Program glitch or discovery?

[ABV]

Base on this theory
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#
Simulator got this glitch.
Is it discovery?
...
Hi Alex,

I'm not sure to understand all what you mean and may be I'm wrong on some
points.
So here is my simulation to clarify the problem (a hard work :
http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com.wm2d

If you have not Workingmodel software (it is free), here is the video (6Mb,
probably long to download):
http://exvacuo.free.fr/div/Sciences/Experiences/Meca/Momentum/com.avi

Here are 3 jpg snapshots. N�0 is the initial position. N�1 is the state just
after the pulse. N�2 is some seconds later.

The speed vector of each rod is displayed in the animation. The impulse force it
not displayed. It provides 100 N during 0.01s between the rods, at points of
same coordinates (= the position of the center of mass of the green rod).

In the red and green small windows, you can monitor the speeds, kinetic energies
and momenta of each rod. The default equations of the program are used.

In the small blue windows, I have put the equations to obtain the angular speed,
the moment of inertia and the angular momentum of each rod, calculated from the
common center of mass of the system.
This center of mass is displayed in the simulation (it is calculated by the
program, it's not a fixed point. It is at rest because there is no external
force acting onto the system. It is the mid-point of the line joining the
centers of mass of the two rods).
To simplify the calculi, the positions of the rods are chosen in order the
common center of mass to be at position x,y = 0,0.

With this simulation, I have discovered that from the common center of mass C,
each rod possesses a "hidden" angular momentum: each rod flies horizontally away
one another but one is above and the other under the horizontal x axis
containing their common center of mass. Thus the angle delimited by the
horizontal x axis and the line joining the centers of mass of the two rods
(crossing at C), decreases when the distance of the rods from the origin
increases. This variation of the angle is to be considered as an angular
velocity. Then from it, we can calculate the moment of inertia of each rod and
its angular momentum. The result is displayed in the small blue windows.

IMPORTANT : We see that the sum of the angular momenta of the two rods,
calculated in the referential frame of their common center of mass, is equal but
with opposite sign, to the angular momentum of the only rotating rod, calculated
in its proper frame.
Thus by adding these two momenta, we find zero. I guess the key of the problem
lies around this, but it is not yet completely clear for me.

Fran�ois

 

[kurros]

There is some highly flawed physics going on at that website, especially in the second experiment. The two objects are initially stationary, and then by some unspecified process they repel each other and one of them starts rotating. He treats the system as closed, yet in a closed system it is impossible for what I just said to happen, since it violates conservation of angular momentum (as he recognises it does). However he is not discovering anything through this, he is simply not treating the system correctly. The force repelling the objects from each other must be externally generated and apply a torque to the system to obtain the desired resultant motion, which then won't conserve energy or momentum so his analysis won't work.

The simulator itself seems based on this flawed understanding too, so really it is neither a glitch or a discovery, it is just bad physics in the first place.

 

[James R]

The situation is not clear for me.

There's an "initial pulse force", which I assume comes from outside the system, but I don't know at what point that force is applied.

Also, I'm not clear on whether the rods are initially connected to each other or not. If they are connected, how do they separate? Is the pulse force applied to both of the rods or just to one?

 

[D H]

There are at least three things wrong here.

1. The geometry is all messed up, Alex.

2. It appears that the forces involved may not be equal but opposite and directed along the same line. That third law reaction forces are equal-but-opposite is a necessary and sufficient condition to yield conservation of linear momentum. It is a necessary but not sufficient condition to yield conservation of angular momentum. In addition, the reaction forces need to be collinear to yield conservation of angular momentum.

3. There is angular momentum due to translation as well as rotation. You are ignoring this.

 

[kurros]

Hmm yes I forgot about the angular momentum from the translational motion too, oops. So maybe what I said isn't strictly correct, but the two objects still won't end up with equal linear velocities, since some of the energy of one is spent on rotating it.

 

[D H]

Assuming the masses are equal the two objects will end up with equal-but-opposite linear velocities when viewed from the system center of mass frame. Linear momentum would not be conserved otherwise.

They won't end up with the same kinetic energy. One of the is spinning, the other is not.

 

[kurros]

You are right of course. Man I'm not doing well with my classical mechanics.
So ok, isn't the program fine then except that they didn't add in the contribution to the angular momentum of the system due to the linear motion of the rods? Except that that seems to be calculated in the blue box and is exactly opposite to the "spin" angular momentum of the red rod, as one would hope, although it drifts out after a while presumably due to numerical error...
I'm not sure what you are worried about with your other two points, well ok so the rods cross through each other but is that all you mean by the geometry is messed up? As far as unequal forces being applied I don't think that is happening. As you say linear momentum has been conserved so I assume equal and opposite impulses have been applied to each rod in a horizontal direction at the point they intersect.

I think I've changed my mind, the program (or at least the avi I have been watching) is fine, hasn't discovered any new physics, and perhaps has just had its results misinterpreted.

 

[ABV]

DH
Thank you for your response

Angular momentum and translation momentums are not same thing.
They not correlate to each other.
If body starts rotating then it won't be change body's translation move.
However, body can start conduct translation and rotation movement from one hit (momentum). Good example - experiment 2.
It possible to make these equations:
P=m2*v2=m1*v1
L=I*w=m1*v1*R
But it won't describe experiment 2. It describes 2 movements. One is translation. Another one is rotation. These 2 movements cannot use same momentum m1*v1. It's wrong. These 2 equations describe 2 unrelated movements and they shoud use 2 independent momentums.
To solve this problem for experiment 2, please look on my site.

hxxp://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#

Thank you

 

[D H]

Alex, you are not capturing all of the angular momentum. The angular momentum of a collection of point masses with respect to some origin is

$${\mathbf L}_{\text{tot}} = \sum_{\iota} m_{\iota} \,{\mathbf r}_{\iota} \times {\mathbf v}_{\iota}$$

For a rigid body of mass this simplifies to

$$\mathbf L = m {\mathbf r}_{\text{cm}} \times {\mathbf v}_{\text{cm}} + {\mathbf I}{\mathbf{\omega}$$

where $${\mathbf r}_{\text{cm}}$$ and $${\mathbf v}_{\text{cm}}$$ are the position and velocity of the body's center of mass with respect to the origin of the frame of interest.

You are ignoring the $$m {\mathbf r}_{\text{cm}} \times {\mathbf v}_{\text{cm}}$$ term, so of course you get the wrong answer.

 

[DRZion]

May be off topic, but for all your information, using law of sines and law of cosines for very small angles will give you large errors.

 

[D H]

May be off topic, but for all your information, using law of sines and law of cosines for very small angles will give you large errors.

Completely off-topic, completely useless, and completely wrong. Home run!

 

[darksidZz]

All this makes me think physics should be used more in medicine :\

"Doctor I'm sick"

"Yes you are, now tell me where or how"

"Everywhere, my body aches"

"Very well let me write out the script for you, it'll take me 2 hours and by then all your pain will be gone"

"Why?"

"Ever heard time cures everything?"

lulz

 

[DRZion]

Completely off-topic, completely useless, and completely wrong. Home run!

How is it off topic??? We are talking about program errors so I thought I'd share my experience and maybe inspire people to rethink certain assumptions.

And here some evidence to seal the deal:
http://en.wikipedia.org/wiki/Law_of_cosines#Applications
"These formulas produce high round-off errors in floating point calculations if the triangle is very acute, i.e., if c is small relative to a and b or γ is small compared to 1."

 

[D H]

No, we are talking about what the OP thinks is a program error. The real problem with the OP is an inadequate knowledge of physics. This has absolutely nothing to do with what you posted.

 

[ABV]

I checked angular velocities equations for this simulation.
w=(V_xS_y+V_yS_x)/(S_y^2+S_x^2)
I'm not sure this is a good equations.


However. I did my own simulations.
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#
The experiment 2 has extra rotation for one rod. This additional torque come into one rod only. This torque doesn't have pair with negative value.
I made another simulation where substitute initial momentum displacement to additional torque. And I got same result.

experiment 1 + additional torque = experiment 2

This experiment 2 show how generate additional torque for Isolated System. The initial pulse force displacement between experiment 1 and 2 should not produce extra torque or angular momentum.

Adding one torque without its own pair is breaking law of angular momentum conservation.

 

[D H]

Physics fail! This is freshman-level physics that you are failing to grasp, Alex. Read a freshman physics text.

 

[DRZion]

Physics fail! This is freshman-level physics that you are failing to grasp, Alex. Read a freshman physics text.

We understand all physics so don't bother considering alternatives and trust what we've got. [the odds are against you ;p]

 

[AlphaNumeric]

We understand all physics so don't bother considering alternatives and trust what we've got. [the odds are against you ;p]

That isn't what DH is saying at all. ABV is making the claim that "Classical mechanics says [something] but I have found it also says [something else] and they don't match. So classical mechanics is wrong."

DH's point is that ABV is getting what classical mechanics say wrong because ABV is a thick clod. Ok, I'm calling him a thick clod but the gist is the same.

 

[D H]

Alex has not discovered a paradox in classical mechanics. He has made several mistakes in the google.docs document cited in the original post, and the simulation done by Alex's friend also appears to be buggy.

Alex, you have a tendency to make overly complicated systems with some poorly defined concepts. This makes it a bit tougher to find where your reasoning has gone awry. So, let's simplify and clarify this system.

Simplification #1: In both experiments a pair of equal-but-opposite forces are applied to the two rods. The force is applied at the center of mass of rod #1 in both experiments.

A question arises: Why bother with a rod for rod #1? There is zero difference in the resulting physics if you use a point mass, so you might as well use a point mass.

Clarification: What is the nature of the force? If you make the energy source that applies the force internal to the system then the total linear momentum and total angular momentum of the system comprising the two objects and the energy source will be conserved. If you make that force a conservative force then energy will be conserved as well. For example, a spring will do the trick. All of the conserved quantities are conserved: energy, linear momentum, and angular momentum.

You can still make the mass+rod+spring system rather complex. For example, the finite amount of time it takes for the spring to fully propel the point mass away from the rod means that the rod rotates while it is ejecting the mass. One way around this:

Simplification #2: Make the spring extremely stiff. Making the spring very stiff makes it so one can ignore this rotation issue. Whether you do this or not, Alex, the bottom line is that energy, linear momentum, and angular momentum will be conserved.

 

[ABV]

Both rods have same mass and moment of inertia.
However, on experiment 2 the rotating rod has extra torque, which easy to simulate. Please look on experiment 3.
This means the simulator apply extra translation force for rotation rod. Which is should not be on real world.
If part of applied force spent for rotation movement (torque) then this portion of force should be removed from translation part of applied force. The sum of these rotation and translation portions of applied force should be equal to applied force for non-rotating rod. (3-rd Newton’s law)
The translation velocities of these rods should be different. However, they are equal on simulator. Because simulator knows nothing about rotation with translation movement and two independent rotation and translation movements cannot describe this experiment 2 correctly.The classical mechanics should include new standalone translation with rotation movement to describe natural phenomenon correctly.

Please look on my site
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#
I added
Steps How to build these experiments on working model demo version.