# What exactly is the reactive centrifugal force (split)

by A.T.
Tags: centrifugal, force, reactive, split
P: 3,178
 Quote by rcgldr 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.
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P: 6,355
 Quote by A.T. 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
P: 3,178
 Quote by DaleSpam 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.

 Quote by DaleSpam 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.
P: 3,178
 Quote by Andrew Mason 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.
 Quote by Andrew Mason There is no outward force.
There are even two in the rotating frame.
 Quote by Andrew Mason 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.
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P: 6,772
 Quote by A.T. 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.
P: 3,178
 Quote by rcgldr 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.
Mentor
P: 15,601
 Quote by A.T. 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.
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P: 6,772
 Quote by rcgldr Wouldn't the sail be reflecting the particles back towards the sun?
 Quote by A.T. 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).
P: 3,178
 Quote by DaleSpam 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.
P: 3,178
 Quote by rcgldr 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.
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P: 6,772
 Quote by A.T. 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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P: 15,601
 Quote by A.T. 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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 Quote by A.T. 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
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P: 15,601
 Quote by Andrew Mason 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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P: 6,355
 Quote by DaleSpam 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.
You can call it what you like. It is due to inertia and it does not accelerate anything away from the centre of rotation.

AM
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P: 15,601
 Quote by Andrew Mason It does not accelerate anything away from the centre of rotation.
Whether it does or not is irrelevant to the naming convention.

But in some cases it clearly does.
 HW Helper Sci Advisor P: 6,355 We have beaten this issue to death. So let me recap where I think we differ. I think we fundamentally disagree on what third law action and reaction pairs are. It is easy in a collision of free bodies analysed from an inertial reference frame. One body accelerates in one direction and the other accelerates in the opposite direction - changes in momentum are equal and opposite. Or, if a rocket sends rocket gas in one direction, the rocket recoils in the other. These are third law action/reaction pairs. No problem. The problem occurs when you do not have a system that defines an inertial reference frame. For example, you say that the reaction to the centripetal force supplied by mechanical means is the force that the body being accelerated applies to the body or part of the body that is supplying the accelerating force. I say that is simply an inertial effect and the real reaction/action pairs are the centripetal force that the rotating body experiences and the centripetal force that this body exerts on the rest of the system as both rotate about their common centre of mass. Similarly, the third-law pair in applying a force to a box with my hand (with me standing on the earth) is the force I exert in the opposite direction on the earth. The change of momentum of the earth that results is equal and opposite to the change of momentum of the box. You seem to be saying that it is the force of the box on my hand. I say that force of the box on my hand is simply an inertial effect - a pseudo force really because it can never be associated with an acceleration. In my view, this is where we disagree. We can argue all we like about Newton's third law but we will continue to disagree because we take somewhat different view on the application of the law to these situations. AM
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P: 15,601
 Quote by Andrew Mason I think we fundamentally disagree on what third law action and reaction pairs are.
I agree that this is the fundamental disagreement. However, this is not simply a difference of equally-valid opinions. You wish to state Newton's 3rd law in terms of changes in momentum instead of forces.

1) You have been unable to produce any textbook or online reference which expresses the third law mathematically in terms of momenta, while I have produced several that express it mathematically in terms of forces.

2) You have also been unable to produce any textbook or online reference which even clearly expresses it verbally in terms of momenta.

3) You have been able to find one reference which is, by your own admission, quite unclear so a small part of it can be construed in a way that at least doesn't contradict your position. However, even that reference contradicts your position when you read the whole reference instead of taking that small part out of context.

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