If someone came along and proved that the first Law of Thermodynamics was faulty, and that energy can be created from nothing, how would the physics masters out there feel?, realizing their not thinking all that much afterall.
Not saying that it will happen, it's just a what if question.
someone will
QUOTE (Dr Quest+Sep 10 2007, 01:07 PM)
how would the physics masters out there feel?, realizing their not thinking all that much afterall.
They are thinking, that much is clear and bear in mind that ideas and theories are crushed all the time in science. That's what progress is about, improving and sometimes replacing older ideas.
Someone would have to be extremely precise about their experiments because it would be an enormous thing to physicists if they could do such a thing. There are a great many effects which might, at first glance, seem like producing energy from nothing (the Casamir effect for instance) but are infact a very complicated and subtle effect due to a phenomena already known about and quantified.
There's also the difference between "Energy from nothing!" and "Energy from a new power source!". Before Einstein, mass and energy were considered seperate, so that noone knew where the Sun was getting it's power from because noone understood it was converting it's mass into energy. But once that link was understood, it was seen that it's not 'power from nothing' but 'power from a previously unknown source'. Is the source of power literally limitless (in which case it's closer to 'energy from nothing') or is there some finite reserve of it which we could then tap?
Finding a previously unknown but finite energy source would not be breaking the 1st Law but just expanding our knowledge about energy and how to change it from one form to another. I imagine that even if someone did find a way to produce energy without limit, it would be seen as a new form of energy we previously didn't consider or know about but will now factor into our equations/models when it's relevent.
They are thinking, that much is clear and bear in mind that ideas and theories are crushed all the time in science. That's what progress is about, improving and sometimes replacing older ideas.
Someone would have to be extremely precise about their experiments because it would be an enormous thing to physicists if they could do such a thing. There are a great many effects which might, at first glance, seem like producing energy from nothing (the Casamir effect for instance) but are infact a very complicated and subtle effect due to a phenomena already known about and quantified.
There's also the difference between "Energy from nothing!" and "Energy from a new power source!". Before Einstein, mass and energy were considered seperate, so that noone knew where the Sun was getting it's power from because noone understood it was converting it's mass into energy. But once that link was understood, it was seen that it's not 'power from nothing' but 'power from a previously unknown source'. Is the source of power literally limitless (in which case it's closer to 'energy from nothing') or is there some finite reserve of it which we could then tap?
Finding a previously unknown but finite energy source would not be breaking the 1st Law but just expanding our knowledge about energy and how to change it from one form to another. I imagine that even if someone did find a way to produce energy without limit, it would be seen as a new form of energy we previously didn't consider or know about but will now factor into our equations/models when it's relevent.
Very simple experiments prove that the Law of Conservation of Energy is false, or they negate Newtonian Physics. One or the other formulas has to be false.
A cylinder that is spun on its central axis can be stopped by tethered masses first held on the cylinders surface, (at 180°) and then the masses are allowed to swing out. As the tethered masses swing out the cylinder will stop. When the tethered masses have all the motion they can conserve only one of the two quantities of motion, kinetic energy (1/2mv²), or Newtonian momentum (mv). The reverse of this process is the ballistics pendulum; it conserves Newtonian momentum.
Such experiments can be assembled for less than $25, and require only slightly sophisticated video capturing and playback equipment. If a few people would repeat this experiment The Law of Conservation of Energy would be a thing of the past.
A concise statement of claims for an energy producing machine-
Objects can be released from circular motion: when they are released they travel tangent to the circle with linear motion that has equal displacement to the previous arc motion.
The circular interaction of arc motion in ballistic pendulums conserves linear momentum.
The experiment seeks to prove that arc motion (in meters per sec) and tangent linear motion are the same and should be covered under the same Laws of Physics (Newton’s Three Laws of Motion; which predicts momentum conservation) and that both should be conserved.
The experiment; A spinning cylinder with spheres embedded at 180° is released to allow the spheres to feed out on the end of tethers. As the spheres feed out they stop the rotation of the cylinder. At this point all the arc motion (tangent linear motion) is held by the spheres. If arc motion is conserved the spheres must be moving faster by a proportion roughly equal to the total mass (spheres and cylinder) divided by the mass of the spheres. If this proportional increase of motion holds, then energy increases.
Cylinder: 364g: 5 inch O.D. 4.5 inch I.D. PVC pipe coupler; and a smaller 4 inch I.D. 4.5 inch O.D. PVC pipe fitted inside the larger pipe to seat the spheres, 40g.
Spheres: 67g each, 1 inch dia. Steel (2)
Another experiment on a frictionless plane; had a
Disk: 284g, ½ inch HDPE 7 in dia.
Pucks: 32.3g each, standard 2.75 inch air pucks.
In both experiments the spheres and pucks stopped the cylinder and disk, respectively. The experimenter saw an increase in energy of over 300%.
The length of the tether varies with models and seems to make no difference, except that at the point when the cylinder is stopped the angle of the tether changes with different lengths of tether. The tether length is typically held to about the length of one radius of the cylinder.
A cylinder that is spun on its central axis can be stopped by tethered masses first held on the cylinders surface, (at 180°) and then the masses are allowed to swing out. As the tethered masses swing out the cylinder will stop. When the tethered masses have all the motion they can conserve only one of the two quantities of motion, kinetic energy (1/2mv²), or Newtonian momentum (mv). The reverse of this process is the ballistics pendulum; it conserves Newtonian momentum.
Such experiments can be assembled for less than $25, and require only slightly sophisticated video capturing and playback equipment. If a few people would repeat this experiment The Law of Conservation of Energy would be a thing of the past.
A concise statement of claims for an energy producing machine-
Objects can be released from circular motion: when they are released they travel tangent to the circle with linear motion that has equal displacement to the previous arc motion.
The circular interaction of arc motion in ballistic pendulums conserves linear momentum.
The experiment seeks to prove that arc motion (in meters per sec) and tangent linear motion are the same and should be covered under the same Laws of Physics (Newton’s Three Laws of Motion; which predicts momentum conservation) and that both should be conserved.
The experiment; A spinning cylinder with spheres embedded at 180° is released to allow the spheres to feed out on the end of tethers. As the spheres feed out they stop the rotation of the cylinder. At this point all the arc motion (tangent linear motion) is held by the spheres. If arc motion is conserved the spheres must be moving faster by a proportion roughly equal to the total mass (spheres and cylinder) divided by the mass of the spheres. If this proportional increase of motion holds, then energy increases.
Cylinder: 364g: 5 inch O.D. 4.5 inch I.D. PVC pipe coupler; and a smaller 4 inch I.D. 4.5 inch O.D. PVC pipe fitted inside the larger pipe to seat the spheres, 40g.
Spheres: 67g each, 1 inch dia. Steel (2)
Another experiment on a frictionless plane; had a
Disk: 284g, ½ inch HDPE 7 in dia.
Pucks: 32.3g each, standard 2.75 inch air pucks.
In both experiments the spheres and pucks stopped the cylinder and disk, respectively. The experimenter saw an increase in energy of over 300%.
The length of the tether varies with models and seems to make no difference, except that at the point when the cylinder is stopped the angle of the tether changes with different lengths of tether. The tether length is typically held to about the length of one radius of the cylinder.
I was wondering when your pucks, tethers, and cylinders would show up again ... hope to draw in a few new marks to foist your mixed-up physics upon?
I'll be watching ...
I'll be watching ...
To question the First Law of Thermodynamic is the most logical position an individual can take. To allocate the loss of 99% of the energy of a ballistic pendulum to frictional heat loss is unimaginable in this modern age of mechanical innovation and electronic sensing equipment.
Control of friction is a multibillion dollar industry: precision ground hardened metals, bearings, wheels, and electromagnetic levitation. Yet no one can reduce even to the slightest amount this huge energy loss to friction in a ballistics pendulum.
Infrared thermometers are now common in test labs, yet no one can measure this heat produced by the ballistic pendulum. Some make the excuse that it quickly dissipates into the air, well remove the air; you are familiar with vacuums I assume.
Galileo used his own heart beat to time objects rolling down an incline, the founding fathers of science had no sufficient means of controlling or measuring this alleged heat in a ballistics pendulum, but those days are gone. To continue this contrived ignorance is a scientific travesty. The heat is simply not there, and your Law of Conservation of Energy is false.
The cylinder and spheres experiment reverses the ballistic pendulum, and if it conserves linear momentum as does the ballistic pendulum, then it makes energy. But people will not repeat this experiment because their little world of physics will fall apart. Imagine what would happen to a professor that even suggested that the Law of Conservation of Energy might be false. The professor could conduct the experiment to prove that linear momentum is not conserved, but that would mean that Newtonian Physics is false, and that would get him fired even quicker.
Control of friction is a multibillion dollar industry: precision ground hardened metals, bearings, wheels, and electromagnetic levitation. Yet no one can reduce even to the slightest amount this huge energy loss to friction in a ballistics pendulum.
Infrared thermometers are now common in test labs, yet no one can measure this heat produced by the ballistic pendulum. Some make the excuse that it quickly dissipates into the air, well remove the air; you are familiar with vacuums I assume.
Galileo used his own heart beat to time objects rolling down an incline, the founding fathers of science had no sufficient means of controlling or measuring this alleged heat in a ballistics pendulum, but those days are gone. To continue this contrived ignorance is a scientific travesty. The heat is simply not there, and your Law of Conservation of Energy is false.
The cylinder and spheres experiment reverses the ballistic pendulum, and if it conserves linear momentum as does the ballistic pendulum, then it makes energy. But people will not repeat this experiment because their little world of physics will fall apart. Imagine what would happen to a professor that even suggested that the Law of Conservation of Energy might be false. The professor could conduct the experiment to prove that linear momentum is not conserved, but that would mean that Newtonian Physics is false, and that would get him fired even quicker.
In my experience, EVERY time someone enters a sentence like "[you/he/someone] can't/won't do this because they're [afraid of being wrong/their world will crumble/they can't see the truth]" they're wrong. You can safely ignore said post at that point.
I should patent that as a law of science forums...
I should patent that as a law of science forums...
LeTUOtter quote“safely ignore said post” In what way do you feel threatened?
The sentence you are complaining about attempts to explain why you won’t do a 25$ experiment, yet you claim to be a scientist. Your reason is certainly not scientific, so maybe I am more correct than you would like to admit.
You can’t handle a fundamental law of physics being in error. But you have no scientific choice. If Newton’s linear momentum conservation is true then The Law of Conservation of Energy is false.
You can only choice to do nothing and hope all your comrades do the same. It is the only way your little world will remain the same.
The sentence you are complaining about attempts to explain why you won’t do a 25$ experiment, yet you claim to be a scientist. Your reason is certainly not scientific, so maybe I am more correct than you would like to admit.
You can’t handle a fundamental law of physics being in error. But you have no scientific choice. If Newton’s linear momentum conservation is true then The Law of Conservation of Energy is false.
You can only choice to do nothing and hope all your comrades do the same. It is the only way your little world will remain the same.
QUOTE (Limon+Sep 12 2007, 01:54 AM)
If Newton’s linear momentum conservation is true then The Law of Conservation of Energy is false.
Too bad you've spent your life disproving linear momentum by employing experiments using angular momentum.
By the way, LeTUOtter, I think what you've observed is somewhere on that "crank" points list ...
So, from the Crackpot Index, I think Limon's last post qualifies to score:
20. 20 points for each favorable comparison of yourself to Newton or claim that classical mechanics is fundamentally misguided (without good evidence).
24. 20 points for each use of the phrase "self-appointed defender of the orthodoxy".(I think we can reasonably score this one out of his last sentence).
And for the post before that we can score the following from the action of "do-nothing professors":
29. 40 points for claiming that the "scientific establishment" is engaged in a "conspiracy" to prevent your work from gaining its well-deserved fame, or suchlike.
8. 5 points for each mention of "Einstien", "Hawking" or "Feynman". (Well, I'm sure these three illuminaries would be glad to include Galileo)
When you consider the entire corpus of Limon's work, I'm sure we can get a few more points assigned to him.
Too bad you've spent your life disproving linear momentum by employing experiments using angular momentum.
By the way, LeTUOtter, I think what you've observed is somewhere on that "crank" points list ...
So, from the Crackpot Index, I think Limon's last post qualifies to score:
20. 20 points for each favorable comparison of yourself to Newton or claim that classical mechanics is fundamentally misguided (without good evidence).
24. 20 points for each use of the phrase "self-appointed defender of the orthodoxy".(I think we can reasonably score this one out of his last sentence).
And for the post before that we can score the following from the action of "do-nothing professors":
29. 40 points for claiming that the "scientific establishment" is engaged in a "conspiracy" to prevent your work from gaining its well-deserved fame, or suchlike.
8. 5 points for each mention of "Einstien", "Hawking" or "Feynman". (Well, I'm sure these three illuminaries would be glad to include Galileo)
When you consider the entire corpus of Limon's work, I'm sure we can get a few more points assigned to him.
To NoCleverName: Obviously you have nothing important to do with your time. Perhaps you should learn to be a more careful reader. I use linear momentum conservation to disprove The Law of Conservation of Energy.
Ignoring Limon's nonsense for a while, i have the following to say to the original poster:
In case it turns out possible "that energy can be created from nothing", all that has to be done in physics equations is to state the energy of that "nothing" is infinite. Then the law of conservation will obviously still hold - infinity = infinity+5 = infinity+10000000000 and such, so it'll always be conserved.
Of course you might object that we'd have absolutely no need for a law that simply says "infinity=infinity" in any situation. And you're right - whenever we're dealing with systems that include the specific "nothing" that has infinite energy, we can just ignore this law.
But the law can still be useful for situations, processes and reactions which do not include that magical "nothing".
Now re:limon's nonsense, i'd love to explain him where he's wrong, but don't want to do his homework for him. Let's first see him provide actual calculations for the energy and momentum of the systems he describes in two different states...
In all likelihood, once he does the actual calculations, he'll realize his mistake without outside help. (hint: linear momentum is a vector quantity, not a scalar, and the law of conservation concerns the vector sum of all objects' momentum, which in the described setups is always 0 for the cylinder/disk by itself, and 0 for the set of spheres/pucks, no matter how much energy they exchange between themselves)
BTW, limon, if you do decide to argue about your point, you'd better start a new topic with the appropriate calculations instead of hijacking this one.
In case it turns out possible "that energy can be created from nothing", all that has to be done in physics equations is to state the energy of that "nothing" is infinite. Then the law of conservation will obviously still hold - infinity = infinity+5 = infinity+10000000000 and such, so it'll always be conserved.
Of course you might object that we'd have absolutely no need for a law that simply says "infinity=infinity" in any situation. And you're right - whenever we're dealing with systems that include the specific "nothing" that has infinite energy, we can just ignore this law.
But the law can still be useful for situations, processes and reactions which do not include that magical "nothing".
Now re:limon's nonsense, i'd love to explain him where he's wrong, but don't want to do his homework for him. Let's first see him provide actual calculations for the energy and momentum of the systems he describes in two different states...
In all likelihood, once he does the actual calculations, he'll realize his mistake without outside help. (hint: linear momentum is a vector quantity, not a scalar, and the law of conservation concerns the vector sum of all objects' momentum, which in the described setups is always 0 for the cylinder/disk by itself, and 0 for the set of spheres/pucks, no matter how much energy they exchange between themselves)
BTW, limon, if you do decide to argue about your point, you'd better start a new topic with the appropriate calculations instead of hijacking this one.
QUOTE (visual+Sep 12 2007, 02:29 PM)
Let's first see him provide actual calculations for the energy and momentum of the systems he describes in two different states...
He won't because he doesn't understand basic Newtonian dynamics. He's been whining about it for months and has been pointed in the direction of the information he needs by a number of people, including myself.
I suggested learning Lagrangian methods, because it would allow him to derive the conservation of energy, momentum and angular momentum for the system he's talking about and see how it all fits together.
But like all cranks, learning is too much for him to handle so he continues to whine about the same things, round and round in circles.
He won't because he doesn't understand basic Newtonian dynamics. He's been whining about it for months and has been pointed in the direction of the information he needs by a number of people, including myself.
I suggested learning Lagrangian methods, because it would allow him to derive the conservation of energy, momentum and angular momentum for the system he's talking about and see how it all fits together.
But like all cranks, learning is too much for him to handle so he continues to whine about the same things, round and round in circles.
Yeah but Limon you are taking energy out of the system as you let the weights out on the tethers. The weights pull on the tethers. You release the tether. Energy is force times distance. So energy in the system will not be conserved.
Tim; the pipe (cylinder) with embedded spheres is on a vertical axis, as the spheres and cylinder are spun and released both are dropped and the pipe and spheres are in free fall. As the spheres swing out on the end of tethers the force in the tether is applied equally to the pipe and to the spheres (Newton’s Third Law of Motion). According to Newtonian Physics (F =ma or the Second Law) whatever momentum the pipe loses the spheres will gain. When the 3 kg cylinder that was moving 1 m/sec stops then that 3 units of momentum have to be transferred to the spheres (F = ma). The spheres that were moving 1 m/sec now have to be moving 4 m/sec. Linear Newtonian momentum is conserved but energy is not, 4 kg moving 1 m/sec is 2 joules and 1 kg moving 4 m/sec is 8 joules. 1/2mv² The 4 kg can rise to .051 m and the 1 kg can rise to .8155 m
There is no Law of Conservation of Kinetic Energy, and no one really expects it to be conserved, but heat can not come to the rescue as it allegedly does in the ballistics pendulum. This time energy is being made not lost.
NoCleverName: Suspend a 5 kg ballistic pendulum block from two strings in a V shape so that a second pendulum bob can swing between the arms of the V and strike the block. Construct the pendulum length for the block on the end of the V to be .5 meters, and construct the length of the pendulum delivering the incoming projectile (bob) to be 5 meters. Let the incoming projectile have a mass of 1 kg and let it drop .5 m before impact. Let the center pendulum string drop away upon impact.
What is the velocity of the combination of the projectile and block after impact? Show what formulas you use.
This is a thought experiment but some ballistic pendulums are actually constructed in this manner.
Don’t waste time by telling me to do the problem myself, I already have, it took 15 seconds. I want to know if you can do it.
To Others: NoCleverName has a conundrum; if he does the problem correctly he will not use angular momentum conservation, because ballistic pendulums do not conserve angular momentum. Even though the initial motion of the incoming projectile is in angular motion and the final motion of the block and projectile is in angular motion, angular momentum is not conserved. What is conserved is linear motion that happens to be moving in a circular path; both Galileo and Newton had no problem with this concept of linear motion moving in a circular path.
What the ballistic pendulum conserves is linear Newtonian momentum. Linear Newtonian momentum conservation is what it think (and have measured) is conserved in my experiments. Angular momentum is not conserved in the lab, nor do I use the concept in my evaluations. Angular momentum conservation was invented for object in space and that is where the concept is properly applied. If NoCleverName gets the problem right he is going to have to conserve linear momentum.
I suspect that NoCleverName will default and refuse to do the problem, and will end his post with a brief evaluation of my intelligence. Maybe LeTUOtter will give it a try?
Ballistic pendulums are also not fooled by improper application of vectors.
There is no Law of Conservation of Kinetic Energy, and no one really expects it to be conserved, but heat can not come to the rescue as it allegedly does in the ballistics pendulum. This time energy is being made not lost.
NoCleverName: Suspend a 5 kg ballistic pendulum block from two strings in a V shape so that a second pendulum bob can swing between the arms of the V and strike the block. Construct the pendulum length for the block on the end of the V to be .5 meters, and construct the length of the pendulum delivering the incoming projectile (bob) to be 5 meters. Let the incoming projectile have a mass of 1 kg and let it drop .5 m before impact. Let the center pendulum string drop away upon impact.
What is the velocity of the combination of the projectile and block after impact? Show what formulas you use.
This is a thought experiment but some ballistic pendulums are actually constructed in this manner.
Don’t waste time by telling me to do the problem myself, I already have, it took 15 seconds. I want to know if you can do it.
To Others: NoCleverName has a conundrum; if he does the problem correctly he will not use angular momentum conservation, because ballistic pendulums do not conserve angular momentum. Even though the initial motion of the incoming projectile is in angular motion and the final motion of the block and projectile is in angular motion, angular momentum is not conserved. What is conserved is linear motion that happens to be moving in a circular path; both Galileo and Newton had no problem with this concept of linear motion moving in a circular path.
What the ballistic pendulum conserves is linear Newtonian momentum. Linear Newtonian momentum conservation is what it think (and have measured) is conserved in my experiments. Angular momentum is not conserved in the lab, nor do I use the concept in my evaluations. Angular momentum conservation was invented for object in space and that is where the concept is properly applied. If NoCleverName gets the problem right he is going to have to conserve linear momentum.
I suspect that NoCleverName will default and refuse to do the problem, and will end his post with a brief evaluation of my intelligence. Maybe LeTUOtter will give it a try?
Ballistic pendulums are also not fooled by improper application of vectors.
QUOTE (lemon+)
I suspect that NoCleverName will default and refuse to do the problem, and will end his post with a brief evaluation of my intelligence.
I too can't be arsed doing the calc. But an evaluation of your brief intelligence is no problem 
1/18 m.
This setup is hard to visualize, but it seems to be an elastic collision between a moving mass of m=1kg and a rest mass of M=5kg. I do not believe the pendulum lengths play any part. The moving block starts with potential energy P.E.=gmh = (1)(9.8)(0.5) which is fully converted to kinetic energy K.E = 1/2mv^2.
Therefore the velocity at impact is v = sqrt(2g h) = sqrt((2)(9.8)(0.5)) = sqrt(9.8) m/s.
Since block 2 is at rest, departure velocity V for block 2 in an elastic collision is
V = [(2*m)*v] / (m+M) = (2)(1)(sqrt(9.8)/(1+5) = sqrt(9.8)/3 m/s.
The big block now has K.E. = 1/2 MV^2 which it converts to P.E. = gMH on its climb. Solving for H:
1/2 MV^2 = gMH
H = V^2/(2g)
= [sqrt(9.8)/3]^2 / [(2)(9.8)] = 1/18 m, hopefully.
No doubt your 15 second answer came out 1/10 m or something like that.
This setup is hard to visualize, but it seems to be an elastic collision between a moving mass of m=1kg and a rest mass of M=5kg. I do not believe the pendulum lengths play any part. The moving block starts with potential energy P.E.=gmh = (1)(9.8)(0.5) which is fully converted to kinetic energy K.E = 1/2mv^2.
Therefore the velocity at impact is v = sqrt(2g h) = sqrt((2)(9.8)(0.5)) = sqrt(9.8) m/s.
Since block 2 is at rest, departure velocity V for block 2 in an elastic collision is
V = [(2*m)*v] / (m+M) = (2)(1)(sqrt(9.8)/(1+5) = sqrt(9.8)/3 m/s.
The big block now has K.E. = 1/2 MV^2 which it converts to P.E. = gMH on its climb. Solving for H:
1/2 MV^2 = gMH
H = V^2/(2g)
= [sqrt(9.8)/3]^2 / [(2)(9.8)] = 1/18 m, hopefully.
No doubt your 15 second answer came out 1/10 m or something like that.
I think I calculated the climbing height when all you wanted was velocities.
I had the big one going at sqrt(9.8)/3 = 1.04 m/s. There is a similar formula for the first block; I won't bother with it.
There is an entirely different formula for an inelastic collision, but I don't see this as applicable. Of course, ypur mileage may vary if there is deformation at impact rather than total elastic or inelastic.
I had the big one going at sqrt(9.8)/3 = 1.04 m/s. There is a similar formula for the first block; I won't bother with it.
There is an entirely different formula for an inelastic collision, but I don't see this as applicable. Of course, ypur mileage may vary if there is deformation at impact rather than total elastic or inelastic.
NoCleverName check the literature, ballistic pendulums do not conserve Kinetic Energy, they conserve linear momentum. You had the initial velocity correct at 3.132 m/sec for an initial linear momentum of 1 * 3.132 = 3.132 divided by the new mass 3.312 / 6 = .5220 m/sec
Quiz for NOM; construct a .3 m ring that is strong and smooth but with low mass, and place it on a smooth hard surface. Attach a 2 kg point mass at 0° (on top) and a 1 kg point mass at 180° (on the bottom) around the ring. In relationship to the table top; tell me the velocity of the 2 kg and 1 kg masses when the 2 kg point mass has rolled to the bottom of the ring.
To vector cancellation proponents: Buy four ballistic pendulums. Place all four facing north; operate them all simultaneously and calculate their total momentum. Next face one north; face one south, one east and one west. Operate them all simultaneously and again calculate their linear momentum. You can’t possibly believe it is now zero, do you? Four independent quantities of momentum canceled each other for no apparent reason; you have got to be kidding. Operate three of the four and you would have more momentum?
Atwood’s machines have nearly half their mass moving up and half the mass moving down. If these momentums canceled F would not equal ma, and that is what an Atwood’s proves.
Vectors are for unbalanced forces that are being applied to each other, if the forces are not being applied to one another or are balanced then the concept of vector addition or cancellation does not apply. A balanced spinning wheel has no unbalanced forces working between the masses; they are all happily minding their own business.
There is this concept that the changing direction of objects in circular motion is viewed as acceleration, but obviously the Moon has been there a long time, and the forces are viewed as balanced. Plus; the forces are between the point of rotation and the object being rotated, not between the point masses of the wheel.
Here is an experiment I conducted once. I fixed a string to an immovable point used for rotation just off of an air table, the other end of the string I affixed to a 84g puck on the air table. I tied a second string to the 84g puck and attached a 32g puck to the other end. I rotated both pucks, trailing the second puck behind the larger puck at 90° to the first string. The 90° position was fixed until release. Upon release the 32 g puck swung out beyond the 84g puck and stopped it, exactly as the spheres stop the cylinder in the cylinder and spheres experiment. The data was similar as well, showing linear momentum conservation.
In the 84 and 32 gram puck experiment there is no opportunity to cancel the motion from the other side using vector cancellation, because there is no other side. If you use linear momentum conservation the results are the same as the cylinder and spheres experiment, the results are different with vector cancellation. It seems like you would want a theory that work consistently like Newton’s does. Besides what good is you vector cancellation concept that tells you nothing; what good is a world of zeros.
Quiz for NOM; construct a .3 m ring that is strong and smooth but with low mass, and place it on a smooth hard surface. Attach a 2 kg point mass at 0° (on top) and a 1 kg point mass at 180° (on the bottom) around the ring. In relationship to the table top; tell me the velocity of the 2 kg and 1 kg masses when the 2 kg point mass has rolled to the bottom of the ring.
To vector cancellation proponents: Buy four ballistic pendulums. Place all four facing north; operate them all simultaneously and calculate their total momentum. Next face one north; face one south, one east and one west. Operate them all simultaneously and again calculate their linear momentum. You can’t possibly believe it is now zero, do you? Four independent quantities of momentum canceled each other for no apparent reason; you have got to be kidding. Operate three of the four and you would have more momentum?
Atwood’s machines have nearly half their mass moving up and half the mass moving down. If these momentums canceled F would not equal ma, and that is what an Atwood’s proves.
Vectors are for unbalanced forces that are being applied to each other, if the forces are not being applied to one another or are balanced then the concept of vector addition or cancellation does not apply. A balanced spinning wheel has no unbalanced forces working between the masses; they are all happily minding their own business.
There is this concept that the changing direction of objects in circular motion is viewed as acceleration, but obviously the Moon has been there a long time, and the forces are viewed as balanced. Plus; the forces are between the point of rotation and the object being rotated, not between the point masses of the wheel.
Here is an experiment I conducted once. I fixed a string to an immovable point used for rotation just off of an air table, the other end of the string I affixed to a 84g puck on the air table. I tied a second string to the 84g puck and attached a 32g puck to the other end. I rotated both pucks, trailing the second puck behind the larger puck at 90° to the first string. The 90° position was fixed until release. Upon release the 32 g puck swung out beyond the 84g puck and stopped it, exactly as the spheres stop the cylinder in the cylinder and spheres experiment. The data was similar as well, showing linear momentum conservation.
In the 84 and 32 gram puck experiment there is no opportunity to cancel the motion from the other side using vector cancellation, because there is no other side. If you use linear momentum conservation the results are the same as the cylinder and spheres experiment, the results are different with vector cancellation. It seems like you would want a theory that work consistently like Newton’s does. Besides what good is you vector cancellation concept that tells you nothing; what good is a world of zeros.
QUOTE (Limon+Sep 13 2007, 10:10 PM)
NoCleverName check the literature, ballistic pendulums do not conserve Kinetic Energy, they conserve linear momentum. You had the initial velocity correct at 3.132 m/sec for an initial linear momentum of 1 * 3.132 = 3.132 divided by the new mass 3.312 / 6 = .5220 m/sec
You, of course, are wrong in this instance because there is no mention of an inelastic collision that would remove kinetic energy.
Now, have the decency to remove yourself from this thread as you have, in typical fashion, hijacked it for your own agenda.
You, of course, are wrong in this instance because there is no mention of an inelastic collision that would remove kinetic energy.
Now, have the decency to remove yourself from this thread as you have, in typical fashion, hijacked it for your own agenda.
I did not know that ballistic pendulums had collisions other than inelastic collisions.
Quote Dr Quest: If someone came along and proved that the first Law of Thermodynamics was faulty,
I am answering his question; I have proved that the first Law of Thermodynamics is faulty.
Quote Dr Quest: If someone came along and proved that the first Law of Thermodynamics was faulty,
I am answering his question; I have proved that the first Law of Thermodynamics is faulty.
QUOTE (Limon+Sep 14 2007, 01:50 AM)
I did not know that ballistic pendulums had collisions other than inelastic collisions.
Quote Dr Quest: If someone came along and proved that the first Law of Thermodynamics was faulty,
I am answering his question; I have proved that the first Law of Thermodynamics is faulty.
While that is classic BP behavior and use and I could have assumed it, there was no specification so I chose the opposite. Unreasonable, maybe. But why shouldn't I indulge myself once in a while?
What you really have proved, time and time again, is the inability to properly model a problem mathematically according to settled principles. I suppose nothing stops you from shouting Eureka when any odd move made by a block occurs. But, really, you have to properly apply mechanics rather than just "take a guess" and inso finding a problem make out as if you have made a wonderous discovery.
There's always that alternative possibilty: that you're wrong. It's generally the most likely one, too.
Quote Dr Quest: If someone came along and proved that the first Law of Thermodynamics was faulty,
I am answering his question; I have proved that the first Law of Thermodynamics is faulty.
While that is classic BP behavior and use and I could have assumed it, there was no specification so I chose the opposite. Unreasonable, maybe. But why shouldn't I indulge myself once in a while?
What you really have proved, time and time again, is the inability to properly model a problem mathematically according to settled principles. I suppose nothing stops you from shouting Eureka when any odd move made by a block occurs. But, really, you have to properly apply mechanics rather than just "take a guess" and inso finding a problem make out as if you have made a wonderous discovery.
There's always that alternative possibilty: that you're wrong. It's generally the most likely one, too.
QUOTE (Limon+Sep 14 2007, 01:50 PM)
I am answering his question; I have proved that the first Law of Thermodynamics is faulty.
Then go ahead and build your energy out of nothing machine and make trillions of dollars by undercutting the energy industry.
What is stopping you?
Um... could it be that your mistaken theories don't work?
Then go ahead and build your energy out of nothing machine and make trillions of dollars by undercutting the energy industry.
What is stopping you?
Um... could it be that your mistaken theories don't work?
To all: I posted this wheel (Quiz for NOM;) for a reason, please evaluate it. Such wheels pass the low point and nearly come back up to the top.
Quiz for NOM; construct a .3 m ring that is strong and smooth but with low mass, and place it on a smooth hard surface. Attach a 2 kg point mass at 0° (on top) and a 1 kg point mass at 180° (on the bottom) around the ring. In relationship to the table top; tell me the velocity of the 2 kg and 1 kg masses when the 2 kg point mass has rolled to the bottom of the ring.
Quiz for NOM; construct a .3 m ring that is strong and smooth but with low mass, and place it on a smooth hard surface. Attach a 2 kg point mass at 0° (on top) and a 1 kg point mass at 180° (on the bottom) around the ring. In relationship to the table top; tell me the velocity of the 2 kg and 1 kg masses when the 2 kg point mass has rolled to the bottom of the ring.
QUOTE (Limon+Sep 14 2007, 11:09 AM)
To all: I posted this wheel (Quiz for NOM;) for a reason, please evaluate it. Such wheels pass the low point and nearly come back up to the top.
Quiz for NOM; construct a .3 m ring that is strong and smooth but with low mass, and place it on a smooth hard surface. Attach a 2 kg point mass at 0° (on top) and a 1 kg point mass at 180° (on the bottom) around the ring. In relationship to the table top; tell me the velocity of the 2 kg and 1 kg masses when the 2 kg point mass has rolled to the bottom of the ring.
(Assume diameter=0.3m)
Initial energy = mgh = (2x9.8x0.3)+(1x9.8x0) = 5.88J
Energy of 2kg mass when at bottom = 0 (zero height and zero speed)
Energy of 1kg mass when at top = 5.88J (from conservation of energy)
5.88J = GPE(1kg mass) + KE(1kg mass) = (1x9.8x0.3)+(0.5mv^2) = (2.94 + 0.5x1xv^2)
v(1kg mass) = [(5.88-2.94)x2]^0.5 = 2.42ms^-1
velocity of ring = 1.21ms^-1
Any obvious errors?
Quiz for NOM; construct a .3 m ring that is strong and smooth but with low mass, and place it on a smooth hard surface. Attach a 2 kg point mass at 0° (on top) and a 1 kg point mass at 180° (on the bottom) around the ring. In relationship to the table top; tell me the velocity of the 2 kg and 1 kg masses when the 2 kg point mass has rolled to the bottom of the ring.
(Assume diameter=0.3m)
Initial energy = mgh = (2x9.8x0.3)+(1x9.8x0) = 5.88J
Energy of 2kg mass when at bottom = 0 (zero height and zero speed)
Energy of 1kg mass when at top = 5.88J (from conservation of energy)
5.88J = GPE(1kg mass) + KE(1kg mass) = (1x9.8x0.3)+(0.5mv^2) = (2.94 + 0.5x1xv^2)
v(1kg mass) = [(5.88-2.94)x2]^0.5 = 2.42ms^-1
velocity of ring = 1.21ms^-1
Any obvious errors?
I'm afraid I must agree with Limon on one point: the definition of a ballistic pendulum definitely does require the collision to be perfectly inelastic. This does not need to be stated explicitly, since it is incorporated into the definition.
The 5m long pendulum with the 5kg mass will have a velocity of sqrt(2*g*h) just before the collision. This can be calculated either from conversion of gravitational potential energy into kinetic energy, or directly from Newton's second law. The result comes out the same in either case, so conservation of energy is NOT required for this stage of the calculation. The numerical value of sqrt(2*g*h) is sqrt(2*9.81*0.5) = 3.132m/s
The impact conserves linear momentum because there are no outside forces in the horizontal direction at the moment of impact (both strings exert only net vertical forces). Therefore, it doesn't matter to the collision whether one or the other of the strings drops away upon impact. This matters only later, in guiding the subsequent motion. Since a 1kg mass strikes a 5kg mass and sticks to it, conservation of momentum gives
mv = (m+M)v', so v' = m/(m+M)*v = 1/6*3.132m/s = 0.522m/s.
Does that meet your needs?
--Stuart Anderson
The 5m long pendulum with the 5kg mass will have a velocity of sqrt(2*g*h) just before the collision. This can be calculated either from conversion of gravitational potential energy into kinetic energy, or directly from Newton's second law. The result comes out the same in either case, so conservation of energy is NOT required for this stage of the calculation. The numerical value of sqrt(2*g*h) is sqrt(2*9.81*0.5) = 3.132m/s
The impact conserves linear momentum because there are no outside forces in the horizontal direction at the moment of impact (both strings exert only net vertical forces). Therefore, it doesn't matter to the collision whether one or the other of the strings drops away upon impact. This matters only later, in guiding the subsequent motion. Since a 1kg mass strikes a 5kg mass and sticks to it, conservation of momentum gives
mv = (m+M)v', so v' = m/(m+M)*v = 1/6*3.132m/s = 0.522m/s.
Does that meet your needs?
--Stuart Anderson
QUOTE (bm1957+Sep 14 2007, 05:08 AM)
velocity of ring = 1.21ms^-1
Any obvious errors?
Looks right to me. Of course, Limon does not accept conservation of energy, so he may want a calculation using only Newton's second law. That calculation would involve finding the acceleration as a function of angle and present angular velocity, and then integrating. It would be very messy but would result of course in the same answer you gave.
QUOTE (Limon+Sep 13 2007, 03:10 PM)
NoCleverName check the literature, ballistic pendulums do not conserve Kinetic Energy, they conserve linear momentum. You had the initial velocity correct at 3.132 m/sec for an initial linear momentum of 1 * 3.132 = 3.132 divided by the new mass 3.312 / 6 = .5220 m/sec
I've responded to this question earlier. Below, I respond to each of your other points
This is simple to compute using conservation of energy, and bm1957 has already done so. I could also calculate this result using just Newton's second law, but it would be much longer. Since I know you have doubts about conservation of energy, it occurs to me that you may not accept an energy based solution to this problem. If you would like to see a purely Newtonian solution, I will post one, but as it would be a lot of work, I'll not do it unless asked.
This is simple to compute using conservation of energy, and bm1957 has already done so. I could also calculate this result using just Newton's second law, but it would be much longer. Since I know you have doubts about conservation of energy, it occurs to me that you may not accept an energy based solution to this problem. If you would like to see a purely Newtonian solution, I will post one, but as it would be a lot of work, I'll not do it unless asked.
To vector cancellation proponents: Buy four ballistic pendulums. Place all four facing north; operate them all simultaneously and calculate their total momentum.
If they are identical, each will have the same momentum P. Since they are aligned, the momenta will sum up to 4P.
Yes, that is exactly what I believe. "Believe" is not quite the right word, though. It should be "know." The issues here seem to be whether it is appropriate to treat momentum as a vector, and whether is is appropriate to combine the momenta of different objects. Is that a fair statement of your criticism?
If so, then let me give my answer here. I will give my answer to the second issue first, because it is more basic. Let me start by asking you a philosophical question: what is an object exactly? It seems to me that it is anything that shows a behavior that allows you to think of it as a single thing instead of as a bunch of separate parts. Consider a baseball. It is made of rubber, hide, horsehair, and thread, many parts. Yet when I throw it, I do not have to separately ask where the thread went and where the hide went; I can treat all those parts as a single object.
Next, consider the spray of shot pellets from a shotgun. The gun fires a cloud of tiny balls, but the whole cloud (although it spreads out as it travels) is fundamentally following the same trajectory. It has a definite total mass and the center of the cloud has a definite velocity, and you can treat the whole cloud of shot as a single object for purposes of calculating whether it will hit a target. This means that it is possible to treat multiple small parts, with no direct connection between them, as a single object. In other words, it is not necessary for the pieces to be actually attached to each other, as in the baseball, in order for them to be treated as a single object for purposes of calculation their motion.
To go the other direction with this reasoning, suppose you take a solid object, such as a brick, and throw it straight up with a speed of 10m/s. Now you repeat the experiment, only you first break the brick in half and place the two halves back together (without any glue) and then CAREFULLY through it upwards so that both halves remain touching and both go up at 10m/s. In each case, there is exactly the same amount of mass going at exactly the same speed in exactly the same direction, so there must be the same momentum. But the two halves of the brick are not attached in any way, so it is certainly possible to think of them separately and compute the momentum of each one. Each has mass m/2 and velocity v, so it has momentum mv/2, and the sum of the momenta of the two pieces is therefore mv/2 + mv/2 = mv, which is the momentum of the whole brick. You could do this with more than two pieces, of course, in fact, as many as you like. This proves that the momenta of several small pieces MUST sum up to the momentum of the entire object.
This should satisfy your first objection.
Now about momentum being a vector, there are two ways to see that this must be true. First, from Newton's second law, F=ma, and "a" by definition is dv/dt, so F = mdv/dt = d(mv)/dt =dp/dt. This means that if you start with an object at rest, and apply a net force F(t) to it for a time T, the momentum will by P = integral from 0 to t (F(t)dt). The F in Newton's second law is the VECTOR sum of all the forces acting on the object, and P is an integral (which is just a fancy sort of sum) of F. Therefore, since F is a vector, P must be a vector.
The second way is more fundamental. The definition of a vector, physically, is that it behaves like displacements. For instance, if I start from my home and go 100m north and then 100m east, I'll reach point A. Then from A, if I go 100m south and then 50m west, I will end up 50m east of my home. This is clear from drawing the path on a map, and also from the idea that adding the x components gives (+100) + (-50) = 50, and adding the y components gives (+100) + (-100) = 0. I have chosen to put x as east and y as north. It is clear that adding the components with proper + and - signs on them gives a correct prediction of where I'll end up. Therefore, displacements add as vectors.
Velocities are just the rates of change of displacements. Suppose point A is fixed, but points B and C are moving. At any given time, I can look at the displacement vector r_BA from A to B (that is, how many meters is B north of A and east of A, with south and west counted as negative). I can do the same for the displacement r_CB from B to C at the same instant in time. (I could take an aerial photograph to do this for instance.) Now I know that the vector r_BA + r_CB = r_CA, because I can see that this is true if I measure them on my photograph. This will be true at ANY time. If I take a new photograph a little later, I'll find B and C in different places than on the first photograph, but I can still measure the vectors and it will still be true that they add up just as before.
This is important, because if an equation is true at all times, then the rate of change of both sides must be equal. It's pretty obvious, because if the sides changed at different rates, they couldn't always stay equal to each other. But the rate of change of the position vector is the velocity, by definition, and so v_BA + v_CB = v_CA, which shows that velocities also behave as vectors. Since momentum is defined to me mass*velocity, it too must be a vector, since it is just a constant times something you already know is a vector. If velocity adds like a vector but momentum doesn't, then p=mv could not always be true, but it is the DEFINITION of momentum, so it cannot be false.
This should satisfy your second objection.
Yes, that is exactly what I believe. "Believe" is not quite the right word, though. It should be "know." The issues here seem to be whether it is appropriate to treat momentum as a vector, and whether is is appropriate to combine the momenta of different objects. Is that a fair statement of your criticism?
If so, then let me give my answer here. I will give my answer to the second issue first, because it is more basic. Let me start by asking you a philosophical question: what is an object exactly? It seems to me that it is anything that shows a behavior that allows you to think of it as a single thing instead of as a bunch of separate parts. Consider a baseball. It is made of rubber, hide, horsehair, and thread, many parts. Yet when I throw it, I do not have to separately ask where the thread went and where the hide went; I can treat all those parts as a single object.
Next, consider the spray of shot pellets from a shotgun. The gun fires a cloud of tiny balls, but the whole cloud (although it spreads out as it travels) is fundamentally following the same trajectory. It has a definite total mass and the center of the cloud has a definite velocity, and you can treat the whole cloud of shot as a single object for purposes of calculating whether it will hit a target. This means that it is possible to treat multiple small parts, with no direct connection between them, as a single object. In other words, it is not necessary for the pieces to be actually attached to each other, as in the baseball, in order for them to be treated as a single object for purposes of calculation their motion.
To go the other direction with this reasoning, suppose you take a solid object, such as a brick, and throw it straight up with a speed of 10m/s. Now you repeat the experiment, only you first break the brick in half and place the two halves back together (without any glue) and then CAREFULLY through it upwards so that both halves remain touching and both go up at 10m/s. In each case, there is exactly the same amount of mass going at exactly the same speed in exactly the same direction, so there must be the same momentum. But the two halves of the brick are not attached in any way, so it is certainly possible to think of them separately and compute the momentum of each one. Each has mass m/2 and velocity v, so it has momentum mv/2, and the sum of the momenta of the two pieces is therefore mv/2 + mv/2 = mv, which is the momentum of the whole brick. You could do this with more than two pieces, of course, in fact, as many as you like. This proves that the momenta of several small pieces MUST sum up to the momentum of the entire object.
This should satisfy your first objection.
Now about momentum being a vector, there are two ways to see that this must be true. First, from Newton's second law, F=ma, and "a" by definition is dv/dt, so F = mdv/dt = d(mv)/dt =dp/dt. This means that if you start with an object at rest, and apply a net force F(t) to it for a time T, the momentum will by P = integral from 0 to t (F(t)dt). The F in Newton's second law is the VECTOR sum of all the forces acting on the object, and P is an integral (which is just a fancy sort of sum) of F. Therefore, since F is a vector, P must be a vector.
The second way is more fundamental. The definition of a vector, physically, is that it behaves like displacements. For instance, if I start from my home and go 100m north and then 100m east, I'll reach point A. Then from A, if I go 100m south and then 50m west, I will end up 50m east of my home. This is clear from drawing the path on a map, and also from the idea that adding the x components gives (+100) + (-50) = 50, and adding the y components gives (+100) + (-100) = 0. I have chosen to put x as east and y as north. It is clear that adding the components with proper + and - signs on them gives a correct prediction of where I'll end up. Therefore, displacements add as vectors.
Velocities are just the rates of change of displacements. Suppose point A is fixed, but points B and C are moving. At any given time, I can look at the displacement vector r_BA from A to B (that is, how many meters is B north of A and east of A, with south and west counted as negative). I can do the same for the displacement r_CB from B to C at the same instant in time. (I could take an aerial photograph to do this for instance.) Now I know that the vector r_BA + r_CB = r_CA, because I can see that this is true if I measure them on my photograph. This will be true at ANY time. If I take a new photograph a little later, I'll find B and C in different places than on the first photograph, but I can still measure the vectors and it will still be true that they add up just as before.
This is important, because if an equation is true at all times, then the rate of change of both sides must be equal. It's pretty obvious, because if the sides changed at different rates, they couldn't always stay equal to each other. But the rate of change of the position vector is the velocity, by definition, and so v_BA + v_CB = v_CA, which shows that velocities also behave as vectors. Since momentum is defined to me mass*velocity, it too must be a vector, since it is just a constant times something you already know is a vector. If velocity adds like a vector but momentum doesn't, then p=mv could not always be true, but it is the DEFINITION of momentum, so it cannot be false.
This should satisfy your second objection.
Four independent quantities of momentum canceled each other for no apparent reason; you have got to be kidding. Operate three of the four and you would have more momentum?
That is correct, and I am not kidding. If the reason why they cancel out is not apparent to you, I'll try to make it apparent:
First, about the word "independent." I fully agree with you that if you regard the 4 pendula as 4 separate objects, then it makes no sense to add their momenta in the first place. That would be like saying 2 midgets who are each 1 meter tall should get together and join a basketball team, because together they are 2 meters tall. It would not make any sense.
However, as I mentioned above, it is possible to consider separate, unconnected objects as parts of a larger composite object, and to ask about the behavior of that large object. The universe does not contain objects, only lots and lots of particles that are attracting and repelling each other with various degrees of force. It is the human mind that cuts the universe up into objects, and I can cut it up any way I choose in order to study and analyze it. The fact that the universe allows us to do this, to make statements about large objects, instead of dealing with each particle individually, is the basic reason why classical physics works at all. If you could not do this, then to find out where a baseball would go when you threw it would require dealing with each particle in it separately, and you would have about 10^24 equations to solve in order to predict its motion. Obviously this is not true, and the reason is that you CAN lump all the parts of an object together into one whole.
The key to this is the concept of the center of mass. It is this concept that allows Newton's laws to apply to large objects instead of just to points. Most introductory physics classes do not emphasize this sufficiently, and some do not mention it at all, so I'll say it here: Newton's laws apply to POINTS, not to real-life large objects. Only by using the center of mass can you extend these laws to large objects. As an example, suppose I asked you to compute the speed of a rolling ball. The correct answer would be to tell me to shut up and stop talking nonsense. A rolling ball doesn't have a velocity. Different parts of it are moving at different speeds; the top is moving fast and the bottom is not moving at all at the instant it touches the ground. A ball has many different velocities at different points. However, the center point of the ball has a nice smooth linear motion, and that's the one that is meant. This is not the whole ball, though, it is the velocity of the center of mass point of the ball.
The calculation of the center of mass shows how this works. The center of mass position is simply the average of the position vectors of each gram of mass in the object. For example, if I have a dumbbell consisting of a mass m_1 at coordinates r_1 = (x_1, y_1, z_1) and a mass m_2 connected to it by a light rod at coordinates r_2=(x_2, y_2, z_2), then the center of mass location is (m_1*r_1 + m_2*r_2)/(m_1+m_2). This is simply the weighted average of the locations of the masses. Of course, for objects consisting of many masses (you could consider various weights hanging at many points on a pegboard, for instance), you average over all the masses, so that the center of mass coordinate vector is R = (sum of m_i*r_i)/(sum of m_i).
Therefore if you consider an object that is made up of smaller parts, each of which has its own little bit of mass and its own position, you can investigate what happens when these individual parts are moving. For instance, think of a swarm of bees flying along. The individual bees follow erratic, random paths, but the whole swarm of bees forms a cloud that stays together, even though the individual bees are not touching each other. What can you say about the whole cloud, supposing you know the precise flight path of each bee?
The location of the center of mass of the swarm is R = (sum of m_i*r_i)/(sum of m_i), so the velocity of this center point is
V =
dR/dt =
d{(sum of m_i*r_i)/(sum of m_i)}/dt = ..... by definition of R
1/(sum of m_i) * d(sum of m_i*r_i)/dt = ..... by factoring out the denominator
1/M * (sum of d(m_i*r_i)/dt) = ..... by naming the total mass M
1/M * (sum of m_i*(dr_i/dt)) = ..... by factoring m_i outside the rate of change
1/M * (sum of m_i*(v_i)) = ..... by definition of velocity
1/M * (sum of p_i)) = ..... by definition of momentum for each bee
P/M ..... by naming the total momentum of the swarm P.
This is a very important fact: The velocity of the center of mass, and the total momentum and mass of a set of moving particles, satisfy P = MV, which is identical to the equation p=mv satisfied by individual point masses. That means that the definition of momentum for point masses is also true for large objects, provided that you use the velocity of the center of mass point to describe the motion of the object. Everything then works just the same as with point masses, but you use the total momentum and total mass of the aggregate object.
Now starting from V = P/M, you could repeat the above calculation again, and everything world work exactly the same way, and you would get at the end
A = dV/dt = F/M. At the point in the calculation where you originally got (sum of m_i*(v_i)), you would get (sum of m_i*(a_i)) instead, because you have started from the velocity and taken one more derivative, so you are seeing acceleration. Then at the next step, instead of replacing m_i*v_i with p_i, you replace m_i*a_i with f_i. This means you use Newton's second law f_i = m_i*a_i on each bee separately. From there, the calculation follows through just like the first time, and you get F=MA, which is identical to the equation f=ma satisfied by individual point masses. That means that if Newton's second law is true for point masses, it is also true for large objects, provided that you use the acceleration of the center of mass point to describe the motion of the object. Everything then works just the same as with point masses, but you use the total force and total mass of the aggregate object.
Now with all this theory set up, it is possible to show why the total momentum of the 4 ballistic pendula is zero. First, collect all of them into a single object. (If this is troublesome, imagine that you have the bases of all 4 of them nailed down to a platform which is floating on a lake on a windless day. Then you are asking, "what is the total momentum of this object?") Let's put the 4 pendula facing outwards at the corners of a square, with their hanging block masses at the corners. Now where is the center of mass of the system? It is obviously at the exact center of the square, since all masses are equal and are symmetrically placed. Just after the collision, each of the 4 blocks is moving outwards at the same speed, and so a short time later, they are still at the corners of a square, but it is a larger square. However, it has expanded equally in all directions, and so the center of the square is exactly where it was before. The center of mass of the object consisting of all 4 ballistic pendula has therefore not moved. If it doesn't move, what is its velocity? Zero! Since the total momentum of the combined object is P =MV, and V=0, it follows that P = 0.
Now suppose you operate only 3 of the 4 pendula. Then 3 of the masses will move outward at a speed of 1m/s and one will stay still. Let's say the south mass stays still and the north, east, and west masses move outward. To make this concrete, let's say that the masses are all 1kg and initially have coordinates 1meter East = (1,0), 1meter West = (-1,0), 1meter North = (0,1), and 1meter South = (0,-1). Then the initial center of mass position is {1*(1,0) + 1*(-1,0) + 1*(0,1) + 1*(0,-1)}/(1+1+1+1) = (1-1+0+0, 0+0+1-1)/4 = (0,0)/4 = (0,0). If each mass except South moves out by 0.01m in 0.01 sec, then the new coordinates of the center of mass point are {1*(1.01,0) + 1*(-1.01,0) + 1*(0,1.01) + 1*(0,-1)}/(1+1+1+1) = (1.01-1.01+0+0, 0+0+1.01-1)/4 = (0,0.01)/4 = (0,0.0025). Therefore, the velocity of the center of mass is (0,0.0025)/0.01 = (0,0.25) = 0.25m/s North. Since the total mass of the combined object is 4kg, the momentum is P=MV = 4kg*(0,0.25m/s) = (0,1kgm/s) = 1kgm/s North. Compare this with what you get by doing it the easy way: The momenta of the three masses are 1kg*(1m/s,0) + 1kg*(-1m/s,0) + 1kg*(0,1m/s) + 1kg*(0,0) = (1-1,1+0) = (0,1) = 1kgm/s North. It is exactly the same either way, which shows that adding the individual momenta exactly predicts the correct motion of the center of mass point. It also shows that you have more momentum with 3 pendula moving than with all 4.
It is because the momenta DO cancel that Atwood's machine satisfies F=ma. To see this, apply F=ma to each of the parts of the system separately, and then to the system as a whole:
Let M be the large mass and m be the small mass. Let the pulley and string be massless and frictionless. Since the pulley is massless, no force is required to turn it, and so the tension on both parts of the string will be the same value, T. Now M will accelerate downward, so the weight Mg must be larger than the tension T of the string. Thus Mg - T = F_net = Ma (downwards). For the smaller mass, the acceleration will be upward, so T must exceed mg. Thus T-mg = F_net = ma (upwards). The upwards and downwards accelerations are equal in size, because the string does not stretch. The pulley does not move up or down, but has the string tension pulling down on it from both sides, so there must be a force N on the axis of the pulley to hold it up. Since the acceleration of the pulley in the vertical direction is zero, the net force on the pulley must be N - 2T = ma = 0, so that N=2T.
Adding the two mass equations together gives
Mg - T = Ma
T - mg = ma
---------------------
(M-m)g =(M+m)a.
Therefore, a = g*(M-m)/M+m). This means that after a time t has passed, M has a velocity t*a downward and therefore a downward momentum p_M = M*t*a=Mtg(M-m)/(M+m). The small mass m has an upward velocity v = t*a and an upward momentum p_m = m*t*a = mtg(M-m)/(M+m). Since the momenta are vectors and the downward momentum is greater (since M is a larger mass but both are moving at the same speed), therefore the net momentum is p_M - p_m downward. This is Mtg(M-m)/(M+m) - mtg(M-m)/(M+m) = tg((M-m)^2)/(M+m). Now since the total momentum is the total mass * center of mass velocity, you can get the velocity of the center of mass by dividing the momentum by (M+m). This gives V = {tg((M-m)^2)/(M+m)}/(M+m) = tg((M-m)/(M+m))^2 downward. Since the acceleration is constant, the velocity of the center of mass must be V = t*A, so all you need to do to find A is to divide V by t: A = g((M-m)/(M+m))^2 downward.
This is the acceleration that Newton's second law will apply to for the whole system. Therefore, it should check out that F_total = M_total*A. You've got A, and M_total is just M+m, so those are all known. What is F_total? For the whole system, it would be the sum of all the forces on all the parts: Two upward copies of T on the masses, two downward copies of T on the pulley, the two weights Mg and mg, both downward, and N upward. Since the acceleration A is known to be downward, let's count downward forces as positive. Therefore, F_total = T+T-T-T+Mg+mg-N = (M+m)g-N. This is good, but what is N? You know N= 2T from the pulley equation, but what is T? You can get that from one of the mass equations (it doesn't matter which, you'll get the same thing, since it was built into the equations that they have the same T anyway).
Let's use the equation for the small mass, and substitute the known acceleration:
T-mg=ma --->
T =
ma+mg = ..... by factoring out m
m(a+g) = ..... by substituting the formula for a
m(g*(M-m)/M+m) + g) = ..... by factoring out g
mg*((M-m)/M+m) + 1) = ..... by using a common denominator
mg*((M-m)/(M+m) + (M+m)/(M+m)) = ..... by summing numerators
mg*(2M/(M+m)) = ..... by factoring out 2M
2mMg/(M+m).
Therefore, N = 2T = 4mMg/(M+m). Now you can substitute this into the total force, to get:
F_total =
(M+m)g - N = ..... by substituting the formula for N
(M+m)g - 4mMg/(M+m) = ..... by factoring out g
g*((M+m) - 4mM/(M+m)) = ..... by using a common denominator
g*(((M+m)^2)/(M+m) - 4mM/(M+m)) = ..... by multiplying out the square
g*((M^2 + 2mM + m^2)/(M+m) - 4mM(M+n)) = ..... by summing numerators
g*(M^2 - 2mM + m^2)/(M+m) = ..... by collecting the square
g*((M-m)^2)/(M+m).
Now, what have you got?
F_total = g*((M-m)^2)/(M+m),
M_total = (M+m),
A = g((M-m)/(M+m))^2.
Multiplying M_total*A gives
M_total*A = ..... by substituting known formulas
(M+m)*g((M-m)/(M+m))^2 = ..... by canceling factors of M+m
g*((M-m)/^2)(M+m) = ..... by comparison with formula for F_total
F_total.
Therefore, if Newton's second law is applied to each part of Atwood's machine, and if the momenta of the moving masses are treated as vectors, so that they partially cancel, then Newton's second law is true for the Atwood's machine as a whole. This shows that, instead of the trouble you describe, where the (partial) cancellation of the momentum vectors due to their opposite directions violates Newton's second law, this vector cancellation is exactly what is needed to PRESERVE Newton's second law for the entire system.
It is because the momenta DO cancel that Atwood's machine satisfies F=ma. To see this, apply F=ma to each of the parts of the system separately, and then to the system as a whole:
Let M be the large mass and m be the small mass. Let the pulley and string be massless and frictionless. Since the pulley is massless, no force is required to turn it, and so the tension on both parts of the string will be the same value, T. Now M will accelerate downward, so the weight Mg must be larger than the tension T of the string. Thus Mg - T = F_net = Ma (downwards). For the smaller mass, the acceleration will be upward, so T must exceed mg. Thus T-mg = F_net = ma (upwards). The upwards and downwards accelerations are equal in size, because the string does not stretch. The pulley does not move up or down, but has the string tension pulling down on it from both sides, so there must be a force N on the axis of the pulley to hold it up. Since the acceleration of the pulley in the vertical direction is zero, the net force on the pulley must be N - 2T = ma = 0, so that N=2T.
Adding the two mass equations together gives
Mg - T = Ma
T - mg = ma
---------------------
(M-m)g =(M+m)a.
Therefore, a = g*(M-m)/M+m). This means that after a time t has passed, M has a velocity t*a downward and therefore a downward momentum p_M = M*t*a=Mtg(M-m)/(M+m). The small mass m has an upward velocity v = t*a and an upward momentum p_m = m*t*a = mtg(M-m)/(M+m). Since the momenta are vectors and the downward momentum is greater (since M is a larger mass but both are moving at the same speed), therefore the net momentum is p_M - p_m downward. This is Mtg(M-m)/(M+m) - mtg(M-m)/(M+m) = tg((M-m)^2)/(M+m). Now since the total momentum is the total mass * center of mass velocity, you can get the velocity of the center of mass by dividing the momentum by (M+m). This gives V = {tg((M-m)^2)/(M+m)}/(M+m) = tg((M-m)/(M+m))^2 downward. Since the acceleration is constant, the velocity of the center of mass must be V = t*A, so all you need to do to find A is to divide V by t: A = g((M-m)/(M+m))^2 downward.
This is the acceleration that Newton's second law will apply to for the whole system. Therefore, it should check out that F_total = M_total*A. You've got A, and M_total is just M+m, so those are all known. What is F_total? For the whole system, it would be the sum of all the forces on all the parts: Two upward copies of T on the masses, two downward copies of T on the pulley, the two weights Mg and mg, both downward, and N upward. Since the acceleration A is known to be downward, let's count downward forces as positive. Therefore, F_total = T+T-T-T+Mg+mg-N = (M+m)g-N. This is good, but what is N? You know N= 2T from the pulley equation, but what is T? You can get that from one of the mass equations (it doesn't matter which, you'll get the same thing, since it was built into the equations that they have the same T anyway).
Let's use the equation for the small mass, and substitute the known acceleration:
T-mg=ma --->
T =
ma+mg = ..... by factoring out m
m(a+g) = ..... by substituting the formula for a
m(g*(M-m)/M+m) + g) = ..... by factoring out g
mg*((M-m)/M+m) + 1) = ..... by using a common denominator
mg*((M-m)/(M+m) + (M+m)/(M+m)) = ..... by summing numerators
mg*(2M/(M+m)) = ..... by factoring out 2M
2mMg/(M+m).
Therefore, N = 2T = 4mMg/(M+m). Now you can substitute this into the total force, to get:
F_total =
(M+m)g - N = ..... by substituting the formula for N
(M+m)g - 4mMg/(M+m) = ..... by factoring out g
g*((M+m) - 4mM/(M+m)) = ..... by using a common denominator
g*(((M+m)^2)/(M+m) - 4mM/(M+m)) = ..... by multiplying out the square
g*((M^2 + 2mM + m^2)/(M+m) - 4mM(M+n)) = ..... by summing numerators
g*(M^2 - 2mM + m^2)/(M+m) = ..... by collecting the square
g*((M-m)^2)/(M+m).
Now, what have you got?
F_total = g*((M-m)^2)/(M+m),
M_total = (M+m),
A = g((M-m)/(M+m))^2.
Multiplying M_total*A gives
M_total*A = ..... by substituting known formulas
(M+m)*g((M-m)/(M+m))^2 = ..... by canceling factors of M+m
g*((M-m)/^2)(M+m) = ..... by comparison with formula for F_total
F_total.
Therefore, if Newton's second law is applied to each part of Atwood's machine, and if the momenta of the moving masses are treated as vectors, so that they partially cancel, then Newton's second law is true for the Atwood's machine as a whole. This shows that, instead of the trouble you describe, where the (partial) cancellation of the momentum vectors due to their opposite directions violates Newton's second law, this vector cancellation is exactly what is needed to PRESERVE Newton's second law for the entire system.
Vectors are for unbalanced forces that are being applied to each other, if the forces are not being applied to one another or are balanced then the concept of vector addition or cancellation does not apply. A balanced spinning wheel has no unbalanced forces working between the masses; they are all happily minding their own business.
I have already talked about the idea that you can consider several separate objects as part of one large object. In that case, forces that are on these separate pieces must be included together as acting on the aggregate object in order to get correct experimental predictions (as with my swarm of bees example). Thus the concept of vector addition applies whenever the objects are being treated as part of a larger object, and does not apply when they are being treated separately.
When you say that if the forces are balanced, then vector addition or cancellation does not apply, I have trouble making sense out of this statement. If the forces are balanced, that means that they push on the same object with equal strengths in opposite directions, right? In that case they are supposed to have zero net effect on the motion of the object. But if the forces are represented by equal and opposite vectors, then these vectors will add up to zero by vector cancellation, which exactly AGREES with the idea that the forces balance out and have no net effect. So in this case, vector cancellation is giving you exactly the result that you expect and know is correct, so where is the problem with it?
About the balanced spinning wheel: If there are no unbalanced forces between the mases, then the following experiment should work. Take a metal wheel and with a chisel, cut a small chip out of the surface. Now carefully straighten out the chip if the chisel bent it, so that it is exactly the same size and shape as the hole it left behind. Place it back into that hole, but do not fasten it in place. The wheel now looks exactly the same as before, with all the same mass in all the same locations. Now spin the wheel. According to your ideas, the chip should stay in place, and not fall off the wheel, no matter how fast the wheel spins. After all, this is the same bit of mass, executing the same exact motion as before. The only difference is that it is not firmly held to the rest of the wheel now. If there were NO unbalanced forces between the masses of the wheel, then there cannot have been any forces holding the chip in place, and therefore cutting it loose must not remove any forces that were needed. Therefore it should stay in place.
On the other hand, if it flies off the wheel, then its behavior is clearly different from what it was before, and so the absence of attachment does make a difference. In that case, the attachment to the wheel must have been necessary. According to Newton's second law, the ONLY thing that affects the motion of a mass is the net force on it. If the motion is different, the net force on the chip must therefore be different. But in the new situation, the net force is clearly zero, because the chip has been cut free of the wheel. If the forces are different, and the new net force is zero, then the old net force must have been different from zero. Therefore, there was a nonzero net force on the chip in the original situation. Therefore, the chip was accelerating, even though the wheel was smoothly turning at a constant speed.
That is not true. The earth's gravity pulls the moon toward the earth. There is no opposing force, and therefore the forces are not balanced. The moon IS accelerating toward the earth. If you needed an opposing force, there would have to be some kind of tower holding the moon up. There obviously is no other force on it besides gravity, which is entirely consistent with Newton's point of view. In fact, the acceleration of the moon toward the earth is one of the fundamental pieces of evidence that Newton presented in favor of his theory of gravitation. If the moon is not accelerating toward the earth, then Newton was simply wrong about gravity.
That is not true. The earth's gravity pulls the moon toward the earth. There is no opposing force, and therefore the forces are not balanced. The moon IS accelerating toward the earth. If you needed an opposing force, there would have to be some kind of tower holding the moon up. There obviously is no other force on it besides gravity, which is entirely consistent with Newton's point of view. In fact, the acceleration of the moon toward the earth is one of the fundamental pieces of evidence that Newton presented in favor of his theory of gravitation. If the moon is not accelerating toward the earth, then Newton was simply wrong about gravity.
Here is an experiment I conducted once. I fixed a string to an immovable point used for rotation just off of an air table, the other end of the string I affixed to a 84g puck on the air table. I tied a second string to the 84g puck and attached a 32g puck to the other end. I rotated both pucks, trailing the second puck behind the larger puck at 90° to the first string. The 90° position was fixed until release. Upon release the 32 g puck swung out beyond the 84g puck and stopped it, exactly as the spheres stop the cylinder in the cylinder and spheres experiment. The data was similar as well, showing linear momentum conservation.
In the 84 and 32 gram puck experiment there is no opportunity to cancel the motion from the other side using vector cancellation, because there is no other side. If you use linear momentum conservation the results are the same as the cylinder and spheres experiment, the results are different with vector cancellation. It seems like you would want a theory that work consistently like Newton’s does. Besides what good is you vector cancellation concept that tells you nothing; what good is a world of zeros.
I'm not sure I fully see the details of this experiment. Do you have a fuller explanation of it available somewhere?
Anyway, I hope my comments have been useful to you.
--Stuart Anderson
I've responded to this question earlier. Below, I respond to each of your other points
QUOTE
Quiz for NOM; construct a .3 m ring that is strong and smooth but with low mass, and place it on a smooth hard surface. Attach a 2 kg point mass at 0° (on top) and a 1 kg point mass at 180° (on the bottom) around the ring. In relationship to the table top; tell me the velocity of the 2 kg and 1 kg masses when the 2 kg point mass has rolled to the bottom of the ring.
This is simple to compute using conservation of energy, and bm1957 has already done so. I could also calculate this result using just Newton's second law, but it would be much longer. Since I know you have doubts about conservation of energy, it occurs to me that you may not accept an energy based solution to this problem. If you would like to see a purely Newtonian solution, I will post one, but as it would be a lot of work, I'll not do it unless asked.
QUOTE (->
| QUOTE |
| Quiz for NOM; construct a .3 m ring that is strong and smooth but with low mass, and place it on a smooth hard surface. Attach a 2 kg point mass at 0° (on top) and a 1 kg point mass at 180° (on the bottom) around the ring. In relationship to the table top; tell me the velocity of the 2 kg and 1 kg masses when the 2 kg point mass has rolled to the bottom of the ring. |
This is simple to compute using conservation of energy, and bm1957 has already done so. I could also calculate this result using just Newton's second law, but it would be much longer. Since I know you have doubts about conservation of energy, it occurs to me that you may not accept an energy based solution to this problem. If you would like to see a purely Newtonian solution, I will post one, but as it would be a lot of work, I'll not do it unless asked.
To vector cancellation proponents: Buy four ballistic pendulums. Place all four facing north; operate them all simultaneously and calculate their total momentum.
If they are identical, each will have the same momentum P. Since they are aligned, the momenta will sum up to 4P.
QUOTE
Next face one north; face one south, one east and one west. Operate them all simultaneously and again calculate their linear momentum. You can’t possibly believe it is now zero, do you?
Yes, that is exactly what I believe. "Believe" is not quite the right word, though. It should be "know." The issues here seem to be whether it is appropriate to treat momentum as a vector, and whether is is appropriate to combine the momenta of different objects. Is that a fair statement of your criticism?
If so, then let me give my answer here. I will give my answer to the second issue first, because it is more basic. Let me start by asking you a philosophical question: what is an object exactly? It seems to me that it is anything that shows a behavior that allows you to think of it as a single thing instead of as a bunch of separate parts. Consider a baseball. It is made of rubber, hide, horsehair, and thread, many parts. Yet when I throw it, I do not have to separately ask where the thread went and where the hide went; I can treat all those parts as a single object.
Next, consider the spray of shot pellets from a shotgun. The gun fires a cloud of tiny balls, but the whole cloud (although it spreads out as it travels) is fundamentally following the same trajectory. It has a definite total mass and the center of the cloud has a definite velocity, and you can treat the whole cloud of shot as a single object for purposes of calculating whether it will hit a target. This means that it is possible to treat multiple small parts, with no direct connection between them, as a single object. In other words, it is not necessary for the pieces to be actually attached to each other, as in the baseball, in order for them to be treated as a single object for purposes of calculation their motion.
To go the other direction with this reasoning, suppose you take a solid object, such as a brick, and throw it straight up with a speed of 10m/s. Now you repeat the experiment, only you first break the brick in half and place the two halves back together (without any glue) and then CAREFULLY through it upwards so that both halves remain touching and both go up at 10m/s. In each case, there is exactly the same amount of mass going at exactly the same speed in exactly the same direction, so there must be the same momentum. But the two halves of the brick are not attached in any way, so it is certainly possible to think of them separately and compute the momentum of each one. Each has mass m/2 and velocity v, so it has momentum mv/2, and the sum of the momenta of the two pieces is therefore mv/2 + mv/2 = mv, which is the momentum of the whole brick. You could do this with more than two pieces, of course, in fact, as many as you like. This proves that the momenta of several small pieces MUST sum up to the momentum of the entire object.
This should satisfy your first objection.
Now about momentum being a vector, there are two ways to see that this must be true. First, from Newton's second law, F=ma, and "a" by definition is dv/dt, so F = mdv/dt = d(mv)/dt =dp/dt. This means that if you start with an object at rest, and apply a net force F(t) to it for a time T, the momentum will by P = integral from 0 to t (F(t)dt). The F in Newton's second law is the VECTOR sum of all the forces acting on the object, and P is an integral (which is just a fancy sort of sum) of F. Therefore, since F is a vector, P must be a vector.
The second way is more fundamental. The definition of a vector, physically, is that it behaves like displacements. For instance, if I start from my home and go 100m north and then 100m east, I'll reach point A. Then from A, if I go 100m south and then 50m west, I will end up 50m east of my home. This is clear from drawing the path on a map, and also from the idea that adding the x components gives (+100) + (-50) = 50, and adding the y components gives (+100) + (-100) = 0. I have chosen to put x as east and y as north. It is clear that adding the components with proper + and - signs on them gives a correct prediction of where I'll end up. Therefore, displacements add as vectors.
Velocities are just the rates of change of displacements. Suppose point A is fixed, but points B and C are moving. At any given time, I can look at the displacement vector r_BA from A to B (that is, how many meters is B north of A and east of A, with south and west counted as negative). I can do the same for the displacement r_CB from B to C at the same instant in time. (I could take an aerial photograph to do this for instance.) Now I know that the vector r_BA + r_CB = r_CA, because I can see that this is true if I measure them on my photograph. This will be true at ANY time. If I take a new photograph a little later, I'll find B and C in different places than on the first photograph, but I can still measure the vectors and it will still be true that they add up just as before.
This is important, because if an equation is true at all times, then the rate of change of both sides must be equal. It's pretty obvious, because if the sides changed at different rates, they couldn't always stay equal to each other. But the rate of change of the position vector is the velocity, by definition, and so v_BA + v_CB = v_CA, which shows that velocities also behave as vectors. Since momentum is defined to me mass*velocity, it too must be a vector, since it is just a constant times something you already know is a vector. If velocity adds like a vector but momentum doesn't, then p=mv could not always be true, but it is the DEFINITION of momentum, so it cannot be false.
This should satisfy your second objection.
QUOTE (->
| QUOTE |
| Next face one north; face one south, one east and one west. Operate them all simultaneously and again calculate their linear momentum. You can’t possibly believe it is now zero, do you? |
Yes, that is exactly what I believe. "Believe" is not quite the right word, though. It should be "know." The issues here seem to be whether it is appropriate to treat momentum as a vector, and whether is is appropriate to combine the momenta of different objects. Is that a fair statement of your criticism?
If so, then let me give my answer here. I will give my answer to the second issue first, because it is more basic. Let me start by asking you a philosophical question: what is an object exactly? It seems to me that it is anything that shows a behavior that allows you to think of it as a single thing instead of as a bunch of separate parts. Consider a baseball. It is made of rubber, hide, horsehair, and thread, many parts. Yet when I throw it, I do not have to separately ask where the thread went and where the hide went; I can treat all those parts as a single object.
Next, consider the spray of shot pellets from a shotgun. The gun fires a cloud of tiny balls, but the whole cloud (although it spreads out as it travels) is fundamentally following the same trajectory. It has a definite total mass and the center of the cloud has a definite velocity, and you can treat the whole cloud of shot as a single object for purposes of calculating whether it will hit a target. This means that it is possible to treat multiple small parts, with no direct connection between them, as a single object. In other words, it is not necessary for the pieces to be actually attached to each other, as in the baseball, in order for them to be treated as a single object for purposes of calculation their motion.
To go the other direction with this reasoning, suppose you take a solid object, such as a brick, and throw it straight up with a speed of 10m/s. Now you repeat the experiment, only you first break the brick in half and place the two halves back together (without any glue) and then CAREFULLY through it upwards so that both halves remain touching and both go up at 10m/s. In each case, there is exactly the same amount of mass going at exactly the same speed in exactly the same direction, so there must be the same momentum. But the two halves of the brick are not attached in any way, so it is certainly possible to think of them separately and compute the momentum of each one. Each has mass m/2 and velocity v, so it has momentum mv/2, and the sum of the momenta of the two pieces is therefore mv/2 + mv/2 = mv, which is the momentum of the whole brick. You could do this with more than two pieces, of course, in fact, as many as you like. This proves that the momenta of several small pieces MUST sum up to the momentum of the entire object.
This should satisfy your first objection.
Now about momentum being a vector, there are two ways to see that this must be true. First, from Newton's second law, F=ma, and "a" by definition is dv/dt, so F = mdv/dt = d(mv)/dt =dp/dt. This means that if you start with an object at rest, and apply a net force F(t) to it for a time T, the momentum will by P = integral from 0 to t (F(t)dt). The F in Newton's second law is the VECTOR sum of all the forces acting on the object, and P is an integral (which is just a fancy sort of sum) of F. Therefore, since F is a vector, P must be a vector.
The second way is more fundamental. The definition of a vector, physically, is that it behaves like displacements. For instance, if I start from my home and go 100m north and then 100m east, I'll reach point A. Then from A, if I go 100m south and then 50m west, I will end up 50m east of my home. This is clear from drawing the path on a map, and also from the idea that adding the x components gives (+100) + (-50) = 50, and adding the y components gives (+100) + (-100) = 0. I have chosen to put x as east and y as north. It is clear that adding the components with proper + and - signs on them gives a correct prediction of where I'll end up. Therefore, displacements add as vectors.
Velocities are just the rates of change of displacements. Suppose point A is fixed, but points B and C are moving. At any given time, I can look at the displacement vector r_BA from A to B (that is, how many meters is B north of A and east of A, with south and west counted as negative). I can do the same for the displacement r_CB from B to C at the same instant in time. (I could take an aerial photograph to do this for instance.) Now I know that the vector r_BA + r_CB = r_CA, because I can see that this is true if I measure them on my photograph. This will be true at ANY time. If I take a new photograph a little later, I'll find B and C in different places than on the first photograph, but I can still measure the vectors and it will still be true that they add up just as before.
This is important, because if an equation is true at all times, then the rate of change of both sides must be equal. It's pretty obvious, because if the sides changed at different rates, they couldn't always stay equal to each other. But the rate of change of the position vector is the velocity, by definition, and so v_BA + v_CB = v_CA, which shows that velocities also behave as vectors. Since momentum is defined to me mass*velocity, it too must be a vector, since it is just a constant times something you already know is a vector. If velocity adds like a vector but momentum doesn't, then p=mv could not always be true, but it is the DEFINITION of momentum, so it cannot be false.
This should satisfy your second objection.
Four independent quantities of momentum canceled each other for no apparent reason; you have got to be kidding. Operate three of the four and you would have more momentum?
That is correct, and I am not kidding. If the reason why they cancel out is not apparent to you, I'll try to make it apparent:
First, about the word "independent." I fully agree with you that if you regard the 4 pendula as 4 separate objects, then it makes no sense to add their momenta in the first place. That would be like saying 2 midgets who are each 1 meter tall should get together and join a basketball team, because together they are 2 meters tall. It would not make any sense.
However, as I mentioned above, it is possible to consider separate, unconnected objects as parts of a larger composite object, and to ask about the behavior of that large object. The universe does not contain objects, only lots and lots of particles that are attracting and repelling each other with various degrees of force. It is the human mind that cuts the universe up into objects, and I can cut it up any way I choose in order to study and analyze it. The fact that the universe allows us to do this, to make statements about large objects, instead of dealing with each particle individually, is the basic reason why classical physics works at all. If you could not do this, then to find out where a baseball would go when you threw it would require dealing with each particle in it separately, and you would have about 10^24 equations to solve in order to predict its motion. Obviously this is not true, and the reason is that you CAN lump all the parts of an object together into one whole.
The key to this is the concept of the center of mass. It is this concept that allows Newton's laws to apply to large objects instead of just to points. Most introductory physics classes do not emphasize this sufficiently, and some do not mention it at all, so I'll say it here: Newton's laws apply to POINTS, not to real-life large objects. Only by using the center of mass can you extend these laws to large objects. As an example, suppose I asked you to compute the speed of a rolling ball. The correct answer would be to tell me to shut up and stop talking nonsense. A rolling ball doesn't have a velocity. Different parts of it are moving at different speeds; the top is moving fast and the bottom is not moving at all at the instant it touches the ground. A ball has many different velocities at different points. However, the center point of the ball has a nice smooth linear motion, and that's the one that is meant. This is not the whole ball, though, it is the velocity of the center of mass point of the ball.
The calculation of the center of mass shows how this works. The center of mass position is simply the average of the position vectors of each gram of mass in the object. For example, if I have a dumbbell consisting of a mass m_1 at coordinates r_1 = (x_1, y_1, z_1) and a mass m_2 connected to it by a light rod at coordinates r_2=(x_2, y_2, z_2), then the center of mass location is (m_1*r_1 + m_2*r_2)/(m_1+m_2). This is simply the weighted average of the locations of the masses. Of course, for objects consisting of many masses (you could consider various weights hanging at many points on a pegboard, for instance), you average over all the masses, so that the center of mass coordinate vector is R = (sum of m_i*r_i)/(sum of m_i).
Therefore if you consider an object that is made up of smaller parts, each of which has its own little bit of mass and its own position, you can investigate what happens when these individual parts are moving. For instance, think of a swarm of bees flying along. The individual bees follow erratic, random paths, but the whole swarm of bees forms a cloud that stays together, even though the individual bees are not touching each other. What can you say about the whole cloud, supposing you know the precise flight path of each bee?
The location of the center of mass of the swarm is R = (sum of m_i*r_i)/(sum of m_i), so the velocity of this center point is
V =
dR/dt =
d{(sum of m_i*r_i)/(sum of m_i)}/dt = ..... by definition of R
1/(sum of m_i) * d(sum of m_i*r_i)/dt = ..... by factoring out the denominator
1/M * (sum of d(m_i*r_i)/dt) = ..... by naming the total mass M
1/M * (sum of m_i*(dr_i/dt)) = ..... by factoring m_i outside the rate of change
1/M * (sum of m_i*(v_i)) = ..... by definition of velocity
1/M * (sum of p_i)) = ..... by definition of momentum for each bee
P/M ..... by naming the total momentum of the swarm P.
This is a very important fact: The velocity of the center of mass, and the total momentum and mass of a set of moving particles, satisfy P = MV, which is identical to the equation p=mv satisfied by individual point masses. That means that the definition of momentum for point masses is also true for large objects, provided that you use the velocity of the center of mass point to describe the motion of the object. Everything then works just the same as with point masses, but you use the total momentum and total mass of the aggregate object.
Now starting from V = P/M, you could repeat the above calculation again, and everything world work exactly the same way, and you would get at the end
A = dV/dt = F/M. At the point in the calculation where you originally got (sum of m_i*(v_i)), you would get (sum of m_i*(a_i)) instead, because you have started from the velocity and taken one more derivative, so you are seeing acceleration. Then at the next step, instead of replacing m_i*v_i with p_i, you replace m_i*a_i with f_i. This means you use Newton's second law f_i = m_i*a_i on each bee separately. From there, the calculation follows through just like the first time, and you get F=MA, which is identical to the equation f=ma satisfied by individual point masses. That means that if Newton's second law is true for point masses, it is also true for large objects, provided that you use the acceleration of the center of mass point to describe the motion of the object. Everything then works just the same as with point masses, but you use the total force and total mass of the aggregate object.
Now with all this theory set up, it is possible to show why the total momentum of the 4 ballistic pendula is zero. First, collect all of them into a single object. (If this is troublesome, imagine that you have the bases of all 4 of them nailed down to a platform which is floating on a lake on a windless day. Then you are asking, "what is the total momentum of this object?") Let's put the 4 pendula facing outwards at the corners of a square, with their hanging block masses at the corners. Now where is the center of mass of the system? It is obviously at the exact center of the square, since all masses are equal and are symmetrically placed. Just after the collision, each of the 4 blocks is moving outwards at the same speed, and so a short time later, they are still at the corners of a square, but it is a larger square. However, it has expanded equally in all directions, and so the center of the square is exactly where it was before. The center of mass of the object consisting of all 4 ballistic pendula has therefore not moved. If it doesn't move, what is its velocity? Zero! Since the total momentum of the combined object is P =MV, and V=0, it follows that P = 0.
Now suppose you operate only 3 of the 4 pendula. Then 3 of the masses will move outward at a speed of 1m/s and one will stay still. Let's say the south mass stays still and the north, east, and west masses move outward. To make this concrete, let's say that the masses are all 1kg and initially have coordinates 1meter East = (1,0), 1meter West = (-1,0), 1meter North = (0,1), and 1meter South = (0,-1). Then the initial center of mass position is {1*(1,0) + 1*(-1,0) + 1*(0,1) + 1*(0,-1)}/(1+1+1+1) = (1-1+0+0, 0+0+1-1)/4 = (0,0)/4 = (0,0). If each mass except South moves out by 0.01m in 0.01 sec, then the new coordinates of the center of mass point are {1*(1.01,0) + 1*(-1.01,0) + 1*(0,1.01) + 1*(0,-1)}/(1+1+1+1) = (1.01-1.01+0+0, 0+0+1.01-1)/4 = (0,0.01)/4 = (0,0.0025). Therefore, the velocity of the center of mass is (0,0.0025)/0.01 = (0,0.25) = 0.25m/s North. Since the total mass of the combined object is 4kg, the momentum is P=MV = 4kg*(0,0.25m/s) = (0,1kgm/s) = 1kgm/s North. Compare this with what you get by doing it the easy way: The momenta of the three masses are 1kg*(1m/s,0) + 1kg*(-1m/s,0) + 1kg*(0,1m/s) + 1kg*(0,0) = (1-1,1+0) = (0,1) = 1kgm/s North. It is exactly the same either way, which shows that adding the individual momenta exactly predicts the correct motion of the center of mass point. It also shows that you have more momentum with 3 pendula moving than with all 4.
QUOTE
Atwood’s machines have nearly half their mass moving up and half the mass moving down. If these momentums canceled F would not equal ma, and that is what an Atwood’s proves.
It is because the momenta DO cancel that Atwood's machine satisfies F=ma. To see this, apply F=ma to each of the parts of the system separately, and then to the system as a whole:
Let M be the large mass and m be the small mass. Let the pulley and string be massless and frictionless. Since the pulley is massless, no force is required to turn it, and so the tension on both parts of the string will be the same value, T. Now M will accelerate downward, so the weight Mg must be larger than the tension T of the string. Thus Mg - T = F_net = Ma (downwards). For the smaller mass, the acceleration will be upward, so T must exceed mg. Thus T-mg = F_net = ma (upwards). The upwards and downwards accelerations are equal in size, because the string does not stretch. The pulley does not move up or down, but has the string tension pulling down on it from both sides, so there must be a force N on the axis of the pulley to hold it up. Since the acceleration of the pulley in the vertical direction is zero, the net force on the pulley must be N - 2T = ma = 0, so that N=2T.
Adding the two mass equations together gives
Mg - T = Ma
T - mg = ma
---------------------
(M-m)g =(M+m)a.
Therefore, a = g*(M-m)/M+m). This means that after a time t has passed, M has a velocity t*a downward and therefore a downward momentum p_M = M*t*a=Mtg(M-m)/(M+m). The small mass m has an upward velocity v = t*a and an upward momentum p_m = m*t*a = mtg(M-m)/(M+m). Since the momenta are vectors and the downward momentum is greater (since M is a larger mass but both are moving at the same speed), therefore the net momentum is p_M - p_m downward. This is Mtg(M-m)/(M+m) - mtg(M-m)/(M+m) = tg((M-m)^2)/(M+m). Now since the total momentum is the total mass * center of mass velocity, you can get the velocity of the center of mass by dividing the momentum by (M+m). This gives V = {tg((M-m)^2)/(M+m)}/(M+m) = tg((M-m)/(M+m))^2 downward. Since the acceleration is constant, the velocity of the center of mass must be V = t*A, so all you need to do to find A is to divide V by t: A = g((M-m)/(M+m))^2 downward.
This is the acceleration that Newton's second law will apply to for the whole system. Therefore, it should check out that F_total = M_total*A. You've got A, and M_total is just M+m, so those are all known. What is F_total? For the whole system, it would be the sum of all the forces on all the parts: Two upward copies of T on the masses, two downward copies of T on the pulley, the two weights Mg and mg, both downward, and N upward. Since the acceleration A is known to be downward, let's count downward forces as positive. Therefore, F_total = T+T-T-T+Mg+mg-N = (M+m)g-N. This is good, but what is N? You know N= 2T from the pulley equation, but what is T? You can get that from one of the mass equations (it doesn't matter which, you'll get the same thing, since it was built into the equations that they have the same T anyway).
Let's use the equation for the small mass, and substitute the known acceleration:
T-mg=ma --->
T =
ma+mg = ..... by factoring out m
m(a+g) = ..... by substituting the formula for a
m(g*(M-m)/M+m) + g) = ..... by factoring out g
mg*((M-m)/M+m) + 1) = ..... by using a common denominator
mg*((M-m)/(M+m) + (M+m)/(M+m)) = ..... by summing numerators
mg*(2M/(M+m)) = ..... by factoring out 2M
2mMg/(M+m).
Therefore, N = 2T = 4mMg/(M+m). Now you can substitute this into the total force, to get:
F_total =
(M+m)g - N = ..... by substituting the formula for N
(M+m)g - 4mMg/(M+m) = ..... by factoring out g
g*((M+m) - 4mM/(M+m)) = ..... by using a common denominator
g*(((M+m)^2)/(M+m) - 4mM/(M+m)) = ..... by multiplying out the square
g*((M^2 + 2mM + m^2)/(M+m) - 4mM(M+n)) = ..... by summing numerators
g*(M^2 - 2mM + m^2)/(M+m) = ..... by collecting the square
g*((M-m)^2)/(M+m).
Now, what have you got?
F_total = g*((M-m)^2)/(M+m),
M_total = (M+m),
A = g((M-m)/(M+m))^2.
Multiplying M_total*A gives
M_total*A = ..... by substituting known formulas
(M+m)*g((M-m)/(M+m))^2 = ..... by canceling factors of M+m
g*((M-m)/^2)(M+m) = ..... by comparison with formula for F_total
F_total.
Therefore, if Newton's second law is applied to each part of Atwood's machine, and if the momenta of the moving masses are treated as vectors, so that they partially cancel, then Newton's second law is true for the Atwood's machine as a whole. This shows that, instead of the trouble you describe, where the (partial) cancellation of the momentum vectors due to their opposite directions violates Newton's second law, this vector cancellation is exactly what is needed to PRESERVE Newton's second law for the entire system.
QUOTE (->
| QUOTE |
| Atwood’s machines have nearly half their mass moving up and half the mass moving down. If these momentums canceled F would not equal ma, and that is what an Atwood’s proves. |
It is because the momenta DO cancel that Atwood's machine satisfies F=ma. To see this, apply F=ma to each of the parts of the system separately, and then to the system as a whole:
Let M be the large mass and m be the small mass. Let the pulley and string be massless and frictionless. Since the pulley is massless, no force is required to turn it, and so the tension on both parts of the string will be the same value, T. Now M will accelerate downward, so the weight Mg must be larger than the tension T of the string. Thus Mg - T = F_net = Ma (downwards). For the smaller mass, the acceleration will be upward, so T must exceed mg. Thus T-mg = F_net = ma (upwards). The upwards and downwards accelerations are equal in size, because the string does not stretch. The pulley does not move up or down, but has the string tension pulling down on it from both sides, so there must be a force N on the axis of the pulley to hold it up. Since the acceleration of the pulley in the vertical direction is zero, the net force on the pulley must be N - 2T = ma = 0, so that N=2T.
Adding the two mass equations together gives
Mg - T = Ma
T - mg = ma
---------------------
(M-m)g =(M+m)a.
Therefore, a = g*(M-m)/M+m). This means that after a time t has passed, M has a velocity t*a downward and therefore a downward momentum p_M = M*t*a=Mtg(M-m)/(M+m). The small mass m has an upward velocity v = t*a and an upward momentum p_m = m*t*a = mtg(M-m)/(M+m). Since the momenta are vectors and the downward momentum is greater (since M is a larger mass but both are moving at the same speed), therefore the net momentum is p_M - p_m downward. This is Mtg(M-m)/(M+m) - mtg(M-m)/(M+m) = tg((M-m)^2)/(M+m). Now since the total momentum is the total mass * center of mass velocity, you can get the velocity of the center of mass by dividing the momentum by (M+m). This gives V = {tg((M-m)^2)/(M+m)}/(M+m) = tg((M-m)/(M+m))^2 downward. Since the acceleration is constant, the velocity of the center of mass must be V = t*A, so all you need to do to find A is to divide V by t: A = g((M-m)/(M+m))^2 downward.
This is the acceleration that Newton's second law will apply to for the whole system. Therefore, it should check out that F_total = M_total*A. You've got A, and M_total is just M+m, so those are all known. What is F_total? For the whole system, it would be the sum of all the forces on all the parts: Two upward copies of T on the masses, two downward copies of T on the pulley, the two weights Mg and mg, both downward, and N upward. Since the acceleration A is known to be downward, let's count downward forces as positive. Therefore, F_total = T+T-T-T+Mg+mg-N = (M+m)g-N. This is good, but what is N? You know N= 2T from the pulley equation, but what is T? You can get that from one of the mass equations (it doesn't matter which, you'll get the same thing, since it was built into the equations that they have the same T anyway).
Let's use the equation for the small mass, and substitute the known acceleration:
T-mg=ma --->
T =
ma+mg = ..... by factoring out m
m(a+g) = ..... by substituting the formula for a
m(g*(M-m)/M+m) + g) = ..... by factoring out g
mg*((M-m)/M+m) + 1) = ..... by using a common denominator
mg*((M-m)/(M+m) + (M+m)/(M+m)) = ..... by summing numerators
mg*(2M/(M+m)) = ..... by factoring out 2M
2mMg/(M+m).
Therefore, N = 2T = 4mMg/(M+m). Now you can substitute this into the total force, to get:
F_total =
(M+m)g - N = ..... by substituting the formula for N
(M+m)g - 4mMg/(M+m) = ..... by factoring out g
g*((M+m) - 4mM/(M+m)) = ..... by using a common denominator
g*(((M+m)^2)/(M+m) - 4mM/(M+m)) = ..... by multiplying out the square
g*((M^2 + 2mM + m^2)/(M+m) - 4mM(M+n)) = ..... by summing numerators
g*(M^2 - 2mM + m^2)/(M+m) = ..... by collecting the square
g*((M-m)^2)/(M+m).
Now, what have you got?
F_total = g*((M-m)^2)/(M+m),
M_total = (M+m),
A = g((M-m)/(M+m))^2.
Multiplying M_total*A gives
M_total*A = ..... by substituting known formulas
(M+m)*g((M-m)/(M+m))^2 = ..... by canceling factors of M+m
g*((M-m)/^2)(M+m) = ..... by comparison with formula for F_total
F_total.
Therefore, if Newton's second law is applied to each part of Atwood's machine, and if the momenta of the moving masses are treated as vectors, so that they partially cancel, then Newton's second law is true for the Atwood's machine as a whole. This shows that, instead of the trouble you describe, where the (partial) cancellation of the momentum vectors due to their opposite directions violates Newton's second law, this vector cancellation is exactly what is needed to PRESERVE Newton's second law for the entire system.
Vectors are for unbalanced forces that are being applied to each other, if the forces are not being applied to one another or are balanced then the concept of vector addition or cancellation does not apply. A balanced spinning wheel has no unbalanced forces working between the masses; they are all happily minding their own business.
I have already talked about the idea that you can consider several separate objects as part of one large object. In that case, forces that are on these separate pieces must be included together as acting on the aggregate object in order to get correct experimental predictions (as with my swarm of bees example). Thus the concept of vector addition applies whenever the objects are being treated as part of a larger object, and does not apply when they are being treated separately.
When you say that if the forces are balanced, then vector addition or cancellation does not apply, I have trouble making sense out of this statement. If the forces are balanced, that means that they push on the same object with equal strengths in opposite directions, right? In that case they are supposed to have zero net effect on the motion of the object. But if the forces are represented by equal and opposite vectors, then these vectors will add up to zero by vector cancellation, which exactly AGREES with the idea that the forces balance out and have no net effect. So in this case, vector cancellation is giving you exactly the result that you expect and know is correct, so where is the problem with it?
About the balanced spinning wheel: If there are no unbalanced forces between the mases, then the following experiment should work. Take a metal wheel and with a chisel, cut a small chip out of the surface. Now carefully straighten out the chip if the chisel bent it, so that it is exactly the same size and shape as the hole it left behind. Place it back into that hole, but do not fasten it in place. The wheel now looks exactly the same as before, with all the same mass in all the same locations. Now spin the wheel. According to your ideas, the chip should stay in place, and not fall off the wheel, no matter how fast the wheel spins. After all, this is the same bit of mass, executing the same exact motion as before. The only difference is that it is not firmly held to the rest of the wheel now. If there were NO unbalanced forces between the masses of the wheel, then there cannot have been any forces holding the chip in place, and therefore cutting it loose must not remove any forces that were needed. Therefore it should stay in place.
On the other hand, if it flies off the wheel, then its behavior is clearly different from what it was before, and so the absence of attachment does make a difference. In that case, the attachment to the wheel must have been necessary. According to Newton's second law, the ONLY thing that affects the motion of a mass is the net force on it. If the motion is different, the net force on the chip must therefore be different. But in the new situation, the net force is clearly zero, because the chip has been cut free of the wheel. If the forces are different, and the new net force is zero, then the old net force must have been different from zero. Therefore, there was a nonzero net force on the chip in the original situation. Therefore, the chip was accelerating, even though the wheel was smoothly turning at a constant speed.
QUOTE
There is this concept that the changing direction of objects in circular motion is viewed as acceleration, but obviously the Moon has been there a long time, and the forces are viewed as balanced. Plus; the forces are between the point of rotation and the object being rotated, not between the point masses of the wheel.
That is not true. The earth's gravity pulls the moon toward the earth. There is no opposing force, and therefore the forces are not balanced. The moon IS accelerating toward the earth. If you needed an opposing force, there would have to be some kind of tower holding the moon up. There obviously is no other force on it besides gravity, which is entirely consistent with Newton's point of view. In fact, the acceleration of the moon toward the earth is one of the fundamental pieces of evidence that Newton presented in favor of his theory of gravitation. If the moon is not accelerating toward the earth, then Newton was simply wrong about gravity.
QUOTE (->
| QUOTE |
| There is this concept that the changing direction of objects in circular motion is viewed as acceleration, but obviously the Moon has been there a long time, and the forces are viewed as balanced. Plus; the forces are between the point of rotation and the object being rotated, not between the point masses of the wheel. |
That is not true. The earth's gravity pulls the moon toward the earth. There is no opposing force, and therefore the forces are not balanced. The moon IS accelerating toward the earth. If you needed an opposing force, there would have to be some kind of tower holding the moon up. There obviously is no other force on it besides gravity, which is entirely consistent with Newton's point of view. In fact, the acceleration of the moon toward the earth is one of the fundamental pieces of evidence that Newton presented in favor of his theory of gravitation. If the moon is not accelerating toward the earth, then Newton was simply wrong about gravity.
Here is an experiment I conducted once. I fixed a string to an immovable point used for rotation just off of an air table, the other end of the string I affixed to a 84g puck on the air table. I tied a second string to the 84g puck and attached a 32g puck to the other end. I rotated both pucks, trailing the second puck behind the larger puck at 90° to the first string. The 90° position was fixed until release. Upon release the 32 g puck swung out beyond the 84g puck and stopped it, exactly as the spheres stop the cylinder in the cylinder and spheres experiment. The data was similar as well, showing linear momentum conservation.
In the 84 and 32 gram puck experiment there is no opportunity to cancel the motion from the other side using vector cancellation, because there is no other side. If you use linear momentum conservation the results are the same as the cylinder and spheres experiment, the results are different with vector cancellation. It seems like you would want a theory that work consistently like Newton’s does. Besides what good is you vector cancellation concept that tells you nothing; what good is a world of zeros.
I'm not sure I fully see the details of this experiment. Do you have a fuller explanation of it available somewhere?
Anyway, I hope my comments have been useful to you.
--Stuart Anderson
Place one end of a .15 m tube on a frictionless bearing point. Hold the tube straight up and place a 1 kg sphere on the other end of the tube. No friction should give the sphere the same max (low point) velocity that a standard pendulum would give it if it had been dropped the same distance. This means that 9.81 newtons have worked on 1 kg of mass over the .3 m distance, even though the portions of the circular arc are different, the drop is the same and the end velocities are the same. In the 3 kg rolling ring the 9.81 newtons would work on 3 kg giving you an acceleration rate of 9.81 / 3 = 3.27m/sec/sec, and a final velocity of √(2*3.27*.3 = 1.4007 m/sec * 3kg = 4.202 units of momentum.
This 1.4007 m/sec is probably correct because the change in the center of mass from the 2 kg being on top to the 2 kg being on the bottom is .1m, and if an object is dropped .1m it will be moving 1.4007 m/sec. In fact if you slide the ring .1 m (vertical drop) on an incline and then force it into a roll, the center of mass would be moving 1.4007 m/sec. You could force a roll by wrapping a string around the ring and having it come tight at the bottom of the incline and at the bottom of the ring. You could also make the ring into a pendulum bob that has the 2kg at 270° in the circle and the 1kg at 90° of the circle. Drop the ring as a bob .1 m and then force a roll, and again the rings center of mass will be rolling at 1.4007 m/sec.
The ring is rolling and in relationship to the table the 2 kg mass is stopped at the bottom and only the center of mass is moving 1.4007 m/sec. The center of mass is 1/3 of the distance from the 2 kg to the 1 kg (at .1 m). This means that the 1 kg (at .3 m) is moving 4.202 m/sec. According to the Law of Conservation of Momentum (F = ma) the total momentum of the system must be maintained, which it is.
The distance formula (1/2v²/a) tells us that the 1 kg moving 4.202 m/sec if released will rise to .9 m above its release point of .3 m. It will be at 1.2 m. Attach a pulley from above and connect a ground level 1kg mass to the kg at 1.2 m, use .3 units of rise to restore the ground level mass to the top of the ring, and you still have .9 m height above the ground, three times higher than when you started. This is 300% the original energy.
This 1.4007 m/sec is probably correct because the change in the center of mass from the 2 kg being on top to the 2 kg being on the bottom is .1m, and if an object is dropped .1m it will be moving 1.4007 m/sec. In fact if you slide the ring .1 m (vertical drop) on an incline and then force it into a roll, the center of mass would be moving 1.4007 m/sec. You could force a roll by wrapping a string around the ring and having it come tight at the bottom of the incline and at the bottom of the ring. You could also make the ring into a pendulum bob that has the 2kg at 270° in the circle and the 1kg at 90° of the circle. Drop the ring as a bob .1 m and then force a roll, and again the rings center of mass will be rolling at 1.4007 m/sec.
The ring is rolling and in relationship to the table the 2 kg mass is stopped at the bottom and only the center of mass is moving 1.4007 m/sec. The center of mass is 1/3 of the distance from the 2 kg to the 1 kg (at .1 m). This means that the 1 kg (at .3 m) is moving 4.202 m/sec. According to the Law of Conservation of Momentum (F = ma) the total momentum of the system must be maintained, which it is.
The distance formula (1/2v²/a) tells us that the 1 kg moving 4.202 m/sec if released will rise to .9 m above its release point of .3 m. It will be at 1.2 m. Attach a pulley from above and connect a ground level 1kg mass to the kg at 1.2 m, use .3 units of rise to restore the ground level mass to the top of the ring, and you still have .9 m height above the ground, three times higher than when you started. This is 300% the original energy.
mr homm: You are correct about the moon it should not have been placed in the discussion of balanced forces. The motion of the moon is maintained because the moon’s preexisting linear momentum carries it away from the earth at the same rate that the unbalance force of gravity pulls it in. I stand corrected; it does not apply to the discussion.
This is vastly different than objects being rotated on the end of a string where centripetal and centrifugal forces are equal. But if these forces are at 90° to the linear motion of the object being rotated they have no affect upon the momentum of the object being rotated.
I strongly disagree with you on the Atwood’s machine however. The extra mass on the one side of the Atwood’s machine has a quantity of force that accelerates both the ascending and descending sides of the Atwood. Let us say that you have 4.5 kg on both sides and an extra 1 kg on one of the sides. That will be 9.81 newtons accelerating 10 kg. After one second the whole 10 kg system is moving .981 m/sec, if the ascending and descending momentums cancel you are left with 1kg only moving .981 m/sec after a one second application of 9.81 newtons of force. F would not equal ma. You can not cancel momentums simply because they are moving in the opposite directions.
This is vastly different than objects being rotated on the end of a string where centripetal and centrifugal forces are equal. But if these forces are at 90° to the linear motion of the object being rotated they have no affect upon the momentum of the object being rotated.
I strongly disagree with you on the Atwood’s machine however. The extra mass on the one side of the Atwood’s machine has a quantity of force that accelerates both the ascending and descending sides of the Atwood. Let us say that you have 4.5 kg on both sides and an extra 1 kg on one of the sides. That will be 9.81 newtons accelerating 10 kg. After one second the whole 10 kg system is moving .981 m/sec, if the ascending and descending momentums cancel you are left with 1kg only moving .981 m/sec after a one second application of 9.81 newtons of force. F would not equal ma. You can not cancel momentums simply because they are moving in the opposite directions.
QUOTE (Limon+Sep 15 2007, 02:10 AM)
In the 3 kg rolling ring the 9.81 newtons would work on 3 kg giving you an acceleration rate of 9.81 / 3 = 3.27m/sec/sec, and a final velocity of √(2*3.27*.3) = 1.4007 m/sec * 3kg = 4.202 units of momentum.
Help me out a bit here. the formula for the final velocity that you use is
V=sqrt(2*acceleration*distance)?
but why do you decide that acceleration is 9.81 / 3? you act as if the force is fixed 9.81 N no matter what mass it acts on, but if you're talking about gravity, it should be proportional to mass. obviously for a 3kg ring we have 3*9.81 N force, and the usual 9.81m/s/s acceleration.
on the other hand, why do you use 0.3, for a distance? you explained in the rest of your post that the center of mass of the ring only moves 0.1m...
what gives? what in the world do you mean with this sqrt(2*3.27*.3) expression?
---
anyway, i see your main error now.
for a rolling ring you can not assume that the end velocity of the center of masses will be the same as for freefall over the given distance. it only works for the pendulum case.
in the pendulum case the reaction to weight in the attachment point is always perpendicular to the motion of the center of mass (it is along the "tube" or string or whatever), so the only force that can affect the velocity of that motion is gravity. this is why you get the same result as if the mass just fell the given distance.
for the rolling ring case though, the reaction to weight at the attachment point is the normal force with which the surface pushes the ring, and it is not always perpendicular to the motion of the ring's center of mass.
so you shouldn't assume the end velocity is the same as in freefall over the given distance.
your approach would be correct for a case where the ring, instead of rolling on a table, is fixed on its central axis. then indeed the setup would be equivalent to your standard pendulum analogy - it's the same as if we have a 3kg object on a 0.05m tube, falling from 0(up) to 180(down) degrees.
then your result for the center of mass velocity will be correct, but of course due to the different "rest point" your calculations for the velocities of the two masses will not apply. if you do them for that case, you'll find energy is conserved and all is well.
Help me out a bit here. the formula for the final velocity that you use is
V=sqrt(2*acceleration*distance)?
but why do you decide that acceleration is 9.81 / 3? you act as if the force is fixed 9.81 N no matter what mass it acts on, but if you're talking about gravity, it should be proportional to mass. obviously for a 3kg ring we have 3*9.81 N force, and the usual 9.81m/s/s acceleration.
on the other hand, why do you use 0.3, for a distance? you explained in the rest of your post that the center of mass of the ring only moves 0.1m...
what gives? what in the world do you mean with this sqrt(2*3.27*.3) expression?
---
anyway, i see your main error now.
for a rolling ring you can not assume that the end velocity of the center of masses will be the same as for freefall over the given distance. it only works for the pendulum case.
in the pendulum case the reaction to weight in the attachment point is always perpendicular to the motion of the center of mass (it is along the "tube" or string or whatever), so the only force that can affect the velocity of that motion is gravity. this is why you get the same result as if the mass just fell the given distance.
for the rolling ring case though, the reaction to weight at the attachment point is the normal force with which the surface pushes the ring, and it is not always perpendicular to the motion of the ring's center of mass.
so you shouldn't assume the end velocity is the same as in freefall over the given distance.
your approach would be correct for a case where the ring, instead of rolling on a table, is fixed on its central axis. then indeed the setup would be equivalent to your standard pendulum analogy - it's the same as if we have a 3kg object on a 0.05m tube, falling from 0(up) to 180(down) degrees.
then your result for the center of mass velocity will be correct, but of course due to the different "rest point" your calculations for the velocities of the two masses will not apply. if you do them for that case, you'll find energy is conserved and all is well.
In the initial position of 2 kg on top of the .3 m ring and 1 kg on the bottom, one of the 1 kg on top and the 1 kg on the bottom are in balance and will show no tendency to rotate. The second kg on top introduces 9.81 newtons (one kilogram of mass has 9.81 newtons of force on the earth’s surface) of force, and it will accelerate all three kg fixed on the ring. F = ma, F/m = a, 9.81/3 = 3.27m/sec/sec.
The sqrt (2*3.27*.3) = v comes from the distance formula, d = 1/2at², t = v/a, t² = v²/a², therefore d = ½ v²/a, and d * 2 * a = v², or sqrt(2 * a* d) = v
Here are a few reasons I think the acceleration of a rolling ring should be very nearly identical to the acceleration of a frictionless sliding mass or a pendulum bob. Drape a string over a frictionless pulley and suspend a mass from it, attach the other end to a block on a frictionless plane, and then later attach it to the top of a ring on a smooth hard surface. Now let us compare the two motions; the center of mass of both the block and ring are moving at the same rate as the falling suspended mass, the individual points of mass are moving at the same rate as the falling mass in the block and very nearly the same rate for each point of mass in the ring. The average rate of motion of the point masses of the ring are moving at the same rate as the falling suspended mass. These similarities seem to show that the roll and the slide accelerate in the same manner. Plus; Galileo collected large quantities of data from items rolling down inclines, that he and Newton used to establish the concept of F = ma. So to think that roll, freefall, pendulums, or sliding accelerate under different rules seems unlikely.
Try to imagine reversing the rolling ring process; you have a .3 m low mass ring with a 2 kg point mass attached that is resting on the table with noting attached to the top. You strike the top of the ring horizontally with a 1 kg projectile that then embeds in the ring. You have to strike it with enough motion to bring the 2 kg back up to the top, to where it had started in the original description.
The incoming motion of the 1 kg is instantly shared with the other 2 kg, ballistic pendulum in all forms show that it is momentum that is conserved. So quickly all 3 kg are moving at the same velocity and their total momentum equals the input momentum.
Now what is going to make the rolling ring stop? There is 1 kg out of balance for 9.81 newtons of force, that force accelerates 3 kg of mass, F = ma = 9.81/3 = 3.27m/sec/sec. The 2 kg needs to rise .3 m so the initial velocity will have to be, sqrt(2*3.27*.3) = v = 1.4007 m/sec. For all 3 kg to have a velocity of 1.4007 m/sec which is 4.202 units of momentum, then the incoming 1 kg must have a velocity of 4.202 m/sec. This is a pendulum bob drop from .9 m.
The sqrt (2*3.27*.3) = v comes from the distance formula, d = 1/2at², t = v/a, t² = v²/a², therefore d = ½ v²/a, and d * 2 * a = v², or sqrt(2 * a* d) = v
Here are a few reasons I think the acceleration of a rolling ring should be very nearly identical to the acceleration of a frictionless sliding mass or a pendulum bob. Drape a string over a frictionless pulley and suspend a mass from it, attach the other end to a block on a frictionless plane, and then later attach it to the top of a ring on a smooth hard surface. Now let us compare the two motions; the center of mass of both the block and ring are moving at the same rate as the falling suspended mass, the individual points of mass are moving at the same rate as the falling mass in the block and very nearly the same rate for each point of mass in the ring. The average rate of motion of the point masses of the ring are moving at the same rate as the falling suspended mass. These similarities seem to show that the roll and the slide accelerate in the same manner. Plus; Galileo collected large quantities of data from items rolling down inclines, that he and Newton used to establish the concept of F = ma. So to think that roll, freefall, pendulums, or sliding accelerate under different rules seems unlikely.
Try to imagine reversing the rolling ring process; you have a .3 m low mass ring with a 2 kg point mass attached that is resting on the table with noting attached to the top. You strike the top of the ring horizontally with a 1 kg projectile that then embeds in the ring. You have to strike it with enough motion to bring the 2 kg back up to the top, to where it had started in the original description.
The incoming motion of the 1 kg is instantly shared with the other 2 kg, ballistic pendulum in all forms show that it is momentum that is conserved. So quickly all 3 kg are moving at the same velocity and their total momentum equals the input momentum.
Now what is going to make the rolling ring stop? There is 1 kg out of balance for 9.81 newtons of force, that force accelerates 3 kg of mass, F = ma = 9.81/3 = 3.27m/sec/sec. The 2 kg needs to rise .3 m so the initial velocity will have to be, sqrt(2*3.27*.3) = v = 1.4007 m/sec. For all 3 kg to have a velocity of 1.4007 m/sec which is 4.202 units of momentum, then the incoming 1 kg must have a velocity of 4.202 m/sec. This is a pendulum bob drop from .9 m.
Learnt any Lagrangian mechanics yet Limon or are you putting all your time into whining and perpetuating your ignorance?
So Lemon. Have you managed to build your energy from nothing machine yet?
look at it this way:
the velocity WONT be the same as freefall, and that's that. if you have the means to accurately measure the ring's velocity, do the experiment. i don't, but i don't need to anyway.
now instead of arguing against this simple FACT, lets just try to understand and explain to ourselves why there's a differency.
i am pretty sure the differency can be explained by the fact that here the forces are not always acting to accelerate the tangential motion of the total mass, but to make it rotate as well. some of the forces effect doesn't end up as linear momentum, but angular momentum.
the velocity WONT be the same as freefall, and that's that. if you have the means to accurately measure the ring's velocity, do the experiment. i don't, but i don't need to anyway.
now instead of arguing against this simple FACT, lets just try to understand and explain to ourselves why there's a differency.
i am pretty sure the differency can be explained by the fact that here the forces are not always acting to accelerate the tangential motion of the total mass, but to make it rotate as well. some of the forces effect doesn't end up as linear momentum, but angular momentum.
QUOTE (visual+Sep 17 2007, 09:10 AM)
look at it this way:
the velocity WONT be the same as freefall, and that's that. if you have the means to accurately measure the ring's velocity, do the experiment. i don't, but i don't need to anyway.
now instead of arguing against this simple FACT, lets just try to understand and explain to ourselves why there's a differency.
i am pretty sure the differency can be explained by the fact that here the forces are not always acting to accelerate the tangential motion of the total mass, but to make it rotate as well. some of the forces effect doesn't end up as linear momentum, but angular momentum.
I didn't bother to read the entire problem since they tend to have exhausting detail in which the flaw generally lies --- but if this is the rolling wheel situation then some of the "up and down" momentum in converted to linear momentum in the direction of the wheel's motion along its track because the path of the weight is what is called "constrained motion".
Now Limon would have you believe that this linear momentum is available for conversion back to "up and down" but it isn't because mechanical "work" has been performed displacing the entire system along the track. In general, work consumes energy and a change of energy manifests a force and forces change momentum.
Now, if you try to pretend work, and thus kinetic energy, don't count, you can prove anything.
the velocity WONT be the same as freefall, and that's that. if you have the means to accurately measure the ring's velocity, do the experiment. i don't, but i don't need to anyway.
now instead of arguing against this simple FACT, lets just try to understand and explain to ourselves why there's a differency.
i am pretty sure the differency can be explained by the fact that here the forces are not always acting to accelerate the tangential motion of the total mass, but to make it rotate as well. some of the forces effect doesn't end up as linear momentum, but angular momentum.
I didn't bother to read the entire problem since they tend to have exhausting detail in which the flaw generally lies --- but if this is the rolling wheel situation then some of the "up and down" momentum in converted to linear momentum in the direction of the wheel's motion along its track because the path of the weight is what is called "constrained motion".
Now Limon would have you believe that this linear momentum is available for conversion back to "up and down" but it isn't because mechanical "work" has been performed displacing the entire system along the track. In general, work consumes energy and a change of energy manifests a force and forces change momentum.
Now, if you try to pretend work, and thus kinetic energy, don't count, you can prove anything.
I am experimenting and I do have the instruments.
I need to make a correction, with the mass draped over a pulley, the string has to be connected to the center of the ring (add a bearing point or use a small shaft) in order for the ring’s center of mass to move at the same rate as the falling mass. The average motion of the individual partials is equal to the motion of the falling mass but the individual motion of each partial of mass is mostly greater or less than the motion of the falling mass.
When the draped string is connected to the top of the ring the motion of the falling mass is twice that of the center of mass. A string at the top would get both the motion of the center of mass and the rotational motion of the ring. Here the motion of the individual partials is roughly equal the motion of the falling mass. Sorry for not catching this before I posted.
But here is something much easier to grasp. If the center of mass of a two point mass ring, rolling down an incline, accelerates according to F = ma then an energy increase can be secured from one of the two point masses as it rolls to the top.
I need to make a correction, with the mass draped over a pulley, the string has to be connected to the center of the ring (add a bearing point or use a small shaft) in order for the ring’s center of mass to move at the same rate as the falling mass. The average motion of the individual partials is equal to the motion of the falling mass but the individual motion of each partial of mass is mostly greater or less than the motion of the falling mass.
When the draped string is connected to the top of the ring the motion of the falling mass is twice that of the center of mass. A string at the top would get both the motion of the center of mass and the rotational motion of the ring. Here the motion of the individual partials is roughly equal the motion of the falling mass. Sorry for not catching this before I posted.
But here is something much easier to grasp. If the center of mass of a two point mass ring, rolling down an incline, accelerates according to F = ma then an energy increase can be secured from one of the two point masses as it rolls to the top.
yes one of the point masses rises and its energy increases, but the other goes lower and its energy decreases.
conservation is a fundamental fact, not just a rule you're encouraged to follow by your teacher
so quit this already, you won't manage to work around it, and you're becoming ridiculous.
well, if you don't, i'm quitting anyway, don't expect more responses.
conservation is a fundamental fact, not just a rule you're encouraged to follow by your teacher
so quit this already, you won't manage to work around it, and you're becoming ridiculous.well, if you don't, i'm quitting anyway, don't expect more responses.
NoCleverName’s Quote: Now Limon would have you believe that this linear momentum is available for conversion back to "up and down" but it isn't because mechanical "work" has been performed displacing the entire system along the track. In general, work consumes energy and a change of energy manifests a force and forces change momentum.
Limon’s Answer: I took a new aluminum automobile rim and taped a 500 gram mass to the concave area of its outer rim. I placed the rim with the 500 g mass in the up position on a flat hard surface and allowed the rim to start rolling. The 500 g mass does not hit the table as it rolls. It did a very beautiful job of rolling down to a point with the 500 g mass at the bottom and then back up again into very nearly the same position on the other end of the counter. The rim didn’t experience the mechanical problems mentioned by NoCleverName.
Just so there are no misunderstandings; here is what I mean. Construct a low mass ring with 1 kg point masses at 0° and 180°, allow this ring to roll down a smooth hard incline surface for a vertical displacement of .051 m. If F = ma then the center of mass of the ring will be moving 1 m/sec as it rolls onto a horizontal surface. When one point mass is touching the surface at the bottom (of the ring) the top mass must be moving 2 m/sec, if released and caught as a pendulum bob it will rise .204 m and the final center of mass will be 2 times higher than when the roll started.
Confirmation of this would involve experiments similar to the ones Galileo preformed some 375 years ago. I have purchased materials for such experiments; I will use photo gate timers etc.
Limon’s Answer: I took a new aluminum automobile rim and taped a 500 gram mass to the concave area of its outer rim. I placed the rim with the 500 g mass in the up position on a flat hard surface and allowed the rim to start rolling. The 500 g mass does not hit the table as it rolls. It did a very beautiful job of rolling down to a point with the 500 g mass at the bottom and then back up again into very nearly the same position on the other end of the counter. The rim didn’t experience the mechanical problems mentioned by NoCleverName.
Just so there are no misunderstandings; here is what I mean. Construct a low mass ring with 1 kg point masses at 0° and 180°, allow this ring to roll down a smooth hard incline surface for a vertical displacement of .051 m. If F = ma then the center of mass of the ring will be moving 1 m/sec as it rolls onto a horizontal surface. When one point mass is touching the surface at the bottom (of the ring) the top mass must be moving 2 m/sec, if released and caught as a pendulum bob it will rise .204 m and the final center of mass will be 2 times higher than when the roll started.
Confirmation of this would involve experiments similar to the ones Galileo preformed some 375 years ago. I have purchased materials for such experiments; I will use photo gate timers etc.
but if you did the measurement, or the right calculation for that matter, you'd see - when one point mass is at the bottom, the top mass will be moving at 1.414m/sec, not 2m/sec. when released and caught as a pendulum bob it will rise .102m, and the final center of mass will be exactly where it was when the roll started.
normally, when each mass falls 0.51m, each will have a velocity of 1m/s.
with the ring, we have one mass end up at rest, as it transfered its energy(note! not momentum) to the other mass.
yes, that means the rolling ring will be falling slower than, say, a sliding object of the same mass.
so far i've tried to explain (to myself and to you) this difference with both the reaction force of the surface and with angular momentum... now i realize these "explanations" are incorrect or incomplete.
but i have a new one for your review.
in order to have a perfect "rolling" of the ring along the surface instead of just sliding - i.e. the point of the ring that is in contact with the surface to always be at rest, you need friction. you can not ignore friction. so i guess exactly that friction force is what counters gravity and causes the end velocity to be less than a "free fall" or "sliding" equivalent.
normally, when each mass falls 0.51m, each will have a velocity of 1m/s.
with the ring, we have one mass end up at rest, as it transfered its energy(note! not momentum) to the other mass.
yes, that means the rolling ring will be falling slower than, say, a sliding object of the same mass.
so far i've tried to explain (to myself and to you) this difference with both the reaction force of the surface and with angular momentum... now i realize these "explanations" are incorrect or incomplete.
but i have a new one for your review.
in order to have a perfect "rolling" of the ring along the surface instead of just sliding - i.e. the point of the ring that is in contact with the surface to always be at rest, you need friction. you can not ignore friction. so i guess exactly that friction force is what counters gravity and causes the end velocity to be less than a "free fall" or "sliding" equivalent.
Let lemon try and build his device. He might then discover that even though he may not understand the mechanics, reality does.
Friction really is not a problem; even for sliding we have the frictionless plane, and in larger machines friction would be minimal. I am rolling 2 inch O.D. bronze sleeve bearings down an incline made of drill rods. That is a lot better than what Galileo had to use, and he overcame the friction problem. And I am sure photo gates are better than heart beats. In any case: all I am proving is that F = ma. I am not nervous about building another energy producing machine, I am sure it can be done because I have built one already (the cylinder and spheres).
Experiments 9-17-07; It is apparent that rings rolled down an incline plane achieve one increment of velocity in one increment of time, and two increments of velocity in two increments of time.
At the end of the first increment of time the ring is moving twice as fast as the average velocity for the first increment, that velocity will carry it two units of first increment distance in the second increment of time.
The added velocity developed in the second increment of time will add another increment of distance equal to the first increment of distance that was crossed in the first increment of time.
One increment of time times acceleration gives you the final velocity for the first increment. 1 * a
½ * (One increment of time * acceleration) give you the average velocity in the first increment of time; or the distance traveled in the first increment.
Time * time gives you the total number of first increment distances. One unit of first increment time gives one first increment distance, 2 units of time gives you 4 first increment distances, 3 units of first increment time give you 9 units of first increment distance.
Or d = 1/2at², this is called the distance formula (one of the most proven formulas in physics) and the rolling ring (hollow cylinder) appears to comply with this formula. And if we know the final distance we can derive the final velocity, which is what we are after.
Because d = 1/2v²/a, since t² = v²/a²
Compliance with the distance formula seems to show that we have a standard Newtonian acceleration in the incline plane experiments.
So what is left to debate, the quantity of acceleration? F = ma? It is obvious that mass does not change, so apparently some are now challenging the quantity of force. The angle to the center of mass of the ring in the incline plane is constant; acceleration also appears to be constant.
On an incline plane F is the force of gravity (9.81 newtons per kilogram) times the sine of the angle of inclination.
Let the incline angel equal 14.47°, sine will be .25, F will be 2.45 newtons times two kilograms = 4.9,
Acceleration will be 4.9 newtons / 2 kg = 2.45 m/sec/sec. F = ma
This is the acceleration down the incline plane and the plane’s length (in this experiment) is .051 / sin14.47° = .204 m.
v = sqrt(2*2.45*.204) = 1 m/sec.
Momentum will be 1.00 * 2 kg = 2 units of momentum.
If the top mass in this 2 kg ring experiment is not moving 2 m/sec then momentum is not conserved and Newtonian Physics is violated. It seems peculiar that you are all so willing to give up on Newtonian Physics. There is no other mass in motion; either the top mass is moving 2 m/sec or Newton’s Three Laws of Motion are false.
Experiments 9-17-07; It is apparent that rings rolled down an incline plane achieve one increment of velocity in one increment of time, and two increments of velocity in two increments of time.
At the end of the first increment of time the ring is moving twice as fast as the average velocity for the first increment, that velocity will carry it two units of first increment distance in the second increment of time.
The added velocity developed in the second increment of time will add another increment of distance equal to the first increment of distance that was crossed in the first increment of time.
One increment of time times acceleration gives you the final velocity for the first increment. 1 * a
½ * (One increment of time * acceleration) give you the average velocity in the first increment of time; or the distance traveled in the first increment.
Time * time gives you the total number of first increment distances. One unit of first increment time gives one first increment distance, 2 units of time gives you 4 first increment distances, 3 units of first increment time give you 9 units of first increment distance.
Or d = 1/2at², this is called the distance formula (one of the most proven formulas in physics) and the rolling ring (hollow cylinder) appears to comply with this formula. And if we know the final distance we can derive the final velocity, which is what we are after.
Because d = 1/2v²/a, since t² = v²/a²
Compliance with the distance formula seems to show that we have a standard Newtonian acceleration in the incline plane experiments.
So what is left to debate, the quantity of acceleration? F = ma? It is obvious that mass does not change, so apparently some are now challenging the quantity of force. The angle to the center of mass of the ring in the incline plane is constant; acceleration also appears to be constant.
On an incline plane F is the force of gravity (9.81 newtons per kilogram) times the sine of the angle of inclination.
Let the incline angel equal 14.47°, sine will be .25, F will be 2.45 newtons times two kilograms = 4.9,
Acceleration will be 4.9 newtons / 2 kg = 2.45 m/sec/sec. F = ma
This is the acceleration down the incline plane and the plane’s length (in this experiment) is .051 / sin14.47° = .204 m.
v = sqrt(2*2.45*.204) = 1 m/sec.
Momentum will be 1.00 * 2 kg = 2 units of momentum.
If the top mass in this 2 kg ring experiment is not moving 2 m/sec then momentum is not conserved and Newtonian Physics is violated. It seems peculiar that you are all so willing to give up on Newtonian Physics. There is no other mass in motion; either the top mass is moving 2 m/sec or Newton’s Three Laws of Motion are false.
... or, as usual, your calculations, observations, and conclusions are false.
Lemme take a wild guess ... it's Alternative 3, isn't it?
Here's the thing. It's no issue with me that you're screwed up. I'm just here to warn anybody gullible enough to listen to you for more than a second to just move on, nothing here to see.
Sure, it would be nice if I could remember enough from way back when to slap down your calculations ... but why make the effort? It would never convince you, so I'll save my breath. You know why I know your calculations are bad? As soon as you say things like "physics is wrong". Can't be your calculations, can it, because they're perfect? Must be those sheep (or is it parrot?) physicists --- can't be you. There hasn't been one time yet that I haven't found the problem --- and since I'm so rusty I have to do massive research and retooling to get the answer. 'bout time you put in some effort.
No doubt in this example you are ignoring the fact that this is rolling, not sliding motion, and we have to calculate using torques and angular momentum and angular veclocities. Not even counting the moment of inertia for a ring. I really don't want to relearn this stuff again just to refute you. Why don't you recast your calculations in the proper model and report back to us later?
Oh, and the cylinders and spheres created energy, did they? So I guess they still must be running then. Wait, you say they stopped soon after the energy was "created"? Guess that created energy just isn't as perpetual as it used to be. Well maybe you created just enough energy to go just a little distance. I wouldn't suppose that you didn't burn any of that "created energy" up resetting the machine for another go --- like I guess getting the thing spinning in the first place must come for free.
The only perpetual anything I see here is your perpetual errors.
Lemme take a wild guess ... it's Alternative 3, isn't it?
Here's the thing. It's no issue with me that you're screwed up. I'm just here to warn anybody gullible enough to listen to you for more than a second to just move on, nothing here to see.
Sure, it would be nice if I could remember enough from way back when to slap down your calculations ... but why make the effort? It would never convince you, so I'll save my breath. You know why I know your calculations are bad? As soon as you say things like "physics is wrong". Can't be your calculations, can it, because they're perfect? Must be those sheep (or is it parrot?) physicists --- can't be you. There hasn't been one time yet that I haven't found the problem --- and since I'm so rusty I have to do massive research and retooling to get the answer. 'bout time you put in some effort.
No doubt in this example you are ignoring the fact that this is rolling, not sliding motion, and we have to calculate using torques and angular momentum and angular veclocities. Not even counting the moment of inertia for a ring. I really don't want to relearn this stuff again just to refute you. Why don't you recast your calculations in the proper model and report back to us later?
Oh, and the cylinders and spheres created energy, did they? So I guess they still must be running then. Wait, you say they stopped soon after the energy was "created"? Guess that created energy just isn't as perpetual as it used to be. Well maybe you created just enough energy to go just a little distance. I wouldn't suppose that you didn't burn any of that "created energy" up resetting the machine for another go --- like I guess getting the thing spinning in the first place must come for free.
The only perpetual anything I see here is your perpetual errors.
Did I forget to mention that the velocity at the "top" of the ring has to be measured normal to the incline, or that it had both a linear component from the center of mass plus a tangential component from the angular motion? Or that, if this is still the eccentrically weighted ring, the center of mass is not smoothly moving down the incline but taking more of a cycloidic path?
There's probably a few other flaws I've missed, but these should keep you busy for a while.
There's probably a few other flaws I've missed, but these should keep you busy for a while.
Quote NoCleverName: Did I forget to mention that the velocity at the "top" of the ring has to be measured normal to the incline,
Limon’s Answer: You could let the ring roll onto a horizontal surface.
Quote NoCleverName:or that it had both a linear component from the center of mass plus a tangential component from the angular motion?
Limon’s Answer: When it is released from the top it has one and only one velocity, one and only one mass, and one and only one direction. Your imaginary friends have to be left at home.
Quote NoCleverName: Or that, if this is still the eccentrically weighted ring, the center of mass is not smoothly moving down the incline but taking more of a cycloidic path?
Limon’s Answer: So; so what if it is cycloid?
But let us keep it simple; let’s use two 1 kg point masses at 0° and 180°. Then the center of mass moves in a straight line as the ring rolls across a horizontal surface.
A low mass rolling ring with two 1 kg point masses (180°,0°) that has been rolled down an incline .051 vertical meters, must have a center of mass moving 1 m/sec to comply with both the Law of Conservation of Momentum and The Law of Conservation of Energy. But if you (by your calculations) have the top mass moving 1.414 m/sec, then the center of mass is only moving .707 m/sec. What do you do when the point masses roll back to 90° and 270°, what is their forward motion when they are on the same horizontal line as the center of mass? And what is the energy of 2 kg moving .707 m/sec? And when you transfer a sliding ring into a roll what happens to 29% of your motion. Friction becomes an imaginary friend: again, doesn’t it. You can cause a roll with a wrapped string, which is a lot of friction (in 1/4 rotation: or less) for an unwinding string.
Limon’s Answer: You could let the ring roll onto a horizontal surface.
Quote NoCleverName:or that it had both a linear component from the center of mass plus a tangential component from the angular motion?
Limon’s Answer: When it is released from the top it has one and only one velocity, one and only one mass, and one and only one direction. Your imaginary friends have to be left at home.
Quote NoCleverName: Or that, if this is still the eccentrically weighted ring, the center of mass is not smoothly moving down the incline but taking more of a cycloidic path?
Limon’s Answer: So; so what if it is cycloid?
But let us keep it simple; let’s use two 1 kg point masses at 0° and 180°. Then the center of mass moves in a straight line as the ring rolls across a horizontal surface.
A low mass rolling ring with two 1 kg point masses (180°,0°) that has been rolled down an incline .051 vertical meters, must have a center of mass moving 1 m/sec to comply with both the Law of Conservation of Momentum and The Law of Conservation of Energy. But if you (by your calculations) have the top mass moving 1.414 m/sec, then the center of mass is only moving .707 m/sec. What do you do when the point masses roll back to 90° and 270°, what is their forward motion when they are on the same horizontal line as the center of mass? And what is the energy of 2 kg moving .707 m/sec? And when you transfer a sliding ring into a roll what happens to 29% of your motion. Friction becomes an imaginary friend: again, doesn’t it. You can cause a roll with a wrapped string, which is a lot of friction (in 1/4 rotation: or less) for an unwinding string.
You didn't answer my question. Afraid to admit you've spent zero time trying to understand mainstream mechanics?
I thought Newton was mainstream mechanics; maybe you are the one that is side tracked.
The point masses have a velocity of translation and a velocity of rotation --- which in turn has instantaneous x and y components if you like. (At the point of contact these vectors will sum to zero velocity, even as the center of mass moves forward).
Just guessing, but I'd imagine a wheel going downhill could be analyzed equally as a wheel on level ground being driven by a torque. Where's the mystery?
The point of the cycloid is that the center of mass does not move continously downhill in a straight line but occasionally steepens and occasionally goes uphill.
Just guessing, but I'd imagine a wheel going downhill could be analyzed equally as a wheel on level ground being driven by a torque. Where's the mystery?
The point of the cycloid is that the center of mass does not move continously downhill in a straight line but occasionally steepens and occasionally goes uphill.
QUOTE (Limon+Sep 20 2007, 11:57 AM)
I thought Newton was mainstream mechanics; maybe you are the one that is side tracked.
So that's a no, you haven't bothered to learn how to succinctly and elegantly describe such dynamical systems as being discussed in this thread.
Can't your professor friend help or were you lying about him?
So that's a no, you haven't bothered to learn how to succinctly and elegantly describe such dynamical systems as being discussed in this thread.
Can't your professor friend help or were you lying about him?
The cycloid motion of the center of mass (when using different mass point masses) can be avoided by using a rim that has its center at the center of mass of the system but with a radius that is the distance to the larger mass. The larger mass can still be rolled into the zero velocity position. This provides a minor engineering challenge. If you use a 3 to 1 mass ration (moving 1 m/sec) in your two point masses, then when the large mass is stationary the small mass might be moving twice as fast or four times as fast. 2m/sec violates the conservation of momentum; 4 m/sec violates the Law of Conservation of Energy.
I just thought I should mention, Limon: nobody cares.
Since Limon seems adament to ignore my question, since he knows his answer wouldn't reflect well on here, here's what I was referring to :
Consider a mass m in a gravitational field created by a mass M, where M>>m. Kinetic energy, T, is 0.5m(r')². Potential energy is V = -GMm/r.
Lagrangian L = T-V = 0.5m(r')²+GM/r
Euler-Lagrange equations of motion are (d/dt)(dL/dq') - dL/dq = 0, where q is a variable in the system, in this case q=r and q=t.
Therefore have 2 equations of motion :
dL/dt = 0
(d/dt)(dL/dr') - dL/dr = 0
By Hamiltonian equations of motion, dL/dt = dH/dt, where H = T+V = Energy of system = E. Thus dE/dt = 0. T+V = 0.5m(r')²-GM/r, so 0.5m(r')²-GM/r = E = constant.
Thus the energy lost in falling in a gravitational field is converted into potental energy.
0 = (d/dt)(dL/dr') - dL/dr = (d/dt)(mr') + GMm/r²
mr'' = -GMm/r²
This is Newton's second law of motion
If we're not used the approximation that the Earth does not move due to the motion of the point mass, we'd have derived conservation of linear momentum.
Limon, can you see how to generalise this to prove conservation of angular momentum?
Do you understand how this shows that a system of point masses obeys both conservation of momentum and of energy?
Consider a mass m in a gravitational field created by a mass M, where M>>m. Kinetic energy, T, is 0.5m(r')². Potential energy is V = -GMm/r.
Lagrangian L = T-V = 0.5m(r')²+GM/r
Euler-Lagrange equations of motion are (d/dt)(dL/dq') - dL/dq = 0, where q is a variable in the system, in this case q=r and q=t.
Therefore have 2 equations of motion :
dL/dt = 0
(d/dt)(dL/dr') - dL/dr = 0
By Hamiltonian equations of motion, dL/dt = dH/dt, where H = T+V = Energy of system = E. Thus dE/dt = 0. T+V = 0.5m(r')²-GM/r, so 0.5m(r')²-GM/r = E = constant.
Thus the energy lost in falling in a gravitational field is converted into potental energy.
0 = (d/dt)(dL/dr') - dL/dr = (d/dt)(mr') + GMm/r²
mr'' = -GMm/r²
This is Newton's second law of motion
If we're not used the approximation that the Earth does not move due to the motion of the point mass, we'd have derived conservation of linear momentum.
Limon, can you see how to generalise this to prove conservation of angular momentum?
Do you understand how this shows that a system of point masses obeys both conservation of momentum and of energy?
Four kilograms are moving one meter per second horizontally in the same direction. This preexisting motion is forced into a roll where three of the kilograms soon stop. All the motion is now in the one kilogram and it is moving in the same original direction. Tell us AlphaNumeric what velocity (a Number) for the 1 kg satisfies all three of these equations; mv, 1/2mv², mr²ω.
A real number v, please. v =
m = 1kg
A real number v, please. v =
m = 1kg
QUOTE (Limon+Sep 25 2007, 12:31 AM)
mv, 1/2mv², mr²ω.
You don't get it do you? I just proved that conservation of energy is NOT "conservation of kinetic energy" but the conservation of the sum of kinetic and potential energy.
The Lagrangian for a rotating system would be :
L = 0.5mr²ω² + GMm/r
Notice you didn't even get angular energy right! Conservation of energy, by Newtonian formalism, will mean that
mr²ω² - GMm/r = E = constant
Conservation of angular momentum will mean that, if you ignore gravity,
mrω² = L = constant
If you include gravity and linear momentum (so that the system is moving as well as rotating) then things are more complex. Can you compute the equations of motion for such a system? Can you see why they do not lead to the 'constants' you claim?
This is evidence you have not bothered to actually learn Newtonian dynamics. You display the classic crank attitude of "If I don't understand something, it can't be my fault!"
You don't get it do you? I just proved that conservation of energy is NOT "conservation of kinetic energy" but the conservation of the sum of kinetic and potential energy.
The Lagrangian for a rotating system would be :
L = 0.5mr²ω² + GMm/r
Notice you didn't even get angular energy right! Conservation of energy, by Newtonian formalism, will mean that
mr²ω² - GMm/r = E = constant
Conservation of angular momentum will mean that, if you ignore gravity,
mrω² = L = constant
If you include gravity and linear momentum (so that the system is moving as well as rotating) then things are more complex. Can you compute the equations of motion for such a system? Can you see why they do not lead to the 'constants' you claim?
This is evidence you have not bothered to actually learn Newtonian dynamics. You display the classic crank attitude of "If I don't understand something, it can't be my fault!"
Newtonian dynamics says; 4 m/sec, how succinct can you get. But the other two formulas are not conserved.
QUOTE (Limon+Sep 25 2007, 01:28 AM)
how succinct can you get
Succinctness is pointless if you give an incorrect answer. I just proved your equations are not what Newton or Leibnitz predict for your system. You didn't address that, you ignored it, probably because you either dont' understand it or you know you're wrong.
Answer my questions :
Can you compute the equations of motion for such a system? Can you see why they do not lead to the 'constants' you claim?
Succinctness is pointless if you give an incorrect answer. I just proved your equations are not what Newton or Leibnitz predict for your system. You didn't address that, you ignored it, probably because you either dont' understand it or you know you're wrong.
Answer my questions :
Can you compute the equations of motion for such a system? Can you see why they do not lead to the 'constants' you claim?
QUOTE (Limon+Sep 24 2007, 11:31 PM)
Four kilograms are moving one meter per second horizontally in the same direction. This preexisting motion is forced into a roll ...
Forced into a roll, eh? How much energy did that take?
Forced into a roll, eh? How much energy did that take?
The same amount lost when an object is caught on the end of a string, none.
Are you unable to answer my questions Limon?
QUOTE (Limon+Sep 25 2007, 06:15 AM)
The same amount lost when an object is caught on the end of a string, none.
Limon, where are you? I will send you my old textbook, "On Mechanics and Heat", if you think it would be of some use to you. To read, not to level your television table.
Limon, where are you? I will send you my old textbook, "On Mechanics and Heat", if you think it would be of some use to you. To read, not to level your television table.
Sapo; Maybe you need to read it.
Ballistic pendulums prove that there is no loss of energy when an object enters a circular path.
Ballistic pendulums force the projectile motion into a circular path, but no loss the energy is attributed to this change. The energy loss is attributed to the projectile sharing its motion with another object (usually a much larger block). The percentage loss is directly related to the block mass divided by the total mass of the projectile and block. The same percent loss occurs on a frictionless plane and there is no motion forced into a circle. You get the same results with or without the circular motion. Momentum is of course conserved in both situations.
Galileo’s incline plane experiments showed that objects that roll down inclines have the capability to roll up a similar incline the same vertical distance that they rolled down, less a small amount of friction.
Hollow cylinders can roll down an incline and then cross a horizontal distance and back up an incline the same vertical distance they came down.
The velocity given the center of mass during the descent is necessary for the accent.
A three kilogram cylinder with the same dimensions as a 1 kg cylinder will roll at the same rate.
All points or sections of the hollow cylinder roll at the same rate.
A 1 kg hollow cylinder with two 1 kg point masses embedded in the cylinder wall at 180° will roll at the same rate as the same cylinder less the point masses.
A 1 kg cylinder with two 1 kg point masses at 180° will continue rolling at the same rate after the two point masses have been simultaneously dropped off. And the cylinder will return to the same height from which it started.
If the two point masses are simultaneously dropped off when one mass is on the horizontal surface then the other point mass will be moving twice as fast as the center of mass of the cylinder.
The velocity of the center of mass of the cylinder was sufficient to take it to its original height; twice that velocity will take the upper point mass to four times that height. At that point the center of mass of the two point masses would be twice as high as when they started and the cylinder would be at the same height. That is 167% the original energy.
Ballistic pendulums prove that there is no loss of energy when an object enters a circular path.
Ballistic pendulums force the projectile motion into a circular path, but no loss the energy is attributed to this change. The energy loss is attributed to the projectile sharing its motion with another object (usually a much larger block). The percentage loss is directly related to the block mass divided by the total mass of the projectile and block. The same percent loss occurs on a frictionless plane and there is no motion forced into a circle. You get the same results with or without the circular motion. Momentum is of course conserved in both situations.
Galileo’s incline plane experiments showed that objects that roll down inclines have the capability to roll up a similar incline the same vertical distance that they rolled down, less a small amount of friction.
Hollow cylinders can roll down an incline and then cross a horizontal distance and back up an incline the same vertical distance they came down.
The velocity given the center of mass during the descent is necessary for the accent.
A three kilogram cylinder with the same dimensions as a 1 kg cylinder will roll at the same rate.
All points or sections of the hollow cylinder roll at the same rate.
A 1 kg hollow cylinder with two 1 kg point masses embedded in the cylinder wall at 180° will roll at the same rate as the same cylinder less the point masses.
A 1 kg cylinder with two 1 kg point masses at 180° will continue rolling at the same rate after the two point masses have been simultaneously dropped off. And the cylinder will return to the same height from which it started.
If the two point masses are simultaneously dropped off when one mass is on the horizontal surface then the other point mass will be moving twice as fast as the center of mass of the cylinder.
The velocity of the center of mass of the cylinder was sufficient to take it to its original height; twice that velocity will take the upper point mass to four times that height. At that point the center of mass of the two point masses would be twice as high as when they started and the cylinder would be at the same height. That is 167% the original energy.
For God's sakes. 
You are fatuous, and impervious to logic and sarcasm. I give up. AlphaNumeric may wish to argue your points with you, but I am not as tenacious.
You are fatuous, and impervious to logic and sarcasm. I give up. AlphaNumeric may wish to argue your points with you, but I am not as tenacious.
Limon, you continue to avoid my question. Care to explain why?
Also, the reason you get an error is that you are incorrect in your simplications.
The system does depend on wether the cylinder is spinning since if it's not, letting the masses go will not do anything, the system continues as normal since it's in an inertial frame. The lower mass experiences no friction, so it just continues moving at speed v.
The only way the lower mass can stuff is if the system is rolling and not skidding (as it does in the frictionless case). Rolling is a very precise constraint on the system, since it means that the lower mass is precisely at rest at the bottom of the cylinder when a mass is there. Any more or less and it's got residual motion. But if it's rolling, you must include angular momentum and a very precise statement about the rate of roll is required.
I was taught such things in my 1st year dynamics course. You obviously never bothered to try to learn such elementary things and continue in your ignorant claims. No doubt you'll ignore this post too since you know you'll have to respond to my question you've been ignoring. Not to mention, you will not understand the explaination I've given since you don't understand Newtonian mechanics.
Also, the reason you get an error is that you are incorrect in your simplications.
The system does depend on wether the cylinder is spinning since if it's not, letting the masses go will not do anything, the system continues as normal since it's in an inertial frame. The lower mass experiences no friction, so it just continues moving at speed v.
The only way the lower mass can stuff is if the system is rolling and not skidding (as it does in the frictionless case). Rolling is a very precise constraint on the system, since it means that the lower mass is precisely at rest at the bottom of the cylinder when a mass is there. Any more or less and it's got residual motion. But if it's rolling, you must include angular momentum and a very precise statement about the rate of roll is required.
I was taught such things in my 1st year dynamics course. You obviously never bothered to try to learn such elementary things and continue in your ignorant claims. No doubt you'll ignore this post too since you know you'll have to respond to my question you've been ignoring. Not to mention, you will not understand the explaination I've given since you don't understand Newtonian mechanics.
Alright, the hell with not allowing my cruel side out. Zephir, I have reported your 'guest' spamming twice today, you miserable coprolite.
QUOTE (Sapo+Sep 26 2007, 09:47 AM)
Alright, the hell with not allowing my cruel side out. Zephir, I have reported your 'guest' spamming twice today, you miserable coprolite.
I was thinking, coprolite - not much of an insult. But that was because I hadn't heard of the word. Quite fitting for Zephir.
I was thinking, coprolite - not much of an insult. But that was because I hadn't heard of the word. Quite fitting for Zephir.
I'm too tired(and lazy) at the moment to go through every detail but I"m curious about a couple things.
@Limon
Do you realize that you don't need friction to create heat from kinetic engery?
ex if you pound a hammer on an anvil it will heat both up.
Also do you realize that you need friction to roll something?
@Limon
Do you realize that you don't need friction to create heat from kinetic engery?
ex if you pound a hammer on an anvil it will heat both up.
Also do you realize that you need friction to roll something?
When a locomotive begins accelerating its cars the coal in the cars does not heat up. Only the friction points of the train experience thermal increases; couplers, brakes, bearings, wheels circumferences, etc.
Hardened polished metal striking hardened polished metal does not create a large amount of heat. The image of heat in your example of the hammer and anvil probably includes a soft metal between the two hardened metals and usually involves a preheating of the softer metal.
You can maintain the same functional arrangement of a hardened polished metal projectile embedding in a hardened polished metal block in a ballistic pendulum. By increasing the mass of the block this alleged heat loss can be 50%, 75%, 90%, 99.9%, even though the two surfaces that struck each other are identical. This heat loss is a ridiculous concept; and I remind you again, no one has ever measure this alleged heat. I believe the heat is not there to be measured.
Cut a circumference groove in the center of a hollow cylinder; wrap a string around the cylinder in this groove. Place the cold cylinder on an incline plane made of dry ice with the string at the bottom of the cylinder and the end of the string up hill, the cylinder will roll down the incline. This proves that friction is not necessary to make an object roll, for if the string were not present the cylinder would slide and not roll.
Hardened polished metal striking hardened polished metal does not create a large amount of heat. The image of heat in your example of the hammer and anvil probably includes a soft metal between the two hardened metals and usually involves a preheating of the softer metal.
You can maintain the same functional arrangement of a hardened polished metal projectile embedding in a hardened polished metal block in a ballistic pendulum. By increasing the mass of the block this alleged heat loss can be 50%, 75%, 90%, 99.9%, even though the two surfaces that struck each other are identical. This heat loss is a ridiculous concept; and I remind you again, no one has ever measure this alleged heat. I believe the heat is not there to be measured.
Cut a circumference groove in the center of a hollow cylinder; wrap a string around the cylinder in this groove. Place the cold cylinder on an incline plane made of dry ice with the string at the bottom of the cylinder and the end of the string up hill, the cylinder will roll down the incline. This proves that friction is not necessary to make an object roll, for if the string were not present the cylinder would slide and not roll.
Limon, I take it you're ignoring my posts correcting your claims because of one or both of the following reasons :
1. You don't understand them
2. You realise they disprove your claims and don't want to admit it.
You'll notice that when expressed in such nice formalism, as I've been telling you for months, it's obvious that gravitational systems as you describe neither conserve kinetic energy or momentum. However, for trivial systems (no gravity, no interactions), both kinetic energy and momentum are conserved. If you're going to model a gravitational system properly (ie to get momentum conservation) you have to consider the effect Newton's 3rd Law will
1. You don't understand them
2. You realise they disprove your claims and don't want to admit it.
You'll notice that when expressed in such nice formalism, as I've been telling you for months, it's obvious that gravitational systems as you describe neither conserve kinetic energy or momentum. However, for trivial systems (no gravity, no interactions), both kinetic energy and momentum are conserved. If you're going to model a gravitational system properly (ie to get momentum conservation) you have to consider the effect Newton's 3rd Law will