Is the concept of reactive centrifugal force valid?

[Doc Al]

Then you admit that the mass of the object that the rope attaches is relevant, because the acceleration of that object due to this force is determined by its mass. a = f/m. Second law. Good. This is progress.

The mass is irrelevant in Dale's scenario since the end of the rope attached to the center of rotation.

If the rope is tied to a 10 kg mass, let's call it C, would you not agree that C would rotate about the centre of mass of this system?

What happened to the fixed pivot? You're changing the scenario once again. Come on.

Would you not agree then that the force on on C would be a centripetal one? Where is the centrifugal force?

1. You are probably (who knows?) using a different definition of 'centrifugal' than others are in this thread. (I just read your definitions.)
2. I'll be glad to consider alternative scenarios once the given ones have been adequately agreed upon.

 

[A.T.]

This a ludicrously confused 'definition'. In the first part it seems to treat centrifugal force as a pseudoforce (what else can they mean by it saying it is 'not a true force'). Then they go on to state how the 'ball is said to exert a centrifugal force on the string'. Laughable.

Yeah: It acts on the string but balances the centripetal force on the ball. But hey, it's from a howstuffworks.com! Much better than engineering books that don't count as references for Andrew.

This is better, but is not the meaning of 'centrifugal force' used in this thread.

Better, but not much better: "An object traveling in a circle behaves as if it is experiencing an outward force." So the centrifugal pseudo force acts in the inertial frame, where the object is traveling in a circle?

 

[A.T.]

Just for fun

Astronauts exerting centrifugal reaction forces on their space ship:

http://www.youtube.com/watch?v=UpiC-KbtLYI#t=12s

It seems that it would have to rotate faster than that (see reflection at 0:20) in order to generate 1g. But by running with the rotation you can increase the centrifugal reaction forces.

 

[Andrew Mason]

People who don't know how to apply Newton's laws to homework-style questions fail freshman physics. You don't know how to apply Newton's laws to homework-style questions.

I am trying to get a physics professor who has written a leading text on classical mechanics to opine on this. I will let him grade my analysis, but thanks. No one can argue that you mince words.

Here is another example of your misunderstanding of Newton's laws. If X exerts a force on Y then the reaction is the force that Y exerts on X. The only force on B is exerted by rope 2, so the reaction force from B is the force on rope 2 exerted by B.

You are assuming that A is pulling on the rope. It isn't. How can it possibly pull on the rope if it is accelerating away from B more slowly than B is accelerating toward A? Simple question? How is that possible?

Btw, I am also still waiting for the two mainstream scientific references. One that supports your claim that Newton's 3rd law doesn't apply to non-isolated systems, and another that supports your claim that Newton's 3rd law doesn't apply to "massless" ropes.

I have never made either claim. Such claims would be false, of course. Newton's 3d law applies to all interactions. Otherwise, the most fundamental law we know about, the conservation of momentum, would be violated. No one is suggesting that.

Newton's law applies to non-isolated systems. But unless you know the motion of the centre of mass how do you determine an inertial frame of reference in which to measure the forces?

Newton's third law applies to massless ropes provided the frame of reference you are measuring the forces in is the same for the forces at both ends of the rope. You are not doing that. You think you are, but you really aren't. That issue will be cleared up shortly.

AM

 

[Andrew Mason]

Astronauts exerting centrifugal reaction forces on their space ship:

http://www.youtube.com/watch?v=UpiC-KbtLYI#t=12s

It seems that it would have to rotate faster than that (see reflection at 0:20) in order to generate 1g. But by running with the rotation you can increase the centrifugal reaction forces.

Are you seriously suggesting that this "centrifugal force" is a real force? (Kubrick took a lot of artistic license in portraying space travel but he did not change physics that much).

By running against the rotation you could become weightless. Is that because the centrifugal and centripetal forces balance or is it because there is no more rotation?

AM

 

[Doc Al]

Are you seriously suggesting that this "centrifugal force" is a real force?

Are you seriously suggesting that the surface of the rotating ship doesn't exert a real force against the feet of the passenger?

 

[RonL]

This is similar to Newton's bucket: the surface of a liquid in a bucket will change (it will form a parabaloid shape) if you hang the bucket from a rope, twist the rope and then let it spin.

What happens in your example, and in Newton's bucket, is that the forces between the molecules of liquid are not strong enough to provide the centripetal acceleration needed to keep the water moving in a circle as the wafer's/bucket's spin increases. In the wafer's frame of reference, the liquid will move outward as if there is a force pushing it out. But it is just the tendency of the water molecules to move in a straight line due to inertia. In the bucket case, water molecules build up on the inside surface of the bucket due to this inertial effect.

AM

Thanks for your answer Andrew,
There is much more going on than your answer would suggest, but I feel the example is a bit of an intrusion to the thread, so I'll leave it alone.
Should anyone up to speed on wafer coating physics see anything useful in the example, maybe the action/reaction can represent a clear and intuitive example of what pushes and pulls the resist (in my mind it has motion in four directions, all as it moves to the edge of the wafer).

I'll get back to my observation post and see if I can learn some more.:shy:

Ron

 

[Andrew Mason]

Thanks for your answer Andrew,
There is much more going on than your answer would suggest, but I feel the example is a bit of an intrusion to the thread, so I'll leave it alone.
Should anyone up to speed on wafer coating physics see anything useful in the example, maybe the action/reaction can represent a clear and intuitive example of what pushes and pulls the resist (in my mind it has motion in four directions, all as it moves to the edge of the wafer).

I'll get back to my observation post and see if I can learn some more.:shy:

Ron

It is a bit complicated. What pushes or pulls the resist (which I gather is a liquid) are the friction forces between the resist and the wafer as well as the forces between the liquid molecules.

These friction forces apply to the liquid at the point where the resist first contacts the wafer and tend to drag the molecules with the wafer in a tangential path. But as soon as that happens, the wafer accelerates away from that tangential path (ie. toward the centre) and tries to pull the resist with it. As the rotational speed of the wafer increases, the tangential speed of the resist increases. But, more important, the friction forces are not strong enough to provide the needed centripetal acceleration. The molecules' inertia keeps them moving in a tangential direction which means that they keep going in whatever direction they were traveling when the friction forces failed, and they leave the wafer. In effect, they are thrown off the wafer. It is not a force that causes them to leave. It is the insufficiency of the force that is required to make them stay that does it.

AM

 

[Andrew Mason]

Are you seriously suggesting that the surface of the rotating ship doesn't exert a real force against the feet of the passenger?

No. Of course it does. The passenger is undergoing centripetal acceleration. We went through this about 200 posts ago. That does not mean that he exerts a centrifugal reaction force on the space station. The reaction force causes the centre of mass of the space station to accelerate toward the centre of rotation. So the reaction force is a centripetal force as well.

AM

 

[Dale]

You are assuming that A is pulling on the rope.

If the rope is pulling on B then B is pulling on the rope in the opposite direction, that is required by Newton's 3rd law. No knowledge of A is necessary although in this case it is true that A is pulling on the rope.

It isn't. How can it possibly pull on the rope if it is accelerating away from B more slowly than B is accelerating toward A? Simple question? How is that possible?

The distance is what is important for determining if a rope is under tension, not the relative acceleration.

There is no requirement that Newton's laws only be applied to isolated systems where all of the forces are internal.

There is if you want to analyse third law pairs of forces.

The rope can't exert a reaction force unless it has mass.

Btw, I am also still waiting for the two mainstream scientific references. One that supports your claim that Newton's 3rd law doesn't apply to non-isolated systems, and another that supports your claim that Newton's 3rd law doesn't apply to "massless" ropes.

I have never made either claim.

Such claims would be false, of course. Newton's 3d law applies to all interactions. Otherwise, the most fundamental law we know about, the conservation of momentum, would be violated. No one is suggesting that.

Newton's law applies to non-isolated systems. ...

Newton's third law applies to massless ropes ...

I am glad that you have changed your mind about these important issues. Given that, do you now agree that my analysis is correct?

 

[Andrew Mason]

If the rope is pulling on B then B is pulling on the rope in the opposite direction, that is required by Newton's 3rd law. No knowledge of A is necessary although in this case it is true that A is pulling on the rope.

What B is "pulling" on depends on what the rope is connected to. If it is not connected to any mass he cannot pull at all. The physics says he must pull on a mass whose centre of mass is on the opposite side of the centre of rotation. That is what B is pulling on. A is not pulling on B. It just appears that way to the non-inertial observer who does not realize that, in fact, B is accelerating toward A.

The relative speed is what is important for determining if a rope is under tension, not the relative acceleration.

My instinct says that relative velocity in the direction of the acceleration is the issue, is it not?

During any given time interval dt, the relative velocity in the direction of the acceleration (ie perpendicular to their tangential velocity) is nil. Perhaps you disagree. I may not have thought that through completely but it seems right.

I am glad that you have changed your mind about these important issues. Given that, do you now agree that my analysis is correct?

Let's see what your answer is to the above first. You may persuade yourself that there is at least an interesting aspect to this that we may have overlooked.

AM

 

[Andrew Mason]

I have no idea what you're talking about. We are viewing things from an inertial frame, so we can apply Newton's 2nd law without modification.

Ok. So if B is accelerating toward A, ie in a direction in which they have no initial relative velocity, how does A move way from B to put tension on the rope so as to have B pull it toward B?

One difference: In Dale's scenario the two masses have different accelerations, but in yours they would have the same acceleration. No problem: Just apply Newton's 2nd law.

Let's do that. 2nd law: a = m\omega^2r => a_B > a_A (toward centre). Do you agree that B is accelerating more rapidly toward A than A is accelerating away from B? In order for A to put tension on the rope, does it not have to stretch the rope a tad. Does that not mean it has to at some point accelerate just a tad in the direction away from B?

AM

 

[Andrew Mason]

Since you keep changing the scenario instead of analyzing the one given, perhaps I mixed up which one you were talking about. Are you talking about your hollow rope scenario? Or your four mass version? Link to the post you are asking about.

The hollow rope scenario, post #252:

There is no force between A and B in the hollow rope scenario.

The tension on the first rope would be the same as in Dale's scenario if B didn't exist in Dale's scenario. It would be 39N + 0.

I asked how Dale's scenario differed from this one in terms of the motions and forces. You have not provided an answer.

I am still waiting for an answer.

This a ludicrously confused 'definition'. In the first part it seems to treat centrifugal force as a pseudoforce (what else can they mean by it saying it is 'not a true force'). Then they go on to state how the 'ball is said to exert a centrifugal force on the string'. Laughable.

Don't blame me. I didn't write them. I tried to find definitions from someone who agreed that such forces exist, but I couldn't find any.

Ok. I will take a first crack at my own definitions:

Centrifugal "force": A force that is applied to a body rotating about a central inertial point that tends to accelerate the center of mass of that body outward or away from the central inertial point.

Centripetal force: A force that is applied to a body that tends to accelerate the center of mass of that body inward or toward a central inertial point causing the mass to prescribe rotational motion about that central inertial point.

How's that?

AM

 

[A.T.]

Astronauts exerting centrifugal reaction forces on their space ship

Are you seriously suggesting that this "centrifugal force" is a real force?

Any force exerted by some object on some other object is a real force.

Your problem of course is that you cannot possibly understand the difference between "real forces" and "inertial forces", because you don't understand:
- Newtons 3rd Law
- The concept of reference frames

 

[Dale]

What B is "pulling" on depends on what the rope is connected to.

Not as far as the 3rd law is concerned. For the 3rd law if the rope is pulling on B then B is pulling on the rope with an equal and opposite force. No other considerations are involved.

Since the only force acting on B is from the rope, the only reaction force from B is the one on the rope.

My instinct says that relative velocity in the direction of the acceleration is the issue, is it not?

Sorry, while you were writing your post I had made an edit to mine. I realized that the thing which determines the tension is the distance: if it is greater than the unstrained length then there is tension.

Are you ready to agree to my analysis or produce one of your own?

 

[Doc Al]

No. Of course it does. The passenger is undergoing centripetal acceleration. We went through this about 200 posts ago. That does not mean that he exerts a centrifugal reaction force on the space station. The reaction force causes the centre of mass of the space station to accelerate toward the centre of rotation. So the reaction force is a centripetal force as well.

Again you seem confused about Newton's 3rd law. If, as you seem to agree, the space station exerts an inward force on the passengers feet, then those feet must exert an outward force on the space station. It's that simple.

 

[Doc Al]

Ok. So if B is accelerating toward A, ie in a direction in which they have no initial relative velocity, how does A move way from B to put tension on the rope so as to have B pull it toward B?

Let's do that. 2nd law: a = m\omega^2r => a_B > a_A (toward centre). Do you agree that B is accelerating more rapidly toward A than A is accelerating away from B? In order for A to put tension on the rope, does it not have to stretch the rope a tad. Does that not mean it has to at some point accelerate just a tad in the direction away from B?

Sure, in order for an actual rope to exert tension it must stretch a bit. But we already have the final configuration of the ropes and masses--any needed stretching is already done.

It seems that you are confused about how circular motion and the resulting acceleration works.

 

[Doc Al]

The hollow rope scenario, post #252:

There is no force between A and B in the hollow rope scenario.

Right. They are essentially unconnected.

The tension on the first rope would be the same as in Dale's scenario if B didn't exist in Dale's scenario. It would be 39N + 0.

OK. Of course in Dale's scenario B is connected to A.

I asked how Dale's scenario differed from this one in terms of the motions and forces. You have not provided an answer.

You just said it yourself! The tension in the rope connecting to A is different in Dale's scenario.
​

I am still waiting for an answer.

Answered.

Don't blame me. I didn't write them. I tried to find definitions from someone who agreed that such forces exist, but I couldn't find any.

I wanted your definitions, not something you copied from a website. How are you using the term in this thread? If you're not sure, why post?

Ok. I will take a first crack at my own definitions:

Centrifugal "force": A force that is applied to a body rotating about a central inertial point that tends to accelerate the center of mass of that body outward or away from the central inertial point.

Quite vague. No mention of pseudoforce, so that 'definition' would also apply to the dreaded 'reactive centrifugal force'. Doesn't seem consistent with your posts. (Seems that you're saying here that 'centrifugal forces' really do exist.)

Centripetal force: A force that is applied to a body that tends to accelerate the center of mass of that body inward or toward a central inertial point causing the mass to prescribe rotational motion about that central inertial point.

As long as you realize that it's the net force that causes the centripetal acceleration.

 

[Dadface]

No, the details of the fixing do not matter. As long as you don't change any of the specified conditions the solution to the problem will not change.

That is correct.

The rotation is about the end of the rope. That was specified in the problem. However you want to apply the force to the end of the rope is up to you, but you cannot change the problem without, uh, changing the problem. If you do not change the problem then you will not alter any of the answers that I posted, regardless of the nature of the set up "to the left" of the end of the rope.

Please feel free to demonstrate that equation. I do not think that your claim here is correct unless you change the problem.

If the center of rotation isn't at the end of the rope then you have changed the problem. There are an infinite number of ways to achieve that, and which one you pick is not important.

I will keep as close to the original problem as is possible.To recap there is a 2m long straight rope of negligible mass.There are two 1 Kg masses attached to the rope,one of them(A) being attached at one end of the rope and the second one(B) being attached to the centre of the rope.The assembley is rotating about the end of the rope which is opposite to A.

What,if anything is at attached to this opposite end has not been specified and as far as any potential question solver is concerned there could be something or there could nothing.If there is something a relevant property of whatever it is,is its mass.In the absence of further information this something (or nothing) will be considered as a mass of mass M where M is unknown and can have any value from zero upwards.

To keep the problem as broad as possible no initial assumptions will be made about the location of the point about which the assembley rotates.Assume that the point of rotation is beween B and M and at a distance x from B.Also,let the tension between A and B be equal to T1 and the tension between B and M be equal to T2.

For circular motion the resultant force needed is given by:

F=mrw^2 ( Here the symbols have their usual meanings)

Applying this equation to each mass in turn we can write:

For A T1=(1+x)w^2

For B T2-T1=xw^2

For M T2=M(1-x)w^2

Solving for x we can write:

x=(M-1)/(2+M)

Plugging in values we show that:

When M=0 x=-0.5 A and B would be rotating about the point midway between them.

When M=1 x=0 A and B would be rotating about B,the mid point of the rope.

As M increases x increases and approaches the end of the rope as and although it can approach the end to a vanishingly small amount it can never quite reach the end and conform exactly to the conditions as specified in the question.A major effect of increasing M is that as M increases then so does the gravitational attraction between the masses.

 

[Dale]

To keep the problem as broad as possible no initial assumptions will be made about the location of the point about which the assembley rotates.

Here is where you are changing the problem. The problem isn't broad, it is specific. The location of the center of rotation isn't something you have to assume, it is something that is given. If you change that then you are changing the problem. Don't do that.

For any mass M there is some length L of extra rope which you can attach to the left end of the existing rope to get the entire assembly to rotate about the given point in isolation, but you must satisfy the constraint of rotation about the given point. Alternatively, you can have whatever is to the left be non-isolated also, or more complicated shape than a simple mass and rope, or actively generating thrust via rocket exhaust or ...

There are an infinite number of such assemblies satisfying the problem. All of them will result in the same answers as given above. The details are irrelevant.

 

[Andrew Mason]

Again you seem confused about Newton's 3rd law. If, as you seem to agree, the space station exerts an inward force on the passengers feet, then those feet must exert an outward force on the space station. It's that simple.

What I am confused about is your interpretation of Newton's third law.

Newton's third law says that the centre of mass of an isolated system cannot accelerate due to the internal forces within the system. The centre of mass of the system has to be the centre of rotation.

The force of the space station on the passenger is toward the centre of rotation because the passenger accelerates toward the centre of rotation.

If the reaction force, the force of the passenger on the space station, caused the centre of mass of the space station to move in any direction other than toward the centre of rotation, this would violate the third law because that would result in the centre of mass of the system accelerating and that would violate Newton's third law.

That is all I am saying. Maybe we are talking at cross-purposes here because I know that you understand this as well.

AM

 

[Andrew Mason]

Right. They are essentially unconnected.


OK. Of course in Dale's scenario B is connected to A.


You just said it yourself! The tension in the rope connecting to A is different in Dale's scenario.​

??! The tensions in all the rope sections are all the same. The rope connecting between A and B is the same tension 79N. Once it is rotating like that, you could tie it to A with a baby hair and the baby hair would not break or even stretch!
In fact, you could tie the hollow rope to the rope inside with baby's hair and there would be no stretching of those the baby's hair. There is no force between A and B in either scenario. There is only force between the centre and each object.

Quite vague. No mention of pseudoforce, so that 'definition' would also apply to the dreaded 'reactive centrifugal force'. Doesn't seem consistent with your posts. (Seems that you're saying here that 'centrifugal forces' really do exist.)

I don't think it is vague. It is general, not vague. It doesn't say whether they exist or not. Newton's third law says they don't exist.

As long as you realize that it's the net force that causes the centripetal acceleration.

Agreed. No problem.

AM​

 

[Doc Al]

What I am confused about is your interpretation of Newton's third law.

Right back at you. You continually dodge the simple statement of Newton's 3rd law: If A pushes B, then B pushes A. That's it in a nutshell, regardless of whatever else you can deduce.

Newton's third law says that the centre of mass of an isolated system cannot accelerate due to the internal forces within the system. The centre of mass of the system has to be the centre of rotation.

That's a true statement, but it is not simply a statement of what "Newton's third law says". But of course you can deduce that result from Newton's laws. It's irrelevant to the point here, of course.

The force of the space station on the passenger is toward the centre of rotation because the passenger accelerates toward the centre of rotation.

Right. Now apply Newton's 3rd law!

If the reaction force, the force of the passenger on the space station, caused the centre of mass of the space station to move in any direction other than toward the centre of rotation, this would violate the third law because that would result in the centre of mass of the system accelerating and that would violate Newton's third law.

The 'reaction force' acts on the portion of the space station directly under the passenger's feet. It's a contact force! Put a scale under the passenger's feet an you'll measure that force. And it works both ways. Newton's 3rd law tells us that you cannot have some mysterious disembodied force that only works from A to B but not from B to A. The space station floor cannot exert a real contact force on the passenger's feet (as you admit that it does) without the passenger's feet also exerting an equal and opposite real contact force on the floor of the space station. That's what Newton's 3rd law tells us!

Obviously the center of mass of the entire system of 'space station + passenger' cannot accelerate, since no external force acts. The net force on the entire system is zero.

Consider the space station and the passenger as separate systems. They clearly exert forces on each other. Thus their separate centers of mass must be accelerating. And they are! They are rotating!

That is all I am saying. Maybe we are talking at cross-purposes here because I know that you understand this as well.

You've made some bizarre statements in this thread that cannot be swept under the rug of 'talking at cross-purposes'.

 

[Doc Al]

??! The tensions in all the rope sections are all the same. The rope connecting between A and B is the same tension 79N.

What? There is no rope connecting A and B in your scenario. Those are two separate ropes!

In your scenario, the tension in the (hollow) rope connecting A to the center is 39.5 N.
In Dale's scenario, the tension in the rope connecting A to the center is 118.5 N.

Sounds pretty different to me!

Once it is rotating like that, you could tie it to A with a baby hair and the baby hair would not break or even stretch!

Sure: They are not connected! All the forces needed to support the accelerations of A and B in your scenario are provided by completely separate ropes. Do you seriously not see the difference?

I don't think it is vague. It is general, not vague. It doesn't say whether they exist or not. Newton's third law says they don't exist.

 

[JeffKoch]

Y'all do realize that this thread is now nearly 20 pages long, and that any entrenched points of view will not be changed even if it drags on to 200 pages.

 

[Andrew Mason]

Here is where you are changing the problem. The problem isn't broad, it is specific. The location of the center of rotation isn't something you have to assume, it is something that is given. If you change that then you are changing the problem. Don't do that.

For any mass M there is some length L of extra rope which you can attach to the left end of the existing rope to get the entire assembly to rotate about the given point in isolation, but you must satisfy the constraint of rotation about the given point. Alternatively, you can have whatever is to the left be non-isolated also, or more complicated shape than a simple mass and rope, or actively generating thrust via rocket exhaust or ...

There are an infinite number of such assemblies satisfying the problem. All of them will result in the same answers as given above. The details are irrelevant.

Dadface's analysis is correct. The frame of reference that is fixed to the end of the rope is not an inertial frame because that point cannot be centre of rotation. It can get arbitrarily close if the mass that the rope is attached to is arbitrarily large, but it cannot be the centre of rotation.

I disagree that you can have the force supplied by a rocket exerting an outward force on the rope. If that was the case, the centre of rotation really would accelerate.

AM

 

[Doc Al]

Dadface's analysis is correct. The frame of reference that is fixed to the end of the rope is not an inertial frame because that point cannot be centre of rotation. It can get arbitrarily close if the mass that the rope is attached to is arbitrarily large, but it cannot be the centre of rotation.

Again, you focus on the nonessential. You seem to think that the rope can exert a centripetal force on the mass and thus must have tension. Yet you don't accept that there must be an outward force on the rope. Amazing that you can have tension in a rope without something pulling on it.

 

[Dale]

Dadface's analysis is correct.

If someone asks you "What is 2+2?" and you respond "2+3 is 5" then the analysis is incorrect. It correctly answers the wrong question. Similarly, if an engineer is asked to build a better moustrap and correctly builds a better bridge then the analysis is incorrect. It correctly solves the wrong problem. Dadface's analysis is incorrect for the same reason, he correctly analyzed a different problem than the one given.

The frame of reference that is fixed to the end of the rope is not an inertial frame because that point cannot be centre of rotation. It can get arbitrarily close if the mass that the rope is attached to is arbitrarily large, but it cannot be the centre of rotation.

Sure it can be the center. I explained how in my response to Dadface.

 

[Dale]

Newton's third law says that the centre of mass of an isolated system cannot accelerate due to the internal forces within the system. The centre of mass of the system has to be the centre of rotation.

Although this is true it is not what Newton's 3rd law says. In any case, the system is not isolated, so it can accelerate and the center of rotation does not need to be the center of mass.

 

[Andrew Mason]

Again, you focus on the nonessential. You seem to think that the rope can exert a centripetal force on the mass and thus must have tension. Yet you don't accept that there must be an outward force on the rope. Amazing that you can have tension in a rope without something pulling on it.

It is pulling on something that Dale will not provide us information on. It has to be an infinite mass. We know that it is NOT a rocket providing the centripetal force.

Who said you can have tension in this rope without two masses pulling on it? Not me. Never. .Of course something is pulling on it. Just not A. The mass at the other end of the rope is pulling on it. A is not pulling on the rope between A and B. That is what I am saying.

AM