Difference between centrifugal force vs reactive centrifugal force?

[Aeronautic Freek]

Difference between centrifugal force vs reactive centrifugal force?
and practic example

 

[A.T.]

Difference between centrifugal force vs reactive centrifugal force?
and practic example

 

[Dale]

Interaction forces (also called “real forces“, but I prefer the term “interaction” as it is less philosophical and more descriptive) arise from interactions between objects. Importantly these interactions follow Newton’s 3rd law and they come in pairs.

Inertial forces (also called “fictitious forces”, but I prefer the term “inertial” as it is less dismissive and more descriptive) arise from the choice of reference frame and disappear in an inertial frame. They do not come in pairs nor follow Newton’s third law.

Importantly, the acceleration produced by inertial forces can not be detected by accelerometers.

Consider a rock on the end of a string being spun in a circle (in the absence of gravity). In an inertial frame the rock is accelerating inward. There is an inward-pointing interaction force from the string acting on the rock, this is the centripetal force. (Edit) An accelerometer on the rock detects the inward acceleration produced by this force.

By Newton’s 3rd law there is an outward-pointing interaction force from the rock acting on the string. This is the reactive centrifugal force. Note that it is an interaction force.

In a co-rotating frame the rock is stationary. It is still acted on by the centripetal interaction force from the rope. In order to explain its motion in this frame we need to invoke an outward pointing inertial force called the inertial centrifugal force. This force does not result from any interaction, it does not have a 3rd law pair, it is not detected by the accelerometer, and it only exists in the rotating reference frame. That said, it is necessary for describing the motion of the rock in the rotating reference frame

 

[Aeronautic Freek]

View attachment 264226

1)is co-rotating frame same name for rotating frame and non-inertial frame?

2)why in co-roating frame we have Frcf?
Isnt Fcp=Ficf or Fcp=Ficf+Frcf ?

I never see in tasks using Frcf for roating frames?

 

[A.T.]

1)is co-rotating frame same name for rotating frame and non-inertial frame?

Co-rotating means rotating with the space station. All 3 bodies are at rest in that frame, so the forces on each of them must balance.

2)why in co-roating frame we have Frcf?

Because it's a frame invariant interaction force. It can be measured with a scale at the wall of the station, and that scale will read the same in every frame.

 

[Aeronautic Freek]

In a co-rotating frame the rock is stationary. It is still acted on by the centripetal interaction force from the rope. In order to explain its motion in this frame we need to invoke an outward pointing inertial force called the inertial centrifugal force. This force does not result from any interaction, it does not have a 3rd law pair, it is not detected by the accelerometer, and it only exists in the rotating reference frame. That said, it is necessary for describing the motion of the rock in the rotating reference frame

Where is reactive centrifugal force in co-rotating frame in your discription, A.T. picture say that reactive centifugal force exist in every frame?

 

[Aeronautic Freek]

Because it's a frame invariant interaction force. It can be measured with a scale at the wall of the station, and that scale will read the same in every frame.

So what is realtion of magnitudes of Ficf and Frcf,are they equal?

 

[A.T.]

So what is realtion of magnitudes of Ficf and Frcf, are they equal?

There is no general relation between their magnitudes, even if they are equal in some cases. Also note that they act on different bodies.

 

[Aeronautic Freek]

There is no general relation between their magnitudes, even if they are equal in some cases. Also note that they act on different bodies.

So what is free body diagram for astronaut in co-roataing frame?
Fcp= Frcf or Fcp=Ficf + Frcf ?

 

[A.T.]

So what is free body diagram for astronaut in co-roataing frame?
Fcp= Frcf or Fcp=Ficf + Frcf ?

Frcf is not acting on the astronaut. The red forces act on the astronaut:
Fcp + Ficf = 0

 

To be honest, I had never actually heard of the term "reactive centrifugal force", though it seems to exist (perhaps this is ignorance on my part...). In any case, it just seems like a confusing terminology.

If there is a body performing some sort of motion in some frame of reference, then the component of the resultant force (which might have contributions from any number of individual forces...) orthogonal to the velocity vector is termed the centripetal force (i.e. the normal component of force in an intrinsic coordinate system). If there is only one force contributing to the centripetal force (i.e. tension in a string, a gravitational force from one other star, etc.) then I suppose the term makes sense: it's just the other force in that force pair.

However, what if there are two strings providing tension forces that contribute to the centripetal force? Then you have a reaction force on each string: which is the "reactive centrifugal force" then?

In any case, it seems like an unnecessary concept. A centrifugal force is a body force $mr\omega^2$ that arises in a rotating frame of reference and a centripetal force is the normal component of the resultant force. Why not just leave it at that, and call the other forces what they really are?

 

[Aeronautic Freek]

Frcf is not acting on the astronaut. The red forces act on the astronaut:
Fcp + Ficf = 0

in my logic Frcf and Ficf are same force and must be same in magnitude ...mass x Vsquare / radius
Fcp is also mass x Vsquare / radius..
so they are all same magnitude with different direction

 

[A.T.]

...it just seems like a confusing terminology...

We already have a 17 pages thread about that terminology:
https://www.physicsforums.com/threads/reactive-centrifugal-force.523212/
We don't need another one.

 

[A.T.]

in my logic Frcf and Ficf are same force ...

They don't even act on the same body.

 

[Aeronautic Freek]

forget about semantics,and just look at real life everywhere around you..

 

[PeroK]

forget about semantics,and just look at real life everywhere around you..

Forget about physics, you mean!

 

[Ibix]

forget about semantics,and just look at real life everywhere around you..

This entire issue is semantics. Nobody is disputing that it's possible to drive on a wall. The only question is about how to name the forces when you do so...

 

[A.T.]

forget about semantics...

Isn't semantics your whole argument? You think two forces acting on different objects are the same force, because some people use similar names for them.

 

[Aeronautic Freek]

This entire issue is semantics. Nobody is disputing that it's possible to drive on a wall. The only question is about how to name the forces when you do so...

CENTRIFUGAL FORCE

(i waiting to someone who will now write that centripetal force keep these car on wall,becuse centrifugal force "dont exist"...)

 

[Doc Al]

(i waiting to someone who will now write that centripetal force keep these car on wall,becuse centrifugal force "dont exist"...)

You do realize that the cars are accelerating, right?

 

[PeroK]

CENTRIFUGAL FORCE

Let me get this right. There is a real centrifugal force on the car, pushing it out. The car pushes against the wall and receives an equal and opposite centripetal (normal) force from the wall. These two real forces on the car balance, leaving the car with a net zero force. The car, therefore, remains at rest or continues with uniform motion in a straight line ...? That can't be right!

Maybe the centripetal force from the wall is not real? But, that's a real contact force with the wall - that must be real. So, it must be that there is no centrifugal force on the car!

So, there is only a centripetal force on the car from the wall. That means the car must be accelerating towards the centre. What do we know about unifrom circular motion? That there is an acceleration towards the centre! Aha, that explains it: centripetal force on the car = acceleration towards the centre = circular motion.

Isn't physics wonderful!

 

The difference between the rotating frame and the inertial frame in this case is just the side of the equation on which you choose to put the $ma$ term . To be specific, in the inertial frame the thing is doing circular motion:$$F_r = ma_r \implies -N = -mr\dot{\theta}^2$$But if you move that $mr\omega^2$ term to the LHS, you end up with $$-N + mr \dot{\theta}^2 = 0$$You will notice that in the rotating frame, the radial acceleration of the car is zero (it's radial coordinate is always $R$). In order to make Newton's laws work right, note that we need to insert an apparent force of $mr\dot{\theta}^2$ away from the axis of rotation!

It might be worth brushing up on circular motion. I personally wouldn't stress so much over the so-called "centrifugal reaction force".

 

[Dale]

Where is reactive centrifugal force in co-rotating frame in your discription, A.T. picture say that reactive centifugal force exist in every frame?

Yes, it is the third law reaction force to the centripetal force. So it is exerted by the rock on the rope in all frames.

 

[Dale]

in my logic Frcf and Ficf are same force and must be same in magnitude

Your logic has some gaps:

1) they act on different objects
2) the reactive force exists in all frames
3) the inertial force exists only in rotating frames
4) the inertial force violates Newton’s 3rd law

 

[Aeronautic Freek]

Your logic has some gaps:

1) they act on different objects
2) the reactive force exists in all frames
3) the inertial force exists only in rotating frames
4) the inertial force violates Newton’s 3rd law

show me equation for magnitude for Ficf and Frcf

 

[A.T.]

show me equation for magnitude for Ficf and Frcf

The magnitude of Frcf depends on the rotation rate of the station.
The magnitude of Ficf depends on the rotation rate of the chosen reference frame.
Their magnitudes are not the same in general, just in some reference frames.

And as already said, they act on different objects.

 

[Aeronautic Freek]

The magnitude of Frcf depends on the rotation rate of the station.
The magnitude of Ficf depends on the rotation rate of the chosen reference frame.
Their magnitudes are not the same in general, just in some reference frames.

And as already said, they act on different objects.

so If we put weight scale between astronaut feet and station wall,scale will show Frcf and it will be
mass x Vsquare / radius ?
(assumption that weight scale is massless)

 

[Ibix]

so If we put weight scale between astronaut feet and station wall,scale will show Frcf and it will be
mass x Vsquare / radius ?
(assumption that weight scale is massless)

It will show $F_{\mathrm{rcf}}=F_{\mathrm{cp}}=m\omega_{\mathrm{station}}^2r=mv_{\mathrm{station}}^2/r$, yes. It will show this whatever frame you choose to use.

 

[A.T.]

so If we put weight scale between astronaut feet and station wall,scale will show Frcf and it will be
mass x Vsquare / radius ?
(assumption that weight scale is massless)

If "V" is the tangential velocity of the astronaut in the inertial frame frame, then yes.

 

[Delta2]

This is probably one of the biggest misconceptions in classical mechanics where i believe our intuition can play nasty tricks on us.
Lets take for example that example of the car driving at wall. The centripetal force is the result of the normal force from wall and from the weight of the car. Because the wall pushes the car with a normal force, the car pushes the wall with an equal and opposite force , that's Newton's 3rd law.

Now we can imagine , as an artifact of our intuition or our imagination, that an invisible hand pushes the car against the wall, with a force equal to centrifugal force, and that's why the car pushes on the wall, but it is all about our intuition misguiding us, there is no invisible hand and no centrifugal force, the car pushes on the wall as the result of Newton's 3rd law, not because an invisible hand is pushing it against the wall.

Centrifugal force is not a real, or more precisely an interaction force, it is just a term that appears when we do the math in a rotating frame of reference. But if we let our intuition mislead us (like my intuition was misleading me when i was at secondary high school 30 years ago) then we might tend to believe that it is a real force.

 

[A.T.]

Centrifugal force is not a real, or more precisely an interaction force, ...

Part of the problem is using the "real" label for interaction forces. It tends to turn technical discussions into philosophical musings.

 

[Aeronautic Freek]

The magnitude of Ficf depends on the rotation rate of the chosen reference frame.
Their magnitudes are not the same in general, just in some reference frames.

I still don't understand why Ficf and Frcf are not same in magnitude...
can you make simple example with numbers..

 

[jbriggs444]

By Newton’s 3rd law there is an outward-pointing interaction force from the rock acting on the string. This is the reactive centrifugal force. Note that it is an interaction force. An accelerometer on the rock detects the inward acceleration produced by this force.

I think that you mis-spoke here, @Dale. The phrase "this force" refers to the reactive centrifugal force from the rock acting on the string. There is no inward acceleration of the rock produced by "this force". First, because it acts outward and second because it acts on the string. The inward acceleration of the rock is, of course, produced by the centripetal force.

 

[A.T.]

I still don't understand why Ficf and Frcf are not same in magnitude...
can you make simple example with numbers..

In the inertial frame Ficf is zero, while Frcf is not.

 

[Dale]

show me equation for magnitude for Ficf and Frcf

A vector has more than just a magnitude. However, here are the equations in an inertial frame:
$|F_{icf}|=0$
$|F_{rcf}|=|F_{cf}|$