Re: Newton e Hooke

Jonathan Thornburg [remove -color to reply] il 17/02/2025 09:19:28 ha
scritto:

In article <m1dng8FkasU1@mid.dfncis.de>, I wrote that if we push on
one end of a spring,

the spring's center-of-mass
acceleration is determined by *all* of the applied force, while at the
same time the spring compresses.

and I worked this out in detail for a simple model system.
>
I neglected to point out that there's actually a much simpler way of
coming to this same conclusion: simply apply conservation of momentum
to the spring.
>
That is, consider the spring's total (horizontal) linear momentum p.
Since there's an external force F pushing the spring to the right, p
must change according to Newton's 2nd law (dp/dt = F), i.e., the spring's
center of mass must be accelerating to the right at an acceleration
a = F/m_total.
>
The fact that the spring also has a bunch of internal dynamics isn't
relevant here -- we ony consider the spring's total linear momentum,
and the (horizontal) external force acting on the spring.
>
In fact, this argument is still true if there's no spring at all, just a
pair of (unconnected) point masses at the ends of the "spring".  That is,
if we apply a force F to one mass, leaving the other mass stationary
(i.e., F is applied to mass #1, and there is no force applied to mass #2),
the center-of-mass of the two masses (at position xc = (x1 + x2)/2)
accelerates with an acceleration
  ac = (a1 + a2)/2
     = (F/m + 0)/2
     = F/(2m)
     = F/m_total
even though there the force is only pushing on one of the two masses.
>
In article <vot1s2$lj72$1@dont-email.me>, Luigi Fortunati wrote

The acceleration of the center of mass cannot be determined by *all*
the applied force because that force does NOT act on the center of
mass!

For the momentum argument I gave above, it doesn't matter where the
applied force acts.  We only care that it's an external force applied
to *somewhere* in the system of interest (the spring).

It is not true that we do not care where the external force is applied,
because it is precisely at that point that the exchange of forces
occurs.

If you exert a force at the end of a body it is not the same as if you
exert it at the center of mass.

In the first case you compress (and cause the reaction), in the second
case you do not!

Note that in accelerated bodies, the internal forces are not zero (and
here we should open another separate discussion).

Luigi also wrote

If you refuse to watch my animation https://www.geogebra.org/m/mrjtyuwk
you cannot notice that the applied force Fa (black) acts on point A"
and, before it can reach C, it must confront the opposing blue force.

>
Forces don't "reach" points or "confront" other forces.  Forces (only)
act on, or are applied to, or push/pull on, objects or points on those
objects.
>

The question is simple: does the opposing blue force in my animation
exist or not?

>
Yes, there is a reaction force of magnitude F pushing left on the hand.
This force is pushing on the hand, not on the car, so it doesn't affect
the car's motion.

It is not true that the blue force acts only on the hand, it also acts
on the car and it is easy to demonstrate this.

If it were as you say, only the force F to the right would act on the
car and no blue force to the left.

But no!

On point A" a blue force to the left also acts and is exerted by the
next particle B (which I have added now and colored in red to highlight
it).

The origin of this force is the following: point A" receives the black
force F to the right and transmits it to the next particle B which
reacts by returning a force to the left to particle A" (action and
reaction).

Here is therefore demonstrated that there are *also* blue forces to the
left exerted by the particles of the car (B) on other particles of the
car (A") and not only on the hand.

And it is not only valid for particles A" and B but also for all the
others up to C, passing from particle to particle.

I point out that in this transmission of force from particle to
particle, in each passage a little force is lost and when it arrives
the center of gravity C, the black force F is no longer the same as
before.

And even less so when it arrives from the opposite side of the car.

Luigi Fortunati