# What exactly is the reactive centrifugal force (split)

by A.T.
Tags: centrifugal, force, reactive, split
Mentor
P: 17,344
 Quote by stevendaryl 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 actually think that this is an excellent example. I too think that the term "fictitious force" is unfortunate. Nonetheless, I recognize that it isn't up to me and that people who use it in the standard way are making correct statements from an authoritative definition. Similarly, you should realize that the definition of "physics", although you think it is unfortunate, it is not up to you and that people who use it in the standard way are making correct statements from an authoritative definition.

If I were having your argument with A.T. I would either reference a definition of physics that included the philosophical bits you like or I would challenge him to reference a definition of physics that excluded them. Many times terms have multiple authoritative definitions, but that doesn't mean that it is good practice to simply go about redefining them yourself, since nobody is likely to consider your definition authoritative.
P: 2,150
 Quote by DaleSpam I actually think that this is an excellent example. I too think that the term "fictitious force" is unfortunate. Nonetheless, I recognize that it isn't up to me and that people who use it in the standard way are making correct statements from an authoritative definition. Similarly, you should realize that the definition of "physics", although you think it is unfortunate, it is not up to you and that people who use it in the standard way are making correct statements from an authoritative definition.
I disagree. There is nothing sacred about terminology, other than the fact that if one is using a technical term that is not standard, then there is time that must be spent explaining the terminology that could possibly be better spent on something else.

When it comes to technical terminology, such as "fictitious force" versus "inertial force" versus what I consider better "connection coefficients", different authors use different terminology. There is no absolute authority about what terms to use.

As for the definition of "physics" itself, I can't see any point whatsoever in worrying about it, or caring what the official definition is. A particular journal can decide to accept or reject a paper based on whatever criteria they choose; it doesn't need to be a universal definition of what counts as "physics".
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P: 6,679
 Quote by DaleSpam 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 a good example because the gravitational force and normal force are not third law pairs. They are not equal and opposite, for one thing. The normal force is always a bit less than the gravitational force except at the poles.

The third law pair to the gravitational force of the earth on a body is necessarily the gravitational force of the body on the earth. Gravitational force supplies the centripetal acceleration required of the body on the surface of the earth. The body supplies the earth with its centripetal acceleration about the body-earth centre of mass (which is obviously extremely close to the earth's centre of mass).

 This is NOT how Newtons third law works. It is the individual forces which are in action reaction pairs, not the net force.
I disagree. Your example illustrates why it can be difficult to apply Newton's third law. Where there is a body experiencing a change in motion there is always some other body experiencing an equal and opposite change in motion. Those are the third law pairs.

AM
P: 4,078
 Quote by stevendaryl 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.
Who said something against making a distinction? I said that I prefer the terms "inertial forces" and "interaction forces". These are two distinct names.

Also note that I used the word "prefer", clearly indicating that it is just personal preference. Contrary to you, who is using terms like "falsehood", "crippled physics" etc. to denounce a naming convention which you don't like.
P: 4,078
 Quote by Andrew Mason The third law pair to the gravitational force of the earth on a body is necessarily the gravitational force of the body on the earth. Gravitational force supplies the centripetal acceleration required of the body on the surface of the earth. The body supplies the earth with its centripetal acceleration about the body-earth centre of mass (which is obviously extremely close to the earth's centre of mass).
What acceleration? Consider a non-rotating planet with you on it. Newtonian gravity. There is no acceleration. It is all static. But there is one interaction (electro-magnetic) between your feet and the ground, with a 3rd law force pair. And there is a second interaction (gravitational) between you and the planet, with a second 3rd law force pair. Two times action & reaction without any change in motion.

 Quote by Andrew Mason Where there is a body experiencing a change in motion there is always some other body experiencing an equal and opposite change in motion. Those are the third law pairs.
Nope, the third law pairs are individual forces, not accelerations.
http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion
 Quote by Wikipedia Third law: When a first body exerts a force F1 on a second body, the second body simultaneously exerts a force F2 = −F1 on the first body.
Your problem is as usual: You think that the most trivial and primitive example (two bodies, one interaction) represents the general case. So you fail to apply the laws correctly to an even slightly more complex scenario (like the one above, with two interactions).
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P: 6,679
 Quote by A.T. You claim that all forces in my scenario are centripetal. That would include Frcf which is obviously centrifugal.
It depends on where the force acts. If one says that a force acts on the centre of mass of the body (ie. the space station), then the force is centripetal because it causes the space station and other astronaut (as the other body) to prescribe a centripetal acceleration towards the centre of rotation.

 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.
It is not wrong. It is just a way of looking at the third law that differs slightly, but not fundamentally, from your view.

The fact is that Newton refers to "forces" in his first and second laws. He does not use the word "force" in the third law.

Your view, I gather, is that his omission of the word "force" and his use of "action" and "reaction" were not deliberate. My view is that this was deliberate.

The third law can be viewed as the law of interacting bodies. If there is a change in motion of a body there must be another equal and opposite change in motion of some other interacting body or bodies. That, it seems to me, is the very essence of the third law. If there are forces but no actions (changes in motion), there is no need for the third law. The first law covers the situation.

AM
P: 2,150
 Quote by A.T. Who said something against making a distinction? I said that I prefer the terms "inertial forces" and "interaction forces". These are two distinct names.
Sure, but "connection coefficient" is what they are, mathematically, and has the advantage of having a precise definition saying how to compute them. Connection coefficients are needed whenever one needs to take a derivative of a vector or tensor field. Calling these terms "fictitious forces" doesn't make sense, because not all uses of derivatives have anything to do with forces. They really don't have anything to do with forces, in general.
Mentor
P: 17,344
 Quote by Andrew Mason the gravitational force and normal force are not third law pairs.
I never said they were. They act on the same body so they cannot be third law pairs. Each of those two forces (gravitational and normal) have their own third law reaction forces acting on the earth.

 Quote by Andrew Mason The normal force is always a bit less than the gravitational force except at the poles.
Fine. Consider the person to be standing on a turntable at the pole such that he is at rest in a Newtonian inertial frame. And consider the planet to be isolated far from any significant gravitational body.

 Quote by Andrew Mason The third law pair to the gravitational force of the earth on a body is necessarily the gravitational force of the body on the earth.
Yes, this is correct. Note that the gravitational force (mg) is not equal to the change in momentum (ma=0). So your correct statement here contradicts your own incorrect statement elsewhere.

 Quote by Andrew Mason Where there is a body experiencing a change in motion there is always some other body experiencing an equal and opposite change in motion. Those are the third law pairs.
Not only is this not how Newtons third law works, it is not even true.

Consider three identical masses, each connected by identical springs of neglible mass to both of the others and starting at rest from a position where the springs are all equally stretched. Each object is experiencing a change in motion and there is no other body experiencing an equal and opposite change in motion.
P: 4,078
 Quote by Andrew Mason It depends on where the force acts.
Look at the picture again. Try to figure out where the blue arrows start. That's where the astronaut applies the force to the space station.

 Quote by Andrew Mason If one says that a force acts on the centre of mass of the body...
...then one obviously has a serious case of cognitive dissonance.

 Quote by Andrew Mason The fact is that Newton refers to "forces" in his first and second laws. He does not use the word "force" in the third law.
Doesn't matter. In today's physics the 3rd law refers to forces. Every physics book will confirm this.
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P: 6,679
 Quote by A.T. What acceleration? Consider a non-rotating planet with you on it. Newtonian gravity. There is no acceleration. It is all static. But there is one interaction (electro-magnetic) between your feet and the ground, with a 3rd law force pair. And there is a second interaction (gravitational) between you and the planet, with a second 3rd law force pair. Two times action & reaction without any change in motion.
I agree. Those are the forces and the interactions. They all sum to 0. But you don't need the third law to analyse that. There is no "action" (ignoring the acceleration relative to the sun and other planets etc.) . Newton made the distinction between force and action. I didn't.

 Nope, the third law pairs are individual forces, not accelerations. http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion
Wikipedia does not state Newton's the third law. It states one interpretation of Newton's third law but it does not use the terms "action" and "reaction" that Newton used. In my view, it implicitly combines the first, second and third laws of Newton so it makes it difficult to see how the third law is distinct from the first and second. If forces are balanced so there is no net force, the first law applies. You don't need the third law to determine that if a body's motion is not changing all the forces must sum to 0. You just need the first.

 Your problem is as usual: You think that the most trivial and primitive example (two bodies, one interaction) represents the general case. So you fail to apply the laws correctly to an even slightly more complex scenario (like the one above, with two interactions).
I am not disagreeing with you on the physics. All I am saying is that if there is no acceleration, you don't need the third law. The first law suffices.

An approach that says that people who take a valid but different view of things are wrong is not particularly helpful.

AM
P: 2,150
 Quote by DaleSpam Consider three identical masses, each connected by identical springs of neglible mass to both of the others and starting at rest from a position where the springs are all equally stretched. Each object is experiencing a change in motion and there is no other body experiencing an equal and opposite change in motion.
I agree with you that thinking of equal and opposite forces works better than thinking of equal and opposite changes of motion, but in the case you are talking about, the change in momentum of one of the masses is equal and opposite to the vectorial sum of the changes of momentum of the other two masses. That might be what AM meant.
P: 4,078
 Quote by Andrew Mason Wikipedia does not state Newton's the third law. It states one interpretation of Newton's third
It is the interpretation of Newton's third that you will find in every modern physics book. It doesn't matter how you call this Law ("3rd law" or "generalized 3rd law"). All your hand waving cannot change the fact that:
- It is a law of physics.
- It states two equal but opposite forces.
- It applies to static cases too.
- It means that my astronauts exert a centrifugal reaction force on the station
P: 2,150
 Quote by Andrew Mason I am not disagreeing with you on the physics. All I am saying is that if there is no acceleration, you don't need the third law. The first law suffices.
Hmm. Let's see if that's true. Suppose you have three objects 1, 2, 3. There are correspondingly 6 forces:
1. $\stackrel{\rightarrow}{F_{12}}$ = force of 1 on 2
2. $\stackrel{\rightarrow}{F_{21}}$ = force of 2 on 1
3. $\stackrel{\rightarrow}{F_{13}}$ = force of 1 on 3
4. $\stackrel{\rightarrow}{F_{31}}$ = force of 3 on 1
5. $\stackrel{\rightarrow}{F_{23}}$ = force of 2 on 3
6. $\stackrel{\rightarrow}{F_{32}}$ = force of 3 on 2

The fact that the situation is static means that the force on any object is zero. So we have:
1. $\stackrel{\rightarrow}{F_{12}} + \stackrel{\rightarrow}{F_{32}} = 0$
2. $\stackrel{\rightarrow}{F_{13}} + \stackrel{\rightarrow}{F_{23}} = 0$
3. $\stackrel{\rightarrow}{F_{21}} + \stackrel{\rightarrow}{F_{31}} = 0$

That's three constraints. I don't see how you can derive that
$\stackrel{\rightarrow}{F_{12}} + \stackrel{\rightarrow}{F_{21}} = 0$
 Mentor P: 17,344 Andrew Mason, please provide a mainstream scientific reference for Newton's 3rd law being as you describe, equal and opposite net forces or equal and opposite changes in momentum.
Mentor
P: 17,344
 Quote by stevendaryl Hmm. Let's see if that's true. Suppose you have three objects 1, 2, 3. There are correspondingly 6 forces:$\stackrel{\rightarrow}{F_{12}}$ = force of 1 on 2 $\stackrel{\rightarrow}{F_{21}}$ = force of 2 on 1 $\stackrel{\rightarrow}{F_{13}}$ = force of 1 on 3 $\stackrel{\rightarrow}{F_{31}}$ = force of 3 on 1 $\stackrel{\rightarrow}{F_{23}}$ = force of 2 on 3 $\stackrel{\rightarrow}{F_{32}}$ = force of 3 on 2 The fact that the situation is static means that the force on any object is zero. So we have:$\stackrel{\rightarrow}{F_{12}} + \stackrel{\rightarrow}{F_{32}} = 0$ $\stackrel{\rightarrow}{F_{13}} + \stackrel{\rightarrow}{F_{23}} = 0$ $\stackrel{\rightarrow}{F_{21}} + \stackrel{\rightarrow}{F_{31}} = 0$ That's three constraints. I don't see how you can derive that $\stackrel{\rightarrow}{F_{12}} + \stackrel{\rightarrow}{F_{21}} = 0$
In fact, you cannot derive it. Here are a set of forces that satisfy Newton's 1st law but not his 3rd:
1. $\stackrel{\rightarrow}{F_{12}}=2$
2. $\stackrel{\rightarrow}{F_{21}}=1$
3. $\stackrel{\rightarrow}{F_{13}}=3$
4. $\stackrel{\rightarrow}{F_{31}}=-1$
5. $\stackrel{\rightarrow}{F_{23}}=-3$
6. $\stackrel{\rightarrow}{F_{32}}=-2$
P: 1
 Quote by hms.tech I think that there is no such thing as centrifugal force . Am I right ? is this force fictitious ?
Yes.
You can study the centrifugal and Coriolis forces here:
http://www.gethome.no/paulba/Spaceship.html

Hit the "slow motion" button.
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P: 6,679
 Quote by stevendaryl But in general, there are more than two objects. So suppose we have three objects, $O_1, O_2$ and $O_3$, with $O_2$ in the center. Object $O_1$ exerts a force only on $O_2$, and $O_2$ exerts forces on $O_1$ and $O_3$. Object $O_3$ exerts a force only on $O_2$ In this setup, it's not the case that the change in momentum of $O_1$ is equal and opposite to the change in momentum of $O_2$, because $O_1$ is not the only object acting on $O_2$. 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 $O_1$ due to $O_2$" and "the force on $O_2$ due to $O_1$"
If there are three bodies interacting, it is much more difficult to analyse forces because the duration of the interactions between bodies may differ. What one can easily determine, however, are the changes in motion that result from the interactions. For three interacting bodies, A, B and C, if body A experiences a change in motion (momentum), bodies B and C must experience a change in motion that, combined, is equal in magnitude to the change in A's motion but opposite in direction.

AM
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P: 6,679
 Quote by DaleSpam Andrew Mason, please provide a mainstream scientific reference for Newton's 3rd law being as you describe, equal and opposite net forces or equal and opposite changes in momentum.
Do you consider Newton to be mainstream?

Newton's explanation of the third law uses both the concept of a force and of change in motion. He begins by talking of a body pressing on another, the latter pressing back on the first (a finger pressing a stone and the stone pressing back on the finger), and then says:
If a body impinges upon another, and by its force changes the motion of the other, that body also (because of the equality of the mutual pressure) will undergo an equal change, in its own motion, toward the contrary part. The changes made by these actions are equal, not in the velocities but in the motions of the bodies; that is to say, if the bodies are not hindered by any other impediments. For, as the motions are equally changed, the changes of the velocities made toward contrary parts are reciprocally proportional to the bodies.
It is interesting that Newton speaks of "bodies ... not hindered by any other impediments". One can argue whether he was equating action with "force on an otherwise unimpeded body", but that is not an unreasonable interpretation.

With respect to the astronaut space-station example, the third law says that a body undergoing centripetal change in motion due to contact with another unimpeded body causes that other body to have an equal change of motion in the opposite direction. So if we are talking about the astronaut undergoing a centripetal change in motion due to contact with the space station/other astronaut (taken as one body), the third law "reaction" is a change in motion of the space station/other astronaut in the opposite direction. That change in motion always points to the centre of rotation.

AM

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