The second principle and inertia

[Luigi Fortunati]

In animation
https://www.geogebra.org/m/myfn2dgd
there are 2 wagons, one full of water and the other not, but they both
have the same overall mass.

[[Mod. note -- Here I have merged in 2 lines of text from the author's
following post -- jt]]

Animation in English
https://www.geogebra.org/m/adsgahwf


When stationary, the rigid wagon has no horizontal force.

Instead, in the other car there is water which also exerts horizontal
forces against the vertical walls of the car (they are the red colored
forces).

But, since these red forces are equal and opposite, they have no
influence on the motion.

However, when we press "Start" to activate the blue F forces (the same
for both wagons), the left red force (that of the wagon with water)
increases and the right one decreases, as you can see from the
different length of the two forces.

And therefore, a net red force is activated contrary to the blue force,
which reduces the resultant of all the horizontal forces (blue and red)
acting on the wagon with the water.

On the other hand, in the rigid wagon, there are no other horizontal
forces that can counteract the blue force.

And then, will the two wagons arrive at the end together (as in the
animation) or will the rigid carriage arrive first that has no red
force to hinder the blue force?

[Richard Livingston]

On Saturday, October 8, 2022 at 12:50:33 PM UTC-5, Luigi Fortunati wrote:
> In animation
...

> Animation in English
> https://www.geogebra.org/m/adsgahwf

You are missing the inertial force ("red") of the "rigid" wagon weight against the
applied "blue" force. The "rigid" "red" force would be greater than the
corresponding red force on the water wagon.

Rich L.

[Luigi Fortunati]

Richard Livingston domenica 09/10/2022 alle ore 17:35:01 ha scritto:
>> In animation https://www.geogebra.org/m/adsgahwf

>
> You are missing the inertial force ("red") of the "rigid" wagon weight against the applied "blue" force.

Is the red force I added to my animation okay?

> The "rigid" "red" force would be greater than the corresponding red force on the water wagon.

Ok.

Is the acceleration of the two vagons the same?

Luigi

[Richard Livingston]

On Monday, October 10, 2022 at 1:18:14 AM UTC-5, Luigi Fortunati wrote:
...

> Is the acceleration of the two vagons the same?
>
> Luigi

No, not initially. The water wagon will initially accelerate faster as
the effective mass is less (because the water is not moving yet). But
when the water sloshes back, towards the rear, it will slow down. The
exact motion depends on the "impedance" of the tractor pulling the
wagons.

Rich L.

[Luigi Fortunati]

Richard Livingston lunedì 10/10/2022 alle ore 05:21:46 ha scritto:
> ...

The red force that at your suggestion I added to the rigid wagon of my animation
https://www.geogebra.org/m/adsgahwf
in your opinion, is there or is it not?

>> Is the acceleration of the two vagons the same?
>

> No, not initially. The water wagon will initially accelerate faster as
> the effective mass is less (because the water is not moving yet). But
> when the water sloshes back, towards the rear, it will slow down. The
> exact motion depends on the "impedance" of the tractor pulling the
> wagons.

The water cannot move back and forth because the tank car is totally
full and there are no empty spaces.

[Richard Livingston]

In that case the two cars will behave exactly the same.

Rich L.

[Luigi Fortunati]

Richard Livingston luned=EC 10/10/2022 alle ore 05:21:46 ha scritto:
> ...

La forza rossa che su tuo suggerimento ho aggiunto al vagone rigido
della mia animazione
https://www.geogebra.org/m/adsgahwf
secondo il tuo parere, c'=E8 o non c'=E8?

>> Is the acceleration of the two vagons the same?
>

> No, not initially. The water wagon will initially accelerate faster as
> the effective mass is less (because the water is not moving yet). But
> when the water sloshes back, towards the rear, it will slow down. The
> exact motion depends on the "impedance" of the tractor pulling the
> wagons.

[Luigi Fortunati]

Richard Livingston martedì 11/10/2022 alle ore 16:04:20 ha scritto:
>> The water cannot move back and forth because the tank car is totally full and there are no empty spaces.
>
> In that case the two cars will behave exactly the same.

And this means that in the rigid vagon there is a horizontal inertial
force during acceleration, as can be seen in my last animation
> https://www.geogebra.org/m/eega7se6

The strange thing is that inertial forces should only "appear" in
accelerated reference frame and not in my animation set in inertial
reference frame.

Luigi

[Luigi Fortunati]

A similar situation to my previous animation is the following:
https://www.geogebra.org/m/gbqzyba2

In the inertial reference frame of the ground, when bodies are still,
there is no centripetal force and there is no centrifugal force either.

During the rotation, the blue centripetal force is activated which is
the only force acting on body A.

Instead, on body B (full of water but with the same overall mass as
body A), in addition to the blue force, the red force of the water
directed outwards (i.e. in the opposite direction to the centripetal
force) also acts.

In your opinion, is it correct that different forces act on body B than
on body A, only because of their different rigidity?

[Richard Livingston]

No, the weight without water has the same "red" reaction force. There
is no difference.

Rich L.

[Jonathan Thornburg [remove -color to reply]]

Luigi Fortunati <fortuna...@gmail.com> wrote:
> A similar situation to my previous animation is the following:
> https://www.geogebra.org/m/gbqzyba2
>
> In the inertial reference frame of the ground, when bodies are still,
> there is no centripetal force and there is no centrifugal force either.
>
> During the rotation, the blue centripetal force is activated which is
> the only force acting on body A.

I would phrase this slightly differently: in the rotating state there
*must* be a "blue" centripetal force acting on each body, otherwise the
body would move in a straight line, not in a circle.

> Instead, on body B (full of water but with the same overall mass as
> body A), in addition to the blue force, the red force of the water
> directed outwards (i.e. in the opposite direction to the centripetal
> force) also acts.

The mass of the empty B must be less than the mass of A (the difference
being the mass of the water). So, the overall centripetal force acting
on the water-filled B must be sum of the centripetal force which would
be required to give the empty B that same centripetal acceleration, plus
the centripetal force which is required to give the water its centripetal
acceleration. (*)

If we approximate the diameter of B as small relative to the radius
of B's orbit around the rotation axis, then the centripetal acceleration
is uniform across the diameter of B.

Let's introduce a few symbols:
M_A = mass of A
M_B = mass of the solid body B without the water
M_W = mass of the water
a = the magnitude of the centripetal acceleration
T = the tension in the string that ties each body to the rotation
axis; this tension is the same for A and B since M_A = M_B + M_W

Then by Newton's 2nd law, the net force (centripetal) acting on A
must be given by $T = M_A a$, and the net force (again centripetal)
acting on the water-filled B must be
T = (M_B + M_W) a = M_B a + M_W a (1)
which is just the statement (*) above.

Ultimately the only thing exerting forces on the water is B, i.e.,
the solid body B must exert an inward (centripetal) net force of
$M_W a$ on the water, and by Newton's 3rd law the water must exert
a corresponding (equal in magnitude, opposite in direction) centrifugal
force $M_W a$ on the solid body B.

As is often the case in Newtonian mechanics, it's instructive to draw
some free-body diagrams. I'll show these for the case where A and B
have each rotated 180 degrees, i.e., they are directly to the right
of the rotation axis, so their centripetal accelerations point left:

Free-body diagram for A:

<-------- A

There is only one force acting on A: the string tension $T$, pointing
inwards towards the rotation axis (to the left). This is shown in blue
in Luigi's animation.

Free-body diagram for the solid body B (note we are considering the water
to be a separate body):

<-------- B ----->

There are two forces acting on the solid body B (again, we are considering
the water to be a separate body):
* the string tension $T$ to the left; this is shown in blue in
Luigi's animation
* the force the water exerts on B, $M_W a$, to the right; this is
shown in red in Luigi's animation, but the animation wrongly shows
the magnitude of this force as being the same as that of the blue
force $T$; it's actually smaller than $T$

The net force on B is the algebraic sum of these two forces, $T - M_W a$
to the left. This is smaller than the net force acting on A, which is just
what we expect, since A and B have the same (centripetal) acceleration $a$,
but the mass of the solid body B is less than the mass of A. In fact,

F_net = T - M_W a (2)
= (M_B + M_W) a - M_W a [by equation (1)] (3)
= M_B a (4)

And finally, here's a free-body diagram for the water:

<----- W

There is only one force acting on the water, which is exerted by B
and which points inwards towards the rotation axis (i.e., to the left).
The magnitude of this force is $M_W a$.

We can also analyze the forces between B and the water in detail, but
that takes a little bit more work (and some elementary calculus):

To keep things simple, I'll assume that the water in B rotates like
a rigid body, with the same angular frequency as B. (If the rotation
of B is uniform and lasts long enough, then the small-but-nonzero
viscosity of water will result in the water's velocity field gradually
approaching this rigid-body-rotation state.)

Assuming this rigid-body rotation, let's consider what force(s) act
on each little (infinitesimal) volume element of water. In particular,
consider an infintesimal volume element of mass $dm$, which is at a
radius $r$ from the rotation axis.

There are two physical processes which can exert forces on this water
element:
(a) there may be a pressure gradient in the water, and
(b) the walls of B may directly exert forces on the water immediately
adjacent to the walls

Consider process (a): if we take our infinitesimal volume element to
of water be a cylinder with flat sides of area dA facing towards & away
from the rotation axis, and to extend from radius-from-rotation-axis r
to r+dr, then we have

dm = dA dr rho (5)

where rho is the density of water (we assume rho to be constant, i.e.,
we're neglecting the compressibility of water). The net force acting
on the volume element of water due to the pressure gradient is then
the difference between the forces acting on the two opposite flat faces
of the volume element, i.e.,

dF_net = p(r+dr) dA - p(r) dA (6)

Also, by Newton's 2nd law (since this volume element of mass dm is moving
with centripetal acceleration $a$), there must be a net force acting on
the volume element of

dF_net = dm a (7)
= dA dr a [here we're using (5)] (8)

Comparing (6) and (8), we then have

dA dr rho a = (p(r+dr) dA - p(r)) dA (9)

where p = p(r) is the pressure as a function of position. Dividing
by dA, this gives

dr rho a = p(r+dr) - p(r) (10)

and hence

rho a = dp/dr (11)

and hence

p(r) = K + rho a r (12)

where $K$ is an arbitrary constant. If we think about the case where
$a = 0$, i.e., there is no rotation and hence no centripetal acceleration,
we can see that the constant $K$ is just the initial water pressure in
this non-rotating state (i.e., the water pressure in B before B started
rotating).

Equation (12) tells us how the water pressure varies with position inside
B. In particular, (12) says that the water pressure is larger on the side
of B which is farther away from the rotation axis, and smaller on the side
of B which is closer to the rotation axis. (If the constant $K$ is small
enough, the pressure may even be negative in some/all of the water;
this corresponds to "tension".)

Now considering process (b) above, at each point on B's inside surface
the water pressure acts perpendicular to B's inside surface, and by
Newton's 3rd law B's inside surface exerts an equal and opposite force
(also perpendicular to B's surface) on the water.

Now let's temporarily assume that $K$ is large and positive, so that
the pressure given by (12) is always positive. Then:

Since the water pressure is larger on the side of B which is farther
away from the rotation axis, where the water exerts a force on B
*away* from the rotation axis, and smaller on the side of B which is
closer to the rotation axis, where the water exerts a force on B
*towards* the rotation axis, it's easy to see that the net force of
the water on B must point *away* from the rotation axis. (We could
work this force out in detail by integrating the radial component of
the right hand side of (12) over B's inner surface.) We saw above
that this force has magnitude $M_W a$.

Similarly, since the pressure B's inside surface exerts on the water
is larger on the side of B which is farther away from the rotation axis,
where this force points *towards* the rotation axis, and smaller on
the side of B which is closer to the rotation axis, where this force
points *away* from the rotation axis, it's easy to see that the net
force of B on the water must point *towards* the rotation axis.
(Again, we could work this out in detail by integrating the radial
component of the right hand side of equation (12) over B's inner
surface.) Again, we saw above that this force has magnitude $M_W a$.

Finally, let's return to our temporary assumption that $K$ is large
and positive, so that the pressure given by (12) is always positive.
Notice that $K$ corresponds to a pressure which is constant everywhere
on B's inside surface. Thus, by symmetry, changing $K$ does *not* change
the net force the water exerts on B, or the net force B exerts on the
water. Thus, our conclusions in the previous two paragraphs about
these net forces don't depend on the value of $K$, and so remain valid
for an arbitrary $K$. That is, we can now drop our temporary assumption
that $K$ was large and positive.

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.th...@gmail-pink.com>
on the west coast of Canada
"Now back when I worked in banking, if someone went to Barclays,
pretended to be me, borrowed UKP10,000 and legged it, that was
`impersonation', and it was the bank's money that had been stolen,
not my identity. How did things change?" -- Ross Anderson

[Jonathan Thornburg [remove -color to reply]]

In article <Y0uTogSz...@iron.bkis-orchard.net>, I wrote:
[[analysis of Luigi's system]]

Two small corrections to my derivation of the water pressure inside B
as a function of r (the distance from the rotation axis):

Correction #1: I wrote:

> The net force acting
> on the volume element of water due to the pressure gradient is then
> the difference between the forces acting on the two opposite flat faces
> of the volume element, i.e.,
>
> dF_net = p(r+dr) dA - p(r) dA (6)
>

Here (right after equation (6)) I should have inserted the phrase

where p = p(r) is the pressure as a function of position.

Correction #2: A few lines later, I wrote:

> Also, by Newton's 2nd law (since this volume element of mass dm is moving
> with centripetal acceleration $a$), there must be a net force acting on
> the volume element of
>
> dF_net = dm a (7)
> = dA dr a [here we're using (5)] (8)

Oops, I left out a factor of rho in the right hand side of (8),
i.e., I should have written

dF_net = dm a (7)

= dA dr rho a [here we're using (5)] (8)

It may also be worth noting that nothing in my analysis made use of the
fact that A and B are actually in *circular* motion about their rotation
axis; I only used the fact that they have a net acceleration (which I
called $a$) in the direction towards that axis, i.e., to the left in my
free-body diagrams. So, my analysis, including those free-body
diagrams, would have been unchanged had the physical situation instead
been that the string had a rocket with thrust $T$ attached to it,
accelerating A and B to the left with respect to an inertial reference
frame.

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.th...@gmail-pink.com>
on the west coast of Canada

"quit When the quit statement is read, the bc processor is terminated,
regardless of where the quit statement is found. For example,
'if (0 == 1) quit' will cause bc to terminate.
-- old Unix manpage for 'bc'"

[Luigi Fortunati]

Jonathan Thornburg [remove -color to reply] domenica 16/10/2022 alle
ore 14:09:31 ha scritto:

> Ultimately the only thing exerting forces on the water is B

Ok.

> The net force on B is the algebraic sum of these two forces

Ok.

You break body B into two parts: the solid body B and the water it
contains.

So, I updated my animation which is now this:
https://www.geogebra.org/m/deqamevm

I added at 180° stop button, where (as per your right suggestion) you
can see that the centrifugal red force (5) exerted by the water on body
B is smaller than the centripetal blue force (10) exerted by the rope
on the body B.

And I also added the blue (centripetal) reaction force of body B on the
water.

Therefore, on the solid structure of the body B acts (as you wrote) a
net force of 5 (10 of the string minus 5 of the water).

And on the water only the blue centripetal force (5) of the solid body
B acts.

Is my animation correct now?

[[Mod. note -- It looks correct to me. -- jt]]

[Luigi Fortunati]

Is this red reaction force (you speak of) directed against body A (like
the red water force of body B) or is it directed against the rope?

Is this red reaction force (you speak of) a real force (like the red
water force of body B) or is it an apparent force?

Luigi