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Re: Multidimensional physics

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  • abelov0927
    The rotated cylinder center mass has slowest linear velocity than cylinder without rotation on repulsion action. The statement of proof. Let s take two
    Message 1 of 3 , Jul 29, 2009
      The rotated cylinder center mass has slowest linear velocity than cylinder without rotation on repulsion action.

      The 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.
       
      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=mfrac{v^2}{2}+Ifrac{omega^2}{2}
      T_1=mfrac{v_1^2}{2}+0(no_._._.rotation)
      T_2=mfrac{v_2^2}{2}+0(no_._._.rotation)
      v_1=v_2Rightarrow T_1=T_2
      As it shown on equation their energies are equal.
      Let's take the derivatives from their parts of energy.
      dot E_t_1 = mv_1
      dot E_t_2 = mv_2
      v_1=v_2Rightarrow mv_1=mv_2
      As it shown on equation their momentums are equal.
      This experiment action is symmetric relatively to 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_1=mfrac{v_1^2}{2}+0(no_._._.rotation)
      T_2=mfrac{v_2^2}{2}+Ifrac{omega^2}{2}(has_._._.rotation)
      v_1=v_2_._._._.omega_1neomega_2Rightarrow T_1ne T_2
      As it shown on equation their energies are not equal.
      Let's take the derivatives from their parts of energy.
      dot E_t_1 = mv_1_._._._._._. dot E_r_1=0
      dot E_t_2=mv_2_._._._._._. dot E_r_2=Iomega_2
      v_1=v_2_._._._.omega_1neomega_2Rightarrow mv_1=mv_2_._._._. 0ne Iomega_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 observer.
      This assumption broke 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.
      These experiments is explaining the translation with rotation movement is standalone natural phenomenon. And law of momentum conservation should cover this movement.
       
      My assumption this movement have a linear and angular momentum together.
      This is the law of momentum conservation for translation with rotation movement.

      [sum P_1_i]^2 +[sum L_1_i] ^2 = [sum P_2_i]^2 +[sum L_2_i] ^2
    • abelov0927
      My assumption this movement have a linear and angular momentums together. This is the law of momentum conservation for the translation with rotation movement:
      Message 2 of 3 , Aug 2, 2009
        My assumption this movement have a linear and angular momentums together.
        This is the law of momentum conservation for the translation with rotation movement:

        sum P_j +[sum L_k]i = Const
         
        This law has a complex number and it has linear and angular momentums.
        One of these parts linear or angular should be imaginary number. It depends on observer which frame of reference he uses.
        Full momentum transfer for this movement has calculation by absolute value:
         
        Z^2=[sum P_j]^2 +[sum L_k]^2

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