I didn't say that the accelerations are equal and opposite in third law pairs! I said that every change in motion gives rise "to offsetting change in motion (ie. of other bodies)".
I don't understand what you mean by "offsetting change in motion". In any case, there is no mainstream scientific reference that you can produce which supports your interpretation of the 3rd law as not applying to static scenarios, so further discussion of the 3rd law in those terms is speculative and not appropriate for PF.

D H said:
I think Dale mean to say exactly what he said, Andrew: What Dale said is exactly right.
No it isn't. If the only forces are a single pair of third law forces between A and B then there is one force on A and one on B. Both accelerate! This is the essence of Newton's third law.

You seem to have a fundamental misunderstanding of Newton's laws.
What I said is correct. You have to read what I said. If there is only a single pair of third law forces, there has to be acceleration. I can give you several examples:

The force between A and B is gravity. A exerts a gravitational force on B. B exerts an equal and opposite gravitational force on A. A and B both accelerate. How can they not unless there are other force (pairs) keeping them apart?

A and B collide. A exerts a force on B. B exerts a force on A. Both experience change in motion.

If you think this is wrong, give me one example of where a single third law force pair DOES NOT produce acceleration.

A lot of students do, too. That is perhaps because there are several implicit assumptions that underlie Newton's laws. Key amongst these underlying assumptions are
• The change in the motion of some object does not immediately depend on the forces that object exerts on other objects. That object A exerts some force on object B has no direct bearing on object A's motion.
• Well it certainly has a bearing. It has to produce acceleration in B if that is the only force on B. And if B's reaction force on A is the only force on A it has to produce an acceleration in A.

Look. No one here misunderstands Newton's three laws. What we are disagreeing with here is how to analyse a situation because there is a certain depth and subtlety involved. But to maintain a civil and productive discussion we have to actually read what is said.

If you think I am still missing something. Tell me again, how does a single pair of third law forces NOT produce acceleration. Give me an example. I can guarantee you that there will always be acceleration.

All that matters with respect to the immediate change in the motion of object A are the forces exerted by object B (and C and D and ...) on object A.
Exactly. So if the only force on A is the force exerted by B then A must accelerate. That was my point. Again, if you disagree then give an example.

This is a source of confusion amongst many students. We repeatedly get students asking how anything can move given that forces are equal but opposite. The answer is that the force object A exerts on object B has nothing to do (directly) with object A's motion.
The point about the third law is that if body A accelerates, something else in the universe has to accelerate. If you disagree, give me an example of where only one body accelerates!

AM

Andrew Mason said:
No it isn't. If the only forces are a single pair of third law forces between A and B then there is one force on A and one on B. Both accelerate! This is the essence of Newton's third law.
No it isn't. That's a consequence of the 2nd law: Whenever you have an net force on an object, it accelerates. Obviously if only a single force acts on a object you'll have acceleration. So what?

In the examples of the the merry-go-round and the spaceship there are multiple forces involved. Deal with it! Don't keep harping on the trivial case where you have only two particles.

For some bizarre reason, you claim that a 'centripetal' force cannot have a 'reaction' that is 'centrifugal' despite having been given several simple examples to the contrary.

Perhaps you're reading more into the words 'centripetal' and 'centrifugal' than warranted. In the context of this thread all they mean are 'towards the center' and 'away from the center', no more significant than saying a force acts 'up' or 'down'.

Andrew Mason said:
The point about the third law is that if body A accelerates, something else in the universe has to accelerate. If you disagree, give me an example of where only one body accelerates!
That is a red herring, Andrew. The third law says absolutely nothing about acceleration. Newton's third law is used in plenty of applications where nothing accelerates. Big chunks of civil and mechanical engineering address bodies that are not accelerating in which Newton's third law nonetheless plays a central role. Just because a bridge or building is not accelerating does not mean that no forces act on it. The static loads that act on a bridge, a building, an aircraft, a boat, etc. can make the object in question unsafe.

DaleSpam said:
I don't understand what you mean by "offsetting change in motion".
If a change in motion of body A is caused by interaction with body B the interaction will produce an equal and opposite change in motion of B only if A and B have the same mass. This is not a problem if we speak about "force". The forces are always equal in magnitude and opposite in direction. But the changes in motion are not (unless the masses are equal).

All I wanted to say is that a change in the motion of body A requires some change in motion of some body somewhere in the universe. And that change in motion "offsets" the change in motion of A so there is no overall change in "momentum". This is a rather subtle but fundamental principle embodied in the third law. The word "offset" is mine. Perhaps it is not the best word to use so if you have some better suggestion let me know.

So when we have an an acceleration we know there is corresponding third law acceleration somewhere and we have to find it. In the case of rotational motion where there is centripetal acceleration, there must be an "offsetting" acceleration, not a static force. That is why, in my view, the reactive centripetal force analysis is flawed.

AM

Doc Al said:
No it isn't. That's a consequence of the 2nd law: Whenever you have an net force on an object, it accelerates. Obviously if only a single force acts on a object you'll have acceleration. So what?
Ok. The accelerations of A and B derive from the first two laws. But the fact that the acceleration of A has to be accompanied by an acceleration of B (if the only interactions between A and the rest of the universe and between B and the rest of the universe are with each other) is found only in the third law.

In the examples of the the merry-go-round and the spaceship there are multiple forces involved. Deal with it! Don't keep harping on the trivial case where you have only two particles.

For some bizarre reason, you claim that a 'centripetal' force cannot have a 'reaction' that is 'centrifugal' despite having been given several simple examples to the contrary.
Those examples are deceptive.

If there are two masses A and B tethered with a rope (of arbitrarily small mass) rotating about their centre of mass I gather that you would agree that the forces are all (ie. both, since there are only two) centripetal. By Newton's third law, each force must be the reaction to the other. If, however, I tether them with two such ropes but fashioned as a sling you appear to be saying that everything changes: the reaction force to the centripetal force on A is a centrifugal force on the sling. I say nothing fundamentally has changed.

AM

D H said:
That is a red herring, Andrew. The third law says absolutely nothing about acceleration. Newton's third law is used in plenty of applications where nothing accelerates. Big chunks of civil and mechanical engineering address bodies that are not accelerating in which Newton's third law nonetheless plays a central role. Just because a bridge or building is not accelerating does not mean that no forces act on it. The static loads that act on a bridge, a building, an aircraft, a boat, etc. can make the object in question unsafe.
I agree. But all I am saying is that if that is all that Newton's third law said, the third law would simply be a corollary to the first two. So the third law is not required to explain static forces. But the third law is not merely a corollary of the first two. You cannot capture the principle of absolute symmetry that is embodied in the third law (and thus derive the law of conservation of momentum) from the first two laws.

AM

Andrew Mason said:
Those examples are deceptive.

If there are two masses A and B tethered with a rope (of arbitrarily small mass) rotating about their centre of mass I gather that you would agree that the forces are all (ie. both, since there are only two) centripetal. By Newton's third law, each force must be the reaction to the other. If, however, I tether them with two such ropes but fashioned as a sling you appear to be saying that everything changes: the reaction force to the centripetal force on A is a centrifugal force on the sling. I say nothing fundamentally has changed.

AM
I agree that nothing fundamental has changed. Since you are using the rope to exert the force between the two masses you cannot just ignore it as you please. In both cases, the rope/sling must be exerting a centripetal force on the masses. Obviously the reaction to those forces is a 'centrifugal' force on the rope/sling. Nice try!

I have no earthly idea why this baffles you.

If you want a clean case where two masses only exert 'centripetal' forces on each other, use a long-range interaction such as gravity. But so what?

And what's the obsession with only two point-like masses? Treat the examples given: the merry-go-round and the spaceship. The extended bodies and the forces involved are much more interesting to analyze.

With all the discussion about the third law, I thought it might be a good idea to provide this excerpt from Newton's Principia: Axioms, or Laws of Motion, Law III:

Isaac Newton said:
"... If a body impinges upon another, and by its force change the motion of the other, that body also (became of the quality of, the mutual pressure) will undergo an equal change, in its own motion, towards the contrary part. The changes made by these actions are equal, not in the velocities but in the motions of bodies; that is to say, if the bodies are not hindered by any other impediments. For, because the motions are equally changed, the changes of the velocities made towards contrary parts are reciprocally proportional to the bodies. This law takes place also in attractions, as will be proved in the next scholium."

AM

Andrew Mason said:
If a change in motion of body A is caused by interaction with body B the interaction will produce an equal and opposite change in motion of B only if A and B have the same mass. This is not a problem if we speak about "force". The forces are always equal in magnitude and opposite in direction. But the changes in motion are not (unless the masses are equal).
So then the third law deals with forces and not changes in motion.

Andrew Mason said:
All I wanted to say is that a change in the motion of body A requires some change in motion of some body somewhere in the universe. And that change in motion "offsets" the change in motion of A so there is no overall change in "momentum".
This is certainly true for isolated systems, but Newton's laws can also be applied to non-isolated systems where the momentum of the system is not conserved. Of course, if you proceed to large enough scales then you should be able to find an isolated system, but it is not necessary for using Newton's laws.

Andrew Mason said:
So when we have an an acceleration we know there is corresponding third law acceleration somewhere and we have to find it. In the case of rotational motion where there is centripetal acceleration, there must be an "offsetting" acceleration, not a static force. That is why, in my view, the reactive centripetal force analysis is flawed.
Sure, for an isolated system nobody disagrees that if one part of the system is undergoing centripetal acceleration about the COM then some other part of the system is also undergoing centripetal acceleration about the COM. Acceleration of each part of the system is due to the net force on the part (which is centripetal). However, that in no way implies that there are no centrifugal forces on the part, only that those centrifugal forces are smaller than the centripetal forces.

Please answer the following. Do you understand the difference between a force and the net force? Do you understand that if a body is being influenced by multiple forces that some of those forces may point in opposing directions? If so, then why is it at all confusing that the sum of a centripetal force and a centrifugal force may be a centripetal net force?

DaleSpam said:
Sure, for an isolated system nobody disagrees that if one part of the system is undergoing centripetal acceleration about the COM then some other part of the system is also undergoing centripetal acceleration about the COM. Acceleration of each part of the system is due to the net force on the part (which is centripetal). However, that in no way implies that there are no centrifugal forces on the part, only that those centrifugal forces are smaller than the centripetal forces.
To interject in this long and sometimes fascinating discussion if I may -- what makes you say that the centrifugal force is smaller than the centripetal force?

Because the acceleration is centripetal, therefore the net force is centripetal, therefore the centrifugal force must be smaller than the centripetal force.

DaleSpam said:
Because the acceleration is centripetal, therefore the net force is centripetal, therefore the centrifugal force must be smaller than the centripetal force.
But if the centripetal acceleration were zero then the centrifugal force would also be zero (not, for example, less than zero). How could they be any different in magnitude?

I mean, if you were to pull a cart by a rope with tension T such that the cart were accelerating, the cart would be pulling back with tension T (not less than T), right?

What am I missing here?

olivermsun said:
But if the centripetal acceleration were zero then the centrifugal force would also be zero (not, for example, less than zero).
If the centripetal acceleration were zero then you wouldn't have uniform circular motion so the designations of centripetal and centrifugal would be rather meaningless IMO. However, if you want to include such cases then just change the "less than" to "less than or equal to".

olivermsun said:
I mean, if you were to pull a cart by a rope with tension T such that the cart were accelerating, the cart would be pulling back with tension T (not less than T), right?
Here you are talking about a third-law pair acting on different bodies. Above I am talking about the second-law net force acting on a single body. AM seems to have a confusion about net forces.

DaleSpam said:
If the centripetal acceleration were zero then you wouldn't have uniform circular motion so the designations of centripetal and centrifugal would be rather meaningless IMO. However, if you want to include such cases then just change the "less than" to "less than or equal to".

Count me "still confused." You were saying the centrifugal force was less than the centripetal force in an isolated two body system, so I am just trying to figure out how they would be related -- would they scale by some factor or be different by a fixed offset?

Doc Al said:
I agree that nothing fundamental has changed. Since you are using the rope to exert the force between the two masses you cannot just ignore it as you please. In both cases, the rope/sling must be exerting a centripetal force on the masses. Obviously the reaction to those forces is a 'centrifugal' force on the rope/sling. Nice try!

I have no earthly idea why this baffles you.
It is not baffling. My point is that it is wrong, and very misleading and confusing, to use the term centrifugal "force". There is certainly a centrifugal effect. But it is not a force. The effect cannot and does not cause, or tend to cause, any outward acceleration of anything, ever. It is that simple.

If you want a clean case where two masses only exert 'centripetal' forces on each other, use a long-range interaction such as gravity. But so what?

And what's the obsession with only two point-like masses? Treat the examples given: the merry-go-round and the spaceship. The extended bodies and the forces involved are much more interesting to analyze.
Let's look at a case where I am really exerting an outward force on the sling (to make it simple, there is no rotation): Ball A and Ball B are connected with a rope sling but there is a spring between A and B. As the rope between them is ratcheted shorter, the sling around A is being pulled toward B, and the sling around B is being pulled toward A, compressing the spring.

I stop shortening the rope. Let's analyse the static force pairs. You could say that the force pairs are the inward force of the rope/sling on A and the outward force of the spring and ball on the sling. The same goes for B. (I would say that the forces pairs are 1. the two outward forces of the spring on A and B, and the inward forces of the rope/sling on A and B, but it is rather arbitrary if the forces are static). Then you would actually have an outward (centrifugal) force on the sling because if I let go of the rope there would be actual outward acceleration for a moment while the spring expanded against the balls.

But in the case of circular rotation, there is no such outward force. The "force pairs" between the sling and the ball are not "forces". That is the subtle but, I think, important distinction.

AM

olivermsun said:
Count me "still confused." You were saying the centrifugal force was less than the centripetal force in an isolated two body system, so I am just trying to figure out how they would be related -- would they scale by some factor or be different by a fixed offset?
I wasn't limiting it to a two body system, simply an isolated system. It would be a fixed offset, the offset being the net force required for the centripetal acceleration of that part of the system.

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