Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss
Centrifugal force (simulation)
53 views
Skip to first unread message
Luigi Fortunati
Dec 14, 2022, 3:54:21 PM12/14/22
to
I have prepared two simulations, in the first one there is a rotating
disc with a tube in which there is a ball bound with a rope.
When we start the rotation with the "Start", *pairs* of opposing forces
are activated: the rope exerts its centripetal force on the ball
(otherwise the ball would not rotate) and the ball exerts its
centrifugal force on the rope (otherwise the tension of the rope would
not exist).
Each pair of contiguous particles (of the ball, rope, and disc) exerts
(on each other) a reciprocal pair of opposing forces (centripetal and
centrifugal).
The opposing forces are both there or both are missing.
Is there (somewhere in my simulation) a centripetal force lacking its
corresponding centrifugal force?
Is there (somewhere in my simulation) a preponderance of centripetal
over centrifugal forces?
It seems to me not.
The simulation is this:
https://www.geogebra.org/m/tnas7mbn
The second simulation has a rotating tube in which there is a ball
which (unlike the previous one) is free to move.
By selecting the appropriate boxes, you can choose normal or slowed
down speed, and you can choose whether to have (or not have) the trace
of the forces applied by the tube on the ball during rotation.
These blue forces (when they arrive on the wall of the ball) are always
orthogonal to the tube and, therefore, tangential.
But what happens to these forces after the impact against the ball's
surface? Do they disappear or does their action go even further by
continuing inside the ball which (in the meantime) is rotating together
with the tube?
My hypothesis is that they do not disappear and, based on this
hypothesis, I chose a single point of the ball (point P) and when,
during the rotation, the tube touches P, I started the continuation of
the force, coloring it in red (to highlight it).
And I noticed that the force does not go in a tangential direction but
(because of the rotation of the tube and the ball) it finds itself
advancing in a centrifugal direction (with respect to the center of
rotation)!
To better observe this centrifugal red force, choose the slow rotation
(or stop the animation midway) and look at the enlargement of the ball
on the right.
Obviously, what is true for point P is true for any other point on the
ball.
The simulation is as follows:
https://www.geogebra.org/m/xcx6rj4s
Can any of you tell me if this is really the case or if I need to make
corrections to my simulations?
I am ready to make any additions or modifications that may be
necessary.
disc with a tube in which there is a ball bound with a rope.
When we start the rotation with the "Start", *pairs* of opposing forces
are activated: the rope exerts its centripetal force on the ball
(otherwise the ball would not rotate) and the ball exerts its
centrifugal force on the rope (otherwise the tension of the rope would
not exist).
Each pair of contiguous particles (of the ball, rope, and disc) exerts
(on each other) a reciprocal pair of opposing forces (centripetal and
centrifugal).
The opposing forces are both there or both are missing.
Is there (somewhere in my simulation) a centripetal force lacking its
corresponding centrifugal force?
Is there (somewhere in my simulation) a preponderance of centripetal
over centrifugal forces?
It seems to me not.
The simulation is this:
https://www.geogebra.org/m/tnas7mbn
The second simulation has a rotating tube in which there is a ball
which (unlike the previous one) is free to move.
By selecting the appropriate boxes, you can choose normal or slowed
down speed, and you can choose whether to have (or not have) the trace
of the forces applied by the tube on the ball during rotation.
These blue forces (when they arrive on the wall of the ball) are always
orthogonal to the tube and, therefore, tangential.
But what happens to these forces after the impact against the ball's
surface? Do they disappear or does their action go even further by
continuing inside the ball which (in the meantime) is rotating together
with the tube?
My hypothesis is that they do not disappear and, based on this
hypothesis, I chose a single point of the ball (point P) and when,
during the rotation, the tube touches P, I started the continuation of
the force, coloring it in red (to highlight it).
And I noticed that the force does not go in a tangential direction but
(because of the rotation of the tube and the ball) it finds itself
advancing in a centrifugal direction (with respect to the center of
rotation)!
To better observe this centrifugal red force, choose the slow rotation
(or stop the animation midway) and look at the enlargement of the ball
on the right.
Obviously, what is true for point P is true for any other point on the
ball.
The simulation is as follows:
https://www.geogebra.org/m/xcx6rj4s
Can any of you tell me if this is really the case or if I need to make
corrections to my simulations?
I am ready to make any additions or modifications that may be
necessary.
Tom Roberts
Dec 15, 2022, 12:54:09 AM12/15/22
to
On 12/14/22 2:52 PM, Luigi Fortunati wrote:
> The simulation is this: https://www.geogebra.org/m/tnas7mbn
> The simulation is this: https://www.geogebra.org/m/tnas7mbn
> When we start the rotation with the "Start", *pairs* of opposing
> forces are activated: the rope exerts its centripetal force on the
> ball (otherwise the ball would not rotate) and the ball exerts its
> centrifugal force on the rope (otherwise the tension of the rope
> would not exist).
This is wrong. Your diagram shows the system in the inertial frame of
> forces are activated: the rope exerts its centripetal force on the
> ball (otherwise the ball would not rotate) and the ball exerts its
> centrifugal force on the rope (otherwise the tension of the rope
> would not exist).
the center. In this frame there is no "centrifugal force". The tension
in the rope exerts a centripetal force on the ball, causing it go move
in a circle. This force is UNOPPOSED, and the ball accelerates around
the circle. (After all, that is what F=ma means.)
> Is there (somewhere in my simulation) a centripetal force lacking
> its corresponding centrifugal force?
center, because in an inertial frame there is no "centrifugal force".
Hint: if the centripetal force was opposed, then
the net force on the ball would be zero, and it
would not accelerate. But is MUST accelerate to
move around in a circle.
> Is there (somewhere in my simulation) a preponderance of centripetal
> over centrifugal forces?
"centrifugal force" everywhere (except the center), including places
with no corresponding centripetal force. Of course in such a diagram the
ball would remain at one place -- it would not accelerate because the
net force on it would be zero ("centrifugal" and centripetal forces
canceling each other).
Plain and simple, "centrifugal force" is a KLUDGE that permits one to
more easily apply Newtonian mechanics in rotating coordinates [#]. It is
purely due to the use of rotating coordinates, and does not correspond
to any natural phenomenon, because it is an unnatural artifact of
rotating coordinates.
[#] In general one must also include "Coriolis and
Euler forces". These are also unnatural artifacts
of rotating coordinates.
That's why we apply the label "fictitious forces" to the three
"centrifugal, Coriolis, and Euler forces" -- they do not correspond to
any natural phenomenon, and are merely unnatural artifacts of using
rotating coordinates.
[In General Relativity, and in geometrized Newtonian
gravitation, those three "fictitious forces" appear
in the geometrical connection, which is explicitly
coordinate dependent. In GR, that's also where one
finds "gravitational force".]
Tom Roberts
Luigi Fortunati
Dec 16, 2022, 1:55:51 PM12/16/22
to
Tom Roberts mercoledě 14/12/2022 alle ore 14:54:04 ha scritto:
> On 12/14/22 2:52 PM, Luigi Fortunati wrote:
>> The simulation is this: https://www.geogebra.org/m/tnas7mbn
>
>> When we start the rotation with the "Start", *pairs* of opposing
>> forces are activated: the rope exerts its centripetal force on the
>> ball (otherwise the ball would not rotate) and the ball exerts its
>> centrifugal force on the rope (otherwise the tension of the rope
>> would not exist).
>
> This is wrong.
If my simulation is wrong, can you help me correct it?
> On 12/14/22 2:52 PM, Luigi Fortunati wrote:
>> The simulation is this: https://www.geogebra.org/m/tnas7mbn
>
>> When we start the rotation with the "Start", *pairs* of opposing
>> forces are activated: the rope exerts its centripetal force on the
>> ball (otherwise the ball would not rotate) and the ball exerts its
>> centrifugal force on the rope (otherwise the tension of the rope
>> would not exist).
>
> This is wrong.
Before you press "Start" everything is still and there are no blue
forces nor red forces.
So far there is nothing to fix, right?
Then I press "Start", the disk starts to rotate and some forces (in the
inertial frame of my simulation) are activated.
> Your diagram shows the system in the inertial frame of
> the center. In this frame there is no "centrifugal force".
centripetal forces are there and the red centrifugal ones are not.
So, I just take out the red forces and everything is fine?
Ok I will.
But first explain something to me: without the red centrifugal force
(exerted by the ball on the string) how is the tension of the string
justified (in the inertial frame of my simulation)?
> The tension in the rope exerts a centripetal force on the ball,
> causing it go move in a circle.
of my simulation, there is no blue centrifugal force exerted on it by
the ball?
Luigi Fortunati
Dec 16, 2022, 1:55:52 PM12/16/22
to
Tom Roberts mercoledì 14/12/2022 alle ore 14:54:04 ha scritto:
> On 12/14/22 2:52 PM, Luigi Fortunati wrote:
>> The simulation is this: https://www.geogebra.org/m/tnas7mbn
>
>> When we start the rotation with the "Start", *pairs* of opposing
>> forces are activated: the rope exerts its blue centripetal force on
> On 12/14/22 2:52 PM, Luigi Fortunati wrote:
>> The simulation is this: https://www.geogebra.org/m/tnas7mbn
>
>> When we start the rotation with the "Start", *pairs* of opposing
>> the ball (otherwise the ball would not rotate) and the ball exerts
>> its red centrifugal force on the rope (otherwise the tension of the
>> rope would not exist).
>
> This is wrong. Your diagram shows the system in the inertial frame of
> the center. In this frame there is no "centrifugal force".
In my simulation the red forces are centrifugal and now you tell me that
>
> This is wrong. Your diagram shows the system in the inertial frame of
> the center. In this frame there is no "centrifugal force".
there are no centrifugal forces.
That is fine.
However, if there are no centrifugal forces, then have I made a mistake
in drawing the red forces and should I eliminate them?
Or, do you mean that the red forces I drew exist (and, therefore, I have
to leave them) but are not centrifugal forces?
> The tension in the rope exerts a centripetal force on the ball,
> causing it go move in a circle.
The tension of the rope exists and it is thanks to it that the ball
moves in a circle.
But there is also the tension of the ball and it is thanks to him that
the rope stretches.
I hope you won't tell me that the rope tension is there and the ball
tension isn't.
The two tensions are specular, if one exists there is also the other, if
one disappears the other disappears too.
I also point out that, during the rotation, EVERYTHING is under tension,
not just the rope.
The ball is under tension and the disc is also under tension.
> This force is UNOPPOSED
In my animation, there isn't a single blue force unopposed by a red
force, nor a red force unopposed by a blue force.
Can you tell me what corrections I need to make to my simulation to
highlight the unopposed force you're talking about?
Luigi Fortunati.
Luigi Fortunati
Dec 23, 2022, 2:24:11 AM12/23/22
to
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?
> 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?
Hendrik van Hees
Dec 23, 2022, 10:09:41 AM12/23/22
to
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/
Julio Di Egidio
Dec 25, 2022, 1:34:18 AM12/25/22
to
On Friday, 23 December 2022 at 16:09:41 UTC+1, Hendrik van Hees wrote:
> On 23/12/2022 08:24, Luigi Fortunati wrote:
<snip>
> On 23/12/2022 08:24, Luigi Fortunati wrote:
> 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.
Please correct me if I am mistaken, but F=ma is not only valid in inertial
> 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.
frames, is it? Indeed, I find the way you are presenting things here still
risks to give the impression that these "fictitious", aka "apparent" forces
are only an artefact of the algebra, while in that sense they are rather a
misnomer for the quite real forces an observer in that non-inertial frame
would feel and measure. Put simply, if one jumps on a merry-go-round,
fighting the centrifugal force is a real and properly physical thing, no?
Julio
[[Mod. note --
1. Just to be clear: the quoted text was written by Hendrik van Hees,
not by Luigi Fortunati.
2. I agree with Hendrik: F=ma (with F only including "real" forces) is only
valid in an inertial reference frame. To do Newtonian dynamics in a
non-inertial reference frame, one must augment F to also include
fictitious forces (such as the Coriolis force).
3. From a Newtonian-dynamics-in-an-inertial-reference-frame perspective,
if I jump on (and hold on to) a merry-go-round, I don't feel a
centrifugal force. Rather, the merry-go-round exerts a *centripetal*
force on me, accelerating me inwards (so that I move in a circle around
the rotation axis). The *centripetal* acceleration is what I feel.
-- jt]]
Luigi Fortunati
Dec 25, 2022, 3:55:11 AM12/25/22
to
Mod. note -- sabato 24/12/2022 alle ore 15:34:14 ha scritto:
> 3. From a Newtonian-dynamics-in-an-inertial-reference-frame perspective,
> if I jump on (and hold on to) a merry-go-round, I don't feel a
> centrifugal force. Rather, the merry-go-round exerts a *centripetal*
> force on me, accelerating me inwards (so that I move in a circle around
> the rotation axis). The *centripetal* acceleration is what I feel.
What you feel is centripetal force (not acceleration), because force is
> 3. From a Newtonian-dynamics-in-an-inertial-reference-frame perspective,
> if I jump on (and hold on to) a merry-go-round, I don't feel a
> centrifugal force. Rather, the merry-go-round exerts a *centripetal*
> force on me, accelerating me inwards (so that I move in a circle around
> the rotation axis). The *centripetal* acceleration is what I feel.
felt, acceleration is not.
And when (jumping on the merry-go-round) you feel the centripetal
force, the pole you cling to feels your centrifugal force!
You could never feel the centripetal force of the pole you cling to if,
at the same time, the pole did not feel your centrifugal force.
And when you jump off the merry-go-round, the two forces disappear
together exactly as they were born together!
The centripetal force of the pole and your centrifugal force travel
*together*: both are there or both are not.
Hendrik van Hees
Dec 25, 2022, 5:00:05 AM12/25/22
to
What I've tried to say is that the equation of motion is always m a = F,
valid for basis-independent vectors, which always describe an inertial
frame of reference. You cannot describe the equations of motion
otherwise, because they must be formulated within the Galilei-Newton
model. The "inertial forces" are not "fictitious" but just occur due to
the time-dependence of the basis ("rotations" if you use Cartesian ones)
and an acceleration of the origin of the non-inertial frame relative to
the inertial frames.
valid for basis-independent vectors, which always describe an inertial
frame of reference. You cannot describe the equations of motion
otherwise, because they must be formulated within the Galilei-Newton
model. The "inertial forces" are not "fictitious" but just occur due to
the time-dependence of the basis ("rotations" if you use Cartesian ones)
and an acceleration of the origin of the non-inertial frame relative to
the inertial frames.
0 new messages