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Re: Paradox of Classical Mechanics

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  • abelov0927
    For this model, à repulsion and à collision comes on different planes. The vertical plane doesn t have any contacts with the rolling body. Otherwise the
    Message 1 of 11 , Jul 11, 2009
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      For this model, à repulsion and à collision comes on different planes.

      The vertical plane doesn't have any contacts with the rolling body.
      Otherwise the horizontal plane always has a physical contact with rolling body.

      From vertical plane point of view, all elements of rolling body have a movement.
      If calculate a rolling body translation momentum with respect to vertical plane then all elements of rolling body must be included.

      Otherwise if look on horizontal plane, one element of rolling body stays on the surface all the time.If calculate a rolling body translation momentum with respect to horizontal plane then one element of rolling body must be included to the mass of horizontal plane.


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

      --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
      >
      > Hi.
      > I think this is a clue for understanding.
      >
      > The element with zero values of linear velocity on the surface is too small. Base on math the geometrical size of this element strives to zero limit. However, this is the physical element, and it has its own geometrical size.
      >
      > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#
      >
      > --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@> wrote:
      > >
      > >
      > > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
      > > qm1l0s4ys/9
      > > <http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1x\
      > > mqm1l0s4ys/9> #
      > >
      > > No objects with mass in translation movement can transfer momentum
      > > without collision action. However, rotation with translation movement
      > > gives unique situation when object touches the surface without momentum
      > > transfer on a rolling body linear movement direction.
      > > The idea is very simple.
      > >
      > >
      > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/cutring2.jpg>
      > >
      > >
      > >
      > > If spit a rolling ring to small parts set of n elements (1,2,3,...,n)
      > > with mass m then each of them conducts linear and circular movement on
      > > surface. Good example is ­Caterpillar tracks (Continuous track).
      > > <http://en.wikipedia.org/wiki/Continuous_track>
      > >
      > > Each piece has constant angular velocity. This is the perfect picture
      > > from UPENN site
      > > <http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsu\
      > > bsection4_1_4_3.html> .
      > >
      > >
      > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/translationwithrotatio\
      > > n.gif>
      > >
      > > Each piece of ring has variable linear velocity at a surface point. Once
      > > per circle each element of ring stop on surface. This (fig. 3) shows the
      > > red bit trajectory. At the surface point, this element has linear
      > > velocity value equal to zero.
      > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/redbit.jpg>
      > > This (fig.2 ) shows all action between ring and surface.
      > >
      > > This action has 3 phases.
      > >
      > >
      > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase1.jpg>
      > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase2.jpg>
      > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase3%20(1).jpg>
      > > 1. The ring and the surface are getting momentums.
      > > The ring has mass n*m
      > > The surface has mass M
      > >
      > >
      > >
      > >
      > >
      > > 2. Cut and hold the red bit on the surface (The red bit has velocity
      > > zero V=0 on surface. The red bit doesn't transfer momentum to the ground
      > > on rings linear movement direction)
      > >
      > >
      > >
      > >
      > > 3. The pseudo chain transfer momentum to the ground. The chain has mass
      > > (n-1)*m
      > > The surface has mass M+m
      > >
      > > As it has shown on (fig. 2), the ring with mass n*m and the surface with
      > > mass M takes momentums on phase 1. On phase 2 the ring must be broken in
      > > one location and one element with zero values of linear velocity holding
      > > by the surface. This is not meaning to stop the whole ring at one time.
      > > This mean stops the red piece and cut the ring at the same time. The
      > > ring transforms to chain and lose the mass to new value (n-1)*m. On
      > > phase 3 the surface with mass M+m holds just one element of ring and
      > > other elements of pseudo chain with mass (n-1)*m is continuing movement
      > > by his own trajectories until they rich the ground. The very important
      > > thing is: phase 1 objects are different from phase 3 objects.
      > >
      > >
      > > What will happen on an end of this action?
      > >
      > > Will this surface return to be initial velocity(V0=0)?
      > >
      > > In problem complexity
      > >
      > > Each element of chain won't stop at the same time. Each element has
      > > different momentum but same mass m. If join each stopped element to the
      > > surface, then the surface mass increase faster than chain return
      > > momentum. This means the surface mass growth will help ground to keep
      > > his own momentum.
      > >
      > > My suggestion:
      > > As it was described at beginning of this text, rotation with translation
      > > movement gives unique situation when object touches the surface without
      > > momentum transfer on a rolling body linear movement direction.If ring
      > > has set of n elements then for the surface point only ring's set of n-1
      > > elements are always moving. However, one element with zero values of
      > > velocity stands down. For this particular case, this set of n-1
      > > elements (broken thin ring or pseudo chain) not equivalent to set of n
      > > elements (initial ring). Base on law of momentum conservation net
      > > momentum for set n-1 elements would have same initial momentum, but
      > > momentum density will change for each element of this set n-1. The
      > > chain will transfer his own momentum to the surface on the end of action
      > > on phase 3. Base on law of momentum conservation:
      > >
      > > P = m*V = const
      > >
      > > m1*V1 = m2*V2
      > >
      > > The surface with new mass will take a velocity V1 from set of elements
      > > n-1. This velocity is different from initial surface velocity V0,
      > > because the surface mass has been changed.
      > >
      > > What is the platform end up velocity equation?
      > >
      > > (Thank you to video_ranger, who provided equation for end up velocity)
      > >
      > > "If the platform is completely free to move (say floating in outer
      > > space) momentum conservation requires that it will end up with a
      > > positive forward velocity:
      > >
      > > V1=(n*m/(M+n*m) )*V
      > >
      > > Kinetic energy is not conserved because as each link slaps down on the
      > > surface some energy is converted to heat.
      > >
      > > For a full ring rolling at constant velocity there's no horizontal force
      > > between the bottom of the ring and the surface but that requires the
      > > ring to be balanced (rotationally symmetric). As links become missing
      > > from the circle that's no longer true so the succeeding links that hit
      > > the surface do have a forward pull on them accelerating the platform
      > > forward."
      > >
      > >
      > >
      > > Follow the law of momentum conservation, an isolated system with
      > > transformed a rolling body may have linear velocity more than zero.
      > > The surface will return to initial velocity V=0 if rest of the chain
      > > could increase momentum on cut action. However, it's nonsense. Base on
      > > law of momentum conservation the chain should keep same initial ring's
      > > momentum.
      > >
      >
    • abelov0927
      Quote from forum. Follow by law of momentum conservation V0x(mx(n-1))=V1x(all mass) If discount one element from the ring then and average speed for (n-1)
      Message 2 of 11 , Jul 14, 2009
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        Quote from forum.
        "Follow by law of momentum conservation
        V0x(mx(n-1))=V1x(all mass)
        If discount one element from the ring then and average speed for (n-1) elements is changing.
        The rest or ring elements have average velocity is V'0=(n/(n-1))xV0!"
         
        This is absolutely correct. The momentum is conserve. The average velocity is changing.
        P=const=mV=mVo=((n-1)m)x(n/(n-1))Vo. Just one little obstacle this model has. One ring's element stays on the ground with velocity zero.
        To solve this problem let reverse frame of reference for this model. The platform moves horizontally with mass M and velocity V. The ring with mass n*m stays and does rotation movement.
        For vertical plane (a wall) the platform has a translation momentum P=MV. However, the translation momentum for horizontal plane (the platform surface) counts ring's element, which is joining to the surface with velocity V. In this case the translation momentum is P=(M+m)V. The average velocity is equal to V.
        How the momentum can be different for these planes? Answer is horizontal and vertical planes have a different frame of reference. To reach same momentum P=MV. The frame reference center should move with velocity V/n.
        WAIT A SECOND!
        Are these vertical and horizontal frames of references move relatively to each other?
        THIS IS THE ONE.
        If action starts from zero velocity on one frame of reference and this action finish with zero velocity for another frame of reference then relativity to the first frame of reference the system will continue to move on the end of action!
         
        The momentum is conserve. The frame of reference is changing. This frame of reference exchange gives the system ability to move.


        --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
        >
        > For this model, à repulsion and à collision comes on different planes.
        >
        > The vertical plane doesn't have any contacts with the rolling body.
        > Otherwise the horizontal plane always has a physical contact with rolling body.
        >
        > From vertical plane point of view, all elements of rolling body have a movement.
        > If calculate a rolling body translation momentum with respect to vertical plane then all elements of rolling body must be included.
        >
        > Otherwise if look on horizontal plane, one element of rolling body stays on the surface all the time.If calculate a rolling body translation momentum with respect to horizontal plane then one element of rolling body must be included to the mass of horizontal plane.
        >
        >
        > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#
        >
        > --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" abelov0927@ wrote:
        > >
        > > Hi.
        > > I think this is a clue for understanding.
        > >
        > > The element with zero values of linear velocity on the surface is too small. Base on math the geometrical size of this element strives to zero limit. However, this is the physical element, and it has its own geometrical size.
        > >
        > > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#
        > >
        > > --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@> wrote:
        > > >
        > > >
        > > > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
        > > > qm1l0s4ys/9
        > > > <http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1x\
        > > > mqm1l0s4ys/9> #
        > > >
        > > > No objects with mass in translation movement can transfer momentum
        > > > without collision action. However, rotation with translation movement
        > > > gives unique situation when object touches the surface without momentum
        > > > transfer on a rolling body linear movement direction.
        > > > The idea is very simple.
        > > >
        > > >
        > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/cutring2.jpg>
        > > >
        > > >
        > > >
        > > > If spit a rolling ring to small parts set of n elements (1,2,3,...,n)
        > > > with mass m then each of them conducts linear and circular movement on
        > > > surface. Good example is ­Caterpillar tracks (Continuous track).
        > > > <http://en.wikipedia.org/wiki/Continuous_track>
        > > >
        > > > Each piece has constant angular velocity. This is the perfect picture
        > > > from UPENN site
        > > > <http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsu\
        > > > bsection4_1_4_3.html> .
        > > >
        > > >
        > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/translationwithrotatio\
        > > > n.gif>
        > > >
        > > > Each piece of ring has variable linear velocity at a surface point. Once
        > > > per circle each element of ring stop on surface. This (fig. 3) shows the
        > > > red bit trajectory. At the surface point, this element has linear
        > > > velocity value equal to zero.
        > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/redbit.jpg>
        > > > This (fig.2 ) shows all action between ring and surface.
        > > >
        > > > This action has 3 phases.
        > > >
        > > >
        > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase1.jpg>
        > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase2.jpg>
        > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase3%20(1).jpg>
        > > > 1. The ring and the surface are getting momentums.
        > > > The ring has mass n*m
        > > > The surface has mass M
        > > >
        > > >
        > > >
        > > >
        > > >
        > > > 2. Cut and hold the red bit on the surface (The red bit has velocity
        > > > zero V=0 on surface. The red bit doesn't transfer momentum to the ground
        > > > on rings linear movement direction)
        > > >
        > > >
        > > >
        > > >
        > > > 3. The pseudo chain transfer momentum to the ground. The chain has mass
        > > > (n-1)*m
        > > > The surface has mass M+m
        > > >
        > > > As it has shown on (fig. 2), the ring with mass n*m and the surface with
        > > > mass M takes momentums on phase 1. On phase 2 the ring must be broken in
        > > > one location and one element with zero values of linear velocity holding
        > > > by the surface. This is not meaning to stop the whole ring at one time.
        > > > This mean stops the red piece and cut the ring at the same time. The
        > > > ring transforms to chain and lose the mass to new value (n-1)*m. On
        > > > phase 3 the surface with mass M+m holds just one element of ring and
        > > > other elements of pseudo chain with mass (n-1)*m is continuing movement
        > > > by his own trajectories until they rich the ground. The very important
        > > > thing is: phase 1 objects are different from phase 3 objects.
        > > >
        > > >
        > > > What will happen on an end of this action?
        > > >
        > > > Will this surface return to be initial velocity(V0=0)?
        > > >
        > > > In problem complexity
        > > >
        > > > Each element of chain won't stop at the same time. Each element has
        > > > different momentum but same mass m. If join each stopped element to the
        > > > surface, then the surface mass increase faster than chain return
        > > > momentum. This means the surface mass growth will help ground to keep
        > > > his own momentum.
        > > >
        > > > My suggestion:
        > > > As it was described at beginning of this text, rotation with translation
        > > > movement gives unique situation when object touches the surface without
        > > > momentum transfer on a rolling body linear movement direction.If ring
        > > > has set of n elements then for the surface point only ring's set of n-1
        > > > elements are always moving. However, one element with zero values of
        > > > velocity stands down. For this particular case, this set of n-1
        > > > elements (broken thin ring or pseudo chain) not equivalent to set of n
        > > > elements (initial ring). Base on law of momentum conservation net
        > > > momentum for set n-1 elements would have same initial momentum, but
        > > > momentum density will change for each element of this set n-1. The
        > > > chain will transfer his own momentum to the surface on the end of action
        > > > on phase 3. Base on law of momentum conservation:
        > > >
        > > > P = m*V = const
        > > >
        > > > m1*V1 = m2*V2
        > > >
        > > > The surface with new mass will take a velocity V1 from set of elements
        > > > n-1. This velocity is different from initial surface velocity V0,
        > > > because the surface mass has been changed.
        > > >
        > > > What is the platform end up velocity equation?
        > > >
        > > > (Thank you to video_ranger, who provided equation for end up velocity)
        > > >
        > > > "If the platform is completely free to move (say floating in outer
        > > > space) momentum conservation requires that it will end up with a
        > > > positive forward velocity:
        > > >
        > > > V1=(n*m/(M+n*m) )*V
        > > >
        > > > Kinetic energy is not conserved because as each link slaps down on the
        > > > surface some energy is converted to heat.
        > > >
        > > > For a full ring rolling at constant velocity there's no horizontal force
        > > > between the bottom of the ring and the surface but that requires the
        > > > ring to be balanced (rotationally symmetric). As links become missing
        > > > from the circle that's no longer true so the succeeding links that hit
        > > > the surface do have a forward pull on them accelerating the platform
        > > > forward."
        > > >
        > > >
        > > >
        > > > Follow the law of momentum conservation, an isolated system with
        > > > transformed a rolling body may have linear velocity more than zero.
        > > > The surface will return to initial velocity V=0 if rest of the chain
        > > > could increase momentum on cut action. However, it's nonsense. Base on
        > > > law of momentum conservation the chain should keep same initial ring's
        > > > momentum.
        > > >
        > >
        >

      • abelov0927
        A few philosophy thinks. Collisions may be classified in two groups. Explicit and implicit. 1. Explicit collisions – happens between objects which conducts
        Message 3 of 11 , Jul 16, 2009
        • 0 Attachment

          A few philosophy thinks.

          Collisions may be classified in two groups. Explicit and implicit.
           
          1. Explicit collisions – happens between objects which conducts a simple movements relativity to each other.
          For example. A collision between rolling body and wall on the surface. Law of momentum conservation is working.
           
          2. Implicit collisions – happens between objects which conducts a complicate movements relativity to each other.
          For example. A collision between a rolling body and a surface (It's kind of weird thing)
          This it may happen in 2 cases.
              a. Between these objects distortions. Rolling friction.
              b. These objects may have shared points to each other during movement action. Coupling through construction elements.
          For ideal model the rolling body is conducting rotation with translation movement without any collisions on straight line surface.  The rolling body with complicated movement has a parallel tangential line to the surface.
          If use same frame of reference for implicit and explicit collisions the momentum will have a different value.
           
          To avoid this law of momentum conservation problem, the model can implement two possibilities.
          a. The model should transform own frame of reference.
          If a platform (the surface) is a center of frame of reference then the calculated average velocity of rolling ring is higher than rolling ring's center mass velocity.
          If reverse this frame of reference and put rolling ring into center then the platform velocity should be reduced for calculations. From the other word this frame of reference should base on law of momentum conservation. Base on this frame of reference momentum measurement other parameters (like velocity) should be readjusted.
          b. The model should include a dark matter.
          This dark matter with own mass gives ability to law of momentum conservation works without frame of reference replacement.
          For rolling ring model can happening two things. The dark matter may reduce the platform mass or be between objects to compensate extra momentum.
           
          =====================================================================
          Are cosmology theories having a same problem?
          Is new frame of reference with law of momentum conservation core can eliminate this problem? 

          --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
          >
          > Quote from forum.
          > "Follow by law of momentum conservation
          > V0x(mx(n-1))=V1x(all mass)
          > If discount one element from the ring then and average speed for (n-1)
          > elements is changing.
          > The rest or ring elements have average velocity is
          > V'0=(n/(n-1))xV0!" This is absolutely correct. The momentum is
          > conserve. The average velocity is changing.
          > P=const=mV=mVo=((n-1)m)x(n/(n-1))Vo. Just one little obstacle this model
          > has. One ring's element stays on the ground with velocity zero. To
          > solve this problem let reverse frame of reference for this model. The
          > platform moves horizontally with mass M and velocity V. The ring with
          > mass n*m stays and does rotation movement.
          > For vertical plane (a wall) the platform has a translation momentum
          > P=MV. However, the translation momentum for horizontal plane (the
          > platform surface) counts ring's element, which is joining to the
          > surface with velocity V. In this case the translation momentum is
          > P=(M+m)V. The average velocity is equal to V.
          > How the momentum can be different for these planes? Answer is horizontal
          > and vertical planes have a different frame of reference. To reach same
          > momentum P=MV. The frame reference center should move with velocity V/n.
          > WAIT A SECOND!
          > Are these vertical and horizontal frames of references move relatively
          > to each other?
          > THIS IS THE ONE.
          > If action starts from zero velocity on one frame of reference and this
          > action finish with zero velocity for another frame of reference then
          > relativity to the first frame of reference the system will continue to
          > move on the end of action! The momentum is conserve. The frame of
          > reference is changing. This frame of reference exchange gives the system
          > ability to move.
          >
          > --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
          > abelov0927@ wrote:
          > >
          > > For this model, à repulsion and à collision comes on different
          > planes.
          > >
          > > The vertical plane doesn't have any contacts with the rolling body.
          > > Otherwise the horizontal plane always has a physical contact with
          > rolling body.
          > >
          > > From vertical plane point of view, all elements of rolling body have a
          > movement.
          > > If calculate a rolling body translation momentum with respect to
          > vertical plane then all elements of rolling body must be included.
          > >
          > > Otherwise if look on horizontal plane, one element of rolling body
          > stays on the surface all the time.If calculate a rolling body
          > translation momentum with respect to horizontal plane then one element
          > of rolling body must be included to the mass of horizontal plane.
          > >
          > >
          > >
          > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
          > qm1l0s4ys/9#
          > >
          > > --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
          > abelov0927@ wrote:
          > > >
          > > > Hi.
          > > > I think this is a clue for understanding.
          > > >
          > > > The element with zero values of linear velocity on the surface is
          > too small. Base on math the geometrical size of this element strives to
          > zero limit. However, this is the physical element, and it has its own
          > geometrical size.
          > > >
          > > >
          > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
          > qm1l0s4ys/9#
          > > >
          > > > --- In gravitationalpropulsionstevenson@yahoogroups.com,
          > "abelov0927" <abelov0927@> wrote:
          > > > >
          > > > >
          > > > >
          > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
          > \
          > > > > qm1l0s4ys/9
          > > > >
          > <http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1x\
          > \
          > > > > mqm1l0s4ys/9> #
          > > > >
          > > > > No objects with mass in translation movement can transfer momentum
          > > > > without collision action. However, rotation with translation
          > movement
          > > > > gives unique situation when object touches the surface without
          > momentum
          > > > > transfer on a rolling body linear movement direction.
          > > > > The idea is very simple.
          > > > >
          > > > >
          > > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/cutring2.jpg>
          > > > >
          > > > >
          > > > >
          > > > > If spit a rolling ring to small parts set of n elements
          > (1,2,3,...,n)
          > > > > with mass m then each of them conducts linear and circular
          > movement on
          > > > > surface. Good example is ­Caterpillar tracks (Continuous
          > track).
          > > > > <http://en.wikipedia.org/wiki/Continuous_track>
          > > > >
          > > > > Each piece has constant angular velocity. This is the perfect
          > picture
          > > > > from UPENN site
          > > > >
          > <http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsu\
          > \
          > > > > bsection4_1_4_3.html> .
          > > > >
          > > > >
          > > > >
          > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/translationwithrotatio\
          > \
          > > > > n.gif>
          > > > >
          > > > > Each piece of ring has variable linear velocity at a surface
          > point. Once
          > > > > per circle each element of ring stop on surface. This (fig. 3)
          > shows the
          > > > > red bit trajectory. At the surface point, this element has linear
          > > > > velocity value equal to zero.
          > > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/redbit.jpg>
          > > > > This (fig.2 ) shows all action between ring and surface.
          > > > >
          > > > > This action has 3 phases.
          > > > >
          > > > >
          > > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase1.jpg>
          > > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase2.jpg>
          > > > >
          > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase3%20(1).jpg>
          > > > > 1. The ring and the surface are getting momentums.
          > > > > The ring has mass n*m
          > > > > The surface has mass M
          > > > >
          > > > >
          > > > >
          > > > >
          > > > >
          > > > > 2. Cut and hold the red bit on the surface (The red bit has
          > velocity
          > > > > zero V=0 on surface. The red bit doesn't transfer momentum to the
          > ground
          > > > > on rings linear movement direction)
          > > > >
          > > > >
          > > > >
          > > > >
          > > > > 3. The pseudo chain transfer momentum to the ground. The chain has
          > mass
          > > > > (n-1)*m
          > > > > The surface has mass M+m
          > > > >
          > > > > As it has shown on (fig. 2), the ring with mass n*m and the
          > surface with
          > > > > mass M takes momentums on phase 1. On phase 2 the ring must be
          > broken in
          > > > > one location and one element with zero values of linear velocity
          > holding
          > > > > by the surface. This is not meaning to stop the whole ring at one
          > time.
          > > > > This mean stops the red piece and cut the ring at the same time.
          > The
          > > > > ring transforms to chain and lose the mass to new value (n-1)*m.
          > On
          > > > > phase 3 the surface with mass M+m holds just one element of ring
          > and
          > > > > other elements of pseudo chain with mass (n-1)*m is continuing
          > movement
          > > > > by his own trajectories until they rich the ground. The very
          > important
          > > > > thing is: phase 1 objects are different from phase 3 objects.
          > > > >
          > > > >
          > > > > What will happen on an end of this action?
          > > > >
          > > > > Will this surface return to be initial velocity(V0=0)?
          > > > >
          > > > > In problem complexity
          > > > >
          > > > > Each element of chain won't stop at the same time. Each element
          > has
          > > > > different momentum but same mass m. If join each stopped element
          > to the
          > > > > surface, then the surface mass increase faster than chain return
          > > > > momentum. This means the surface mass growth will help ground to
          > keep
          > > > > his own momentum.
          > > > >
          > > > > My suggestion:
          > > > > As it was described at beginning of this text, rotation with
          > translation
          > > > > movement gives unique situation when object touches the surface
          > without
          > > > > momentum transfer on a rolling body linear movement direction.If
          > ring
          > > > > has set of n elements then for the surface point only ring's set
          > of n-1
          > > > > elements are always moving. However, one element with zero values
          > of
          > > > > velocity stands down. For this particular case, this set of n-1
          > > > > elements (broken thin ring or pseudo chain) not equivalent to set
          > of n
          > > > > elements (initial ring). Base on law of momentum conservation net
          > > > > momentum for set n-1 elements would have same initial momentum,
          > but
          > > > > momentum density will change for each element of this set n-1. The
          > > > > chain will transfer his own momentum to the surface on the end of
          > action
          > > > > on phase 3. Base on law of momentum conservation:
          > > > >
          > > > > P = m*V = const
          > > > >
          > > > > m1*V1 = m2*V2
          > > > >
          > > > > The surface with new mass will take a velocity V1 from set of
          > elements
          > > > > n-1. This velocity is different from initial surface velocity V0,
          > > > > because the surface mass has been changed.
          > > > >
          > > > > What is the platform end up velocity equation?
          > > > >
          > > > > (Thank you to video_ranger, who provided equation for end up
          > velocity)
          > > > >
          > > > > "If the platform is completely free to move (say floating in outer
          > > > > space) momentum conservation requires that it will end up with a
          > > > > positive forward velocity:
          > > > >
          > > > > V1=(n*m/(M+n*m) )*V
          > > > >
          > > > > Kinetic energy is not conserved because as each link slaps down on
          > the
          > > > > surface some energy is converted to heat.
          > > > >
          > > > > For a full ring rolling at constant velocity there's no horizontal
          > force
          > > > > between the bottom of the ring and the surface but that requires
          > the
          > > > > ring to be balanced (rotationally symmetric). As links become
          > missing
          > > > > from the circle that's no longer true so the succeeding links that
          > hit
          > > > > the surface do have a forward pull on them accelerating the
          > platform
          > > > > forward."
          > > > >
          > > > >
          > > > >
          > > > > Follow the law of momentum conservation, an isolated system with
          > > > > transformed a rolling body may have linear velocity more than
          > zero.
          > > > > The surface will return to initial velocity V=0 if rest of the
          > chain
          > > > > could increase momentum on cut action. However, it's nonsense.
          > Base on
          > > > > law of momentum conservation the chain should keep same initial
          > ring's
          > > > > momentum.
          > > > >
          > > >
          > >
          >
        • abelov0927
          Let imagine observer and two objects (A and B). One object A with simple movement can move only into one plane. The observer in own frame of reference sees
          Message 4 of 11 , Jul 16, 2009
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            Let imagine observer and two objects (A and B). One object A with simple movement can move only into one plane. The observer in own frame of reference sees only this plane also.
            Another object B with complicated movement can move in two planes and observer can see only one of these planes.
            The object A has position without movement for observer. The object B has a movement. The object B has component velocity for each plane. However, observer can see only one component velocity of objects B . The object A takes momentum from object B after collision and objects A velocity may be higher than inspected.
            a. The observer can assume the object A has a huge mass. Or if observer knows the object B mass then he can assume include dark matter for collision process.
            b. If frame of reference base on law of momentum conservation then using all know parameters easy to calculate another plane velocity component of object B which is invisible for observer. This velocity component for observer has imaginary character. From the other words, the frame of reference use law of momentum conservation to readjust another object parameters.


            A few words about movement without external forces. The object can rotate on own center mass without external forces. However, if alternate simple and complicated movement on object collisions then for observer frame of references it may possible exchange rotation to translation movement.
            Everything Is Relative.
          • abelov0927
            This is a good example is shown transfer between angular and linear momentums. Unfortunately the simulator does not allow giving different angular
            Message 5 of 11 , Jul 19, 2009
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               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.



              --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
              >
              > A few philosophy thinks. Collisions may be classified in two groups.
              > Explicit and implicit. 1. Explicit collisions – happens between
              > objects which conducts a simple movements relativity to each other.
              > For example. A collision between rolling body and wall on the surface.
              > Law of momentum conservation is working. 2. Implicit collisions –
              > happens between objects which conducts a complicate movements relativity
              > to each other.
              > For example. A collision between a rolling body and a surface (It's
              > kind of weird thing)
              > This it may happen in 2 cases.
              > a. Between these objects distortions. Rolling friction.
              > b. These objects may have shared points to each other during
              > movement action. Coupling through construction elements. For ideal model
              > the rolling body is conducting rotation with translation movement
              > without any collisions on straight line surface. The rolling body with
              > complicated movement has a parallel tangential line to the surface.
              > If use same frame of reference for implicit and explicit collisions the
              > momentum will have a different value. To avoid this law of momentum
              > conservation problem, the model can implement two possibilities.
              > a. The model should transform own frame of reference.
              > If a platform (the surface) is a center of frame of reference then the
              > calculated average velocity of rolling ring is higher than rolling
              > ring's center mass velocity.
              > If reverse this frame of reference and put rolling ring into center then
              > the platform velocity should be reduced for calculations. From the other
              > word this frame of reference should base on law of momentum
              > conservation. Base on this frame of reference momentum measurement other
              > parameters (like velocity) should be readjusted. b. The model should
              > include a dark matter. This dark matter with own mass gives ability to
              > law of momentum conservation works without frame of reference
              > replacement.
              > For rolling ring model can happening two things. The dark matter may
              > reduce the platform mass or be between objects to compensate extra
              > momentum.
              > =====================================================================
              > Are cosmology theories having a same problem?
              > Is new frame of reference with law of momentum conservation core can
              > eliminate this problem?
              > --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
              > abelov0927@ wrote:
              > >
              > > Quote from forum.
              > > "Follow by law of momentum conservation
              > > V0x(mx(n-1))=V1x(all mass)
              > > If discount one element from the ring then and average speed for (n-1)
              > > elements is changing.
              > > The rest or ring elements have average velocity is
              > > V'0=(n/(n-1))xV0!" This is absolutely correct. The momentum is
              > > conserve. The average velocity is changing.
              > > P=const=mV=mVo=((n-1)m)x(n/(n-1))Vo. Just one little obstacle this
              > model
              > > has. One ring's element stays on the ground with velocity zero. To
              > > solve this problem let reverse frame of reference for this model. The
              > > platform moves horizontally with mass M and velocity V. The ring with
              > > mass n*m stays and does rotation movement.
              > > For vertical plane (a wall) the platform has a translation momentum
              > > P=MV. However, the translation momentum for horizontal plane (the
              > > platform surface) counts ring's element, which is joining to the
              > > surface with velocity V. In this case the translation momentum is
              > > P=(M+m)V. The average velocity is equal to V.
              > > How the momentum can be different for these planes? Answer is
              > horizontal
              > > and vertical planes have a different frame of reference. To reach same
              > > momentum P=MV. The frame reference center should move with velocity
              > V/n.
              > > WAIT A SECOND!
              > > Are these vertical and horizontal frames of references move relatively
              > > to each other?
              > > THIS IS THE ONE.
              > > If action starts from zero velocity on one frame of reference and this
              > > action finish with zero velocity for another frame of reference then
              > > relativity to the first frame of reference the system will continue to
              > > move on the end of action! The momentum is conserve. The frame of
              > > reference is changing. This frame of reference exchange gives the
              > system
              > > ability to move.
              > >
              > > --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
              > > abelov0927@ wrote:
              > > >
              > > > For this model, à repulsion and à collision comes on different
              > > planes.
              > > >
              > > > The vertical plane doesn't have any contacts with the rolling body.
              > > > Otherwise the horizontal plane always has a physical contact with
              > > rolling body.
              > > >
              > > > From vertical plane point of view, all elements of rolling body have
              > a
              > > movement.
              > > > If calculate a rolling body translation momentum with respect to
              > > vertical plane then all elements of rolling body must be included.
              > > >
              > > > Otherwise if look on horizontal plane, one element of rolling body
              > > stays on the surface all the time.If calculate a rolling body
              > > translation momentum with respect to horizontal plane then one element
              > > of rolling body must be included to the mass of horizontal plane.
              > > >
              > > >
              > > >
              > >
              > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
              > \
              > > qm1l0s4ys/9#
              > > >
              > > > --- In gravitationalpropulsionstevenson@yahoogroups.com,
              > "abelov0927"
              > > abelov0927@ wrote:
              > > > >
              > > > > Hi.
              > > > > I think this is a clue for understanding.
              > > > >
              > > > > The element with zero values of linear velocity on the surface is
              > > too small. Base on math the geometrical size of this element strives
              > to
              > > zero limit. However, this is the physical element, and it has its own
              > > geometrical size.
              > > > >
              > > > >
              > >
              > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
              > \
              > > qm1l0s4ys/9#
              > > > >
              > > > > --- In gravitationalpropulsionstevenson@yahoogroups.com,
              > > "abelov0927" <abelov0927@> wrote:
              > > > > >
              > > > > >
              > > > > >
              > >
              > http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
              > \
              > > \
              > > > > > qm1l0s4ys/9
              > > > > >
              > >
              > <http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1x\
              > \
              > > \
              > > > > > mqm1l0s4ys/9> #
              > > > > >
              > > > > > No objects with mass in translation movement can transfer
              > momentum
              > > > > > without collision action. However, rotation with translation
              > > movement
              > > > > > gives unique situation when object touches the surface without
              > > momentum
              > > > > > transfer on a rolling body linear movement direction.
              > > > > > The idea is very simple.
              > > > > >
              > > > > >
              > > > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/cutring2.jpg>
              > > > > >
              > > > > >
              > > > > >
              > > > > > If spit a rolling ring to small parts set of n elements
              > > (1,2,3,...,n)
              > > > > > with mass m then each of them conducts linear and circular
              > > movement on
              > > > > > surface. Good example is ­Caterpillar tracks (Continuous
              > > track).
              > > > > > <http://en.wikipedia.org/wiki/Continuous_track>
              > > > > >
              > > > > > Each piece has constant angular velocity. This is the perfect
              > > picture
              > > > > > from UPENN site
              > > > > >
              > >
              > <http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsu\
              > \
              > > \
              > > > > > bsection4_1_4_3.html> .
              > > > > >
              > > > > >
              > > > > >
              > >
              > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/translationwithrotatio\
              > \
              > > \
              > > > > > n.gif>
              > > > > >
              > > > > > Each piece of ring has variable linear velocity at a surface
              > > point. Once
              > > > > > per circle each element of ring stop on surface. This (fig. 3)
              > > shows the
              > > > > > red bit trajectory. At the surface point, this element has
              > linear
              > > > > > velocity value equal to zero.
              > > > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/redbit.jpg>
              > > > > > This (fig.2 ) shows all action between ring and surface.
              > > > > >
              > > > > > This action has 3 phases.
              > > > > >
              > > > > >
              > > > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase1.jpg>
              > > > > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase2.jpg>
              > > > > >
              > > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase3%20(1).jpg>
              > > > > > 1. The ring and the surface are getting momentums.
              > > > > > The ring has mass n*m
              > > > > > The surface has mass M
              > > > > >
              > > > > >
              > > > > >
              > > > > >
              > > > > >
              > > > > > 2. Cut and hold the red bit on the surface (The red bit has
              > > velocity
              > > > > > zero V=0 on surface. The red bit doesn't transfer momentum to
              > the
              > > ground
              > > > > > on rings linear movement direction)
              > > > > >
              > > > > >
              > > > > >
              > > > > >
              > > > > > 3. The pseudo chain transfer momentum to the ground. The chain
              > has
              > > mass
              > > > > > (n-1)*m
              > > > > > The surface has mass M+m
              > > > > >
              > > > > > As it has shown on (fig. 2), the ring with mass n*m and the
              > > surface with
              > > > > > mass M takes momentums on phase 1. On phase 2 the ring must be
              > > broken in
              > > > > > one location and one element with zero values of linear velocity
              > > holding
              > > > > > by the surface. This is not meaning to stop the whole ring at
              > one
              > > time.
              > > > > > This mean stops the red piece and cut the ring at the same time.
              > > The
              > > > > > ring transforms to chain and lose the mass to new value (n-1)*m.
              > > On
              > > > > > phase 3 the surface with mass M+m holds just one element of ring
              > > and
              > > > > > other elements of pseudo chain with mass (n-1)*m is continuing
              > > movement
              > > > > > by his own trajectories until they rich the ground. The very
              > > important
              > > > > > thing is: phase 1 objects are different from phase 3 objects.
              > > > > >
              > > > > >
              > > > > > What will happen on an end of this action?
              > > > > >
              > > > > > Will this surface return to be initial velocity(V0=0)?
              > > > > >
              > > > > > In problem complexity
              > > > > >
              > > > > > Each element of chain won't stop at the same time. Each element
              > > has
              > > > > > different momentum but same mass m. If join each stopped element
              > > to the
              > > > > > surface, then the surface mass increase faster than chain return
              > > > > > momentum. This means the surface mass growth will help ground to
              > > keep
              > > > > > his own momentum.
              > > > > >
              > > > > > My suggestion:
              > > > > > As it was described at beginning of this text, rotation with
              > > translation
              > > > > > movement gives unique situation when object touches the surface
              > > without
              > > > > > momentum transfer on a rolling body linear movement direction.If
              > > ring
              > > > > > has set of n elements then for the surface point only ring's set
              > > of n-1
              > > > > > elements are always moving. However, one element with zero
              > values
              > > of
              > > > > > velocity stands down. For this particular case, this set of n-1
              > > > > > elements (broken thin ring or pseudo chain) not equivalent to
              > set
              > > of n
              > > > > > elements (initial ring). Base on law of momentum conservation
              > net
              > > > > > momentum for set n-1 elements would have same initial momentum,
              > > but
              > > > > > momentum density will change for each element of this set n-1.
              > The
              > > > > > chain will transfer his own momentum to the surface on the end
              > of
              > > action
              > > > > > on phase 3. Base on law of momentum conservation:
              > > > > >
              > > > > > P = m*V = const
              > > > > >
              > > > > > m1*V1 = m2*V2
              > > > > >
              > > > > > The surface with new mass will take a velocity V1 from set of
              > > elements
              > > > > > n-1. This velocity is different from initial surface velocity
              > V0,
              > > > > > because the surface mass has been changed.
              > > > > >
              > > > > > What is the platform end up velocity equation?
              > > > > >
              > > > > > (Thank you to video_ranger, who provided equation for end up
              > > velocity)
              > > > > >
              > > > > > "If the platform is completely free to move (say floating in
              > outer
              > > > > > space) momentum conservation requires that it will end up with a
              > > > > > positive forward velocity:
              > > > > >
              > > > > > V1=(n*m/(M+n*m) )*V
              > > > > >
              > > > > > Kinetic energy is not conserved because as each link slaps down
              > on
              > > the
              > > > > > surface some energy is converted to heat.
              > > > > >
              > > > > > For a full ring rolling at constant velocity there's no
              > horizontal
              > > force
              > > > > > between the bottom of the ring and the surface but that requires
              > > the
              > > > > > ring to be balanced (rotationally symmetric). As links become
              > > missing
              > > > > > from the circle that's no longer true so the succeeding links
              > that
              > > hit
              > > > > > the surface do have a forward pull on them accelerating the
              > > platform
              > > > > > forward."
              > > > > >
              > > > > >
              > > > > >
              > > > > > Follow the law of momentum conservation, an isolated system with
              > > > > > transformed a rolling body may have linear velocity more than
              > > zero.
              > > > > > The surface will return to initial velocity V=0 if rest of the
              > > chain
              > > > > > could increase momentum on cut action. However, it's nonsense.
              > > Base on
              > > > > > law of momentum conservation the chain should keep same initial
              > > ring's
              > > > > > momentum.
              > > > > >
              > > > >
              > > >
              > >
              >
            • abelov0927
              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
              Message 6 of 11 , Jul 21, 2009
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                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.
              • 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 7 of 11 , Jul 23, 2009
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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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