Is the concept of reactive centrifugal force valid?

[Dale]

The center of mass of the rotating space station is orbiting about the space station + astronaut center of mass. The curvature is inward, not outward. The space station is therefore subject to a centripetal force, just as is the astronaut.

Your first two sentences are correct, but the third sentence does not follow. The first two sentences correctly establish the direction of the force, but to determine whether or not a given force is centripetal or centrifugal requires not only specification of the direction of the force, but also it's point of application.

Assuming that the center of rotation is the origin and that a force is pointing in the positive x direction then the force is centrifugal if applied on the positive x axis, centripetal if applied on the negative x axis, and tangential if applied on the y axis.

In this scenario the force is applied at the location of the astronaut and winds up being centrifugal, despite the fact that it is causing uniform circular motion of the center of mass. I.e. It is not applied at the center of mass so the motion of the center of mass is not sufficient.

 

[Andrew Mason]

I submit that there is such a thing as a reactive centrifugal force. The equal but opposite reaction to a real centrifugal force is a real centrifugal force. For example, the interactions between two positively charged objects are equal but opposite centrifugal forces.

I agree. That is an example of a real centrifugal force - and it does not require, or have anything really to do with, rotation.

This however is not what that nonsense wikipedia article is talking about.

That was my take - complete nonsense. For students of classical mechanics, better to stick to a good textbook. But even good textbooks can be misleading or confusing. Even Newton appears to be misleading (incorrect) in this aspect of rotational motion. The animated discussion on this thread highlights the challenge and difficulty in analysing the physics of rotation.

AM

 

[Andrew Mason]

Sure, the net force is zero. So what? Third law pairs are actual, individual forces, not 'net forces'.

The fact is that there is acceleration always occurring. These are not balanced static forces. The third law pairs are forces that are each causing acceleration.

Answer this question: Does the section of the merry-go-round in contact with the rider exert a force on the rider?

And this: What direction is that force?

The merry-go-round exerts a force on the rider and it is toward the centre of rotation.

And this: Do you agree that Newton's 3rd law applies?

Of course.

AM

 

[A.T.]

The third law pairs are forces that are each causing acceleration.

Wrong. There is nothing in Newtons 3rd Law about the acceleration of the interacting objects. When I lean against the wall, there are two equal opposite forces acting on me and the wall respectively. But neither me nor the wall is accelerating.


That is Newtons 3rd Law:

The merry-go-round exerts a force on the rider and it is toward the centre of rotation, ...

...while the rider exerts a force on the merry-go-round and it is away from the centre of rotation.

 

[A.T.]

Forget the astronaut for a moment.

Or make it 2 astronauts, on opposite sides. Now we have no confusion about center of mass vs. center of rotation and momentum conservation

 

[Doc Al]

Or make it 2 astronauts, on opposite sides. Now we have no confusion about center of mass vs. center of rotation and momentum conservation

Nicely done.

 

[Andrew Mason]

Just put a second rider of equal mass on the opposite side. Both riders will exert reactive centrifugal forces on their seats. These forces point away from :
- the center of rotation
- the center of mass

There is a centrifugal effect of the rider on the chair. But it is not a force. If the chair breaks loose from the merry-go-round there is no acceleration supplied by the rider to the chair. They both would just move in uniform motion in a direction tangential to the merry-go-round.

While the rider rotates, the merry-go-round as a whole, not just the seat, exerts a centripetal force on him. He, in turn, exerts a (reaction) force on the whole merry-go-round as a whole, not just the seat. The rider opposite asserts a force on the whole merry-go-round as a whole, not just his seat and that force is exactly equal and opposite to the force that the first rider exerts on the merry-go-round. Those are true forces.

These two riders could stay on the merry-go-round by letting go of poles or seats and just holding a rope between them. There would be no forces between the riders and the merry-go-round at all. Would you then argue that the riders are exerting a centrifugal force on each other?

AM

 

[Doc Al]

There is a centrifugal effect of the rider on the chair. But it is not a force.

So the chair can exert a force on the rider without the rider exerting a force on the chair? Wow!

While the rider rotates, the merry-go-round as a whole, not just the seat, exerts a centripetal force on him.

Mysterious non-local forces? Wow!

 

[Andrew Mason]

Wrong. There is nothing in Newtons 3rd Law about the acceleration of the interacting objects. When I lean against the wall, there are two equal opposite forces acting on me and the wall respectively. But neither me nor the wall is accelerating.

Fine. These are just balanced forces so neither I nor the wall accelerates. But these are not really the action/reaction pairs of forces that the third law speaks about. Those are dynamic: they occur when masses experience accelerations eg in collisions of matter.

Suppose I am standing on a floor. My weight is balanced by the normal force of the floor. But if the floor gives way, the acceleration that the pieces of the floor experience is not from my weight pushing on them. It is from gravity. The reaction force to the gravitational force on me is my gravitational pull on the earth. The same goes for the floor: the reaction force to the gravitational acceleration of the floor pieces is their gravitational pull on the earth.

AM

 

[Andrew Mason]

So the chair can exert a force on the rider without the rider exerting a force on the chair? Wow!

The rider exerts a force on the chair only because the chair is connected to the merry-go-round. If the chair is not connected, the rider exerts no force on it. So it is incorrect to say that the rider exerts a force on just the chair.

When you analyse collisions between billiard balls, do you say that the balls only exert forces on the molecules that they are in contact with? Can you ignore the fact that the molecules of each ball are all connected? If you do, you will not be able to explain the changes in motion that occur.

AM

 

[Dale]

There is nothing in Newtons 3rd Law about the acceleration of the interacting objects. When I lean against the wall, there are two equal opposite forces acting on me and the wall respectively.

These are just balanced forces so neither I nor the wall accelerates. But these are not really the action/reaction pairs of forces that the third law speaks about.

A.T. Is correct. These are exactly the action/reaction pairs the third law speaks about. The third law is not restricted to collisions nor to accelerating objects.

 

[Andrew Mason]

Or make it 2 astronauts, on opposite sides. Now we have no confusion about center of mass vs. center of rotation and momentum conservation

These drawings are very well done but are not correct. In the inertial frame the only forces are centripetal forces. The astronauts and the space station are undergoing net acceleration so there is a net (centripetal) force on each astronaut and on each part of the space station. The reaction centripetal forces on each element of mass are conducted through the space station structure to provide the centripetal force on the mass opposite. To see this, remove one of the astronauts and see what happens to the centre of rotation.

In the rotating frame, the "forces" need to balance (ie sum to 0) as the acceleration is not apparent. The non-inertial astronaut postulates a centrifugal force to balance the force he feels pressing into him from the space station.

AM

 

[D H]

Wrong. There is nothing in Newtons 3rd Law about the acceleration of the interacting objects. When I lean against the wall, there are two equal opposite forces acting on me and the wall respectively. But neither me nor the wall is accelerating.

Fine. These are just balanced forces so neither I nor the wall accelerates. But these are not really the action/reaction pairs of forces that the third law speaks about. Those are dynamic: they occur when masses experience accelerations eg in collisions of matter.

A.T. is correct here, Andrew. The force that the wall exerts on A.T. and the force that A.T. exerts on the wall are exactly the kind of pair-wise forces that Newton's third law addresses. Force is subject to the superposition principle. The F in F=ma is the net force acting on the object (or on A.T. in this case). The reason A.T. doesn't accelerate when he leans against the wall is because other forces are in play.

A.T. isn't accelerating because the floor is exerting both a normal vertical force that counteracts gravity and a frictional horizontal force that counteracts the normal force the wall exerts on A.T. Should A.T. try to lean against a frictionless wall while standing on a frictionless floor he will accelerate. (You can see a demonstration of this principle when a newbie ice skater tries to lean against the skating rink wall.)

Equal-but-opposite is not sufficient to qualify as a third law forces. For example, the force that the wall exerts on A.T. and the (horizontal) frictional force that the floor exerts on A.T., although equal-but-opposite, are not third law forces.

 

[A.T.]

There is a centrifugal effect of the rider on the chair. But it is not a force.

Of course it is a force. Anything else would violate Newtons 3rd Law, because the chair exerts a force on the rider. And "centrifugal effect" is just vague gibberish.

If the chair breaks loose from the merry-go-round there is no acceleration supplied by the rider to the chair. They both would just move in uniform motion in a direction tangential to the merry-go-round.

Completely irrelevant for the argument above.

While the rider rotates, the merry-go-round as a whole, not just the seat, exerts a centripetal force on him.

No, the interaction is local. The point of application of both forces (centripetal on the rider, and reactive centrifugal on the seat) is at the contact area between rider and seat. As DaleSpam already explained: You have to take this application point and the direction of the force into account, to determine if it's centrifugal or centripetal.

 

[A.T.]

Or make it 2 astronauts, on opposite sides. Now we have no confusion about center of mass vs. center of rotation and momentum conservation

These drawings are very well done but are not correct. In the inertial frame the only forces are centripetal forces.

Wrong. You are confusing all forces with all net forces on each part.

The astronauts and the space station are undergoing net acceleration so there is a net (centripetal) force on each astronaut and on each part of the space station.

Correct, but irrelevant. See above. There can still be centrifugal forces acting, even if all the net forces on individual parts are centripetal.

To see this, remove one of the astronauts and see what happens to the centre of rotation.

I made it balanced so you don't get confused. But you insist on getting confused? Removing the other astronaut will not change the fact that:

F_rcf has the same direction, as the vector from rotation center to the point of application of F_rcf. Therefore F_rcf is a centrifugal force

The net acceleration of the space station's COM is irrelevant here. But I made it zero, deal with it.

In the rotating frame, the "forces" need to balance (ie sum to 0) as the acceleration is not apparent. The non-inertial astronaut postulates a centrifugal force to balance the force he feels pressing into him from the space station.

Correct. You understand the inertial centrifugal force well. You should be able to tell it apart from the reactive centrifugal force based on the differences I listed in the legend.

 

[D H]

I made it balanced so you don't get confused.

I disagree. Your making it balanced is an obfuscating complication. We're talking Newtonian mechanics here, and ultimately Newtonian mechanics is about a set of point masses (aka particles) that interact via third law pairs.

So let's make it simple rather than complex. Consider just a pair of particles that interact via some central force. If the force is attractive both particles undergo either straight line or centripetal motion when viewed from the perspective of an inertial observer. The particles undergo either straight line or centrifugal motion if the force is repulsive.

The concept of a reactive centrifugal force to a centripetal force does not jibe with Newtonian mechanics in the case of point masses. So when does the concept of reactive centrifugal force to a centripetal force apply? When you over-complicate and obfuscate things.

 

[Dale]

These drawings are very well done but are not correct. In the inertial frame the only forces are centripetal forces.

The drawings are correct, this analysis is not correct. The 3rd law pairs are, as A.T. drew them, between each astronaut and the space station. This can be seen by placing scales or pressure plates at the foot of each astronaut. Each scale registers a 3rd law pair, so there are four forces.

In the rotating frame, the "forces" need to balance (ie sum to 0) as the acceleration is not apparent. The non-inertial astronaut postulates a centrifugal force to balance the force he feels pressing into him from the space station.

This is correct, and agrees with A.T.'s drawing.

 

[Doc Al]

So let's make it simple rather than complex. Consider just a pair of particles that interact via some central force. If the force is attractive both particles undergo either straight line or centripetal motion when viewed from the perspective of an inertial observer. The particles undergo either straight line or centrifugal motion if the force is repulsive.

How about sticking to the example of the merry-go-round that A.T. nicely illustrated?

Obviously, in the special case of the two particles attracting at a distance the only forces happen to be centripetal. So what? Stick to the example.

 

[Dale]

I disagree. Your making it balanced is an obfuscating complication.

Whether it is complicated or not, it is still a legitimate scenario that Newton's laws can analyze.

We're talking Newtonian mechanics here, and ultimately Newtonian mechanics is about a set of point masses (aka particles) that interact via third law pairs.

I disagree with this in general. Newtonian mechanics does just fine with extended bodies and continuums. However, for the sake of argument let's analyze the point particle of a space-station floor atom directly underneath the astronaut's boot. There are two forces acting on this atom. One is the contact force from the boot, and the other is the strain force from the neighboring floor atoms. These two forces are not equal, so there is a net force acting on the atom, and the net force is centripetal.

The strain force acts in the same direction as the net force, so it is also centripetal, but the contact force acts in the direction opposite the net force, so it is centrifugal.

Looking at the interacting atoms, the reaction force to the centrifugal contact force acting on the floor atom is a centripetal contact force acting on the boot atom, and the reaction force to the centripetal strain force acting on the floor atom is a centrifugal strain force acting on the neighboring atoms.

So let's make it simple rather than complex. Consider just a pair of particles that interact via some central force. If the force is attractive both particles undergo either straight line or centripetal motion when viewed from the perspective of an inertial observer. The particles undergo either straight line or centrifugal motion if the force is repulsive.

I agree, in an isolated two particle central force scenario the third-law pair of forces will be the same (centrifugal or centripetal). But Newton's laws can be used to analyze situations more complicated than two isolated point particles acting via some central force, and in such situations it is indeed possible to have the 3rd law pairs being different (centrifugal or centripetal).

The concept of a reactive centrifugal force to a centripetal force does not jibe with Newtonian mechanics in the case of point masses.

No, it only doesn't jibe in the case of 2 isolated point masses. It jibes with more general cases of point masses.

 

[A.T.]

A.T. Is correct. These are exactly the action/reaction pairs the third law speaks about. The third law is not restricted to collisions nor to accelerating objects.

In fact usually static examples are used introducing Newton's 3rd Law:

From
http://www.grc.nasa.gov/WWW/k-12/WindTunnel/Activities/third_law_motion.html

The book lying on the table is exerting a downward force on the table, while the table is exerting an upward reaction force on the book. Because the forces are equal and opposite, the book remains at rest. Notice also that the table legs are in contact with the floor and exert a force downward on it, while the floor in turn exerts an equal and opposite force upward.

The third law pairs are forces that are each causing acceleration.

I have never come across a book limiting Newtons 3rd to net forces which are directly related to acceleration. Can you provide us with a reference that supports your statement?

 

[Doc Al]

In fact usually static examples are used introducing Newton's 3rd Law:

Unfortunately that reference is very badly worded:

The book lying on the table is exerting a downward force on the table, while the table is exerting an upward reaction force on the book. Because the forces are equal and opposite, the book remains at rest.

The first sentence is describing Newton's 3rd law pairs. The second sentence implies that it is those third law pairs that produce equilibrium. Yikes!

 

[A.T.]

I disagree. Your making it balanced is an obfuscating complication.

Putting aside the question why a well balanced space station is "more complicated" than a wobbling one, let me point out that the balancing is not needed to prove my point. Even in the unbalanced case it would still be true that:

Frcf has the same direction, as the vector from rotation center to the point of application of Frcf. Therefore Frcf is a centrifugal force

I didn't make it balanced because I had to, but because it leaves no room to weasel around. That's why you try to change the scenario:

Consider... [Some scenario. Where is the reactive centrifugal force now?]

This is the same flawed line of argument Andrew exhausted pages ago. Nobody ever claimed that the reactive centrifugal force exists in every possible scenario.

The concept of a reactive centrifugal force to a centripetal force does not jibe with Newtonian mechanics in the case of point masses.

Wrong as DaleSpam already explained.

So when does the concept of reactive centrifugal force to a centripetal force apply? When you over-complicate and obfuscate things.

Well, if any scenario more complex that two point masses is "over-complicated" for you, then indeed you should not bother with reactive centrifugal forces.

 

[A.T.]

The first sentence is describing Newton's 3rd law pairs. The second sentence implies that it is those third law pairs that produce equilibrium. Yikes!

Indeed. That formulation can lead to the wrong understanding I pointed out to harrylin few pages ago. Ironically their question 3 is exactly about that potential flaw in reasoning they seem to imply themselves.

But my point was simply that this static scenario is a typical example of Newton's 3rd Law, used in many books.

 

[Andrew Mason]

A.T. is correct here, Andrew. The force that the wall exerts on A.T. and the force that A.T. exerts on the wall are exactly the kind of pair-wise forces that Newton's third law addresses. Force is subject to the superposition principle. The F in F=ma is the net force acting on the object (or on A.T. in this case). The reason A.T. doesn't accelerate when he leans against the wall is because other forces are in play.

If there is no acceleration (by having sufficient friction between AT and the floor and wall), all forces are balanced and inter-connected. It seems to me somewhat arbitrary to choose which force pairs are third law pairs and which are not. But I think I see your point: if the other forces are removed (ie if there was no/insufficient static friction (on the wall or floor) the wall (along with the Earth it is connected to) and AT would exert equal and opposite forces on each other each would accelerate away from each other in a horizontal direction. Those forces then would definitely be third law force pairs.

AM

 

[harrylin]

..Once you get to the non-inertial frame there are fictitious forces. You cannot remove those fictitious forces and correctly determine the trajectory.

I'm sorry if you cannot determine the trajectory without them, but I already referred to and indicated how to do that without introducing anything fictitious. If you want to get more explanation then please start a topic on it.

 

[A.T.]

If there is no acceleration (by having sufficient friction between AT and the floor and wall), all forces are balanced and inter-connected. It seems to me somewhat arbitrary to choose which force pairs are third law pairs and which are not.

What is arbitrary for Newton's 3rd, is the definition of objects (e.g. me, wall ,floor). This arbitrariness is a great feature, because it allows to use the law on different levels of abstraction.

But I think I see your point: if the other forces are removed (ie if there was no/insufficient static friction (on the wall or floor) the wall (along with the Earth it is connected to) and AT would exert equal and opposite forces on each other each would accelerate away from each other in a horizontal direction. Those forces then would definitely be third law force pairs.

Provide a reference that demands acceleration as a result from the Newtons 3rd force pair.

 

[Dale]

I'm sorry if you cannot determine the trajectory without them, but I already referred to and indicated how to do that without introducing anything fictitious.

Then you misunderstand the method you used. Your method is exactly the method which introduces the appropriate fictitious forces for a given non-inertial coordinate system. It introduces them whether you call them fictitious forces or not.

 

[A.T.]

The concept of a reactive centrifugal force...

It occurs to me, that the problem you and Andrew might be having, is simply exaggerated expectations. You seem to think that "reactive centrifugal force" is supposed to be some general concept, like the inertial centrifugal force which appears in every rotating frame.

But in fact it just a trivial naming convention applicable in a subset of cases involving rotation, where the local reaction to the centripetal force points away from the rotation center. I would not even call it a "concept". The concept here is Newtons 3rd.

 

[Andrew Mason]

The drawings are correct, this analysis is not correct. The 3rd law pairs are, as A.T. drew them, between each astronaut and the space station. This can be seen by placing scales or pressure plates at the foot of each astronaut. Each scale registers a 3rd law pair, so there are four forces.

There are gazillions of forces if you want to look at each atom and the forces between them. [There is the force of gravity between the atoms, as well, which we are ignoring. There are also the tidal forces between the astronaut and the space station because of differences in their centripetal acceleration due to slight difference in their radii of rotation, which we are ignoring].

What you have to do is ask: what is accelerating? In this case, all the molecules of the space station and astronauts are accelerating toward the centre of rotation. So all forces on a molecule must sum to a net force on each molecule equal to the centripetal force it is experiencing due to its centripetal acceleration. So we don't have to worry about the intermolecular forces. We just need to worry about the net force that causes acceleration. So the diagram should show the net force vector for each molecule. That will be a vector pointing toward the centre whose magnitude is mv^2/r.

Let's suppose that there are two space stations rotating about a common axis line but not mechanically connected. Initially, the first one has no astronauts and the second has the two astronauts shown. One astronaut then jumps onto the other space station. What happens? Do the stresses between the molecules in the second space station increase? The answer is: no they don't. The centre of rotation just changes.

If both astronauts jump at the same time, do the stresses increase? The answer is: yes they do. Now if you want to suggest that the additional stresses occur because of a centrifugal force that the astronaut exerts on the space station, why does it only occur if both astronauts are on the space station but not if only one is present?

AM

 

[A.T.]

There are gazillions of forces if you want to look at each atom and the forces between them.

Yes, but who wants that? As I already said: What is arbitrary for Newton's 3rd, is the definition of objects (e.g. me, wall ,floor). This arbitrariness is a great feature, because it allows to use the law on different levels of abstraction.

What you have to do is ask: what is accelerating?

Is this still about the "reactive centrifugal force"? Because I still haven't seen that reference stating that Newtons 3rd Law applies only to net forces resulting in acceleration.

So the diagram should show the net force vector for each molecule.

I don't know if that's a good idea. D H thinks it is already over-complicated.

 

[rcgldr]

One astronaut then jumps onto the other space station. What happens? Do the stresses between the molecules in the second space station increase?

You've just increased the mass of a rotating system, so the stresses increase overall, or locally increase in the vicinity of the single astronaut (and decrease elsewhere), depending if angular velocity or angular momentum was preservered during the transfer.

 

[Dale]

There are gazillions of forces if you want to look at each atom and the forces between them.

Sure, I did a similar analysis in my post 119 to D_H. In such a situation there are also clearly reactive centrifugal forces for many of the atoms, as I described above.

However, that does not detract from the correctness of A.T.'s drawing. Given the partitioning of the system into 3 objects (two astronauts and the space station) then the forces are correct as drawn. You can certainly do a different partitioning, but that does not make A.T.'s diagram incorrect.

So all forces on a molecule must sum to a net force on each molecule equal to the centripetal force it is experiencing due to its centripetal acceleration.

Certainly, nobody has ever said otherwise. We are not saying that the net force is centrifugal, simply that some of the individual forces are.

So the diagram should show the net force vector for each molecule.

Have you never done a free-body diagram before? Net forces are not drawn on a free-body diagram, and you only draw the external forces for each object. The drawing is correct.

 

[Andrew Mason]

Sure, I did a similar analysis in my post 119 to D_H. In such a situation there are also clearly reactive centrifugal forces for many of the atoms, as I described above.

However, that does not detract from the correctness of A.T.'s drawing. Given the partitioning of the system into 3 objects (two astronauts and the space station) then the forces are correct as drawn. You can certainly do a different partitioning, but that does not make A.T.'s diagram incorrect.

Certainly, nobody has ever said otherwise. We are not saying that the net force is centrifugal, simply that some of the individual forces are.

Have you never done a free-body diagram before? Net forces are not drawn on a free-body diagram, and you only draw the external forces for each object. The drawing is correct.

?? But if the bodies are accelerating you have to show the forces on them that cause acceleration. All the molecules in space station are accelerating, along with the astronauts. What forces produce that acceleration?

Eliminate the astronauts. Does that eliminate the acceleration of the matter in the space station? Where is the centrifugal force on the molecules in the space station then? Answer: there is no centrifugal force.

AM

 

[Dale]

?? But if the bodies are accelerating you have to show the forces on them that cause acceleration.

Which A.T. did.

Note that the space station is not accelerating. This can be seen from the fact that the forces on the space station (labeled Frcf on A.T.'s diagram) are balanced.

All the molecules in space station are accelerating, along with the astronauts. What forces produce that acceleration?

Did you not read my post 119 to D_H? I laid that out very carefully.

Eliminate the astronauts. ...

That is a different scenario and therefore a different diagram.

 

[A.T.]

Eliminate the astronauts. (...)

And the space station too. That should finally eliminate those pesky centrifugal reaction forces.

Where is the centrifugal force on the molecules in the space station then?

All over the place. Some molecules of the space station will still experience individual interaction forces from other molecules of the space station, which have a centrifugal component.

 

[Andrew Mason]

Which A.T. did.

Note that the space station is not accelerating. This can be seen from the fact that the forces on the space station (labeled Frcf on A.T.'s diagram) are balanced.

The space station is accelerating.

Every molecule in the space station is accelerating. They are just not all accelerating in the same direction. The centre of mass is not accelerating but the centre of mass is a mathematical abstraction.

You could say the same thing about two billard balls when they are colliding. The centre of mass does not accelerate. But that does not mean there is no acceleration. The physics are very different during the collision than before and after, even though there is absolutely no change to the motion of the centre of mass.

AM

 

[Andrew Mason]

And the space station too. That should finally eliminate those pesky centrifugal reaction forces.

All over the place. Some molecules of the space station will still experience individual interaction forces from other molecules of the space station, which have a centrifugal component.

Reduce the space station to a circular string of molecules. What object of matter is causing this centrifugal "force" on a given molecule? What is exerting it and what is that object exerting the centrifugal force on?

AM

 

[Dale]

Reduce the space station to a circular string of molecules. What object of matter is causing this centrifugal "force" on a given molecule? What is exerting it and what is that object exerting the centrifugal force on?

You still haven't read post 119, have you?

 

[Dale]

The space station is accelerating.

No, it isn't. The net external force on the space station is 0, so by Newton's second law it is not accelerating.

Every molecule in the space station is accelerating. They are just not all accelerating in the same direction. The centre of mass is not accelerating but the centre of mass is a mathematical abstraction.

Everything in physics is a mathematical abstraction, that is hardly a counter-argument. When you use Newton's 2nd law to determine the acceleration of some system the thing which is accelerating is the system's center of mass. Are you attempting to reject Newton's 2nd law since it describes the behavior of a mathematical abstraction?

 

[D H]

But in fact it just a trivial naming convention applicable in a subset of cases involving rotation, where the local reaction to the centripetal force points away from the rotation center. I would not even call it a "concept". The concept here is Newtons 3rd.

I disagree. Then again, I might be being a bit irrational; 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?

For the sake of argument, let's drag it into the mix. For point masses and when viewed from the perspective of an inertial frame, the reaction to a centripetal force is also centripetal while the reaction to a centrifugal force is also centrifugal. I don't see how you can argue it is anything but.

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.

 

[Dale]

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.

 

[Andrew Mason]

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

 

[A.T.]

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.

 

[olivermsun]

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?

 

[Dale]

I have. I and have said I disagree with that analysis.

Sorry, I missed that. What was your disagreement with my analysis in 119?

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.

 

[A.T.]

I truly despise the name "reactionary centrifugal force".

Why make it political?

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?

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

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.

 

[A.T.]

...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?

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.

 

[Andrew Mason]

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

 

[Doc Al]

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.

 

[Dale]

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."