All rolling motions are a resultant of contractive pulling-in motions.

Rybo
On Jul 9, 2009, at 9:22 PM, abelov0927 wrote:

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> A rolling disk on a straight line surface.
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> The diagram has shown a rolling disk on straight line surface. This
> movement has rolling and static frictions. This disk contains n
> sectors with mass m. The (centre mass) CM disk has an initial
> translation velocity V. The red sector has linear velocity zero.
> Base on previous explanations, only n-1 sector on the rolling disk
> has velocity more then zero. These moving sectors transfer linear
> momentumdP to the surface with mass M+m per time frame dt.
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> If use classical model: The disk with linear velocity V transfer
> momentum to the surface. The total velocity on the end of action is:
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> (equation 1)
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> If discount the red sector with zero linear velocity: The total
> velocity on the end of action is:
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> (equation 2)
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> The velocity difference between these two models is:
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> (equation 3)
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> If sectors geometrical size strives to zero, then sectors mass
> strive to zero also. For these velocity difference equation, it
> gives a zero result.
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> (equation 4)
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> However, the physical elements have its own geometrical size and
> end up velocity may be count by (equation 2)
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> Velocity difference between classical and this modern models is:
> (equastion 3)
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