Is the concept of reactive centrifugal force valid?

[A.T.]

This addresses the point that AT made, which I have not yet properly answered, questioning whether Newton's third law is only about accelerations.

No it doesn't. A proper reference supporting your version of Newton's 3rd Law would address it.

Static forces are, by definition, equal and opposite - otherwise they would not be static.

Sounds like the common flawed reasoning, that the two equal and opposite forces are balancing each other.

I think our discussion shows that we really run into problems if we try to apply Newton's third law to static forces.

There are no problems. Static cases are classic examples of Newtons 3rd Law.

It becomes really difficult, if not arbitrary, to select third-law force pairs.

To people interested in problem solving this is not difficult. It is a feature, which makes the Law applicable on different levels of abstraction. It might be a problem to armchair philosophers though.

If we ask, where is the acceleration that is the third law reaction...

The third law is not about acceleration.

The reaction to a centripetal acceleration...

The third law is not about acceleration.

 

[A.T.]

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

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

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

 

[Andrew Mason]

It sounds like you do not understand basic Newtonian mechanics. Newtons third law has no problem with static forces. Newtons third law is used extensively in statics, and it is absolutely required for analyzing the stresses in a beam or structural member.

Are you suggesting that Newtonian mechanics is simple! I think discussion shows that it is far from simple.

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

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

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

AM

 

[Dale]

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

This shows that you have the usual misunderstanding about Newton's 3rd law. If object A is static then all of the forces on A sum to zero, as required by Newton's 1st law. Newton's 3rd law describes the relationship of the force on A due to B and the force on B due to A. You can never sum a 3rd law pair of forces to get a net force because the forces act on different objects. The 1st and 3rd laws say completely different things.

The essence of the third law, it seems to me, is that it establishes the principle that every change in motion gives rise, instantaneously, to offsetting change in motion (ie. of other bodies). ... The reaction to any acceleration is another acceleration ie of another body. That is what the third law says.

This is also not correct, although at least it is an unusual misunderstanding. The third law says nothing about acceleration. It strictly deals with forces. There are many cases where the accelerations due to third-law pairs are unequal, and other cases where they are both zero but still interacting.

 

[rcgldr]

A bit off topic:

It is not that Newton's third law has a problem with static forces. It is just that it does not say anything new about static forces.

That static (and all) forces only exist in equal and opposing pairs?

Wiki article:

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

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

exceptional case for rotation and centrifugal acceleration ...

I think we all agree that rotation does not create a centrifugal acceleration.

What about the previously described case of an object sliding inside a frictionless cylinder that rotates end over end? This is an exceptional case though.

back on topic ...

suggested here that rotation - centripetal acceleration, gives rise (sometimes) to a static centrifugal force.

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

 

[Andrew Mason]

This shows that you have the usual misunderstanding about Newton's 3rd law. If object A is static then all of the forces on A sum to zero, as required by Newton's 1st law. Newton's 3rd law describes the relationship of the force on A due to B and the force on B due to A.

You have to read what I wrote. I think I made this point very clear in my first post.

You can never sum a 3rd law pair of forces to get a net force because the forces act on different objects. The 1st and 3rd laws say completely different things.

I don't think you meant to say that. I think you meant to say that you could not get a single third law pair forces NOT to produce a net force. You can never have a single pair of third law forces that do NOT produce acceleration.

This is also not correct, although at least it is an unusual misunderstanding. The third law says nothing about acceleration. It strictly deals with forces. There are many cases where the accelerations due to third-law pairs are unequal, and other cases where they are both zero but still interacting.

I didn't say that the accelerations are equal and opposite in third law pairs! I said that every change in motion gives rise "to offsetting change in motion (ie. of other bodies)".

AM

 

[A.T.]

You can never have a single pair of third law forces that do NOT produce acceleration.

Acceleration is "produced" by net forces, not by third law force pairs. Most scenarios involve more than one single interaction, so in general you cannot say anything about acceleration, based solely on single pair of third law forces.

 

[D H]

You can never sum a 3rd law pair of forces to get a net force because the forces act on different objects. The 1st and 3rd laws say completely different things.

I don't think you meant to say that. I think you meant to say that you could not get a single third law pair forces NOT to produce a net force. You can never have a single pair of third law forces that do NOT produce acceleration.

I think Dale mean to say exactly what he said, Andrew: What Dale said is exactly right.

You seem to have a fundamental misunderstanding of Newton's laws. A lot of students do, too. That is perhaps because there are several implicit assumptions that underlie Newton's laws. Key amongst these underlying assumptions are

• The change in the motion of some object does not immediately depend on the forces that object exerts on other objects. That object A exerts some force on object B has no direct bearing on object A's motion. All that matters with respect to the immediate change in the motion of object A are the forces exerted by object B (and C and D and ...) on object A.
  
  This is a source of confusion amongst many students. We repeatedly get students asking how anything can move given that forces are equal but opposite. The answer is that the force object A exerts on object B has nothing to do (directly) with object A's motion.
• "Force" is subject to the superposition principle. Newton's second law is a statement about the net force, or the superposition of all of the forces, acting on some body. It says nothing about the response of some body to an individual force acting on the body. Newton's third is a statement about an individual force between a pair of bodies. It too says nothing about the acceleration that results from this force.

 

[Dale]

You have to read what I wrote. I think I made this point very clear in my first post.

Yes, you have been inconsistent in the correct application of the 3rd law. The OP was correct, your more recent posts have not been.

I didn't say that the accelerations are equal and opposite in third law pairs! I said that every change in motion gives rise "to offsetting change in motion (ie. of other bodies)".

I don't understand what you mean by "offsetting change in motion". In any case, there is no mainstream scientific reference that you can produce which supports your interpretation of the 3rd law as not applying to static scenarios, so further discussion of the 3rd law in those terms is speculative and not appropriate for PF.

 

[Andrew Mason]

I think Dale mean to say exactly what he said, Andrew: What Dale said is exactly right.

No it isn't. If the only forces are a single pair of third law forces between A and B then there is one force on A and one on B. Both accelerate! This is the essence of Newton's third law.

You seem to have a fundamental misunderstanding of Newton's laws.

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

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

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

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

A lot of students do, too. That is perhaps because there are several implicit assumptions that underlie Newton's laws. Key amongst these underlying assumptions are

• The change in the motion of some object does not immediately depend on the forces that object exerts on other objects. That object A exerts some force on object B has no direct bearing on object A's motion.

• Well it certainly has a bearing. It has to produce acceleration in B if that is the only force on B. And if B's reaction force on A is the only force on A it has to produce an acceleration in A.
  
  Look. No one here misunderstands Newton's three laws. What we are disagreeing with here is how to analyse a situation because there is a certain depth and subtlety involved. But to maintain a civil and productive discussion we have to actually read what is said.
  
  If you think I am still missing something. Tell me again, how does a single pair of third law forces NOT produce acceleration. Give me an example. I can guarantee you that there will always be acceleration.
  
  

  All that matters with respect to the immediate change in the motion of object A are the forces exerted by object B (and C and D and ...) on object A.

  Exactly. So if the only force on A is the force exerted by B then A must accelerate. That was my point. Again, if you disagree then give an example.
  
  

  This is a source of confusion amongst many students. We repeatedly get students asking how anything can move given that forces are equal but opposite. The answer is that the force object A exerts on object B has nothing to do (directly) with object A's motion.

  The point about the third law is that if body A accelerates, something else in the universe has to accelerate. If you disagree, give me an example of where only one body accelerates!
  
  AM

 

[Doc Al]

No it isn't. If the only forces are a single pair of third law forces between A and B then there is one force on A and one on B. Both accelerate! This is the essence of Newton's third law.

No it isn't. That's a consequence of the 2nd law: Whenever you have an net force on an object, it accelerates. Obviously if only a single force acts on a object you'll have acceleration. So what?

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

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

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

 

[D H]

The point about the third law is that if body A accelerates, something else in the universe has to accelerate. If you disagree, give me an example of where only one body accelerates!

That is a red herring, Andrew. The third law says absolutely nothing about acceleration. Newton's third law is used in plenty of applications where nothing accelerates. Big chunks of civil and mechanical engineering address bodies that are not accelerating in which Newton's third law nonetheless plays a central role. Just because a bridge or building is not accelerating does not mean that no forces act on it. The static loads that act on a bridge, a building, an aircraft, a boat, etc. can make the object in question unsafe.

 

[Andrew Mason]

I don't understand what you mean by "offsetting change in motion".

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

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

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

AM

 

[Andrew Mason]

No it isn't. That's a consequence of the 2nd law: Whenever you have an net force on an object, it accelerates. Obviously if only a single force acts on a object you'll have acceleration. So what?

Ok. The accelerations of A and B derive from the first two laws. But the fact that the acceleration of A has to be accompanied by an acceleration of B (if the only interactions between A and the rest of the universe and between B and the rest of the universe are with each other) is found only in the third law.

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

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

Those examples are deceptive.

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

AM

 

[Andrew Mason]

That is a red herring, Andrew. The third law says absolutely nothing about acceleration. Newton's third law is used in plenty of applications where nothing accelerates. Big chunks of civil and mechanical engineering address bodies that are not accelerating in which Newton's third law nonetheless plays a central role. Just because a bridge or building is not accelerating does not mean that no forces act on it. The static loads that act on a bridge, a building, an aircraft, a boat, etc. can make the object in question unsafe.

I agree. But all I am saying is that if that is all that Newton's third law said, the third law would simply be a corollary to the first two. So the third law is not required to explain static forces. But the third law is not merely a corollary of the first two. You cannot capture the principle of absolute symmetry that is embodied in the third law (and thus derive the law of conservation of momentum) from the first two laws.

AM

 

[Doc Al]

Those examples are deceptive.

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

AM

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

I have no earthly idea why this baffles you.

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

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

 

[Andrew Mason]

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

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

AM

 

[Dale]

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

So then the third law deals with forces and not changes in motion.

All I wanted to say is that a change in the motion of body A requires some change in motion of some body somewhere in the universe. And that change in motion "offsets" the change in motion of A so there is no overall change in "momentum".

This is certainly true for isolated systems, but Newton's laws can also be applied to non-isolated systems where the momentum of the system is not conserved. Of course, if you proceed to large enough scales then you should be able to find an isolated system, but it is not necessary for using Newton's laws.

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

Sure, for an isolated system nobody disagrees that if one part of the system is undergoing centripetal acceleration about the COM then some other part of the system is also undergoing centripetal acceleration about the COM. Acceleration of each part of the system is due to the net force on the part (which is centripetal). However, that in no way implies that there are no centrifugal forces on the part, only that those centrifugal forces are smaller than the centripetal forces.

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

 

[olivermsun]

Sure, for an isolated system nobody disagrees that if one part of the system is undergoing centripetal acceleration about the COM then some other part of the system is also undergoing centripetal acceleration about the COM. Acceleration of each part of the system is due to the net force on the part (which is centripetal). However, that in no way implies that there are no centrifugal forces on the part, only that those centrifugal forces are smaller than the centripetal forces.

To interject in this long and sometimes fascinating discussion if I may -- what makes you say that the centrifugal force is smaller than the centripetal force?

 

[Dale]

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

 

[olivermsun]

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

But if the centripetal acceleration were zero then the centrifugal force would also be zero (not, for example, less than zero). How could they be any different in magnitude?

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

What am I missing here?

 

[Dale]

But if the centripetal acceleration were zero then the centrifugal force would also be zero (not, for example, less than zero).

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

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

Here you are talking about a third-law pair acting on different bodies. Above I am talking about the second-law net force acting on a single body. AM seems to have a confusion about net forces.

 

[olivermsun]

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

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

 

[Andrew Mason]

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

I have no earthly idea why this baffles you.

It is not baffling. My point is that it is wrong, and very misleading and confusing, to use the term centrifugal "force". There is certainly a centrifugal effect. But it is not a force. The effect cannot and does not cause, or tend to cause, any outward acceleration of anything, ever. It is that simple.

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

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

Let's look at a case where I am really exerting an outward force on the sling (to make it simple, there is no rotation): Ball A and Ball B are connected with a rope sling but there is a spring between A and B. As the rope between them is ratcheted shorter, the sling around A is being pulled toward B, and the sling around B is being pulled toward A, compressing the spring.

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

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

AM

 

[Dale]

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

I wasn't limiting it to a two body system, simply an isolated system. It would be a fixed offset, the offset being the net force required for the centripetal acceleration of that part of the system.

 

[Andrew Mason]

So then the third law deals with forces and not changes in motion.

Not according to Newton - see post above #167. He speaks about changes in motion.

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

Sure. An example would be if you put a compressed spring in the middle of the rotating bodies. The sum of the inward tension forces and the outward spring forces would equal the centripetal forces. But if you take away the spring, there is no centrifugal force at all. None.

AM

 

[Dale]

My point is that it is wrong, and very misleading and confusing, to use the term centrifugal "force". There is certainly a centrifugal effect. But it is not a force. The effect cannot and does not cause, or tend to cause, any outward acceleration of anything, ever. It is that simple.

This is incorrect. The "effect" obeys Newton's laws, so it is definitely a force.

Not according to Newton - see post above #167. He speaks about changes in motion.

You missed his important qualifier: "if the bodies are not hindered by any other impediments".

Sure. An example would be if you put a compressed spring in the middle of the rotating bodies. The sum of the inward tension forces and the outward spring forces would equal the centripetal forces.

That outward spring force is a centrifugal force simply because it is pointing outward. That is all "centrifugal" means.

 

[olivermsun]

I wasn't limiting it to a two body system, simply an isolated system. It would be a fixed offset, the offset being the net force required for the centripetal acceleration of that part of the system.

Well, let me limit it to just 2 bodies for the moment to try and get this straight, since I don't really see why this is so different from the application of the 3rd law in the rope-and-cart example.

Suppose you have the regular old problem, two balls connected by a rope, twirling around a common center of mass. Pick any old inertial reference frame, e.g., the magic one in which the COM is not moving. Now it seems to me that the entire centripetal acceleration Fr on Ball 1 is exerted by Ball 2 through tension on the string. Conversely, Ball 2 must pull back on the string with exactly the same force, just like the cart pulling back on the rope in the earlier example. This pull is outward along the radius, hence this is the "reactive" centrifugal force, and it must be exactly equal to -Fr or else the COM would start to move.

So where is the constant offset?

 

[olivermsun]

My point is that it is wrong, and very misleading and confusing, to use the term centrifugal "force". There is certainly a centrifugal effect. But it is not a force. The effect cannot and does not cause, or tend to cause, any outward acceleration of anything, ever. It is that simple.

Gravity pushes me down toward the ground. I infer that the ground pushes back, more or less, since I am not currently sinking into the ground. I don't happen to believe that it misleading and wrong to say the ground is exerting an upward "force" on me, even if the force only exists as long as I am here to push down on it (it won't suddenly fling other, lighter, objects into space, for example).

Let's look at a case where I am really exerting an outward force on the sling (to make it simple, there is no rotation): Ball A and Ball B are connected with a rope sling but there is a spring between A and B. As the rope between them is ratcheted shorter, the sling around A is being pulled toward B, and the sling around B is being pulled toward A, compressing the spring.

Ok, so you are saying A and B are held apart by a spring, but a rope is used to impose an additional force which compresses the spring and moves A and B closer together. So far, so good.

I let go of the rope there would be actual outward acceleration for a moment while the spring expanded against the balls.

But in the case of circular rotation, there is no such outward force.

Of course there are outward forces in the rotating system. The rope or whatever is holding the balls together is pulling with the amount of inward, "centripetal" force (through tension) required to maintain the circular motion of each ball. An outward, "centrifugal" force (as viewed from the inertial, non rotating frame) is exerted by each ball on the line and keeps the line taut. Otherwise the balls would just collapse inward and the line would crumple. It also happens that the outward force of one ball also equals the inward force on the other, since they're connected by the same line, after all.

 

[Doc Al]

Try this: Imagine the ball on the end of rope being divided into an inner and an outer half (the inner half being the half connected to the rope). The ball is twirling around. Analyze the forces on the inner half. There are two: The force of the rope pulling inward (a centripetal force) and the force of the outer half pulling outward (a 'centrifugal' force). The net force on that inner half must equal its mass times its centripetal acceleration.

I think this difference between the net force on a body causing centripetal acceleration and the actual centripetal force acting on it is the 'offset' that Dale was talking about.

 

[olivermsun]

Try this: Imagine the ball on the end of rope being divided into an inner and an outer half (the inner half being the half connected to the rope).

Great example.

One could also imagine a Ferris wheel or a roller coaster or something, where there's the car which goes in a circle, and then there's you in the car (being pushed inward by the seat or the floor of the the car). Or a merry-go-round where there's the horsey and then there's you hanging on to the horsey... (wait, that example has been done already in this thread, right?)

 

[Doc Al]

One could also imagine a Ferris wheel or a roller coaster or something, where there's the car which goes in a circle, and then there's you in the car (being pushed inward by the seat or the floor of the the car). Or a merry-go-round where there's the horsey and then there's you hanging on to the horsey... (wait, that example has been done already in this thread, right?)

Exactly. The merry-go-round and rotating spaceship examples have been discussed, but their analyses keep getting ignored (by some) in favor of less interesting examples.

 

[Dale]

Well, let me limit it to just 2 bodies for the moment to try and get this straight, since I don't really see why this is so different from the application of the 3rd law in the rope-and-cart example.

It is no different. The third law is the same regardless of the situation. Andrew Mason is incorrect in his assertions that there is something different in rotational motion which violates or modifies Newton's laws.

Suppose you have the regular old problem, two balls connected by a rope, twirling around a common center of mass. Pick any old inertial reference frame, e.g., the magic one in which the COM is not moving. Now it seems to me that the entire centripetal acceleration Fr on Ball 1 is exerted by Ball 2 through tension on the string. Conversely, Ball 2 must pull back on the string with exactly the same force, just like the cart pulling back on the rope in the earlier example. This pull is outward along the radius, hence this is the "reactive" centrifugal force, and it must be exactly equal to -Fr or else the COM would start to move.

So where is the constant offset?

Again, the "offset" is for the 2nd law on a single body, not the 3rd law between two bodies. If a single body undergoing uniform circular motion is experiencing a centripetal force Fi (inwards) and a centrifugal force Fo (outwards) then by Newton's 2nd law Fi-Fo = ma where a is directed inwards. So the offset between Fi and Fo is given by Fi = Fo+ma which follows directly from the 2nd law.

Are you clear on the fact that the forces involved in uniform circular motion obey Newton's laws?

 

[Andrew Mason]

Gravity pushes me down toward the ground. I infer that the ground pushes back, more or less, since I am not currently sinking into the ground. I don't happen to believe that it misleading and wrong to say the ground is exerting an upward "force" on me, even if the force only exists as long as I am here to push down on it (it won't suddenly fling other, lighter, objects into space, for example).

There is absolutely nothing wrong with calling the normal force that the Earth exerts on you an outward or force. You can call it centrifugal if you like but that usually refers to rotation.

So long as you understand that the "centrifugal" or normal force is NOT in any way shape or form a reaction to the centripetal acceleration that you experience due to the Earth's rotation, you can call it a centrifugal force. The centripetal acceleration is provided by gravity. The reaction to the centripetal force on you is the attractive gravitational force that you exert on the earth.AM

 

[Andrew Mason]

Ok, so you are saying A and B are held apart by a spring, but a rope is used to impose an additional force which compresses the spring and moves A and B closer together. So far, so good.


Of course there are outward forces in the rotating system. The rope or whatever is holding the balls together is pulling with the amount of inward, "centripetal" force (through tension) required to maintain the circular motion of each ball. An outward, "centrifugal" force (as viewed from the inertial, non rotating frame) is exerted by each ball on the line and keeps the line taut. Otherwise the balls would just collapse inward and the line would crumple. It also happens that the outward force of one ball also equals the inward force on the other, since they're connected by the same line, after all.

Think of the rope as a spring. The spring stretches to provide the necessary centripetal force (F = -kx). The only way that spring can exert an inward centripetal force is if it is also pulling inward on the other ball with the same force. So the spring is pulling inward.

[Note: We assume the rope is of negligible mass (otherwise you have to factor in the centripetal force on each part of the rope, which complicates things), the net force on each part of the rope is 0. So the net force is between the ends.]

There is nothing pushing outward. The spring stretches in order to supply the centripetal force that is required to maintain rotational motion. If it is not strong enough, the inertial motion of the balls will carry them away from the centre. They would not crumple to the centre.

By the way, you are talking about the centrifugal force that everyone agrees is fictitious, not the reactive centrifugal force.

AM

 

[Andrew Mason]

It is no different. The third law is the same regardless of the situation.

Agreed.

Andrew Mason is incorrect in his assertions that there is something different in rotational motion which violates or modifies Newton's laws.

I said that?! I certainly assert that there is something about rotational motion that is different than linear motion or linear acceleration: ie acceleration that is constantly changing direction. But it does not violate or require modification to the third law to explain.

Again, the "offset" is for the 2nd law on a single body, not the 3rd law between two bodies. If a single body undergoing uniform circular motion is experiencing a centripetal force Fi (inwards) and a centrifugal force Fo (outwards) then by Newton's 2nd law Fi-Fo = ma where a is directed inwards. So the offset between Fi and Fo is given by Fi = Fo+ma which follows directly from the 2nd law.

Are you clear on the fact that the forces involved in uniform circular motion obey Newton's laws?

You have to be careful here. You may confuse people into making the same mistake that you incorrectly accused me of making.

There is no centrifugal force that has to be subtracted here to determine its centripetal acceleration. That, in fact, is how you prove there is no centrifugal force on the body - by measuring the centripetal force and showing that it is equal to its centripetal acceleration times the mass of the body.

Rather the "centrifugal reaction force" is (said to be) on the matter that is in contact with the rotating mass. We all agree that there is a reaction force to the centripetal acceleration. I am just saying it is the force of one rotating mass on the other and both are inward. The others, except for DH it seems, says that there is an outward component of force that is a reaction to the force of one rotating body on the other.

AM

 

[A.T.]

There is certainly a centrifugal effect.

"Centrifugal effect" is still just vague gibberish.

But it is not a force.

Calling forces "effects" doesn't change physics. It is just pointless obfuscation.

 

[olivermsun]

Again, the "offset" is for the 2nd law on a single body, not the 3rd law between two bodies.

I guess it wasn't completely clear to me due to the title and the ongoing discussion about finding the 3rd law "reaction" to the centripetal force. But thanks.

Are you clear on the fact that the forces involved in uniform circular motion obey Newton's laws?

No need to condescend. It's the shifting language and incessant bickering about definitions that confuses me, not Newtonian mechanics properly expressed.

By the way, you are talking about the centrifugal force that everyone agrees is fictitious, not the reactive centrifugal force.

Actually, I was talking about the reactive centrifugal force. But thanks anyway.

 

[Dale]

No need to condescend. It's the shifting language and incessant bickering about definitions that confuses me, not Newtonian mechanics properly expressed.

I was not intending to be condescending, but only making sure that you were clear on that point since it is extremely important. It appears to be a point of confusion with others in this thread.

 

[Dale]

I said that?! I certainly assert that there is something about rotational motion that is different than linear motion or linear acceleration: ie acceleration that is constantly changing direction. But it does not violate or require modification to the third law to explain.

Yes, when you say that a centrifugal force is an effect and not a force you are saying that Newton's laws don't apply or that they need modifications.

There is no centrifugal force that has to be subtracted here to determine its centripetal acceleration. That, in fact, is how you prove there is no centrifugal force on the body - by measuring the centripetal force and showing that it is equal to its centripetal acceleration times the mass of the body.

This is a good suggestion. Let's take the following example: In the absence of gravity, two balls are attached to two ropes as shown. Each ball has a mass of 1 kg. Each rope is massless with a length of 1 m and a gauge attached to measure the tension in the rope. The whole assembly is swung about the left end of the rope in a uniform circular motion at a rate of 1 revolution per second. Using Newton's laws of motion find:
1) the acceleration of masses A and B
2) the net forces acting on masses A and B and their directions
3) the measured tensions in gauges 1 and 2
4) all forces acting on masses A and B and their directions

 

[Andrew Mason]

The whole assembly is swung about the left end of the rope in a uniform circular motion at a rate of 1 revolution per second.

I think that means that the masses are not rotating about their centre of mass. If so, you have to add in the acceleration and mass or moment of inertia of whatever it is that it is tied to. So we need more information. You can avoid this problem by having the two tethered balls rotate in space about their centre of mass.

To demonstrate the third law you need objects that are "not hindered by any other impediments".

AM

 

[Dale]

I think that means that the masses are not rotating about their centre of mass. If so, you have to add in the acceleration and mass or moment of inertia of whatever it is that it is tied to. So we need more information.

No further information is needed. The system is not isolated, but Newton's laws still apply and can be used without modification or additional information.

 

[Andrew Mason]

No further information is needed. The system is not isolated, but Newton's laws still apply and can be used without modification or additional information.

Of course Newton's laws still apply. But you need to know what other mass is involved. A mass cannot just rotate in space. It can only rotate about a centre of mass of a rotating system. The rope cannot rotate all by itself.

The common example of a centrifugal reactive force that engineers use is a car turning on a road. The centripetal force supplied by the road results in a reaction force away from the car, it is said. So they conclude that the reaction force is a centrifugal force. It appears that way but it isn't. This is a very subtle and understandable mistake.

The reason: it is not the road that is supplying the centripetal force - it is the earth. If the road was not connected to the earth, the car would not turn. So the reaction force to the centripetal force of the Earth on the car is a force of the car on the earth. To see whether it is centripetal or centrifugal you have to determine the centre of mass of the car/earth system and determine how the Earth centre of mass moves in response to the car's acceleration. It always moves toward the centre of mass of the car/earth system.

Of course it is impossible to see or measure because of the difference in mass between the Earth and car. But we don't have to do that if we accept Newton's third law. We just do the physics.

Anyone who thinks that classical mechanics is simple or easy has never tried to work out the physics of a precessing and nutating spinning top. It is difficult. It is counter-intuitive.

So what we have to do is put our intuition aside, quit arguing with words and apply the laws strictly and correctly.

AM

 

[rcgldr]

The common example of a centrifugal reactive force that engineers use is a car turning on a road. ... To see whether it is centripetal or centrifugal you have to determine the centre of mass of the car/earth system and determine how the Earth centre of mass moves in response to the car's acceleration.

Much of the reaction at the surface of the Earth would be angular (versus linear) acceleration where the surface of the Earth would accelerate "outwards" from the center of mass.

This is getting off-topic. Earlier in this thread, there have been examples where there is is a reactive centrifugal force and others were there is not. Just because there are cases where there isn't reactive centrifugal force doesn't invalidate the usage of that term to describe the cases where it does apply.

 

[A.T.]

To demonstrate the third law you need objects that are "not hindered by any other impediments".

Not true. You could use soft objects in a static case to demonstrate the deformation on both, as a result of the equal opposite forces. My space station can be covered with soft foam on the inside, that will deform and demonstrate the centrifugal force exerted on it by the astronaut.

Acceleration is just easier to visualize. That's why the special case "no other impediments" is used as an example. But you are stuck at this simplistic special case and fail to understand the general meaning of the law.

We just do the physics.

Well, you don't. Any high school student could "just do the physics" and solve the problem posed by DaleSpam without any further information. But you apparently can't. You demand more information to answer questions that were never asked.

So what we have to do is put our intuition aside, quit arguing with words and apply the laws strictly and correctly.

Yes that is what YOU have to do. Note that DaleSpam did not ask you about the names of all the forces. You can just name them F₁, F₂ ... F_n. But instead of "just do the physics" you continue to argue with words.

 

[A.T.]

To see whether it is centripetal or centrifugal you have to ... acceleration...

You have obviously your private definition of "centripetal" & "centrifugal" in respect to forces, and that's fine. But the more common definition, referred to in Wikipedia, doesn't depend on acceleration. It depends only on:

- center of rotation
- point of application of the individual force (not some net force!)
- direction of the individual force (not some net force!)

Centripetal : Force direction is the same as the direction from the point of force application to the center of rotation.
Centrifugal : Force direction is the same as the direction from the center of rotation to the point of force application, or center of rotation and point of force application are the same point.

There is no point in arguing which definition is correct. Definitions are neither false not correct. It is just a convention. You should just be aware that many people use the convention above, so you don't get confused when they use those terms.

 

[Dale]

Of course Newton's laws still apply. But you need to know what other mass is involved.

No, you don't. There obviously is another mass involved since the system is not isolated, but none of the details of that other mass affect the answers to the questions asked. The problem is fully and completely specified, no additional information is required.

So what we have to do is put our intuition aside, quit arguing with words and apply the laws strictly and correctly.

I agree completely. If you cannot apply the laws strictly and correctly to this freshman-level problem, then perhaps you should listen to and learn from those who can.

I encourage you to work the problem yourself before peeking at the answer:

Spoiler

1)For uniform circular motion the acceleration is r\omega^2 where r is the radius and w is the angular frequency. From the problem \omega=2\pi s^{-1} = 6.28 s^{-1}. So
a_A = (1 m)(6.28 s^{-1})^2 = 39.5 m/s^2 \text{inwards}
a_B = (2 m)(6.28 s^{-1})^2 = 79.0 m/s^2 \text{inwards}

2) From Newton's 2nd law F_{net} = ma. So
F_{A\,net}=(1 kg)(39.5 m/s^2)=39.5 N \text{inwards}
F_{B\,net}=(1 kg)(79.0 m/s^2)=79.0 N \text{inwards}

3) The force from rope 2 on mass B F_{2B} is the only force acting on B, so
F_{2B}=F_{B\,net}=79.0 N \text{inwards}
By Newton's 3rd law the force from mass B on rope 2 F_{B2}=-F_{2B} so the tension in rope 2 is
T_2=|F_{B2}|=79.0 N

Since the rope is massless, by Newton's 2nd law F_{A2}+F_{B2}=ma=0 so by Newton's 3rd law F_{2A}=-F_{A2}=F_{B2}=-F_{2B}=79.0 N \text{outwards}. The forces on mass A are the force from rope 1 and the force from rope 2 F_{1A}+F_{2A}=F_{A\,net}. So F_{1A}=F_{A\,net}-F_{2A}=(39.5 N \text{inwards}) - (79.0 N \text{outwards}) = 118.5 N \text{inwards}. By Newton's 3rd law the force from mass A on rope 1 F_{A1}=-F_{1A} so the tension in rope 2 is
T_1=|F_{A1}|=118.5 N

4) So the forces acting on B are:
F_{2B}=79.0 N \text{inwards}

And the forces acting on A are:
F_{2A}=79.0 N \text{outwards}
F_{1A}=118.5 N \text{inwards}

 

[Andrew Mason]

Much of the reaction at the surface of the Earth would be angular (versus linear) acceleration where the surface of the Earth would accelerate "outwards" from the center of mass.

Lets use centripetal and centrifugal because that is the real issue here. The question is not the direction that the surface moves. It is the direction that the Earth moves as required by Newton's third law.

That direction in relation to the centre of rotation determines whether the reaction to the centripetal force on the car is centripetal or centrifugal. That is the problem that we seem to be having here. It is not a problem in the simplest case where everyone agrees that the reaction to a centripetal force/acceleration is another centripetal force/acceleration eg: gravitational orbit.

Let's say that the road is a huge piece of steel plate mounted on a sheet of ice. As I drive on the plate in a circle, the centre of mass of the steel plate prescribes a rotation. The car and the centre of mass of the steel plate both rotate about the centre of mass of the car/plate system which remains fixed. As measured in the inertial frame of reference of the system centre of mass, the reaction force of the car on the plate is the centripetal force on the plate. That is all I am saying.

This is getting off-topic. Earlier in this thread, there have been examples where there is is a reactive centrifugal force and others were there is not. Just because there are cases where there isn't reactive centrifugal force doesn't invalidate the usage of that term to describe the cases where it does apply.

I say simply that the third law reaction to a centripetal force causing centripetal acceleration is always an equal and opposite centripetal force causing centripetal acceleration as measured in an inertial frame of reference. I would point out that when something is rotating on or around the earth, the centre of mass of the Earth is not an inertial frame of reference.

AM

 

[olivermsun]

It is not a problem in the simplest case where everyone agrees that the reaction to a centripetal force/acceleration is another centripetal force/acceleration eg: gravitational orbit.

Let's say that the road is a huge piece of steel plate mounted on a sheet of ice. As I drive on the plate in a circle, the centre of mass of the steel plate prescribes a rotation.

What's the essential difference between this and the other examples that have already posed, e.g., gravitational orbits, masses connected by strings, merry-go-rounds, etc.?

 

[Andrew Mason]

No, you don't. There obviously is another mass involved since the system is not isolated, but none of the details of that other mass affect the answers to the questions asked. The problem is fully and completely specified, no additional information is required.

The whole point of this exercise is to determine the magnitude and direction of the reaction force to the centripetal force. The reaction to the centripetal force is not on that rope. It is on what is connected to the "fixed" end which, by the way, cannot possibly be "fixed" to an inertial reference frame. If you refuse to tell us what it is fixed to so we can choose an inertial reference frame you cannot analyse the problem. End of story.

If you cannot apply the laws strictly and correctly to this freshman-level problem, then perhaps you should listen to and learn from those who can.

I would not put it in such a condescending fashion but I would suggest that you apply that to yourself. Some things that appear to be "freshman-level problems" raise profound issues. Feynman's freshman level lectures will be studied and discussed for a long time.

I encourage you to work the problem yourself before peeking at the answer:

You are measuring the forces in from a non-inertial frame and you are not including the other half of the rotating system (ie the Earth or what ever it is that you won't say the rope is connected to).

So let's do it right. We have two such ropes and masses and we let them rotate about the centre of mass. We measure all forces in the non-rotating inertial reference frame of the centre of mass/rotation. We will call the balls A1, B1 and A2, B2 and the respective ropes the same. Here is my analysis:

1)Accelerations:
a_{A1} = (1 m)(6.28 s^{-1})^2 = 39.5 m/s^2 \text{centripetal}
a_{B1} = (2 m)(6.28 s^{-1})^2 = 79.0 m/s^2 \text{centripetal}
a_{A2} = (1 m)(6.28 s^{-1})^2 = 39.5 m/s^2 \text{centripetal}
a_{B2} = (2 m)(6.28 s^{-1})^2 = 79.0 m/s^2 \text{centripetal}2) Forces:
F_{A1}=(1 kg)(39.5 m/s^2)=39.5 N \text{centripetal}
F_{B1}=(1 kg)(79.0 m/s^2)=79.0 N \text{centripetal}
F_{A2}=(1 kg)(39.5 m/s^2)=39.5 N \text{centripetal}
F_{B2}=(1 kg)(79.0 m/s^2)=79.0 N \text{centripetal}

3) The rope tensions supply the centripetal forces. So the tension in ropes A1 and in A2 is the sum of these centripetal forces on masses A1 and B1: 118.5 N. The tension in ropes B1 and B2 provide only the centripetal force on masses B1 and B2 respectively: 79N.

The force acting on A1 is equal and opposite to the force acting on A2 and is the difference in rope tensions: 39.5N. Both are centripetal.

Similarly the force acting on B1 is equal and opposite to the force on B2 and is 79 N. Both are centripetal.

Those are the action and reaction pairs and they are the only forces.

AM