Hello
Could someone please help me with the following basic physics class scenario?
Scenario
I have a cart on wheels, with a mass of 20 lb.
On it is a 6ft. long vertical pole, with a mass of 10lb. (mass uniformly distributed)
The pole is attached by a hinge to the cart.
The cart/pole assembly is moving horizontally at 5 mph.
The cart hits a stationary object and is brought to a sudden stop.
Question
Assuming 100% momentum transfer, how fast will the centre of mass be travelling and how fast will the upper end be travelling.
Will the upper end accelerate beyond the 5 mph that it was before the collision?
The momentum of the cart will not be transferred to the pole, the cart is stopped and the pole did not cause the cart to stop, as the spheres caused the cylinder to stop in the cylinder and spheres experiments.
The momentum of the pole will be conserved by its center of mass proceeding at its original velocity, assuming no friction in the hinge. The bottom of the pole will be stopped, the center will be moving 5 miles per hour, and the top will be moving 10 miles per hour. Gravity will accelerate the pole as it falls but I think we are ignoring that part of the motion.
The momentum of the pole is conserved but the energy of the pole is not conserved.
The momentum of the pole will be conserved by its center of mass proceeding at its original velocity, assuming no friction in the hinge. The bottom of the pole will be stopped, the center will be moving 5 miles per hour, and the top will be moving 10 miles per hour. Gravity will accelerate the pole as it falls but I think we are ignoring that part of the motion.
The momentum of the pole is conserved but the energy of the pole is not conserved.
Thanks Limon. As you may have guessed, I'm no physicist. I do have a bit more confusion in my head about this though. Since the sudden stop is due to a reaction force, would there not be some sort of "effect" on the cart/pole assembly? What I mean is, if it were just the cart, wouldn't it be subject to deformation of some sort? It would break or compress/crumple. Given that the pole is attached and is free (somewhat) to move, would it not be the "weak link" and therefore act like a "crumple zone" in a car and be affected by the reaction force through the cart?
By the way, I really appreciate the help.
By the way, I really appreciate the help.
QUOTE (Darrell+Feb 14 2008, 12:03 PM)
Since the sudden stop is due to a reaction force, would there not be some sort of "effect" on the cart/pole assembly? What I mean is, if it were just the cart, wouldn't it be subject to deformation of some sort? It would break or compress/crumple. Given that the pole is attached and is free (somewhat) to move, would it not be the "weak link" and therefore act like a "crumple zone" in a car and be affected by the reaction force through the cart?
Any further thoughts?
Any further thoughts?
You are right that there must be a force on the base of the pole when the cart stops. Otherwise, the pole would continue moving forward with no change in its motion, which contradicts the fact that the hinge must stop. To stop the bottom end of the pole, there must be a horizontal impact force directed backward along the cart's path. This force is exerted by the hinge on the bottom of the pole.
Now since there IS a force on the pole, its linear momentum is not conserved. This force will also exert a torque around the center of the pole, so its angular momentum around the center is not conserved either, which makes it seem as if the problem will be hard to solve. It can be solved based on four facts:
1: the base of the pole stops,
2: the speed of the center = (angular velocity) * (1/2 of length),
3: (the applied force) * time = mass * (change of center velocity), and
4: (the applied force) * (1/2 of length) * time = mass/12 * (length)^2 * (change of angular velocity).
Fact 1 is given, fact 2 is the kinematic relation v = v_axis + omega*r for rotational motion (where the center of mass is of course at a distance length/2 from the base), fact 3 is the impulse momentum theorem, and fact 4 is the angular impulse angular momentum theorem, where the moment of intertia of the pole is (mL^2)/12.
By fact 1, v_axis = 0, so by fact 2, v_c = -wL/2, where v_c is the center of mass velocity, w is the angular speed, and L is the length. The minus sign is because a negative (i.e. clockwise) angular velocity will give a forward ( i.e. positive) velocity for the center of mass. By fact 3, Ft = m(v_c - v_0), where v_0 is the original speed. Therefore, by the previous equation, Ft = m(-wL/2 - v_0). Now fact 4 says FtL/2 = (mL^2)/12 * (w-0), because the initial value of w is 0. This equation simplifies to Ft = mwL/6. Now there are two equations for Ft, so their right sides must be equal, therefore m(-wL/2 - v_0) = mwL/6. Cancelling m and solving for w gives w = -(3/2)*(v_0/L). Therefore, the center of mass velocity of the rod after the collision is given by fact 2 as v_c = -wL/2 = (3/4)*v_0. The top of the rod will be going twice this fast, which is (3/2)*v_0, so the top of the rod actually does go faster after the collision, although the center of the rod slows down. The rod loses exactly 1/4 of its original linear momentum.
There is also a shortcut method: look at the angular momentum of the pole around the hinge instead of around its center. Since the force that stops the bottom end is exerted at the hinge, it makes no torque around the hinge, so the angular momentum around that axis IS conserved. (That's an important thing to understand: angular momentum may be conserved around one axis but not another for the same object.) Before the collision, the angular momentum of the pole is (its momentum) * (distance from axis to center of mass), which is -mv_0L/2. The minus sign is because the center of mass passes the hinge while moving tangent to a clockwise circle, which is the negative direction for rotation. After the collision, the motion is now a rotation around the hinge, so the angular momentum is now (1/3)mL^2*w. (This uses the moment of inertia around the end of the rod, instead of the center, so it is (1/3)mL^2.) Since these must be equal, you get -mv_0L/2 = mL^2w/3, which gives w = -(3/2)v_0/L, just as before.
By the way, Limon will disagree with me here, because he has a different view of conservation of angular momentum than most of us do. You will have to decide whom to believe. Personally, I would conduct the experiment and then believe the POLE. You can't go wrong with reality.
Hope that helps!
--Stuart Anderson
Now since there IS a force on the pole, its linear momentum is not conserved. This force will also exert a torque around the center of the pole, so its angular momentum around the center is not conserved either, which makes it seem as if the problem will be hard to solve. It can be solved based on four facts:
1: the base of the pole stops,
2: the speed of the center = (angular velocity) * (1/2 of length),
3: (the applied force) * time = mass * (change of center velocity), and
4: (the applied force) * (1/2 of length) * time = mass/12 * (length)^2 * (change of angular velocity).
Fact 1 is given, fact 2 is the kinematic relation v = v_axis + omega*r for rotational motion (where the center of mass is of course at a distance length/2 from the base), fact 3 is the impulse momentum theorem, and fact 4 is the angular impulse angular momentum theorem, where the moment of intertia of the pole is (mL^2)/12.
By fact 1, v_axis = 0, so by fact 2, v_c = -wL/2, where v_c is the center of mass velocity, w is the angular speed, and L is the length. The minus sign is because a negative (i.e. clockwise) angular velocity will give a forward ( i.e. positive) velocity for the center of mass. By fact 3, Ft = m(v_c - v_0), where v_0 is the original speed. Therefore, by the previous equation, Ft = m(-wL/2 - v_0). Now fact 4 says FtL/2 = (mL^2)/12 * (w-0), because the initial value of w is 0. This equation simplifies to Ft = mwL/6. Now there are two equations for Ft, so their right sides must be equal, therefore m(-wL/2 - v_0) = mwL/6. Cancelling m and solving for w gives w = -(3/2)*(v_0/L). Therefore, the center of mass velocity of the rod after the collision is given by fact 2 as v_c = -wL/2 = (3/4)*v_0. The top of the rod will be going twice this fast, which is (3/2)*v_0, so the top of the rod actually does go faster after the collision, although the center of the rod slows down. The rod loses exactly 1/4 of its original linear momentum.
There is also a shortcut method: look at the angular momentum of the pole around the hinge instead of around its center. Since the force that stops the bottom end is exerted at the hinge, it makes no torque around the hinge, so the angular momentum around that axis IS conserved. (That's an important thing to understand: angular momentum may be conserved around one axis but not another for the same object.) Before the collision, the angular momentum of the pole is (its momentum) * (distance from axis to center of mass), which is -mv_0L/2. The minus sign is because the center of mass passes the hinge while moving tangent to a clockwise circle, which is the negative direction for rotation. After the collision, the motion is now a rotation around the hinge, so the angular momentum is now (1/3)mL^2*w. (This uses the moment of inertia around the end of the rod, instead of the center, so it is (1/3)mL^2.) Since these must be equal, you get -mv_0L/2 = mL^2w/3, which gives w = -(3/2)v_0/L, just as before.
By the way, Limon will disagree with me here, because he has a different view of conservation of angular momentum than most of us do. You will have to decide whom to believe. Personally, I would conduct the experiment and then believe the POLE. You can't go wrong with reality.
Hope that helps!
--Stuart Anderson
Swing a 1 kg sphere down .4587 meters on the end of a twenty meter pendulum string, it will be moving 3 m/sec. Have it embed in a 2 kilogram block that is suspended on the end of a one meter pendulum string. The twenty meter string releases; and the 3 kilogram combination will be moving 1 meter per second in a 1 meter pendulum.
Linear Newtonian Momentum is conserved.
Is angular momentum conserved?
Is kinetic energy conserved?
I am the one that does the experiments and my experiments show that the cylinder and spheres device makes energy. This is not a violation of Newtonian Physics.
And homm you are the one that is afraid to do experiments. It is awfully hard for you to prove a theory that rarely (keep the pendulum length the same, then angular momentum conservation might work) works in the lab.
You hold onto your theories with no evidence that they actually work. Don’t give examples in space where gravity changes linear momentum, find examples where angular momentum conservation works in the lab; you will have a hard time of it.
Linear Newtonian Momentum is conserved.
Is angular momentum conserved?
Is kinetic energy conserved?
I am the one that does the experiments and my experiments show that the cylinder and spheres device makes energy. This is not a violation of Newtonian Physics.
And homm you are the one that is afraid to do experiments. It is awfully hard for you to prove a theory that rarely (keep the pendulum length the same, then angular momentum conservation might work) works in the lab.
You hold onto your theories with no evidence that they actually work. Don’t give examples in space where gravity changes linear momentum, find examples where angular momentum conservation works in the lab; you will have a hard time of it.
QUOTE (Lemon+)
You hold onto your theories with no evidence that they actually work. Don’t give examples in space where gravity changes linear momentum, find examples where angular momentum conservation works in the lab; you will have a hard time of it.
Only by applying your lack of understanding to the problem
Only by applying your lack of understanding to the problem
@Limon
Since you have given me the speed after the collision, I assume that I may use this data in my calculations, since it is the result of your experiment.
Since you have given me the speed after the collision, I assume that I may use this data in my calculations, since it is the result of your experiment.
Linear Newtonian Momentum is conserved.
Let's check:
Before, P = 1kg*3m/s = 3kgm/s; after, P = 3kg*1m/s = 3kgm/s. These are indeed the same, so yes, linear momentum is conserved.
Around which axis? The large pendulum is swinging around an axis 20m above the point of impact, and the small pendulum is swinging around an axis 1m above the point of impact.
Let's check:
Around the higher axis, before, L = 20m*1kg*3m/s = 60kgm^2/s; after, L = 20m*3kg*1m/s = 60kgm^2/s. These are indeed the same, so yes, the angular momentum around the higher axis is conserved.
Around the lower axis, before, L = 1m*1kg*3m/s = 3kgm^2/s; after, L = 1m*3kg*1m/s = 3kgm^/2. These are indeed the same, so yes, the angular momentum around the lower axis is conserved.
Around which axis? The large pendulum is swinging around an axis 20m above the point of impact, and the small pendulum is swinging around an axis 1m above the point of impact.
Let's check:
Around the higher axis, before, L = 20m*1kg*3m/s = 60kgm^2/s; after, L = 20m*3kg*1m/s = 60kgm^2/s. These are indeed the same, so yes, the angular momentum around the higher axis is conserved.
Around the lower axis, before, L = 1m*1kg*3m/s = 3kgm^2/s; after, L = 1m*3kg*1m/s = 3kgm^/2. These are indeed the same, so yes, the angular momentum around the lower axis is conserved.
Is kinetic energy conserved?
Let's check:
Before, KE = 1/2*1kg*(3m/s)^2 = 4.5J; after, KE = 1/2*3kg*(1m/s)^2 = 1.5J. These are different, and the final KE is only 1/3 of the initial KE. Therefore, 2/3 of the KE is lost to heat, sound, and vibration, and it is not conserved.
Since I used data that you provided, you cannot question their experimental basis without questioning your own experiment. Application of the definition of angular momentum around an axis shows that for each of the two obvious axes, the angular momentum is exactly the same before and after the collision. In fact, I could place the axis at ANY height, and the calculation would clearly come out the same before and after the collision. The idea of conservation of angular momentum never said that the angular momentum around two different axes must be the same, only that if you use the SAME axis before and after the collision, you will get the same value, and you can see from my calculation that you clearly do.
You must remember that the definition of angular momentum is r X p, where r is the vector from the axis location to the mass. For a large mass, you sum up the angular momentum of the mass at each location within it. The axis can be placed anywhere in the universe, and for each axis, you will get a different angular momentum. So there is not just ONE angular momentum for a given object; there is a different angular momentum for each axis you choose to compute around. This is directly from the definition of angular momentum. Conservation of angular momentum is always referred to a specific axis, and if the value of L around that same axis is equal before and after the collision, then we say that L is conserved around that axis. It may at the same time be not conserved around some other axis, because the whole conservation/nonconservation question is to be answered on a per-axis basis. Again, this is part of the definition of L.
Therefore, your own data proves that L is conserved around each of the pendulum axes. The only way you could think it is not conserved is if you use a different axis after the collision than you used before the collision. But that contradicts the definition of L, so if you do that, you aren't even comparing the same L's because they are around two different axes. Then you are comparing apples and oranges, and if you want to say they're not equal, well of course I'll agree with that. The L around the upper axis wasn't equal to the L around the lower axis even before the collision, so why should they be equal afterwards? It's perfectly true that they are unequal, it simply has nothing to do with conservation of L, that's all.
Hope that helps.
--Stuart Anderson
QUOTE
Swing a 1 kg sphere down .4587 meters on the end of a twenty meter pendulum string, it will be moving 3 m/sec. Have it embed in a 2 kilogram block that is suspended on the end of a one meter pendulum string. The twenty meter string releases; and the 3 kilogram combination will be moving 1 meter per second in a 1 meter pendulum.
Since you have given me the speed after the collision, I assume that I may use this data in my calculations, since it is the result of your experiment.
QUOTE (->
| QUOTE |
| Swing a 1 kg sphere down .4587 meters on the end of a twenty meter pendulum string, it will be moving 3 m/sec. Have it embed in a 2 kilogram block that is suspended on the end of a one meter pendulum string. The twenty meter string releases; and the 3 kilogram combination will be moving 1 meter per second in a 1 meter pendulum. |
Since you have given me the speed after the collision, I assume that I may use this data in my calculations, since it is the result of your experiment.
Linear Newtonian Momentum is conserved.
Let's check:
Before, P = 1kg*3m/s = 3kgm/s; after, P = 3kg*1m/s = 3kgm/s. These are indeed the same, so yes, linear momentum is conserved.
QUOTE
Is angular momentum conserved?
Around which axis? The large pendulum is swinging around an axis 20m above the point of impact, and the small pendulum is swinging around an axis 1m above the point of impact.
Let's check:
Around the higher axis, before, L = 20m*1kg*3m/s = 60kgm^2/s; after, L = 20m*3kg*1m/s = 60kgm^2/s. These are indeed the same, so yes, the angular momentum around the higher axis is conserved.
Around the lower axis, before, L = 1m*1kg*3m/s = 3kgm^2/s; after, L = 1m*3kg*1m/s = 3kgm^/2. These are indeed the same, so yes, the angular momentum around the lower axis is conserved.
QUOTE (->
| QUOTE |
| Is angular momentum conserved? |
Around which axis? The large pendulum is swinging around an axis 20m above the point of impact, and the small pendulum is swinging around an axis 1m above the point of impact.
Let's check:
Around the higher axis, before, L = 20m*1kg*3m/s = 60kgm^2/s; after, L = 20m*3kg*1m/s = 60kgm^2/s. These are indeed the same, so yes, the angular momentum around the higher axis is conserved.
Around the lower axis, before, L = 1m*1kg*3m/s = 3kgm^2/s; after, L = 1m*3kg*1m/s = 3kgm^/2. These are indeed the same, so yes, the angular momentum around the lower axis is conserved.
Is kinetic energy conserved?
Let's check:
Before, KE = 1/2*1kg*(3m/s)^2 = 4.5J; after, KE = 1/2*3kg*(1m/s)^2 = 1.5J. These are different, and the final KE is only 1/3 of the initial KE. Therefore, 2/3 of the KE is lost to heat, sound, and vibration, and it is not conserved.
Since I used data that you provided, you cannot question their experimental basis without questioning your own experiment. Application of the definition of angular momentum around an axis shows that for each of the two obvious axes, the angular momentum is exactly the same before and after the collision. In fact, I could place the axis at ANY height, and the calculation would clearly come out the same before and after the collision. The idea of conservation of angular momentum never said that the angular momentum around two different axes must be the same, only that if you use the SAME axis before and after the collision, you will get the same value, and you can see from my calculation that you clearly do.
You must remember that the definition of angular momentum is r X p, where r is the vector from the axis location to the mass. For a large mass, you sum up the angular momentum of the mass at each location within it. The axis can be placed anywhere in the universe, and for each axis, you will get a different angular momentum. So there is not just ONE angular momentum for a given object; there is a different angular momentum for each axis you choose to compute around. This is directly from the definition of angular momentum. Conservation of angular momentum is always referred to a specific axis, and if the value of L around that same axis is equal before and after the collision, then we say that L is conserved around that axis. It may at the same time be not conserved around some other axis, because the whole conservation/nonconservation question is to be answered on a per-axis basis. Again, this is part of the definition of L.
Therefore, your own data proves that L is conserved around each of the pendulum axes. The only way you could think it is not conserved is if you use a different axis after the collision than you used before the collision. But that contradicts the definition of L, so if you do that, you aren't even comparing the same L's because they are around two different axes. Then you are comparing apples and oranges, and if you want to say they're not equal, well of course I'll agree with that. The L around the upper axis wasn't equal to the L around the lower axis even before the collision, so why should they be equal afterwards? It's perfectly true that they are unequal, it simply has nothing to do with conservation of L, that's all.
Hope that helps.
--Stuart Anderson
You have got to be kidding.
You pretend that a one meter pendulum has an axis twenty meters from the bob, or a twenty meter radius. Can’t you see all the ridicules garbage you have to come up with to make your phony math work? A one meter pendulum has a one meter radius. This isn’t never never land.
A twenty meter pendulum doesn’t have an axis one meter away or a one meter radius; it is a twenty meter radius. Rather than admit that angular momentum conservation is a false concept you come up with all this outrageous nonsense.
Homm quote: so if you do that, you aren't even comparing the same L's because they are around two different axes.
Well: that happens to be exactly what is happening, they are two different radii. Just because your math doesn’t work; don’t go inventing imaginary axes. Why didn’t you pick the moon for an axis? The bob on the end of a one meter pendulum does not know from whence it came, pick three, pick seven, pick the sun for an axis. You picked twenty because that makes you false theory look good.
Try not to get angry with me; I am not angry with you. I know this isn’t your theory per say, if you did not argue in favor of it someone else would. But angular momentum conservation in the lab is just a totally preposterous concept.
I took a 4.2 cm (1.5 in. inside diameter) outside diameter pipe and placed it on its side on one end of a cart. I also placed a 17.8 cm O.D. (13.9 cm I.D.) ring on the same end of the cart. I pointed both the ring and the pipe in a, soon to be, down hill direction. I then lifted that end of the cart, and the ring struck the other end of the cart a short time before the pipe.
Let’s say they hit at the same time and they were both rolling 1 m/sec, and let’s say they both had a mass of 1 kilogram. I could cut off a few inches of the pipe and they would have the same mass, and of course the short pipe would roll at the same rate as the long pipe.
What is the angular momentum of the pipe and the ring when they hit the other end of the cart?
And what does your answer do to F = ma?
F = ma does not make or conserve angular momentum it makes and conserves linear momentum.
You pretend that a one meter pendulum has an axis twenty meters from the bob, or a twenty meter radius. Can’t you see all the ridicules garbage you have to come up with to make your phony math work? A one meter pendulum has a one meter radius. This isn’t never never land.
A twenty meter pendulum doesn’t have an axis one meter away or a one meter radius; it is a twenty meter radius. Rather than admit that angular momentum conservation is a false concept you come up with all this outrageous nonsense.
Homm quote: so if you do that, you aren't even comparing the same L's because they are around two different axes.
Well: that happens to be exactly what is happening, they are two different radii. Just because your math doesn’t work; don’t go inventing imaginary axes. Why didn’t you pick the moon for an axis? The bob on the end of a one meter pendulum does not know from whence it came, pick three, pick seven, pick the sun for an axis. You picked twenty because that makes you false theory look good.
Try not to get angry with me; I am not angry with you. I know this isn’t your theory per say, if you did not argue in favor of it someone else would. But angular momentum conservation in the lab is just a totally preposterous concept.
I took a 4.2 cm (1.5 in. inside diameter) outside diameter pipe and placed it on its side on one end of a cart. I also placed a 17.8 cm O.D. (13.9 cm I.D.) ring on the same end of the cart. I pointed both the ring and the pipe in a, soon to be, down hill direction. I then lifted that end of the cart, and the ring struck the other end of the cart a short time before the pipe.
Let’s say they hit at the same time and they were both rolling 1 m/sec, and let’s say they both had a mass of 1 kilogram. I could cut off a few inches of the pipe and they would have the same mass, and of course the short pipe would roll at the same rate as the long pipe.
What is the angular momentum of the pipe and the ring when they hit the other end of the cart?
And what does your answer do to F = ma?
F = ma does not make or conserve angular momentum it makes and conserves linear momentum.
If you use the same twenty meter and one meter pendulums, but you increase the mass of the block by 2 kilograms the missing energy would be 80% instead of 66%. You would have the same collision surfaces and bearing but a difference of 14%. Energy conservation is a ridiculous concept as well. An imaginary unmeasured quantity of heat covers for the phony concept. Increase the mass of the block to 40 kilogram and you lose 97.5% of the energy, still using the same bearing point and the same contact surfaces.
In all the situations above, of 97.5%, 80%, and 66% energy loss, linear Newtonian Momentum is conserved. Angular momentum is not conserved useless you are willing to use an axis other than the one the bob is using, what rules of logic cover this mathematical use of a physically unused axis is beyond me.
Belief in unmeasured heat and arbitrary placement of the axis should be classified as dogma not science. And of course you can only cling onto your dogma by refusing to do experiments.
In all the situations above, of 97.5%, 80%, and 66% energy loss, linear Newtonian Momentum is conserved. Angular momentum is not conserved useless you are willing to use an axis other than the one the bob is using, what rules of logic cover this mathematical use of a physically unused axis is beyond me.
Belief in unmeasured heat and arbitrary placement of the axis should be classified as dogma not science. And of course you can only cling onto your dogma by refusing to do experiments.
QUOTE (Limon+Feb 27 2008, 01:04 AM)
And what does your answer do to F = ma?
F = ma does not make or conserve angular momentum it makes and conserves linear momentum.
Actually, it works for all kinds of momentum. You just have to be able to do multi-dimensional integrals so that you can work out the effect of F=ma on each 'element' of extended objects like wheels.
When you do the integrals, you end up with τ = Iθ'', which is the rotational version of F=ma, torque is equal to the moment of inertia of an object multipled by the acceleration of the angle about the axis used to define I.
This is completely in line with what Mr Homm said, that the equation depends on the axis you define your rotation about, but the conservation of angular momentum is always there, when you factor in the angular momentum of the Earth, which is producing the torque and thus having a torque applied on it (as inline with Newton's Third Law, but the rotational version).
As I said to you months ago, doing experiments is pointless because you don't know the theory which you are trying to disprove. You don't know how to put the values into the equations, so you put them in incorrectly and then find things don't match up. Rather than consider we have 350 years of experimental evidence for Newton's work, and things like cars work on torque principles, you think everyone else must be wrong and you're right.
Have you ever worked through a textbook on Newtonian mechanics? I am certain you haven't. And yet you think you understand a topic you have obviously never taken the time to grasp.
You're a staggeringly moronic fool.
F = ma does not make or conserve angular momentum it makes and conserves linear momentum.
Actually, it works for all kinds of momentum. You just have to be able to do multi-dimensional integrals so that you can work out the effect of F=ma on each 'element' of extended objects like wheels.
When you do the integrals, you end up with τ = Iθ'', which is the rotational version of F=ma, torque is equal to the moment of inertia of an object multipled by the acceleration of the angle about the axis used to define I.
This is completely in line with what Mr Homm said, that the equation depends on the axis you define your rotation about, but the conservation of angular momentum is always there, when you factor in the angular momentum of the Earth, which is producing the torque and thus having a torque applied on it (as inline with Newton's Third Law, but the rotational version).
As I said to you months ago, doing experiments is pointless because you don't know the theory which you are trying to disprove. You don't know how to put the values into the equations, so you put them in incorrectly and then find things don't match up. Rather than consider we have 350 years of experimental evidence for Newton's work, and things like cars work on torque principles, you think everyone else must be wrong and you're right.
Have you ever worked through a textbook on Newtonian mechanics? I am certain you haven't. And yet you think you understand a topic you have obviously never taken the time to grasp.
You're a staggeringly moronic fool.
First off AlphaNumeric you don’t impress anyone by calling others names, it is not part of an intelligent argument.
Mr Homm gave a clear and concise argument for the theories that are held by the scientific community; concerning angular momentum conservation and energy conservation. I appreciate his time and effort. I have the highest regard for his math ability and his intelligence. If anyone thought that I directed my arguments at him personally; I apologize.
Mathematics and mathematical formulas are to be used to describe the events in nature; the descriptions are to be concise enough that one even is distinct from another even.
If you assign an angular momentum number to a one meter pendulum whose one kilogram bob is moving one meter per second, then that number must be distinct to that particular pendulum. A different number must be assigned if the length changes and mass and velocity remain the same. A different number must be assigned if mass changes and length and velocity remain the same. A different number must be assigned if velocity changes and length and mass remain the same. Otherwise you can not define one pendulum from another.
So lets say you enter a room in a laboratory and you find a 1 meter pendulum swinging one meter per second (at the down swing) that has a bob with a mass of 3 kilograms. One and only one number (angular momentum number) defines this pendulum. If you then pretend that the length changes to twenty then you are talking about a different pendulum. Mr Homm then compares this different pendulum to the other twenty meter pendulum. But it is not the original one meter pendulum.
If this is the kind of thing you need to do to support a theory then you should consider that the theory may be incorrect.
Mr Homm gave a clear and concise argument for the theories that are held by the scientific community; concerning angular momentum conservation and energy conservation. I appreciate his time and effort. I have the highest regard for his math ability and his intelligence. If anyone thought that I directed my arguments at him personally; I apologize.
Mathematics and mathematical formulas are to be used to describe the events in nature; the descriptions are to be concise enough that one even is distinct from another even.
If you assign an angular momentum number to a one meter pendulum whose one kilogram bob is moving one meter per second, then that number must be distinct to that particular pendulum. A different number must be assigned if the length changes and mass and velocity remain the same. A different number must be assigned if mass changes and length and velocity remain the same. A different number must be assigned if velocity changes and length and mass remain the same. Otherwise you can not define one pendulum from another.
So lets say you enter a room in a laboratory and you find a 1 meter pendulum swinging one meter per second (at the down swing) that has a bob with a mass of 3 kilograms. One and only one number (angular momentum number) defines this pendulum. If you then pretend that the length changes to twenty then you are talking about a different pendulum. Mr Homm then compares this different pendulum to the other twenty meter pendulum. But it is not the original one meter pendulum.
If this is the kind of thing you need to do to support a theory then you should consider that the theory may be incorrect.
QUOTE (Limon+Feb 29 2008, 12:10 AM)
First off AlphaNumeric you don’t impress anyone by calling others names, it is not part of an intelligent argument.
No, it's just an added bonus to an otherwise intelligent argument from myself.
Where's your intelligent argument? You couldn't counter my point that if you don't understand the equations, experiments are pointless because you cannot see how the equations fit your experimental results together.
You do the experiment, bugger up the equations and then whine that it's everyone else's fault you couldn't do the equations properly.
Let's see you answer question 4. Just to prove to everyone you can do the equations needed to compute the moment of inertia of simple objects about things other than their centre of mass (and even then things like a cone aren't simple).
Where's your intelligent argument? You couldn't counter my point that if you don't understand the equations, experiments are pointless because you cannot see how the equations fit your experimental results together.
You do the experiment, bugger up the equations and then whine that it's everyone else's fault you couldn't do the equations properly.
Let's see you answer question 4. Just to prove to everyone you can do the equations needed to compute the moment of inertia of simple objects about things other than their centre of mass (and even then things like a cone aren't simple).
QUOTE (Limon+Feb 29 2008, 12:10 AM)
If this is the kind of thing you need to do to support a theory then you should consider that the theory may be incorrect.
You should learn the perpendicular axis theorem. But then you should learn a lot of things.
It doesn't matter where you define your axis of rotation, angular momentum is ALWAYS conserved, provided you keep the axis the same before and after your calculation. Things like the perpendicular axis theorem allow you to move your axis of rotation about in space. Anyone whose studied angular momentum knows this.
You don't.
You should learn the perpendicular axis theorem. But then you should learn a lot of things.
It doesn't matter where you define your axis of rotation, angular momentum is ALWAYS conserved, provided you keep the axis the same before and after your calculation. Things like the perpendicular axis theorem allow you to move your axis of rotation about in space. Anyone whose studied angular momentum knows this.
You don't.
QUOTE (Limon+Feb 27 2008, 10:44 PM)
what rules of logic cover this mathematical use of a physically unused axis is beyond me.
It's quite clear that this is the underlying problem.
It's quite clear that this is the underlying problem.
Okay; you have a one meter pendulum with a one kilogram bob moving one meter per second. How many numbers do you get to describe this event if you are allowed to pick an axis (any axis) other than the one that the bob is using? It would be an infinite number wouldn’t it? You should be able to see this as a logical fallacy.
If you can give one event any number then of course you can make it equal to any other event that itself can be given any number.
If one even is equal to anything then anything is equal to it. Angular momentum conservation, in the lab, is a logical fallacy.
If you can give one event any number then of course you can make it equal to any other event that itself can be given any number.
If one even is equal to anything then anything is equal to it. Angular momentum conservation, in the lab, is a logical fallacy.
QUOTE (Limon+Mar 1 2008, 04:15 AM)
..... is a logical fallacy.
It's quite logical to assume you're a phallus.
It's quite logical to assume you're a phallus.
QUOTE (Limon+Mar 1 2008, 05:15 AM)
Okay; you have a one meter pendulum with a one kilogram bob moving one meter per second. How many numbers do you get to describe this event if you are allowed to pick an axis (any axis) other than the one that the bob is using? It would be an infinite number wouldn’t it? You should be able to see this as a logical fallacy.
If you can give one event any number then of course you can make it equal to any other event that itself can be given any number.
You obviously have never actually done any calculations.
Linear momentum is easy to define, since it doesn't require a reference point. Angular momentum requires an axis about which you define your rotation.
Let's take your example : A 1 kg mass moving at 1m/s in a straight line. What is it's angular momentum?
Any ideas?
Any at all?
Well it depends on where your axis is. You can put it anyway. In doing that, you'll give your blob some amount of angular momentum. This is no different to having the freedom for linear momentum to pick your notion of 'zero velocity', since speed is relative.* So just as boosting to a different inertial frame gives your 1kg mass a different velocity, moving my axis of rotation gives it a different angular momentum.
* Infact, speed boosts and rotations are the two parts of Lorentz transforms!
Is this a problem? No, of course not. Why? Because the conservation of linear and angular momentum still applies. The linear momentum before and the linear momentum after some interaction are still equal. The angular momentum before and the angular momentum afterwards are still equal.
Suppose I've got two objects, Objects 1 and 2. In some inertial frame Object 1 has x amount of linear momentum, Object 2 has y Total linear momentum is x+y. I now boost to a different inertial frame where Object 1 has X amount of momentum, 2 has Y. Total amount X+Y. The objects then interact. In the original inertial frame the total linear momentum afterwards will be x+y. In the second inertial frame it's X+Y.
Despite them having different momenta in different frames, conservation still occurs.
Picking a different axis of rotation is exactly the same for angular momentum, the values are different but the total before is still the same as the total afterwards.
This demonstrates you don't even grasp the concepts of momentum, linear or angular.
I notice you completely ignored my challenge you answer some relatively simple questions expected of students. And you, again, ignored my question about putting in numbers incorrectly to equations renders the experiments pointless.
I guess we all know why, you don't want to admit you cannot do the equations so all your whining is moronic. Your display of ignorance just puts another nail in your coffin. Why do you waste so much time avoiding learning any of the physics you spend so much time trying to discuss? It's like living in France and deliberately trying to avoid French people and not learning the language.
If you can give one event any number then of course you can make it equal to any other event that itself can be given any number.
You obviously have never actually done any calculations.
Linear momentum is easy to define, since it doesn't require a reference point. Angular momentum requires an axis about which you define your rotation.
Let's take your example : A 1 kg mass moving at 1m/s in a straight line. What is it's angular momentum?
Any ideas?
Any at all?
Well it depends on where your axis is. You can put it anyway. In doing that, you'll give your blob some amount of angular momentum. This is no different to having the freedom for linear momentum to pick your notion of 'zero velocity', since speed is relative.* So just as boosting to a different inertial frame gives your 1kg mass a different velocity, moving my axis of rotation gives it a different angular momentum.
* Infact, speed boosts and rotations are the two parts of Lorentz transforms!
Is this a problem? No, of course not. Why? Because the conservation of linear and angular momentum still applies. The linear momentum before and the linear momentum after some interaction are still equal. The angular momentum before and the angular momentum afterwards are still equal.
Suppose I've got two objects, Objects 1 and 2. In some inertial frame Object 1 has x amount of linear momentum, Object 2 has y Total linear momentum is x+y. I now boost to a different inertial frame where Object 1 has X amount of momentum, 2 has Y. Total amount X+Y. The objects then interact. In the original inertial frame the total linear momentum afterwards will be x+y. In the second inertial frame it's X+Y.
Despite them having different momenta in different frames, conservation still occurs.
Picking a different axis of rotation is exactly the same for angular momentum, the values are different but the total before is still the same as the total afterwards.
This demonstrates you don't even grasp the concepts of momentum, linear or angular.
I notice you completely ignored my challenge you answer some relatively simple questions expected of students. And you, again, ignored my question about putting in numbers incorrectly to equations renders the experiments pointless.
I guess we all know why, you don't want to admit you cannot do the equations so all your whining is moronic. Your display of ignorance just puts another nail in your coffin. Why do you waste so much time avoiding learning any of the physics you spend so much time trying to discuss? It's like living in France and deliberately trying to avoid French people and not learning the language.
AlphaNumeric quote: “Let's take your example : A 1 kg mass moving at 1m/s in a straight line. What is it's angular momentum?
Well it depends on where your axis is. You can put it anyway. In doing that, you'll give your blob some amount of angular momentum. ------ moving my axis of rotation gives it a different angular momentum.”
Yes assigning any axis gives you a different number (angular momentum).
So: A 1 kg mass moving at 1m/s in a straight line can have an infinite number of angular momentums because you can give it an infinite number of axes.
So to you a one newton force working for one second on a one kilogram mass will give you an infinite number of angular momentums.
A 1 kg mass moving at 1m/s in a straight line, is the motion caused by one newton working for one second. And you can give it an infinite number of angular momentums.
A certain Force working for a certain time can give you one and only one Momentum. F = ma
You can not pretend that some special something is conserved when you can assign one particular event an infinite number of angular momentums.
Well it depends on where your axis is. You can put it anyway. In doing that, you'll give your blob some amount of angular momentum. ------ moving my axis of rotation gives it a different angular momentum.”
Yes assigning any axis gives you a different number (angular momentum).
So: A 1 kg mass moving at 1m/s in a straight line can have an infinite number of angular momentums because you can give it an infinite number of axes.
So to you a one newton force working for one second on a one kilogram mass will give you an infinite number of angular momentums.
A 1 kg mass moving at 1m/s in a straight line, is the motion caused by one newton working for one second. And you can give it an infinite number of angular momentums.
A certain Force working for a certain time can give you one and only one Momentum. F = ma
You can not pretend that some special something is conserved when you can assign one particular event an infinite number of angular momentums.
QUOTE (Limon+Mar 2 2008, 05:03 AM)
A certain Force working for a certain time can give you one and only one Momentum. F = ma
No, it'll give you one and only one change in momentum, since F = dp/dt.
It depends entirely on what velocity it's moving at initially. So there's infinitely many momenta it could have.
However, the total momentum before and the total momentum after, when you consider the entire system, will be the same.
Just as I explained in my previous post and precisely what applies to angular momentum too. You didn't bother to try to understand, did you? Or did you try and you're just too stupid to grasp it?
No, it'll give you one and only one change in momentum, since F = dp/dt.
It depends entirely on what velocity it's moving at initially. So there's infinitely many momenta it could have.
However, the total momentum before and the total momentum after, when you consider the entire system, will be the same.
Just as I explained in my previous post and precisely what applies to angular momentum too. You didn't bother to try to understand, did you? Or did you try and you're just too stupid to grasp it?
QUOTE (Limon+Mar 2 2008, 05:03 AM)
You can not pretend that some special something is conserved when you can assign one particular event an infinite number of angular momentums.
You pick an axis and then you compute the angular momentum about that axis before and after. Just as you pick a frame for linear momentum and compute the linear momentum in that frame before and after. In both cases the relevant momentum is conserved.
Come on, at least try to understand. All you're doing it making yourself look thick as ****.
And you, again, ignored the questions I asked you. Are you too scared to admit you cannot do them? If you ignore me again, I'll take it that you do accept you're too scared to do it.
Come on, at least try to understand. All you're doing it making yourself look thick as ****.
And you, again, ignored the questions I asked you. Are you too scared to admit you cannot do them? If you ignore me again, I'll take it that you do accept you're too scared to do it.
Momentum is defined as mass times velocity. Momentum occurs when a certain force is applied to a mass for a period of time. Angular momentum does not conserve this quantity in the lab. Therefore the concepts of angular momentum conservation are false.
It follows that angular momentum conservation can not be used to discredit experiments like the cylinder and spheres experiment, by claiming that the cylinder and spheres experiment must comply with the false law of angular momentum conservation.
Newton clearly states in his discussion of his First Law of Motion that rotational motion preserves linear motion.
It follows that angular momentum conservation can not be used to discredit experiments like the cylinder and spheres experiment, by claiming that the cylinder and spheres experiment must comply with the false law of angular momentum conservation.
Newton clearly states in his discussion of his First Law of Motion that rotational motion preserves linear motion.
QUOTE (Limon+Mar 3 2008, 06:49 AM)
Momentum is defined as mass times velocity. Momentum occurs when a certain force is applied to a mass for a period of time. Angular momentum does not conserve this quantity in the lab. Therefore the concepts of angular momentum conservation are false.
The fact that text books sight the skater should tell you something, why did they pick an irregular mass with such a complex motion, and muscles involved no less? Why not something simple like a 20 meter pendulum striking a 1 meter pendulum?
Here is an experiment that will prove that the angular momentum theory is false.
Suspend a pendulum bob from a sled horizontally mounted on a roller track; accelerate both the sled and the bob to one meter per second keeping the string in a vertical position. Have the sled impact a stationary block. The bob should rise .051 meters. Do the same experiment with one, two, and three meter strings for the pendulum lengths, the bob should rise .051 meters in all situations.
Velocity (before rise) of the bob is maintained in the three different pendulum lengths.
Linear momentum of the bob is maintained in the three different pendulum lengths.
Angular momentum is not maintained in the three different pendulum lengths.
It is true that linear momentum is conserved under different frames of reference. Let’s say a one kilogram mass moving 1 m/sec on a frictionless plane is struck from behind by a 3 kilogram mass moving 3 m/sec. You can use the frame of reference of any point on the air table; the frame of reference of any point in the one kilogram mass, or the frame of reference of any point in the 3 kilogram mass and momentum is always conserved from that vantage point.
This is the exact same event seen from the vantage points of three different objects, two of the vantage points are in motion and one vantage point is outside the motion. There is nothing imaginary about the frames of references in the 3 and 1 kilogram pucks experiment.
But when you change your axis all axes are using the same point of reference as that used by the air table, it is not a change in reference it is merely a means of pretending that the radius has not changed, which of course has to be avoided because if the radius changes your conservation of angular momentum is lost. Further it is not the same event; a twenty meter pendulum is not a one meter pendulum. One or the other pendulums is a figment of the imagination. There is nothing imaginary about the frames of references in the 3 and 1 kilogram pucks experiment.
Here is an experiment that will prove that the angular momentum theory is false.
Suspend a pendulum bob from a sled horizontally mounted on a roller track; accelerate both the sled and the bob to one meter per second keeping the string in a vertical position. Have the sled impact a stationary block. The bob should rise .051 meters. Do the same experiment with one, two, and three meter strings for the pendulum lengths, the bob should rise .051 meters in all situations.
Velocity (before rise) of the bob is maintained in the three different pendulum lengths.
Linear momentum of the bob is maintained in the three different pendulum lengths.
Angular momentum is not maintained in the three different pendulum lengths.
It is true that linear momentum is conserved under different frames of reference. Let’s say a one kilogram mass moving 1 m/sec on a frictionless plane is struck from behind by a 3 kilogram mass moving 3 m/sec. You can use the frame of reference of any point on the air table; the frame of reference of any point in the one kilogram mass, or the frame of reference of any point in the 3 kilogram mass and momentum is always conserved from that vantage point.
This is the exact same event seen from the vantage points of three different objects, two of the vantage points are in motion and one vantage point is outside the motion. There is nothing imaginary about the frames of references in the 3 and 1 kilogram pucks experiment.
But when you change your axis all axes are using the same point of reference as that used by the air table, it is not a change in reference it is merely a means of pretending that the radius has not changed, which of course has to be avoided because if the radius changes your conservation of angular momentum is lost. Further it is not the same event; a twenty meter pendulum is not a one meter pendulum. One or the other pendulums is a figment of the imagination. There is nothing imaginary about the frames of references in the 3 and 1 kilogram pucks experiment.
QUOTE (AlphaNumeric to Lemon+)
You didn't bother to try to understand, did you? Or did you try and you're just too stupid to grasp it?
I'm voting for "too Stupid"
I'm voting for "too Stupid"
QUOTE (N O M+Mar 4 2008, 03:09 AM)
QUOTE (AlphaNumeric to Lemon+)
You didn't bother to try to understand, did you? Or did you try and you're just too stupid to grasp it?
I'm voting for "too Stupid"
My money's on "exceedingly far too stupid, in the extreme"
I'm voting for "too Stupid"
My money's on "exceedingly far too stupid, in the extreme"
Angular momentum is composed of three quantities, mass, linear velocity, and radius. Radius appears in the formula three times.
The distance traveled around the circumference of the circle in a unit period of time is equal to linear velocity. This linear velocity is then divided by radius to obtain radians per sec. The product of mass and radians (linear velocity/radius) per sec is then multiplied by radius two more times. Since radians are linear velocity divided by radius and we multiply by radius two more times this makes two radii drop out of the equation. This leaves us with mass times linear velocity times radius; one radius beyond Newtonian Physics.
This is the funny thing about angular momentum, why not just drop the one extra radius and everything would work just fine. The reason the third radius is not dropped is that angular momentum was developed by Kepler for use with satellites (comets, planets, and etc.) and the extra radius was necessary to compensate for a huge increase in linear velocity caused by gravity. For satellites a small radius means huge linear velocity, and a huge radius mean small linear velocity, because of gravitational acceleration.
This brings us to yet another proof that angular momentum conservation will not work in the lab. It should be obvious to any mathematician that a formula that works for satellites where you have huge increases in linear velocity caused by gravity can not possibly work in situations where there is no increase in linear velocity caused by gravity.
The distance traveled around the circumference of the circle in a unit period of time is equal to linear velocity. This linear velocity is then divided by radius to obtain radians per sec. The product of mass and radians (linear velocity/radius) per sec is then multiplied by radius two more times. Since radians are linear velocity divided by radius and we multiply by radius two more times this makes two radii drop out of the equation. This leaves us with mass times linear velocity times radius; one radius beyond Newtonian Physics.
This is the funny thing about angular momentum, why not just drop the one extra radius and everything would work just fine. The reason the third radius is not dropped is that angular momentum was developed by Kepler for use with satellites (comets, planets, and etc.) and the extra radius was necessary to compensate for a huge increase in linear velocity caused by gravity. For satellites a small radius means huge linear velocity, and a huge radius mean small linear velocity, because of gravitational acceleration.
This brings us to yet another proof that angular momentum conservation will not work in the lab. It should be obvious to any mathematician that a formula that works for satellites where you have huge increases in linear velocity caused by gravity can not possibly work in situations where there is no increase in linear velocity caused by gravity.
A thought experiment that should clarify the impossibility of angular momentum conservation in the lab would be to put a pin at the position of the Sun and run a string to the comet at apogee. The comet will continue to rotate around the pin at a distance of apogee.
Place a second pin at a distance from the comet that would have been equal to perigee. Place this second pin between the comet and the first pin so that the string comes in contact with the second pin and the comet then begins orbiting the second pin at a distance of perigee. No linear velocity change has occurred because there is no gravitational acceleration caused by the Sun.
The angular momentum formula is: M ω R² where ω equals radians per sec or linear velocity / r.
M * (linear velocity / r ) * r * r = M * linear velocity * r
The two formulas that have to be equal to each other if angular momentum conservation is to be true are:
M * linear velocity * apogee = M * linear velocity * perigee With the use of pins mass and linear velocity remained constant; and the equation is false because apogee does not equal perigee. The two equations are equal only if gravity causes a huge increase in linear velocity.
With the Sun back in place; linear velocity at apogee / linear velocity at perigee = perigee / apogee, and the equation (M * linear velocity at apogee * apogee = M * linear velocity at perigee * perigee) is true.
This two pin experiment can be performed on a frictionless plane; only the scale is changed but the principle remains the same. If there is no gravitational acceleration angular momentum conservation does not work.
Kepler needed no frame of reference, he got it right. Your frame of reference only pretends that the experiment was not conducted, by pretending that the radius remains the same.
Place a second pin at a distance from the comet that would have been equal to perigee. Place this second pin between the comet and the first pin so that the string comes in contact with the second pin and the comet then begins orbiting the second pin at a distance of perigee. No linear velocity change has occurred because there is no gravitational acceleration caused by the Sun.
The angular momentum formula is: M ω R² where ω equals radians per sec or linear velocity / r.
M * (linear velocity / r ) * r * r = M * linear velocity * r
The two formulas that have to be equal to each other if angular momentum conservation is to be true are:
M * linear velocity * apogee = M * linear velocity * perigee With the use of pins mass and linear velocity remained constant; and the equation is false because apogee does not equal perigee. The two equations are equal only if gravity causes a huge increase in linear velocity.
With the Sun back in place; linear velocity at apogee / linear velocity at perigee = perigee / apogee, and the equation (M * linear velocity at apogee * apogee = M * linear velocity at perigee * perigee) is true.
This two pin experiment can be performed on a frictionless plane; only the scale is changed but the principle remains the same. If there is no gravitational acceleration angular momentum conservation does not work.
Kepler needed no frame of reference, he got it right. Your frame of reference only pretends that the experiment was not conducted, by pretending that the radius remains the same.
Limon, why do you continue to ignore my questions?
1. Do you know anything about the parallel and perpendicular axis theorems?
2. Question 4
3. Do you not think that experiments are pointless if you cannot do the relevent equations? Why use is having numbers to put into equations when you cannot do the equations?
Come on, simple direct questions. Try to answer them directly and relevently.
1. Do you know anything about the parallel and perpendicular axis theorems?
2. Question 4
3. Do you not think that experiments are pointless if you cannot do the relevent equations? Why use is having numbers to put into equations when you cannot do the equations?
Come on, simple direct questions. Try to answer them directly and relevently.
You could release the comet orbiting pin #2 in the direction of pin #1. Have the comet come within a perigee distance of pin #1 and recapture it on the end of a string. Now your point of reference that you want to use is the same point around which the comet is circling, except now the radius is diminished. So it is not an issue of position is space it is only that you don’t want to use the real radius for your angular momentum calculations. And no wonder you don’t want to use the real radius because that will prove that angular momentum conservation is false.
You could build a cylinder and spheres device and attempt to prove that your angular momentum conservation overrides linear momentum conservation, the spheres can have only one velocity in the open position. Only 2 times the original velocity satisfies angular momentum conservation, but it will take 4 times the original velocity to satisfy linear momentum conservation. This is in a 3 to 1 mass relationship between the cylinder and spheres. The experiment should be awarded the Nobel Prize in Physics.
When objects rotating on the end of a string wrap around a thick center post it does not have a change in linear velocity. If a one kilogram object moving one meter per second starts wrapping its one meter string around a stationary post it will be moving at exactly the same speed when the string shortens to .5 meters or .25 meter. This means that angular momentum is diminishing. Also; if the circling object is unwrapping from a post there will be no change in linear velocity, you could release the circling object and direct it to move in the same direction with the same linear momentum at 10 meters radius as well as at 2.5 meters radius.
Proponents of angular momentum will use the frame of reference notion to keep the radius the same. But by doing this they basically ignore the experiment.
Real scientists don’t ignore reality: the radius is really changing and velocity is not. The best way to deal with this reality is to accept Newton’s view; (mv) radius has nothing to do with momentum.
I can change the string length on the cylinder and spheres and it will not change the final maximum velocity of the spheres (according to my measurements). Newtonian Physics works: no frame of reference is necessary.
Being able to plug numbers into a formula does not mean the formula is correct, nor does it mean that the one who is doing the plugging knows what the formula is saying or why the formula should work. It may only mean that the student can emulate the behavior required by the professor.
You could build a cylinder and spheres device and attempt to prove that your angular momentum conservation overrides linear momentum conservation, the spheres can have only one velocity in the open position. Only 2 times the original velocity satisfies angular momentum conservation, but it will take 4 times the original velocity to satisfy linear momentum conservation. This is in a 3 to 1 mass relationship between the cylinder and spheres. The experiment should be awarded the Nobel Prize in Physics.
When objects rotating on the end of a string wrap around a thick center post it does not have a change in linear velocity. If a one kilogram object moving one meter per second starts wrapping its one meter string around a stationary post it will be moving at exactly the same speed when the string shortens to .5 meters or .25 meter. This means that angular momentum is diminishing. Also; if the circling object is unwrapping from a post there will be no change in linear velocity, you could release the circling object and direct it to move in the same direction with the same linear momentum at 10 meters radius as well as at 2.5 meters radius.
Proponents of angular momentum will use the frame of reference notion to keep the radius the same. But by doing this they basically ignore the experiment.
Real scientists don’t ignore reality: the radius is really changing and velocity is not. The best way to deal with this reality is to accept Newton’s view; (mv) radius has nothing to do with momentum.
I can change the string length on the cylinder and spheres and it will not change the final maximum velocity of the spheres (according to my measurements). Newtonian Physics works: no frame of reference is necessary.
Being able to plug numbers into a formula does not mean the formula is correct, nor does it mean that the one who is doing the plugging knows what the formula is saying or why the formula should work. It may only mean that the student can emulate the behavior required by the professor.
QUOTE (Limon+Mar 7 2008, 11:43 PM)
Being able to plug numbers into a formula does not mean the formula is correct
No, but you can demonstrate it isn't immediately wrong.
You claim that if you do an accurate, well conducted, experiment and then put those obtained numbers into the equations for angular momentum, it doesn't work.
This means either the equations are wrong, your experimentation is poor or you cannot grasp how to turn your experimental results into the right things to put into the equation.
You think it's the equations which are wrong and you cannot accept that it might be that you don't know how to manipulate/apply the equations properly.
Consider F=ma. If you measure the force to be 10N, the mass of an object to be 10kg, F=ma says the object will be accelerated at 1m/s/s. But if youo put into the equation that m=20, not 10, then you'll get that a=0.5, yet you'll measure it to be 1. Is it the equation is wrong or your use of the equation?
As has been repeatedly demonstrated, you don't know how to compute the moment of inertia of objects.
I have asked you three direct questions. You ignored them all. Why was that? Are you unable to answer them?
No, but you can demonstrate it isn't immediately wrong.
You claim that if you do an accurate, well conducted, experiment and then put those obtained numbers into the equations for angular momentum, it doesn't work.
This means either the equations are wrong, your experimentation is poor or you cannot grasp how to turn your experimental results into the right things to put into the equation.
You think it's the equations which are wrong and you cannot accept that it might be that you don't know how to manipulate/apply the equations properly.
Consider F=ma. If you measure the force to be 10N, the mass of an object to be 10kg, F=ma says the object will be accelerated at 1m/s/s. But if youo put into the equation that m=20, not 10, then you'll get that a=0.5, yet you'll measure it to be 1. Is it the equation is wrong or your use of the equation?
As has been repeatedly demonstrated, you don't know how to compute the moment of inertia of objects.
I have asked you three direct questions. You ignored them all. Why was that? Are you unable to answer them?
The cylinder and spheres experiment takes the motion of four units of mass and places all that motion in one unit of mass. This transfer of motion guarantees that one or more of these equations is false. mv, 1/2mv², and or MωR²
Who claims that kinetic energy is conserved? That's obviously not true. Two massive objects seperated in space but not moving relative to one another will accelerate towards one another, their kinetic energy increasing all the time at the expensve of potential energy. Their momenta will always sum to zero. And if they have tangential motion relative to the line which joins them, they have non-zero angular momentum and that is conserved.
You demonstrate you don't know how to do the equations.
Maybe that explains why you have so much trouble answering three direct questions.
Either address the questions I asked you or admit you cannot do them.
Ignoring them just demonstrates how BS you realise your position is but you cannot admit it to yourself.
You demonstrate you don't know how to do the equations.
Maybe that explains why you have so much trouble answering three direct questions.
Either address the questions I asked you or admit you cannot do them.
Ignoring them just demonstrates how BS you realise your position is but you cannot admit it to yourself.
Even the comet thought experiment could be done on a frictionless plane, I have done many experiments like these mentioned. Why don’t you pick one and evaluate it? How about the roller track sled experiment, why is angular momentum not conserved when all the input motions are the same?
QUOTE (AlphaNumeric+Mar 8 2008, 09:29 PM)
Either address the questions I asked you or admit you cannot do them.
Ignoring them just demonstrates how BS you realise your position is but you cannot admit it to yourself.
Again.
Ignoring them just demonstrates how BS you realise your position is but you cannot admit it to yourself.
Again.
alphanumeric-já que oce adora a lingua portuguesa( sua tradução é absurda e ridícula).SUAS PALAVRAS SAO PURAS BOBAGENS.
Both Galileo* and Newton² made clear that you should not use a description of an event that is more complex than is sufficient to explain the event. This axiom would include any mathematical expressions.
Let’s explain this event. A 10 kilogram balanced rim has a mass that is rotating in a horizontal plane at 1.1 m/sec. It has a high quality center bearing, and an attached string is being threaded around its circumference. On the other end of the string is a one kilogram mass resting on an air table. Assume that the string does not stretch or break. AlphaNumeric or others: What happens when the string comes taut?
_______________
* Galileo Galilei: from Naturally Accelerated Motion. “Finally, in the investigation of naturally accelerated motion we are led, by hand as it were, in following the habit and custom of nature herself, in all her various other processes, to employ only those means which are most common, simple and easy.-------When therefore, I observe a stone initially at rest falling from an elevated position and continually acquiring new increments of speed, why should I not believe that such increases take place in a manner which is exceedingly simple and rather obvious to everyone.”
² Isaac Newton: from Rules of Reasoning in Philosophy, Rule I, We are to admit no more causes of natural things than such as are both true and sufficient to explain their appearances.
Let’s explain this event. A 10 kilogram balanced rim has a mass that is rotating in a horizontal plane at 1.1 m/sec. It has a high quality center bearing, and an attached string is being threaded around its circumference. On the other end of the string is a one kilogram mass resting on an air table. Assume that the string does not stretch or break. AlphaNumeric or others: What happens when the string comes taut?
_______________
* Galileo Galilei: from Naturally Accelerated Motion. “Finally, in the investigation of naturally accelerated motion we are led, by hand as it were, in following the habit and custom of nature herself, in all her various other processes, to employ only those means which are most common, simple and easy.-------When therefore, I observe a stone initially at rest falling from an elevated position and continually acquiring new increments of speed, why should I not believe that such increases take place in a manner which is exceedingly simple and rather obvious to everyone.”
² Isaac Newton: from Rules of Reasoning in Philosophy, Rule I, We are to admit no more causes of natural things than such as are both true and sufficient to explain their appearances.
QUOTE (Limon+Mar 12 2008, 12:48 AM)
Let’s explain this event. A 10 kilogram balanced rim has a mass that is rotating in a horizontal plane at 1.1 m/sec. It has a high quality center bearing, and an attached string is being threaded around its circumference. On the other end of the string is a one kilogram mass resting on an air table. Assume that the string does not stretch or break. AlphaNumeric or others: What happens when the string comes taut?
First I'd like to you see you work out the angular momentum of the system. I want to see how you answer it, since you seem to be ignoring my questions :
Either address the questions I asked you or admit you cannot do them.
Ignoring them just demonstrates how BS you realise your position is but you cannot admit it to yourself.
Funny how you'll read the wordy works of Newton and Galileo but you can't actually do their mathematical work. I have asked you several times to compute the angular momentum of a few shapes, to show you know how to translate the Newtonian dynamics into proper physical predictions and you ignore me.
Are you so pathetic you cannot admit you cannot do it?
And Mott, my Portuguese is as poor as your English. If you are insulted by my use of Google Translate, too bad.
Either address the questions I asked you or admit you cannot do them.
Ignoring them just demonstrates how BS you realise your position is but you cannot admit it to yourself.
Funny how you'll read the wordy works of Newton and Galileo but you can't actually do their mathematical work. I have asked you several times to compute the angular momentum of a few shapes, to show you know how to translate the Newtonian dynamics into proper physical predictions and you ignore me.
Are you so pathetic you cannot admit you cannot do it?
And Mott, my Portuguese is as poor as your English. If you are insulted by my use of Google Translate, too bad.
Algebra is sufficient to explain the event.
10 kg moving 1.1 m/sec = 11 initial units of momentum.
Final velocity will be; 11kg * v = 11 units of momentum, which is 1 m/sec.
This experiment kills two birds with one stone.
First; not only do we not need AlphaNumeric’s more complex formula, it is a violation of scientific procedure to use it. Both Galileo and Newton knew that truth can be lost in complexity; people will pretend to know and understand more complex formulas in an attempt to impress others, when the formula itself and/or their understanding of the formula is faulty.
Second; the balanced rim has mass moving south and an equal amount of mass moving north, if you use vectors to let the north momentum cancel the south momentum (and momentum toward the east cancels westerly momentum, and so forth around the rim) then how can the wheel give momentum to the one kilogram mass on the frictionless plane when you say it has no momentum?
You will probably say “well it has energy but not momentum”. That is funny because it is conserving momentum not energy. So you say it conserves something it does not have and can not conserve that which you assume it does have. This is the kind of nonsense that Galileo and Newton where warning against.
10 kg moving 1.1 m/sec = 11 initial units of momentum.
Final velocity will be; 11kg * v = 11 units of momentum, which is 1 m/sec.
This experiment kills two birds with one stone.
First; not only do we not need AlphaNumeric’s more complex formula, it is a violation of scientific procedure to use it. Both Galileo and Newton knew that truth can be lost in complexity; people will pretend to know and understand more complex formulas in an attempt to impress others, when the formula itself and/or their understanding of the formula is faulty.
Second; the balanced rim has mass moving south and an equal amount of mass moving north, if you use vectors to let the north momentum cancel the south momentum (and momentum toward the east cancels westerly momentum, and so forth around the rim) then how can the wheel give momentum to the one kilogram mass on the frictionless plane when you say it has no momentum?
You will probably say “well it has energy but not momentum”. That is funny because it is conserving momentum not energy. So you say it conserves something it does not have and can not conserve that which you assume it does have. This is the kind of nonsense that Galileo and Newton where warning against.
QUOTE (Limon+Mar 12 2008, 09:21 PM)
then how can the wheel give momentum to the one kilogram mass on the frictionless plane when you say it has no momentum?
You will probably say “well it has energy but not momentum”. That is funny because it is conserving momentum not energy. So you say it conserves something it does not have and can not conserve that which you assume it does have. This is the kind of nonsense that Galileo and Newton where warning against.
Obviously it gets the momentum from the object the wheel is ultimately attached to - the Earth.
Looks like you missed both birds.
You will probably say “well it has energy but not momentum”. That is funny because it is conserving momentum not energy. So you say it conserves something it does not have and can not conserve that which you assume it does have. This is the kind of nonsense that Galileo and Newton where warning against.
Obviously it gets the momentum from the object the wheel is ultimately attached to - the Earth.
Looks like you missed both birds.
This is the kind of nonsense that Galileo and Newton were warning against.
If it got the motion from the earth then why does the wheel slow down? Even the Coriolis effect does not speed up a pendulum it only changes its direction.
If it got the motion from the earth then why does the wheel slow down? Even the Coriolis effect does not speed up a pendulum it only changes its direction.
QUOTE (Limon+Mar 12 2008, 10:21 PM)
First; not only do we not need AlphaNumeric’s more complex formula, it is a violation of scientific procedure to use it.
The formulae I talk about are not 'complex' on the scale of things. It's just too complex for you.
The formulae I talk about are not 'complex' on the scale of things. It's just too complex for you.
QUOTE (Limon+Mar 12 2008, 10:21 PM)
Both Galileo and Newton knew that truth can be lost in complexity; people will pretend to know and understand more complex formulas in an attempt to impress others, when the formula itself and/or their understanding of the formula is faulty.
This is a very pathetic and very transparent excuse for why you haven't bothered to learn any mechanics. You don't know any and now you're trying to use your ignorance as a defence.
It's not a matter of impressing people. I'm deeply unimpressed that for all your whining about angular momentum you don't even know how to compute it!
Newton saw the power of calculus in accurately describing things. He invented it, after all! So rather than waving your arms, crunch some numbers other than "10 = 11 is wrong".
Remember how you couldn't grasp momentum conservation still exists despite the initial momentum being definable as anything you like? You were inconsistent with your application of "Choice of frame" (or rather you just failed to understand it at all) when talking about angular and linear momenta. This came about because you've never done any calculations. And then you assume that because you can't get the algebra to work, it's the algebra's fault. No, it's yours. Prove me wrong. Show you can actually compute the angular momenta of things. Answer just one of those questions I asked you. If you're willing to do hours of experiments to try to show you're right is 5 minutes of algebra too much to ask?
Or are you like all the other cranks here, someone who whines about something you don't understand, delude yourself into thinking you can do said area of physics but then refuse to at every turn?
Also, if our grasp of torque, forces and momenta was as poor as you make out, how do we build cars? They involve forces, angular momenta, acceleration, all done in precise ways to make cars as efficient and powerful as possible.
It's not a matter of impressing people. I'm deeply unimpressed that for all your whining about angular momentum you don't even know how to compute it!
Newton saw the power of calculus in accurately describing things. He invented it, after all! So rather than waving your arms, crunch some numbers other than "10 = 11 is wrong".
Remember how you couldn't grasp momentum conservation still exists despite the initial momentum being definable as anything you like? You were inconsistent with your application of "Choice of frame" (or rather you just failed to understand it at all) when talking about angular and linear momenta. This came about because you've never done any calculations. And then you assume that because you can't get the algebra to work, it's the algebra's fault. No, it's yours. Prove me wrong. Show you can actually compute the angular momenta of things. Answer just one of those questions I asked you. If you're willing to do hours of experiments to try to show you're right is 5 minutes of algebra too much to ask?
Or are you like all the other cranks here, someone who whines about something you don't understand, delude yourself into thinking you can do said area of physics but then refuse to at every turn?
Also, if our grasp of torque, forces and momenta was as poor as you make out, how do we build cars? They involve forces, angular momenta, acceleration, all done in precise ways to make cars as efficient and powerful as possible.
It is 10kg* 1.1 m/sec = 11 units of momentum.
Specifically which algebra problem, did I get wrong?
I don’t remember you actually doing a problem; I can remember you telling us what a skilled mathematician you are. To bad we can’t see some of it in action. Explain Galileo’s pendulum using both algebra and calculus. Make the left side of the pendulum 5 meters in length and the right side 10 meters in length. Explain how angular momentum is conserved even though the pendulum lengths are constantly changing back and forth from 5 to 10. Let the bob have a maximum velocity of one meter per second.
I really don’t quite remember what problems you presented that I would not do, could you re-post those problems. I hope they are real problems not something taken out of a book that no one has ever done. Galileo’s pendulum is real.
Specifically which algebra problem, did I get wrong?
I don’t remember you actually doing a problem; I can remember you telling us what a skilled mathematician you are. To bad we can’t see some of it in action. Explain Galileo’s pendulum using both algebra and calculus. Make the left side of the pendulum 5 meters in length and the right side 10 meters in length. Explain how angular momentum is conserved even though the pendulum lengths are constantly changing back and forth from 5 to 10. Let the bob have a maximum velocity of one meter per second.
I really don’t quite remember what problems you presented that I would not do, could you re-post those problems. I hope they are real problems not something taken out of a book that no one has ever done. Galileo’s pendulum is real.
QUOTE (Limon+Mar 13 2008, 11:50 PM)
Specifically which algebra problem, did I get wrong
I have yet to see you attempt to do any algebra in a way where you setup the Newtonian equations of motion and solve them. Can you show me where you've done that?
You continue to claim that some physicists think that 0.5mv^2 is preserved when it's quite obvious it isn't. But you blame that on physicists, not your lack of understanding.
I have yet to see you attempt to do any algebra in a way where you setup the Newtonian equations of motion and solve them. Can you show me where you've done that?
You continue to claim that some physicists think that 0.5mv^2 is preserved when it's quite obvious it isn't. But you blame that on physicists, not your lack of understanding.
QUOTE (Limon+Mar 13 2008, 11:50 PM)
I don’t remember you actually doing a problem; I can remember you telling us what a skilled mathematician you are.
I have shown my competance in this thread and others, including doing specific maths questions on relativity and quantum field theory.
QUOTE (Limon+Mar 13 2008, 11:50 PM)
Explain Galileo’s pendulum using both algebra and calculus. Make the left side of the pendulum 5 meters in length and the right side 10 meters in length. Explain how angular momentum is conserved even though the pendulum lengths are constantly changing back and forth from 5 to 10. Let the bob have a maximum velocity of one meter per second.
Very nice, but very obvious, attempt to turn this around onto me. You're the one claiming to be competant enough at this physics to have spotted a mistake, so you're the one whose claims should be scrutinized. As a matter of course, it's important you demonstrate you have a firm grasp of the area you deride otherwise how can you be sure you have done the calculations properly?
As usual, you dodge the questions.
I asked you them here. By now, even someone as incompetant as you could have found the answers on Google! They are very standard questions (well the first two parts of Q4 anyway).
And you didn't answer my question about why, if the error is so glaring, do things like cars work so well, when they make enormous use of notions of torque, angular acceleration and momentum? If our understanding was out by the 10~20% you seem to think it is, why do all the machines we build based on those principles work?
As usual, you dodge the questions.
I asked you them here. By now, even someone as incompetant as you could have found the answers on Google! They are very standard questions (well the first two parts of Q4 anyway).
And you didn't answer my question about why, if the error is so glaring, do things like cars work so well, when they make enormous use of notions of torque, angular acceleration and momentum? If our understanding was out by the 10~20% you seem to think it is, why do all the machines we build based on those principles work?
Because angular momentum conservation is not used to build machines.
QUOTE (Limon+Mar 12 2008, 11:11 PM)
This is the kind of nonsense that Galileo and Newton were warning against.
If it got the motion from the earth then why does the wheel slow down? Even the Coriolis effect does not speed up a pendulum it only changes its direction.
If you were specifically asking about angular momentum then you should have used the word "angular." I thought you were talking about the conservation of momentum which also must be conserved.
If it got the motion from the earth then why does the wheel slow down? Even the Coriolis effect does not speed up a pendulum it only changes its direction.
If you were specifically asking about angular momentum then you should have used the word "angular." I thought you were talking about the conservation of momentum which also must be conserved.
QUOTE (Limon+Mar 14 2008, 02:12 AM)
Because angular momentum conservation is not used to build machines.
So things which spin and involve energy and force, like all the turbines used in power generation, just ignore angular momentum?
And I notice you couldn't spend 3 minutes answering just one of those questions.
So things which spin and involve energy and force, like all the turbines used in power generation, just ignore angular momentum?
And I notice you couldn't spend 3 minutes answering just one of those questions.
Question 4
i solid sphere through diameter I = 2/5MR²
ii hollow sphere through diameter I = 2/3 MR²
solid cylinder or disk I = 1/2MR² rotating on the center of mass around a line parallel to the cylinder side.
hollow cylinder or ring I = MR² rotating on the center of mass around a line parallel to the cylinder side.
These are moments of inertia to get angular momentum you multiply these by radians per second.
Radians per second (ω) is linear velocity (of a cylinder or point) divided by radius.
The only shapes I have used to make energy are the disk (solid cylinder), the sphere (on the end of a string; and treated as a point) and the hollow cylinder.
A point (comet) has a moment of inertia the same as a hollow cylinder I = MR².
This is the formula (angular momentum = MωR²) I used to prove that angular momentum conservation works in space as Kepler introduced it, but angular momentum conservation is a false concept for the laboratory.
i solid sphere through diameter I = 2/5MR²
ii hollow sphere through diameter I = 2/3 MR²
solid cylinder or disk I = 1/2MR² rotating on the center of mass around a line parallel to the cylinder side.
hollow cylinder or ring I = MR² rotating on the center of mass around a line parallel to the cylinder side.
These are moments of inertia to get angular momentum you multiply these by radians per second.
Radians per second (ω) is linear velocity (of a cylinder or point) divided by radius.
The only shapes I have used to make energy are the disk (solid cylinder), the sphere (on the end of a string; and treated as a point) and the hollow cylinder.
A point (comet) has a moment of inertia the same as a hollow cylinder I = MR².
This is the formula (angular momentum = MωR²) I used to prove that angular momentum conservation works in space as Kepler introduced it, but angular momentum conservation is a false concept for the laboratory.
I see you managed to use Google. You avoided answering the one about a cone because that one isn't on things like the Wikipedia list of moments of inertia, you'd have had to grasp how to apply the perpendicular and parallel axis theorems, which were other things I asked you about and which directly relate to computing the angular momentum of objects about arbitrary axes.
Do you understand what those theorems allow physicists to do?
So let's see you do something with them. Question 3
Even deriving the second order PDE will be enough. It's very little work, it's almost immediately obvious once you formulate the system mathematically.
Yet again.
Do you understand what those theorems allow physicists to do?
So let's see you do something with them. Question 3
Even deriving the second order PDE will be enough. It's very little work, it's almost immediately obvious once you formulate the system mathematically.
QUOTE
This is the formula (angular momentum = MωR²) I used to prove that angular momentum conservation works in space as Kepler introduced it, but angular momentum conservation is a false concept for the laboratory.
Well considering that such a formula would only apply to a point mass and not the extended objects you just listed, it demonstrate you don't grasp the concepts of angular momentum.Yet again.
I use a hollow cylinder and spheres as point masses, their momentums are both calculated exactly the same.
I knew you would keep hiding your inability to think outside the box by introducing more and more useless math problems. I got right the math I need to get right.
You can’t assume you understand the motion of a cone when you can’t even explain Galileo’s pendulum; with a point mass no less. You pretend to know so much but you can explain nothing.
When the spheres have all the motion, of the cylinder and spheres experiment, they can conserve only one of these two formulas; mv or 1/2mv² not both. And angular momentum conservation in the lab is a hoax. So what choices do you have left but the conservation of Newtonian Linear Momentum? Why conserving Newtonian momentum seems so strange to you is difficult to understand.
I knew you would keep hiding your inability to think outside the box by introducing more and more useless math problems. I got right the math I need to get right.
You can’t assume you understand the motion of a cone when you can’t even explain Galileo’s pendulum; with a point mass no less. You pretend to know so much but you can explain nothing.
When the spheres have all the motion, of the cylinder and spheres experiment, they can conserve only one of these two formulas; mv or 1/2mv² not both. And angular momentum conservation in the lab is a hoax. So what choices do you have left but the conservation of Newtonian Linear Momentum? Why conserving Newtonian momentum seems so strange to you is difficult to understand.
QUOTE (Limon+Mar 15 2008, 01:50 AM)
I use a hollow cylinder and spheres as point masses, their momentums are both calculated exactly the same.
I knew you would keep hiding your inability to think outside the box by introducing more and more useless math problems. I got right the math I need to get right.
How can they be point masses when their moments of inertia are not that of a point mass? That if you consider them rotating about anything other than their axes of symmetry their moments of inertia become very complicated?
I'm not hiding behind anything. I have yet to see you demonstrate you can do anything other than use Google, after days of prompting.
I knew you would keep hiding your inability to think outside the box by introducing more and more useless math problems. I got right the math I need to get right.
How can they be point masses when their moments of inertia are not that of a point mass? That if you consider them rotating about anything other than their axes of symmetry their moments of inertia become very complicated?
I'm not hiding behind anything. I have yet to see you demonstrate you can do anything other than use Google, after days of prompting.
QUOTE (Limon+Mar 15 2008, 01:50 AM)
You pretend to know so much but you can explain nothing.
I've explained many times in this thread and others why you do not even understand the concepts of Newtonian dynamics.
You couldn't even grasp how linear momentum conservation is possible irrespective of the frame you're moving in relative to the system you're considering. You couldn't even see how your understanding was inconsistent. Despite me explaining it several times.
In looking at moments of inertia formulas it becomes obvious that they are hugely axis specific. If the axis is 5 meters from the center of a sphere you use MR², if the axis is on the edge of the sphere you use 7/5MR², if the axis is in the center of the sphere you use 2/5MR². These formula changes do not compensate for the changing radius R; they indicate the distribution of mass about the axis. The sphere where the axis is a tangent line to the surface of the sphere does not have the same angular momentum as a sphere with an axis twenty meters away.
Your major argument to protect the angular momentum conservation theory is that you can use any frame of reference and move the axis wherever you want. This is eminently false; moving the axis even changes the formula that you use to determine angular momentum.
Your change of axis isn’t changing the axis at all it is keeping the axis where it was. When a twenty meter pendulum strikes a pin at the down swing position and becomes a one meter pendulum, (as in Galileo’s pendulum) you pretend the axis remained in place. You make believe that the experiment has not been conducted; it is the only way for you protect your faulty concept of angular momentum conservation in the lab.
AlphaNumeric quote; No one claims 1/2mv² is conserved. Do you understand the difference between 'total energy' and 'kinetic energy'?
Pequaide; Well; you can not pretend that the cylinder and spheres absorbs heat from the environment, kinetic energy is all there is. Either it is conserved or it is not, it can not have a mystical friend (heat) as you attribute to the ballistic pendulum.
You couldn't even grasp how linear momentum conservation is possible irrespective of the frame you're moving in relative to the system you're considering. You couldn't even see how your understanding was inconsistent. Despite me explaining it several times.
QUOTE (Limon+Mar 15 2008, 01:50 AM)
When the spheres have all the motion, of the cylinder and spheres experiment, they can conserve only one of these two formulas; mv or 1/2mv² not both. And angular momentum conservation in the lab is a hoax. So what choices do you have left but the conservation of Newtonian Linear Momentum? Why conserving Newtonian momentum seems so strange to you is difficult to understand.
Noone claims 1/2mv² is conserved. Do you understand the difference between 'total energy' and 'kinetic energy'?
This is your problem. You attack claims noone has made. The error is in your understanding and despite numerous people explaining it to you, you don't want to listen.
This is your problem. You attack claims noone has made. The error is in your understanding and despite numerous people explaining it to you, you don't want to listen.
In looking at moments of inertia formulas it becomes obvious that they are hugely axis specific. If the axis is 5 meters from the center of a sphere you use MR², if the axis is on the edge of the sphere you use 7/5MR², if the axis is in the center of the sphere you use 2/5MR². These formula changes do not compensate for the changing radius R; they indicate the distribution of mass about the axis. The sphere where the axis is a tangent line to the surface of the sphere does not have the same angular momentum as a sphere with an axis twenty meters away.
Your major argument to protect the angular momentum conservation theory is that you can use any frame of reference and move the axis wherever you want. This is eminently false; moving the axis even changes the formula that you use to determine angular momentum.
Your change of axis isn’t changing the axis at all it is keeping the axis where it was. When a twenty meter pendulum strikes a pin at the down swing position and becomes a one meter pendulum, (as in Galileo’s pendulum) you pretend the axis remained in place. You make believe that the experiment has not been conducted; it is the only way for you protect your faulty concept of angular momentum conservation in the lab.
AlphaNumeric quote; No one claims 1/2mv² is conserved. Do you understand the difference between 'total energy' and 'kinetic energy'?
Pequaide; Well; you can not pretend that the cylinder and spheres absorbs heat from the environment, kinetic energy is all there is. Either it is conserved or it is not, it can not have a mystical friend (heat) as you attribute to the ballistic pendulum.
ah, Limon. I see you're back with the old "conservation of angular momentum conspiracy theory". Got a whole bunch of new lab results, maybe?
With oil at over $100/bbl, your perpetual motion and free energy from space will be real valuable. Publish now and strike it rich!
Got a whole bunch of new lab results, maybe?With oil at over $100/bbl, your perpetual motion and free energy from space will be real valuable. Publish now and strike it rich!
Galileo’s pendulum should be sufficient proof that angular momentum conservation does not work in the lab. At least for those of us that did not swallow everything our teachers and professors taught us.
Yes; $110 a barrel my inspire some to spend $25 to make an energy producing machine.
It is published; and it goes all around the world, the internet.
Yes; $110 a barrel my inspire some to spend $25 to make an energy producing machine.
It is published; and it goes all around the world, the internet.
QUOTE (Limon+Mar 18 2008, 05:45 AM)
Yes; $110 a barrel my inspire some to spend $25 to make an energy producing machine.
It is published; and it goes all around the world, the internet.
Oh, I want one.
Where can I get the plans?
Arthur
It is published; and it goes all around the world, the internet.
Oh, I want one.
Where can I get the plans?
Arthur
QUOTE (Limon+Mar 18 2008, 02:13 AM)
Your major argument to protect the angular momentum conservation theory is that you can use any frame of reference and move the axis wherever you want. This is eminently false; moving the axis even changes the formula that you use to determine angular momentum.
Just as changing your own motion relative to a system changes the expressions for linear momentum but the momentum before is still the momentum after an interaction. It's just what you measure that total to be is different.
In one frame you'll have X total momentum before and after.
In another frame you'll have Y total momentum before and after.
Similarly, picking a different axis gives you :
About one axis you have A total angular momentum before and after.
About another axis you have B total angular momentum before and after.
Angular momentum is conserved.
This seems too hard for you to grasp, since you've had it explained to you many times.
Limon: It is true that linear momentum is conserved under different frames of reference. Let’s say a one kilogram mass moving 1 m/sec on a frictionless plane is struck from behind by a 3 kilogram mass moving 3 m/sec. You can use the frame of reference of any point on; the air table, the one kilogram mass, or the 3 kilogram mass and momentum is always conserved from that frame of reference.
But when you change your axis all axes are using the same point of reference as that used by the air table, it is not a change in reference it is merely a means of pretending that the radius has not changed, which of course has to be avoided because if the radius changes your conservation of angular momentum is lost.
AlphaNumeric; About one axis you have A total angular momentum before and after.
About another axis you have B total angular momentum before and after.
The term axis does not appear in the angular momentum formula, so what you really are saying is radius, radius does appear in the formula.
"About one radius you have A total angular momentum before and after.
About another radius you have B total angular momentum before and after."
And again you have not conducted the experiment. You ignore that the radius has changed.
Apply your A B theory to where angular momentum does work (for satellites) and see if you’re A B theory works then.
About one radius (apogee) you have A total angular momentum before and after (when after you have a huge increase in linear velocity at perigee you still keep the radius as if it were at apogee). Is angular momentum conserved? No it is a false statement.
About another radius (perigee) you have B total angular momentum before and after (when after you have a huge decrease in linear velocity at apogee you still keep the radius as if it were at perigee). Is angular momentum conserved? No it is a false statement. Your A B axis concept is false.
When you use the ‘pick an axis any axis’ excuse: you hide the fact that angular momentum conservation in the laboratory is a false concept.
AlphaNumeric quote: You'll find that such examples are common place in textbooks.
Limon: So; all the textbooks are teaching it wrong. And this is not the only thing they haven't got right.
Just as changing your own motion relative to a system changes the expressions for linear momentum but the momentum before is still the momentum after an interaction. It's just what you measure that total to be is different.
In one frame you'll have X total momentum before and after.
In another frame you'll have Y total momentum before and after.
Similarly, picking a different axis gives you :
About one axis you have A total angular momentum before and after.
About another axis you have B total angular momentum before and after.
Angular momentum is conserved.
This seems too hard for you to grasp, since you've had it explained to you many times.
QUOTE (Limon+Mar 18 2008, 02:13 AM)
Your change of axis isn’t changing the axis at all it is keeping the axis where it was. When a twenty meter pendulum strikes a pin at the down swing position and becomes a one meter pendulum, (as in Galileo’s pendulum) you pretend the axis remained in place. You make believe that the experiment has not been conducted; it is the only way for you protect your faulty concept of angular momentum conservation in the lab.
You'll find that such examples are common place in textbooks. You'll also find that if you're doing everything properly you have to consider the resultant angular momentum of the 'wall' holding the pin in place. If you don't, it's easy to prove all kinds of things. For instance, if you consider a ball, released from rest, accelerating in Earth's gravity, it seems to violate linear momentum, it's speeding up! But then you have to remember you should include the fact the Earth is also rising up ever so slightly to meet the ball, as per Newton's equations. Momentum is conserved.
But as usual, you demonstrate you don't grasp all the factors to consider and as always, you think your ignorance is someone else's fault. It always has and by the looks of it, always will be your fault.
Do you think these experiments haven't been done? Of course they have, the dynamics Newton came up with were extensively tested for more than 250 years.
But as usual, you demonstrate you don't grasp all the factors to consider and as always, you think your ignorance is someone else's fault. It always has and by the looks of it, always will be your fault.
Do you think these experiments haven't been done? Of course they have, the dynamics Newton came up with were extensively tested for more than 250 years.
Limon: It is true that linear momentum is conserved under different frames of reference. Let’s say a one kilogram mass moving 1 m/sec on a frictionless plane is struck from behind by a 3 kilogram mass moving 3 m/sec. You can use the frame of reference of any point on; the air table, the one kilogram mass, or the 3 kilogram mass and momentum is always conserved from that frame of reference.
But when you change your axis all axes are using the same point of reference as that used by the air table, it is not a change in reference it is merely a means of pretending that the radius has not changed, which of course has to be avoided because if the radius changes your conservation of angular momentum is lost.
AlphaNumeric; About one axis you have A total angular momentum before and after.
About another axis you have B total angular momentum before and after.
The term axis does not appear in the angular momentum formula, so what you really are saying is radius, radius does appear in the formula.
"About one radius you have A total angular momentum before and after.
About another radius you have B total angular momentum before and after."
And again you have not conducted the experiment. You ignore that the radius has changed.
Apply your A B theory to where angular momentum does work (for satellites) and see if you’re A B theory works then.
About one radius (apogee) you have A total angular momentum before and after (when after you have a huge increase in linear velocity at perigee you still keep the radius as if it were at apogee). Is angular momentum conserved? No it is a false statement.
About another radius (perigee) you have B total angular momentum before and after (when after you have a huge decrease in linear velocity at apogee you still keep the radius as if it were at perigee). Is angular momentum conserved? No it is a false statement. Your A B axis concept is false.
When you use the ‘pick an axis any axis’ excuse: you hide the fact that angular momentum conservation in the laboratory is a false concept.
AlphaNumeric quote: You'll find that such examples are common place in textbooks.
Limon: So; all the textbooks are teaching it wrong. And this is not the only thing they haven't got right.
QUOTE (Limon+Mar 18 2008, 06:46 PM)
Limon: So; all the textbooks are teaching it wrong. And this is not the only thing they haven't got right.
So when are you going to publish your findings and become FAMOUS.
(I'd love to be able to say "I knew him when he was first posting on Physorg....)
Arthur
PS, I'm still waiting on my plans for that Energy Making machine.
So when are you going to publish your findings and become FAMOUS.
(I'd love to be able to say "I knew him when he was first posting on Physorg....)
Arthur
PS, I'm still waiting on my plans for that Energy Making machine.
So, Limon, you are in the curious position of holding that the "one true path" is the straight-line Eulidian path and that all other paths "leak" momentum. However, I find it hard to reconcile your dislike of the curved path since all a curve is is a series of very, very small straight-line paths chained together. Then, too, what would you say to the denizens of a spherical shell who perceive many paths as straight although we would see them as curved? Do they never observe momentum conservation?
You got some big problems to overcome, my man.
You got some big problems to overcome, my man.
adoucette:A one kilogram object dropped one meter has a velocity of 4.423 m/sec.
A one kilogram object only needs 4.423 units of momentum to rise one meter. A pendulum bob with a low point velocity of 4.423 m/sec will rise 1 meter.
A one kilogram object dropped one meter on a 9 kilogram (ten kilograms total) Attwood’s Machine will developed 14.007 units of momentum. √(2*.981*1) = 1.40 m/sec* 10 kg = 14.0 Only the one extra (over balanced) kilogram has been drop 1 meter, the center of mass of the 9 kg has not changed.
The pulley of the Attwood’s Machine can be placed in a horizontal plane by using frictionless bearing points for the pulley itself and the draped masses.
The horizontal pulley or wheel can be used as a cylinder and spheres machine.
The cylinder and spheres machine can give the motion of a large spinning object (Attwood) to small portions of itself. If the cylinder and spheres machine gives all the 14.007 units of momentum to one kilogram it will rise 10 meters. Ten times higher than what it was dropped.
All data being collected from the cylinder and spheres experiments confirm Newtonian linear momentum conservation, and therefore confirm the production of energy.
The only portion of this set of steps that is not well accepted physics is the cylinder and spheres machine. The fact that the 90° (the cylinder stops when the spheres are at full extention) stop only occurs with a precise momentum is being retained proves that the cylinder and spheres experiments conserves momentum. And again: it has been timed with strobe light film photography; photo gates; and frame by frame stop action video tapes.
So if $25 is not too much to have free energy, I would start building these simple cylinder and spheres experiments, or disk and pucks experiments.
A one kilogram object only needs 4.423 units of momentum to rise one meter. A pendulum bob with a low point velocity of 4.423 m/sec will rise 1 meter.
A one kilogram object dropped one meter on a 9 kilogram (ten kilograms total) Attwood’s Machine will developed 14.007 units of momentum. √(2*.981*1) = 1.40 m/sec* 10 kg = 14.0 Only the one extra (over balanced) kilogram has been drop 1 meter, the center of mass of the 9 kg has not changed.
The pulley of the Attwood’s Machine can be placed in a horizontal plane by using frictionless bearing points for the pulley itself and the draped masses.
The horizontal pulley or wheel can be used as a cylinder and spheres machine.
The cylinder and spheres machine can give the motion of a large spinning object (Attwood) to small portions of itself. If the cylinder and spheres machine gives all the 14.007 units of momentum to one kilogram it will rise 10 meters. Ten times higher than what it was dropped.
All data being collected from the cylinder and spheres experiments confirm Newtonian linear momentum conservation, and therefore confirm the production of energy.
The only portion of this set of steps that is not well accepted physics is the cylinder and spheres machine. The fact that the 90° (the cylinder stops when the spheres are at full extention) stop only occurs with a precise momentum is being retained proves that the cylinder and spheres experiments conserves momentum. And again: it has been timed with strobe light film photography; photo gates; and frame by frame stop action video tapes.
So if $25 is not too much to have free energy, I would start building these simple cylinder and spheres experiments, or disk and pucks experiments.
Go ahead and build your free energy machine then Lemon. Prove you aren't the idiot we all think you are. Free energy will solve the oil crisis, it will solve global warming, it will even solve much of the Middle-East crisis as oil won't be worh as much.
Go on Lemon, become the richest person in the world. We will all have to eat our words.
Go on.
Go on Lemon, become the richest person in the world. We will all have to eat our words.
Go on.
QUOTE (Limon+Mar 19 2008, 06:25 PM)
adoucette:A one kilogram object dropped one meter has a velocity of 4.423 m/sec.
.....
I notice you didn't actually answer my question.
Oh, and I'm still waiting for you to post a set of plans for this FABULOUS machine.
Arthur
.....
I notice you didn't actually answer my question.
QUOTE
when are you going to publish your findings and become FAMOUS?
Oh, and I'm still waiting for you to post a set of plans for this FABULOUS machine.
Arthur
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