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Newton e Hooke

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  Newton e Hooke Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-01-26 08:51 -0800
    Re: Newton e Hooke Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-01-30 08:42 +0000
      Re: Newton e Hooke Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-02-04 23:47 -0800
        Re: Newton e Hooke Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-02-10 22:25 -0800
    Re: Newton e Hooke "Jonathan Thornburg [remove color- to reply]" <dr.j.thornburg@pink-gmail.com> - 2025-02-12 08:02 +0000
      Re: Newton e Hooke Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-02-14 08:09 +0000
        Re: Newton e Hooke "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com> - 2025-02-16 08:55 +0000
          Re: Newton e Hooke Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-02-16 10:42 -0800
            Re: Newton e Hooke "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com> - 2025-02-17 08:19 +0000
              Re: Newton e Hooke Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-02-17 22:45 -0800
            Re: Newton e Hooke Tom Roberts <tjoberts137@sbcglobal.net> - 2025-02-18 08:23 +0000
        Re: Newton e Hooke Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-02-16 10:42 -0800
          Re: Newton e Hooke "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com> - 2025-02-17 08:19 +0000
            Re: Newton e Hooke "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com> - 2025-02-18 08:23 +0000
              Re: Newton e Hooke Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-02-20 07:56 +0100
                Re: Newton e Hooke Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-02-22 18:13 +0100
                  inelastic collision (was: Re: Newton e Hooke) "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-red.com> - 2025-02-26 08:02 +0000
                    Re: inelastic collision (was: Re: Newton e Hooke) Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-02-28 14:44 -0800
                      Re: inelastic collision (was: Re: Newton e Hooke) "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com> - 2025-03-01 09:18 +0000
                        Re: inelastic collision (was: Re: Newton e Hooke) Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-03-03 08:09 +0000
                          Re: inelastic collision (was: Re: Newton e Hooke) Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-03-12 12:37 +0000
                            Re: inelastic collision (was: Re: Newton e Hooke) Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-03-16 09:28 +0000
                              Re: inelastic collision (was: Re: Newton e Hooke) Luigi Fortunati <fortunati.luigi@gmail.com> - 2025-03-16 21:28 -0700
                      Re: inelastic collision (was: Re: Newton e Hooke) "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com> - 2025-03-02 09:14 +0000

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#11730 — Re: inelastic collision (was: Re: Newton e Hooke)

The animation https://www.geogebra.org/classic/hxvcaphh shows how two 
rigid bodies interact when they collide head-on.

When body A comes into contact with body B, an often imperceptible 
contraction occurs (first phase) and a subsequent elastic return to the 
initial form (second phase).

The quantitative model is the following, before the collision the 
masses are both equal to 1 (m_A=1 and m_B=1), the initial velocities 
are vi_A=+1 and vi_B=-1, and the momentum quantities are pi_A=+1 and 
p_Bi=-1.

In the first phase of the collision, a compression zone and a pair of 
action and reaction forces F1 and F2 are activated simultaneously in 
the contact area (the compression is the cause, the opposite forces are 
the effects).

The force F1 slows the motion of body B to the left until it stops and 
the force F2 does the same, slowing the motion of body A to the right 
until it stops (the forces are the cause, the decelerations are the 
effects).

During this phase, the changes in the value of the two opposing blue 
and red action and reaction forces are displayed.

When the two bodies have stopped, the contraction is at maximum (1), 
the velocities have become zero and the opposing forces measure F1=+1 
and F2=-1.

You can stop the motion at this moment with the "Max compression" 
button.

In the second phase of the impact, the opposing forces F1 and F2 
accelerate the two bodies A and B in the opposite direction until the 
initial position of contact where both the compression, the forces and 
the acceleration cease to exist.

From this point on, the motion of the two bodies returns to being 
inertial, the velocities have been inverted:
body A from vi_A=+1 to vf_A=-1 and body B from vi_B=-1 to vf_B=+1
and the quantities of motion as well:
body A from pi_A=+1 to pf_A=-1 and body B from pi_B=-1 to pf_B=+1.


All this is very simple because the 2 bodies have the same mass but 
things get complicated when we choose to increase (with the appropriate 
button) the mass of body A from m_A=1 to m_A=2 because, to the main 
contraction between the particles A1 and B1, the secondary compression 
between A2 and A1 is added whose action on the left concerns 
exclusively body A, while that on the right concerns both body A and 
body B.

But I will talk about this in the next post because here I have gone on 
too long.

Luigi Fortunati

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#11731 — Re: inelastic collision (was: Re: Newton e Hooke)

Luigi Fortunati il 12/03/2025 13:37:09 ha scritto:
> The animation https://www.geogebra.org/classic/hxvcaphh shows how two 
> rigid bodies interact when they collide head-on.
>
> When body A comes into contact with body B, an often imperceptible 
> contraction occurs (first phase) and a subsequent elastic return to the 
> initial form (second phase).
>
> The quantitative model is the following, before the collision the 
> masses are both equal to 1 (m_A=1 and m_B=1), the initial velocities 
> are vi_A=+1 and vi_B=-1, and the momentum quantities are pi_A=+1 and 
> p_Bi=-1.
>
> In the first phase of the collision, a compression zone and a pair of 
> action and reaction forces F1 and F2 are activated simultaneously in 
> the contact area (the compression is the cause, the opposite forces are 
> the effects).
>
> The force F1 slows the motion of body B to the left until it stops and 
> the force F2 does the same, slowing the motion of body A to the right 
> until it stops (the forces are the cause, the decelerations are the 
> effects).
>
> During this phase, the changes in the value of the two opposing blue 
> and red action and reaction forces are displayed.
>
> When the two bodies have stopped, the contraction is at maximum (1), 
> the velocities have become zero and the opposing forces measure F1=+1 
> and F2=-1.
>
> You can stop the motion at this moment with the "Max compression" 
> button.
>
> In the second phase of the impact, the opposing forces F1 and F2 
> accelerate the two bodies A and B in the opposite direction until the 
> initial position of contact where both the compression, the forces and 
> the acceleration cease to exist.
>
> From this point on, the motion of the two bodies returns to being 
> inertial, the velocities have been inverted:
> body A from vi_A=+1 to vf_A=-1 and body B from vi_B=-1 to vf_B=+1
> and the quantities of motion as well:
> body A from pi_A=+1 to pf_A=-1 and body B from pi_B=-1 to pf_B=+1.
>
>
> All this is very simple because the 2 bodies have the same mass but 
> things get complicated when we choose to increase (with the appropriate 
> button) the mass of body A from m_A=1 to m_A=2 because, to the main 
> contraction between the particles A1 and B1, the secondary compression 
> between A2 and A1 is added whose action on the left concerns 
> exclusively body A, while that on the right concerns both body A and 
> body B.
>
> But I will talk about this in the next post because here I have gone on 
> too long.

In the case of body A in my animation with mass m_A=2, the forces 
generated by the main compression are F1 and F2, and those generated by 
the secondary compression are the forces F3 and F4, both are pairs of 
equal and opposite forces that guarantee the conservation of the 
quantity of motion.

Therefore, the forces that push to the right (F1+F3) are exactly equal 
and opposite to the forces that push to the left (F2+F4).

But Newton's third law does not say what we all know, that is, that the 
forces that push to the right are equal and opposite to the forces that 
push to the left, but says something completely different: it says that 
the forces that body A exerts on body B are equal and opposite to those 
that body B exerts on body A.

So I ask: is it body B that exerts the force F4 that pushes body A to 
the left? No!

The force F4 is exerted by particle A1 against particle A2, that is, it 
is exerted by body A against itself, it is an *internal* force that 
body A uses to give itself a push to the right against body B: the 
force F3 directed towards body B that would not exist without the 
internal force F4!

The force F3 is released to the right against body B and the force F4 
is released to the left against the same body A.

Body A exerts two pushes on body B (F1+F3) and body B exerts only one 
push on body A (F2), while the other push F4 remains inside body A.

Therefore, the action of body A on body B is *greater* than the 
reaction of body B on body A.

But greater than how much? I had already asked myself this for a long 
time but I never dwelt on it too much and, instead, it was essential to 
do so.

Now, on the Italian physics newsgroup, an interlocutor (rightly) 
pointed out to me that if a law is not good, it must be replaced by a 
new one that takes its place.

And so, I said to myself that, if the action is not equal to the 
reaction, I absolutely had to find what relationship links the action 
to the reaction, knowing full well that it had to depend on the two 
masses.

I did some calculations and found that the force F_AB exerted by body A 
on body B and the force F_BA exerted by body B on body A are linked by 
this equation:
FA_B=-F_BA(1+(mA-mB)^2/(mA+mB))

I verified its accuracy on the numerical data of my animation with body 
A of mass m_A=1 and then with that of mass M_A=2, and it works 
perfectly, just as it works with all the other combinations of masses 
of bodies A and B.

So, the new law is defined as follows: "For every action there is a 
corresponding opposite reaction depending on the ratio between the two 
masses expressed by the equation FA_B=-F_BA(1+(mA-mB)^2/(mA+mB))".

Luigi Fortunati

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#11732 — Re: inelastic collision (was: Re: Newton e Hooke)

Luigi Fortunati il 16/03/2025 10:28:35 ha scritto:
>> The animation https://www.geogebra.org/classic/hxvcaphh shows how two 
>> rigid bodies interact when they collide head-on.
>>
>> ... 
>
> So, the new law is defined as follows: "For every action there is a 
> corresponding opposite reaction depending on the ratio between the two 
> masses expressed by the equation FA_B=-F_BA(1+(mA-mB)^2/(mA+mB))".

Surprisingly and referring to inertia, Newton said something very 
similar: "The vis insita, or innate force of matter, is a power of 
resisting, by which every body, as much as it lies, endeavors to 
persevere in its present state, whether it be of rest, or of moving 
uniformly forward in a right line. This force is __proportional__ to 
the body whose force it is...".

Proportional and not equal!

In the collision this is exactly what happens: each of the bodies A and 
B endeavors to persevere in its current state of motion and rebels with 
a force proportional to the other body that tries to modify it.

Now that I think I have found this proportionality I realize that I 
still have a long way to go.

First of all, why does the proportionality depend on 
(1+(mA-mB)^2/(mA+mB)) and not on anything else?

I found it indirectly starting from the conservation of momentum and 
knowing how the velocities of bodies vary after the collision but I 
would need to compare myself with someone who is better than me to find 
the direct explanation, to check that everything is correct, to fix 
every imperfection or error and to transcribe everything with the 
correct scientific language suitable for an official publication.

Is there anyone willing to collaborate with me to fine-tune everything 
together or to find any errors that I may have missed?

Luigi Fortunati

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#11728 — Re: inelastic collision (was: Re: Newton e Hooke)

In article <vpd0kg$23r6$1@dont-email.me>, Luigi Fortunati writes:
> In my animation https://www.geogebra.org/m/rs4cfxzg body A of 5 particles
> collides inelastically with body B of 3 particles.

This animation shows a specific set of motions of the 5 bodies (shown
as spheres), with a specific set of deformations (the spheres overlap)
at various times.

This animation might be what actually happens in an inelastic collision,
or it might not be.  How can we find out?  That is, how can we apply the
known laws of Newtonian mechanics to work out what happens?

To do this, we need a *quantitative* model of body A and of body B, and
of how they interact during the collison.

Earlier in this thread (article <m1dng8FkasU1@mid.dfncis.de>) I
suggested a model consisting of a set of point particles interconnected
with "springs with friction".  That model is reasonable for elastic or
partly-elastic collisions, but as we'll see, we need to modify it a bit
to handle fully-inelastic collisions.

An example of a fully-inelastic collision is crunching snowballs
(or lumps of modelling clay) together.  Notice that if you compress a
snowball (or a lump of clay), then let go, the snowball (or clay) stays
compressed: it doesn't rebound to its original size.  In contrast, in
our earlier model the springs can (and would) re-expand after being
compressed.

So, to model a fully-inelastic collision, let's change our model a bit:

We still have a set of point particles (5 for body A, 3 for body B),
equally spaced along a line.  But, now the particles are interconnected
with "struts", which behave like springs-with-friction *except* that
they don't re-expand after being compressed.  More precisely, if L is
the (time-dependent) current length of a strut, and L_eq is the original
equilibrium length of a strut, then

  If dL/dt < 0, i.e., if the strut's length is currently decreasing,
  then the strut acts like a spring-with-friction, exerting a restoring
  force R = -k (L - L_eq) - mu dL/dt, where k is the elastic spring
  constant, mu is the spring friction coefficient, and our sign convention
  is that R > 0 pushes "outwards" (i.e., tends to increase L).

  Otherwise, the strut exerts no restoring force (i.e., R = 0) AND
  we reset L_eq to the current value of L.

For example, if we imagine compressing the strut a bit, then removing
the applied compression force, then after we remove the applied
compression force there's no restoring force, so the strut doesn't
re-expand.

The reason we reset L_eq to the current (compressed) L is so that the
system behaves reasonably if there's a further compression.  If we didn't
reset L_eq, then even if there were only a very small further compression,
the restoring force R would immediately jump to a large value, which seems
physically implausible.  By resetting L_eq, we ensure that if there's
only a small further compression, there will only be a small restoring
force R, which seems more reasonable.

We also need to say what happens in our model when a pair of particles
collide (as do A1 and B1 at the very beginning of the collision). A
reasonable answer is to say that the particles then stick together,
moving as a single particle with mass 2.  (This means that the system
of equations changes -- there is now one fewer particle, and not all
the particles have the same masses.)

Now we can use Newton's laws to construct a full-fledged dynamical
model of the collision.  Each point particle moves according to Newton's
2nd law,
  m_i d^2 x_i/dt^2 = F_net_i
where m_i and x_i are the mass and position of particle i, and F_net_i
is the net force acting on particle i, i.e., the algebraic sum of the
restoring force R_left_i from the strut to the left of the particle
(if there is a strut there) and the restoring force R_right_i from
the strut to the right of the particle (if there is a strut there),
  F_net_i = R_left_i - R_right_i



So, what happens in the collision?  To solve the model numerically and
find out, during the simulation we have to
(a) detect any time(s) at which any strut stops compressing (so that
    we can reset L_eq to the current value of L), and
(b) detect any time(s) at which any pair of particles collide (so that
    we can replace them by a single particle with their total mass, and
    change the system of equations accordingly).

<<begin slight digression on numerical software design>>
The standard design for ODE integration routines (e.g., the Gnu Scientific
Library routines I used in the program I posted earlier in this thread)
solves the equations
  du(t)/dt = F(t,u(t))   where F is a user-provided subroutine
  u(t=0) = u_0           where u_0 are user-provided initial data
where u_0, u(t), and the result of F are all N-dimensional real vectors.
You write code to specify a sequence of "goal" times t_goal (these are
usually just the times at which you want to print the solution), and
for each goal time the ODE integration routines take as many internal
time steps as are needed to advance the solution vector to or beyond
the goal time, then (if the integration actually went beyond the goal
time) time-interpolate to determine the solution vector at the goal time.

This design usually works well, and it's very efficient because it lets
the ODE integration routine figure out the right internal time steps
based on the behavior of the solution.

But for our inelastic-collision problem, the standard design is a bit
clumsy: because we don't know if or when condition (a) and/or (b) might
occur, we have to instead specify each goal time to be only a tiny step
ahead of the current time, so that we can check for (a) and/or (b) before
calling the ODE-integration routine again to further advance the solution.
That is, we have to bypass all the fancy "automatically choose the most
efficient time step" mechanisms in the ODE integration routine.  The
result is extra work for us (the programmer) and a much less efficient
program.

An alternative approach is to use fancier ODE integration software
which can monitor "auxiliary conditions" automatically.  That is, ODE
integration routines like LSODAR (<https://www.netlib.org/odepack/>) or
its successor SUNDIALS (<https://computing.llnl.gov/projects/sundials/>)
can solve the equations
  du(t)/dt = F(t,u(t))   where F is a user-provided subroutine
  u(t=0) = u_0           where u_0 are user-provided initial data
where u_0, u(t), and the result of F are all N-dimensional real vectors,
AND they can (optionally) also concurrently monitor a set of M auxiliary
functions G(t,u(t)) (where G is a user-provided subroutine returning
an M-dimensional vector), so that if any component of G(t,u(t)) = 0,
then the integration will stop and the ODE integration routine will
report the time and which component of G is zero.

This make it much easier to program our dynamical model, because we can
now easily check for (a) by having one of the components of G be dL/dt,
and we can check for (b) by having one of the components of G be the
distance between an adjacent pair of particles.  And, with this type of
programming, we can return to specifying only the goal times when we want
to print the solution, and let the ODE integration routine automatically
choose efficient internal step sizes.

So, at the cost of having to use a somewhat more complicated set of
ODE integration software, we get both easier programming *and* a much
more efficient program.
<<end slight digression on numerical software design>>

If I find the time, I'll write a program implementing the above and
post some of the output.

ciao,
-- 
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
   (he/him; currently on the west coast of Canada)
   "The ideal subject of totalitarian rule is not the convinced Nazi or
    the convinced Communist, but people for whom the distinction between
    fact and fiction (i.e., the reality of experience) and the distinction
    between true and false (i.e., the standards of thought) no longer exist."
                             -- Hannah Arendt, The Origins of Totalitarianism

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