ABV

I agree. Real rolling resistance depends on many factors. But I disagree if we talk about ideal models. Anyway rolling resistance is just a force. Even if increase force, acting time for deceleration period going to be short. But deceleration acting time depends from rolling body kinetic energy, which will be spend to this deceleration (i.e work). In this case no matter what value of force will apply to rolling body, net momentum will depend from work which rolling body does at deceleration period.

2 questions. Is rolling resistance depends on rolling body shape?(thin ring and solid disk only) Is rolling resistance depends on acting time?

This is link to wikipedia. http://en.wikipedia.org/wiki/Rolling_resistance

Will it helps? These diagrams will be the same for thin ring and solid disk. But thin ring acting time for deceleration period will be higher than same deceleration acting time for solid disk.

Thank you for answer. But if look on this Isolated System, we can see same straight tracks inside. Newton laws should not be change. And law of conservation mast work for Isolated System also. Paradox. This is I try to explain.

First. You try to reverse calculation. This is incorrect. Second. thin ring has more energy than solid disk at same velocity. And this thin ring energy gains more momentum to the track than solid disk does. The rolling movement is more complicated than linear, but try to understand rolling body convert energy to work by rolling friction ONLY. Base on spent energy value, rolling body gains momentum to track. Energy cannot disappear without any reason, same cannot come from nowhere. Merged post follows: Consecutive posts merged One more question. Will thin ring get longer ride than solid disk with same mass and velocity? Yes or No? (please use same rolling resistance on a straight track)

What do you mean. Could you explain in details please. And what? You mean momentum and energy does not relate to each other? P.S. I didn't get answer about extra energy. Where it goes?

All equations are present inside Isolated System. The momentum is depending from value of energy. All kinetic energy would discharge and convert to work, with 2 values - resistance force and acting time. Momentum is just multiplication of force and acting time values. Correct? Which way it should? Where extra kinetic energy goes?

There is no magic – difference capacity of kinetic energy. Rolling bogy kinetic energy depend o 2 parts - linear and angular. The thin ring has angular part of kinetic energy bigger than the same part of energy for solid disk. Rolling body with highest energy gets longer ride than same body with lower energy. This longer ride with same rolling resistance gains higher momentum to the track.

The law of momentum conservation works for Isolated System Internally. But to make it work, Isolated System must move into one direction. Please pay attention to equation between thin ring and solid disk momentums on curve part of track. And momentum difference on deceleration part of track (has better explanation than acceleration part). If you agree with that, what will happen with this Isolated System on end of rolling body moving circle? Merged post follows: Consecutive posts mergedActually one thing will be different. The thin ring Acceleration/deceleration acting time will be longer than solid disk acting time for same parts of track. It means Full circle time for rolling thin ring and solid disk with same mass and velocities on curve part of track will be different.

I didn’t break any physics laws. I’m using physics laws for explanation. Please do not mix Newton's third law and law of momentum conservation. I’m using both of them there.

Are any external forces present there? This Isolated System has 4 stages. I may be devide image to all of them, But it won't be helpful. (But I'll think about it) All forces have a direction. Did I miss the direction for any foce there? Please read text. "Net momentum which rolling body gains to track from these forces will be a parallel to acceleration/deceleration momentums too" I disagree. No any law of forces conservation is present on Classical Mechanics. If we talk about law of linear momentum conservation, I showed may examples on previous posts and explained how it works. no comments

Some implementation things? Magnet gun, percussion device, turned wheel, and much, much more. All these devices should stand on oval track. There is no error. This is paradox. I you see some errors on force directions or any kind of things, let me know. If it’s real error – I fix it. But again. This is not logical error -This is paradox. We can discuss about each part of this presentation. I agree. Isolated System should stand on one position. At least, modern physics say about it. The law of momentum conservation is working there, inside Isolated System. But I see equation conflict between momentums from curve and straight parts of track. A few questions. Will extra energy allow to rolling body moves longer? Which rolling body, thin ring or solid disk, with same mass and velocity moves longer? (Let’s use same rolling resistant in both cases) Are Newton’s laws will change from track geometry? P.S. The experiments may be not so clear, but shows whole test dynamic in comparing view.

Let's accelerate thin ring inside u-track without external forces. In this case thin ring gains momentum to track. Correct? Thin ring will cause centripetal force on curve part of track. Track gains another momentum with opposite on this period of time. Correct? During deceleration time the thin ring gains momentum to the track again. Direction for acceleration/deceleration momentums on u-track will be identical. Correct? For thin ring Net momentums curve part of track will be less than sum of acceleration/deceleration momentums. For solid disk they will be equal to each other. Why? Because the thin ring has more energy than solid disk has. And during deceleration time (let's talk about that part. I didn't describe how to accelerate thin ring) thin ring gains bigger momentum to track than solid disk can do. Inside Isolated system law of conservation will work. But for external viewers Isolated System will oscillate and move into one direction on each circle. BTW. I updated site. I did a lot of changes. Please take a look.

Yes. I know it should not. But in this particular case, I see where law of conservation works internally and allow Isolated System moves. Merged post follows: Consecutive posts mergedI would agree about, this Isolated System won’t move anywhere, if the thin ring rolling resistance acting time has same value for solid disk too. But the rolling thin ring kinetic energy value much bigger than rolling solid disk kinetic energy. This thin ring “extra” energy part with same value of rolling friction force gains “extra” momentum to track. This is normal behavior of rolling bodies, which easy to see on straight line track (Isolated System also). Why rules should change to ellipse track? Inside ellipse acceleration/deceleration parts of track use straight line also.

It does not break any laws of conservations. It just makes condition, which moves Isolated System without any external forces and law of conservation works internally. Anyway, thank you for you posts.

This is the rolling body. This body has 2 movements. Yes, this 2 momentums are perpendicular to each other and impossible add them together except rolling friction for deceleration (let just talk about this part). On rolling friction this 2 momentums work together. Linear part momentum will gains momentum to track, but extra angular energy gives more rolling capacity. This rolling capacity increase net momentum, which rolling body gains to the track. I absolutely agree. Isolated System should not move without any external forces. But. I proposed whole document about it. In this particular combination, Isolated System has unusual behavior.

Just keep in mind, from same momentum the thin ring won’t reach same velocity as the solid disk. It’s rolling body which has movement on 2 directions, linear path and circular path. If thin ring and solid disk will get same acceleration/decoration momentums, velocity on curve part of track will be a different for them. Otherwise, to keep same velocity on curve part of track, the thin ring and solid disk should have different acceleration/deceleration momentums. In thin ring case, additional energy comes from the angular part of energy. You may create very thin ring with huge radius and energy comparing to solid disk wish small radius and energy. This will work just for rolling bodies. And it doesn’t work for linear moves

This is the next step of implementation part. You would probably saw that on a bottom of document. Even if the flywheel can change velocity without external forces, during acceleration/deceleration time the track causes acceleration/deceleration force. This is result of difference between a new angular velocity of flywheel and old flywheel linear velocity. This will equate through these forces. Assuming the ellipse track with solid disk won’t move anywhere, because all momentums from centripetal, acceleration/deceleration forces give us zero on result. But the thin ring has more kinetic energy comparing to the solid disk. And this extra energy will work between thin ring and track during acceleration/deceleration time only. The thin ring centripetal forces will be the same as for disks forces. Both Isolated Systems should have same velocities everywhere. The difference only is acting time on acceleration/deceleration part of track. This difference will bring different acceleration/deceleration momentums, which will move Isolated System.

Yes, I agree. It changes the magnitude and direction of the centripetal force. I’ll create a new picture where 2 bodies running synchronously in opposite direction. This will compensate all other centripetal force directions. Yes. And all laws work. But if you closely look into the Isolated System, you’ll see acceleration and deceleration part have same direction (not opposite). So, if thing ring gain more energy from the track on acceleration part. This kinetic energy will push back to track from thin ring on deceleration part. Both these parts have same direction. Just centripetal force will opposite to them. And this centripetal force won’t change, even if rolling body has a different moment of inertia. Other words, following by law of momentum conservation, Isolated System will move into one direction

4 forces are present on the oval: 2 centripetal acceleration and deceleration. For solid disk the oval won’t move to anywhere and net momentum will be zero. P(s_disk)(centripetal_1)= P(s_disk) (acceleration) + P(s_disk) (deceleration) + P(s_disk) (centripetal_2) But the thin ring has more energy than solid disk. Even if acceleration/deceleration forces are equal to same solid disk forces, to reach same velocity, acceleration/deceleration acting time for thin ring will higher, than the same solid disk action. P(t_ring)(centripetal_1) <> P(t_ring)(acceleration) + P(t_ring)(deceleration) + P(t_ring)(centripetal_2) The law of momentum conservation works locally. During acceleration action extra momentum passes to extra momentum deceleration action. This extra momentum won’t affect momentum from centripetal force. Because curve part of track centripetal force depends on mass only. And do nothing to the rolling body moment of inertia. P(t_ring)(centripetal_1) = P(s_disk) (centripetal_1) P(t_ring)(centripetal_2) = P(s_disk) (centripetal_2) P(t_ring)(acceleration) = P(s_disk) (acceleration) + P(extra) P(t_ring)(deceleration) = P(s_disk) (deceleration) + P(extra) These two extra momentums have opposite direction on strait track. And sum of these momentums will be a zero. But these momentums have same direction on U-track. And sum of these extra momentums not equal zero.

No any new physic laws. This is correct. But the paradox is isolated system moves without any external forces. And all law of conservation is working. All momentums are present. The Ring just passes some “extra” kinetic energy thought curve part of track. Acceleration force charges high energy to thin ring. Isolated System causes high momentum. And Isolated System will cause same high momentum on friction force acting. Forces have a same direction on acceleration and friction acting time. But System will cause low momentum on centripetal force action time. Just this momentum will opposite to others 2 momentums. P(centripetal) << P(acceleration)+P(friction)

This is correct. This is explaining why centripetal force doesn’t depend on bodies shape. For 2 cases, rolling disk and ring mass m – const, velocity for curve part of track V – const kinetic energy E depends on shape. The thin ring has a big radius R. Even if ring and disk has same mass and velocity, the ring kinetic energy will be higher than disk kinetic energy. Centripetal force for curve part of track won’t change on both cases. m - const, for curve part of track V – const, R(ring)>>R(disk), E(ring)>>E(disk), F(centripetal) = F(ring) = F(disk) - const Even if ring rolling friction equal to disk rolling friction. F(fr)(ring)==F(fr)(disk) Ring momentum from rolling friction will be higher than disk momentum. Because ring kinetic energy higher than disk kinetic energy. E(ring)>>E(disk) P(fr)(ring)>>P(fr)(disk) This means U-track with thin ring will move into one direction, but the law of momentum conservation will work locally.

The idea is very simple. Centripetal force doesn’t care about moment of inertia. It means, body with shape causes same centripetal force on curve trajectory. Centripetal force just depends on mass, velocity and curve radius. But bodies with different shapes but same velocity and mass have different kinetic energy. From the other words we have to spend DIFFERENT amount of energy to reach same velocity for ring and disk. But bodies will cause SAME centripetal force on same curve trajectory. So, in U-track case acceleration and break energy will have same direction, but amount of them will be different for disk and ring. Otherwise centripetal forces for disk and ring are equal. If use only rolling friction for break bodies, vector of momentum will be different. (Not all break energy will be spent to internal works) This is “two words” explanation. Sorry about my English. I just try to bring a brilliant idea to world Merged post follows: Consecutive posts mergedFull kinetic E = m*V*V/2+J*W*W/2 The Disk moment of inertia J= m*R*R/2. The Ring moment of inertia J = m*(R1*R1+R2*R2)/2. If internal and external radius of ring close to each other R2-R1 ->0, then Ring radius is infinite R1->oo, R2->oo. And the ring full kinetic energy E->oo. Other words, the ring with same velocity and mass will get a huge kinetic energy. But this won’t affect curve part of track, because centripetal force doesn’t depend on moment of inertia. Both disk and ring with different kinetic energy will cause same centripetal force on curve part of track F=m*V*V/R. But need complete different chunk of energy to accelerate them to same velocity V. Or reduce their velocity V to 0.

Are these mistakes corrupt the main idea?