energy producing experiments

[Tarsier79]

A piece of lead surely absorbs a lot of energy during the squishing process, does the same momentum conservation occur with a projectile and block that doesnt deform?

[daanopperman]

Hi pequaide ,

: The camera could take 1000 frames per second;

Mit camera 1 trillion f/s

[broli]

Why does the heat need to be concentrated at the spot of impact? Have you ever heard of Surface phonons? At impact a shockwave travels throughout the entire material at the speed of sound causing the atoms to vibrate collectively.

Here's a good explanation on this: http://news.mit.edu/2010/explained-phonons-0706

This vibration does heat up the entire block. So if we use the following exercise:

http://farside.ph.utexas.edu/teaching/3 ... ode81.html

and use a heat capacity of 1700 J /kg °C for wood (http://www.engineeringtoolbox.com/speci ... d_391.html).

What we end up with is 3263.9J that heats up the wood. Given the heat capacity the 5.2kg block of wood will heat up 0.36°C or (0.65°F). I wouldn't put my hazmat suit on for that.

[pequaide]

Well: lets do the experiment and find out. At your own risk.

Take a 5.188 kilogram cylinder and place a long string through one of its diameters; by drilling two holes. On the two ends the long strings, that now extends out from each side of the cylinder, place 6 grams. Tightly wrap the equal length strings around the circumference of the cylinder, until you can hold the two 6 gram masses up against the opposite sides of the cylinder.

Now: spin the side of the cylinder at about 1.7 m/sec and release the 6 gram masses.

Wear eye protection and have no one and nothing in the vicinity of the speeding 6 gram masses.

If this were done in a vacuum the two 6 grams masses would return to their “original� 738 m/sec velocity.

[pequaide]

The slow motion video of the cylinder and spheres with a 4.5 (total mass) to 1 (sphere mass) is making its rounds. The latest evaluation was from a cautious business owner.

He stated that there is very little slow down from start to finish; he compared it to the slowdown of a pendulum.  And you need instruments to detect the, one period, slow down of a pendulum.  So I thought that was a pretty good way to explain it.

Okay let’s review this cylinder and spheres: there is the starting rotational speed of the system; then a quick stop of the cylinder; then a restart of the cylinder that occurs at full extension of the tethers and spheres; Then there is another cylinder slow down to a full second stop of the cylinder; followed by a second full restart of the cylinder. The end rotational speed is equal (within experimental measurement) to the starting rotational motion.

Your theory requires that 53% of the motion is lost (as heat) in the first restart; and 53% of that remaining motion must be lost (as heat) in the second restart. This leaves you with 22% of the original motion.  But the business owner saw no reduction of motion.

The mass ratio of the 5.2 kilogram (to 12 grams) cylinder is about the same as the dawn mission (1400 kilogram to 3 kilograms) mass ratio. And from NASA own words they got an 8% back spin with only 5%, by their theory, of the energy remaining available. Which itself disproves the energy conservation theory.  So I guess we don’t need to build that experiment; NASA disproved energy conservation for us.
 
Suppose you have a 100 newton rocket engine; you attach the rocket engine to a one kilogram mass and fire the engine for one second. After the firing of the engine you will have a one kilogram mass moving 100 m/sec. This will be 100 units of momentum and 5000 joules of energy.
 
Suppose you have a 100 newton rocket engine; you attach the rocket engine to a ten kilogram mass and fire the engine for one second. After the firing of the engine you will have a ten kilogram mass moving 10 m/sec. This will be 100 units of momentum and 500 joules of energy.
 
Suppose you have a 100 newton rocket engine; you attach the rocket engine to a one kilogram mass and fire the engine for one second. After the firing of the engine you will have a one kilogram mass moving 100 m/sec. You then collide the 1 kilogram with 9 kilograms and you have 10 kilograms moving 10 meter per second. This, according to your theory, will give you 500 joules of motion and 4500 joules of heat. 
 
So (from your theory) the same engine fired for the same time can give you:

5000 joule of motion,

Or 500 joules of motion,

Or 500 joules of motion and 4500 joules of heat.  One free kcal; really?

The same three arrangements in Newtonian Physics would yield: 100 units of momentum; and 100 units of momentum, and 100 units of momentum.

[ME]

Okay let’s review this cylinder and spheres: there is the starting rotational speed of the system; then a quick stop of the cylinder; then a restart of the cylinder that occurs at full extension of the tethers and spheres; Then there is another cylinder slow down to a full second stop of the cylinder; followed by a second full restart of the cylinder. The end rotational speed is equal (within experimental measurement) to the starting rotational motion.

Just out of curiosity:

How is the restoration of motion NOT proving the conservation of energy ??
It is exactly in line with expectation and the known formula's.

[pequaide]

You cannot lose heat and get it back. And the only way that energy can be conserved, when a small mass shares its motion with a large mass, is to lose heat.

There are three times in the video (one cycle of the 4.5 to 1 cylinder and sphere) when the rotational velocity of the system is measured to be the same. This is where 4.5 units of mass is moving with 1 m/sec velocity.

There are two times in the video when the spheres have all the motion. For Newtonian momentum to be conserved the sphere velocity must be 4.5 m/sec. For energy to be conserved the spheres velocity must be 2.12 m/sec for the first stop.

To move from shared motion to sphere motion the maximum velocity of the sphere can only be the square root of 4.5 m/sec, 2.12m/sec; in order that energy is conserved. And you will only have 2.12 units of momentum to get back to the 4.5 units of momentum of the first restart.

For momentum to be conserved the velocity of the spheres must be 4.5 m/sec. And you will have 4.5 units of momentum to get back to the 4.5 units of momentum of the first restart.

Only momentum can be transferred from small (spheres) to large (cylinder and spheres) so you only have 2.12 units left for the second stop; if energy had been conserved. And you remove 31% (the square root of 2.12 = 1.456 m/sec) more from 2.12 m/sec for the second stop. This leaves you with only 1.456 m/sec for the velocity of the spheres for the second time they have all the motion. This will be 1.456 units of momentum given to 4.5 units of mass; for .3235 m/sec for the final rotational velocity of the system; when the original rotational velocity was 1 m/sec.

This exposes an error; I had used a percentage (53%) of loss for the second stop. It is not a percentage loss; it is a square root loss. The square root of a smaller number is a larger percentage, of that number, than that of a large number. 2.12 of 4.5 is 47% but 1.456 of 2.12 is 67% This leaves the final required motion for energy conservation at 32% rather than 22%. But of course no percentage loss is observed.

In Newtonian physics it is 4.5 m/sec for both the point where the spheres have all the motion. And it is 1 m/sec, through out, for the original velocity and the two restarts.

A more extreme number can be found for the 3 kilograms in the Dawn Mission. NASA predicts that only 5% of the momentum remains. You can not return the motion to the satellite if you have lost 95% of your momentum. The square root of 400 is only 20.

The motion only fulfills the expectations of momentum formulas.

[pequaide]

Oops: sorry; it is .222 m/sec for the second restart of the 4.5 to 1 model of the cylinder and spheres. Because it is not the square root of the sphere velocity or the energy. These applies to last part of paragraph 6 and paragraph 7 of the previous post.

The remaining motion will be the square root of the mass difference over the mass difference, between the total mass and the sphere mass. 2.12 over 4.5 = 47% The percent loss would then of course be that which remains from 100% ; 53%

The mass difference remains the same for the second shared position and therefore the % loss is the same at 53%. Removing 53% of the motion twice leaves you with 22% of the original motion.

If energy were conserved; it would take 18 frames for the spinning cylinder to move the distance of the black square.  But all are noticing that the video shows that it only takes 4 frames.  There is no perceived loss of motion.

Intuitive mistakes may be avoided if you do the math on paper rather than in your head; as was done in my errant correction of .3235 m/sec.

[pequaide]

The squares are clearer in the video.

This is another 4.5 to 1 mass ratio; with the cylinder having a mass of 1080g.

I discover that by using a transparent film for a marker on the monitor you can watch both the rotation and the side of the cylinder at the same time. As you go frame by frame you can mark the rotation of the square and the side shift of the cylinder (if it has any).

This cylinder is at the second stop; about 16 frames later it will be fully restarted. The final rate of rotation is equal to the initial rate of rotation. It is not 22% of the initial rotation rate.

So we have 450% of the initial energy when the cylinder is stopped; twice.

I am distributing videos of these 4.5 to 1 cylinder and spheres machines.

[pequaide]

second try

[pequaide]

Even with a short tether length of 1.89 r (cylinder r) the 304 grams of spheres can stop a cylinder of 2276 grams . This is 8.487 to 1 (2580 total grams, to 304 g). For a Newtonian velocity of the spheres, when the spheres have all the motion, of 8.487 time the initial velocity. And if energy is conserved the velocity would be only 2.91 time the initial. This 2.91 is only 34% of the needed momentum to restore the motion of the cylinder.

The motion of the cylinder is fully restored before it strikes the ground (foam pads). Five frames and five frames to cross the black square.

This is 848% of the original energy.

[ME]

Roughly calculated, you found this:

M = Mass cylinder
R = Radius cylinder
m = combined mass spheres
r = length of tether
(note: spheres don't actually start it the cylinder's rim due to some diameter)

Conservation of angular momentum: L=I·ω

r/R = Sqrt( M/m + 1 )-1
r/R = Sqrt( 2276/304 + 1)-1
r/R = 1.9132

Conservation of Energy: E=½I·ω²

<--> ½M·R²·&#969;² + ½m·R²·&#969;² = ½m·(R+r)²·&#969;²
<--> M·R² + m·R² = m·(R+r)²
<--> R² ( M+m)/m = (R+r)²
<--> R·Sqrt( M/m+1 ) = (R+r)
<--> Sqrt( ( M+m)/m ) = 1+r/R
r/R = Sqrt( M/m + 1 )-1
...idem...
r/R = 1.9132

note: (M/m+1) = (2276/304+1)=8.487

Speed when cylinder at full speed: v=&#969;·R
Speed of sphere when fully extended: v=&#969;·(R+r)

[pequaide]

Ballistic pendulums do not conserve angular momentum or energy. Half of this experiment is a ballistic pendulum and the motion is conserved. That eliminates two of your formulas.

[ME]

eeh ok.. Perhaps I miss the part being a ballistic pendulum, I thought you did a yo-yo-despin....

[pequaide]

Apparently you did; NASA did the yo-yo despin and then they mislead you about the velocity of the released masses (missiles). Don't you think it strange that they don't just come out and say the velocity was x, when they surly have the ability to measure it. They are so sure of their concept that experiments were not necessary.

I do not release the missiles and then make outlandish speculations about the spheres' velocity.

I leave the spheres attached and force them to show us how much motion they have.

By leaving the tethers and spheres attached after full extension the second half of this experiment becomes a ballistic pendulum experiment.

A ballistic pendulum experiment is where a small mass gives all its motion to a large mass; and that is what is done in the second half of the experiment. Only the Newtonian momentum is given to the large object; the other two formulas fail.

The spheres return all of the motion back to the cylinder. That is a return of all the momentum. So there is no other solution possible; the spheres have all the momentum when they have all the motion.

NASA speculated that the masses would have 5% of the initial momentum; but how can 5% return 100%?

I am going to add 156 grams to the cylinder; and then I will lengthen the tether to get back to a prefect stop.  This will allow us to discuss a 9 to 1 mass ratio. The motion (when the spheres have all the motion) needed to conserve energy will then be only 33% of the motion need to conserve Newtonian momentum.

The spheres will have a mass of 304 grams and the total mass will be 2736 grams. When the spheres have all the motion; a velocity 3 times the initial velocity will conserve energy, and a velocity of nine times the initial velocity will be needed to conserve Newtonian Momentum. This rounds the numbers of the experiment above (8.48 and 2.91).

With an initial velocity of one meter per second the sphere velocity will have to be 9 m/sec in order to have enough momentum to return the cylinder to the original rate of rotation.