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## Re: How to build math model for that?

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• A rolling disk on a straight line surface. The diagram has shown a rolling disk on
Message 1 of 13 , Jul 9, 2009
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).
> >
>

[Non-text portions of this message have been removed]
• For this model, à repulsion and à collision comes on different planes. The vertical plane doesn t have any contacts with the rolling body. Otherwise the
Message 2 of 13 , Jul 11, 2009
For this model, à repulsion and à collision comes on different planes.

The vertical plane doesn't have any contacts with the rolling body.
Otherwise the horizontal plane always has a physical contact with rolling body.

From vertical plane point of view, all elements of rolling body have a movement.
If calculate a rolling body translation momentum with respect to vertical plane then all elements of rolling body must be included.

Otherwise if look on horizontal plane, one element of rolling body stays on the surface all the time.If calculate a rolling body translation momentum with respect to horizontal plane then one element of rolling body must be included to the mass of horizontal plane.

http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#

--- In mathforfun@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
>
> 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).
> > >
> >
>
>
>
> [Non-text portions of this message have been removed]
>
• Quote from forum. Follow by law of momentum conservation V0x(mx(n-1))=V1x(all mass) If discount one element from the ring then and average speed for (n-1)
Message 3 of 13 , Jul 14, 2009
Quote from forum.
"Follow by law of momentum conservation
V0x(mx(n-1))=V1x(all mass)
If discount one element from the ring then and average speed for (n-1)
elements is changing.
The rest or ring elements have average velocity is
V'0=(n/(n-1))xV0!" This is absolutely correct. The momentum is
conserve. The average velocity is changing.
P=const=mV=mVo=((n-1)m)x(n/(n-1))Vo. Just one little obstacle this model
has. One ring's element stays on the ground with velocity zero. To
solve this problem let reverse frame of reference for this model. The
platform moves horizontally with mass M and velocity V. The ring with
mass n*m stays and does rotation movement.
For vertical plane (a wall) the platform has a translation momentum
P=MV. However, the translation momentum for horizontal plane (the
platform surface) counts ring's element, which is joining to the
surface with velocity V. In this case the translation momentum is
P=(M+m)V. The average velocity is equal to V.
How the momentum can be different for these planes? Answer is horizontal
and vertical planes have a different frame of reference. To reach same
momentum P=MV. The frame reference center should move with velocity V/n.
WAIT A SECOND!
Are these vertical and horizontal frames of references move relatively
to each other?
THIS IS THE ONE.
If action starts from zero velocity on one frame of reference and this
action finish with zero velocity for another frame of reference then
relativity to the first frame of reference the system will continue to
move on the end of action! The momentum is conserve. The frame of
reference is changing. This frame of reference exchange gives the system
ability to move.

--- In mathforfun@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
>
> For this model, à repulsion and à collision comes on different
planes.
>
> The vertical plane doesn't have any contacts with the rolling body.
> Otherwise the horizontal plane always has a physical contact with
rolling body.
>
> From vertical plane point of view, all elements of rolling body have a
movement.
> If calculate a rolling body translation momentum with respect to
vertical plane then all elements of rolling body must be included.
>
> Otherwise if look on horizontal plane, one element of rolling body
stays on the surface all the time.If calculate a rolling body
translation momentum with respect to horizontal plane then one element
of rolling body must be included to the mass of horizontal plane.
>
>
>
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
qm1l0s4ys/9#
>
> --- In mathforfun@yahoogroups.com, "abelov0927" abelov0927@ wrote:
> >
> > 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).
> > > >
> > >
> >
> >
> >
> > [Non-text portions of this message have been removed]
> >
>

[Non-text portions of this message have been removed]
• A few philosophy thinks. Collisions may be classified in two groups. Explicit and implicit. 1. Explicit collisions – happens between objects which conducts
Message 4 of 13 , Jul 16, 2009
A few philosophy thinks. Collisions may be classified in two groups.
Explicit and implicit. 1. Explicit collisions  happens between
objects which conducts a simple movements relativity to each other.
For example. A collision between rolling body and wall on the surface.
Law of momentum conservation is working. 2. Implicit collisions 
happens between objects which conducts a complicate movements relativity
to each other.
For example. A collision between a rolling body and a surface (It's
kind of weird thing)
This it may happen in 2 cases.
a. Between these objects distortions. Rolling friction.
b. These objects may have shared points to each other during
movement action. Coupling through construction elements. For ideal model
the rolling body is conducting rotation with translation movement
without any collisions on straight line surface. The rolling body with
complicated movement has a parallel tangential line to the surface.
If use same frame of reference for implicit and explicit collisions the
momentum will have a different value. To avoid this law of momentum
conservation problem, the model can implement two possibilities.
a. The model should transform own frame of reference.
If a platform (the surface) is a center of frame of reference then the
calculated average velocity of rolling ring is higher than rolling
ring's center mass velocity.
If reverse this frame of reference and put rolling ring into center then
the platform velocity should be reduced for calculations. From the other
word this frame of reference should base on law of momentum
conservation. Base on this frame of reference momentum measurement other
parameters (like velocity) should be readjusted. b. The model should
include a dark matter. This dark matter with own mass gives ability to
law of momentum conservation works without frame of reference
replacement.
For rolling ring model can happening two things. The dark matter may
reduce the platform mass or be between objects to compensate extra
momentum.
=====================================================================
Are cosmology theories having a same problem?
Is new frame of reference with law of momentum conservation core can
eliminate this problem?
--- In mathforfun@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
>
> Quote from forum.
> "Follow by law of momentum conservation
> V0x(mx(n-1))=V1x(all mass)
> If discount one element from the ring then and average speed for (n-1)
> elements is changing.
> The rest or ring elements have average velocity is
> V'0=(n/(n-1))xV0!" This is absolutely correct. The momentum is
> conserve. The average velocity is changing.
> P=const=mV=mVo=((n-1)m)x(n/(n-1))Vo. Just one little obstacle this
model
> has. One ring's element stays on the ground with velocity zero. To
> solve this problem let reverse frame of reference for this model. The
> platform moves horizontally with mass M and velocity V. The ring with
> mass n*m stays and does rotation movement.
> For vertical plane (a wall) the platform has a translation momentum
> P=MV. However, the translation momentum for horizontal plane (the
> platform surface) counts ring's element, which is joining to the
> surface with velocity V. In this case the translation momentum is
> P=(M+m)V. The average velocity is equal to V.
> How the momentum can be different for these planes? Answer is
horizontal
> and vertical planes have a different frame of reference. To reach same
> momentum P=MV. The frame reference center should move with velocity
V/n.
> WAIT A SECOND!
> Are these vertical and horizontal frames of references move relatively
> to each other?
> THIS IS THE ONE.
> If action starts from zero velocity on one frame of reference and this
> action finish with zero velocity for another frame of reference then
> relativity to the first frame of reference the system will continue to
> move on the end of action! The momentum is conserve. The frame of
> reference is changing. This frame of reference exchange gives the
system
> ability to move.
>
> --- In mathforfun@yahoogroups.com, "abelov0927" abelov0927@ wrote:
> >
> > For this model, à repulsion and à collision comes on different
> planes.
> >
> > The vertical plane doesn't have any contacts with the rolling body.
> > Otherwise the horizontal plane always has a physical contact with
> rolling body.
> >
> > From vertical plane point of view, all elements of rolling body have
a
> movement.
> > If calculate a rolling body translation momentum with respect to
> vertical plane then all elements of rolling body must be included.
> >
> > Otherwise if look on horizontal plane, one element of rolling body
> stays on the surface all the time.If calculate a rolling body
> translation momentum with respect to horizontal plane then one element
> of rolling body must be included to the mass of horizontal plane.
> >
> >
> >
>
http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
\
> qm1l0s4ys/9#
> >
> > --- In mathforfun@yahoogroups.com, "abelov0927" abelov0927@ wrote:
> > >
> > > 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).
> > > > >
> > > >
> > >
> > >
> > >
> > > [Non-text portions of this message have been removed]
> > >
> >
>
>
>
>
> [Non-text portions of this message have been removed]
>

[Non-text portions of this message have been removed]
• Let imagine observer and two objects (A and B). One object A with simple movement can move only into one plane. The observer in own frame of reference sees
Message 5 of 13 , Jul 16, 2009
Let imagine observer and two objects (A and B). One object A with simple movement can move only into one plane. The observer in own frame of reference sees only this plane also.
Another object B with complicated movement can move in two planes and observer can see only one of these planes.
The object A has position without movement for observer. The object B has a movement. The object B has component velocity for each plane. However, observer can see only one component velocity of objects B . The object A takes momentum from object B after collision and objects A velocity may be higher than inspected.
a. The observer can assume the object A has a huge mass. Or if observer knows the object B mass then he can assume include dark matter for collision process.
b. If frame of reference base on law of momentum conservation then using all know parameters easy to calculate another plane velocity component of object B which is invisible for observer. This velocity component for observer has imaginary character. From the other words, the frame of reference use law of momentum conservation to readjust another object parameters.

A few words about movement without external forces. The object can rotate on own center mass without external forces. However, if alternate simple and complicated movement on object collisions then for observer frame of references it may possible exchange rotation to translation movement.
Everything Is Relative.

--- In mathforfun@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
>
> A few philosophy thinks. Collisions may be classified in two groups.
> Explicit and implicit. 1. Explicit collisions  happens between
> objects which conducts a simple movements relativity to each other.
> For example. A collision between rolling body and wall on the surface.
> Law of momentum conservation is working. 2. Implicit collisions 
> happens between objects which conducts a complicate movements relativity
> to each other.
> For example. A collision between a rolling body and a surface (It's
> kind of weird thing)
> This it may happen in 2 cases.
> a. Between these objects distortions. Rolling friction.
> b. These objects may have shared points to each other during
> movement action. Coupling through construction elements. For ideal model
> the rolling body is conducting rotation with translation movement
> without any collisions on straight line surface. The rolling body with
> complicated movement has a parallel tangential line to the surface.
> If use same frame of reference for implicit and explicit collisions the
> momentum will have a different value. To avoid this law of momentum
> conservation problem, the model can implement two possibilities.
> a. The model should transform own frame of reference.
> If a platform (the surface) is a center of frame of reference then the
> calculated average velocity of rolling ring is higher than rolling
> ring's center mass velocity.
> If reverse this frame of reference and put rolling ring into center then
> the platform velocity should be reduced for calculations. From the other
> word this frame of reference should base on law of momentum
> conservation. Base on this frame of reference momentum measurement other
> parameters (like velocity) should be readjusted. b. The model should
> include a dark matter. This dark matter with own mass gives ability to
> law of momentum conservation works without frame of reference
> replacement.
> For rolling ring model can happening two things. The dark matter may
> reduce the platform mass or be between objects to compensate extra
> momentum.
> =====================================================================
> Are cosmology theories having a same problem?
> Is new frame of reference with law of momentum conservation core can
> eliminate this problem?
> --- In mathforfun@yahoogroups.com, "abelov0927" <abelov0927@> wrote:
> >
> > Quote from forum.
> > "Follow by law of momentum conservation
> > V0x(mx(n-1))=V1x(all mass)
> > If discount one element from the ring then and average speed for (n-1)
> > elements is changing.
> > The rest or ring elements have average velocity is
> > V'0=(n/(n-1))xV0!" This is absolutely correct. The momentum is
> > conserve. The average velocity is changing.
> > P=const=mV=mVo=((n-1)m)x(n/(n-1))Vo. Just one little obstacle this
> model
> > has. One ring's element stays on the ground with velocity zero. To
> > solve this problem let reverse frame of reference for this model. The
> > platform moves horizontally with mass M and velocity V. The ring with
> > mass n*m stays and does rotation movement.
> > For vertical plane (a wall) the platform has a translation momentum
> > P=MV. However, the translation momentum for horizontal plane (the
> > platform surface) counts ring's element, which is joining to the
> > surface with velocity V. In this case the translation momentum is
> > P=(M+m)V. The average velocity is equal to V.
> > How the momentum can be different for these planes? Answer is
> horizontal
> > and vertical planes have a different frame of reference. To reach same
> > momentum P=MV. The frame reference center should move with velocity
> V/n.
> > WAIT A SECOND!
> > Are these vertical and horizontal frames of references move relatively
> > to each other?
> > THIS IS THE ONE.
> > If action starts from zero velocity on one frame of reference and this
> > action finish with zero velocity for another frame of reference then
> > relativity to the first frame of reference the system will continue to
> > move on the end of action! The momentum is conserve. The frame of
> > reference is changing. This frame of reference exchange gives the
> system
> > ability to move.
> >
> > --- In mathforfun@yahoogroups.com, "abelov0927" abelov0927@ wrote:
> > >
> > > For this model, à repulsion and à collision comes on different
> > planes.
> > >
> > > The vertical plane doesn't have any contacts with the rolling body.
> > > Otherwise the horizontal plane always has a physical contact with
> > rolling body.
> > >
> > > From vertical plane point of view, all elements of rolling body have
> a
> > movement.
> > > If calculate a rolling body translation momentum with respect to
> > vertical plane then all elements of rolling body must be included.
> > >
> > > Otherwise if look on horizontal plane, one element of rolling body
> > stays on the surface all the time.If calculate a rolling body
> > translation momentum with respect to horizontal plane then one element
> > of rolling body must be included to the mass of horizontal plane.
> > >
> > >
> > >
> >
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
> \
> > qm1l0s4ys/9#
> > >
> > > --- In mathforfun@yahoogroups.com, "abelov0927" abelov0927@ wrote:
> > > >
> > > > 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).
> > > > > >
> > > > >
> > > >
> > > >
> > > >
> > > > [Non-text portions of this message have been removed]
> > > >
> > >
> >
> >
> >
> >
> > [Non-text portions of this message have been removed]
> >
>
>
>
> [Non-text portions of this message have been removed]
>
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