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Re: Paradox of Classical Mechanics

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  • abelov0927
    Let s solve simple problem. Given a long thin cylinder with length L, radius R and mass M. Case 1. This cylinder takes momentum P into his center mass. Case 2.
    Message 1 of 11 , Jul 23 4:29 PM
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      Let's solve simple problem.
      Given a long thin cylinder with length L, radius R and mass M.
      Case 1.
      This cylinder takes momentum P into his center mass.
      Case 2.
      This cylinder takes momentum P on a some distance from his center mass.
      The cylinder start rotation with translation movement.
       
      How many times the translation velocity for case 2 is lower than translation velocity for case 1?
      ======================================================================
       
      If calculate this problem using classical mechanics laws, the linear momentum should be conserve. However, these cases take different energies. The cylinder with translation and rotation movements will take two parts (translation and rotation) of energies. The cylinder with translation takes only one( translation) part. These translation kinetic energy parts should be equal.
      Where this rotation kinetic energy come from? From nowhere? What about angular momentum? How it starts rotating if all momentum transfers to translation part?
       
      If assume for case 2 the incoming momentum vector dividing for two parts initially then it will explain everything. The first part – translation momentum gives ability the cylinder moves. The second part – angular momentum gives ability the cylinder rotates. The sum of these parts kinetic energies is equal to full translation kinetic energy from case1. The case 2 cylinder translation velocity lower than case 1 translation cylinder velocity.
      The equation translation velocity is:
       
      V_c_a_s_e_2=V_c_a_s_e_1\sqrt{\frac{E_t}{E_t+E_r}}
      Et - kinetic energy translation part
      Er - kinetic energy rotation part
       
      Let's make a simple experiment with pencil.
      Case 1: Let's hit the pencil on his middle – The pencil fly away on a long distance.
      Case 2: Let's hit the pencil on his edge – The pencil starts rotation and fly away not so far.
      The classical mechanics solution says – For both cases the pencil should fly away on a long distance with same high translation velocity. No matter how fast it is rotating.
       
      Is it really true? I don't see this result.

      ---
      --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
      >
      > Another good example:
      > Let's repulse a long cylinder inside Isolated System. The repulse should hit the cylinder away from object center mass. This hit energy splits for two object movements. This cylinder starts rotation with translation movement. The Isolated System starts move to opposite cylinder translation movement direction. During this object movement let transform this cylinder to a sphere. This sphere is rotating faster than cylinder because this object moment of inertia less than cylinder moment of inertia. However this rotation does not make any sense for translation momentum transfer. The sphere translation momentum is equal to cylinder translation momentum. However from cylinder repulse action the system takes higher translation momentum than sphere has. The sphere translation momentum won't be enough to stop the Isolated System.
      > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#
      >
      >
      > --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" abelov0927@ wrote:
      > This is a good example is shown transfer between angular and linear momentums. Unfortunately the simulator does not allow giving different angular velocities for squares.
      > Anyway, on this model easy to understand these squares with different angular velocities will take a different linear velocities on the end of action.
      > Base on my theory about frame of reference conversion for complicated movement, the law of momentum conservation works, however for correct calculation need use imaginary part of velocity.
      > The momentum is conserve. The frame of reference is changing. This frame of reference exchange gives the system ability to move.
      >
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