These thoughts can be used for a rolling body.

 

--- 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.
>