Connetion between centrifugal force and weight force

[Gabriele Pinna]

Can someone explain me what is the connetion between centrifugal force and weight force thanks ? In which situation should I use the centrifugal force (that is a fake force) ?

 

[jamie.j1989]

Centrifugal force isn't a fake force, tie a mass to the end of a string and spin around, the string gets taught and stops the mass from flying off, the string applies a force in the radial direction towards you, this is centrifugal force. Weight is a gravitational force.

 

[mathman]

Centrifugal force isn't a fake force, tie a mass to the end of a string and spin around, the string gets taught and stops the mass from flying off, the string applies a force in the radial direction towards you, this is centrifugal force. Weight is a gravitational force.

The force in the radial direction is call centripetal. Centrifugal "force" is the inertia which is trying to keep the object going in a straight line.

 

[jbriggs444]

The force in the radial direction is call centripetal. Centrifugal "force" is the inertia which is trying to keep the object going in a straight line.

What I believe jamie.j1989 is referring to is the reactive centrifugal force. https://en.wikipedia.org/wiki/Reactive_centrifugal_force. What Gabriele Pinna is talking about is the inertial centrifugal force. https://en.wikipedia.org/wiki/Centrifugal_force.

 

[hackhard]

weight force = net gravitational force on body
normal force(apparent weight) = mass * gforce = projection of contact force on body in dir normal to contact surface
centrifugal force = (-1) * (centripetal force)
=(-1) * (mass) * (projection of resultant acceleration in dir normal to dir of resultant velocity of body )
= (-1) * ( (radial proj of normal force) + (radial proj of force 2) + (radial proj of force 3) ... )
op must be referring to normal force ---
there is no special connection between centrifugal force and normal force
op must be referring to case where normal force or gravitational force is acting only in radial dir and no other force acts
gravitional force - supppose satallite orbits Earth . net force acting is gravity towards centre . so centrifugal force on sat = mag of gravity away from centre
normal force - suppose lolipop spins about axis . net force on ball of lolipop is normal force (or tension) due to stick inwards. so centrifugal force on ball of lolipop = mag of tension outwards

In which situation should I use the centrifugal force (that is a fake force) ?

add centrifugal force when in rotating frame
you (at rest) see that lolipop spinning (net force = centripetal force = normal force !=0)
. ant on lolipop (at rest ) says lolipop at rest .both correct
so ant adds centrifugal force (due to change of reference frame)(net force = centrifugal force + centripetal = normal force + (- normal force) = 0

 

[vanhees71]

I'd rather call the centrifugal force an inertial force rather than a fictitious force, because then it's immediately clear that it occurs exclusively in non-inertial (rotating) reference frames.

 

[Andrew Mason]

What I believe jamie.j1989 is referring to is the reactive centrifugal force. https://en.wikipedia.org/wiki/Reactive_centrifugal_force. What Gabriele Pinna is talking about is the inertial centrifugal force. https://en.wikipedia.org/wiki/Centrifugal_force.

I'd rather call the centrifugal force an inertial force rather than a fictitious force, because then it's immediately clear that it occurs exclusively in non-inertial (rotating) reference frames.

I think this kind of terminology just makes it more confusing. There is no centrifugal "force" in the Newtonian sense. There is a tension that exists if there is a mechanical connection but that does not mean that there is a centrifugal "force". Nothing can be accelerated outward with such an "effect". If the mechanical connection fails, there is not even a centrifugal motion - there is uniform motion of the fleeing mass in a direction that is at right angles to the radial direction when the connection is lost.

The introduction of the concept of a "centrifugal reaction force" makes this very confusing. It makes it very difficult for students to distinguish between a "centrifugal reaction force" and a centrifugal force (the non-Newtonian force or inertial effect). Just look at the Wikipedia discussion on Reactive Centrifugal Force.

The problem is with the term "centrifugal". Every real force has a reaction force to which it is paired. Centripetal forces are not excepted. That is just Newton's third law. But it is confusing to refer to this reaction force as "centrifugal". It is not necessarily, if ever, centrifugal. If the centripetal force on a rotating body is supplied by a force acting at a distance - e.g. gravity, electro-magnetic force - all forces are centripetal: the reaction to the centripetal force force on one body is a centripetal force on the other bod(y)(ies). In the case where the force between masses is supplied by a mechanical connection, there are tensions in two (eg. tethered rotating mass) or many directions (eg. a rotating rigid body) and all masses experience centripetal accelerations. In a rotating system, the integral of all point masses multiplied by their accelerations sum to 0. So all net forces are centripetal. All tensions are bi- or multi-directional.

In any event, nothing can ever accelerate outward in a radial direction by virtue of a rotation. So I am not sure why anyone wants to refer to a centrifugal real force as a reaction to a centripetal force. In my view, it is clearer and much less confusing to say that in any rotating system, all net forces are centripetal and all forces (which includes "reaction" forces) are either purely centripetal forces and, in cases where there are mechanical connections, a combination of centripetal net forces and bi or multi-directional tensions.

AM

 

[A.T.]

In my view, it is clearer and much less confusing to say that in any rotating system, all net forces are centripetal and all forces (which includes "reaction" forces) are either purely centripetal forces and, in cases where there are mechanical connections, a combination of centripetal net forces and bi or multi-directional tensions.

Now, this sounds very unclear and confusing to me. In my view, it is better not to obsess so much about the "centrifugal"/"centripetal" terminology, or always provide a precise mathematical definition of those terms (which you didn't).

 

[Andrew Mason]

Now, this sounds very unclear and confusing to me. In my view, it is better not to obsess so much about the "centrifugal"/"centripetal" terminology, or always provide a precise mathematical definition of those terms (which you didn't).

No one is "obsessing" about anything. It is a matter of being clear on the physics. If one must talk about centrifugal "force", it is the apparent force that appears in a non-inertial (rotating) frame of reference. To suggest that the reaction force to a centripetal force should be called "the centrifugal reaction force" is just wrong in many cases and confusing where one is referring to tensions that necessarily always operate in at least two directions.

As far as mathematical treatment goes, the mathematical treatment of centripetal force, of course is trivial: $\vec{F_c} = -m\omega^2r \hat r$. For a rotating mass element dm subject to tensions, $d\vec{F} = \sum T_i = -dm\omega^2r \hat r$. How to apply to a particular case depends on the geometry of the system being considered.

AM

 

[Andrew Mason]

Centrifugal force isn't a fake force, tie a mass to the end of a string and spin around, the string gets taught and stops the mass from flying off, the string applies a force in the radial direction towards you, this is centrifugal force. Weight is a gravitational force.

Just to follow up on my last couple of posts, this is a perfect example of why it is so confusing to speak of the "reactive centrifugal force". The string applies a force to both the rotating mass and to "you". The real forces acting on the rotating mass and "you" are centripetal, not centrifugal. In the non-inertial frame of "you", there seems to be some force pulling the mass away from you so it seems centrifugal. That is not a real (Newtonian) force. "You" (the person who is holding the other end) are actually rotating (kind of wobbling) about an inertial point somewhere in between you and the rotating mass. The rope is supplying a centripetal force to both (the rotating) you and the (rotating) mass. So the string or rope is supplying centripetal forces at both ends.

Even if the mass is rotating about a post fixed to the earth, the centre of the post is not really an inertial point. It, and the Earth to which it is fixed, has to be actually rotating or wobbling a tiny bit about a point (very close to the centre of the post) in between.

AM

 

[A.T.]

The real forces acting on the rotating mass and "you" are centripetal, not centrifugal.

What about the real forces on the string?

 

[Andrew Mason]

What about the real forces on the string?

What about them?

The real force on any string element dm is the sum of the tensions acting on it: $\vec{dF} = \sum T_i = -dm\omega^2 r \hat r$. It is always toward an inertial point, which is the centre of rotation. r is the distance from the mass element to that centre of rotation. For the string as a whole, the tensions vary along the length of the string itself but the total net force is determined by integrating $\vec{F} =\int \vec{dF}= -\omega^2 \hat r \int r dm$. That is equivalent the centripetal force on a point mass located at the centre of mass of the rope.

AM

 

[A.T.]

The real force on any string element dm is the sum of the tensions acting on it: →dF=∑Ti=−dmω2r^rdF→=∑Ti=−dmω2rr^\vec{dF} = \sum T_i = -dm\omega^2 r \hat r. It is always toward an inertial point, which is the centre of rotation. r is the distance from the mass element to that centre of rotation.

That is the net force on a string element. I'm asking about the force exerted by the mass on the string.

 

[jtbell]

"You" (the person who is holding the other end) are actually rotating (kind of wobbling) about an inertial point somewhere in between you and the rotating mass. The rope is supplying a centripetal force to both (the rotating) you and the (rotating) mass. So the string or rope is supplying centripetal forces at both ends.

To make this more obvious, suppose you and a friend (with roughly equal mass) are holding opposite ends of the rope, and "orbiting" around each other while ice-skating or riding frictionless air-carts.

 

[jbriggs444]

To make this less obvious, consider a hub, as on a carnival ride with chains leading to the seats swinging around in a circle. The force exerted on the hub by the chains is in the centrifugal direction.

 

[Andrew Mason]

That is the net force on a string element. I'm asking about the force exerted by the mass on the string.

It is equal and opposite to the force of the string on the mass.

Whether that force is toward or away from the centre of rotation depends on where the centre of rotation is and what you mean by the string.

If you are asking: "what is the direction of the force that the man is applying to the string as a whole" you have to consider where the centre of mass of the string is in relation to the centre of rotation.

If you are asking: "what is the direction of the force that the man is applying to the mass element of the string of length dL at the end of the string" you have to consider where the centre of mass of the string element is in relation to the centre of rotation. If the centre of rotation is not within the string itself, the force is toward the centre of rotation. For example, where a 100 kg man is swinging a 1 kg rope and mass with his outstretched arms holding the rope, the centre of rotation may well be somewhere between the man's centre of mass and his outstretched hands. In that case, the force on the rope by the man is toward the centre of rotation no matter how you analyse it. And the rope provides a centripetal force to the man as a whole (i.e his centre of mass), even though the tension feels like it is trying to rip his arms away from his body.

If the man ties the rope around his waist and let's the rope swing around him, the force direction depends on which part of the rope you are speaking about. But the only reason he is able to exert any force at all on the rope is because he is accelerating the rope as a whole, and the mass attached to it, centripetally about an inertial point.

AM

 

[Andrew Mason]

To make this less obvious, consider a hub, as on a carnival ride with chains leading to the seats swinging around in a circle. The force exerted on the hub by the chains is in the centrifugal direction.

Only if you think the hub centre is the inertial centre of rotation. To determine the inertial centre of rotation you have to take into account the mass of the earth. The hub/earth actually rotates about that point. So the chain supplies tension to the hub, which distributes that tension throughout the apparatus and the earth, causing the hub/earth to rotate or wobble (albeit imperceptibly) around that inertial point and accelerate centripetally toward that point.

This is why it is not a good idea to talk about the direction of the "reaction" force to a centripetal force supplied by tension as always having a particular direction (eg. "centrifugal"). It depends on the geometry and mass distribution in the rotating system and what parts you are considering. It also makes it very difficult to distinguish the real reaction force from the pseudo (non-Newtonian) force (apparent only in the non-inertial frame of reference) if you call both "centrifugal".

AM

 

[jbriggs444]

Only if you think the hub centre is the inertial centre of rotation. To determine the inertial centre of rotation you have to take into account the mass of the earth.

As long as the center of rotation is anywhere in the interior of the hub, the force of the chains on the hub will be at least primarily in the centrifugal (aka outward) direction.

Yes, the unbalanced component of seats and riders on the carnival ride and the ride and Earth are mutually orbitting a common barycenter and are all undergoing centripetal acceleration around that hypothetical point. But that's hardly a useful or even a particularly accurate description of the situation.

 

[A.T.]

It is equal and opposite to the force of the string on the mass.

So it's opposite to the "centripetal" force, and has the same point of application (just acts on a different object).

Whether that force is toward or away from the centre of rotation depends on where the centre of rotation is and what you mean by the string.

It's the same center that the centripetal force point towards. Since the two forces have the same point of application, but opposite directions, they can hardly both point toward the center of rotation.

 

[Andrew Mason]

As long as the center of rotation is anywhere in the interior of the hub, the force of the chains on the hub will be at least primarily in the centrifugal (aka outward) direction.

Yes, the unbalanced component of seats and riders on the carnival ride and the ride and Earth are mutually orbitting a common barycenter and are all undergoing centripetal acceleration around that hypothetical point. But that's hardly a useful or even a particularly accurate description of the situation.

So how useful is it to always refer to the reaction force to a centripetal force as centrifugal if it is sometimes not centrifugal (even when the force is tensile) and, in the case of force acting at a distance, never centrifugal? And, regardless of how the tension may be directed, that tension can never accelerate a mass away from the centre.

AM

 

[jbriggs444]

So how useful is it to always refer to the reaction force to a centripetal force as centrifugal if it is sometimes not centrifugal (even when the force is tensile) and, in the case of force acting at a distance, never centrifugal?

I dislike the term "reactive centrifugal force" and try not to use it. However, in the case I've described we have a force acting at a distance which is most certainly centrifugal in direction.

 

[A.T.]

So how useful is it to always refer to the reaction force to a centripetal force as centrifugal

Nobody does that always. Some people do it only when it actually points away from the center.

 

[Andrew Mason]

So it's opposite to the "centripetal" force, and has the same point of application (just acts on a different object).

Yes. But the direction of a force is defined as the direction which the force, acting alone on the body, would accelerate the centre of mass of the body, which I believe fits the Newtonian definition of force. The force and its "reaction" force act on different bodies, and therefore on different points. So it is not necessarily the case that the reaction force to a force in the centripetal direction will be in a non-centripetal direction.

It's the same center that the centripetal force point towards. Since the two forces have the same point of application, but opposite directions, they can hardly both point toward the center of rotation.

Why not? Since they act on different bodies at different points, they could both be, and often are, both centripetal forces. The point of application does not determine the direction of the force unless you are only dealing with the mass elements whose centres of mass coincide with the point of application.

If you are dealing with forces at a distance between macroscopic bodies, such as gravity, it becomes a bit clearer. The gravitational forces are treated as applying to the centres of mass of each body. There are tidal forces within each body that result in tensions within each body. We don't refer to them as centrifugal or centripetal because they go in all sorts of directions relative to the non-inertial centre of mass of the respective body. In any event, these tensions produce only centripetal accelerations.

AM

 

[Andrew Mason]

I dislike the term "reactive centrifugal force" and try not to use it. However, in the case I've described we have a force acting at a distance which is most certainly centrifugal in direction.

! In the example you gave, the centripetal forces were mechanical, not "acting at a distance". Where the forces between two bodies are acting at a distance (eg. gravity) both are always centripetal.

AM

 

[jbriggs444]

! In the example you gave, the centripetal forces were mechanical, not "acting at a distance". Where the forces between two bodies are acting at a distance (eg. gravity) both are always centripetal.

The chain can be replaced with a force at a distance without changing the situation in any material way.

Your latter claim is simply false. It holds for the special cases of gravity on point masses and for gravity on spherically symmetric objects simplified by pretending that it acts at their centers of mass. It does not hold for general forces at a distance where the point of application of the force need not coincide with the target object's center of mass.

 

[A.T.]

But the direction of a force is defined as the direction which the force, acting alone on the body, would accelerate the centre of mass of the body, which I believe fits the Newtonian definition of force.

This sense of "direction" alone tells you nothing about pointing towards or away from the center. You have to also consider the point of application.

act on different bodies, and therefore on different points.

Does not follow. They act at the same geometrical point (connection between string and mass) and are opposite in direction. So they cannot both point towards the center.

 

[Andrew Mason]

The chain can be replaced with a force at a distance without changing the situation in any material way.

How would that work?

Your latter claim is simply false. It holds for the special cases of gravity on point masses and for gravity on spherically symmetric objects simplified by pretending that it acts at their centers of mass. It does not hold for general forces at a distance where the point of application of the force need not coincide with the target object's center of mass.

I don't think it is false to say that in a rotating rigid system, all net forces are centripetal. As I stated, there are tensions within rigid systems that can go in many directions. Exactly where a force acts on a rigid body can be a philosophical debate. What matters is how the body acts in response to the force. In the case of two rigid bodies in mutual gravitational orbit about their common barycentre, the centre of mass of each body accelerates in the direction of that barycentre. That does not depend on how the mass is distributed within the body.

If you disagree with how I have stated things, pehaps you could help us by pointing out where Prof. Scott is wrong in this article:

G. David Scott (1957). "Centrifugal Forces and Newton's Laws of Motion" 25. American Journal of Physics. p. 325.

A mass A is moving in a circle; a centripetal force must therefore be acting on A. What is the reaction to this force? The reaction in general will be the force which causes mass B to move in a circle. For example the reaction to the force which move the electron around the proton in the hydrogen atom is the force which moves the proton around their common centre of gravity. The same situation is revealed in the case of the moon and the earth, or the figure skater and her partner as they whirl about together. The reaction to a centripetal force is not a centrifugal force but another centripetal force. Though Newton's third law is used in statics, its real significance is in dynamics. When a force acts to accelerate one mass, then there is always a reaction which is accelerating another mass in the opposite direction.

Practical problems can always be solved using the strict Newtonian approach, but it is often more realistic and intuitively simpler to use a rotating reference frame and introduce the necessary centrifugal forces. The centrifugal forces must be recognized for what they are - non-Newtonian forces acting on masses at rest in a rotating frame.

AM

 

[jbriggs444]

How would that work?

Magnets, charged spheres, big blobs of mass, idealized cords. It does not matter. We have an object over here exerting a force-at-a-distance with a point of application over there.

I don't think it is false to say that in a rotating system, all net forces are centripetal.

That's not the false statement you made.

 

[A.T.]

As I stated, there are tensions within rigid systems that can go in many directions

Tensions act within bodies. But there are forces between the bodies, which can go in many directions, including away from the center.

Also, rigid systems are just a special case, so you cannot derive general statements from it.

 

[Andrew Mason]

This sense of "direction" alone tells you nothing about pointing towards or away from the center. You have to also consider the point of application.

How do you determine the point of application of a force? Does it the point of application really matter - or is this just a semantic debate? Where do the Earth and moon apply their forces to each other?

Does not follow. They act at the same geometrical point (connection between string and mass) and are opposite in direction. So they cannot both point towards the center.

Since the forces act on different bodies, why can they not be both centripetal if the bodies are on opposite sides of the inertial centre of rotation?

AM

 

[Andrew Mason]

Magnets, charged spheres, big blobs of mass, idealized cords. It does not matter. We have an object over here exerting a force-at-a-distance with a point of application over there.

Ok. Put the ride in space and have the sufficiently negatively charged chair circling the sufficiently positively charged hub due to the electric forces between them. It would seem to me that both undergo centripetal acceleration about their common centre of mass.

That's not the false statement you made.

So what is the false part of what I said?

My point is that using the term "centrifugal force"to denote a phenomenon other than the non-Newtonian pseudo-force that appears in non-inertial frames creates a great deal of confusion for anyone trying to learn about the physics of rotating bodies.

AM

 

[A.T.]

Does it the point of application really matter - or is this just a semantic debate?

The point of force application does matter in physics.

if the bodies are on opposite sides of the inertial centre of rotation

Special case, that cannot be generalized.

 

[A.T.]

My point is that using the term "centrifugal force"to denote a phenomenon other than the non-Newtonian pseudo-force that appears in non-inertial frames creates a great deal of confusion for anyone trying to learn about the physics of rotating bodies.

If that was your only point, no one would object. But you make a lot of invalid generalizations and geometrically absurd statements.

 

[jbriggs444]

Ok. Put the ride in space and have the sufficiently negatively charged chair circling the sufficiently positively charged hub due to the electric forces between them. It would seem to me that both undergo centripetal acceleration about their common centre of mass.

Please stop changing the setup to fit your pet scenario. It can be a charged blob on the hub, not a charged hub.

So what is the false part of what I said?

Where the forces between two bodies are acting at a distance (eg. gravity) both are always centripetal.

You did not say "net", and did not mention a point of application.

My point is that using the term "centrifugal force"to denote a phenomenon other than the non-Newtonian pseudo-force that appears in non-inertial frames creates a great deal of confusion for anyone trying to learn about the physics of rotating bodies.

A.T.'s response to this was entirely on point.

 

[Andrew Mason]

If that was your only point, no one would object. But you make a lot of invalid generalizations and geometrically absurd statements.

So tell us where my "invalid generalizations" differ from those of Prof. Scott in his article.

AM

 

[Andrew Mason]

Please stop changing the setup to fit your pet scenario. It can be a charged blob on the hub, not a charged hub.

I don't have a pet scenario. I was trying to imagine how your scenario could work and provide centrifugal forces.

The geometrical distribution of the charge on the hub should not matter. What does matter is that the charge is rigidly connected to the hub.

You did not say "net", and did not mention a point of application.

Except for the tensions within the rotating bodies due to tidal forces, it seems to me that all forces ARE centripetal - If we can ignore tidal forces, there are no forces other than net forces and they are all centripetal (the rings of Saturn, for example)..A.T.'s response to this was entirely on point.[/QUOTE]So tell us where Prof. Scott is wrong. http://www.spmlaw.ca/PF/Scott_Centrifugal_Forces_Newtons_Laws_1957_full.pdf:

AM

 

[A.T.]

So tell us where my "invalid generalizations" differ from those of Prof. Scott in his article.

They don't. He also considers the simplest possible special case of just two bodies as "general" (if that's the correct interpretation of the phrase "in general").

 

[jbriggs444]

I don't have a pet scenario. I was trying to imagine how your scenario could work and provide centrifugal forces.

Your pet scenario is the Earth and Moon. You keep slipping it in at every opportunity, apparently imagining that the only sort of force at a distance is one that acts at the center of mass of the interacting bodies. Your only notion of interacting bodies seems to involve either point particles or featureless spheres.

The geometrical distribution of the charge on the hub should not matter. What does matter is that the charge is rigidly connected to the hub.

The geometrical distribution of the charge on the hub affects the point of application of the force from electrostatic attraction.

 

[Andrew Mason]

Your pet scenario is the Earth and Moon. You keep slipping it in at every opportunity, apparently imagining that the only sort of force at a distance is one that acts at the center of mass of the interacting bodies. Your only notion of interacting bodies seems to involve either point particles or featureless spheres.

I mentioned the earth-moon exactly once, by the way, and it was in form of a question because I wanted to use the simplest example possible. I did not mention anything about featureless spheres or point particles.

AM

 

[vanhees71]

It's amazing, how many words one can make about some not that complicated issues like the socalled "fictitious forces" or, as I prefer to call them "inertial froces" in Newtonian physics. To see, what's really going on, one should do the math. Here, I'll concentrate on the special case of a reference frame rotating, relative to an inertial frame. Examples are the motion on Earth, where the rotation is (in some crude approximation) around a fixed axis or the body frame of a rigid body rotating around its center of mass or around another fixed point (spinning free and heavy top).

We look at the most simple case of a single point particle, moving due to some external force (e.g., in the gravitational field of the Earth or a charged particle in an electric field). Let $\vec{x}$ denote the Cartesian righthanded coordinates in an inertial reference frame and $\vec{x}'$ that with respect to the rotating frame. Then there's a time-dependent matrix $\hat{D}(t) \in \mathrm{SO}(3)$ such that
$$\vec{x}=\hat{D} \vec{x}'.$$
The motion in the inertial frame follows Newton's 2nd Law
$$m \ddot{\vec{x}}=\vec{F}(t,\vec{x},\dot{\vec{x}}).$$
Here $\vec{F}$ is a "true force", i.e., caused by some interaction of the particle with some other particles or due to fields.

Now we simply want to know, how and observer in the rotating system describes the motion, i.e., we must calculate the 2nd time derivative in terms of $\vec{x}'$. Obviously we have
$$\dot{\vec{x}} = \dot{\hat{D}} \vec{x}'+\hat{D} \dot{\vec{x}}'.$$
Multiplying with $\hat{D}^{-1}$ from the left, we find
$$\hat{D}^{-1} \dot{\vec{x}}=\hat{D}^{-1} \dot{\hat{D}} \vec{x}'+\dot{\vec{x}}'.$$
Now since $\hat{D}^{-1} = \hat{D}^{T}$ and thus $\hat{D} \hat{D}^T=1$ and thus through taking the time derivative of this equation
$$\hat{D}^{-1} \dot{\hat{D}} = \hat{D}^{T} \dot{\hat{D}}=-\dot{\hat{D}}^T \hat{D} =-(\hat{D}^t \dot{\hat{D}})^T,$$
we see that $\hat{D}^{-1} \dot{\hat{D}}$ is an antisymmetric matrix and thus we can define an axial vector $\vec{\omega}'$, the components of the momentaneous angular velocity of the rotation of the reference frames in the rotating frame such that
$$\hat{D}^{-1} \dot{\vec{x}}=\dot{\vec{x}}'+\vec{\omega}' \times \vec{x}'.$$
Thus we have
$$\dot{\vec{x}}=\hat{D} (\dot{\vec{x}}'+\vec{\omega}' \times \vec{x}'.$$
Now we can do the same calculation again to find
$$\ddot{\vec{x}}=\hat{D} [\ddot{\vec{x}}'+2\vec{\omega}' \times \dot{\vec{x}}' + \dot{\vec{\omega}}' \times \dot{\vec{x}}' + \vec{\omega}' \times (\vec{\omega}' \times \vec{x}')]$$
Plugging this into the Equation of motion and multiplying finally again with $\hat{D}^{-1}$ from the left leads to
$$m [\ddot{\vec{x}}'+2\vec{\omega}' \times \dot{\vec{x}}' + \dot{\vec{\omega}}' \times \dot{\vec{x}}' + \vec{\omega}' \times (\vec{\omega}' \times \vec{x}')]=\vec{F}'.$$
To bring this in the form of the usual Newtonian equation of motion, one brings all terms on the left-hand side of the equation to the right-hand side except the first term.

Thus according to the rotating observer the particle feels the external force, expressed in hin coordinates and a bunch of "inertial forces", among them
$$\vec{F}_C'=-2 m \vec{\omega}' \times \dot{\vec{x}}' \qquad \text{(Coriolis force)},$$
$$\vec{F}_Z'=-m \vec{\omega}' \times (\vec{\omega}' \times \vec{x}') \qquad \text{(centrifugal force)}.$$
Then there is a force due to the change of the momentaneous rotation axis, which I don't know whether it has a special name,
$$\vec{F}_A'=-m \dot{\vec{\omega}}' \times \vec{x}'.$$
These forces are "ficititious" in the sense that they are not due to interactions with other particles or fields but just due to the description of the motion in a rotating reference frame, and they can thus be eliminated by doing the opposite transformation back to the inertial frame. That's it. Nothing mysterious.

 

[A.T.]

I mentioned the earth-moon exactly once

In this and previous threads, where you made the same generalizations, you tend to ignore counter examples and keep bringing up two isolated bodies orbiting their common center of mass.

I wanted to use the simplest example possible.

You cannot generalize the simplest example possible, to all possible cases.

 

[Andrew Mason]

You cannot generalize the simplest example possible, to all possible cases.

It depends on whether there is something about the forces that is common to all rotating rigid systems. It seems to me that they all have the following in common with regard to the Newtonian forces:

1. There are no net centrifugal forces. All net forces are centripetal. All accelerations are centripetal.

2. There are tensions whose directions depend on the geometry of the system. But these tensions never result in centrifugal accelerations.

I think I will just leave it at that.

AM

 

[A.T.]

You cannot generalize the simplest example possible, to all possible cases.

It depends on whether there is something about the forces that is common to all rotating rigid systems.

By demanding rigidity you have already lost generality, which confirms my quote above, that it doesn't apply to all cases involving rotation.

1. There are no net centrifugal forces.

For the special case of a rigid system trivially true and never contested . What is being contested is when you leave out that "net" bit.

2. There are tensions whose directions depend on the geometry of the system.

There can also be forces between the bodies, which are pointing away from the center. This is what you like to deny, and again fail to acknowledge.

 

[A.T.]

Where the forces between two bodies are acting at a distance (eg. gravity) both are always centripetal.

This is also false. Put two repelling magnets into the closed end of a pipe and spin it. The inner magnet exerts a magnetic force at a distance onto the outer magnet, which points away from the center.