A rolling disk on a straight line surface.



<http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/rollingfriction.jpg>
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:
<http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/d1.jpg>
<http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/d1.jpg> (equation 1)
If discount the red sector with zero linear velocity: The total velocity
on the end of action is:
<http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/d2.jpg>
<http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/d2.jpg> (equation 2)
The velocity difference between these two models is:
<http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/d3.jpg>
<http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/d3.jpg> (equation 3)
If sectors geometrical size strives to zero, then sectors mass strive to
zero also. For these velocity difference equation, it gives a zero
result.
<http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/d4.jpg>
<http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/d4.jpg> (equation 4)
However, the physical elements have its own geometrical size and end up
velocity may be count by (equation 2)
<http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/d2.jpg>
<http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/d2.jpg> Velocity
difference between classical and this modern models is:
(equastion 3)
<http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/d3.jpg>
--- In mathforfun@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
>
> Thank you for this explanation. It's good.
> I'm trying to find out the platform finish line velocity. I'm sure
it'll close to zero, but the platform may not stop. Why? Because even if
follow law of momentum conservation, the bodies on phases 1 and 3 have
different mass.
> Thanks anyway. I'll continue work with this problem and I'll put more
details on knoll site.
>
> --- In mathforfun@yahoogroups.com, "video_ranger" video_ranger@ wrote:
> >
> > --- In mathforfun@yahoogroups.com, "abelov0927" <abelov0927@> wrote:
> > >
> > > By 'model', I just want to describe behavior of this system. All
laws of classical mechanic should be preserved.
> > > Initial platform velocity is zero V0=0. Will platform return to
this initial velocity after all? This is what I'm looking for.
> > > Is it physically possible calculating this on math modeling?
> > >
> > > Thank you
> > >
> > >
> >
> > 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=((M+nm)/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.
> >
> > But offhand I'm not sure how to calculate the detailed dynamics that
describe how the platform goes from 0 velocity to its final velocity
(the system has n degrees of freedom so it's more complicated than a
simple rolling ring with 1 degree of freedom).
> >
>



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