A rolling disk on a straight line surface.

 

--- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:
>
> These thoughts can be used for a rolling body.
>
>
>
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/rollingfriction.jpg>
> 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 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:
> 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 gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927"
> abelov0927@ wrote:
> >
> >
> >
> http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xm\
> \
> > qm1l0s4ys/9
> >
> <http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1x\
> \
> > mqm1l0s4ys/9> #
> >
> > No objects with mass in translation movement can transfer momentum
> > without collision action. However, rotation with translation movement
> > gives unique situation when object touches the surface without
> momentum
> > transfer on a rolling body linear movement direction.
> > The idea is very simple.
> >
> >
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/cutring2.jpg>
> >
> >
> >
> > If spit a rolling ring to small parts set of n elements (1,2,3,...,n)
> > with mass m then each of them conducts linear and circular movement on
> > surface. Good example is ­Caterpillar tracks (Continuous track).
> > <http://en.wikipedia.org/wiki/Continuous_track>
> >
> > Each piece has constant angular velocity. This is the perfect picture
> > from UPENN site
> >
> <http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsu\
> \
> > bsection4_1_4_3.html> .
> >
> >
> >
> <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/translationwithrotatio\
> \
> > n.gif>
> >
> > Each piece of ring has variable linear velocity at a surface point.
> Once
> > per circle each element of ring stop on surface. This (fig. 3) shows
> the
> > red bit trajectory. At the surface point, this element has linear
> > velocity value equal to zero.
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/redbit.jpg>
> > This (fig.2 ) shows all action between ring and surface.
> >
> > This action has 3 phases.
> >
> >
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase1.jpg>
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase2.jpg>
> > <http://knol.google.com/k/-/-/1xmqm1l0s4ys/h6o9ht/phase3%20(1).jpg>
> > 1. The ring and the surface are getting momentums.
> > The ring has mass n*m
> > The surface has mass M
> >
> >
> >
> >
> >
> > 2. Cut and hold the red bit on the surface (The red bit has velocity
> > zero V=0 on surface. The red bit doesn't transfer momentum to the
> ground
> > on rings linear movement direction)
> >
> >
> >
> >
> > 3. The pseudo chain transfer momentum to the ground. The chain has
> mass
> > (n-1)*m
> > The surface has mass M+m
> >
> > As it has shown on (fig. 2), the ring with mass n*m and the surface
> with
> > mass M takes momentums on phase 1. On phase 2 the ring must be broken
> in
> > one location and one element with zero values of linear velocity
> holding
> > by the surface. This is not meaning to stop the whole ring at one
> time.
> > This mean stops the red piece and cut the ring at the same time. The
> > ring transforms to chain and lose the mass to new value (n-1)*m. On
> > phase 3 the surface with mass M+m holds just one element of ring and
> > other elements of pseudo chain with mass (n-1)*m is continuing
> movement
> > by his own trajectories until they rich the ground. The very important
> > thing is: phase 1 objects are different from phase 3 objects.
> >
> >
> > What will happen on an end of this action?
> >
> > Will this surface return to be initial velocity(V0=0)?
> >
> > In problem complexity
> >
> > Each element of chain won't stop at the same time. Each element has
> > different momentum but same mass m. If join each stopped element to
> the
> > surface, then the surface mass increase faster than chain return
> > momentum. This means the surface mass growth will help ground to keep
> > his own momentum.
> >
> > My suggestion:
> > As it was described at beginning of this text, rotation with
> translation
> > movement gives unique situation when object touches the surface
> without
> > momentum transfer on a rolling body linear movement direction.If ring
> > has set of n elements then for the surface point only ring's set of
> n-1
> > elements are always moving. However, one element with zero values of
> > velocity stands down. For this particular case, this set of n-1
> > elements (broken thin ring or pseudo chain) not equivalent to set of n
> > elements (initial ring). Base on law of momentum conservation net
> > momentum for set n-1 elements would have same initial momentum, but
> > momentum density will change for each element of this set n-1. The
> > chain will transfer his own momentum to the surface on the end of
> action
> > on phase 3. Base on law of momentum conservation:
> >
> > P = m*V = const
> >
> > m1*V1 = m2*V2
> >
> > The surface with new mass will take a velocity V1 from set of elements
> > n-1. This velocity is different from initial surface velocity V0,
> > because the surface mass has been changed.
> >
> > What is the platform end up velocity equation?
> >
> > (Thank you to video_ranger, who provided equation for end up velocity)
> >
> > "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:
> >
> > V1=(n*m/(M+n*m) )*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."
> >
> >
> >
> > Follow the law of momentum conservation, an isolated system with
> > transformed a rolling body may have linear velocity more than zero.
> > The surface will return to initial velocity V=0 if rest of the chain
> > could increase momentum on cut action. However, it's nonsense. Base on
> > law of momentum conservation the chain should keep same initial ring's
> > momentum.
> >
>