John Cuthber 3705 Posted April 30, 2014 another answer to John Cuthber, (on 30 Apr 2014 - 10:53 AM), take another approach to the problem. non conservative force like friction does not conserve mechanical energy. mechanical energy is PE and KE. consider a case in horizontal direction, then PE is out of the equation KE=1/2 mv^2= 1/2pv so if KE is not conserved then p is not conserved. If you are in objection with this . please explain with an example or give good reference link that explains this( it will be useful to many readers including me) You have posted irrelevant stuff about work, i.e. energy in relation to my question about the conservation of momentum. Also you say "so if KE is not conserved then p is not conserved." Well, I have news for you; p is strictly conserved. Would you please explain why you think you can ignore the conservation of momentum. Specifically, if the craft accelerates without some sort of "projectile" it breaches the conservation of momentum (That's essentially the bit about Newton's laws I cited earlier and which you have not addressed.) You need to provide an answer of the form "My idea can break the conservation of momentum because..." or something like that. 0 Share this post Link to post Share on other sites

dijinj 4 Posted May 1, 2014 Since you have only one v I am here going to assume you are only considering the energy of the ship. Yes, the energy and momentum of the ship do change during the friction phase. However, the total energy and total momentum of the ship+ball system does not change assuming we also take into account heat energy in the ship that is being generated due to the friction. Whenever the ball is transferring momentum to the ship, the ball loses the very same momentum. This happens both in the friction area, the area where you accelerate the ball, and the curves (while the absolute value of linear momentum does not change here, the direction does - the vectorial difference is transferred to the ship). As long as the force accelerating the ball is not external to the entire contraption, the total momentum of ball+ship will not change. It is very possible that the momentum transferred during the friction segment is not of the same magnitude as that transferred during the acceleration segment. However, this difference will be exactly the same, but with opposite sign, as the difference of the momentum transferred in the two different curves. The end result will simply be generating heat in the device due to the friction. You can clearly see that path integral of work is arc length and force (a function, if force is not constant). Here pi is arc length of semicircle and 2 is force. Regular integral of work is length of chord and force. If length of straight section, where electromagnetic force is given; is 3.14 and force is 2; then work done in y direction is 6.18. Momentum transfer is in y direction only. Then in upper semicircular arc of radius 1 and arc length 3.14; constant friction force of 2 is given then work done is also 6.18. Since both x and y component of that tangential friction force contribute; there will be a momentum transfer in oblique direction. So this initial and final momentum transfer may not cancel each other. You have posted irrelevant stuff about work, i.e. energy in relation to my question about the conservation of momentum. Also you say "so if KE is not conserved then p is not conserved." Well, I have news for you; p is strictly conserved. Would you please explain why you think you can ignore the conservation of momentum. Specifically, if the craft accelerates without some sort of "projectile" it breaches the conservation of momentum (That's essentially the bit about Newton's laws I cited earlier and which you have not addressed.) You need to provide an answer of the form "My idea can break the conservation of momentum because..." or something like that. I would not say say my device will breach conservation of momentum since i don't have a working prototype. but either one of momentum or path dependence has to be extended. or there has to be a " beauty of an illustration" to show that both are true. I would show a experiment where KE and both Momentum are conserved. see this youtube video here there are two bars of mass 1, red bar has initial velocity 1 and blue bar is stationary no force act on both bars. after collision red bar has velocity 0.7..something and blue bar has velocity 0.2..something. both these velocity add up to 1 so linear momentum is conserved. but collision causes rotation of same speed and same direction to viewers about CG of both bars respectively. but you know that direction of angular momentum is perpendicular to plane of rotation. one of bars angular Momentum vector is pointing toward viewer and vice verse for other bar so angular momentum too conserved. KE lost in squaring the new velocity of both bar is used for generating angular momentum of both bars. so KE too is conserved. see the beauty of this Computer simulation all laws are not breached still one crazy output of rotation of same speed and same direction to viewers about CG of both bars respectively; this could be useful in some crazy devices. That's why I am asking for a dynamic simulation soft ware that uses line integral for calculation. help me in this regard 0 Share this post Link to post Share on other sites

John Cuthber 3705 Posted May 1, 2014 (edited) "I would not say say my device will breach conservation of momentum "Yes you did. "line integral of a closed path to be zero for just wiggle back and to.stays in original position. but friction does do some work in closed path so there must be equal and opposite reaction of that work resulting in thrust of space craft in any direction." That thrust- with nothing to push against, is a breach of the conservation of momentum. So either explain why the laws of physics don't apply or accept that you are wrong. You need to provide an answer of the form"My idea can break the conservation of momentum because..."or something like that. Edited May 1, 2014 by John Cuthber 0 Share this post Link to post Share on other sites

swansont 7203 Posted May 1, 2014 You can clearly see that path integral of work is arc length and force (a function, if force is not constant). But momentum transfer, which is the quantity of interest, is not. And since speed is a function of position, the time spent exerting the force is a different function of the path, in such a way that the path dependence cancels. Here pi is arc length of semicircle and 2 is force. Regular integral of work is length of chord and force. If length of straight section, where electromagnetic force is given; is 3.14 and force is 2; then work done in y direction is 6.18. Momentum transfer is in y direction only. Then in upper semicircular arc of radius 1 and arc length 3.14; constant friction force of 2 is given then work done is also 6.18. Since both x and y component of that tangential friction force contribute; there will be a momentum transfer in oblique direction. So this initial and final momentum transfer may not cancel each other. But they will. I would not say say my device will breach conservation of momentum Your claim depends on this being true. That's why I am asking for a dynamic simulation soft ware that uses line integral for calculation. help me in this regard For the Nth time: the momentum will not depend on a line integral of work. It will depend on the time integral of the force. You should be able to solve a simple case analytically and show that momentum is conserved, because all you'd really be doing with software is a series of small linear solutions. And if each element (or any arbitrary element) conserves momentum, then there is no thrust. Put another way, instead of the complicated case show where exactly that momentum is not conserved, i.e. where the thrust comes from. 0 Share this post Link to post Share on other sites

Orodruin 15 Posted May 1, 2014 You do not need to do a simulation or do any consideration of energies to see that your overall thrust is zero. For continuous "propulsion" your accelerated mass would have to have the same velocity every time it arrives at the accelerating zone. If this is not true, it will just mean that your device wobbles more and more every time the mass passes. So, assume that this velocity is v0 and that the accelerated velocity is v1. This gives us the following cycle: During the acceleration, the mass has gained momentum m(v1-v0) which has been transferred to the ship. The ship has thus received the momentum m(v0-v1). In the turn, the mass has velocity v1. Since it is turning around completely and direction of momentum matters, the mass gains momentum -2m v1, since it is going from velocity v1 to -v1 and therefore from momentum m v1 to -m v1. This momentum has been transferred to the ship which now has momentum m(v0-v1)+2m v1 = m(v0+v1). During the friction phase, the mass slows down to v0 again simply because this is the only place where it can slow down and it needs to be at this speed when it comes back to the acceleration phase. However, it is moving in the negative direction so its linear momentum goes from -m v1 to - m v0, meaning that the momentum change is -m(v0 - v1). Due to conservation of momentum, the ship receives the opposite momentum and now has momentum m(v0+v1)+m(v0-v1) = 2m v0. During the second turn, the mass momentum changes from -m v0 to m v0 and the momentum change is thus 2m v0. The ship receives the opposite momentum change and its total momentum is therefore 2m v0 - 2m v0 = 0. Conclusion: No net momentum is gained by the ship throughout one cycle of operation of your device. No amount of simulation can change this and if you want to argue otherwise you must find exactly where my argumentation is wrong. Since my argumentation is solely based on the conservation of momentum, finding something wrong with it is equivalent to arguing that momentum is not conserved. You may rewrite integrals as much as you like, but momentum conservation will remain and give you exactly the argumentation as presented above. 0 Share this post Link to post Share on other sites

dijinj 4 Posted May 2, 2014 (edited) Okay Momentum is conserved. What about non-conservative force friction and its Path integral? does path integral not required in calculation. if yes can you suggest good simulation software that uses path integral for calculating friction. I only want to make ball at initial position and stationary with respect to space craft. space craft and ball could have some velocity with respect to earths reference frame. For KE and v^2 v.v is dot product. since cos(0) = 1 so only magnitude of v is required for calculation of KE. so is p.v dot product definition is given below Edited May 2, 2014 by dijinj 0 Share this post Link to post Share on other sites

moth 128 Posted May 2, 2014 Sorry for being so dense, but are you trying to transform some of the momentum to heat through friction? Winding up with a net force. 0 Share this post Link to post Share on other sites

John Cuthber 3705 Posted May 2, 2014 " I only want to make ball at initial position and stationary with respect to space craft. space craft and ball could have some velocity with respect to earths reference frame." That's easy. Put the ball in a cupboard on the space ship and then launch it in the usual way. However, it has nothing to do with the original post which is still impossible. 0 Share this post Link to post Share on other sites

Orodruin 15 Posted May 2, 2014 (edited) [math]p =\int Fdt[/math] [math]F =dp/dt[/math] [math]W =\int Fds[/math] [math]W =\int F*vdt[/math] [math]W =F*v*t[/math] [math]W =v*t*dp/dt[/math] [math]W =s*dp/dt[/math] This is how momentum and work are related(correct me if I am wrong). for Path dependence velocity v will be instantaneous velocity and delta change in arc length I did not bother with this before, but it is also wrong. You are always assuming that s = vt and that Fv is constant during the acceleration. Since the mass is accelerating, this is definitely wrong. Instead we have [math]W =\int Fv dt[/math] [math]\frac{dW}{dt} = Fv[/math] [math]\frac{dW}{dt} =v \frac{dp}{dt}[/math] [math]v^{-1} \frac{dW}{dt} = \frac{dp}{dt}[/math] Now by the chain rule [math]v^{-1} \frac{dW}{dt} = \frac{dt}{ds} \frac{dW}{dt} = \frac{dW}{ds} = F = \frac{dp}{dt}[/math] So you end up right where you started. Okay Momentum is conserved. What about non-conservative force friction and its Path integral? does path integral not required in calculation. if yes can you suggest good simulation software that uses path integral for calculating friction. I only want to make ball at initial position and stationary with respect to space craft. space craft and ball could have some velocity with respect to earths reference frame. For KE and v^2 v.v is dot product. since cos(0) = 1 so only magnitude of v is required for calculation of KE. so is p.v dot product definition is given below Again you are just ignoring the facts. The ship must gain momentum if it is to be accelerated. By conservation of momentum, this ends up somewhere else. Unless you break conservation of momentum, the ship will not accelerate. You do not need simulations to end up with this conclusion. You have agreed on the second point and I sincerely hope that you realize the first is also true. So, unless you can show us how point 3 is not true (which you just agreed that it is), point 4 is moot and nobody will lift a finger to investigate it further. Edited May 2, 2014 by Orodruin 0 Share this post Link to post Share on other sites

dijinj 4 Posted May 2, 2014 I know conservation of momentum and external force is required to change velocity and thus momentum . but I think by using conservative and non conservative force I can breach it. here is an example where walls and balls has equal mass 1 and both ball has initial velocity 1 in +ve x direction no external force act on any thing. green walls and balls are perfectly elastic and red wall is damping or inelastic. the ball and wall in upper side loses velocity and thus momentum without external force. but bottom one does not loses velocity (non-elastic material stress is also a non-conservative). can you explain what happens here You should be able to solve a simple case analytically and show that momentum is conserved, because all you'd really be doing with software is a series of small linear solutions. And if each element (or any arbitrary element) conserves momentum, then there is no thrust.Pm. I small linear solutions are tried the y direction will cancel each other and you have left with x direction only as i shown in below calculation earlier if you can not view the image of calculation go to bottom of link https://sites.google.com/site/nonmassexpulsiontypethruster/contact the work will be -4 since we are integrating over +1 to -1 0 Share this post Link to post Share on other sites

swansont 7203 Posted May 2, 2014 I know conservation of momentum and external force is required to change velocity and thus momentum . but I think by using conservative and non conservative force I can breach it. here is an example where walls and balls has equal mass 1 and both ball has initial velocity 1 in +ve x direction no external force act on any thing. green walls and balls are perfectly elastic and red wall is damping or inelastic. the ball and wall in upper side loses velocity and thus momentum without external force. but bottom one does not loses velocity (non-elastic material stress is also a non-conservative). can you explain what happens here Well, you can animate anything you want; the top system appears to be violating the laws of physics. If the ball is slowing from the friction, why isn't it transferring momentum to the box while it's doing so? That's unphysical. You should end up with the ball at rest wrt the box, and the box moving to the right. Assuming a constant F it should transfer a little on the first pass going backward, but after the rebound it should transfer even more in the forward direction, since it's going slower and the force acts for a longer time. Thus in the first cycle the box should be moving forward even when the ball is (which is not what's shown), unlike the elastic example, where the box is at rest. 0 Share this post Link to post Share on other sites

dijinj 4 Posted May 3, 2014 Well, you can animate anything you want; the top system appears to be violating the laws of physics. If the ball is slowing from the friction, why isn't it transferring momentum to the box while it's doing so? That's unphysical. You should end up with the ball at rest wrt the box, and the box moving to the right. Assuming a constant F it should transfer a little on the first pass going backward, but after the rebound it should transfer even more in the forward direction, since it's going slower and the force acts for a longer time. Thus in the first cycle the box should be moving forward even when the ball is (which is not what's shown), unlike the elastic example, where the box is at rest. Its not an animation its a dynamic simulation. its non elastic collision( damping) is happening at red wall (not friction). there is no external force only same initial velocity for both blue balls. can you explain what happens in shock absorbers (dampers) in cars. If dampers conserve momentum the cars should oscillate indefinitely(perpetually); but that's not happening in cars, the oscillation is damped or suppressed sometime after hitting a bump. 0 Share this post Link to post Share on other sites

Orodruin 15 Posted May 3, 2014 Momentum is still conserved in cars with shock absorbers and cars in general. It is also conserved when you jump. There is a force between the car and the ground, resulting in a change of momentum of both but their combined momentum is still the same. So why dont you notice the Earth's change in velocity? Well ... It is pretty heavy ... 0 Share this post Link to post Share on other sites

John Cuthber 3705 Posted May 3, 2014 Its not an animation its a dynamic simulation. its non elastic collision... The important thing which it is, is wrong. Momentum is conserved. 0 Share this post Link to post Share on other sites

dijinj 4 Posted May 3, 2014 (edited) The important thing which it is, is wrong. Momentum is conserved. can somebody has access to dynamic simulation software like ADAMS do this simple simulation( with elastic and nonelastic collision) and post their results here . Momentum is still conserved in cars with shock absorbers and cars in general. It is also conserved when you jump. There is a force between the car and the ground, resulting in a change of momentum of both but their combined momentum is still the same. So why dont you notice the Earth's change in velocity? Well ... It is pretty heavy ... then why a car without dampers but only springs oscillate for more time. you are challenging basic theory of physics without any supporting evidence. viscosity, drag, non elastic stress and friction etc are non conservative ( path dependent) forces There is nothing wrong with my simulation it is correct take a simple experiment take a rubber ball(elastic) and a concrete ball(Nonelastic) . bounce it on a same floor from same height and see which ball bounces high and think why! Edited May 3, 2014 by dijinj -1 Share this post Link to post Share on other sites

John Cuthber 3705 Posted May 3, 2014 If your animation suggests that you can get trust from an isolated system it is wrong because it breaches the law of conservation of momentum. You are the one "challenging basic theory of physics without any supporting evidence" 0 Share this post Link to post Share on other sites

swansont 7203 Posted May 3, 2014 Its not an animation its a dynamic simulation. its non elastic collision( damping) is happening at red wall (not friction). there is no external force only same initial velocity for both blue balls. can you explain what happens in shock absorbers (dampers) in cars. If dampers conserve momentum the cars should oscillate indefinitely(perpetually); but that's not happening in cars, the oscillation is damped or suppressed sometime after hitting a bump. Dynamic simulation, as in the solution to the kinematics equations? Then why, as I asked, is momentum not being properly conserved? can somebody has access to dynamic simulation software like ADAMS do this simple simulation( with elastic and nonelastic collision) and post their results here . then why a car without dampers but only springs oscillate for more time. you are challenging basic theory of physics without any supporting evidence. viscosity, drag, non elastic stress and friction etc are non conservative ( path dependent) forces There is nothing wrong with my simulation it is correct If your simulation is an actual solution to the equations, why do you need other software? 0 Share this post Link to post Share on other sites

Orodruin 15 Posted May 3, 2014 A non-damped car oscillates for longer because it has less damping and therefore it exchanges momentum back and forth with the Earth for quite some time before settling. And there *is* something wrong with your simulation. Since it violates conservation of momentum it cannot be built upon newtonian mechanics. Most likely you have forgotten that the friction force acts on *both* bodies but with opposite direction. 0 Share this post Link to post Share on other sites

dijinj 4 Posted May 4, 2014 If your simulation is an actual solution to the equations, why do you need other software? in the case of a straight line path dependence has no role in case of curved lines path dependence has a role that's why I am asking software that uses line integral 0 Share this post Link to post Share on other sites

Orodruin 15 Posted May 4, 2014 can somebody has access to dynamic simulation software like ADAMS do this simple simulation( with elastic and nonelastic collision) and post their results here . then why a car without dampers but only springs oscillate for more time. you are challenging basic theory of physics without any supporting evidence. viscosity, drag, non elastic stress and friction etc are non conservative ( path dependent) forces There is nothing wrong with my simulation it is correct take a simple experiment take a rubber ball(elastic) and a concrete ball(Nonelastic) . bounce it on a same floor from same height and see which ball bounces high and think why! Since you continue to ignore the facts, let me tell you exactly where your video with the bouncing balls goes wrong: You assume that only the moving object is affected by the frictional force. According to Newton's third law, when one body exerts a force on another, the other body exerts the same but opposite force on the first one. This is not the situation in your animation and as I said in my previous post, you are forgetting about one of the forces. The moving object is subjected to a frictional force from the stationary one, but the stationary one is subject to a frictional force from the moving one (and thus should start moving, which is not the case in your simulation). To answer your last sentence, you need to think a bit more about it. The reason they do not bounce to the same height is that the rubber ball transfers more of its momentum to the Earth because the collision is elastic. Now, the momentum transfer in both cases is so small that you will not notice the change in the Earth's velocity, but if you were able to measure it, the Earth's velocity change would be larger in the elastic case. 0 Share this post Link to post Share on other sites

Sensei 975 Posted May 4, 2014 (edited) viscosity, drag, non elastic stress and friction etc are non conservative ( path dependent) forces They are conserved but at level you can't see - energy. Friction causes heating of material. Kinetic energy of one body is changed to temperature of other body. And first body stops moving, and second body has higher temperature. There was even such experiment - scientists were shooting to water with thermometer and measured change of temperature of water caused by hundred or thousands objects hitting water. Edited May 4, 2014 by Sensei 0 Share this post Link to post Share on other sites

dijinj 4 Posted May 4, 2014 To answer your last sentence, you need to think a bit more about it. The reason they do not bounce to the same height is that the rubber ball transfers more of its momentum to the Earth because the collision is elastic. Now, the momentum transfer in both cases is so small that you will not notice the change in the Earth's velocity, but if you were able to measure it, the Earth's velocity change would be larger in the elastic case. you were wrong; take a case first ball moving and a second stationary. if two ball's mass were equal moving ball stops after collision and stationary ball starts moving with same velocity of first ball. if second ball's mass is less than first ball. then both ball will move in first ball's direction. if first ball's mass is less than second ball. the second ball will move in first ball's direction but first ball will reflect back after collision.(that's why mass less photon reflects back from a surface with original magnitude of velocity) all this happens while conserving momentum if collision is elastic. in non elastic collision stress( a non conserving force) external to ball will change its momentum first ( reduce the momentum since stress force will act against velocity) and then if any momentum left that will be transferred and conserved between those. And kinetic energy lost will be converted to heat energy. They are conserved but at level you can't see - energy. Friction causes heating of material. Kinetic energy of one body is changed to temperature of other body. And first body stops moving, and second body has higher temperature. There was even such experiment - scientists were shooting to water with thermometer and measured change of temperature of water caused by hundred or thousands objects hitting water. I agree to the above 0 Share this post Link to post Share on other sites

imatfaal 2480 Posted May 4, 2014 you were wrong; take a case first ball moving and a second stationary. if two ball's mass were equal moving ball stops after collision and stationary ball starts moving with same velocity of first ball. if second ball's mass is less than first ball. then both ball will move in first ball's direction. if first ball's mass is less than second ball. the second ball will move in first ball's direction but first ball will reflect back after collision.(that's why mass less photon reflects back from a surface with original magnitude of velocity) all this happens while conserving momentum if collision is elastic. in non elastic collision stress( a non conserving force) external to ball will change its momentum first ( reduce the momentum since stress force will act against velocity) and then if any momentum left that will be transferred and conserved between those. And kinetic energy lost will be converted to heat energy. You seem to be hinting that elastic and inelastic collisions vary in how they conserve momentum - this is false. The difference between them is the conservation of kinetic energy - they BOTH conserve momentum http://en.wikipedia.org/wiki/Inelastic_collision Inelastic collisions may not conserve kinetic energy, but they do obey conservation of momentum http://en.wikipedia.org/wiki/Elastic_collision The conservation of the total momentum demands that the total momentum before the collision is the same as the total momentum after the collision, and is expressed by the equation . We cannot help you get to the bottom of why you do not understand this if we do not know what umbers and equations you are putting into your simulation You really need to stop making assertions and put some equations down and actual do the hard yards - if you do this I absolutely guarantee you will not continue to misunderstand; this area of physics is known, consistent and non-contradictory thus thought experiments do not have any chance of overthrowing current thinking. You are making assumptions which are incorrect - this leads to result which violate the conservation laws; it is you who are wrong, if you will permit us we will find out the details of your mistakes and help you correct them 0 Share this post Link to post Share on other sites

Orodruin 15 Posted May 4, 2014 you were wrong; take a case first ball moving and a second stationary. if two ball's mass were equal moving ball stops after collision and stationary ball starts moving with same velocity of first ball. if second ball's mass is less than first ball. then both ball will move in first ball's direction. if first ball's mass is less than second ball. the second ball will move in first ball's direction but first ball will reflect back after collision.(that's why mass less photon reflects back from a surface with original magnitude of velocity) all this happens while conserving momentum if collision is elastic. in non elastic collision stress( a non conserving force) external to ball will change its momentum first ( reduce the momentum since stress force will act against velocity) and then if any momentum left that will be transferred and conserved between those. And kinetic energy lost will be converted to heat energy. So I believe I have identified your misconception about Newtonian mechanics. Momentum is always conserved. What is not conserved in an inelastic collision is the total kinetic energy. Elastic collision of two objects with mass m (one dimension): Before the collision: p1 = m v1 E1 = m v1^2 / 2 p2 = 0m = 0 E2 = 0 P = p1 + p2 = m v1 E = E1 + E2 = m v1^2/2 After collision: p1 = 0m = 0 E1 = 0 p2 = m v1 E2 = m v1^2/2 P = m v1 E = m v1^2/2 Now for the totally inelastic collision: Before: p1 = m v1 E1 = m v1^2/2 p2 = 0 E2 = 0 P = m v1 E = m v1^2/2 After: Equal velocity of both objects gives a velocity v. Momentum is conserved (as a result of Newton's third law), i.e. P = m v1 = (m+m)v = 2m v This gives v = v1/2 and thus E =(2m) v^2/2 = m v1^2 / 4 < m v1^2 /2 Thus, the total kinetic energy after the collision is smaller but the total momentum is still the same. Do not confuse conservation of momentum with conservation of kinetic energy. 0 Share this post Link to post Share on other sites

moth 128 Posted May 4, 2014 (edited) A mass attached to a linear bearing with a launcher on one end of the bearing, and a spring on the otherHalf of the mass of the entire setup is in the linear bearing,launcher, and spring. The other half is in the weight mounted to the linear bearing.The launcher end and the mounted mass are initially together, and the launcher accelerates the movable mass with enough momentum to bounce off the spring end and return to the start of the cycle.Since the masses are equal the mounted mass and the launcher end of the bearing assembly move away from each other until the mounted mass hits the spring.The spring looses some of the momentum to friction and returns the mounted mass to the launcher end.The launcher end produces enough force to complete a full cycleThe spring end looses some of that energy to heat.It seems unbalanced, but the spring releases photons that have inertial mass and that balances/accounts for the conservation of momentum.It is a net force, but not enough to overcome thermal noise in the molecules of the contraption.but I'm just using intuition here. Edit: Please ignore my intrusion. i just realized what Imatfaal said about the difference between momentum and kinetic energy. sorry 'bout that. Edited May 4, 2014 by moth 0 Share this post Link to post Share on other sites