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## What exactly is the reactive centrifugal force (split)

 Quote by A.T. That is not relevant of the circular motion. In the plane of the circle there is a centripetal force and a centrifugal reaction, which accelerates the particles outwards.
In order to accelerate the particles outwards, it seems the solar ship would have to travel faster than orbital speed (v^2 > G M / r) in a circular path so that a centripetal force (in addition to gravity) would be required to maintain the circular path, and use some sort of collector and accelerator, which could be an idealized sail, in order to end up accelerating particles outwards.

Mentor
 Quote by A.T. If I understand him correctly he wants L = const for the body on which the centripetal force is exerted, which is the rocket and the remaining fuel.
Yuck, that makes it messy. With this definition you need a tangential thrust in order to have centripetal motion. I don't see any reason for that definition.

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 Quote by A.T. That's mumbo jumbo. You have to use forces acting on the outer block to explain the outer block's outwards acceleration in the rotating frame. There is a fictions centrifugal force acting on the outer block in the rotating frame, but it is not sufficient to overcome friction. Only because of the real centrifugal force from the inner block the outer block accelerates outwards in the rotating frame.
Are you talking about real forces or forces that appear in the non-interial reference frame of the rotating turntable?

If you are talking about real forces, the turntable is trying to grab the outer block and change its velocity. The surface of the turntable is accelerating inward underneath the block. If the friction between the outer block and turntable is not enough to provide the force needed to accelerate both blocks to keep them on the turntable, the turntable surface accelerates away from blocks which continue their inertial motion. They would be accelerated somewhat by the kinetic friction force between the turntable and block until they left the surface.

The only real force that is apparently outward is the reaction force of the block on the turntable. If the turntable and spindle was sitting on a frictionless you would be able to see the block and spindle both rotate about a vertical axis through their common centre of mass. The block would exert a force on the turntable that causes the centre of mass of the turntable to rotate about the common centre of mass (which is between the spindle and the block). Since it is fixed to the earth, the reaction force of the block is exerted on the earth which undergoes a much smaller rotation.

 So there is a centripetal component, and a centrifugal reaction component.
Which means that the force is not radial: r x F ≠ 0

AM

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 Quote by Andrew Mason The only real force that is apparently outward is the reaction force of the block on the turntable.
And the reaction force of the inner block on the outer. That is a real force, not fictitious, and s required to explain the movement in all frames.

This line of conversation is still irrelevant. Regardless of the validity of your reasons for disliking it, the term exists. You cannot wish it away, nor can you reason it away.

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 Quote by DaleSpam And the reaction force of the inner block on the outer. That is a real force, not fictitious, and s required to explain the movement in all frames. This line of conversation is still irrelevant. Regardless of the validity of your reasons for disliking it, the term exists. You cannot wish it away, nor can you reason it away.
You can call it what you like. It is due to inertia and it does not accelerate anything away from the centre of rotation.

AM

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 Quote by Andrew Mason It does not accelerate anything away from the centre of rotation.
Whether it does or not is irrelevant to the naming convention.

But in some cases it clearly does.
 Recognitions: Homework Help Science Advisor We have beaten this issue to death. So let me recap where I think we differ. I think we fundamentally disagree on what third law action and reaction pairs are. It is easy in a collision of free bodies analysed from an inertial reference frame. One body accelerates in one direction and the other accelerates in the opposite direction - changes in momentum are equal and opposite. Or, if a rocket sends rocket gas in one direction, the rocket recoils in the other. These are third law action/reaction pairs. No problem. The problem occurs when you do not have a system that defines an inertial reference frame. For example, you say that the reaction to the centripetal force supplied by mechanical means is the force that the body being accelerated applies to the body or part of the body that is supplying the accelerating force. I say that is simply an inertial effect and the real reaction/action pairs are the centripetal force that the rotating body experiences and the centripetal force that this body exerts on the rest of the system as both rotate about their common centre of mass. Similarly, the third-law pair in applying a force to a box with my hand (with me standing on the earth) is the force I exert in the opposite direction on the earth. The change of momentum of the earth that results is equal and opposite to the change of momentum of the box. You seem to be saying that it is the force of the box on my hand. I say that force of the box on my hand is simply an inertial effect - a pseudo force really because it can never be associated with an acceleration. In my view, this is where we disagree. We can argue all we like about Newton's third law but we will continue to disagree because we take somewhat different view on the application of the law to these situations. AM

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 Quote by Andrew Mason I think we fundamentally disagree on what third law action and reaction pairs are.
I agree that this is the fundamental disagreement. However, this is not simply a difference of equally-valid opinions. You wish to state Newton's 3rd law in terms of changes in momentum instead of forces.

1) You have been unable to produce any textbook or online reference which expresses the third law mathematically in terms of momenta, while I have produced several that express it mathematically in terms of forces.

2) You have also been unable to produce any textbook or online reference which even clearly expresses it verbally in terms of momenta.

3) You have been able to find one reference which is, by your own admission, quite unclear so a small part of it can be construed in a way that at least doesn't contradict your position. However, even that reference contradicts your position when you read the whole reference instead of taking that small part out of context.

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 Quote by DaleSpam However, this is not simply a difference of equally-valid opinions. You wish to state Newton's 3rd law in terms of changes in momentum instead of forces.
What Newton meant by "Newton's third law" is perfectly clear. First he defines
 Quote by Principia Definition IV An impressed force is an action exerted upon a body...
In other words, "actions" are "forces".
 Quote by Principia Law III For every action there is always opposed an equal reaction: ...
The examples following the definition are clearly about forces, not momentum. For the example of a horse drawing a stone tied to a rope, he is talking about the equal and opposite forces the rope exerts on the horse and the stone. By any reasonable interpretation of that scenario, the changes of momentum of the horse and the stone are not equal and opposite, in general.

The explanation of law III continues
 If a body impinge upon another, and by its force changes the momentum of the other, the change in momentum of that body will be equal and opposite; that is to say, if the (movement of the) bodies are not hindered by any other impediments.
(My bold).

To summarize, action and reaction forces are always equal and opposite in pairs, but they don't necessarily cause equal and opposite changes of momentum if the motion of the two bodies is constrained in some other way.

That explains why the changes in momentum of the horse and the stone are not equal and opposite, of course.

Reformulating classical mechanics in terms of "Mason's laws of motion" is fine by me, but don't mis-attribute Mason's laws to Newton where they are different.

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 Quote by AlephZero What Newton meant by "Newton's third law" is perfectly clear. ..... In other words, "actions" are "forces".
If it was "perfectly clear" there would be no discussion of how confusing it is, as in this quote that I referred to earlier from a physicist , Mark Hammond:
"Newton’s explanation starts out talking about forces and pressures, speaks of obstructing and advancing “progress,” and finally ends up talking about what appears to be momentum. I’ll say right here that I am unimpressed with the clarity of Newton’s explanation. I go back and forth between thinking action/reaction should be read as the change in motion as opposed to being read as the thing that changes the motion. But rather than pick nits (and criticize an author who can’t defend himself), let’s zero in on what Newton seems most intent upon telling us in this, his third law of motion: there is a specific relationship between the changes in the motions of two interacting objects. Hence, this is a law of motion."

 To summarize, action and reaction forces are always equal and opposite in pairs, but they don't necessarily cause equal and opposite changes of momentum if the motion of the two bodies is constrained in some other way.
They do cause equal and opposite changes of momentum if you look at the motions of the body and the rest of the inertial system the body is interacting with. You can always do that.
 That explains why the changes in momentum of the horse and the stone are not equal and opposite, of course. Reformulating classical mechanics in terms of "Mason's laws of motion" is fine by me, but don't mis-attribute Mason's laws to Newton where they are different.
It is not that the principle of reciprocity of forces between contact points is wrong (it is correct in most cases). But this reciprocity is demonstrated by the changes in momentum that the forces create as measured by the changes in motion of the centres of mass of the two interacting parts of an inertial system. If you want to explain it in terms of the reciprocity of forces, it is explained by the forces as measured by the accelerations of the centres of mass of the interacting parts of an inertial system. It is neither illuminating nor practical to apply Newton's third law in terms of the reciprocity of forces between the contact points. So I have difficulty believing that is what Newton fundamentally intended to convey in the third law.

In the simple situation of an astronaut pushing off from a space ship with his legs, the third law is demonstrated by the equal and opposite changes of momentum that result to the two parts of this inertial system (ie. astronaut and space ship). If you just apply the reciprocity of forces between the astronaut's feet and the surface of the space station, this does not explain the physics. You have to take into account what the feet and space station surface are connected to. You then have to take into account the masses of those bodies and the motions of the centres of mass of those bodies. Those are legitimate third law action/reaction pairs. They do not lose that quality simply because the centres of mass are not touching. There are a myriad of other forces (between different parts/atoms etc.) that we simply ignore because they are not important.

AM

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 Quote by Andrew Mason It is neither illuminating nor practical to apply Newton's third law in terms of the reciprocity of forces between the contact points. So I have difficulty believing that is what Newton fundamentally intended to convey in the third law.
Then find a good reference that supports your interpretation. Otherwise you are simply engaging in personal speculation.

In context, Newton's Principia does not support your approach, and I have yet to see any other scientific reference that does either.