ABV

Question to you. Why? May be I wrong with something, but I'm talking about physic. Beside you, who just keep silence and move thread without any reason. I need explanation first. Otherwise, I would mention you virtual name as a person who mislead others without any reason. Merged post follows: Consecutive posts mergedOriginally Posted by ABV But it's nonsense. You mean, law of momentum conservation doesn't work? Or you didn't understand the post? Merged post follows: Consecutive posts mergedOk. This is what other says about this model. "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=(nm/(M+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." P.S. Please check this out for details on knol site again. I did a lot of changes there. http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9# Merged post follows: Consecutive posts mergedJust keep in mind. This model has 3 phases. Bodies at phase1 is different from bodies at phase3. The law of momentum conservation works, but it gives different result for bodies with different mass.

If someone looks closely to this problem, it's easy to see difference between how to accelerate ring and how this ring stop. During acceleration time whole ring moves on surface and rich const rotation and translation velocities. During collision time the pseudo chain return momentum to the surface through one chains element (red bit on picture). Is it makes any sense? Yes. Because this red bit caught with zero value of velocity V=0 and it didn’t transfer any momentum to the ground. Otherwise rest of the chain will return whole initial ring’s momentum. But those elements of chain return momentums to the surface with new mass. The red bit element was already stopped at begin of collision time. It means the surface join the red bit before the rest of chain returned the momentum and ground didn’t change own velocity. If apply momentum to the surface with new mass then the surface will conduct movement after chain stop. Merged post follows: Consecutive posts mergedRotation with translation movement without slipping on the surface - each piece of the ring has velocity zeto V=0 on the ground. 3 actions. 1. Ring and surface are getting momentums. 2. Cut and hold red bit on surface (red bit has velocity zero V=0 on surface. Red bit don't transfer momentum to the ground) 3. The pseudo chain transfer momentum to the ground. These 3 actions happen in different time.

The ground links(red bit) momentum transfers to the links on the top of ring. The momentum is always constant for ring. Even it cut (and no push between links) rest of the chain (n-1), the chain will have initial ring's momentum. This means links on the chain will increase velocity after cut to keeping same net momentum. But even if apply this momentum to the surface with new mass, it on't return surface to initial velocity. Because the surface velocity wasn't change during the surface apply ground link(red one) of chain. --- The surface will return back to initial velocity V=0 if the rest of chain could increase momentum. But it's nonsense. Base on law of momentum concervation chain should keep same initial ring's momentum.

I think it's pretty simple. The rolling ring has n*m mass and momentum P and surface with mass M and momentum -P After stop(catch by surface) red element and cut the ring. The chain has a mass (n-1)*m and surface with mass M+m The surface would stop if it keep same initial mass M, but the surface mass with given velocity has been changed to M(surface)+m(red element) P.S. Thank you for your posts anyway. I think it's a good problem for brainwashing

The ring with mass n*m has a movement with some linear momentum. After conversion without extra linear momentum. The chain with mass (n-1)*m has a movement with some linear momentum. Both of them part othe isolated system. But this is not means if little human pushh tje ring system will stay. System will start move to oppsite direction. If little human catch the ring system will stop. Push and catch this ring to opposite direction will return system to initial point. Here is complete different behavior with ring and chain. The little human push the ring with one mass, but chain is stopping with other mass. Even if momentums are same, the surface mass will change at conversion moment. The red element stop without extra momentum.

The Chain n-1 will have a ring net momentum. But the surface won't return to initial velocity because the surface mass has been changed. The red element already on surface.

Yeah, but little human push rolling body with mass n*m. And surface accept momentum back from chain with mass (n-1)*m. In this case chain return back momentum to surface with different mass M(surface)+m(red element). No body turn red element to linear velocity zero. No extra momentum for this action. It means the surface won't return to initial velocity after all. ---- If red element takes mass zero then the surface return back to initial velocity. But on real world each element has a mass.

1. Just static friction should be there. 2. Translation with rotation motion gives different linear velocity at surface point. Red piece shows V=0. 2V(not shown) on the top of this ring. image from upenn: http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsubsection4_1_4_3.html 3. The piece with zelo linear velocity must be join to the surface as solid material at stop point (just for modeling). The Idea is stop red piece and cut the ring at the same time. Rest of the ring return whole momentum to the surface back. But surface mass(M(surface) + m(red piece) is different from initial mass M(surface) at this time. The surface velocity V1<>0 will be different from initial velocity V0=0 because momentum returns back to different platform (the surface mass has been changed). P=mV=const m1V1=m2V2

Please look on this picture for better understanding.

My suggestion: If ring has set of n elements then for surface point only ring's set of n-1 elements is always move. But one element with zero value of velocity stand down and it will be always part to surface. For this particular case, this set of n-1 elements (broken thin ring or 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 for each element of this set n-1 will change. The surface will take all elements momentums back when they’ll stop. Base on law of momentum conservation, from same momentum the body with higher mass will take lower velocity then body with lower mass. 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. Follow the law of momentum conservation, even all elements of broken ring will stop on the surface, this surface will have linear velocity more than zero.

Correct, but even and odd sets are a big difference. These sets (in generic) can't covert to each other without extra transformation (loses). This means set with one less element won't return whole momentum back. Just keep in mind, mass of chain and surface changed. Transformation of this chain started when ring stopped and cut the thin. This transformation move net momentum to n-1 set. Other words, (I'm not sure about it) if a thin ring do geometry transformation during riding time then this thin ring will stop without any extra forces. Would be some math specialists can help?

Hi. First. I apologize if this thread breaks any rules of this forum. I’m working on rolling body (thin ring) problem and found interesting thing. Main question is: Could be even and odd sets transform to each other? If answer is no then it means broken rolling ring to chain would loose momentum on surface by this insufficient sets transformation. This is the main problem: The idea is very simple. If spit a rolling ring to small parts set of n elements (1,2,3,...,n) with mass m then each of them conduct linear and circular movement on surface. Each piece has constant angular velocity. And each piece has variable linear velocity at surface point. Once per circle each element stop on surface. At surface point this element has linear velocity value equal to zero at this time. If break chain then all of these elements except one is continue to move. If calculate net linear momentum of these element this should be equal to this ring initial linear momentum. But one of these elements is stop already and net momentum will be for n-1 elements. In this case one element attached to the surface and mass is M(suface) + m(element). Set of elements has a mass equal to (n-1)*m now. It’s change initial condition. The surface is still keeping same momentum and increase own mass. Set of elements n-1 should hold same momentum, but lost one piece with mass m. Is this net of linear momentums for set of elements n-1 with net mass (n-1)*m is equal to the ring initial linear momentum? My suggestion: If ring has set of n elements then for surface point only ring's set of n-1 elements is always move. But one element with zero value of velocity stand down and it will be always part to surface. For this particular case for this broken thin ring or chain, the set of n-1 elements not equivalent to set of n elements. It means net momentum of set of n elements won’t be equal to net momentum of set of n-1 elements. The original problem location: http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9# There also review and other solution by RPenner.

In this case one element attached to the surface and mass is M(suface) + m(element). Array of elements has a mass equal to (n-1)*m. It’s change initial condition. The surface is still keeping same momentum and increase own mass. Array of elements should hold same momentum, but lost one piece with mass m. A few elements will return back small value momentums to surface and increase surface mass. Will other elements with own momentums compensate surface momentum with new mass value?

Idea is very simple. If divide a ring to small parts of n elements array with mass m then each of them conduct linear and circular movement. Each piece has constant angular velocity and variable linear velocity. Once per circle each element stop on surface. Value of linear velocity of this element is zero at this time. And if break chain then all of these elements except one is continue to move. If calculate net linear moments of these element this should be equal to this ring initial linear momentum. But one of these elements is stop already and net momentum will be for n-1 elements. Is this net of linear momentums for n-1 elements array is equal to the ring initial linear momentum?

THANK YOU SO MUCH to all. This is very good answer for me. But anyway I have another IDEA. If a rolling body got linear momentum, we have to put same linear momentum to stop it. This is correct. But Let's look on this one. Let's break rolling body. What will happen? http://knol.google.com/k/alex-belov/-/1xmqm1l0s4ys/9# The conditions: A System has static friction. No rolling friction. A gravity force works as well too. Would this system break the law?

Thank you for kinematic equation. But why [math] \dot v = -\,\frac {F_r} {M+2I/r^2} [/math] after [math]\vec F_t = \vec F_r + \vec F_c = -\left(F_r + \frac I {r^2} \dot v\right) \hat x [/math] it would be [math] \dot v = -\,\frac {F_r} {M+I/r^2} [/math]

Without slippery? Diagram is simple. The little human stopping a rolling body with force Ffr. This little human is transferring a mometum from rolling body to the ground. Why I'm asking? Because an extra moment of inertia of rolling body will gain extra momentum through static friction.

The rolling bodies have same mass, initial velocity and different momet of inertia. same force different time a momentum from rolling bodies to the ground.

In case with a little humans, same rolling bodies masses, velocities and radiuses. The transferred momentum through little human will be a different for rolling bodies with different moment of inertia. Correct?

Thank you for explanation. Let's look a bit close to momentum, which human transfer from rolling body to surface. Let's say a little human base on cart and push a rolling body. Base on law of momentum conservation, the human takes same momentum as a rolling body. So, if the human will push another rolling body with highest moment of inertia the previous one, he will take highest momentum to. Correct?

Easy. You would look on my site: http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics/1xmqm1l0s4ys/3# Let say one rolling body concentrate mass into disk center with small radius r and other part of disk to external R is weightless. Another opposite rolling body is a thin ring, which concentrate a mass on external radius only.

This is very important for me. If I take 2 rolling bodies with different moment of inertia but same mass and radius. And try to accelerate them with same force to same velocity, but different acting time and distance. I will transfer a different momentum to the track, because rolling bodies have a different full kinetic energy. Correct? I meant dV=V1-V0 for example. I out case 0-V = -V deceleration. But it was extra condition.

No, no extra energy as source The bottom line is, a little human transfers a different momentum for rolling bodies with different (moment of inertia/full kinetic energy). Even if mass, initial velocity, and veocity frame are equal. Correct?

The tread disappeared at first time and I opened another. Sorry about it. But I would like to see/get some notifications.

Well, I'll remove it. I didn't understand it may confuse you.