John Cuthber Posted April 27, 2014 Share Posted April 27, 2014 Firstly, we know what path independence means. Secondly, you are still trying to argue against the known laws of physics. Why are you wasting your time doing this? Also, the diagram doesn't show up, but it doesn't matter. The diagram can't explain why your idea works, because the idea doesn't work. Link to comment Share on other sites More sharing options...
swansont Posted April 27, 2014 Share Posted April 27, 2014 Yes, work is path dependent. You are trying to tie this into momentum somehow, but there is no connection there. Momentum transfer is not work, and is not path dependent. If an object comes to rest under some force, it will have transferred its momentum to the second system, the one that exerted the force. It will not depend on the path. All of the momentum will be transferred. Every time. Link to comment Share on other sites More sharing options...
dijinj Posted April 28, 2014 Author Share Posted April 28, 2014 (edited) I assume that you can at-least agree that same magnitude of force in the direction of motion is required to start moving a ball from initial point to another point and that same Magnitude of force in opposite direction to motion is required to move ball from that point to stop the ball at original initial position. Since friction force will act against motion in opposite direction to that motion, at every point along that curved path friction force will act against motion and reduce velocity and there by momentum of the ball. Normal force in upper semicircular path will cause change in angular momentum. That angular momentum will be counter balanced by normal force given earlier in bottom semicircular path. If conservation of angular momentum is true in this case too. If an object comes to rest under some force, it will have transferred its momentum to the second system, the one that exerted the force. It will not depend on the path. All of the momentum will be transferred. Every time. this will be dependent on path, if friction force is applied to stop the ball. the momentum transfer will be tangential along the path since it is against the direction of force exerted by second body. force is rate of change of momentum. friction force does negative work in most of the reference frames that's how they are connected Edited April 28, 2014 by dijinj Link to comment Share on other sites More sharing options...
John Cuthber Posted April 28, 2014 Share Posted April 28, 2014 No. The momentum transfer is not path dependent. All the momentum added to the ball is subtracted from the rest of the apparatus. The net change in momentum is zero. Why are you trying to argue with the laws of physics? Link to comment Share on other sites More sharing options...
swansont Posted April 28, 2014 Share Posted April 28, 2014 I assume that you can at-least agree that same magnitude of force in the direction of motion is required to start moving a ball from initial point to another point and that same Magnitude of force in opposite direction to motion is required to move ball from that point to stop the ball at original initial position. Since friction force will act against motion in opposite direction to that motion, at every point along that curved path friction force will act against motion and reduce velocity and there by momentum of the ball. Normal force in upper semicircular path will cause change in angular momentum. That angular momentum will be counter balanced by normal force given earlier in bottom semicircular path. If conservation of angular momentum is true in this case too. this will be dependent on path, if friction force is applied to stop the ball. the momentum transfer will be tangential along the path since it is against the direction of force exerted by second body. force is rate of change of momentum. friction force does negative work in most of the reference frames that's how they are connected No. Regardless of the path, if the ball starts with some momentum P and comes to rest, the object that was exerting the force will have a momentum P. There's no way to change the path and get a different outcome within those boundary conditions. Link to comment Share on other sites More sharing options...
imatfaal Posted April 28, 2014 Share Posted April 28, 2014 ... Since friction force will act against motion in opposite direction to that motion, at every point along that curved path friction force will act against motion and reduce velocity and there by momentum of the ball. Normal force in upper semicircular path will cause change in angular momentum. That angular momentum will be counter balanced by normal force given earlier in bottom semicircular path. If conservation of angular momentum is true in this case too. ... Agree with SwansonT and JohnC above; have snipped bit of quote dealt with. Re the action of friction (which you still havent spelled out how you think it is quantified) - in a path such as you have described what you will observe is that the action of friction will be to mostly impart angular momentum. the outside of the ball where it touches the inside of the wall is under a force opposite to that of motion - this generates a torque; if the ball is travelling anti-clockwise around the circuit then the ball itself will roll in a clockwise rotation. This will be the same at the top and at the bottom curve. The ball will spin and so will the craft; even though the net angular momentum of the system will remain constant. Link to comment Share on other sites More sharing options...
dijinj Posted April 28, 2014 Author Share Posted April 28, 2014 (edited) in a semi circular path, normal force will only change angular momentum. tangential force will change instantaneous velocity; Instantaneous velocity is always tangential to trajectory. Slope of tangent of position or displacement time graph is instantaneous velocity and its slope of chord is average velocity. Here is where path dependence and line integral comes; for path independent force, resultant force along chord and length of chord will give work done. since momentum is mass X velocity. that tangential force(friction) will reduce linear momentum. since friction will act against direction of motion, friction force is tangential along a curved path. Edited April 28, 2014 by dijinj Link to comment Share on other sites More sharing options...
Endy0816 Posted April 28, 2014 Share Posted April 28, 2014 (edited) Please comment your views on feasibility of the above concept It is not feasible. Friction is conservative if you do the full calculations. Typically we just don't bother. Edited April 28, 2014 by Endy0816 Link to comment Share on other sites More sharing options...
swansont Posted April 28, 2014 Share Posted April 28, 2014 in a semi circular path, normal force will only change angular momentum. tangential force will change instantaneous velocity; Instantaneous velocity is always tangential to trajectory. Slope of tangent of position or displacement time graph is instantaneous velocity and its slope of chord is average velocity. Here is where path dependence and line integral comes; for path independent force, resultant force along chord and length of chord will give work done. since momentum is mass X velocity. that tangential force(friction) will reduce linear momentum. since friction will act against direction of motion, friction force is tangential along a curved path. And how does that matter for the momentum transfer? The two examples are not equivalent to each other; if the ball does not come to rest the momentum change is different than if it does. We know this, because the tangential dp/dt is zero for one case and not for the other. You're not comparing two different paths, you're comparing two completely different systems. Link to comment Share on other sites More sharing options...
dijinj Posted April 29, 2014 Author Share Posted April 29, 2014 If constant path dependent force given along curved path, change in magnitude of velocity will be equal to change in magnitude of velocity by a same amount of path independent force along a straight line of length equal to curved path. However for path dependent force, x and y component of that tangential path dependent friction force will contribute in change in magnitude of velocity. So the x and y component (y component of less magnitude than that initial electromagnetic force, x component will be new without any cancelling initial force) will cause an equal and opposite reaction force on space craft. This will result in in some motion of space craft in oblique direction. Link to comment Share on other sites More sharing options...
John Cuthber Posted April 29, 2014 Share Posted April 29, 2014 (edited) If constant path dependent force given along curved path, change in magnitude of velocity will be equal to change in magnitude of velocity by a same amount of path independent force along a straight line of length equal to curved path. However for path dependent force, x and y component of that tangential path dependent friction force will contribute in change in magnitude of velocity. So the x and y component (y component of less magnitude than that initial electromagnetic force, x component will be new without any cancelling initial force) will cause an equal and opposite reaction force on space craft. This will result in in some motion of space craft in oblique direction. But, since it doesn't, it doesn't. Edited April 29, 2014 by John Cuthber Link to comment Share on other sites More sharing options...
CaptainPanic Posted April 29, 2014 Share Posted April 29, 2014 ! Moderator Note dijinj, This thread is now over 30 posts long, and all this time you've been arguing mostly against a number of experts in physics, about a rather basic physics topic. It could be that you are trying to learn, and that you are just debating to find out where you are wrong. While this can be a very useful technique to learn a lesson, it is incredibly frustrating for all the other participants, because it seems you are just ignoring them. Please show that you try to understand the points of all the other participants in the thread. And if you disagree with them, please explain why. However, if you do not disagree with them, but you still don't see why your own idea is wrong, please don't only rephrase your own idea. Make a short comment about agreeing/disagreeing with the other posts (and why). Right now, your debating style comes across as soapboxing, and that is against our rules (section 2.8). Link to comment Share on other sites More sharing options...
swansont Posted April 29, 2014 Share Posted April 29, 2014 It boils down to F = dp/dt, or [math]p =\int Fdt[/math] There is no path integral in time. There is just the force, which, by the third law, is equal and opposite for the two parts involved. Link to comment Share on other sites More sharing options...
dijinj Posted April 29, 2014 Author Share Posted April 29, 2014 Friction is conservative if you do the full calculations. Typically we just don't bother. friction is non-conservative or path dependent. if there is only conservative force why bother to call them conservative force not just force.The term conservative force comes from the fact that when a conservative force exists, it conserves mechanical energy. The most familiar conservative forces are gravity, the electric force (in a time-independent magnetic field), and spring force. A force field F, defined everywhere in space (or within a simply-connected volume of space), is called a conservative force or conservative vector field if it meets any of these three equivalent conditions: 1. The curl of F is zero: 2. There is zero net work (W) done by the force when moving a particle through a trajectory that starts and ends in the same place: 3. The force can be written as the negative gradient of a potential Many forces (particularly those that depend on velocity) are not force fields. In these cases, the above three conditions are not mathematically equivalent. For example, the magnetic force satisfies condition 2 (since the work done by a magnetic field on a charged particle is always zero), but does not satisfy condition 3, and condition 1 is not even defined (the force is not a vector field, so one cannot evaluate its curl). Accordingly, some authors classify the magnetic force as conservative, while others do not. The magnetic force is an unusual case; most velocity-dependent forces, such as friction, do not satisfy any of the three conditions, and therefore are unambiguously nonconservative. It boils down to F = dp/dt, or [math]p =\int Fdt[/math] There is no path integral in time. There is just the force, which, by the third law, is equal and opposite for the two parts involved. I agree with F = dp/dt, or [math]p =\int Fdt[/math] (force is rate of change of momentum) But you can see that displacement (ds) can be changed to (vdt) by change of variables in integration, now the integration for calculating W will be with respect to time.Path integral is for calculating the work done by a non-conservative force Link to comment Share on other sites More sharing options...
swansont Posted April 29, 2014 Share Posted April 29, 2014 I agree with F = dp/dt, or [math]p =\int Fdt[/math] (force is rate of change of momentum) But you can see that displacement (ds) can be changed to (vdt) by change of variables in integration, now the integration for calculating W will be with respect to time.Path integral is for calculating the work done by a non-conservative force Calculating W has nothing to do with momentum, and thrust requires momentum to not be constant. Bringing up work and path dependence is a red herring. Link to comment Share on other sites More sharing options...
dijinj Posted April 29, 2014 Author Share Posted April 29, 2014 (edited) Calculating W has nothing to do with momentum, and thrust requires momentum to not be constant. Bringing up work and path dependence is a red herring. [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 Edited April 29, 2014 by dijinj Link to comment Share on other sites More sharing options...
swansont Posted April 29, 2014 Share Posted April 29, 2014 [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 You have solved for work, not momentum. The value of momentum doesn't show up there, the derivative does. There's no path integral for momentum. Momentum is not a path-dependent quantity. Link to comment Share on other sites More sharing options...
dijinj Posted April 30, 2014 Author Share Posted April 30, 2014 (edited) You have solved for work, not momentum. The value of momentum doesn't show up there, the derivative does. [math]W =s*dp/dt[/math] rearranging [math]dp =(W/s)dt[/math] integrating both side [math]p =\int (W/s)dt[/math] Edited April 30, 2014 by dijinj Link to comment Share on other sites More sharing options...
John Cuthber Posted April 30, 2014 Share Posted April 30, 2014 For a start, it reminds me of this joke. http://cauchy.math.ttu.edu/jokes/joke-086.html Secondly, you have not explained why you don't think the conservation laws apply to your idea. I have asked repeatedly. Your failure to reply is a breach of the rules. Link to comment Share on other sites More sharing options...
dijinj Posted April 30, 2014 Author Share Posted April 30, 2014 (edited) If you can not see the image of calculation got link https://sites.google.com/site/nonmassexpulsiontypethruster/contact Correct me if I am wrong Secondly, you have not explained why you don't think the conservation laws apply to your idea. I have asked repeatedly. Your failure to reply is a breach of the rules. if you can not view the image of calculation go to bottom of link https://sites.google.com/site/nonmassexpulsiontypethruster/contact correct me if I am wrong Edited April 30, 2014 by dijinj Link to comment Share on other sites More sharing options...
swansont Posted April 30, 2014 Share Posted April 30, 2014 [math]W =s*dp/dt[/math] rearranging [math]dp =(W/s)dt[/math] integrating both side [math]p =\int (W/s)dt[/math] And where is the path integral? All that's happened here is you've replaced F with W/s. Work normalized to the path length is not the work. Both are path dependent, and that cancels out. Link to comment Share on other sites More sharing options...
dijinj Posted April 30, 2014 Author Share Posted April 30, 2014 (edited) And where is the path integral? All that's happened here is you've replaced F with W/s. Work normalized to the path length is not the work. Both are path dependent, and that cancels out. path integral is used for calculating work done by path dependent force like friction, so W can be replaced with path integral in that equation. for the difference in calculating path integral and regular integral see the below calculation if you can not view the image of calculation go to bottom of link https://sites.google.com/site/nonmassexpulsiontypethruster/contact Secondly, you have not explained why you don't think the conservation laws apply to your idea. 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) Edited April 30, 2014 by dijinj Link to comment Share on other sites More sharing options...
Orodruin Posted April 30, 2014 Share Posted April 30, 2014 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) 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. Link to comment Share on other sites More sharing options...
swansont Posted April 30, 2014 Share Posted April 30, 2014 path integral is used for calculating work done by path dependent force like friction, so W can be replaced with path integral in that equation. for the difference in calculating path integral and regular integral see the below calculation 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) One thing about this is it's a good lesson in not blindly applying an equation where you don't know what assumptions were made in deriving it. That's important here because KE is a scalar, while momentum is a vector (as is velocity). So the question becomes how did you arrive at the equation and what assumptions were made in doing so? It's a simple substitution of variables, so there is no accommodation for any sort of interaction — it's for a single particle. The equation does not hold for more than 1 particle. And it's true that if you change KE for a single particle, i.e. it's not conserved, that the magnitude of the momentum will change — there must be a force, which changes the momentum and does work. Link to comment Share on other sites More sharing options...
imatfaal Posted April 30, 2014 Share Posted April 30, 2014 path integral is used for calculating work done by path dependent force like friction, so W can be replaced with path integral in that equation. for the difference in calculating path integral and regular integral see the below calculation 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) The object in your latter example is slowing down on a frictional surface - so clearly the momentum of that object is not conserved. But whatever is slowing it down by exerting a frictional force is by newtons third law feeling an equal and opposite force. This 3rd law force on the object will accelerate that object and cause it to have a momentum. If you add the momentum of your initial object to the momentum of the surface then you will always get the same answer. Link to comment Share on other sites More sharing options...
Recommended Posts