On 23/12/2022 08:24, Luigi Fortunati wrote:
In my simulation
https://www.geogebra.org/m/tnas7mbn
I have highlighted, coloring them in red, some centrifugal forces.
These forces exist because they are the reaction to the corresponding centripetal forces (third principle).
These forces are centrifugal because they point in the opposite direction to the corresponding centripetal forces.
Every centripetal force has its corresponding centrifugal force (and vice versa).
I'm not discussing outlandish or far-fetched theories but pure Newtonian physics.
And, as I said, if there are any errors in my simulation, I'm ready to make whatever corrections are necessary.
So why is nobody willing to clarify the matter? Why so much reticence?
Although I have let this posting through as one of the moderators of
this newsgroup, I must say that this point of view is full of
misconceptions about the meaning of "inertial forces" (sometimes also
called "fictitious forces", but I don't like this name).
First of all, it must be very clear that "inertial forces", like the centrifugal and Coriolis forces in reference frames rotating with
respect to the inertial reference frames, only occur in non-inertial
frames of reference.
Second, they are not the "reactio" of the "actio" in Newton's Lex Prima.
These are interaction forces occuring in inertial frames of reference,
and the statement is that if there is an interaction force between two
bodies (labelled by "1" and "2") and the force on particle 1 is
\vec{F}_{12}, then the the force on particle 2 is
\vec{F}_{21}=-\vec{F}_{12}, i.e., the "actio-reactio pair of forces" act
on different particles, never at the same particle.
The equations of motion of Newtonian mechanics for point particles with constant mass are always defined with reference to an inertial frame of reference, and you get the equations describing these Newtonian within a non-inertial reference frame simply by writing the coordinates of the
point particles wrt. an inertial frame in terms of coordinates referring
to a non-inertial frame.
This leads to additional terms when taking the time derivatives. If you
have a rotating reference frame, i.e., a frame of reference whose
(Cartesian) basis is rotating wrt. the Cartesian basis of the inertial reference frame, then the time-derivative of the components of an
arbitrary vector with respect to the rotating basis reads
D_t \vec{V}'=d_t \vec{V}'+\vec{\omega}' \times \vec{V}',
where d_t is the usual time derivative of these components, and
\vec{\omega}' are the components of the momentary angular velocity of
the rotating basis wrt. the inertial basis.
The equations of motion are the same as in the inertial frame of
reference, but when expressed in terms of the non-inertial components of
the position vector, you have to use the covariant time derivative twice
to calculate the (components of the) acceleration of your particle in
terms of the (components of the) position vector:
m D_t^2 \vec{x}'=\vec{F}'(\vec{x}'),
where \vec{F}' are the components of the "true force" (e.g., the
gravitational force of the Earth on the particle) wrt. the rotating
basis. Working out the time derivatives yields
m (d_t^2 \vec{x}' + 2 \vec{\omega}' \times d_t \vec{x}'
+ \vec{\omega}' \times (\vec{\omega}' \times \vec{r}')
+ d_t \vec{\omega}' \times d_t \vec{x}'
=\vec{F}'(\vec{x}')
Now you bring all the additional terms to the right-hand side and
reinterpret them as "inertial forces".
As you see from this derivation, indeed these "inertial forces" only
occur in the non-inertial reference frame and have nothing to do with
reaction forces, which would be not forces on the same particle but to
other particles interacting with it. In your Geogebra example the
reaction force is on the rope not on the particle.
--
Hendrik van Hees
Goethe University (Institute for Theoretical Physics)
D-60438 Frankfurt am Main
http://itp.uni-frankfurt.de/~hees/
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