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Natural Science Forum / Physics / Research / July 2009How to calculate this?
problem link http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9# 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 conduct linear and circular movement on surface. Each piece has constant angular velocity. Each piece of ring has variable linear velocity at surface point. Once per circle each element of ring stop on surface. At surface point this element has linear velocity value equal to zero. The ring must be broken in one location and one element of chain which has a zero value of linear velocity holding by the surface. This is not mean to stop the whole ring at this time. This mean stop the red piece and cut the ring at the same time. The surface holds just one element of ring and other elements of chain is continuing movement by own trajectories. If calculate net linear momentum of these elements then this net should be equal to this ring initial linear momentum. But one of these elements is stop already and net momentum will be for n-1 elements. In this case one element has been join to the surface and mass is M (suface) + m(element). Set of elements has a mass equal to (n-1)*m now. It=92s change initial condition. The surface is still keeping same momentum and increase own mass. But chain (set of elements n-1) should hold same ring initial momentum. Is this net of linear momentums for set of elements n-1 with net mass (n-1)*m is equal to the ring initial linear momentum? Would set of n-1 elements return whole ring momentum back to surface? Will this surface return to initial velocity? > problem linkhttp://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/... > [quoted text clipped - 3 lines] > with mass m then each of them conduct linear and circular movement on > surface. So, our system has mass nm, angular momentum L = \Sum_i r_i \times mv_i, and Linear Momentum P = \Sum_i mv_i > Each piece has constant angular velocity. Each piece of ring has > variable linear velocity at surface point. Once per circle each > element of ring stop on surface. At surface point this element has > linear velocity value equal to zero. Incorrect. A ROLLING ring has angular momenta, not linear. Thus,p = 0. > The ring must be broken in one location and one element of chain which > has a zero value of linear velocity holding by the surface. This is [quoted text clipped - 11 lines] > momentum and increase own mass. But chain (set of elements n-1) should > hold same ring initial momentum. The initial linear momentum is zero, and the ring loses mass m when you cut a piece out, thus costing your ring momentum. > Is this net of linear momentums for set of elements n-1 with net mass > (n-1)*m is equal to the ring initial linear momentum? > > Would set of n-1 elements return whole ring momentum back to surface? > > Will this surface return to initial velocity? If you cut the ring once, nothing changes. If you cut a section of the ring out, then, by conservation of momentum, it loses angular mometum r \times mv, and gains the same in linear momenta of the broken off piece. Thus, the piece goes flying away, and the ring keeps rolling, albeit at a slower roll. Just keep in mind. This model has 3 phases. Bodies at phase1 is different from bodies at phase3. The law of momentum conservation works, but it gives different result for bodies with different mass. This is the comment from other forum. "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 V=(nm/(M+nm) )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." Please read and look on knol site for details. http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9# |
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