Experimental Proof That Energy Is Not A Conserved Quantity.

[exchemist]

Absolutely total rubbish.

 

 

 

Oh, you “suppose” so? Well, you supposed incorrectly.

 

In an elastic collision, the velocity of the 400 kg mass will be just under 0.1 m/s or more accurately, 0.09975 m/s. The final velocity of the 1 kg mass will be -19.9 m/s. where the minus sign indicates a reversal in direction (the 1 kg mass bounces off the 400 kg mass)

 

There is no “166 % increase in momentum” and certainly no 2000%! Nobody, except you has ever said there was.

 

In this elastic collision, both momentum and energy are conserved:

 

Initial momentum is:

(1 kg) (20 m/s) = 20 kg m/s

 

Initial energy is:

(1 kg) (20 m/s)^2 = 400 Joules

 

Final momentum is:

(1 kg) (-19.9 m/s) + 400 kg (0.09975 m/s) = 20 kg m/s 

 

Final energy is:

(1 kg) (-19.9 m/s)^2 + (400 kg) (0.09975 m/s)^2 = 400 Joules

 

My advice to you is do not suppose anything when you can do the math!

 

 

 

No, let’s not even discuss rotational dynamics until you can demonstrate an understanding of simple problems involving linear velocities.

Nice one! 

 

Actually, I've found this tutorial about 15 mins long on YouTube, which explains and contrasts the calculations for the 2 extreme cases of a perfectly inelastic collision and a perfectly elastic one. I don't generally approve of referring people to YouTube videos on a discussion forum but in this case it may help our poster to understand from a neutral 3rd party source, the difference between the two:

 

I've also looked up the ballistic pendulum, which he was trying to interest me in earlier.  Again this is a close-to-perfectly inelastic example.

 

I am fairly sure he has decided to live in a cosy (and imaginary) world, in which all linear collisions are perfectly inelastic, enabling him to propose that energy conservation is a myth and thus he can make ludicrous claims about objects in rotary motion.  

Edited January 17, 2017 by exchemist

[DelburtPhend]

I think he said momentum is always conserved and kinetic energy is rarely conserved.

 

The spheres return all of the momentum back to the cylinder, so they have to have all of the original momentum. Momentum is always conserved; These are real experiments and they always conserve momentum.      

[DelburtPhend]

One of the interesting experiments that I saw was two equal mass sleds on an air track. The masses of the sleds were powerful bar magnets. On the ends of the air track were other bar magnets so that the direction of the sled was reversed without a change in speed.

 

Now in your youtube video those cars could have just as easily been magnets mounted on sleds; just as in the other experiment. Then these two sleds would bounce apart in a perfectly elastic collision and both would head for the ends of the track.

 

What if one of these sleds reached the end of the track before the other; and the one had its direction reversed. Now you have both sleds moving in the same direction. Now you have no choice but to add their momentum instead of subtracting them.

 

Why were you subtracting them in the first place; should they ever be subtracted?

[OceanBreeze]

I think he said momentum is always conserved and kinetic energy is rarely conserved.

 

The spheres return all of the momentum back to the cylinder, so they have to have all of the original momentum. Momentum is always conserved; These are real experiments and they always conserve momentum.      

 

Yes, Kinetic Energy is not conserved unless the collision is perfectly elastic, which is rare. But Total energy is always conserved. When the collision is inelastic much of the initial kinetic energy is converted into other forms. For example, there will always be heat energy as well as sound energy (which is a type of mechanical oscillatory energy) and there may even be light energy given off if the collision is severe and there is energy lost to deformation.

 

The yo-yo de-spin also conserves rotational kinetic energy, or at least it comes very close to doing so. There will always be some small amount of energy lost to friction of the tethers being pulled out to full extension but there is no need to consider that because it is so small.

[DelburtPhend]

The Dawn Mission goes from 400 kilograms moving 1 m/sec to one kilogram moving v m/sec. How do you conserve the 400 units of momentum? It takes the same force to make a rim spin as it would to make the same mass move in a straight line.

[OceanBreeze]

One of the interesting experiments that I saw was two equal mass sleds on an air track. The masses of the sleds were powerful bar magnets. On the ends of the air track were other bar magnets so that the direction of the sled was reversed without a change in speed.

 

Now in your youtube video those cars could have just as easily been magnets mounted on sleds; just as in the other experiment. Then these two sleds would bounce apart in a perfectly elastic collision and both would head for the ends of the track.

 

What if one of these sleds reached the end of the track before the other; and the one had its direction reversed. Now you have both sleds moving in the same direction. Now you have no choice but to add their momentum instead of subtracting them.

 

Why were you subtracting them in the first place; should they ever be subtracted?

 

 

 

As has been explained to you countless times, momentum is a vector. That is, it has both an amplitude and a direction and are added according to the rules for vector addition. We have been discussing the simplest cases here, where all the directions are along a single straight line. When the two momentum vectors are in exactly opposite directions, they are subtractive.

 

 

 

 

Please don’t tell me you didn’t know that!

[OceanBreeze]

One of the interesting experiments that I saw was two equal mass sleds on an air track. The masses of the sleds were powerful bar magnets. On the ends of the air track were other bar magnets so that the direction of the sled was reversed without a change in speed.

 

Now in your youtube video those cars could have just as easily been magnets mounted on sleds; just as in the other experiment. Then these two sleds would bounce apart in a perfectly elastic collision and both would head for the ends of the track.

 

What if one of these sleds reached the end of the track before the other; and the one had its direction reversed. Now you have both sleds moving in the same direction. Now you have no choice but to add their momentum instead of subtracting them.

 

Why were you subtracting them in the first place; should they ever be subtracted?

 

Well, at least your trolling is starting to get slightly more interesting!

 

If one of the sleds hits the end magnet and changes direction before the other, momentum will still be conserved. The symbol for momentum is p.

 

Say they are both moving in opposite directions to start with, sled A with +1 p of momentum and sled B with -1 p. The total momentum to be conserved is 0. When they collide at the center point, if the collision is perfectly elastic they both change directions so now sled A has – 1 p,  and sled B has +1 p conserving momentum.

 

If sled A reaches the end of the track before sled B (maybe because the track lengths are not the same), it will have another elastic collision there, and sled A’s momentum must be conserved. That is, Sled A reverses direction with +1 p of momentum and gives -2 p of momentum to the magnet it hits. +1 -2 = -1 conserving sled A’s momentum. The total momentum is also conserved, -2 to the end magnet, +1 to sled A and +1 to sled B = 0, which is what the initial value was and it is conserved.

 

Just out of curiosity, Del o'l buddy, how much longer do you think you can keep this troll going?

[OceanBreeze]

The Dawn Mission goes from 400 kilograms moving 1 m/sec to one kilogram moving v m/sec. How do you conserve the 400 units of momentum? It takes the same force to make a rim spin as it would to make the same mass move in a straight line.

 

I have already shown that, but I won't go over it again until you demonstrate that you can handle the simple linear examples.

 

And you have both the mass and the velocity wrong.

Edited January 17, 2017 by OceanBreeze

[DelburtPhend]

https://youtu.be/YaUmzekdxTQ

[OceanBreeze]

https://youtu.be/YaUmzekdxTQ

 

Do you know what you are watching? It is Conservation of angular momentum, NOT linear momentum, as you claimed. Rotational kinetic energy is also conserved if the de-spin is lossless.

 

https://www.youtube.com/watch?v=-zJXRjG7DK0

[exchemist]

Yes, Kinetic Energy is not conserved unless the collision is perfectly elastic, which is rare. But Total energy is always conserved. When the collision is inelastic much of the initial kinetic energy is converted into other forms. For example, there will always be heat energy as well as sound energy (which is a type of mechanical oscillatory energy) and there may even be light energy given off if the collision is severe and there is energy lost to deformation.

 

The yo-yo de-spin also conserves rotational kinetic energy, or at least it comes very close to doing so. There will always be some small amount of energy lost to friction of the tethers being pulled out to full extension but there is no need to consider that because it is so small.

Yes and indeed it is perhaps in manoeuvres in space that we find systems that most closely approximate the complete conservation of kinetic energy alone, due to the absence of the friction with the air and the ground that complicates our terrestrial lives. :)

 

I suspect our experimenter has done no more than prove to himself that kinetic energy is not conserved, in most terrestrial interactions. Which, er, we all sort of knew. 

Edited January 18, 2017 by exchemist

[DelburtPhend]

Note that the full restoration of rotational motion occurs after the spheres are at full extension.

 

The proportional masses for the Dawn Mission would be a 400 kilogram thin walled cylinder (spheres included) and a thrown (spheres) mass of 1 kilogram. We have been using an arc velocity of one m/sec.

 

This is 400 kilograms moving one meter per second and that is going to cost you 400 newtons of force working for one second.

 

Kinetic energy conservation is satisfied with a 20 m/sec velocity for the spheres at full extension. That is 20 newtons working for one second.

 

Angular momentum is satisfied with between 14.14 and 16 m/sec. This range of 14 to 16 is because we do not know the exact length of the radius. But this is 15 newton seconds.

 

We know from experimentation that if the tethers were left attached that the full restoration of rotational motion would be returned to the satellite.

 

So the question is how do you get 400 newton seconds out of 15 or 20 newton seconds?

 

The answer is you can't: that is why NASA never collected or reported any data.

 

This math quote is informative: of 400 moving .09975 and the one kilogram rebounding 19.9 m/sec.

 

 

The best thing I see about this math is that you acknowledge that you can’t get 400 units of momentum out of 20. If the spheres have only 20 units of momentum you can’t return all of the motion back to the cylinder. But the experiment forces you to account for the fact that the motion does return to the cylinder. You might consider that the spheres do not have twenty; but that they have 400 units of momentum.

 

The second thing I see in this math is that you are generating 39.9 units of momentum from a consumption of only (20 -19.9) = .1 units of momentum. You think that this is okay because by subtracting you return to the original quantity. No that is not how it works.

 

If this theory worked you could repeat the production of 39.9, 39.8?, 39.6?, 39.2?  units of momentum again and again.

 

After the 400 kilogram moves away with (400 kg * .09975 m/sec) 39.9 units of momentum; the one kilogram that is rebounding at 19.9 m/sec can move to the end of the air track and bounce back from the end of the track and return to a second 400 kilograms, at rest. It would give the second 400 kilograms another 39.8 units of momentum and now you would have 79.7? units of momentum. You could do this again and again and again and you can get hundreds of units of momentum from the 20. But no that is not how it works.

 

The 39.9 units of momentum is 39.9 newton seconds of applied force; and that is what it is going to cost you to make it.  It is not going to be free just because you plan to subtract from it later.

 

You can't make 39.9 units of momentum out of .1 units.

[exchemist]

Oh dear oh dear.

 

Here is a simple illustration of the reason why angular momentum is an essential concept for analysing rotary motion correctly. 

 

Suppose you have a 1kg wheel, "wheel A", 1m in radius, with a tangential, or circumferential, speed at the rim of 1m/sec. One might naively consider its momentum to be 1kg x 1m/sec = 1 kg-m/sec.

 

If you had the same thing with another wheel, "B", 0.5m in radius (i.e. still 1kg mass at the rim, circumferential speed 1m/sec, you would still say its momentum was 1kg-m/sec.  

 

However, if one thinks about leverage, it will be apparent that the mass in case A is further from the pivot point (the axis of rotation) than in case B and so the moving mass at that distance has more leverage than that in case B. So you will have to try harder to stop it spinning than you will for wheel B. Right? 

 

It is also immediately apparent that you cannot account for this by thinking only in terms of linear momentum, p.

 

You need angular momentum, L, which is the product of moment of inertia, I (which has SI units of kg-m²), and angular speed, ω (which has SI units of 1/sec - in fact "radians"/sec but radians are dimensionless). So instead of p = mv you have L = Iω. These quantities are thus the rotational analogues of m and v.

 

In this simple case I = mr², so we have Ia = 1.1² = 1kg-m²; and Ib = 1.(1/2)² = 1/4 kg-m². So B has only a quarter of the moment of inertia of A.

 

The angular speeds of the two wheels are different since the 1m radius one only completes one revolution (2πr) half as quickly as the 0.5m radius one. The relation between circumferential speed and angular speed is ω=v/r. So we have ω(a) = 1/1 = 1 rad/sec, while  ω(b )  = 1/(0.5) = 2 rad/sec.

 

So now we have obtained the rotational equivalents of mass and velocity for both wheels, we can work out the angular momentum, La and Lb and you can guess what we will find. La = Ia.ω(a) = 1.1 = 1kg-m²/sec, but Lb = 1/4 . 2 = 1/2 kg-m²/sec.

 

So as expected, the angular momentum of B is half that of A, reflecting the lesser degree of leverage of its moving mass.

 

YOU CANNOT ANALYSE PROBLEMS IN ROTATING SYSTEMS WITH LINEAR MOMENTUM. If you try, you will f*** up.  

 

P.S. if anyone likes to read the "Discussion" section of this Wiki article, it expands a bit on what I have been saying above: https://en.wikipedia.org/wiki/Angular_momentum

Edited January 19, 2017 by exchemist

[OceanBreeze]

 

 

 

So the question is how do you get 400 newton seconds out of 15 or 20 newton seconds?

 

The answer is you can't: that is why NASA never collected or reported any data.

 

 

 

The best thing I see about this math is that you acknowledge that you can’t get 400 units of momentum out of 20.

 

The second thing I see in this math is that you are generating 39.9 units of momentum from a consumption of only (20 -19.9) = .1 units of momentum. You think that this is okay because by subtracting you return to the original quantity. No that is not how it works.

 

If this theory worked you could repeat the production of 39.9, 39.8?, 39.6?, 39.2?  units of momentum again and again.

 

After the 400 kilogram moves away with (400 kg * .09975 m/sec) 39.9 units of momentum; the one kilogram that is rebounding at 19.9 m/sec can move to the end of the air track and bounce back from the end of the track and return to a second 400 kilograms, at rest. It would give the second 400 kilograms another 39.8 units of momentum and now you would have 79.7? units of momentum. You could do this again and again and again and you can get hundreds of units of momentum from the 20. But no that is not how it works.

 

The 39.9 units of momentum is 39.9 newton seconds of applied force; and that is what it is going to cost you to make it.  It is not going to be free just because you plan to subtract from it later.

 

You can't make 39.9 units of momentum out of .1 units.

 

Trollin, trollin, trollin...............................

 

[exchemist]

Trollin, trollin, trollin...............................

 

Yes I expect so. He knows enough about all this. It's all a pose, I suspect. That's why I did not reply to his post. 

 

But I thought that, as we said previously, it would do no harm to help the understanding of any other readers by going through the principles. :)

 

Note added later: Thinking some more about this, and trying to be as fair as possible to our poster, one can I suppose use linear momentum in a limited way in this example:- 

 

One needs double the angular impulse (Torque x time) to stop wheel A compared to wheel B. But if one applies a force of 1N at the rim of each wheel, say, then the torque this force exerts on wheel A (1N-m) is double that exerted on B (0.5N-m), because the arm of the force is double.

 

So this torque doubling cancels the effect of the double angular momentum, such that after 1sec both wheels are brought to a halt. So one would still get the right answer by thinking solely in terms of linear momentum in this case. So possibly our friend thinks that, on this basis, all this angular stuff is needlessly complex ballocks.

 

However, where this naive approach would cease to work is with a more realistic mass distribution - in say a spacecraft - where the moment of inertia is the integral of r² dm, rather than having all the mass at the same radius. Such moments of inertia are not  trivial to calculate. And then it is quite impossible to apply the linear momentum approach - unless one resorts to integrals of successive shells of mass and tangential velocity, I suppose. 

Edited January 19, 2017 by exchemist

[OceanBreeze]

Yes I expect so. He knows enough about all this. It's all a pose, I suspect. That's why I did not reply to his post. 

 

But I thought that, as we said previously, it would do no harm to help the understanding of any other readers by going through the principles. :)

 

Note added later: Thinking some more about this, and trying to be as fair as possible to our poster, one can I suppose use linear momentum in a limited way in this example:- 

 

One needs double the angular impulse (Torque x time) to stop wheel A compared to wheel B. But if one applies a force of 1N at the rim of each wheel, say, then the torque this force exerts on wheel A (1N-m) is double that exerted on B (0.5N-m), because the arm of the force is double.

 

So this torque doubling cancels the effect of the double angular momentum, such that after 1sec both wheels are brought to a halt. So one would still get the right answer by thinking solely in terms of linear momentum in this case. So possibly our friend thinks that, on this basis, all this angular stuff is needlessly complex ballocks.

 

However, where this naive approach would cease to work is with a more realistic mass distribution - in say a spacecraft - where the moment of inertia is the integral of r² dm, rather than having all the mass at the same radius. Such moments of inertia are not  trivial to calculate. And then it is quite impossible to apply the linear momentum approach - unless one resorts to integrals of successive shells of mass and tangential velocity, I suppose. 

 

 

 

A linear analysis can be used in your example because you know both of the radii and you are taking that into consideration.

In his case, he never calculates the radius of the extended spheres, either because he doesn’t know how, (even though he has been shown) or he doesn’t want to, (more likely) because it will expose his alleged conflict between momentum and energy.

 

At this point, I see nothing to be gained from replying to him anymore. His nonsense has been refuted a number of times and further refutations are redundant and a waste of time.

 

 I do have to wonder why anyone would bother to troll a science site? What can possibly be gained from it? Does he really believe his ideas about basic mechanics are infallible, and the rest of the world, including NASA has it all wrong?

 

Or, as I suspect, he is trolling because he gets a laugh out of having others react to his nonsense?

 

Either way, there is no value in this any more (if there ever was). I think I will take the advice he gave to DrKrettin, and hit the “unfollow button”

[exchemist]

A linear analysis can be used in your example because you know both of the radii and you are taking that into consideration.

In his case, he never calculates the radius of the extended spheres, either because he doesn’t know how, (even though he has been shown) or he doesn’t want to, (more likely) because it will expose his alleged conflict between momentum and energy.

 

At this point, I see nothing to be gained from replying to him anymore. His nonsense has been refuted a number of times and further refutations are redundant and a waste of time.

 

 I do have to wonder why anyone would bother to troll a science site? What can possibly be gained from it? Does he really believe his ideas about basic mechanics are infallible, and the rest of the world, including NASA has it all wrong?

 

Or, as I suspect, he is trolling because he gets a laugh out of having others react to his nonsense?

 

Either way, there is no value in this any more (if there ever was). I think I will take the advice he gave to DrKrettin, and hit the “unfollow button”

Agree it seems very odd. Anyway his trolling - if that's what it is - has failed, insofar as it doesn't make me angry. I have rather enjoyed the trip down memory lane and it's always good to see how simply one can try to explain these things. I was in fact quite gratified to see that something very close to my own example existed as a tutorial on YouPube!

 

In fact my experience of these science fora is that it is often the cranks that give rise to the more interesting discussions, generally by accident rather than design. 

 

All good stuff. Well....until next time - assuming he doesn't come back with something reasonable.