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A rolling disk on a straight line surface.
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 momentum dP to the surface with mass M+m per time frame dt.
If use classical model: The disk with linear velocity V transfer momentum to the surface. The total velocity on the end of action is:
If discount the red sector with zero linear velocity: The total velocity on the end of action is:
The velocity difference between these two models is:
If sectors geometrical size strives to zero, then sectors mass strive to zero also. For these velocity difference equation, it gives a zero result.
However, the physical elements have its own geometrical size and end up velocity may be count by (equation 2)
Velocity difference between classical and this modern models is:
(equastion 3)
--- In TheoryOfEverything@yahoogroups.com, rybo6 <rybo6@...> wrote:
>
> All rolling motions, are a resultant of contractive, pulling inward
> forces.
>
> All pushing out motions, are a resultant of contractive, pulling-
> inward forces.
>
> All pulling outward motions, are a resultant of contractive, pulling-
> inward forces.
>
> Rybo
> On Jul 7, 2009, at 6:27 PM, abelov0927 wrote:
>
> >
> >
> >
> > These thoughts can be used for a rolling body.
> >
> >
> >
> >
> >
> > The diagram has shown a rolling disk which contains n sectors with
> > mass m. This (centre mass) CM disk has a 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 momentum dPto the
> > surface with mass M+m per time frame dt.
> > If use classical model: The disk with linear velocity V transfer
> > momentum to the surface. The total velocity on the end of action is:
> >
> > V1=((n*m)/(n*m+M))*V
> >
> > If discount the red sector with zero linear velocity: The total
> > velocity on the end of action is:
> > V1'=(((n-1)*m)/((n-1)*m+M+m))*V.
> >
> > The velocity difference between these two models is:
> > V1-V1'=m/(n*m+M)*V
> >
> > If sectors geometrical size strives to zero, then sectors mass
> > strive to zero also. For velocity difference equation, it gives a
> > zero result.
> > V1-V1'=0.
> >
> > At a math point of view with a very small sector with zero linear
> > velocity the surface takes equation
> > V1=((n*m)/(n*m+M))*V.
> >
> > However, the physical elements have its own geometrical size and
> > end up velocity for this action is:
> >
> > V1'=(((n-1)*m)/((n-1)*m+M+m))*V
> >
> > Velocity difference is:
> > V1-V1'=m/(n*m+M)*V
> >
> >
> > --- In TheoryOfEverything@yahoogroups.com, "abelov0927"
> > abelov0927@ wrote:
> > >
> > > Would law of momentum conservation helps to move isolated system
> > with transformed rolling body?
> > >
> > > Please look on this site
> > >
> > > http://knol.google.com/k/alex-belov/paradox-of-classical-
> > mechanics-2/1xmqm1l0s4ys/9#
> > >
> >
> >
> >
>
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