# 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 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.

> >

>

**Let's solve simple problem.**Given a long thin cylinder with length L, radius R and mass M.Case 1.

This cylinder takes momentum P into his center mass.Case 2.

This cylinder takes momentum P on a some distance from his center mass.

The cylinder start rotation with translation movement.How many times the translation velocity for case 2 is lower than translation velocity for case 1?======================================================================__If calculate this problem using classical mechanics laws__, the linear momentum should be conserve. However, these cases take different energies. The cylinder with translation and rotation movements will take two parts (translation and rotation) of energies. The cylinder with translation takes only one( translation) part. These translation kinetic energy parts should be equal.

Where this rotation kinetic energy come from? From nowhere? What about angular momentum? How it starts rotating if all momentum transfers to translation part?__If assume for case 2 the__then it will explain everything. The first part translation momentum gives ability the cylinder moves. The second part angular momentum gives ability the cylinder rotates. The sum of these parts kinetic energies is equal to full translation kinetic energy from case1. The case 2 cylinder translation velocity lower than case 1 translation cylinder velocity.**incoming momentum****vector dividing for two parts initially**

The equation translation velocity is:

Et - kinetic energy translation partEr - kinetic energy rotation part__Let's make a simple experiment with pencil.__Case 1: Let's hit the pencil on his middle The pencil fly away on a long distance.

Case 2: Let's hit the pencil on his edge The pencil starts rotation and fly away not so far.__The classical mechanics solution says For both cases the pencil should fly away on a long distance with same high translation velocity. No matter how fast it is rotating.____Is it really true? I don't see this result.__

------ In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:

>

> Another good example:

> Let's repulse a long cylinder inside Isolated System. The repulse should hit the cylinder away from object center mass. This hit energy splits for two object movements. This cylinder starts rotation with translation movement. The Isolated System starts move to opposite cylinder translation movement direction. During this object movement let transform this cylinder to a sphere. This sphere is rotating faster than cylinder because this object moment of inertia less than cylinder moment of inertia. However this rotation does not make any sense for translation momentum transfer. The sphere translation momentum is equal to cylinder translation momentum. However from cylinder repulse action the system takes higher translation momentum than sphere has. The sphere translation momentum won't be enough to stop the Isolated System.

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

>

>

> --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" abelov0927@ wrote:

> This is a good example is shown transfer between angular and linear momentums. Unfortunately the simulator does not allow giving different angular velocities for squares.

> Anyway, on this model easy to understand these squares with different angular velocities will take a different linear velocities on the end of action.

> Base on my theory about frame of reference conversion for complicated movement, the law of momentum conservation works, however for correct calculation need use imaginary part of velocity.

> The momentum is conserve. The frame of reference is changing. This frame of reference exchange gives the system ability to move.

>