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Experimental Proof That Energy Is Not A Conserved Quantity.


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#1 DelburtPhend

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Posted 22 December 2016 - 03:46 PM

I would like to give you a real world experiment that proves that energy is not a conserved quantity.

 

I have built a Dawn Mission type yo-yo de-spin device. I question if NASA is correct in their assumption that the weights conserve the same energy as the spinning space craft. They claim that when the 3 kilogram weights have the 1400 kg spinning rocket stopped; that the weight have the same kinetic energy as the original spin of the satellite body.

 

Just for a point of reference I change the satellite into a one meter diameter hollow cylinder; rotating at one meter per second around the arc of the circle. I think this 1200 kilograms is a ballpark figure; so we have twelve hundred kilograms moving one meter per second.

 

For the satellite energy we have ½ * 1200 kg * 1 m/sec * 1 m/sec = 600 joules

 

For energy conservation of the 3 kilogram masses on the end of the tethers we would need to have ½ * 3 kg * 20 m/sec * 20 m/sec = 600 joules. This 20 m/sec is half the speed of the baseball in the majors.

 

For the satellite momentum we have 1200 kg * 1 m/sec = 1200 units.

 

For momentum conservation of the 3 kilogram masses on the end of the tethers we have 3 kg * 400 m/sec = 1200 units. This 400 m/sec is 20% above the speed of sound.

 

Quite a difference 20 or 400; which one is correct?

 

I built a yo-yo de-spin device to find out. Because only momentum can be transferred from small objects to large objects; and conservation of energy only gives use 20 m/sec * 3 kg = 60 units of momentum to return 1200 units. If the 3 kilogram masses only have 60 units of momentum they can not restart the spin of the satellite.

 

So now for the experimental answer that I obtained from a small de-spin device. After a complete stop of the cylinder the tethered spheres completely restore the spinning motion of the cylinder. The original and final spin rates of the cylinder are the same.

 

If the Dawn Mission tethers were not released the 3 kilogram masses would have rewound and restarted the spin of the satellite. For a complete restoration of motion. Momentum is conserved not energy.

 

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Edited by DelburtPhend, 22 December 2016 - 06:44 PM.


#2 OceanBreeze

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Posted 23 December 2016 - 03:10 AM

I would like to give you a real world experiment that proves that energy is not a conserved quantity.

 

I have built a Dawn Mission type yo-yo de-spin device. I question if NASA is correct in their assumption that the weights conserve the same energy as the spinning space craft. They claim that when the 3 kilogram weights have the 1400 kg spinning rocket stopped; that the weight have the same kinetic energy as the original spin of the satellite body.

 

Just for a point of reference I change the satellite into a one meter diameter hollow cylinder; rotating at one meter per second around the arc of the circle. I think this 1200 kilograms is a ballpark figure; so we have twelve hundred kilograms moving one meter per second.

 

For the satellite energy we have ½ * 1200 kg * 1 m/sec * 1 m/sec = 600 joules

 

For energy conservation of the 3 kilogram masses on the end of the tethers we would need to have ½ * 3 kg * 20 m/sec * 20 m/sec = 600 joules. This 20 m/sec is half the speed of the baseball in the majors.

 

For the satellite momentum we have 1200 kg * 1 m/sec = 1200 units.

 

For momentum conservation of the 3 kilogram masses on the end of the tethers we have 3 kg * 400 m/sec = 1200 units. This 400 m/sec is 20% above the speed of sound.

 

Quite a difference 20 or 400; which one is correct?

 

I built a yo-yo de-spin device to find out. Because only momentum can be transferred from small objects to large objects; and conservation of energy only gives use 20 m/sec * 3 kg = 60 units of momentum to return 1200 units. If the 3 kilogram masses only have 60 units of momentum they can not restart the spin of the satellite.

 

So now for the experimental answer that I obtained from a small de-spin device. After a complete stop of the cylinder the tethered spheres completely restore the spinning motion of the cylinder. The original and final spin rates of the cylinder are the same.

 

If the Dawn Mission tethers were not released the 3 kilogram masses would have rewound and restarted the spin of the satellite. For a complete restoration of motion. Momentum is conserved not energy.

 

 

Before you conclude that you have disproven conservation of energy, you might consider you are not working the problem correctly.

 

First of all, you are solving the problem as if it was a linear collision rather than one involving a spinning mass, where conservation of angular momentum and rotational kinetic energy is more appropriate approach.

 

But, staying with your approach for the moment, in a linear collision, if a 1200 kg mass moving at 1 m/s were to collide with a 3 kg mass, do you suppose the 1200 kg mass will stop dead in its tracks and all the momentum and kinetic energy will be transferred to the 3 kg mass?

What I suggest you do is try solving the momentum and energy equations simultaneously instead of independently.

 

But the simplest approach is to consider the two masses will stick together, as in an inelastic collision. I think that approach is legitimate in this case since, in fact, they are stuck together by the tether.

 

In that case, 1200 kg x 1 m/s = 1203 kg x (…) m/s solve that and you will have a velocity that works for both momentum and energy for the linear collision that you are considering. But that still isn't the answer for the tethered satellite system!

 

If you really want to understand the physics of a tethered satellite system (TSS) well, here is a paper that might interest you. As you can see, the problem is a little more complicated than you thought!



#3 DelburtPhend

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Posted 23 December 2016 - 08:49 PM

I still see this as being very simple. I checked the box in one recent thread for high school level math; B for beginner. But simple; does not mean you do not know what you are doing. In fact I think it is the opposite; it proves that you do know what you are doing.

 

The force F is placed upon the satellite so as to make it rotate. This force F works for a period of time t so as to cause momentum (mv). This mv is stored in the rational motion of the satellite; this mv can not be abated. This is from F = ma; a is v/t Therefore Ft = mv. Please remember that in ALL interactions in the lab linear Newtonian momentum is ALWAYS conserved. The arc motion around the circle is entirely identical to straight line linear motion; by that I mean the same formulas can be used. This linear and arc equality is another part of it being very simple.

 

I have one model de-spin device that starts with an arc velocity of 1.2 m/sec. The cylinder stops and the spheres have all the motion. Soon (.067 sec) the spheres return the 1.2 m/sec motion back to the cylinder. And then (because of the proper mass ratio) the spheres stop the cylinder again. And then the spheres return the same 1.2 m/sec back to the cylinder again. So you have two stops of the cylinder and then two full restarts. Only momentum can return the momentum from small mass to large mass; so only momentum can be causing this back and forth transfer of motion. Simply all other solutions are ruled out.

 

Place a one half kilogram hard rubber ball on the end of a 3 m string. Swing the ball around the room with an arc velocity of 5 meters per second. Have someone hold up a rod so that it interrupts the string at .75 meters from the ball. The arc velocity will remain the same at 5 meters per second.

 

The real momentum (linear or arc) will remain the same. .5 kg * 5 m/sec = 2.5 units

 

The angular momentum will change from 7.5 to 1.875.  Angular momentum can’t even conserve the motion of one single ball, how could it be used to conserve the motion of three or ten. It is a useless concept.

 

Kepler’s Law is for space; it cannot be applied to objects in the lab.



#4 OceanBreeze

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Posted 24 December 2016 - 02:20 AM

I still see this as being very simple. I checked the box in one recent thread for high school level math; B for beginner. But simple; does not mean you do not know what you are doing. In fact I think it is the opposite; it proves that you do know what you are doing.

 

 

Unfortunately, in your case that is exactly what it means. You cannot solve a problem in rotational dynamics using linear momentum and translational kinetic energy. You will need to use angular momentum and rotational kinetic energy, and both are conserved in a lossless system.

Not only are you using the wrong approach to this problem, but your solution using translational momentum and kinetic energy is also done completely wrong. You need to solve the equations for momentum and kinetic energy simultaneously. You cannot end up with two different velocities and then ask “which one”.

Finally, your conclusion that energy is not a conserved quantity and angular momentum is a useless concept, is absurd.

In the spirit of the season, and good will, I have tried to be responsive to your posts, even though your “physics” falls into the category of “not even wrong”. But, if you are going to insist you are right there is no point in continuing to respond to you.


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#5 DelburtPhend

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Posted 24 December 2016 - 07:33 AM

Merry Christmas to you also.

 

 

In space the force quads if the distance halves. But the velocity only doubles if the distance halves. If the radius doubles then the velocity halves. This is linear or arc velocity; and this change in linear velocity is required if angular momentum is to be conserved. But gravity does not double the velocity when the radius is halved in the lab. So how can angular momentum conservation work in the lab? 

 

Remember: I have experiments that prove that I am correct.



#6 DrKrettin

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Posted 24 December 2016 - 07:53 AM

 

Remember: I have experiments that prove that I am correct.

 

Does it not bother you that if you are correct, there is a Nobel Prize waiting for you? Does that not make you question why nobody has done this before?


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#7 OceanBreeze

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Posted 24 December 2016 - 09:08 AM

Merry Christmas to you also.

 

 

In space the force quads if the distance halves. But the velocity only doubles if the distance halves. If the radius doubles then the velocity halves. This is linear or arc velocity; and this change in linear velocity is required if angular momentum is to be conserved. But gravity does not double the velocity when the radius is halved in the lab. So how can angular momentum conservation work in the lab? 

 

Remember: I have experiments that prove that I am correct.

 

By the same argument, why should linear momentum work in the lab? Sure, gravity has some effect on any experiment you do here on the surface of the earth, but that isn’t the problem you are having!

The problem you are having is apples and oranges, or maybe a better analogy is English and Chinese.

There are two mathematical languages that are used to describe your experiments. One is the language of translational motion, and the other is the language of rotational motion.

The factor that ties them both together is r, where r is the radius of rotation, something that you are not even taking into consideration.

For example, the linear velocity you are using is properly called the tangential velocity, v and the angular velocity ω is just v/r. And there are corresponding relationships in the rotational system to the linear system for force, acceleration and momentum as well. The big difference between the rotational system and the linear system, is that every bit of mass needs to be accounted for in the rotational system, depending on where it is. That is accounted for in the moment of inertia I, which is different for different types of bodies. Here is a list of I, and you can see it is different for different types of bodies. Your cylinder does not have the same I as the masses on the tethers! Unless you take that into consideration, you will never understand your experiments.

If you use the rotational system for your experiments instead of the linear system you will find that both rotational kinetic energy (assuming no losses) and angular (rotational) momentum are conserved.

What I find a bit strange, and a little sad, is that you have spent time and trouble to design and run experiments (which shows a lot of interest and initiative) but you have not bothered to learn about rotational dynamics that govern the behavior of those experiments, and it really isn’t that difficult.

 

Merry Christmas and Happy New Year!



#8 DelburtPhend

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Posted 24 December 2016 - 12:54 PM

The thin cylindrical shell has the same moment of inertia as a point mass on the end of a tether. The PVC pipe is being treated as a thin cylindrical shell and the spheres are being treated as point masses.

 

In my calculations (for the radius of the cylinder) I used a point a little over half way from the inside to the outside of the cylinder wall. In the 3 inch inside diameter PVC pipe; that would make the radius about 1.65 inches. This has to be reasonably close to your moment of inertia

.

The spheres are treated as point masses and this is reasonably close to the moment of inertia mentioned; especially when the spheres are at full extension.

 

This does not mean that I agree with angular momentum theories; I fully disagree with the use of angular momentum in the lab. Angular momentum is for deep space where gravity causes massive changes in linear velocity. Please note that the velocity used for angular momentum (when it is properly used in deep space) is in m/sec; this terminology (m/sec) for velocity is fully linear.

 

If you are swinging a point mass on the end of a string you can at any time burn through the string and the mass will proceed in a straight line tangent to the former circle of motion. This is why it is refereed to as tangent velocity. I am sure you know this; I am just doing a little tutorial reminder to all and myself.

 

If the arc motion was 5 m/sec then the tangent velocity will be 5 m/sec; and if released it will travel in a straight line at 5 m/sec. But note that radius is not mentioned; because it is not needed. I do not mention radius often because it is not needed.

 

DrKrettin; No; for several reasons. I have very good experiments that prove I am correct, These are done we high speed cameras; photo gates, etc.

 

Second: I see the obvious errors that are almost universally accepted; such as Angular Momentum working with gravity accelerating the mass and gravity not accelerating the mass. That is impossible on the surface.

 

Third: Anger is not science. I got booted from one thread because I think outside the box. Well: Columbus crossed an ocean with angry sailors that thought they were going to fall off of the horizon; I think my task is a little less daunting than that.



#9 DrKrettin

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Posted 24 December 2016 - 02:22 PM

 

DrKrettin; 

Third: Anger is not science. I got booted from one thread because I think outside the box. 

 

I am not remotely angry. What you must realize is that what you claim is an extraordinary claim, and as such requires extraordinary evidence. This evidence is obfuscated by signs that you really do not even begin to understand applied mathematics which has been accepted for centuries by mainstream science. If you are so convinced of your findings then you should consider submitting them to a scientific journal rather than a mere internet forum.

 

Just ask yourself why nobody has come up with the same conclusions as you have during the past two centuries. Do you really think that the whole corpus of scientific research has failed to notice this monumental error in their assumptions?



#10 DelburtPhend

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Posted 24 December 2016 - 06:43 PM

Yes: they missed it. I have proven that they missed it. Newtonian Physics began to be replaced by Leibnizian physics soon after their death. Conservation of energy has no experimental foundation. They wanted so badly for it to be true they required no experimental proof.

 

What I am saying is not outside conventional physics; at least not Newtonian physics. A one kilogram object moving 20 m/sec will give 399 kilograms (at rest) a velocity of .05 m/sec: this is what will happen on a frictionless plane; this .05 m/sec will happen on an air track; and this is what will happen in a ballistics pendulum, even if both masses are pendulum bobs. This 5 cm per second will happen in any imaginable arrangement?

 

NASA is predicting that the 3 kilograms in the Dawn Mission are only moving 20 m/sec at full extension; but the same initial rotational motion will be returned to the satellite if the 3 kilograms are allowed to re-wrap. So the 3 kilograms with only 60 units of momentum is expected to give the satellite 1200 units of momentum. This 60 making 1200 is what is outside Newtonian physics.

 

I have demonstrated that the full rotational motion does return and that eliminate conservation of energy as a possible velocity for the extended masses.

 

I was not implying that you in particular were angry. But the language used for me is not consistent with my somewhat professional presentation of facts. I think great emotion is felt when people have their core beliefs challenged; and this emotion ends up as an expression of anger toward the person that has unsettled them.



#11 OceanBreeze

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Posted 25 December 2016 - 01:53 AM

Yes: they missed it. I have proven that they missed it. Newtonian Physics began to be replaced by Leibnizian physics soon after their death. Conservation of energy has no experimental foundation. They wanted so badly for it to be true they required no experimental proof.

 

What I am saying is not outside conventional physics; at least not Newtonian physics. A one kilogram object moving 20 m/sec will give 399 kilograms (at rest) a velocity of .05 m/sec: this is what will happen on a frictionless plane; this .05 m/sec will happen on an air track; and this is what will happen in a ballistics pendulum, even if both masses are pendulum bobs. This 5 cm per second will happen in any imaginable arrangement?

 

NASA is predicting that the 3 kilograms in the Dawn Mission are only moving 20 m/sec at full extension; but the same initial rotational motion will be returned to the satellite if the 3 kilograms are allowed to re-wrap. So the 3 kilograms with only 60 units of momentum is expected to give the satellite 1200 units of momentum. This 60 making 1200 is what is outside Newtonian physics.

 

I have demonstrated that the full rotational motion does return and that eliminate conservation of energy as a possible velocity for the extended masses.

 

I was not implying that you in particular were angry. But the language used for me is not consistent with my somewhat professional presentation of facts. I think great emotion is felt when people have their core beliefs challenged; and this emotion ends up as an expression of anger toward the person that has unsettled them.

 

 

 

No, they didn’t miss anything. You are the one who is missing a great deal.

First of all, you are missing an education in physics, but you still feel compelled to tell the whole world, including NASA, that we have it all wrong.

The second thing you are missing is a bit of humility, in thinking you are the only one on the planet who has it right. Lose some of your ego and you might actually learn something.

 

Take a look back at this problem you posed:

 

 

 

Place a one half kilogram hard rubber ball on the end of a 3 m string. Swing the ball around the room with an arc velocity of 5 meters per second. Have someone hold up a rod so that it interrupts the string at .75 meters from the ball. The arc velocity will remain the same at 5 meters per second.

 

The real momentum (linear or arc) will remain the same. .5 kg * 5 m/sec = 2.5 units

 

The angular momentum will change from 7.5 to 1.875.  Angular momentum can’t even conserve the motion of one single ball, how could it be used to conserve the motion of three or ten. It is a useless concept.

 

Kepler’s Law is for space; it cannot be applied to objects in the lab.

 

Amazingly enough, you do have the correct numbers for the initial and final angular momentum, L, as 7.5 and 1.875 kg m 2/s.

 

But, what you are totally missing is the fact that the system has changed from just a ball on a string with a radius of 3 m, to a new system of a ball on a string with a radius of 0.75 m AND an very BIG ball called the Earth, on a string with a radius of 2.25 m. When you take both of these rotating balls into consideration, you find the angular momentum of 7.5 kg m 2/s is perfectly conserved.

 

How? The final I (moment of inertia of the ball) is now 0.28125 kg. m^2, and the I earth is 3.027E25 kg. m^2. The final angular velocity of the ball, ω = v/r = 5/0.75 = 6.66666….rad/s

 

So, 7.5 = (0.28135) (6.6666…) + (3.027E25) (ω earth)

 

ω earth = 5.625 / 3.027E25 = 1.858E-25 rad/s a very small rotation but because of the large mass of the earth it is just enough to conserve the angular momentum of the system.

 

In a similar way, you can find that the rotational kinetic energy of 6.25 J that the ball started out with, is also perfectly conserved. Very little of that goes to the earth, something like 1E-24 J, but it does conserve kinetic energy.

 

So, you see, Mr hot shot fizzikist, you have a lot to learn and you are fortunate that you have not been lampooned and laughed off of this forum already.

 

Show us that you are willing to learn some actual physics and maybe someone will continue to respond to you, but I am done wasting my time with you.


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#12 DelburtPhend

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Posted 27 December 2016 - 12:21 PM

The different moments of inertia for different shapes is only because of the distribution of mass about the many different radii. For any certain radius to any certain point of mass; the moments of inertia is mr². Angular velocity (in radians) is linear (arc) velocity divided by radius.

 

Therefore L = (m * r * r) * v / r : Two r's drop out and you have L = mrv where v is linear velocity.

 

Now why would you go back to the more complex form (L = (m * r * r) * v / r) when you already know that I got the problem correct: L = 7.5 for the 3 m circle and 1.875 for the .75 m circle.

 

What if you started with a .75 m circle and went to a 3 m circle: would you tell mother earth to subtract some for you instead of add.

 

Experiments that have no earth movement excuse is the Dawn Mission space craft; and cylinder and spheres experiments here on the planet. They are not attached to the earth. So we can forget the earth motion stuff.

 

When we video tape the cylinder and spheres at the different stages of extension: and angular momentum is not conserved; then it is not conserved. These are real experiments not imaginary excuses.

 

Lets take up another imaginary excuse: You could take a 2 kilogram block and cover it with a thin wall of Styrofoam. You could leave only a hole for the entry of a 100 grams sphere. A 100 gram sphere could enter through the hole at 20 m/sec and the alleged heat that it would produce would remain inside the foam for a reasonable long period of time. Thermal photographs could be taken of the metal inside of the foam through the hole.

 

As soon as this foamed over block achieves the speed, that satisfies linear momentum conservation, of .952 m/sec you could stop the block and carefully evaluate the heat inside the block. The heat will not change simply because the block is stopped. This experiment has never been required from those that propose this imaginary heat. This is what I mean by this theory being accepted without proof.

 

The cylinder and sphere are real experiments that prove there is no heat and there is no earth absorption of motion.



#13 exchemist

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Posted 27 December 2016 - 01:20 PM

The different moments of inertia for different shapes is only because of the distribution of mass about the many different radii. For any certain radius to any certain point of mass; the moments of inertia is mr². Angular velocity (in radians) is linear (arc) velocity divided by radius.

 

Therefore L = (m * r * r) * v / r : Two r's drop out and you have L = mrv where v is linear velocity.

 

Now why would you go back to the more complex form (L = (m * r * r) * v / r) when you already know that I got the problem correct: L = 7.5 for the 3 m circle and 1.875 for the .75 m circle.

 

What if you started with a .75 m circle and went to a 3 m circle: would you tell mother earth to subtract some for you instead of add.

 

Experiments that have no earth movement excuse is the Dawn Mission space craft; and cylinder and spheres experiments here on the planet. They are not attached to the earth. So we can forget the earth motion stuff.

 

When we video tape the cylinder and spheres at the different stages of extension: and angular momentum is not conserved; then it is not conserved. These are real experiments not imaginary excuses.

 

Lets take up another imaginary excuse: You could take a 2 kilogram block and cover it with a thin wall of Styrofoam. You could leave only a hole for the entry of a 100 grams sphere. A 100 gram sphere could enter through the hole at 20 m/sec and the alleged heat that it would produce would remain inside the foam for a reasonable long period of time. Thermal photographs could be taken of the metal inside of the foam through the hole.

 

As soon as this foamed over block achieves the speed, that satisfies linear momentum conservation, of .952 m/sec you could stop the block and carefully evaluate the heat inside the block. The heat will not change simply because the block is stopped. This experiment has never been required from those that propose this imaginary heat. This is what I mean by this theory being accepted without proof.

 

The cylinder and sphere are real experiments that prove there is no heat and there is no earth absorption of motion.

You are wrong. If you "stop" the block, you have to apply a braking effect. Brakes get hot when they are applied, don't they? So some heat will be generated somewhere if the block is stopped.

 

You are proposing an inelastic collision in which the final k.e. is 1/2 . 2.1 . (0.952) squared =  0.951J, while the initial k. e. of the sphere was 1/2 . 0.1 . (20) squared = 20J. So most but not all of the k.e. has gone into heat and the remainder will go to heat when you stop the block.  



#14 DelburtPhend

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Posted 28 December 2016 - 11:08 AM

Yes: there would be friction; or change of shape; or rise: but the friction used for stopping would be on the outside of the Styrofoam shell. Lets say the outside shell hits a cushion of cotton; this friction would be insulated from the inside metal. And according to the theory there would not be much heat left anyway. I am assuming that the foam shell and the inside metal block were snugly fitted and the stopping friction will be on the outside.

 

Or you could let it rise those 4.6 cm and then thermally photograph it. But this heat should stay in the foam for a reasonable amount of time.

 

But here is an experiment that shows why there can not be a massive loss of motion. This alleged quantity of heat is negligible anyway but the alleged loss of motion is huge.

 

This real experiment starts with two tethered spheres up against a cylinder with an arc motion of 1.2 m/sec.

 

The tethered spheres with a total mass of 132 grams soon have a 396 gram cylinder stopped between them; they quickly start the cylinder up again to full rotation. This full rotation is the fastest speed for the cylinder in the experiment. This fastest speed is determined by counting how many video frames are needed to cross a 20 mm black square on the surface of the cylinder. In this experiment the number of frames needed to cross is 4; this is 4/240 of a second for a velocity (20 mm times 240 / 4) of 1.2 m/sec.

 

The spheres will stop the cylinder again and then restart the cylinder back up to 1.2 m/sec; and the spheres will be up against the cylinder body as at the very start.

 

At two places in the experiment there is a small mass giving motion to a larger mass. And there are two places where a large mass give all its motion to a small mass. There can not be a loss of motion in any of these events because the initial motion and the final motion are identical.

 

If you where to graph the quantity of motion you would have a straight horizontal line with (equal distance between) points on the ends and one in the middle. Half of that straight line would be solid because we know that linear Newtonian momentum is conserved when a small mass give its motion to a large mass (this would be when the spheres give motion back to the cylinder). So half way between the three points would be a solid line extending back from the middle point and the end point.

 

Momentum has no why of inventing itself; nor is it ever lost. The empty spots in this line must also have the same linear Newtonian momentum.

 

This experiment proves that there is no heat lost; and there is no motion grabbed by the earth; there is no conservation of angular momentum; and there is no conservation of energy. The quantity of energy when the spheres have all the motion is about 450% of the original energy.



#15 exchemist

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Posted 28 December 2016 - 11:41 AM

Yes: there would be friction; or change of shape; or rise: but the friction used for stopping would be on the outside of the Styrofoam shell. Lets say the outside shell hits a cushion of cotton; this friction would be insulated from the inside metal. And according to the theory there would not be much heat left anyway. I am assuming that the foam shell and the inside metal block were snugly fitted and the stopping friction will be on the outside.

 

Or you could let it rise those 4.6 cm and then thermally photograph it. But this heat should stay in the foam for a reasonable amount of time.

 

[snip]

Hang on, let's deal with one problem at a time. In your scenario with the 2kg block and the inelastic collision with the 100g sphere, what are you claiming?

 

I am claiming that most of the initial 20 Joules of kinetic energy is converted to heat in the collision, apart from 0.951 Joules which remains in the motion of the combined block/sphere system after the collision. If an external influence then stops the block, this last bit of energy will also be converted to heat. Obviously it will not appear in the styrofoam you are using as a calorimeter, since the braking effect does not occur there. If you care to specify how you wish the braking effect to be achieved, we can determine where this heat will be generated.  

 

By the way, 20 Joules is not a lot of heat. It is about 5 calories, i.e. enough to raise the temperature of 5cc of water by one deg C. So your calorimeter will need to be quite sensitive to detect it.

 

Alternatively, you could read this link, which describes a classroom demonstration of the heat generated at the point of contact between a colliding pair of larger spheres: http://blog.teachers...-steel-spheres/ This demo gets over the calorimetry difficulty by exploiting the small size of the point of contact between the spheres, as a result of which the modest amount of heat produced is concentrated over a tiny area and thus leads to a much more dramatic rise in local temperature. 

 

But I am fascinated: what are you saying happens in this case? That this heat does not appear? That the conversion of kinetic energy to other forms of energy is an illusion?  Don't forget this is how hybrid cars and electric locomotives that make use of regenerative braking, save energy. Do you think that is an illusion, too?


Edited by exchemist, 28 December 2016 - 12:33 PM.


#16 DelburtPhend

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Posted 28 December 2016 - 08:14 PM

When a brake gets hot the car or bus slows down; the heat is not possible with no change in momentum.

 

Energy can be stored by cars or locomotives because the motion of the car or locomotive slows down. If the car or locomotive did not slow down it could not store energy.

 

There is no loss of motion in the back and forth motion of the cylinder and spheres; therefore heat can not be generated. If the cylinder and spheres does not slow down you can not make heat.

 

Look at it this was: You have a device that can give all of the motion of the 2.1 kilograms block back to the 100 grams. That is the 2.1 kg block moving .952 m/sec does give a 100 gram sphere a velocity of 20 m/sec.

 

The cylinder and spheres does this: it returns all of the motion back to the 100 gram mass.

 

If the starting point is 100 g * 20 m/sec : and your end point is 100 grams * 20 m/sec then how can you have lost 95% of your motion in between the two points? You can not generate something if you have not lost anything.

 

NASA proved that all of the motion in the Dawn Mission can be transferred to the 3 kilogram spheres. If they would have left the spheres attached they would have found that all the rotational motion would have come back to the satellite. By predicting energy conservation NASA predicted that 95% of the momentum would be lost. You can not bring all the motion back if you lose 95% of your momentum; because only linear Newtonian momentum can be transferred from small masses to large masses.



#17 exchemist

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Posted 29 December 2016 - 12:02 PM

When a brake gets hot the car or bus slows down; the heat is not possible with no change in momentum.

 

Energy can be stored by cars or locomotives because the motion of the car or locomotive slows down. If the car or locomotive did not slow down it could not store energy.

 

[snip]

 

Look at it this was: You have a device that can give all of the motion of the 2.1 kilograms block back to the 100 grams. That is the 2.1 kg block moving .952 m/sec does give a 100 gram sphere a velocity of 20 m/sec.

 

[snip]

Again staying with your example of the block and the sphere, until we have bottomed it out properly:

 

I think where you go wrong is in thinking that an inelastic collision is reversible: it is not. Energy has been converted to heat and you cannot get all of it back into kinetic energy again. This is basic thermodynamics: remember the Carnot cycle? You cannot  get 100% conversion of heat to work. 

 

You thus do not "have a device that can give all of the motion" of the block/sphere combination back to the sphere. I defy you to invent any scenario in which the block and sphere can spontaneously alter such that you have a block at rest and a sphere moving at 20m/sec. 

 

Now, if you had an elastic collision, which means one with no heat generation, that would be reversible. But in that case the sphere would rebound from the block.

 

 

 

With a braking vehicle, again momentum is conserved. However it is hard to measure, because what happens is momentum of the vehicle is transferred to the surface on which it is travelling, which is in most cases connected rigidly to the Earth. (This becomes more obvious if the rails are not bolted down correctly - if they are not, the rails will shoot forward when the brakes are applied, will they not?) In practice -  on a competently maintained railway -  when a train stops the motion of the Earth is very slightly altered. Momentum is lost by the vehicle but gained by the Earth.