What exactly is the reactive centrifugal force (split)

A.T.

That is true in this case. You have been given other examples, where the force is centrifugal in an inertial frame of reference.
No, that is nonsense in any case. A fictitious force is never part of a 3rd law pair.

DaleSpam

Are you talking about quantum mechanics here? I don't think that is relevant to the discussion which is essentially classical.

And my point is that this is not generally true. Often you cannot assume perfect rigidity and often you will want to analyze the internal stresses of a rotating body, for instance, in turbine blade design. These internal stresses are well-characterized and understood despite your unsupported assertions to the contrary. I can provide a list of references for this if you wish, essentially any statics textbook, but it is simply not true that we can only know about forces which cause accelerations.

Point well taken. "Clearly" is a matter of opinion. I should have said that it was clear to me, at least the part about the ball. I agree that the gravity example is wrong, but the ball example is correct and clear.

A human is an extended body, even if the center of rotation is between the COM of the human and the COM of the ball the reaction force on the hand will still be centrifugal. At least, that is how it has been any time I have swung a ball (or any other object) using my hand.

Andrew Mason

No mainstream texts, I see. The Mook and Vargish example of the planet and sun is just wrong.

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.

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. The centres of mass of the astronaut and the space ship both move at constant velocity in relation to an inertial point (although both would still rotate about their respective centres of mass).

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.

I don't see how that is a centrifugal effect at all. If it is not opposed there would be no centripetal force at all. There would be no rotation.

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. I don't see any forces directed away from the centre of mass. Total forces add to 0. I give myself and the box, respectively, equal and opposite changes in momentum. Both changes in momentum are toward the centre of mass.

Now, if I make the box more massive by ΔM, and apply the same force to the box as before (F), the box accelerates toward the centre of mass (which is now a different point) but with less acceleration (a' = F/(M+ΔM). I accelerate toward the centre of mass at the same rate as before (a=F/m). But the changes in momentum are the same for the box and me. So what seems like an effect directed away from the centre of mass if you only look only at the reduced acceleration of the large box, is not a reduction in the change of momentum of the box. There is no "effect" that is in a direction away from the centre of mass.

If you apply that to centripetal forces due to rotation, in the first case the less massive box and I rotate on the frictionless surface with acceleration toward the centre of mass: [itex]F_{c-box} = m_{box}r_{box}ω^2[/itex] and [itex]F_{c-me} = m_{me}r_{me}ω^2[/itex]

With the more massive box but with the same pulling force between me and the box, the box and I will rotate about a centre of mass of the system that is closer to the centre of mass of the box by Δr. The box will have less centripetal acceleration and I will have greater centripetal acceleration toward the centre of mass of the system (longer radius).[itex]F_{c-box} = m_{box}(r_{box}-\Delta r)ω^2[/itex] and [itex]F_{c-me} = m_{me}(r_{me}+\Delta r)ω^2[/itex]. If the centripetal forces do not change (i.e. my pull force on the box does not change), ω will have to be less so that my acceleration remains the same. Again, there is no centrifugal effect at all.

AM

DaleSpam

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.

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.

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.

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.

Andrew Mason

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 ma_c = 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

A.T.

It is a dead simple, minimal scenario. If your proposed naming reasoning gets complicated here already, then I doubt it will catch on.

There is no inertial space station in the rocket scenario. Here it is again:

A space ship 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.

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.

DaleSpam

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.

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.

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

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

jbriggs444

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

Ok. Fair enough.

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.

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.

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

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.

I note that yet again you are unable to produce any reference supporting one of your claims when asked to do so.

Andrew Mason

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.

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

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.

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

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 it is propagating laterally, how can it be the reaction to centripetal acceleration, which is entirely radial?

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^2r_com). 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

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.

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

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.

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

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.

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.