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I'm not quite sure what you mean about the rolling bodies, but don't forget the floor will take an asymmetrical force pushing it in one direction.
What's stopping you doing a decisive experiment? Some equipment you don't have?




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What do you mean asymmetrical forces? Rolling resistance force does not depend on velocity. Just from mass, radius, material. In my case, radius, mass and material are equivalent. The reaction force for both bodies must be identical, because they repulse from each other. The spring has identical forces on both sides. All these symmetric forces are covering by Newtons third law.
I don't have a high precision equipment and vacuum room for clear experiment. I don't think my experiment is unique case for nature. The rolling bodies on surface with same mass and different moment of inertia is pretty similar case. It would be easy to reproduce. However, even on kinematic equation I see different translational velocities for rolling bodies relatively to repulsing point. What is it? Mistake or paradox?




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What do you mean asymmetrical forces? Rolling The roller with the higher rotational inertia will transmit a greater force to the floor through its friction. So the floor will be pushed in that direction. If you calculate the total linear momentum of the system, including the floor (and earth), it'll be zero. Same for angular momentum. High precision equipment won't help because you haven't made a quantified, testable hypothesis (that I know of). If you do it in an evacuated room with frictionless parts and high speed cameras, and the results are consistent with computer simulations, then what will that tell you? Maybe the effect you hope for is too small to detect with that equipment. So it's a probablypointless experiment.
Last edited by kallog; 10/05/10 11:30 AM.




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The roller with the higher rotational inertia will transmit a greater force to the floor through its friction. So the floor will be pushed in that direction. Here is rolling resistance force explanation http://en.wikipedia.org/wiki/Rolling_resistanceNo any dependencies from velocity. If you calculate the total linear momentum of the system, including the floor (and earth), it'll be zero. Same for angular momentum. I used normal kinematic equation from physics book. Great example of this is a rolling body along incline. http://cnx.org/content/m14312/latest/I don't see any errors on my kinematic equation yet. I would appreciate if you find one. High precision equipment won't help because you haven't made a quantified, testable hypothesis (that I know of). If you do it in an evacuated room with frictionless parts and high speed cameras, and the results are consistent with computer simulations, then what will that tell you? Maybe the effect you hope for is too small to detect with that equipment. So it's a probablypointless experiment.
I made hypothesis about standalone natural phenomenon as rotational and translational motion. http://knol.google.com/k/paradoxofclassicalmechanics2#I don't think effect is too small. I think, the main problem is simplification of physics calculation.




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Ok. Back to model for rolling bodies. This model is not equivalent to experiment on my site, because rolling bodies contact to surface all the time and both have a rotation. The rolling body with higher moment of inertial will have bigger reaction force to the surface. Which should be compensated by spring through surface. However the surface has it's own mass and will keep reaction forces for rolling bodies difference. Unfortunately I have to disregard my model from my site. However, I'm still working for exact model for my experiment and will publish it in a future.




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Maybe you haven't expressed it in the way you're thinking of it. But I don't think the hypothesis "The Rotational and Translational motion is standalone natural phenomenon." is testable. You should specify quantitatively what result of the experiment would support it, and what result would disprove it. That'll tell you how much accuracy is required. It may show that a homemade experiment is adequate, or it may show that a fancy vacuumroom experiment is still not enough. However it sounds like you're not expecting any results to differ from the classical predictions. In that case no experiment can be of any use, other than to confirm the existing theory. But again, how accurately do you want to confirm it? If you get too accurate you'll run into relativistic or quantum effects which disprove the classical theory, so in a way you're forced to do it only roughly.
Last edited by kallog; 10/06/10 03:19 AM.




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This site shows the physics problem. http://knol.google.com/k/alexbelov/thewheels/1xmqm1l0s4ys/18#Two wheels locate on weightless platforms. These platforms repulsing from each other through weightless spring. These wheels start conduct rotational and translational motion without friction on it's own platforms. These wheels don't have a rolling resistance also. Using force Fs1 spring repulse platform with wheel 1 with translational acceleration a3 relatively to repulsing point. Wheel 1 with mass m, moment inertia I1 and radius R has translational acceleration a1 relatively to repulsing point and angular acceleration alpha 1. Using force Fs2 spring repulse platform with wheel 2 with translational acceleration a4 relatively to repulsing point. Wheel 2 with mass m, moment inertia I2 and radius R has translational acceleration a2 relatively to repulsing point and angular acceleration alpha 2. These wheels have same mass and radius. Find out kinematic equations of motions for wheel 1 and wheel 2. Compare theirs translational and angular accelerations.




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I assume you mean there is nonslipping friction between the platform and wheel 2? Like a lossless rack and pinion. Find out kinematic equations of motions for wheel 1 and wheel 2. Compare theirs translational and angular accelerations.
Two issues: 1. I don't like this equation: Fs2 = Fl + Fr It suggests that only part of Fs2 contributes to translational motion, but in fact all of it does. However I might be misunderstanding your meaning. 2. You seem to have neglected the angular acceleration of object 1. To balance angular momentums you have to sum all the angular momentums in the system, taken about the same axis. Translational motion of wheel 1: Fs1 = m * a1 Translational motion of wheel 2: Fs2 = m * a2 Rotational motion of wheel 2 about its center: Moment = I * angular acceleration: Fs2 * R = I * alpha Rotational motion of wheel 1 about the same point. +ve is clockwise: Here I_1 means rotational inertia of object 1 about the center of object 2 for the type of motion it has, which includes no rotation of itself, so it's own 'I' must be ignored: I_1 = m * R^2 Moment = I_1 * angular acceleration: Fs1 * R = I_1 * alpha_1 Fs1 * R = m * R^2 * alpha_1 What's alpha_1? Depends on the linear acceleration: alpha_1 = a1 / R Fs1 * R = m * R^2 * a1 / R Fs1 = m * a1 This agrees with the linear motion equation, so no problem. Qualitatively: Translational: Linear momentums easily cancel out. Rotational: If m is very high compared to I, then object 2 will spin up fast and get some angular momentum (high alpha, low I). The objects will drift apart slowly because of their large masses. Object 1's angular momentum about the center of object 2 is also the same (low alpha, high I). I is high because it's proportional to the high m. Alpha is low because it's proportional to the low linear velocity. So the two angular momentums can cancel out.
Last edited by kallog; 10/07/10 05:03 AM.




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Thank you for answer with equations. Let's look closelly on it. 1. I don't like this equation: Fs2 = Fl + Fr It suggests that only part of Fs2 contributes to translational motion, but in fact all of it does. However I might be misunderstanding your meaning.
Why? Please look on standard problem  a rolling body on incline. http://cnx.org/content/m14312/latest/There you'll see 2 forces 2. You seem to have neglected the angular acceleration of object 1. To balance angular momentums you have to sum all the angular momentums in the system, taken about the same axis.
I don't need to do it, because it's simple wheel, which has standard equation. Kind of flywheel. http://en.wikipedia.org/wiki/Flywheel Translational motion of wheel 1: Fs1 = m * a1
Translational motion of wheel 2: Fs2 = m * a2
I disagree, because just part Fs is using for translational motion. It would be Fs1_part1 = m * a1 Fs2_part1 = m * a2 Rotational motion of wheel 2 about its center: Moment = I * angular acceleration: Fs2 * R = I * alpha
Same thing. Fs1_part2 * R = I1 * alpha (wheel 1) Fs2_part2 * R = I2 * alpha (wheel 2) Rotational motion of wheel 1 about the same point. +ve is clockwise: Here I_1 means rotational inertia of object 1 about the center of object 2 for the type of motion it has, which includes no rotation of itself, so it's own 'I' must be ignored: I_1 = m * R^2
I'm not follow this suggestion. However, nothing should be ignored. Extra rotation around center mass of isolated system may be compensated by gyroscope or symmetrical process. I would divide this problem to two simple ones. If look perspective from platforms, their accelerations can be described like virtual gravity. Objects on these platforms will start experience kind of gravity force. After that easy to see these objects will have simple kinematics equations, which was described before. Here's my solution: http://knol.google.com/k/alexbelov/thewheels/1xmqm1l0s4ys/18#It's good question about center of mass. I'll think about it. Thank you.
Last edited by ABV; 10/08/10 03:55 PM.




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Here is no problem with center mass. System won't move till any forces are applying to the system. For example if something initially move inside isolated system(empty sphere for example) without applied forces then system change own center of mass but system is not moving till object hit the wall of isolated system (for example). So, I don't see any problems with that.




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It's quite a different problem. However the entire force of gravity is applied to the roller, then there are reaction forces pushing it other ways too. I treat your roller on the platform as having a horizontal force applied to its center (equal to spring force), and additionally a moment applied about its center. That's fine. You can do that. Who says you can't? I don't need to do it, because it's simple wheel, which has standard equation. Kind of flywheel.
You need it if you're summing angular momentums to apply the law of conservation of momentum. When you do that you have to include the angular momentum of every part, all measured about the _same_ axis. The simple flywheel formula only applies to angular momentum about the wheel's own axis. Even tho it's not rotating it still has angular momentum about the other wheel's center, and it has a different rotational inertia about that axis too. I think this is a crucial part which needs to be incorporated. You can't just ignore the angular momentum of wheel 1 because it's not rotating. I disagree, because just part Fs is using for translational motion. It would be
Imagine you break the connection between wheel 2 and the platform. Then install a lever fixed to the center of the wheel, and its other end is pinjointed to the platform. From this it's obvious that the entire force is transmitted to the center of the wheel  there's nowhere else it can go. It also shows there's an additional moment applied about the center of the wheel. Hmm I kind of got a bit lost, sorry.




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It's quite a different problem. However the entire force of gravity is applied to the roller, then there are reaction forces pushing it other ways too.
This is the same classics mechanics kinematic problem. The difference is gravity force substituted by force which come from platform acceleration. The solution is very close to rolling body along incline. I treat your roller on the platform as having a horizontal force applied to its center (equal to spring force), and additionally a moment applied about its center. That's fine. You can do that. Who says you can't?
The spring doesn't push rolling object to theirs center of mass. Therefore the force of spring should no be fully spend to object translational motion. The spring force push platforms with rolling objects. Here's different case. You need it if you're summing angular momentums to apply the law of conservation of momentum. When you do that you have to include the angular momentum of every part, all measured about the _same_ axis. The simple flywheel formula only applies to angular momentum about the wheel's own axis. Even tho it's not rotating it still has angular momentum about the other wheel's center, and it has a different rotational inertia about that axis too.
Please look on solution of problem a rolling body along incline. There is just operate with forces which should equate to projection of gravity force. One oh them is translation motion force. Another came from torque with radius multiplication. I think this is a crucial part which needs to be incorporated. You can't just ignore the angular momentum of wheel 1 because it's not rotating. What do you mean is not rotating? The spring is connecting to platforms. These platforms have acceleration and wheel is free moving on them. Both wheels have reverse motion relatively to platforms. Imagine you break the connection between wheel 2 and the platform. Then install a lever fixed to the center of the wheel, and its other end is pinjointed to the platform. From this it's obvious that the entire force is transmitted to the center of the wheel  there's nowhere else it can go. It also shows there's an additional moment applied about the center of the wheel. You're correct with your model. However, the wheels have a free move on platforms. It means part of spring force with spend to translational motion of rolling object. Another part of force will spend to rotate this object which equate to torque and radius of object multiplication. Hmm I kind of got a bit lost, sorry. It's fine It's hard to understand from first look P.S. Thank you for you meaningful answers. I know a lot of forums where opponents know physic but don't understand that. This forum is complete different. Thank you again. Alex
Last edited by ABV; 10/12/10 01:47 PM.




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A few words about frames of reference. Newton's Laws hold only with respect to a certain set of frames of reference called Newtonian or inertial reference frames.[1] The first Newton's law is: "Every body remains in a state of rest or uniform motion (constant velocity) unless it is acted upon by an external unbalanced force. This means that in the absence of a nonzeronet force, the center of mass of a body either remains at rest, or moves at a constant speed in a straight line.[1] However Newton's laws don't deny free move center of mass of isolated system. Second and third laws describes bodies forces interaction. Otherwise, bodies knows nothing about each other and center of mass of isolated system is meaningless without bodies forces interaction. Base on symmetric bodies forces during interaction, the center of mass of isolated system should hold same position. It's true for simple motions, where force has simple meaning. However, for rotational and translational motion force can have two components from simple motions. In this case, net force may achieve same value by different components variation. For example 3+4=7 and 4+3=7. Where first number is translational force component and second number is rotational angular force by radius projection component. Therefore, center of mass of system for bodies forces interaction in rotational and translational motion can move. Otherwise, bodies during interaction should get additional extra forces from nowhere which will help to hold center of mass of isolated system on same position. Energy for these additional extra forces should come from nowhere too. Unfortunately, the modern classical mechanics equalize holding same position of center of mass of isolated system with symmetric forces for any cases of bodies forces interaction, because rotational and translational motion is a product of sum of two simple motions. This solution will follow free move center of mass of isolated system for single standalone rotational and translational motion, because no strong description about it in Newton's laws. This solution won't include any additional extra forces to helping to hold center of mass of isolated system on same position. Refernce: [1]http://en.wikipedia.org/wiki/Newton's_laws_of_motion http://knol.google.com/k/alexbelov/thewheels/1xmqm1l0s4ys/18#




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However Newton's laws don't deny free move center of mass of isolated system.
Perhaps, but the linear momentum conservation law does deny that possibility, even when there's rotation. radius projection component. Therefore, center of mass of system for bodies forces interaction in rotational and translational motion can move. Otherwise, bodies during
Now you're talking about reactionless propulsion. This is certainly impossible, partly because many people have tried many times, and all completely failed to show any result. And partly because it leads to paradoxes.




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Now you're talking about reactionless propulsion. This is certainly impossible, partly because many people have tried many times, and all completely failed to show any result. And partly because it leads to paradoxes.
Here's nothing about reactionless propulsion. I'm trying to explain where modern physic got mistake. If repulse rotated and nonrotated objects with same mass then base on modern physics these objects will have same translational velocity after repulsing action. Because rotational and translational motion is a sum of two simple motion. Whole force during repulsing should create a translational motion for rotated and nonrotated objects. Base on this force induct internal forces inside rotated object which will bring it's rotation during repulsing action. How come, Huh? Internal forces for rotation. From where this force? Where energy for this force came from? This is modern classical mechanics now. You don't believe me? You could ask any physics scientist My opinion, I don't believe to any magical internal forces inside rotated object. Otherwise, object should loose temperature, because internal forces are getting energy from object. The nature is doing simple thing. The repulse force inside rotated objects split for two forces for two motions of this object. Therefore net of these forces for two motions is equal to force which is applying to non rotated object during repulsing action(Third Newton's law). However it's impossible for modern physics now, because rotational and translational motion is sum of simple two motion for modern classical mechanic now which each of these motions must execute it's own law of momentum conservation. My solution is postulate rotation and translational motion as standalone motion with it's own law of momentum conservation. http://knol.google.com/k/alexbelov/paradoxofclassicalmechanics2# The model of rotated object on weightless platform fully describes rotated object during repulsing object. http://knol.google.com/k/alexbelov/thewheels/1xmqm1l0s4ys/18#Therefore, the classical mechanic has just one generic rotational and translational motion. Simple translational and rotational motions just a trivial cases of one main motion only. Sir Newton described both trivial cases of one main motion which simplified understanding about motion. However,excluding main rotational and translational motion brings mistake on nature motion description. I hope my experiment reproduction will prove it. I hope it helps. P.S. I don't think, I'm a first one who see this paradox. A few reasons could be don't discover it now. You can imagine
Last edited by ABV; 10/16/10 05:27 PM.




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I think, I found more simplest solution for this rotational and translational motion explanation. Rotational and translational motion description should include rule: Base on modern classical mechanic where rotational and translational motion is a product of sum of two simple motions as rotational motion and translational motion, each of these simple motions must have own force which induct this kind of motion. Then F=F1+F2. where F  full force, F1,F2  forces for inducting simple motions. Each of these motion will follow it's own law of momentum conservation where product of sum of these momentum will equal to full momentum which applied to this object. Then P=P1+P2. where P  full momentum, P1,P2  momentums for inducting simple motions. After that translational motion will follow F1 and P1 and rotational motion will follow F2*R and P2*R. That's it.




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I understood what was happened. Modern physics use static model for forces which applied to an object. This is the same model if calculate loads in civil engineering. If you this logic for physics problem "a rolling body along incline" then object has two reaction forces. In static mode each of this forces equal to gravity force projection on incline. However, solution of this problem use dynamic model where rotational and translational motion is a sum of two motion and sum of forces which conducted certain type of motion equal to net force(i.e. gravity force projection on incline) The physics problem for rotational and translational motion in free move doesn't use dynamic model and substitute physics process by static model where torque as addon(not part of it) for force in linear direction. This is mistake. The model should be the same as modern physics use for rolling objects on surface




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Rotational and translational motion can be described as a sum of 2 simple motions, which conducted in two separate events. Then for translational motion equation is: F_1=ma Work for this type of motion which equal objects kinetic energy translational part is E_k_t=mv^2/2 Base on third Newtons law symmetric force F_1 is present for this event Same for rotational motion with fixed axis equation is F_2R=Ia Work for this type of motion which equal objects kinetic energy rotational part is E_k_r=Iw^2/2 Base on third Newtons law symmetric force F_2 is present for this event ==== Base on sum of two motions the full kinetic energy which is work of 2 forces is equal: E_k_f=mv^2/2+Iw^2/2 Therefore sum of two symmetric forces for each simple motion necessary to initiate rotational and translational motion: F=(F_1)+(F_2) === However, the modern classical mechanics says if force applied away from center mass of object then just force [tex]F_1[/tex] with symmetrical force F_1 is enough to initiate rotational and translational motion with energy: E_k_f=mv^2/2+Iw^2/2 How this part of force do this part of work Iw^2/2  mystery. This paradox base on suggestion what law of momentum conservation works always and adding symmetric force F_2 is crashing this law. If use little rule where each motion law of momentum conservation should adequate force which induct this certain type of motion and all law of momentum conservation works locally, then not necessary hide force F_2. Center mass of isolated system is not the same as center of mass of objects and doesn't follow same rule. Dark matter is not a mystery anymore. Just a miscalculation http://knol.google.com/k/paradoxofclassicalmechanics2http://knol.google.com/k/alexbelov/thewheels/1xmqm1l0s4ys/18




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The third Newton’s law and law of momentum conservation.
The third Newton’s law declares what each reaction of forces should be symmetrical. F=F where: F  reaction forces of objects
The consequence of this is law of momentum conservation. F=F mv*t=mv*t mv=mv P=P
where: F  reaction forces of objects, m  mass of objects, v  velocity of objects, t  time trame of action, P  momentums of objects.
This consequence should be used for case where objects induce same identical motions and this does not cover the case where objects induce different type of motions. In experiment 2 thin cylinders induce different type of motions and this simple consequence should not cover this objects repulcing action. Therefore, for correct explanation of experiment 2 needs to use prime third Newton’s law and identify all reaction forces by this law. The consequence law of momentum conservation where objects have symmetrical behavior should not be used for this objects repulsing case.




