Jump to content


Photo
- - - - -

Experimental Proof That Energy Is Not A Conserved Quantity.


  • Please log in to reply
169 replies to this topic

#35 DelburtPhend

DelburtPhend

    Advanced Member

  • Members
  • PipPipPip
  • 80 posts

Posted 02 January 2017 - 11:28 AM

Not even the concept of positive and negative (cancellation of momentum) can be attributed to clockwise and counterclockwise. It is the magnitude of the arc motion; and how they interact; not the direction. When you have two fly wheels that are touching or geared to each other one is going clockwise and the other is moving counterclockwise; yet if you try to stop one they will work together to resist the applied force. It takes a careful evaluation to properly apply the concept of vectors.

 

The 4.5 (total mass) to 1 (sphere) mass cylinder and spheres has a cylinder motion of 1.2 m/sec clockwise at the end of the experiment and at the beginning of the experiment; but in the middle the motion (3.5 kg * 1.2 m/sec) is stored counterclockwise. The spheres have a continuous motion clockwise; but they will stop the clockwise motion of the cylinder and then restore all of its motion counterclockwise. The spheres will then stop that counterclockwise motion and fully restore the cylinder motion clockwise. At the end you have exactly what you had at the beginning; 4.5 kg moving clockwise at 1.2 m/sec.

Attached Thumbnails

  • DSC07357.JPG
  • DSC07360.JPG


#36 DelburtPhend

DelburtPhend

    Advanced Member

  • Members
  • PipPipPip
  • 80 posts

Posted 02 January 2017 - 01:40 PM

The cylinder and spheres were released 1/15 of a second (16/240 frames per second) before the picture on the left. This frame shows the cylinder with no rotation.

 

One fifteenth of a second after this picture is the frame (on the right) that shows the cylinder with full restoration of motion (1.2 m/sec): counterclockwise. It is also seen in this second photograph that the spheres are fully extended. Because the spheres are moving to the other side of the cylinder (from the point where the tether is fixed) the force they exert on the cylinder switches direction. This force no longer causes the cylinder to accelerate counterclockwise; the cylinder is now accelerating clockwise. Soon (in about another 1/15 of a second) the spheres will have the cylinder stopped; again.

 

After this stop the spheres will force the cylinder back to its original speed of 1.2 m/sec: and that motion will be clockwise.



#37 DelburtPhend

DelburtPhend

    Advanced Member

  • Members
  • PipPipPip
  • 80 posts

Posted 02 January 2017 - 03:48 PM

The second stop and the second restart.

Attached Thumbnails

  • DSC07363.JPG
  • DSC07366.JPG


#38 DelburtPhend

DelburtPhend

    Advanced Member

  • Members
  • PipPipPip
  • 80 posts

Posted 02 January 2017 - 04:30 PM

At the first stop: the arc velocity required for momentum conservation would be 1.2 *4.5 = 5.4 m/sec.

This is the momentum required to restore all of the motion back to the cylinder: and only momentum is transferred from small to large.

 

The arc velocity required for energy conservation would be only 2.54 m/sec; this is only 47% of the need motion to restore the momentum back to the cylinder; for only momentum can be transferred from the small spheres back to the larger mass cylinder.

 

This same experiment also eliminate improper vector cancellations.

 

And one mass on the end of one string can't conserve angular momentum. These angular momentum conservation velocities are a little slower that energy conservation.

 

This particular experiment has an energy increase of 450%.

 

If anyone would like a slow motion video of one of these experiments; I have several available. Free



#39 sanctus

sanctus

    Resident Diabolist

  • Administrators
  • 4225 posts

Posted 03 January 2017 - 06:54 AM

No formulas needed to show how wrong you are, just an experiment with the feeling in your head and you love experiments :-)

So if angular momentum is not relevant in the lab then try this in your "lab": -)

https://www.youtube....h?v=qa2EzqNwREU

And then sit on you chair and push yourself off the lab-wall as hard as you can.

Then tell me whether your head feels the same..


  • OceanBreeze likes this

#40 exchemist

exchemist

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2771 posts

Posted 03 January 2017 - 11:06 AM

What we have here is a very characteristic phenomenon on science forums. We have a free energy crank who occupies himself in setting up scenarios in mechanics that exceed his competence to analyse - and thus he claims energy is not conserved, all of science has got it wrong, from Newton to Emmy Noether, and only he (it is never a she - female egos are almost never huge enough) is right. :)


Edited by exchemist, 03 January 2017 - 11:08 AM.


#41 DrKrettin

DrKrettin

    Understanding

  • Members
  • PipPipPipPipPip
  • 349 posts

Posted 03 January 2017 - 11:16 AM

What we have here is a very characteristic phenomenon on science forums. We have a free energy crank who occupies himself in setting up scenarios in mechanics that exceed his competence to analyse - and thus he claims energy is not conserved, all of science has got it wrong, from Newton to Emmy Noether, and only he (it is never a she - female egos are almost never huge enough) is right. :)

 

The last time I mentioned the Dunning-Kruger effect, I got into trouble with a moderator, so I won't mention it, other than wonder at what point it might be appropriate. 


  • OceanBreeze likes this

#42 exchemist

exchemist

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2771 posts

Posted 03 January 2017 - 03:09 PM

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.

No, I want you to answer my challenge to this proposed scenario of yours that we have been discussed, not come forward with alternatives that evade the issue.

 

I do not believe you can get a system with a mass of 2.1kg, moving at 0.951m/sec to accelerate a 100g mass to 20m/sec. Do you contend you CAN do so? How?  

 

Because I say that there is not enough kinetic energy in the 2.1kg mass to do it. 

 

How do you work this out?



#43 DelburtPhend

DelburtPhend

    Advanced Member

  • Members
  • PipPipPip
  • 80 posts

Posted 03 January 2017 - 04:31 PM

Newton rejected the concept of energy (mv²); he did not believe in the concept of energy and certainly would have never argued for its conservation.

 

The leaf blower shows me what level of non-scientists I am dealing with.

 

Formulas and equations are only of value when they match experimental facts.

 

All the motion in the cylinder and spheres remains after two motion transfers (collisions); and you are right energy can not to that; only linear Newtonian momentum can do that.

 

I don't need an ego to know I am correct. Your formulas are so bad that your errors are obvious. One claimed to proved I was wrong and he did not even know tether length. And for experiments you sight leaf blowers and ice skaters. Where my experiments are excellent. Four frames covering 20 mm; that is entirely different than flailing arms. No I don't need an ego; just a little common sense.



#44 CraigD

CraigD

    Creating

  • Administrators
  • 8034 posts

Posted 03 January 2017 - 06:21 PM

I’m a late joiner, so apologies for beginning with a “there’s your problem” reference to the first post, possibly overlooking someone’s post saying the same thing, but,

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

There’s your problem! ;)

Momentum is a vector quantity, meaning it has both magnitude and direction. Let’s simplify your 1200 kg hollow cylinder even more, and say it has nearly all of its mass concentrated in a couple of balls at its 12 and 6 o-clock positions, 600 kg at each, the rest of made of something so low-mass we can ignore it.

You’re saying the system’s momentum is
600 kg * 1 m/s + 600 kg * 1 m/s = 1200 kg m/s.
But because momentum is a vector quantity, we have to include a direction for each ball. So we should write something like
(600 kg m/s,0,0) + (-600 kg m/s,0,0) = (0,0,0)

Adding the pair of 3 kg balls at the 3 and 9 o-clock positions, the system’s momentum remains zero, and we’d write that:
(600 kg m/s,0,0) + (-600 kg m/s,0,0) + (0,3 kg m/s,0) + (0,-3 kg m/s,0) = (0,0,0)

There are an infinite number of solutions for momentum of this system, so to find a single solution for the posed yo-yo de-spin problem, we have to use its kinetic energy.

For its initial, all-balls-on-the-cylinder state, we write:
600 kg * (1 m/s)2 + 600 kg * (1 m/s)2 + 3 kg * (1 m/s)2 + 3 kg * (1 m/s)2 = 1206 J

For its big-balls-standing-still state, we write:
600 kg * (0 m/s)2 + 600 kg * (0 m/s)2 + 3 kg * v2 + 3 kg * v2 = 1206 J
Where v= [math]\sqrt{201}[/math] m/s =~ 14.18 m/s

It’s momentum, of course, remains zero, and might be written:
(0,0,0) + (0,0,0) + (0,42.54 kg m/s,0) + (0,-42.54 kg m/s,0) = (0,0,0)

#45 DelburtPhend

DelburtPhend

    Advanced Member

  • Members
  • PipPipPip
  • 80 posts

Posted 03 January 2017 - 06:58 PM

But real experiments show that one and only one form of motion is conserved: linear Newtonian Momentum.

 

But thank you: because 14.18 m/sec could never return the motion back to the cylinder; but the real experiments show that the spheres do return all the motion.

 

Look at real collision experiments only and 'always only' linear Newtonian momentum is conserved.


Edited by DelburtPhend, 03 January 2017 - 07:11 PM.


#46 DelburtPhend

DelburtPhend

    Advanced Member

  • Members
  • PipPipPip
  • 80 posts

Posted 03 January 2017 - 08:19 PM

Turn the 1200 kg back into a rim mass flywheel moving 1 m/sec around the arc of the circle. Now tie a rope to the circumference of the flywheel and tie the other end of the rope onto a 1200 kilogram mass on a frictionless plane. When the rope comes tight I can tell you exactly how fast the 1200 kg on the plane is moving.

 

1200 kg * 1 kg = 2400 kg * .5m/sec

 

It will not be moving.7071 m/sec as required by ½ mv²

 

You have improperly used vectors because the momentum of a flywheel can be calculated and that momentum is useful.



#47 sanctus

sanctus

    Resident Diabolist

  • Administrators
  • 4225 posts

Posted 04 January 2017 - 01:29 AM

Did you try my experiment? Feedback? :-)



#48 exchemist

exchemist

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2771 posts

Posted 04 January 2017 - 04:15 AM

Turn the 1200 kg back into a rim mass flywheel moving 1 m/sec around the arc of the circle. Now tie a rope to the circumference of the flywheel and tie the other end of the rope onto a 1200 kilogram mass on a frictionless plane. When the rope comes tight I can tell you exactly how fast the 1200 kg on the plane is moving.

 

1200 kg * 1 kg = 2400 kg * .5m/sec

 

It will not be moving.7071 m/sec as required by ½ mv²

 

You have improperly used vectors because the momentum of a flywheel can be calculated and that momentum is useful.

By the way, do you by any chance have views on why the electron does not fall into the nucleus of the atom? 

 

Just asking........



#49 OceanBreeze

OceanBreeze

    Creating

  • Moderators
  • 1060 posts

Posted 04 January 2017 - 04:50 AM

I’m a late joiner, so apologies for beginning with a “there’s your problem” reference to the first post, possibly overlooking someone’s post saying the same thing, but,
There’s your problem! ;)

Momentum is a vector quantity, meaning it has both magnitude and direction. Let’s simplify your 1200 kg hollow cylinder even more, and say it has nearly all of its mass concentrated in a couple of balls at its 12 and 6 o-clock positions, 600 kg at each, the rest of made of something so low-mass we can ignore it.

You’re saying the system’s momentum is
600 kg * 1 m/s + 600 kg * 1 m/s = 1200 kg m/s.
But because momentum is a vector quantity, we have to include a direction for each ball. So we should write something like
(600 kg m/s,0,0) + (-600 kg m/s,0,0) = (0,0,0)

Adding the pair of 3 kg balls at the 3 and 9 o-clock positions, the system’s momentum remains zero, and we’d write that:
(600 kg m/s,0,0) + (-600 kg m/s,0,0) + (0,3 kg m/s,0) + (0,-3 kg m/s,0) = (0,0,0)

 

 

That is the reason why we should use angular momentum and rotational kinetic energy in solving this type of problem.

 

There are an infinite number of solutions for momentum of this system, so to find a single solution for the posed yo-yo de-spin problem, we have to use its kinetic energy.

For its initial, all-balls-on-the-cylinder state, we write:
600 kg * (1 m/s)2 + 600 kg * (1 m/s)2 + 3 kg * (1 m/s)2 + 3 kg * (1 m/s)2 = 1206 J

For its big-balls-standing-still state, we write:
600 kg * (0 m/s)2 + 600 kg * (0 m/s)2 + 3 kg * v2 + 3 kg * v2 = 1206 J
Where v= [math]\sqrt{201}[/math] m/s =~ 14.18 m/s

It’s momentum, of course, remains zero, and might be written:
(0,0,0) + (0,0,0) + (0,42.54 kg m/s,0) + (0,-42.54 kg m/s,0) = (0,0,0)

 

 

Yes, you can use that as a quick and easy way to arrive at the tangential velocity of the spheres, but this approach will not tell you anything about the rotational dynamics, which is the heart of the problem.

 

The Dawn spacecraft de-spin is an interesting problem in rotational dynamics. I will spare you all the full analysis here, and just discuss it briefly.

 

If we treat the spacecraft as a cylinder with a mass of 1200 kg and a radius of 0.8 m.

The moment of inertia I = mr^2/2 = 384 kg m^2

 

(actually, the spacecraft is box-shaped, but this is a good approximation for I)

 

If we accept a tangential velocity of 1 m/s, the angular velocity ω = v/r, ω = 1.25 rad/s

 

(actually, the Dawn spacecraft was spinning at 36 rpm, or 3.8 rad/s but I will stay with the problem as it was written here)

 

The Rotational KE = 1/2 I ω^2 = 300 Joules which is conserved

 

In order to de-spin it with a total tethered mass of 3 kg, (divided into two spheres) the tethered mass should have the same moment of inertia as the spacecraft of 384 kg m^2

 

In this case, I = mr^2, r = 11.3 m

 

Interestingly, NASA actually used tethers that were 12 meters long and overshot the de-spin, sending the spacecraft spinning the opposite way at 3 rpm, or 0.314159 rad/s

 

Maybe they were trying to get π/10? :innocent:

 

But don’t think any of this matters to Delbert The Troll. He admitted in this thread, post 8, to being kicked off another forum, so we are wasting our time on him.



#50 OceanBreeze

OceanBreeze

    Creating

  • Moderators
  • 1060 posts

Posted 04 January 2017 - 04:51 AM

Did you try my experiment? Feedback? :-)

 

Maybe some dizziness would be an improvement?



#51 CraigD

CraigD

    Creating

  • Administrators
  • 8034 posts

Posted 04 January 2017 - 07:53 PM

Yes, you can use that as a quick and easy way to arrive at the tangential velocity of the spheres, but this approach will not tell you anything about the rotational dynamics, which is the heart of the problem.

I’ve found that one of the beauties of physics is that quick and easy is often also good and useful. For answering the question “what is the speed of the released weights in a yo-yo despin system when the cylinder is rotating at rate X?”, the simple kinetic energy calculation I showed is all you need. It’s a much simpler system than even something like the elastic collision of 2 balls, where you must solve for both momentum and kinetic energy.

Answering a question like “what is the minimum length of the tether to change the cylinder’s rotation from X to y?” or “when should the tether be released ...?” or “at what angle is the force on the tether’s attachment point when it should be released ...?” are much harder. I’ll confess that, when I first read about a yo-yo despin system, I was surprised that it was possible to stop or reverse something’s spin this way. My intuition told me that, like a skater extending her arms, you could only slow, not stop or reverse it. I would have never thought to invent such a system.

I’d be awed is someone could answer either of the above questions, or describe all the systems positions, velocities, and forces as a function of time, with exact algebra and calculus. If I tried it, I’d likely chicken out, and write computer simulation to get an approximate solution – though even that would be a lot of work, as simulating a weight on a tether unwinding from a spinning cylinders is complicated.

Moderator PS: I banned DelburtPhend, because he wasn’t follow the site rule, and in general acting trollishly. If anyone thinks I acted rashly, and would like him un-banned, please say so.