Is the concept of reactive centrifugal force valid?

[Andrew Mason]

Would you not agree that the merry-go-round must exert a centripetal force on the rider? If so, then the rider exerts and equal and opposite (in this case, outward) force on the merry-go-round. Simple as that. (Call that force what you will.)

All the molecules in the merry-go-round and rider exert forces on each other. But the net force on each particle is a centripetal force. Those centripetal forces sum to zero if the mass is distributed symmetrically in the merry-go-round (ie. the centre of rotation is the centre of mass). (They do not sum to zero if the centre of rotation is not the centre of mass, in which case there is a net force on the Earth which has to be included.) But from Newton's third law, we know that all action/reaction pairs must sum to zero. Therefore, the centripetal forces on all the molecules in the merry-go-round must include all action and reaction pairs. So the reaction force to a centripetal force must be another centripetal force. There can be no centrifugal reaction force. If you add in a "real" centrifugal force they do not sum to zero.

The confusion arises if the merry-go-round is not balanced so that the (apparent) centripetal forces about the centre of the merry-go-round do not sum to zero. In order to make these apparent centripetal forces sum to 0 one has to postulate a centrifugal force on the centre of rotation. But that is a mistake because the centre of the merry-go-round is not an inertial frame in that case. There is a force on the Earth that accelerates the earth, and the merry-go-round centre, towards the centre of mass of the earth/merry-go-round system. In order to see this, one has to think of the Earth as being much less massive than it is.

AM

 

[Doc Al]

All the molecules in the merry-go-round and rider exert forces on each other.

Well, the ones in contact do, but OK.

But the net force on each particle is a centripetal force.

True.

Those centripetal forces sum to zero if the mass is distributed symmetrically in the merry-go-round (ie. the centre of rotation is the centre of mass). (They do not sum to zero if the centre of rotation is not the centre of mass, in which case there is a net force on the Earth which has to be included.) The forces include all action and reaction pairs.

Sure, the net force is zero. So what? Third law pairs are actual, individual forces, not 'net forces'.

So the reaction force to a centripetal force must be another centripetal force.

Nonsense.

There can be no centrifugal reaction force.

Answer this question: Does the section of the merry-go-round in contact with the rider exert a force on the rider?

And this: What direction is that force?

And this: Do you agree that Newton's 3rd law applies?

 

[D H]

But my point is that the reaction force to the centripetal force that is applied to the merry-go-round rider is not a centrigufal reaction force.

Would you not agree that the merry-go-round must exert a centripetal force on the rider? If so, then the rider exerts and equal and opposite (in this case, outward) force on the merry-go-round. Simple as that. (Call that force what you will.)

I agree with Andrew here: The equal but opposite reaction to a real centripetal force is a real centripetal force. It is directed inward, not outward. You have to look at it all squinty-eyed and such to see the reaction force as outward.

A different example: Consider a binary star system; call the two stars A and B. The gravitational force exerted on Star A by star B is directed toward star B and thus toward the system center of mass. This force makes star A's trajectory curve inward toward the center of mass. It is a centripetal force. The gravitational force exerted on Star B by star A is an equal but opposite force and causes star B's trajectory curve inward toward the center of mass. It is also a centripetal force.

The same goes for your merry-go-round. The force on the merry-go-round when viewed from the perspective of an inertial frame is centripetal, not centrifugal.

I submit that there is such a thing as a reactive centrifugal force. The equal but opposite reaction to a real centrifugal force is a real centrifugal force. For example, the interactions between two positively charged objects are equal but opposite centrifugal forces.

This however is not what that nonsense wikipedia article is talking about.

 

[Doc Al]

The same goes for your merry-go-round. The force on the merry-go-round when viewed from the perspective of an inertial frame is centripetal, not centrifugal.

Obviously the net force on some segment of the merry-go-round must be centripetal. But the reaction to the centripetal force that the merry-go-round exerts on the rider must be an outward 'centrifugal' force on the merry-go-round.

No squinting involved!

 

[D H]

Obviously the net force on some segment of the merry-go-round must be centripetal. But the reaction to the centripetal force that the merry-go-round exerts on the rider must be an outward 'centrifugal' force on the merry-go-round.

No squinting involved!

Lots of squinting involved!

Rhetorical question: What is it that makes the merry-go-round+rider situation different from the binary star situation? The answer is that the merry-go-round is attached to the Earth. That attachment is just a distraction. So, let's put the merry-go-round out in space. The result is an astronaut in a rotating space station. The astronaut feels something very much like a gravitational force due to the rotation of the space station.

What's happening to the rotating space station? The answer is that it too is undergoing a uniform circular motion. The center of mass of the rotating space station is orbiting about the space station + astronaut center of mass. The curvature is inward, not outward. The space station is therefore subject to a centripetal force, just as is the astronaut.

 

[Doc Al]

What's happening to the rotating space station? The answer is that it too is undergoing a uniform circular motion. The center of mass of the rotating space station is orbiting about the space station + astronaut center of mass. The curvature is inward, not outward. The space station is therefore subject to a centripetal force, just as is the astronaut.

Again you confuse the net force on an element of the space station with the force on it due to the astronaut.

 

[D H]

Again you confuse the net force on an element of the space station with the force on it due to the astronaut.

What net force? Assume the space station is in deep space; the only force acting on it is the force exerted on it by the astronaut.

 

[A.T.]

But the net force on each particle is a centripetal force.

We are not interested in the net force on the merry-go-round-particle. We are interested in the force that rider-particles exert on merry-go-round-particles. That force is called "reactive centrifugal force".

The fact that the net force on the merry-go-round-particles is centripetal does not change the fact that the force from rider-particles on merry-go-round-particles is centrifugal.

But that is a mistake because the centre of the merry-go-round is not an inertial frame in that case. There is a force on the Earth that accelerates the earth,

Nonsense. I told you already how to balance the merry-go-round:

Just put a second rider of equal mass on the opposite side. Both riders will exert reactive centrifugal forces on their seats. These forces point away from :
- the center of rotation
- the center of mass

You can put the whole thing into space. Considering the interaction with the Earth is not needed here, unless you want to continue to confuse yourself.

 

[Doc Al]

What net force? Assume the space station is in deep space; the only force acting on it is the force exerted on it by the astronaut.

Forget the astronaut for a moment. You have a gigantic space station spinning around its center of mass. Do you not agree that an element of the space station has forces acting on it due to the surrounding material?

 

[RonL]

This thread has taught me one important thing..., not to be embarrassed so much, when I'm not able to understand some simple physics 101 principles.
I wonder if a thought example might help me get a clear picture of what seems to be the point of such confusion.

If on one of the spokes of the merry-go-round, anywhere toward the outer edge, a vertical stand post (rigid) that has a pivot point for a bar or tube which is perfectly balanced and free to rotate (like a prop) assume no wind interference. If a strain gauge was in position to measure any pressure top or bottom, inside and outside, what measurement would show and where ?
If some out of balance be needed, where would it need to be placed to show any thing being discussed ?

Ron

P.S. Just noticed this is post 666

 

[Dale]

The center of mass of the rotating space station is orbiting about the space station + astronaut center of mass. The curvature is inward, not outward. The space station is therefore subject to a centripetal force, just as is the astronaut.

Your first two sentences are correct, but the third sentence does not follow. The first two sentences correctly establish the direction of the force, but to determine whether or not a given force is centripetal or centrifugal requires not only specification of the direction of the force, but also it's point of application.

Assuming that the center of rotation is the origin and that a force is pointing in the positive x direction then the force is centrifugal if applied on the positive x axis, centripetal if applied on the negative x axis, and tangential if applied on the y axis.

In this scenario the force is applied at the location of the astronaut and winds up being centrifugal, despite the fact that it is causing uniform circular motion of the center of mass. I.e. It is not applied at the center of mass so the motion of the center of mass is not sufficient.

 

[Andrew Mason]

I submit that there is such a thing as a reactive centrifugal force. The equal but opposite reaction to a real centrifugal force is a real centrifugal force. For example, the interactions between two positively charged objects are equal but opposite centrifugal forces.

I agree. That is an example of a real centrifugal force - and it does not require, or have anything really to do with, rotation.

This however is not what that nonsense wikipedia article is talking about.

That was my take - complete nonsense. For students of classical mechanics, better to stick to a good textbook. But even good textbooks can be misleading or confusing. Even Newton appears to be misleading (incorrect) in this aspect of rotational motion. The animated discussion on this thread highlights the challenge and difficulty in analysing the physics of rotation.

AM

 

[Andrew Mason]

Sure, the net force is zero. So what? Third law pairs are actual, individual forces, not 'net forces'.

The fact is that there is acceleration always occurring. These are not balanced static forces. The third law pairs are forces that are each causing acceleration.

Answer this question: Does the section of the merry-go-round in contact with the rider exert a force on the rider?

And this: What direction is that force?

The merry-go-round exerts a force on the rider and it is toward the centre of rotation.

And this: Do you agree that Newton's 3rd law applies?

Of course.

AM

 

[A.T.]

The third law pairs are forces that are each causing acceleration.

Wrong. There is nothing in Newtons 3rd Law about the acceleration of the interacting objects. When I lean against the wall, there are two equal opposite forces acting on me and the wall respectively. But neither me nor the wall is accelerating.


That is Newtons 3rd Law:

The merry-go-round exerts a force on the rider and it is toward the centre of rotation, ...

...while the rider exerts a force on the merry-go-round and it is away from the centre of rotation.

 

[A.T.]

Forget the astronaut for a moment.

Or make it 2 astronauts, on opposite sides. Now we have no confusion about center of mass vs. center of rotation and momentum conservation

 

[Doc Al]

Or make it 2 astronauts, on opposite sides. Now we have no confusion about center of mass vs. center of rotation and momentum conservation

Nicely done.

 

[Andrew Mason]

Just put a second rider of equal mass on the opposite side. Both riders will exert reactive centrifugal forces on their seats. These forces point away from :
- the center of rotation
- the center of mass

There is a centrifugal effect of the rider on the chair. But it is not a force. If the chair breaks loose from the merry-go-round there is no acceleration supplied by the rider to the chair. They both would just move in uniform motion in a direction tangential to the merry-go-round.

While the rider rotates, the merry-go-round as a whole, not just the seat, exerts a centripetal force on him. He, in turn, exerts a (reaction) force on the whole merry-go-round as a whole, not just the seat. The rider opposite asserts a force on the whole merry-go-round as a whole, not just his seat and that force is exactly equal and opposite to the force that the first rider exerts on the merry-go-round. Those are true forces.

These two riders could stay on the merry-go-round by letting go of poles or seats and just holding a rope between them. There would be no forces between the riders and the merry-go-round at all. Would you then argue that the riders are exerting a centrifugal force on each other?

AM

 

[Doc Al]

There is a centrifugal effect of the rider on the chair. But it is not a force.

So the chair can exert a force on the rider without the rider exerting a force on the chair? Wow!

While the rider rotates, the merry-go-round as a whole, not just the seat, exerts a centripetal force on him.

Mysterious non-local forces? Wow!

 

[Andrew Mason]

Wrong. There is nothing in Newtons 3rd Law about the acceleration of the interacting objects. When I lean against the wall, there are two equal opposite forces acting on me and the wall respectively. But neither me nor the wall is accelerating.

Fine. These are just balanced forces so neither I nor the wall accelerates. But these are not really the action/reaction pairs of forces that the third law speaks about. Those are dynamic: they occur when masses experience accelerations eg in collisions of matter.

Suppose I am standing on a floor. My weight is balanced by the normal force of the floor. But if the floor gives way, the acceleration that the pieces of the floor experience is not from my weight pushing on them. It is from gravity. The reaction force to the gravitational force on me is my gravitational pull on the earth. The same goes for the floor: the reaction force to the gravitational acceleration of the floor pieces is their gravitational pull on the earth.

AM

 

[Andrew Mason]

So the chair can exert a force on the rider without the rider exerting a force on the chair? Wow!

The rider exerts a force on the chair only because the chair is connected to the merry-go-round. If the chair is not connected, the rider exerts no force on it. So it is incorrect to say that the rider exerts a force on just the chair.

When you analyse collisions between billiard balls, do you say that the balls only exert forces on the molecules that they are in contact with? Can you ignore the fact that the molecules of each ball are all connected? If you do, you will not be able to explain the changes in motion that occur.

AM

 

[Dale]

There is nothing in Newtons 3rd Law about the acceleration of the interacting objects. When I lean against the wall, there are two equal opposite forces acting on me and the wall respectively.

These are just balanced forces so neither I nor the wall accelerates. But these are not really the action/reaction pairs of forces that the third law speaks about.

A.T. Is correct. These are exactly the action/reaction pairs the third law speaks about. The third law is not restricted to collisions nor to accelerating objects.

 

[Andrew Mason]

Or make it 2 astronauts, on opposite sides. Now we have no confusion about center of mass vs. center of rotation and momentum conservation

These drawings are very well done but are not correct. In the inertial frame the only forces are centripetal forces. The astronauts and the space station are undergoing net acceleration so there is a net (centripetal) force on each astronaut and on each part of the space station. The reaction centripetal forces on each element of mass are conducted through the space station structure to provide the centripetal force on the mass opposite. To see this, remove one of the astronauts and see what happens to the centre of rotation.

In the rotating frame, the "forces" need to balance (ie sum to 0) as the acceleration is not apparent. The non-inertial astronaut postulates a centrifugal force to balance the force he feels pressing into him from the space station.

AM

 

[D H]

Wrong. There is nothing in Newtons 3rd Law about the acceleration of the interacting objects. When I lean against the wall, there are two equal opposite forces acting on me and the wall respectively. But neither me nor the wall is accelerating.

Fine. These are just balanced forces so neither I nor the wall accelerates. But these are not really the action/reaction pairs of forces that the third law speaks about. Those are dynamic: they occur when masses experience accelerations eg in collisions of matter.

A.T. is correct here, Andrew. The force that the wall exerts on A.T. and the force that A.T. exerts on the wall are exactly the kind of pair-wise forces that Newton's third law addresses. Force is subject to the superposition principle. The F in F=ma is the net force acting on the object (or on A.T. in this case). The reason A.T. doesn't accelerate when he leans against the wall is because other forces are in play.

A.T. isn't accelerating because the floor is exerting both a normal vertical force that counteracts gravity and a frictional horizontal force that counteracts the normal force the wall exerts on A.T. Should A.T. try to lean against a frictionless wall while standing on a frictionless floor he will accelerate. (You can see a demonstration of this principle when a newbie ice skater tries to lean against the skating rink wall.)

Equal-but-opposite is not sufficient to qualify as a third law forces. For example, the force that the wall exerts on A.T. and the (horizontal) frictional force that the floor exerts on A.T., although equal-but-opposite, are not third law forces.

 

[A.T.]

There is a centrifugal effect of the rider on the chair. But it is not a force.

Of course it is a force. Anything else would violate Newtons 3rd Law, because the chair exerts a force on the rider. And "centrifugal effect" is just vague gibberish.

If the chair breaks loose from the merry-go-round there is no acceleration supplied by the rider to the chair. They both would just move in uniform motion in a direction tangential to the merry-go-round.

Completely irrelevant for the argument above.

While the rider rotates, the merry-go-round as a whole, not just the seat, exerts a centripetal force on him.

No, the interaction is local. The point of application of both forces (centripetal on the rider, and reactive centrifugal on the seat) is at the contact area between rider and seat. As DaleSpam already explained: You have to take this application point and the direction of the force into account, to determine if it's centrifugal or centripetal.

 

[A.T.]

Or make it 2 astronauts, on opposite sides. Now we have no confusion about center of mass vs. center of rotation and momentum conservation

These drawings are very well done but are not correct. In the inertial frame the only forces are centripetal forces.

Wrong. You are confusing all forces with all net forces on each part.

The astronauts and the space station are undergoing net acceleration so there is a net (centripetal) force on each astronaut and on each part of the space station.

Correct, but irrelevant. See above. There can still be centrifugal forces acting, even if all the net forces on individual parts are centripetal.

To see this, remove one of the astronauts and see what happens to the centre of rotation.

I made it balanced so you don't get confused. But you insist on getting confused? Removing the other astronaut will not change the fact that:

F_rcf has the same direction, as the vector from rotation center to the point of application of F_rcf. Therefore F_rcf is a centrifugal force

The net acceleration of the space station's COM is irrelevant here. But I made it zero, deal with it.

In the rotating frame, the "forces" need to balance (ie sum to 0) as the acceleration is not apparent. The non-inertial astronaut postulates a centrifugal force to balance the force he feels pressing into him from the space station.

Correct. You understand the inertial centrifugal force well. You should be able to tell it apart from the reactive centrifugal force based on the differences I listed in the legend.

 

[D H]

I made it balanced so you don't get confused.

I disagree. Your making it balanced is an obfuscating complication. We're talking Newtonian mechanics here, and ultimately Newtonian mechanics is about a set of point masses (aka particles) that interact via third law pairs.

So let's make it simple rather than complex. Consider just a pair of particles that interact via some central force. If the force is attractive both particles undergo either straight line or centripetal motion when viewed from the perspective of an inertial observer. The particles undergo either straight line or centrifugal motion if the force is repulsive.

The concept of a reactive centrifugal force to a centripetal force does not jibe with Newtonian mechanics in the case of point masses. So when does the concept of reactive centrifugal force to a centripetal force apply? When you over-complicate and obfuscate things.

 

[Dale]

These drawings are very well done but are not correct. In the inertial frame the only forces are centripetal forces.

The drawings are correct, this analysis is not correct. The 3rd law pairs are, as A.T. drew them, between each astronaut and the space station. This can be seen by placing scales or pressure plates at the foot of each astronaut. Each scale registers a 3rd law pair, so there are four forces.

In the rotating frame, the "forces" need to balance (ie sum to 0) as the acceleration is not apparent. The non-inertial astronaut postulates a centrifugal force to balance the force he feels pressing into him from the space station.

This is correct, and agrees with A.T.'s drawing.

 

[Doc Al]

So let's make it simple rather than complex. Consider just a pair of particles that interact via some central force. If the force is attractive both particles undergo either straight line or centripetal motion when viewed from the perspective of an inertial observer. The particles undergo either straight line or centrifugal motion if the force is repulsive.

How about sticking to the example of the merry-go-round that A.T. nicely illustrated?

Obviously, in the special case of the two particles attracting at a distance the only forces happen to be centripetal. So what? Stick to the example.

 

[Dale]

I disagree. Your making it balanced is an obfuscating complication.

Whether it is complicated or not, it is still a legitimate scenario that Newton's laws can analyze.

We're talking Newtonian mechanics here, and ultimately Newtonian mechanics is about a set of point masses (aka particles) that interact via third law pairs.

I disagree with this in general. Newtonian mechanics does just fine with extended bodies and continuums. However, for the sake of argument let's analyze the point particle of a space-station floor atom directly underneath the astronaut's boot. There are two forces acting on this atom. One is the contact force from the boot, and the other is the strain force from the neighboring floor atoms. These two forces are not equal, so there is a net force acting on the atom, and the net force is centripetal.

The strain force acts in the same direction as the net force, so it is also centripetal, but the contact force acts in the direction opposite the net force, so it is centrifugal.

Looking at the interacting atoms, the reaction force to the centrifugal contact force acting on the floor atom is a centripetal contact force acting on the boot atom, and the reaction force to the centripetal strain force acting on the floor atom is a centrifugal strain force acting on the neighboring atoms.

So let's make it simple rather than complex. Consider just a pair of particles that interact via some central force. If the force is attractive both particles undergo either straight line or centripetal motion when viewed from the perspective of an inertial observer. The particles undergo either straight line or centrifugal motion if the force is repulsive.

I agree, in an isolated two particle central force scenario the third-law pair of forces will be the same (centrifugal or centripetal). But Newton's laws can be used to analyze situations more complicated than two isolated point particles acting via some central force, and in such situations it is indeed possible to have the 3rd law pairs being different (centrifugal or centripetal).

The concept of a reactive centrifugal force to a centripetal force does not jibe with Newtonian mechanics in the case of point masses.

No, it only doesn't jibe in the case of 2 isolated point masses. It jibes with more general cases of point masses.

 

[A.T.]

A.T. Is correct. These are exactly the action/reaction pairs the third law speaks about. The third law is not restricted to collisions nor to accelerating objects.

In fact usually static examples are used introducing Newton's 3rd Law:

From
http://www.grc.nasa.gov/WWW/k-12/WindTunnel/Activities/third_law_motion.html

The book lying on the table is exerting a downward force on the table, while the table is exerting an upward reaction force on the book. Because the forces are equal and opposite, the book remains at rest. Notice also that the table legs are in contact with the floor and exert a force downward on it, while the floor in turn exerts an equal and opposite force upward.

The third law pairs are forces that are each causing acceleration.

I have never come across a book limiting Newtons 3rd to net forces which are directly related to acceleration. Can you provide us with a reference that supports your statement?