gravitationalpropulsionstevenson : Message: A Rotation with Translation Movement is standalone natural phenomenon(version 2)

A Rotation with Translation Movement is standalone natural phenomenon.

Statement of proof.

Let's take two identical thin cylinders which stay initially relatively to an observer. These cylinders will repulse to each other.
Let's make two experiments.

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

The first experiment.

The cylinders repulse to each other from their center of mass. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation, their linear velocities are identical relatively to the observer. Let's measure their kinetic energies. The full kinetic energy is equal to:

$T=E_t+E_r=m\frac{v^2}{2}+I\frac{\omega^2}{2}$

$T_1=m\frac{v_1^2}{2}+0(\text{no rotation})$

$T_2=m\frac{v_2^2}{2}+0(\text{no rotation})$

$v_1=v_2\Rightarrow (m\frac{v_1^2}{2}+0) = (m\frac{v_2^2}{2}+0) \text{ , }T_1=T_2$

As it shown on equation their energies are equal.
Let's compare their translation and angular momentums.

$P_1 = mv_1 \text{ } L_1=I\omega_1$

$P_2 = mv_2 \text{ } L_2=I\omega_2$

$v_1=v_2\text{ , }\omega_1=\omega_2=0$

$mv_1=mv_2\Rightarrow P_1=P_2\text{ , }0=0\Rightarrow L_1=L_2$

As it shown on equation their momentums are equal.
For this experiment these cylinders have symmetric action as translation movement relatively to initial event (repulsing) and observer.

The second experiment.

The cylinders repulse to each other from their different parts. One cylinder is repulsing form his center mass. Another cylinder is repulsing from his edge. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation let's assume their linear velocities are identical relatively to the observer. Only one of these cylinders rotates. Let's measure their kinetic energies. The full kinetic energy is equal to:

$T=E_t+E_r=m\frac{v^2}{2}+I\frac{\omega^2}{2}$

$T_1=m\frac{v_1^2}{2}+0(\text{no rotation})$

$T_2=m\frac{v_2^2}{2}+I\frac{\omega_2^2}{2}(\text{has rotation})$

$v_1=v_2\text{ , }0\ne\omega_2\text{ }\Rightarrow \text{ } (m\frac{v_1^2}{2}+0) \ne (m\frac{v_2^2}{2}+I\frac{\omega_2^2}{2}) \text{ , }T_1\ne T_2$

As it shown on equation their energies are not equal.
Let's compare their translation and angular momentums.

$P_1 = mv_1 \text{ } L_1=I\omega_1$

$P_2=mv_2\text{ } L_2=I\omega_2$

$v_1=v_2\text{ , }0\ne\omega_2$

$mv_1=mv_2\Rightarrow P_1=P_2 \text{ , }0\ne I\omega_2\Rightarrow L_1 \ne L_2$

As it shown on equation their linear momentums are equal. However the angular momentums are not equal. Only one of these cylinders has angular momentum.
This experiment action is not symmetric relatively to initial event (repulsing) and observer.

This assumption broke the law of angular momentum conservation. The body can't start own rotating without symmetrical action.

This mean the assumption about identical linear velocity was wrong.

One even can't reproduce two type of movements together for one object. One event can reproduce only one type of movement per object.
On experiment 1 both objects have only one translation movement.
The experiment 2 shows only one type of movements for one object and two types of movements for other object. However these objects taked only one event(repulsing). It means it should be one type of movements per object and translation with rotation movements for this experiment should be described as new type of moments.
The experiment 2 shows situation where the translation with rotation movement as a standalone natural phenomenon. This movement should have own momentum and law of momentum conservation.

My assumption this movement have a linear and angular momentums together.

The full momentum of rotation with translation movement is:

$P_f= \sum P_j +\frac{1}{R_u}[\sum L_k]i$

note: Pj - linear momentum Lk - angular momentum Ru - unit radius

This is the law of momentum conservation for the translation with rotation movement:

$\sum P_j +\frac{1}{R_u}[\sum L_k]i = Const$

This law has a complex number and it has linear and angular momentums.

Full momentum transfer for this movement has calculation by absolute value:

$Z^2=[\sum P_j]^2 +[\frac{1}{R_u}\sum L_k]^2$

For example: if reverse time back for experiment 2 then one cylinders angular L2 and translation P2 momentums compensated by another cylinders only translation momentum P1.
For this case equation between momentums is:

$(P_1)^2=(P_2)^2 +(\frac{1}{R_u}L_2)^2$

Full kinetic energy of rotation with translation movement is:

$E=m\frac{v^2}{2}+I\frac{w^2}{2}$

Follow law of momentum conservation and using complex number for translation with rotation momentum, the cylinders translation velocities have a different value on experiment 2.

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Base on previous statement it possible to say the translation with rotation movement is covering translation and rotation movements together. One of these movements is the simple version of main complicated movement.

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A Rotation with Translation Movement is standalone natural phenomenon.

Statement of proof.

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A Rotation with Translation Movement is standalone natural phenomenon.

Statement of proof.

email: abelov0927@...