What exactly is the reactive centrifugal force (split)

In summary: I think that this is the root of my misunderstanding.In summary, the conversation is discussing the use of the term "centrifugal force" in relation to the reactive centrifugal force and the inertial centrifugal force. One person argues that the term is confusing and should not be used, while the other argues that it depends on the reference frame and the direction of the force. They also discuss Newton's third law and its application in determining the reaction force in a system. Ultimately, they agree that the third law is not about tensions, but about equal and opposite forces and changes in momentum.
  • #106
Andrew Mason said:
I would suggest that the reason mainstream texts do not use the term "centrifugal reaction force" to describe the reaction force to a centripetal force is because it obscures the physics. If the direction of the force is determined by the direction of the acceleration of the centre of mass of the body to which the force is applied, which I would suggest is the standard convention, all forces are all centripetal.
Then please provide some references which demonstrate that this is the standard convention. Note, that being a standard convention is a much stronger claim than merely that it is one opinion or that it is an alternate terminology. I don't know how you can possibly support this, but then again, you haven't supported any of your claims.

Andrew Mason said:
These are both good examples of why these do NOT cause centrifugal acceleration. We have discussed the bolt cutting example and all the bolt cutting does is STOP centripetal acceleration. It does not cause any acceleration away from the centre.
Yes, it does. Between the sudden cutting of the bolts and the propagation of the shear wave to the feet of the astronaut there is still a centrifugal reaction force and this centrifugal reaction force accelerates the plate away from the center. I already explained this in detail, since you did not respond further I thought that you had understood it.

Andrew Mason said:
The rocket is quite a bit more complicated. I will need some time to think about it some more. But the situation described seems incomplete - there must be other forces involved in order for the centre of mass of the rocket to rotate about the centre of the space station with the centre of mass of the space station remaining inertial.
No other force is needed. Please analyze it in depth. If a rocket is spinning and burning its engines to produce a constant magnitude of acceleration then there exists some inertial reference frame where it is in uniform circular motion.

Andrew Mason said:
When I pull on a box on a frictionless surface with a force F, the box pulls back on me with force F. Both forces are directed toward the same inertial point - the centre of mass.
Where did you get this idea? The direction of the force is not automatically towards the COM. If you pull horizontally then the force is only directed towards the COM if it is vertically aligned with the COM. Otherwise the force could be directed above or below the COM.

Andrew, this continued discussion is pointless. The terminology exists, is well defined, and commonly accepted. You are absolutely correct that the centrifugal reaction force can cause centripetal acceleration in some circumstances, but that doesn't change a thing.
 
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  • #107
DaleSpam said:
Yes, it does. Between the sudden cutting of the bolts and the propagation of the shear wave to the feet of the astronaut there is still a centrifugal reaction force and this centrifugal reaction force accelerates the plate away from the center. I already explained this in detail, since you did not respond further I thought that you had understood it.
Here is what you said:

"I specified that the bolts were "suddenly" cut for a very important reason. As the astronaut is standing on the floor the floor is under stress with centripetal forces from the bolts and a centrifugal reaction force from the astronaut. The centripetal force is greater than the centrifugal reaction force so there is a net acceleration towards the center.

When the bolts are suddenly cut the stress is relieved from the outside of the section of floor, but the inner part of the floor (where the astronaut is standing) is still under stress. This sets up a shear wave where the floor material transitions from stress to stress-free. During the time between when the bolts are suddenly cut and when that shear wave reaches the feet of the astronaut the centrifugal reaction force still exists, the feet and floor are still in contact, and the floor is accelerating in a direction away from the center. It may help to think of the floor as being made of a stretchy rubber material.

The centrifugal force is every bit as "centrifugal" as the centripetal force is "centripetal". The centrifugal force points away from the center, the centripetal points towards the center. If either is unbalanced then it will result in acceleration in the corresponding direction. If there are other forces involved then the actual acceleration depends on the net force, per Newton's 2nd law.

Let's deal with the first paragraph:

"I specified that the bolts were "suddenly" cut for a very important reason. As the astronaut is standing on the floor the floor is under stress with centripetal forces from the bolts and a centrifugal reaction force from the astronaut. The centripetal force is greater than the centrifugal reaction force so there is a net acceleration towards the center."
This makes no sense to me. How can the centripetal force (presumably by the force exerted by the floor on the astronaut) be greater than the "centrifugal reaction force" (the force exerted by the astronaut on the floor, I assume) if they are equal and opposite 3rd law pairs? There is something wrong here because the centripetal force is not opposed by any force. The centripetal force, by definition, is mac = ω2r. What force is opposing it?

Now your second paragraph:

"When the bolts are suddenly cut the stress is relieved from the outside of the section of floor, but the inner part of the floor (where the astronaut is standing) is still under stress. This sets up a shear wave where the floor material transitions from stress to stress-free. During the time between when the bolts are suddenly cut and when that shear wave reaches the feet of the astronaut the centrifugal reaction force still exists, the feet and floor are still in contact, and the floor is accelerating in a direction away from the center. It may help to think of the floor as being made of a stretchy rubber material.​

You appear to be saying that the relaxation of the stress forces within the floor will cause the floor to expand against the astronaut (and, presumably the relaxation of similar tensions in the astronaut will cause the astronaut to expand against the floor section that has been liberated from the space station). But once the bolts are cut the centre of mass of the astronaut and floor section (taken together so long as they are exerting forces on each other) defines an inertial reference frame - no external forces are acting on them. So there is no acceleration of the centre of mass of the floor/astronaut. The astronaut and floor section will briefly (very briefly) push off against each other as the tensions are relaxed, but that is not a reaction to the centripetal force which is zero at that moment.

Finally, the third paragraph:

"The centrifugal force is every bit as "centrifugal" as the centripetal force is "centripetal". The centrifugal force points away from the center, the centripetal points towards the center. If either is unbalanced then it will result in acceleration in the corresponding direction. If there are other forces involved then the actual acceleration depends on the net force, per Newton's 2nd law."​

Again, the centripetal force is always unbalanced because it is defined as centripetal acceleration x mass. There is never centrifugal acceleration (unless you are in an accelerating frame of reference) even if one were to accept the idea of calling the reaction force to the force causing centripetal acceleration "centrifugal".

AM
 
  • #108
Andrew Mason said:
The rocket is quite a bit more complicated.
It is a dead simple, minimal scenario. If your proposed naming reasoning gets complicated here already, then I doubt it will catch on.

Andrew Mason said:
But the situation described seems incomplete - there must be other forces involved in order for the centre of mass of the rocket to rotate about the centre of the space station with the centre of mass of the space station remaining inertial.
There is no inertial space station in the rocket scenario. Here it is again:

A spaceship is moving on a circular path as seen from an inertial frame, by firing its engine to provide the centripetal acceleration. The burned fuel is exerting a centripetal force on the ship, which causes a centripetal acceleration of the ship. The ship is exerting a centrifugal force on the burned fuel, which causes a centrifugal acceleration of the burned fuel.

Andrew Mason said:
Again, the centripetal force is always unbalanced because it is defined as centripetal acceleration x mass.
That is the definition of "net force", not of "centripetal force". The net force can be centripetal but so can individual forces, which are different from the net force.
 
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  • #109
Andrew Mason said:
Let's deal with the first paragraph:

"I specified that the bolts were "suddenly" cut for a very important reason. As the astronaut is standing on the floor the floor is under stress with centripetal forces from the bolts and a centrifugal reaction force from the astronaut. The centripetal force is greater than the centrifugal reaction force so there is a net acceleration towards the center."
This makes no sense to me. How can the centripetal force (presumably by the force exerted by the floor on the astronaut) be greater than the "centrifugal reaction force" (the force exerted by the astronaut on the floor, I assume) if they are equal and opposite 3rd law pairs? There is something wrong here because the centripetal force is not opposed by any force. The centripetal force, by definition, is mac = ω2r. What force is opposing it?
I don't know why you would say "the centripetal force (presumably by the force exerted by the floor on the astronaut)" when I explicitly said "the floor is under stress with centripetal forces from the bolts". Please draw a free-body diagram for the floor. There are two sets of forces on the floor, one from the bolts, one from the astronaut. The forces from the bolts are directed centripetally. The forces from the astronaut are directed centrifugally. The centripetal force from the bolts is greater than the centrifugal reaction force from the astronaut, giving a net centripetal force as required by Newton's 2nd law. They are not a third law pair since (1) they act on the same object (the floor) and (2) they are between two different pairs of objects.

Andrew Mason said:
"When the bolts are suddenly cut the stress is relieved from the outside of the section of floor, but the inner part of the floor (where the astronaut is standing) is still under stress. This sets up a shear wave where the floor material transitions from stress to stress-free. During the time between when the bolts are suddenly cut and when that shear wave reaches the feet of the astronaut the centrifugal reaction force still exists, the feet and floor are still in contact, and the floor is accelerating in a direction away from the center. It may help to think of the floor as being made of a stretchy rubber material.​

You appear to be saying that the relaxation of the stress forces within the floor will cause the floor to expand against the astronaut (and, presumably the relaxation of similar tensions in the astronaut will cause the astronaut to expand against the floor section that has been liberated from the space station). But once the bolts are cut the centre of mass of the astronaut and floor section (taken together so long as they are exerting forces on each other) defines an inertial reference frame - no external forces are acting on them. So there is no acceleration of the centre of mass of the floor/astronaut. The astronaut and floor section will briefly (very briefly) push off against each other as the tensions are relaxed, but that is not a reaction to the centripetal force which is zero at that moment.
But I am not taking them together. I am analyzing the section of floor separately, which is perfectly legitimate. Taking it separately, there remains a "reaction" force on the floor from the astronaut and that force causes the floor to accelerate centrifugally. Likewise, there remains an "action" force on the astronaut from the floor which causes the astronaut to accelerate centripetally. There is simply no avoiding that.

Andrew Mason said:
Finally, the third paragraph:

"The centrifugal force is every bit as "centrifugal" as the centripetal force is "centripetal". The centrifugal force points away from the center, the centripetal points towards the center. If either is unbalanced then it will result in acceleration in the corresponding direction. If there are other forces involved then the actual acceleration depends on the net force, per Newton's 2nd law."​

Again, the centripetal force is always unbalanced because it is defined as centripetal acceleration x mass. There is never centrifugal acceleration (unless you are in an accelerating frame of reference) even if one were to accept the idea of calling the reaction force to the force causing centripetal acceleration "centrifugal".
Can you provide a reference for this? I.e. a reference which states that, in a situation with multiple forces acting on an object, the centripetal force is the net force? I had always taken the centripetal force to be the individual force which was acting centripetally, not the net force, but perhaps I was wrong in this.

In any case, that doesn't change the analysis for the astronaut and floor during the shear wave propagation. The floor is exerting an unbalanced centripetal force on the astronaut which is accelerating centripetally. The reaction to that force is the centrifugal force on the floor which is also unbalanced and therefore the floor is accelerating centrifugally.
 
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  • #110


Andrew Mason said:
This is not a good example because the gravitational force and normal force are not third law pairs. They are not equal and opposite, for one thing. The normal force is always a bit less than the gravitational force except at the poles.
AM


Why does the normal force is always a bit less than the gravitational force?
 
  • #111


GT1 said:
Why does the normal force is always a bit less than the gravitational force?

The Earth is rotating. If you are "standing still" on your bathroom scale, you are actually traveling in a circular path at a rate of about one rotation per 24 hours.

If you add the inward gravitational force and the outward normal force together, the resultant is the small centripetal force required to keep you on this circular path.

If normal force were equal to gravitational force then you would not be accelerating. You would find yourself following a straight line path tangent to the rotating Earth and would quickly find yourself in outer space.
 
  • #112
DaleSpam said:
I don't know why you would say "the centripetal force (presumably by the force exerted by the floor on the astronaut)" when I explicitly said "the floor is under stress with centripetal forces from the bolts". Please draw a free-body diagram for the floor. There are two sets of forces on the floor, one from the bolts, one from the astronaut. The forces from the bolts are directed centripetally. The forces from the astronaut are directed centrifugally. The centripetal force from the bolts is greater than the centrifugal reaction force from the astronaut, giving a net centripetal force as required by Newton's 2nd law. They are not a third law pair since (1) they act on the same object (the floor) and (2) they are between two different pairs of objects.
Ok. Fair enough.

But I am not taking them together. I am analyzing the section of floor separately, which is perfectly legitimate. Taking it separately, there remains a "reaction" force on the floor from the astronaut and that force causes the floor to accelerate centrifugally. Likewise, there remains an "action" force on the astronaut from the floor which causes the astronaut to accelerate centripetally. There is simply no avoiding that.
Once the bolts are cut, the centre of mass of the astronaut/floor stops accelerating. While the astronaut and floor are still in contact, the centres of mass of bodies rotate about their common centre of mass. The relaxation of tensions just causes them to accelerate away from each other (briefly) ie. they push away on each other. That force is not a centripetal force because it is not constantly directed to a central point.

Can you provide a reference for this? I.e. a reference which states that, in a situation with multiple forces acting on an object, the centripetal force is the net force? I had always taken the centripetal force to be the individual force which was acting centripetally, not the net force, but perhaps I was wrong in this.
Centripetal just means always pointing to the centre, of course. And it is created naturally by rotation of a rigid or constrained body.

You could increase the radial tension artificially. For example in a bicycle wheel, you could tighten the spokes. This would have no bearing on the centripetal acceleration experienced by the rotating bicycle wheel. You would then have a centripetal force due to rotation plus a static radial tension.

In any case, that doesn't change the analysis for the astronaut and floor during the shear wave propagation. The floor is exerting an unbalanced centripetal force on the astronaut which is accelerating centripetally. The reaction to that force is the centrifugal force on the floor which is also unbalanced and therefore the floor is accelerating centrifugally.
It is not directed constantly toward a central point. The astronaut and floor are no longer rotating with the space station when the bolts are cut. They are only rotating about their centre of mass which is inertial (not accelerating). The push against each other is a linear force separating the two parts. It is not directed to the same point. I don't see how it could be centripetal.

AM
 
  • #113
Andrew Mason said:
That force is not a centripetal force because it is not constantly directed to a central point.
Yes, it is. Remember, the floor continues to rotate and the normal force continues to be perpendicular to the floor. For the parts that have not been reached by the shear wave, everything continues as normal.

Andrew Mason said:
Centripetal just means always pointing to the centre, of course. And it is created naturally by rotation of a rigid or constrained body.
I note that yet again you are unable to produce any reference supporting one of your claims when asked to do so.
 
  • #114
DaleSpam said:
Yes, it is. Remember, the floor continues to rotate and the normal force continues to be perpendicular to the floor. For the parts that have not been reached by the shear wave, everything continues as normal.
What is the source of the force that causes centripetal acceleration of the centre of mass of the floor after the bolts are cut?

After the bolts are cut, the floor continues to rotate about its centre of mass, not the centre of mass of the space station. When it is rotating with the space station, its centre of mass is accelerating but when it is freed (bolts cut) its centre of mass stops accelerating.

I note that yet again you are unable to produce any reference supporting one of your claims when asked to do so.
See any good physics text on the derivation of centripetal force: Alonso Finn, Physics, Addison Wesley p. 120-121;Barger and Olssen, Classical Mechanics, p 112, you will see the derivation of central force, defined as the force on a rotating body where dL/dt = 0:

[itex]\vec{L} = \vec{r}\times\vec{p}[/itex]

[itex]\frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt}\times\vec{p} + \vec{r}\times\frac{d\vec{p}}{dt} [/itex]

[itex]\frac{d\vec{L}}{dt} = \vec{v}\times m\vec{v} + \vec{r}\times\frac{d\vec{p}}{dt} [/itex]

[itex]\frac{d\vec{L}}{dt} = 0 + \vec{r}\times\frac{d\vec{p}}{dt} = \vec{r}\times\vec{F}[/itex]

If [itex]\frac{d\vec{L}}{dt}=0[/itex] then [itex]\vec{F}[/itex] is entirely radial: [itex]\vec{r}\times\vec{F}=0[/itex] and this is called the centripetal force.

AM
 
  • #115
Andrew Mason said:
What is the source of the force that causes centripetal acceleration of the centre of mass of the floor after the bolts are cut?
The center of mass of the floor accelerates centrifugally immediately after the bolts are cut. The source of the force is the centrifugal reaction force from the astronaut.

Andrew Mason said:
After the bolts are cut, the floor continues to rotate about its centre of mass, not the centre of mass of the space station. When it is rotating with the space station, its centre of mass is accelerating but when it is freed (bolts cut) its centre of mass stops accelerating.
It isn't "freed" when the bolts are cut. It is "freed" when the shear wave reaches the astronaut's feet and they lose contact. During the time between the bolts being cut and the shear wave reaching the astronaut's feet the floor is in non-rigid body motion. Overall, the COM is accelerating centrifugally under the influence of the centrifugal force, outside the wavefront the floor is accelerating centrifugally, and inside the wavefront the floor continues to accelerate centripetally.

If you are not so familiar with stress and strain then perhaps it will help to think of the astronaut as standing on a spring. While the spring is decompressing the "top" continues to accelerate centripetally while the "bottom" and the COM accelerate centrifugally. It isn't quite the same since the shear wave is progressing laterally while the spring is expanding radially, but both are governed by Hooke's law, so they are closely related.

Again, the term "reactive centrifugal force" is common and you should be familiar with it. You have valid reasons for disliking it, but those reasons don't make the term go away. I sympathize with your position since I have similar objections to other terms that just won't go away, no matter how I wish they would (e.g. "relativistic mass").
 
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  • #116
DaleSpam said:
The center of mass of the floor accelerates centrifugally immediately after the bolts are cut. The source of the force is the centrifugal reaction force from the astronaut.
That reaction force ends as soon the centripetal acceleration of the astronaut ends.

It seems to me that the source of the "force" is the relaxation of the tension, so there is no net force on the centre of mass of either the astronaut or the floor. The relaxation of the tension creates a compression wave that propagates through the floor material much the same way that a stretched spring would oscillate if you were swinging it around and then let it go. The duration of those oscillations would depend on the damping forces within the spring. Are you suggesting that the centre of mass of the spring would still be undergoing centripetal acceleration while the oscillations continued?

If you are not so familiar with stress and strain then perhaps it will help to think of the astronaut as standing on a spring. While the spring is decompressing the "top" continues to accelerate centripetally while the "bottom" and the COM accelerate centrifugally. It isn't quite the same since the shear wave is progressing laterally while the spring is expanding radially, but both are governed by Hooke's law, so they are closely related.
If it is propagating laterally, how can it be the reaction to centripetal acceleration, which is entirely radial?

Again, the term "reactive centrifugal force" is common and you should be familiar with it. You have valid reasons for disliking it, but those reasons don't make the term go away. I sympathize with your position since I have similar objections to other terms that just won't go away, no matter how I wish they would (e.g. "relativistic mass").
Within any extended body rotating about a central body and whose orientation with respect to the radial vector from the centre of rotation does not change, there are necessarily tension forces that will arise within the body. This is due to the fact that different parts of the body are at a different distances from the centre of rotation (ie. centripetal acceleration of the part is different than ω^2rcom). These tensions are all directed radially from the com of the extended body. When the com of the extended body stops accelerating, these tensions continue because the extended body continues to rotate about its centre of mass. So there is no relaxation of these tensions.

AM
 
  • #117
Andrew Mason said:
That reaction force ends as soon the centripetal acceleration of the astronaut ends.
Yes, clearly.

Andrew Mason said:
It seems to me that the source of the "force" is the relaxation of the tension, so there is no net force on the centre of mass of either the astronaut or the floor.
Why wouldn't there be a net force? There is the force between the floor and the astronaut and there are no other forces on either to balance that force, so clearly there is a net force on each.

Andrew Mason said:
If it is propagating laterally, how can it be the reaction to centripetal acceleration, which is entirely radial?
Shear waves are transverse waves. The wave propagates laterally and the forces are transverse to the wave, i.e. the forces are radial.

That is the key difference between the floor example and the spring example. They both follow Hooke's law, but for the spring example, the wave would be a compression wave propagating radially rather than a shear wave propagating laterally. In both cases, the forces are radial. If it is easier for you to grasp the spring example then we can do that.
 
  • #118
DaleSpam said:
Yes, clearly.

Why wouldn't there be a net force? There is the force between the floor and the astronaut and there are no other forces on either to balance that force, so clearly there is a net force on each.
If they are in contact with each other they can have equal and opposite forces on each other. I meant there would be no centripetal acceleration of the centre of mass of either the astronaut or the liberated floor section toward the centre of rotation of the space station.

Shear waves are transverse waves. The wave propagates laterally and the forces are transverse to the wave, i.e. the forces are radial.

That is the key difference between the floor example and the spring example. They both follow Hooke's law, but for the spring example, the wave would be a compression wave propagating radially rather than a shear wave propagating laterally. In both cases, the forces are radial. If it is easier for you to grasp the spring example then we can do that.
The physics here cannot depend on whether the wave is a shear wave or a compression wave. I don't really see much of a shear wave here because the difference in centripetal force across the thickness of the floor is very small, assuming the thickness of the floor is small compared to its distance from the centre of rotation.

It is very easy to create an acceleration that is outward from the centre of rotation at a given moment. You would just have to have an extended object rotating about a central point at speed ω and also rotating about its centre of mass at speed ω+Δω. This is what happens in some amusement rides (eg. Tilt A Whirl). You can set up compressed springs and then release them, sending mass outward. Of course, these are not third law pairs to the centripetal force.

AM
 
  • #119
Andrew Mason said:
The physics here cannot depend on whether the wave is a shear wave or a compression wave
This is a very weird comment. Shear waves and compression waves are physically different waves. As I have described, we can set up a scenario with either kind of wave, but they are physically different scenarios.

So if you have trouble understanding shear waves we can do compression waves. But I don't know why you would say that the physics cannot depend on the nature of the wave.

Andrew Mason said:
I don't really see much of a shear wave here because the difference in centripetal force across the thickness of the floor is very small
What does the difference in centripetal force across the thickness of the floor have to do with shear? I don't think you understand shear stress at all.

Do you have access to a statics textbook? I am glad to help you with shear, but if you prefer we can switch to the compression example. Just let me know your preference.
 
  • #120
DaleSpam said:
This is a very weird comment. Shear waves and compression waves are physically different waves. As I have described, we can set up a scenario with either kind of wave, but they are physically different scenarios.

So if you have trouble understanding shear waves we can do compression waves. But I don't know why you would say that the physics cannot depend on the nature of the wave.

What does the difference in centripetal force across the thickness of the floor have to do with shear? I don't think you understand shear stress at all.
Well, I was assuming you were talking about a stress that is the result of two opposing forces that act on the same body in a way that tends to shear - make parts of the body that are close together move in opposite directions. Not sure how that applies here so if I am missing something, feel free to enlighten me.

Do you have access to a statics textbook? I am glad to help you with shear, but if you prefer we can switch to the compression example. Just let me know your preference.
No mention of shear stress in any of my physics texts. Sounds like engineering.

AM
 
  • #121
DaleSpam said:
So if you have trouble understanding shear waves we can do compression waves.
I really think this is overkill.

If you want a scenario where the centrifugal reaction causes centrifugal acceleration in the rotating frame, then take a turntable where a mass on rollers pushes a sliding mass outwards .

If you want a scenario where the centrifugal reaction causes centrifugal acceleration in the inertial frame, then take the rocket on a circular path.

Of course, none of this is really important, because Newton's 3rd applies regardless of acceleration.
 
  • #122
Andrew Mason said:
Well, I was assuming you were talking about a stress that is the result of two opposing forces that act on the same body in a way that tends to shear - make parts of the body that are close together move in opposite directions. Not sure how that applies here so if I am missing something, feel free to enlighten me.
You are correct, the forces in question are the forces from the bolts and the force from the astronaut's feet. Those forces are in opposite directions and cause the shear stress in the floor. As you correctly indicate, parts of the floor which are close together tend to move in opposite directions, the part closer to the feet tends to move out while the part closer to the bolts tends to move in. This is shear stress.

When the bolts are cut the shear stress in the floor next to the bolts is relieved, but not in the rest of the floor. A shear wave ripples through the material at the speed of sound in the material relieving the shear stress as it goes. Until the shear wave reaches it, the material is in the same state of stress as it was before the bolts were cut.

I can go into more detail if you like, but A.T. is right, none of it is important. The centrifugal reaction force depends on the point of application of Newton's 3rd law, not any acceleration. You can argue with me and A.T. if you like, but even if you do convince us that it is bad terminology, that wouldn't make the bad terminology go away.
 
  • #123
A.T. said:
I really think this is overkill.

If you want a scenario where the centrifugal reaction causes centrifugal acceleration in the rotating frame, then take a turntable where a mass on rollers pushes a sliding mass outwards .
But that is a perfect example of the fictitious centrifugal force, not the reaction to the centripetal force on the mass. This confusion is the real problem.

If you want a scenario where the centrifugal reaction causes centrifugal acceleration in the inertial frame, then take the rocket on a circular path.
That is an interesting scenario but I don't think that the force on the rocket qualfies as a centripetal force.

Centripetal force on a body defined as is dp/dt of the body where dL/dt of the body = 0. But if dL/dt for the rocket + fuel = 0 the rocket would have to constantly be increasing its rotational speed because it is constantly losing mass.

Of course, none of this is really important, because Newton's 3rd applies regardless of acceleration.
This is correct. And in an inertial frame where dL/dt = 0, all dp/dt is centripetal.

AM
 
  • #124
Andrew Mason said:
But that is a perfect example of the fictitious centrifugal force, not the reaction to the centripetal force on the mass. This confusion is the real problem.
Yes, confusing fictitious and real forces is your real problem. In my scenario the sliding block can be very light and have large friction, which by far outweighs the sliding block's fictitious centrifugal force. What moves the sliding block outwards is the real centrifugal force from the inner heavier low friction block.
Andrew Mason said:
That is an interesting scenario but I don't think that the force on the rocket qualfies as a centripetal force.
Based on your criteria most names would apply only to the simple idealized cases, that are found at the begin of a book chapter, where the name is first introduced. Most people who have some abstraction skill prefer more generally applicable names.
Andrew Mason said:
Centripetal force on a body defined as is dp/dt of the body where dL/dt of the body = 0.
That is just the simplest possible case, not the general case.
Andrew Mason said:
rocket would have to constantly be increasing its rotational speed
Fine, if it makes you happy let the speed increase. Or replace the rocket with a plane, flying in a circle. It accelerates air centrifugally, to have a centripetal force.
 
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  • #125
A.T. said:
Fine, if it makes you happy let the speed increase. Or replace the rocket with a plane, flying in a circle. It accelerates air centrifugally, to have a centripetal force.
Or you can simply reduce the thrust in proportion to the reduction of mass. I do like the airplane idea also.
 
  • #126
A.T. said:
... or replace the rocket with a plane, flying in a circle. It accelerates air centrifugally, to have a centripetal force.
In the case of a rocket or aircraft, aren't both the centripetal and centrifugal forces "reaction" forces? For the aircraft, the centripetal force is the reaction (or coexistance) to the air being accelerated outwards, and the centrifugal force is the reaction (or coexistance) to the aircraft being accelerated inwards.
 
  • #127
rcgldr said:
In the case of a rocket or aircraft, aren't both the centripetal and centrifugal forces "reaction" forces? For the aircraft, the centripetal force is the reaction to the air being accelerated outwards, and the centrifugal force is the reaction to the aircraft being accelerated inwards.
The "action/reaction" label is meaningless. It just indicates that it is a 3rd law force pair.
 
  • #128
A.T. said:
Yes, confusing fictitious and real forces is your real problem. In my scenario the sliding block can be very light and have large friction, which by far outweighs the sliding block's fictitious centrifugal force. What moves the sliding block outwards is the real centrifugal force from the inner heavier low friction block.
What moves the block outward is inertia. The inability of the friction force between the turntable and the block to provide the centripetal acceleration required to keep both blocks on the turntable, results in the blocks not accelerating sufficiently to stay on the turntable. There is no outward force.

That is just the simplest possible case, not the general case.
It is the only case in which the force is centripetal - that is constantly directed toward the same inertial point. If dL/dt ≠ 0 the force is not centripetal - it has centripetal and tangential components so it is not pointing toward the same central point.

AM
 
  • #129
DaleSpam said:
Or you can simply reduce the thrust in proportion to the reduction of mass.
That was my original idea, but Andrew wants L = const. So let's speed up to compensate for mass reduction.

DaleSpam said:
I do like the airplane idea also.

The same in space would be a ship with a solar sail, doing small circles at constant distance to the sun. The radial lift component cancels gravity. The other component provides centripetal acceleration. The reaction to the centripetal force accelerates particles centrifugally.
 
  • #130
Andrew Mason said:
What moves the block outward is inertia.
That's mumbo jumbo. You have to use forces acting on the outer block to explain the outer block's outwards acceleration in the rotating frame. There is a fictions centrifugal force acting on the outer block in the rotating frame, but it is not sufficient to overcome friction. Only because of the real centrifugal force from the inner block the outer block accelerates outwards in the rotating frame.
Andrew Mason said:
There is no outward force.
There are even two in the rotating frame.
Andrew Mason said:
If dL/dt ≠ 0 the force is not centripetal - it has centripetal and tangential components so it is not pointing toward the same central point.
So there is a centripetal component, and a centrifugal reaction component.
 
  • #131
A.T. said:
The same in space would be a ship with a solar sail, doing small circles at constant distance to the sun. The radial lift component cancels gravity. The other component provides centripetal acceleration. The reaction to the centripetal force accelerates particles centrifugally.
I assume the sail reflects the particles back towards the sun, so the sail exerts a centripetal force (and acceleration) onto the particles (in addition to gravity) and the particles exert a reactive centrifugal force onto the sail.
 
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  • #132
rcgldr said:
Wouldn't the sail be reflecting the particles back towards the sun?
The ship makes small circles in a plane that is perpendicular to the solar wind and gravity. The sail is at an 45° angle outwards to the solar wind and gravity. The particles are diverted by 90° and fly off centrifugally within the plane of the circular path.
 
  • #133
A.T. said:
That was my original idea, but Andrew wants L = const. So let's speed up to compensate for mass reduction.
For a rocket and its exhaust dL/dt=0 regardless of how fast or slow or in which direction you point the exhaust. It is a great example for that reason.
 
  • #134
rcgldr said:
Wouldn't the sail be reflecting the particles back towards the sun?

A.T. said:
The ship makes small circles in a plane that is perpendicular to the solar wind and gravity. The sail is at an 45° angle outwards to the solar wind and gravity. The particles are diverted by 90° and fly off centrifugally within the plane of the circular path.
OK, but even if the particles are diverted 90°, the direction of acceleration of those particles includes a component towards the sun (radial outwards velocity is reduced to zero), and a tangental component (at that instant in time).
 
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  • #135
DaleSpam said:
For a rocket and its exhaust dL/dt=0
If I understand him correctly he wants L = const for the body on which the centripetal force is exerted, which is the rocket and the remaining fuel. But the example can be adjusted to work either way.
 
  • #136
rcgldr said:
OK, but even if the particles are diverted 90°, the direction of acceleration of those particles includes a component towards the sun
That is not relevant of the circular motion. In the plane of the circle there is a centripetal force and a centrifugal reaction, which accelerates the particles outwards.
 
  • #137
A.T. said:
That is not relevant of the circular motion. In the plane of the circle there is a centripetal force and a centrifugal reaction, which accelerates the particles outwards.
In order to accelerate the particles outwards, it seems the solar ship would have to travel faster than orbital speed (v^2 > G M / r) in a circular path so that a centripetal force (in addition to gravity) would be required to maintain the circular path, and use some sort of collector and accelerator, which could be an idealized sail, in order to end up accelerating particles outwards.
 
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  • #138
A.T. said:
If I understand him correctly he wants L = const for the body on which the centripetal force is exerted, which is the rocket and the remaining fuel.
Yuck, that makes it messy. With this definition you need a tangential thrust in order to have centripetal motion. I don't see any reason for that definition.
 
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  • #139
A.T. said:
That's mumbo jumbo. You have to use forces acting on the outer block to explain the outer block's outwards acceleration in the rotating frame. There is a fictions centrifugal force acting on the outer block in the rotating frame, but it is not sufficient to overcome friction. Only because of the real centrifugal force from the inner block the outer block accelerates outwards in the rotating frame.
Are you talking about real forces or forces that appear in the non-interial reference frame of the rotating turntable?

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

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

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

AM
 
  • #140
Andrew Mason said:
The only real force that is apparently outward is the reaction force of the block on the turntable.
And the reaction force of the inner block on the outer. That is a real force, not fictitious, and s required to explain the movement in all frames.

This line of conversation is still irrelevant. Regardless of the validity of your reasons for disliking it, the term exists. You cannot wish it away, nor can you reason it away.
 
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<h2>1. What is the reactive centrifugal force (split)?</h2><p>The reactive centrifugal force (split) is a fictitious force that appears to act on an object moving in a circular path. It is a result of the object's inertia and its tendency to resist changes in its motion.</p><h2>2. How is the reactive centrifugal force (split) different from the centrifugal force?</h2><p>The centrifugal force is a real force that acts on an object moving in a circular path, pushing the object away from the center of rotation. The reactive centrifugal force (split) is a fictitious force that appears to act on the object in the opposite direction, towards the center of rotation.</p><h2>3. What causes the reactive centrifugal force (split) to appear?</h2><p>The reactive centrifugal force (split) is a result of the object's inertia and its tendency to resist changes in its motion. As the object moves in a circular path, its inertia causes it to continue moving in a straight line, resulting in the appearance of the reactive centrifugal force (split).</p><h2>4. Is the reactive centrifugal force (split) a real force?</h2><p>No, the reactive centrifugal force (split) is a fictitious force and does not have a physical origin. It is simply a mathematical construct used to explain the motion of objects in circular paths.</p><h2>5. How is the reactive centrifugal force (split) related to Newton's laws of motion?</h2><p>The reactive centrifugal force (split) is a result of the first law of motion, also known as the law of inertia. It explains how objects tend to resist changes in their motion and how this resistance results in the appearance of fictitious forces, such as the reactive centrifugal force (split).</p>

1. What is the reactive centrifugal force (split)?

The reactive centrifugal force (split) is a fictitious force that appears to act on an object moving in a circular path. It is a result of the object's inertia and its tendency to resist changes in its motion.

2. How is the reactive centrifugal force (split) different from the centrifugal force?

The centrifugal force is a real force that acts on an object moving in a circular path, pushing the object away from the center of rotation. The reactive centrifugal force (split) is a fictitious force that appears to act on the object in the opposite direction, towards the center of rotation.

3. What causes the reactive centrifugal force (split) to appear?

The reactive centrifugal force (split) is a result of the object's inertia and its tendency to resist changes in its motion. As the object moves in a circular path, its inertia causes it to continue moving in a straight line, resulting in the appearance of the reactive centrifugal force (split).

4. Is the reactive centrifugal force (split) a real force?

No, the reactive centrifugal force (split) is a fictitious force and does not have a physical origin. It is simply a mathematical construct used to explain the motion of objects in circular paths.

5. How is the reactive centrifugal force (split) related to Newton's laws of motion?

The reactive centrifugal force (split) is a result of the first law of motion, also known as the law of inertia. It explains how objects tend to resist changes in their motion and how this resistance results in the appearance of fictitious forces, such as the reactive centrifugal force (split).

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