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http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/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.
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).
Each piece has constant angular velocity. This is the perfect picture from UPENN site.
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.
This (fig.2 ) shows all action between ring and surface.
This action has 3 phases.
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1. The ring and the surface are getting momentums.
The ring has mass n*m
The surface has mass 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:
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.
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