What exactly is the reactive centrifugal force (split)


by A.T.
Tags: centrifugal, force, reactive, split
A.T.
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Jan29-13, 09:43 AM
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Mod note: This thread on the reactive centrifugal force was split from this thread on the (fictitious) centrifugal force.

Quote Quote by sophiecentaur View Post
Using the term 'Centrifugal Force' when I was at school was verboten.
It is really not that confusing if you make clear which reference frame you consider.

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Jan29-13, 12:54 PM
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Quote Quote by A.T. View Post
It is really not that confusing if you make clear which reference frame you consider.

There is only one force on each astronaut provided by the space station. That is the centripetal force provided by the tensions in space station structure. Each astronaut provides a force on the space station that creates tensions within the space station structure and result in a centripetal force on the other astronaut.

So one could say that the reaction force to the centripetal force on one astronaut is that astronaut's (centripetal) force on the other astronaut arising from the structural tensions within the space station structure. In that sense, all the forces are centripetal.

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A.T.
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Jan30-13, 04:03 AM
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Quote Quote by Andrew Mason View Post
Each astronaut provides a force on the space station...
Which points outwards, away from the center of rotation. That's why it is called "centrifugal".

Quote Quote by Andrew Mason View Post
that creates tensions within the space station structure and result in a centripetal force on the other astronaut.
Irrelevant. There is a direct local interaction between astronaut and station with two equal/opposite forces: one is centripetal, one centrifugal.

Quote Quote by Andrew Mason View Post
So one could say that the reaction force to the centripetal force on one astronaut is that astronaut's (centripetal) force on the other astronaut arising from the structural tensions within the space station structure.
No, this is not how Newtons 3rd Law is applied. There is no direct interaction between the two astronauts. So there is no 3rd Law force pair acting on both.

Quote Quote by Andrew Mason View Post
In that sense, all the forces are centripetal.
Then "that sense" is obviously nonsense. Everyone can see that Frcf is pointing outwards, away from the center of rotation. So it is centrifugal.

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Jan30-13, 08:45 AM
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What exactly is the reactive centrifugal force (split)


Quote Quote by A.T. View Post
Which points outwards, away from the center of rotation. That's why it is called "centrifugal".
It depends on how you look at it. The force applied by astronaut1 causes astronaut2, who is mechanically connected to the space station, to accelerate toward the centre of the station. If astronaut1 stopped applying this force, astronaut2's acceleration (and the acceleration of the rest of the space station) toward the centre of the space station would decrease. So if you want to associate the direction of a force with the acceleration that it causes, the direction of the astronaut's force is centripetal.

No, this is not how Newtons 3rd Law is applied. There is no direct interaction between the two astronauts. So there is no 3rd Law force pair acting on both.


Then "that sense" is obviously nonsense. Everyone can see that Frcf is pointing outwards, away from the center of rotation. So it is centrifugal.
Suppose we have two spherical moons orbiting a spherical planet and both moons are directly opposite each other (ie. a line through the moons' centres passes through the planet's centre) on identical orbits. Would you say that that the reaction forces of each moon on the planet are centrifugal?

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sophiecentaur
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Jan30-13, 08:48 AM
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You guys are going to have to agree to differ. The 'other guy' is not actually wrong, in this case. Can't you both see that?
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Jan30-13, 09:16 AM
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Quote Quote by Andrew Mason View Post
...causes ....acceleration
Irrelevant. The term "centrifugal" has nothing to do with causing acceleration. It simply means that the force points away from the center of rotation. Acceleration is a matter of net force, not just one centrifugal force. So a force can be centrifugal without causing centrifugal acceleration.
Quote Quote by Andrew Mason View Post
So if you want to associate the direction of a force with the acceleration that it causes,
I don't. For the reason I state above.
Quote Quote by Andrew Mason View Post
Suppose...
No thanks. I gave an example where a centrifugal reaction force exists. Pointing out other cases where there is no centrifugal reaction force proves nothing. Nobody claimed it is always there.
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Quote Quote by A.T. View Post
I don't see how you can confuse the two. One is an interaction force that exists in every frame, the other is an inertial force that exist only in rotating frames. The diferences are listed in the table here:
http://en.wikipedia.org/wiki/Reactiv...trifugal_force

Calling it "centripetal" as you suggest, despite the fact that it points away from the center, that would be confusing.
We are not in disagreement over the direction of the reaction force. We disagree on the body that is experiencing the reaction. Whether a "reaction" points toward or away from the centre depends on the location of the body experiencing the reaction.

Newton's third law does not talk about tensions. Newton talks about "actions" and "reactions which mean "changes in motion". The third law says "To every action there is always opposed an equal reaction; or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts." In other words, changes in motion cannot occur in just one body. The change in motion (momentum) of one body requires an opposing change in motion of another body. The "reaction" to any "action" is an equal and opposing change in momentum of another.

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stevendaryl
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Jan30-13, 09:27 PM
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Quote Quote by Andrew Mason View Post
Newton's third law does not talk about tensions. Newton talks about "actions" and "reactions which mean "changes in motion". The third law says "To every action there is always opposed an equal reaction; or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts." In other words, changes in motion cannot occur in just one body. The change in motion (momentum) of one body requires an opposing change in motion of another body. The "reaction" to any "action" is an equal and opposing change in momentum of another.

AM
Hmm. I guess I can see what you're saying, but the way that I've seen Newton's third law used in actual problems was in terms of equal and opposite forces rather than momentum changes. In practice, it worked something like this:

Let [itex]\stackrel{\rightarrow}{F_I}[/itex] be the force on object [itex]I[/itex]. Then, if the only forces are two-body forces (which is the case for many problems), we can write:

[itex]\stackrel{\rightarrow}{F_I} = \sum_J \stackrel{\rightarrow}{F_{IJ}}[/itex] where [itex]\stackrel{\rightarrow}{F_{IJ}}[/itex] is the force on object [itex]I[/itex] due to object [itex]J[/itex].

The assumption of equal and opposite forces is just the assumption that

[itex]\stackrel{\rightarrow}{F_{IJ}} = - \stackrel{\rightarrow}{F_{JI}}[/itex]

Ultimately, it's certainly true that changes of momenta have to cancel, because the total momentum can't change due to internal forces. However, it seems to me that accounting for the changes in momenta is more systematic by using pairwise forces, and using the principle of equal and opposite forces.
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Quote Quote by stevendaryl View Post
Hmm. I guess I can see what you're saying, but the way that I've seen Newton's third law used in actual problems was in terms of equal and opposite forces rather than momentum changes.
The forces and changes in momentum are necessarily directly proportional so changes in momentum must be equal and opposite.

Since, in non-relativistic interactions, the duration of interaction is the same for both interacting bodies, the changes in momentum of the interacting bodies are equal and opposite: Body 1: Force exerted by body 1 on body 2 = F1 = dp2/dt ⇔∫F1dt = ∫dp2 = Δp2; For body 2: F2 = - F1 ⇔ ∫F2dt = Δp1 = -∫F1dt = -Δp2

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Jan30-13, 09:59 PM
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Quote Quote by Andrew Mason View Post
The "reaction" to any "action" is an equal and opposing change in momentum of another.
Nonsense. The change in momentum is a net force. It is the individual forces which have action reaction pairs.

Consider a person standing on the ground. There is a gravitational force down and a normal force up, the net force is zero and therefore there is no change in momentum. So according to you there is no normal reaction force on the ground and also no gravitational reaction force since there is no change in momentum.

This is NOT how Newtons third law works. It is the individual forces which are in action reaction pairs, not the net force.
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Jan30-13, 10:02 PM
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Quote Quote by Andrew Mason View Post
The forces and changes in momentum are necessarily directly proportional.
No, they are not. Standing on the ground, my change in momentum is zero, but the force of the ground on my feet is nonzero, as is the reaction force of my feet on the ground.

Your statement is a misrepresentation of Newtons third law.
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Jan30-13, 10:18 PM
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Quote Quote by Andrew Mason View Post
The forces and changes in momentum are necessarily directly proportional so changes in momentum must be equal and opposite.

Since, in non-relativistic interactions, the duration of interaction is the same for both interacting bodies, the changes in momentum of the interacting bodies are equal and opposite: Body 1: Force exerted by body 1 on body 2 = F1 = dp2/dt ⇔∫F1dt = ∫dp2 = Δp2; For body 2: F2 = - F1 ⇔ ∫F2dt = Δp1 = -∫F1dt = -Δp2

AM
But in general, there are more than two objects. So suppose we have three objects, [itex]O_1, O_2[/itex] and [itex]O_3[/itex], with [itex]O_2[/itex] in the center. Object [itex]O_1[/itex] exerts a force only on [itex]O_2[/itex], and [itex]O_2[/itex] exerts forces on [itex]O_1[/itex] and [itex]O_3[/itex]. Object [itex]O_3[/itex] exerts a force only on [itex]O_2[/itex]

In this setup, it's not the case that the change in momentum of [itex]O_1[/itex] is equal and opposite to the change in momentum of [itex]O_2[/itex], because [itex]O_1[/itex] is not the only object acting on [itex]O_2[/itex].

It's certainly true that the total change in momentum is zero, if you add up the changes for all three objects. But they don't come in action/reaction pairs. To me, what comes in pairs is "the force on [itex]O_1[/itex] due to [itex]O_2[/itex]" and "the force on [itex]O_2[/itex] due to [itex]O_1[/itex]"
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Jan31-13, 03:12 AM
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Quote Quote by stevendaryl View Post
Indirectly, it is.
No, it's not affecting the prediction. As you said yourself, it is just rearranging the initial equation and naming some of the terms. To get a prediction you have to rearrange equation anyway to solve for certain terms, and the solutions are mathematically equivalent.

Quote Quote by stevendaryl View Post
Come on. You know what the word "understand" means.
I know what it means to me in the context of physics: Understanding the world = Having a model of the world that can make quantitative predictions. But you seem to have a different definition of "understanding the world" in the context of physics, because you see a difference between the two.

I also don't understand what you mean by the gap between understanding and truth here:
Claiming that the goal of physics is understanding doesn't imply how close our current understanding is to the truth.
To me, in the context of physics, this would mean the gap between quantitative prediction and measurement. But since to you understanding is not about quantitative prediction, I have no idea what truth could mean here.
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Jan31-13, 03:27 AM
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Quote Quote by Andrew Mason View Post
We are not in disagreement over the direction of the reaction force.
You claim that all forces in my scenario are centripetal. That would include Frcf which is obviously centrifugal.



Quote Quote by Andrew Mason View Post
Newton talks about "actions" and "reactions which mean "changes in motion".
Wrong. Action & reaction are forces, not changes in motion. Newton 3rd Law applies to static cases as well, where there are force pairs but no acceleration.
stevendaryl
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Jan31-13, 05:36 AM
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Quote Quote by A.T. View Post
No, it's not affecting the prediction. As you said yourself, it is just rearranging the initial equation and naming some of the terms. To get a prediction you have to rearrange equation anyway to solve for certain terms, and the solutions are mathematically equivalent.
AT, one of the first posts you made to this thread, you said:
"Fictitious force" is just an unfortunate name choice. It gives people the wrong idea that some forces are more "real" than others. But that is just philosophy, irrelevant to physics. That is why I prefer the terms "inertial forces" and "interaction forces".
So, way back then, you seem to agree that a choice of terminology can be "unfortunate" and that it can give people the "wrong idea", which I assume means that it can be misleading. Have you changed your mind about the possibility of terminology being unfortunate and misleading?

If you still believe in the possibility that terminology can be unfortunate and can give people the wrong idea, then why do you object so strenuously when I suggest that YOUR terminology can be unfortunate and can give people the wrong idea?
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Jan31-13, 05:52 AM
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Quote Quote by A.T. View Post
I know what it means to me in the context of physics: Understanding the world = Having a model of the world that can make quantitative predictions. But you seem to have a different definition of "understanding the world" in the context of physics, because you see a difference between the two.
They are certainly two different activities: Constructing, refining, simplifying, clarifying concepts of models is one activity. Comparing a model's predictions with experiment is another activity. "Understanding" can apply to either a model, or to experimental results.

As I pointed out, one of your first posts made complaints about what you considered a bad way to think about models, when you said that the phrase "fictitious forces" is unfortunate, and can give people the wrong idea.
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Jan31-13, 06:23 AM
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Quote Quote by stevendaryl View Post
So, way back then, you seem to agree that a choice of terminology can be "unfortunate" and that it can give people the "wrong idea", which I assume means that it can be misleading. Have you changed your mind about the possibility of terminology being unfortunate and misleading?
I admit that "wrong idea" is not the best way to put it. As I explain later in the post you mentioned, I meant that it leads to pointless philosophizing about what is "real", which is irrelevant to physics.
Quote Quote by stevendaryl View Post
If you still believe in the possibility that terminology can be unfortunate and can give people the wrong idea, then why do you object so strenuously when I suggest that YOUR terminology can be unfortunate and can give people the wrong idea?
Because your justification for calling a certain terminology unfortunate, is exactly the pointless philosophizing about what is "real" or "the truth", which I don't see as relevant.
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Jan31-13, 06:45 AM
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Quote Quote by A.T. View Post
I admit that "wrong idea" is not the best way to put it. As I explain later in the post you mentioned, I meant that it leads to pointless philosophizing about what is "real", which is irrelevant to physics.
In this particular case, we can certainly make the distinction between vectors, whose components are covariant under coordinate transformations, and terms that can be made to vanish by choosing a particular coordinate system. That's a mathematical, not a philosophical distinction, and it's an important distinction for working with the equations of motion.


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