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Experimental Proof That Energy Is Not A Conserved Quantity.


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#18 DelburtPhend

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Posted 29 December 2016 - 01:32 PM

Exchemist quote “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.”

 

I do it all the time; I have dozen of such devices. The 20 m/sec would be a 20 to 1 cylinder to spheres mass ratio. That mass ratio would be in the cart wheel class. I use a 2.6 kilogram cart wheel to throw two 66 gram spheres. But these are not scenarios; they are real working devices.

 

I argue that heat can not be produced because you get 100% back.

 

You argue that you can not get 100% back because heat is produced.

 

The following is a real experiment.

 

I have a cylinder and spheres experiment where it takes 4 frames (from a camera taking 240 frames per second) for the cylinder surface to cross 20 mm at the beginning of the experiment. And then the spheres completely stop the cylinder: the cylinder surface is stopped.

 

The spheres then restart the cylinder and: it takes four frames for the cylinder surface to cross 20 mm in the middle of the experiment. And then the spheres completely stop the cylinder.

 

The spheres then restart the cylinder and: it takes four frames for the cylinder surface to cross 20 mm at the end of the experiment.

 

This is a real experiment not a scenario. There are two complete restorations of motion after two complete stops. This is exactly the experiment you are requesting. I can mail you a video of the experiment on a flash drive.

 

This is another experiment that might help;

 

Take a 2 kg PVC pipe (3 in. I.D.) and drill a hole through its diameter about an inch down from the top. Place a 60 pound test fluorocarbon string through the hole and place 50 gram spheres on the ends of the string. Affix the string so that the tethered spheres have equal lengths of about 1.5 times the pipe diameter . Wrap the tethers around the cylinder and hold the spheres up against the cylinder. Spin the cylinder and spheres at a rotational rate of .952 m/sec. And then release the spheres.

 

Now: We have 2.1 kilogram moving .952 meter per sec. This is .9524 joules of energy: and NASA predicts that when the 100 grams are at full extension (and the cylinder is stopped) the 100 grams will have .952 joules of energy. So; ½ * .100 kg *v * v = .9524 J; v = 4.3643 m/sec ; NASA predicts that the spheres will be moving 4.36 m/sec.

 

From a multitude of collision experiments we know that: When a small mass of .100 kilograms moving 4.36 m/sec collides with a larger mass of 2 kilogram the final velocity will be (.100 kg * 4.36 m/sec = 2.1 kg * v) v = .2076 m/sec. Therefore the origin velocity of .952 m/sec can not be obtained from the velocity NASA predicts.

 

Newton would predict that the .100 kilograms would be moving: 2.1 kg * .952 m/sec = .1 kg * v; v = 20 m/sec. This 20 m/sec would allow for a complete restoration of motion at the end of the experiment; and that is exactly what happens.

 

I think I could set up this experiment in about 20 minutes: but I would have to change the 100 grams into 132 grams because a tapped steel spheres have a mass of 66 grams each. And that would make me change the 2 kilograms pipe into a 132/100 * 2 = 2.64 kg pipe. Then I would have to mount the high speed camera: and of course the outcome would be the same as I have done dozens of times already. The data collected would show that Newton is correct and linear momentum is conserved.



#19 exchemist

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Posted 29 December 2016 - 01:39 PM

Exchemist quote “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.”

 

I do it all the time; I have dozen of such devices. The 20 m/sec would be a 20 to 1 cylinder to spheres mass ratio. That mass ratio would be in the cart wheel class. I use a 2.6 kilogram cart wheel to throw two 66 gram spheres. But these are not scenarios; they are real working devices.

 

I argue that heat can not be produced because you get 100% back.

 

You argue that you can not get 100% back because heat is produced.

 

The following is a real experiment.

 

I have a cylinder and spheres experiment where it takes 4 frames (from a camera taking 240 frames per second) for the cylinder surface to cross 20 mm at the beginning of the experiment. And then the spheres completely stop the cylinder: the cylinder surface is stopped.

 

The spheres then restart the cylinder and: it takes four frames for the cylinder surface to cross 20 mm in the middle of the experiment. And then the spheres completely stop the cylinder.

 

The spheres then restart the cylinder and: it takes four frames for the cylinder surface to cross 20 mm at the end of the experiment.

 

This is a real experiment not a scenario. There are two complete restorations of motion after two complete stops. This is exactly the experiment you are requesting. I can mail you a video of the experiment on a flash drive.

 

This is another experiment that might help;

 

Take a 2 kg PVC pipe (3 in. I.D.) and drill a hole through its diameter about an inch down from the top. Place a 60 pound test fluorocarbon string through the hole and place 50 gram spheres on the ends of the string. Affix the string so that the tethered spheres have equal lengths of about 1.5 times the pipe diameter . Wrap the tethers around the cylinder and hold the spheres up against the cylinder. Spin the cylinder and spheres at a rotational rate of .952 m/sec. And then release the spheres.

 

Now: We have 2.1 kilogram moving .952 meter per sec. This is .9524 joules of energy: and NASA predicts that when the 100 grams are at full extension (and the cylinder is stopped) the 100 grams will have .952 joules of energy. So; ½ * .100 kg *v * v = .9524 J; v = 4.3643 m/sec ; NASA predicts that the spheres will be moving 4.36 m/sec.

 

From a multitude of collision experiments we know that: When a small mass of .100 kilograms moving 4.36 m/sec collides with a larger mass of 2 kilogram the final velocity will be (.100 kg * 4.36 m/sec = 2.1 kg * v) v = .2076 m/sec. Therefore the origin velocity of .952 m/sec can not be obtained from the velocity NASA predicts.

 

Newton would predict that the .100 kilograms would be moving: 2.1 kg * .952 m/sec = .1 kg * v; v = 20 m/sec. This 20 m/sec would allow for a complete restoration of motion at the end of the experiment; and that is exactly what happens.

 

I think I could set up this experiment in about 20 minutes: but I would have to change the 100 grams into 132 grams because a tapped steel spheres have a mass of 66 grams each. And that would make me change the 2 kilograms pipe into a 132/100 * 2 = 2.64 kg pipe. Then I would have to mount the high speed camera: and of course the outcome would be the same as I have done dozens of times already. The data collected would show that Newton is correct and linear momentum is conserved.

No, please give me a scenario in which the system we have been discussing (the block and sphere combination) is able to return to a state in which the block is at rest and the sphere moving at 20m/sec. That is what you said could happen and that is what I am challenging.

 

I really want us to get the mechanics of this simple example of yours straight, before moving on to other setups. One thing at a time. 


Edited by exchemist, 29 December 2016 - 01:39 PM.


#20 DelburtPhend

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Posted 29 December 2016 - 07:53 PM

Arc motion and linear motion are absolutely the same thing. A one kilogram mass moving 10 m/sec in a straight line will move 10 m/sec around the arc of the circle. And a one kilogram mass moving in an arc at ten meters per second will travel 10 m/sec if released into a straight line. There is no significance; one from the other.



#21 DelburtPhend

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Posted 30 December 2016 - 09:53 AM

You could tie a string to a 20 kilogram mass and then wrap the string around a light wheel with a bearing. You could then accelerate the twenty kilograms mass to 1 m/sec. The wheel would also be moving one meter per second and the wheel could throw a 1 kilogram mass 20 m/sec.

 

I used this design once; because it can be designed to not restart. If you design the 20 kg to stop at full extension the sphere can not restart the 20 kilograms because the string, to the 20 kg, goes slack. In most designs a restart will occur if you don't release the sphere.

 

The start is linear and the end is linear; if you release the sphere.



#22 OceanBreeze

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Posted 31 December 2016 - 03:14 AM

 

This is another experiment that might help;

 

Take a 2 kg PVC pipe (3 in. I.D.) and drill a hole through its diameter about an inch down from the top. Place a 60 pound test fluorocarbon string through the hole and place 50 gram spheres on the ends of the string. Affix the string so that the tethered spheres have equal lengths of about 1.5 times the pipe diameter . Wrap the tethers around the cylinder and hold the spheres up against the cylinder. Spin the cylinder and spheres at a rotational rate of .952 m/sec. And then release the spheres.

 

 

 

OK so far………

 

 

Now: We have 2.1 kilogram moving .952 meter per sec.

 

 

No, we do not! We have a 2.1 kg mass rotating with a tangential velocity of 0.952 m/s. The center of mass is staying in one place, so there is no linear velocity that can be assigned to the 2.1 kg mass.

 

 

This is .9524 joules of energy

 

 

Absolutely not! With no translational velocity, there is no translational kinetic energy. You cannot calculate the kinetic energy with a linear calculation. The reason is: for every point on the circumference of the cylindrical body that is moving in a tangential direction, there is another point directly across (through the diameter) that is moving in the exact opposite direction! There is only rotational kinetic energy.

 

 

and NASA predicts that when the 100 grams are at full extension (and the cylinder is stopped) the 100 grams will have .952 joules of energy. So; ½ * .100 kg *v * v = .9524 J; v = 4.3643 m/sec ; NASA predicts that the spheres will be moving 4.36 m/sec.

 

 

NASA predicts no such thing; only you are making this prediction, and it is totally wrong, based on a totally wrong understanding of physics. You are doing a calculation for a linear inelastic collision which is not applicable to this situation. If NASA were to do the calculations, they would use rotational kinetic energy and they will find the spheres will have approximately 0.45 joules of energy, the same as the cylinder has when it is spinning, and that energy is conserved.

 

From a multitude of collision experiments we know that: When a small mass of .100 kilograms moving 4.36 m/sec collides with a larger mass of 2 kilogram the final velocity will be (.100 kg * 4.36 m/sec = 2.1 kg * v) v = .2076 m/sec. Therefore the origin velocity of .952 m/sec can not be obtained from the velocity NASA predicts.

 

Newton would predict that the .100 kilograms would be moving: 2.1 kg * .952 m/sec = .1 kg * v; v = 20 m/sec. This 20 m/sec would allow for a complete restoration of motion at the end of the experiment; and that is exactly what happens.

 

 

No, Newton would be appalled at what you are doing, and so would NASA and so am I.  Now you doing a calculation for linear momentum  in an elastic collision, and that also does not apply here because this is a problem in rotational dynamics.

 

I think I could set up this experiment in about 20 minutes: but I would have to change the 100 grams into 132 grams because a tapped steel spheres have a mass of 66 grams each. And that would make me change the 2 kilograms pipe into a 132/100 * 2 = 2.64 kg pipe. Then I would have to mount the high speed camera: and of course the outcome would be the same as I have done dozens of times already. The data collected would show that Newton is correct and linear momentum is conserved.

 

 

It doesn’t matter how quickly you can do the experiment, if you do not understand what the experiment means, and you do not use the correct mathematical analysis to interpret your results.

 

If you do the calculations correctly, using rotational analysis, the calculations will show that angular momentum (L) and rotational kinetic energy (KE rotational) are both conserved.

 

I will post the correct calculations later.



#23 OceanBreeze

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Posted 31 December 2016 - 05:43 AM

 

 

The 3” ID PVC has an OD of ~ 3.5” or ~ .05m Radius (if you don’t mind, I will use civilized units from the metric system)

 

 

The Moment of Inertia, I for the cylinder is :

[math]{ I }_{ c }\quad =\quad \frac { m{ r }^{ 2 } }{ 2 } \quad =\quad \frac { 2\quad kg\quad { .05m }^{ 2 } }{ 2 } \quad =\quad 0.0025\quad kg\quad { m }^{ 2 }\quad[/math]

For the spheres to cause the cylinder to stop spinning, the Moment of Inertia of the spheres must be the same as for the cylinder:

[math]Is\quad =\quad { m }_{ 1 }{ r }^{ 2 }+\quad { m }_{ 2 }{ r }^{ 2 }\quad =\quad .0025[/math]

Since m1 = m2,  [math](2m){ r }^{ 2 }\quad =\quad .0025\quad ,\quad r\quad =\quad .158m[/math]

Since:  [math]\omega \quad =\quad \frac { v }{ { r }_{ c } } \quad =\quad \frac { 0.952\quad { m }/{ 2 } }{ .05m } \quad \approx \quad 19\quad { r }/s\quad[/math]

The quantities that must be conserved are: [math]{ KE }_{ rotational\quad  }and\quad Angular\quad Momentum\quad L\quad[/math]

[math]{ KE }_{ rotational }\quad =\quad \frac { 1 }{ 2 } { I }_{ c }{ \omega  }^{ 2 }=\quad \frac { 1 }{ 2 } (.0025){ (19) }^{ 2 }\quad =\quad .451\quad Joules\quad[/math]

[math]Angular\quad Momentum\quad =\quad { I }_{ c }{ \omega  }_{ c\quad  }=\quad .0465\quad kg-{ m }^{ 2 }/\quad sec\quad[/math]

Since the Moment of Inertias are the same for the cylinder and the spheres, it will be self-evident to anyone familiar with rotational mechanics, that both angular momentum of .0465 and rotational kinetic energy of 0.451 will be conserved, but I will go through the mathematics to show how it is done.

Momentum says: [math]{ I }_{ c }{ \omega  }_{ ci }\quad +\quad { I }_{ s }{ \omega  }_{ si }\quad =\quad { I }_{ c }{ \omega  }_{ cf }\quad +\quad { I }_{ s }{ \omega  }_{ sf }\quad[/math]

 

[math](.0025)(19)\quad +\quad 0\quad =\quad (.0025){ \omega  }_{ cf }\quad +\quad (.0025){ \omega  }_{ sf }\quad[/math]

 

Solving for [math]{ \omega  }_{ sf }\quad[/math] in terms of the other variables yields:

[math]{ \omega  }_{ sf }\quad =\quad \frac { .0475\quad -\quad .0025\quad { \omega  }_{ cf } }{ .0025 }[/math]

 

Energy says: [math]{ I }_{ c }{ \omega  }_{ ci }^{ 2 }\quad +\quad { I }_{ s }{ \omega  }_{ si }^{ 2 }\quad =\quad { I }_{ c }{ \omega  }_{ cf }^{ 2 }\quad +\quad { I }_{ s }{ \omega  }_{ cf }^{ 2 }\quad[/math]

 

Simplifying and making the substitution for [math]{ \omega  }_{ sf }\quad[/math] into the energy equation:

[math].9025\quad =\quad .0025{ \omega  }_{ cf }^{ 2 }\quad +\quad .0025\quad { \left[ \frac { .0475\quad -\quad .0025{ \omega  }_{ cf } }{ .0025 }  \right]  }^{ 2 }[/math]

 

[math](.0025){ (19) }^{ 2 }\quad =\quad (.0025){ \omega  }_{ cf }^{ 2 }\quad +\quad (.0025){ \omega  }_{ sf }^{ 2 }\quad[/math]

Further simplification and algebraic solution yields the solution of a quadratic eq in [math]{ \omega  }_{ cf }[/math] :

 

[math]{ \omega  }_{ cf }^{ 2 }\quad -\quad 19\quad { \omega  }_{ cf\quad  }=\quad 0\quad[/math]

 

The correct choice is: [math]\quad { \omega  }_{ cf\quad  }=\quad 0\quad[/math]

 

What this means is the final angular velocity of the cylinder is 0. It has been stopped.

That of course means the final angular velocity of the spheres is 19 radians/s

And that of course means that both the initial angular momentum and rotational KE of the system is conserved.

 

  



#24 DelburtPhend

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Posted 31 December 2016 - 07:01 AM

I don't think you even know the length of the tether.

 

All the mass of a spinning rim can be released at the same angular position (at the top for example); and all the mass would be moving in a straight line with the same momentum it had when it was spinning. You have to start thinking beyond what they told you in High School.



#25 DrKrettin

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Posted 31 December 2016 - 07:49 AM

 You have to start thinking beyond what they told you in High School.

 

I really don't know why OceanBreeze goes to such effort, only to have an unbelievably patronizing response like that. It is obvious to me that he did indeed go to High School, and it is also obvious that you did not.



#26 OceanBreeze

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Posted 31 December 2016 - 11:23 AM

I don't think you even know the length of the tether.

 

All the mass of a spinning rim can be released at the same angular position (at the top for example); and all the mass would be moving in a straight line with the same momentum it had when it was spinning. You have to start thinking beyond what they told you in High School.

 

Exactly the response I expected.



#27 DelburtPhend

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Posted 31 December 2016 - 11:26 AM

"thinking beyond what they told you"  is not accusing him of being uneducated; And this is precisely what they teach you in High School. I don't think the statement is offensive; and I think it is absolutely true.

 

The concept that a spinning rim mass wheel has zero momentum comes from the erroneous application of vectors. The thinking of course is that momentum is a vector. Did I see a statement that the momentum(s) cancel; or they subtract from one another. That the mass going east cancels the mass going west; and that is the error.

 

The mass on the north side of the wheel going east causes the wheel to rotate clockwise.

 

The mass on the south side of the wheel going west causes the wheel to rotate clockwise.

 

The mass on the east side of the wheel going south causes the wheel to rotate clockwise.

 

The mass on the west side of the wheel going north causes the wheel to rotate clockwise.

 

Lets say each of these (N,S,E,W) is a one kilogram mass moving one meter per second around the arc of the circle. If you then add an additional one kilogram to the rim of the wheel: the whole wheel will be moving .8 m/sec clockwise. So you see the wheel did have momentum and it still does. If you design the wheel to throw it will throw a one kilogram mass 4 m/sec. The Force times time that causes the (N,S,E,W) mass to rotate is the same FT that would cause the same straight line motion.



#28 OceanBreeze

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Posted 31 December 2016 - 11:27 AM

I really don't know why OceanBreeze goes to such effort, only to have an unbelievably patronizing response like that. It is obvious to me that he did indeed go to High School, and it is also obvious that you did not.

 

I enjoy physics. I had no expectations at all of getting through to the troll. :estrange:



#29 OceanBreeze

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Posted 31 December 2016 - 11:36 AM

 

The concept that a spinning rim mass wheel has zero momentum comes from the erroneous application of vectors. The thinking of course is that momentum is a vector. Did I see a statement that the momentum(s) cancel; or they subtract from one another. That the mass going east cancels the mass going west; and that is the error.

 

The mass on the north side of the wheel going east causes the wheel to rotate clockwise.

 

The mass on the south side of the wheel going west causes the wheel to rotate clockwise.

 

The mass on the east side of the wheel going south causes the wheel to rotate clockwise.

 

The mass on the west side of the wheel going north causes the wheel to rotate clockwise.

 

Lets say each of these (N,S,E,W) is a one kilogram mass moving one meter per second around the arc of the circle. If you then add an additional one kilogram to the rim of the wheel: the whole wheel will be moving .8 m/sec clockwise. So you see the wheel did have momentum and it still does. If you design the wheel to throw it will throw a one kilogram mass 4 m/sec. The Force times time that causes the (N,S,E,W) mass to rotate is the same FT that would cause the same straight line motion.

 

So momentum is not a vector?

 

How interesting!

 

Let me add that to your list of troll physics:

 

1). Energy is not a conserved quantity.

 

2). Angular momentum is a useless concept.

 

3). Momentum is not a vector.

 

This just keeps getting better and better!

 

I can't wait to see what you will come up with next!



#30 DelburtPhend

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Posted 31 December 2016 - 12:11 PM

2). Angular momentum is a useless concept​ "in the lab".

 

3). I said erroneous application of vectors; But most would agree with your interpretation.

 

Other wise: Thank You



#31 DrKrettin

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Posted 31 December 2016 - 12:18 PM

I enjoy physics. I had no expectations at all of getting through to the troll. :estrange:

 

DNFTT  :shocked:



#32 DelburtPhend

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Posted 31 December 2016 - 01:20 PM

I am not sure what that means but I think you should check the box at the top and join the 'unfollow this topic'.


Edited by DelburtPhend, 31 December 2016 - 01:29 PM.


#33 DrKrettin

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Posted 31 December 2016 - 02:07 PM

I am not sure what that means but I think you should check the box at the top and join the 'unfollow this topic'.

 

There is always the possibility to search the internet, and within a microsecond the Urban Dictionary will tell you that it means "Do Not Feed The Troll". I'm just intrigued as to why OB continues to take the time to try and explain something which you are clearly not capable of understanding. That, or you are a troll. The concepts are so simple that I suspect the latter.



#34 DelburtPhend

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Posted 31 December 2016 - 05:00 PM

Please hit the unfollow button; thank you.