Register to reply

What exactly is the reactive centrifugal force (split)

by A.T.
Tags: centrifugal, force, reactive, split
Share this thread:
DaleSpam
#109
Feb8-13, 07:58 AM
Mentor
P: 17,541
Quote Quote by Andrew Mason View Post
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.

Quote Quote by Andrew Mason View Post
"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.

Quote Quote by Andrew Mason View Post
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.
GT1
#110
Feb8-13, 09:58 AM
P: 121
Quote Quote by Andrew Mason View Post
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?
jbriggs444
#111
Feb8-13, 10:14 AM
P: 999
Quote Quote by GT1 View Post
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 travelling 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.
Andrew Mason
#112
Feb8-13, 07:32 PM
Sci Advisor
HW Helper
P: 6,684
Quote Quote by DaleSpam View Post
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
DaleSpam
#113
Feb8-13, 09:26 PM
Mentor
P: 17,541
Quote Quote by Andrew Mason View Post
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.

Quote Quote by Andrew Mason View Post
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.
Andrew Mason
#114
Feb9-13, 06:19 AM
Sci Advisor
HW Helper
P: 6,684
Quote Quote by DaleSpam View Post
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
DaleSpam
#115
Feb9-13, 06:58 AM
Mentor
P: 17,541
Quote Quote by Andrew Mason View Post
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.

Quote Quote by Andrew Mason View Post
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").
Andrew Mason
#116
Feb9-13, 02:31 PM
Sci Advisor
HW Helper
P: 6,684
Quote Quote by DaleSpam View Post
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
DaleSpam
#117
Feb9-13, 04:55 PM
Mentor
P: 17,541
Quote Quote by Andrew Mason View Post
That reaction force ends as soon the centripetal acceleration of the astronaut ends.
Yes, clearly.

Quote Quote by Andrew Mason View Post
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.

Quote Quote by Andrew Mason View Post
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.
Andrew Mason
#118
Feb9-13, 08:56 PM
Sci Advisor
HW Helper
P: 6,684
Quote Quote by DaleSpam View Post
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
DaleSpam
#119
Feb9-13, 09:59 PM
Mentor
P: 17,541
Quote Quote by Andrew Mason View Post
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.

Quote Quote by Andrew Mason View Post
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.
Andrew Mason
#120
Feb9-13, 11:06 PM
Sci Advisor
HW Helper
P: 6,684
Quote Quote by DaleSpam View Post
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
A.T.
#121
Feb10-13, 03:41 AM
P: 4,232
Quote Quote by DaleSpam View Post
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.
DaleSpam
#122
Feb10-13, 07:46 AM
Mentor
P: 17,541
Quote Quote by Andrew Mason View Post
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.
Andrew Mason
#123
Feb10-13, 08:37 AM
Sci Advisor
HW Helper
P: 6,684
Quote Quote by A.T. View Post
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
A.T.
#124
Feb10-13, 09:53 AM
P: 4,232
Quote Quote by Andrew Mason View Post
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.
Quote Quote by Andrew Mason View Post
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.
Quote Quote by Andrew Mason View Post
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.
Quote Quote by Andrew Mason View Post
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.
DaleSpam
#125
Feb10-13, 03:41 PM
Mentor
P: 17,541
Quote Quote by A.T. View Post
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.
rcgldr
#126
Feb10-13, 03:50 PM
HW Helper
P: 7,180
Quote Quote by A.T. View Post
... 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.


Register to reply

Related Discussions
G-Force and (possibly) Centrifugal force applied to the rotation of a car Introductory Physics Homework 0
Reactive force on radiation Classical Physics 5
Reactive Centrifugal Force Classical Physics 403
Reactive centifugal force. Classical Physics 9
Can gravitational force be reactive? General Physics 5