Connetion between centrifugal force and weight force

[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.