Experimental Proof That Energy Is Not A Conserved Quantity.

[DelburtPhend]

At the full restart the arrow will move from one side of the black square to the other side: in four frames. This is equal to the original quantity of motion. I was ready with two pictures but I guess I am out of space.

[DelburtPhend]

Craig's Question;

“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.”

 

The tether length needed to stop the cylinder is dependent upon the mass ratio between the hollow cylinder and the spheres. If the mass of the cylinder is 3.5 and the spheres mass is 1 then it could be stopped with a tether length of about 1 r length (of the cylinder). If the cylinder mass is 20 and the spheres mass is 1 then it could be stopped with a tether length of about 1/2 circumference length (about 3 r of the cylinder). These numbers are just from memory so they would have to be verified. You can either adjust the tether length or the mass ratio. I usually do both. If you seat the sphere and the side wall hole is fixed then you have to adjust the mass ratio.

 

It is easiest to release at full extension: here a loop of the tether merely slips off of a pin. By adjusting mass and/or tether length you can get a full dead stop at full extension. If the tether is left attached after full extension the tethers will accelerate the cylinder in the original direction.

 

If the cylinder is stopped before full extension then the tethers will accelerate it backwards. If the tether length is longer than that which is needed to stop the cylinder then the force in the tether will reverse the motion of the cylinder. The 4.5 total mass to spheres mass model reverses it twice. Is there some place I could send a video; it is very interesting.

 

Some believe that 3 kilograms moving 14.14 m/sec (I did not add 3 to 1200 for simplicity) will cause 1200 kilograms to move 1 m/sec. This is 42.4 units of linear Newtonian momentum creating 1200 units of linear Newtonian momentum. This is only 3.5% of the needed Newtonian momentum. This gentleman is simply not going to happen. You can not throw 3 kg moving 14.14 m/sec behind 1200 kilogram on a frictionless plane and expect the 1200 kg to move one meter per second. You can not throw 3 kg moving 14.14 m/sec onto the edge of a 1200 kilogram flywheel and expect the wheel to move one meter per second around the arc of the circle. Placing 3 kilograms on the end of a long string does not give it fantastic capabilities: it is still just 3 kg moving just 14.14 m/sec no mater what length the string.

 

Suppose we were to reduce both masses to1/1000^th; that would give use a 1.2 kilogram cylinder (this is in the range of the cylinders I build) and the spheres would have a mass of only 3 grams. Now I build this stuff and 3 grams will not work; 3 grams would be moving way way to fast. For an end motion of 14.14 m/sec for the spheres; the spheres would need to have masses about 30 times that much; Newtonian calculations say 85 grams; I am sure he is right.

 

Okay lets set one up: I don't have (85/2) 42.5 gram spheres (but I have one inch steel; 66 gram (each)spheres). We can change the 85 grams of spheres to 132 grams steel spheres. For the same mass relationship of 1.2 kilogram to .085 kilogram we will have to change the 1.2 to 1.86 kilograms; but this is a total mass so the cylinder itself will have a mass of 1.73 kilograms.

 

So we will have a 1.73 kilogram cylinder (spun at an arc speed of 1 m/sec) throwing 132 grams (66 g each) steel spheres into the sky 10 meters. Ya; I have done stuff like that; it will work.

 

So if the 1.2 kilogram cylinder and spheres will throw 85 grams at 14.14 m/sec; then what will it do to 3 g? Well it will throw it so fast : you won't be able to find them again, or you will get injured, or the tether length would be too long for the lab. But the point is 85 grams works; and 3 grams will not work.

 

Now going back to the Dawn Mission analogy with masses of 1200 kg hollow cylinder and 3 kg spheres. 85 kilograms moving 14.14 m/sec will capture all of the momentum and stop the spinning of the satellite; so 3 kg will have to be moving much much faster. In fact 85/3 faster: 400 m/sec.

Edited January 7, 2017 by DelburtPhend

[DelburtPhend]

I noticed some confusion about an earlier post 4 Jan 17.

 

“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.”

 

I used the term flywheel; which does not make clear (my mistake) that I intend that a bearing is in the center of the wheel. This is not a spinning cylinder floating in space or falling from a release point.

 

The bearing requires that the 1200 kg block moves toward the edge of the 1200 kg wheel on a tangent line. The bearing negates the two objects finding or rotating about the center of mass. Someone changed the 1200 to 1 kg so lets go with that.

 

When the one kilogram flywheel moving one meter per second jerks the one kilogram block they will both be moving ½ m/sec. The real momentum of the flywheel at the start is 1 (1kg * 1 m/sec), radius is not important. After the momentum is shared the wheel (rim) and the block will be moving ½ m/sec; the real momentum still 1 (2 * 1/2).

 

The block will approach the rim of the flywheel at a speed of ½ m/sec and this will be in a tangent line. Half the motion is in a straight line; there is no radius and no angular momentum for half of the motion.

 

When the block touches the circumference of the flywheel the string will be warped around the wheel ahead of the block; this would be the same warping as the Dawn Mission or the cylinder and spheres experiments.

 

At the point of touching the flywheel the block will want to continue in a straight line. But the string is now arranged to pull the block around the circle. As the block mass attempts to continue in a straight line the mass will lift off of the flywheel. Tension (force in both directions) will return to the tether line because the block mass want to travel in a straight line and the wheel mass wants to make it go in a circle. This force will slow the rotation of the wheel and accelerate the block mass. Because of the mass ratio the block will very quickly stop the rotation of the wheel. You could cut the string at this point and the block would depart in a straight line with all the motion; and it would be traveling 1 m/sec.

 

I don't know how you can evaluate this with angular momentum conservation when the motion starts with all the motion being in a circle; and then half the motion is in a circle; and then none of the motion is in a circle.

 

Kinetic energy conservation would start as ½ * 1 kg * 1 m/sec * 1 m/sec = .5 joules. Then it would be ½ * 2 kg * ½ m/sec * ½ m/sec = .25 joules; and then back to .5 joules.

 

Linear Newtonian momentum conservation would be 1 kg * 1 m/sec = 1 ; and then 2 kg * ½ m/sec = 1; and then 1 kg * 1 m/sec = 1.

[DelburtPhend]

Craig; here are some tether lengths.

 

With a total cylinder and spheres mass of 598 grams over a spheres mass of 130 grams a 1.2 radius tether stops the cylinder at  90° (to tangent). The spheres will then return the full starting rotation rate of the cylinder. The 1.2 r tether stops a 4.5 (total mass) to 1, cylinder and spheres. The r is the cylinder radius.

 

The same (4.5 to 1 mass ratio) cylinder and spheres can have a double stop if the tether length is pi r.

[DelburtPhend]

A light pulley with 1.75 kg suspended from each side; and an extra 1 kilogram added to the right side, is a 4.5 to 1 Atwood's.

 

The acceleration rate would be 1/ 4.5 * 9.81 m/sec/sec = 2.18 m/sec/sec

 

All of the 4.5 kilogram Atwood's will be moving 1 m/sec after the extra one kilograms has dropped .2294 meters.

 

The momentum of this Atwood's is 4.5 units.

 

One kilogram with 4.5 units of momentum would be moving 4.5 m/sec.

 

A 4.5 to 1 cylinder and spheres moving 1 meter per second can throw the 1 kilogram mass spheres at 4.5 m/sec.

 

A one kilogram mass moving 4.5 m/sec will rise to; 1.032 meters.

 

So a mass dropped 22.94 cm can rise to 103.2 cm.

 

An increase in energy to 450%

[exchemist]

Suggest you read this, to see where you are going off the track: http://www.lasalle.edu/~blum/p105wks/pl105_Energy.htm

[exchemist]

A light pulley with 1.75 kg suspended from each side; and an extra 1 kilogram added to the right side, is a 4.5 to 1 Atwood's.

 

The acceleration rate would be 1/ 4.5 * 9.81 m/sec/sec = 2.18 m/sec/sec

 

All of the 4.5 kilogram Atwood's will be moving 1 m/sec after the extra one kilograms has dropped .2294 meters.

 

The momentum of this Atwood's is 4.5 units.

 

One kilogram with 4.5 units of momentum would be moving 4.5 m/sec.

 

A 4.5 to 1 cylinder and spheres moving 1 meter per second can throw the 1 kilogram mass spheres at 4.5 m/sec.

 

A one kilogram mass moving 4.5 m/sec will rise to; 1.032 meters.

 

So a mass dropped 22.94 cm can rise to 103.2 cm.

 

An increase in energy to 450%

 

OK, curiosity has got the better of me: let’s have a go....

 

The acceleration of the combined system is 1/4.5 x g = 2.18m/sec².

 

At the point you mention, the downward velocity of the heavier mass is 1m/sec and that of the lighter is -1m/sec (i.e. upward).

 

The time at which this occurs is given by v=at, or t=v/a, i.e. 1/2.18 sec.

 

The downward and upward  distances travelled by the masses are given by S=1/2 at² i.e. 2.18 . (1/2.18)²/2 = 1/(4.36) m.

I am leaving this as a fraction to avoid rounding errors later.

 

The kinetic energy gained by the system is the sum of that of the two masses, because k.e., 1/2 mv² is not a vector quantity: 1/2 . 1.75 . 1  + 1/2 . 2.75 . 1 = 4.5/2 = 2.25 J. 

 

To check this answer, consider the net work (Fd) done by gravity on the system. F due to gravity =mg for each mass. The work done on the large mass is Fd  = 2.75 . 9.81 . 1/4.36, while the work done on the small one is minus 1.75 . 9.81 . 1/4.36, as it is lifted, i.e. it gains PE rather than losing it. So we have (2.75 - 1.75) 9.81 x 1/4.36 = 2.25J.

 

The two match, showing that energy is conserved in this process.  

 

2.25J will give a 1kg mass a velocity given by 1/2 mv² , i.e.  v² =(2.25 . 2)/1 = 4.5, so v = 2.12m/sec. This is the speed to which the kinetic energy in your system will be able to accelerate 1kg if energy conversion is 100%. Not 4.5m/sec. 

 

If you add the momenta of two bodies moving in opposite directions, you will get the momentum of the centre of gravity of the combined system. This would be zero if the masses were equal. In this case it will be 2.75 - 1.75 - 1kgm/sec. The CG is therefore (at that instant) moving downwards at a speed of 1/4.5 = 0.222 m/sec, while the two components recede from each other at 2m/sec.  

 

N.B. It makes no sense to add momenta ignoring the directions, because momentum is a vector, as others have already pointed out.

 

Often with problems such as this, using energy to analyse what is happening is the easiest and most powerful method, which is why I have used it. Note that I have no need to assume conservation of energy here, it just follows from applying the standard formulae for kinetic energy and mechanical work. 

Edited January 15, 2017 by exchemist

[DelburtPhend]

I copied this from wikipedia:

 

“The Atwood machine (or Atwood's machine) was invented in 1784 by the English mathematician George Atwood as a laboratory experiment to verify the mechanical laws of motion with constant acceleration. Atwood's machine is a common classroom demonstration used to illustrate principles of classical mechanics.”

 

The mechanical laws of motion; are Newton's Three Laws of Motion

 

“Newton's Second Law of Motion; F = ma: In an inertial reference frame, the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration a of the object: F = ma.”

 

As you can see if the momentum of the masses are treated as vectors the vector sum must include the mass going up and the mass going down; the two vectors must be added together for the correct answer of F = ma.

 

In your method for example: In a 20 to 1 Atwood's (after a drop of one meter) the 9.5 kg going up would cancel the 9.5 kg going down; and the only mass you would have left is 1 kilogram going down at .9904 m/sec. This would give you F = ma: 9.81 N = 1 kg * .4905 m/sec/sec. Which is incorrect.

 

You obviously must accelerate the whole 20 kilograms. Now the equation is correct 9.81 N = 20 kg * .4905 m/sec/sec.

 

We also know from Newton's Three Laws: that the motion caused must be in the same direction as the applied Force. We have no choice but to intellectually agree that the motion up is in the same direction as the motion down. Otherwise the equation F = ma does not work; and we all know that F = ma is the most experimentally proven equation in all of physics. We can be comforted by the concept of the pulley itself. The purpose of the pulley is to change the direction of force.

 

I set up five Atwood’s in Excel; the first had zero balanced masses on the left and right sides, the second had balanced masses of 1.75 kg on both side; the next had 4.5 kilogram each side; then 9.5 each, and 99.5 kg both sides. Each of these five Atwood’s had an added mass of 1 kilogram on the right side. If we refer to them as; accelerating mass over total mass: we would have a 1/1 Atwood’s, a 1/4.5 Atwood’s, 1/10; 1/20 1/200.

 

The extra mass (1 kg) was dropped one meter for each Atwood’s; the center of mass of the balanced mass does not change.

 

We shall list each Atwood’s characteristics; in order

 

Name: 1 / 1; 1 / 4.5; 1 / 10; 1 / 20; 1 /200

 

Acceleration: 9.81 m/sec/sec; 2.18 m/sec/sec; .981 m/sec/sec; .4905 m/sec/sec; .04905 m/sec/sec

 

Final velocities: 4.429 m/sec; 2.088 m/sec; 1.4007 m/sec; .9904 m/sec; .3132 m/sec

 

Final KE are all: 9.81 joules

 

Final momentum: 4.43; 9.396; 14.007; 19.8; 62.64 units

 

When you give all this momentum back to the one kilogram that was dropped; they will have KEs of: 9.81; 44.145; 98.1; 196.2; 1962 joules.

 

This is an energy increase of: none; to 450%; 1000%; 2000%; 20,000%

 

 

 

The one kilogram will rise: 1.00 m; 4.5 m; 10 m; 20 m; 200 m

[DelburtPhend]

Okay: so you think 2.12 units of momentum can produce 4.5 units of momentum. I do not think there are any conditions under which this can occur. 2.12 units of momentum takes 2.12 Ns and it takes 4.5 newton seconds to produce 4.5 units of momentum. A newton of force will not come from nowhere and apply itself for 2.38 seconds.

  

 

If we start with a simple rear end collision of 1 kg moving 2.12 m/sec striking a 3.5 kg mass (at rest) we will get 4.5 kg moving .47 m/sec.  We know we will be .47 m/sec because of a vast quantity of experiments.

 

 

Your claim is that if you put 1 kg moving 2.12 m/sec on the end of a string and wrap the string around a 3.5 kg mass you can get 213% the original momentum. So you say that kinetic energy is conserved and momentum comes along like a puppy on a leash

.

 

In the Dawn Mission you say (and everybody else says; including NASA) 3 kg moving 14.14 m/sec (42.42 units) returns 1200 units of momentum. This is 1157.57 newtons applied for one second and it just come from nowhere.

 

 

I think it is the other way around; linear momentum is conserved and energy floats.

 

So let’s go to the experiments and see who is right;

 

 

I was searching the video archives and I found a video of a cylinder and spheres experiment that had a tether length of 3 or 4 cylinder circumferences. After a release of the spheres; and a quick stop of the cylinder; and a full restart of the cylinder; the tether just kept feeding out and feeding out until the cylinder and spheres hit the ground.

 

 

The reason the tether does not re-wrap is because the tether never crossed the 90° to tangent. The tether is too long for coming to the 90° to tangent and the cylinder just continues its reverse spin; and the tether just keeps feeding out and feeding out. But the magnitude of the original spin and the final spin are the same.

 

 

If the cylinder has a mass nine times that of the spheres then; by the energy conservation theory, 68.4 % of the momentum is lost at the stop of the cylinder.  But you cannot return to the original momentum if 68.4 % of the original momentum is lost. And this cylinder returns to its original rate of spin. It starts with four frames to cross the black square and ends with four frames to cross the black square. If 68.4% of the motion had been lost it would take 12.7 frames to cross the black square. This 12.7 frame crossing is very slow and this is not what happens.

 

So this experiment is another that proves that energy is not a conserved quantity.

Edited January 13, 2017 by DelburtPhend

[exchemist]

Okay: so you think 2.12 units of momentum can produce 4.5 units of momentum. I do not think there are any conditions under which this can occur. 2.12 units of momentum takes 2.12 Ns and it takes 4.5 newton seconds to produce 4.5 units of momentum. A newton of force will not come from nowhere and apply itself for 2.38 seconds.

  

 

If we start with a simple rear end collision of 1 kg moving 2.12 m/sec striking a 3.5 kg mass (at rest) we will get 4.5 kg moving .47 m/sec.  We know we will be .47 m/sec because of a vast quantity of experiments.

 

 

Your claim is that if you put 1 kg moving 2.12 m/sec on the end of a string and wrap the string around a 3.5 kg mass you can get 213% the original momentum. So you say that kinetic energy is conserved and momentum comes along like a puppy on a leash

.

 

In the Dawn Mission you say (and everybody else says; including NASA) 3 kg moving 14.14 m/sec (42.42 units) returns 1200 units of momentum. This is 1157.57 newtons applied for one second and it just come from nowhere.

 

 

I think it is the other way around; linear momentum is conserved and energy floats.

 

So let’s go to the experiments and see who is right;

 

 

I was searching the video archives and I found a video of a cylinder and spheres experiment that had a tether length of 3 or 4 cylinder circumferences. After a release of the spheres; and a quick stop of the cylinder; and a full restart of the cylinder; the tether just kept feeding out and feeding out until the cylinder and spheres hit the ground.

 

 

The reason the tether does not re-wrap is because the tether never crossed the 90° to tangent. The tether is too long for coming to the 90° to tangent and the cylinder just continues its reverse spin; and the tether just keeps feeding out and feeding out. But the magnitude of the original spin and the final spin are the same.

 

 

If the cylinder has a mass nine times that of the spheres then; by the energy conservation theory, 68.4 % of the momentum is lost at the stop of the cylinder.  But you cannot return to the original momentum if 68.4 % of the original momentum is lost. And this cylinder returns to its original rate of spin. It starts with four frames to cross the black square and ends with four frames to cross the black square. If 68.4% of the motion had been lost it would take 12.7 frames to cross the black square. This 12.7 frame crossing is very slow and this is not what happens.

 

So this experiment is another that proves that energy is not a conserved quantity.

I said no such thing, as will be obvious to any reader with half a brain. 

[DelburtPhend]

exchemist quote:

 

The two match, showing that energy is conserved in this process.  

 

2.25J will give a 1kg mass a velocity given by 1/2 mv² , i.e.  v² =(2.25 . 2)/1 = 4.5, so v = 2.12m/sec. This is the speed to which the kinetic energy in your system will be able to accelerate 1kg if energy conversion is 100%. Not 4.5m/sec. "

[exchemist]

 

exchemist quote:

 

The two match, showing that energy is conserved in this process.  

 

2.25J will give a 1kg mass a velocity given by 1/2 mv² , i.e.  v² =(2.25 . 2)/1 = 4.5, so v = 2.12m/sec. This is the speed to which the kinetic energy in your system will be able to accelerate 1kg if energy conversion is 100%. Not 4.5m/sec. "

 

Right. But that is NOT saying that 2.12 "units of momentum" (in fact kg m/sec) gives rise to 4.5 kg m/sec. I explicitly pointed out that you cannot add momenta ignoring their direction. The figure of 4.5 kg m/sec is your figure, not mine and, as I told you, it is wrong. You can't do that. 

 

You are also making another, even more fundamental, wrong assumption, namely that you can choose to convert all the momentum from one body to a second, in an interaction. Whether this takes place or not depends on energy considerations as well: you do not have freedom to specify this.

 

Consider the simplest of examples: a 1kg ball (1) moving at 1m/sec hits a stationary 2kg ball (2) in a frictionless, perfectly elastic collision. Momentum is 1kg-m/sec before and after, but momentum alone does not tell you the velocities - call them V1 and V2 - of the two balls after the impact. All you know is the sum of their momenta is 1 kgm/sec: i.e. 1x V1 + 2xV2 = 1 kg-m/sec.

 

You need a second independent equation to tell you the answer and energy provides it: 1/2 . 1 . 1²  = 1/2J is the k.e. before the collision. The k.e. after is the sum of that in both balls: 1/2 . 1 . V1² + 1/2 . 2 . V2² = 1/2J (by conservation of energy)

 

The momentum equation can be rewritten as V1 = 1-2V2, so we substitute for V1 in the energy equation and we get: 

1/2 (1-2V2)² + 1/2 . 2V2² = 1/2. Multiplying both side by 2, expanding the bracket and collecting like terms we get:

6V2² - 4V2 +1 = 1, from which we can see there are 2 roots: either V2 = 0, in which case V1 is 1, or V2 = 2/3, in which case V1 is minus 1/3.

 

The first corresponds to the situation before the collision.

 

The second shows that, to conserve both energy and momentum, the only possible outcome is that the 1kg ball rebounds from the 2kg one and returns at -1/3m/sec, while the 2kg one acquires a velocity of 2/3m/sec. You do NOT have the freedom to decide that all the momentum goes into the 2kg ball.

 

If that were to happen it would violate the conservation of one or the other and THIS DOES NOT OCCUR. 

 

All your examples obscure this issue. It is a basic misunderstanding of conservation of momentum. You are not free to partition momentum among bodies how you like. How it will be partitioned will be determined by the simultaneous conservation of BOTH momentum AND energy. Both are fundamental conserved quantities in mechanics.

Edited January 15, 2017 by exchemist

[OceanBreeze]

[exchemist]

Yes I don't know why I bother really, but it seems to me possibly useful for some readers to take this right back to basics, in order to clarify it.

 

It is always the same with free energy nuts: devise a system that exceeds their capacity to analyse, analyse it wrongly and, hey presto, the whole of physics for 200 years is wrong, up to and including Noether's Theorem. Brilliant, Nobel Prizes all round!  :)

[DelburtPhend]

Exchemist quote:  "The second shows that, to conserve both energy and momentum, the only possible outcome is that the 1kg ball rebounds from the 2kg one and returns at -1/3m/sec, while the 2kg one acquires a velocity of 2/3m/sec. You do NOT have the freedom to decide that all the momentum goes into the 2kg ball."

 

The experiment decides; and all of the momentum goes into the spheres.

 

Exchemist quote: If that were to happen it would violate the conservation of one or the other and THIS DOES NOT OCCUR.

 

 It does occur when the experiments are inelastic; which is why you avoid them.

 

Exchemist quote: All your examples obscure this issue. It is a basic misunderstanding of conservation of momentum. You are not free to partition momentum among bodies how you like. How it will be partitioned will be determined by the simultaneous conservation of BOTH momentum AND energy. Both are fundamental conserved quantities in mechanics.

 

Again: the experiment decides how the motion is partitioned; and inelastic collisions can not conserve both energy and momentum.

 

Exchemist quote: “Right. But that is NOT saying that 2.12 "units of momentum" (in fact kg m/sec) gives rise to 4.5 kg m/sec. I explicitly pointed out that you cannot add momenta ignoring their direction. The figure of 4.5 kg m/sec is your figure, not mine and, as I told you, it is wrong. You can't do that.”

 

The 4.5 kg m/sec comes from an experiment. The experiment ends (with the same four frames) with the same 4.5 kg m/sec motion as it started. You are saying that in the middle of that experiment it can drop to 2.12 kg m/sec; I am saying that it can not. It must be 4.5 kg m/sec through out.

 

Atwood added 'up' and 'down' motion and he was correct in doing so.

 

Exchemist quote: “You are also making another, even more fundamental, wrong assumption, namely that you can choose to convert all the momentum from one body to a second, in an interaction. Whether this takes place or not depends on energy considerations as well: you do not have freedom to specify this.”

 

You can choose what experiment you are to conduct; some experiments convert all the motion and some don't. If the cylinder is stopped rotating then all the rotational motion is now in the spheres.

 

The slow motion version of the cylinder and spheres experiment takes 7 seconds; talk you administrators into giving me 7 second space and I will post the experiment in question. It has the 4.5 kg  m/sec three times: and in between are two stops of the cylinder.

[exchemist]

Exchemist quote:  "The second shows that, to conserve both energy and momentum, the only possible outcome is that the 1kg ball rebounds from the 2kg one and returns at -1/3m/sec, while the 2kg one acquires a velocity of 2/3m/sec. You do NOT have the freedom to decide that all the momentum goes into the 2kg ball."

 

The experiment decides; and all of the momentum goes into the spheres.

 

Exchemist quote: If that were to happen it would violate the conservation of one or the other and THIS DOES NOT OCCUR.

 

 It does occur when the experiments are inelastic; which is why you avoid them.

 

Exchemist quote: All your examples obscure this issue. It is a basic misunderstanding of conservation of momentum. You are not free to partition momentum among bodies how you like. How it will be partitioned will be determined by the simultaneous conservation of BOTH momentum AND energy. Both are fundamental conserved quantities in mechanics.

 

Again: the experiment decides how the motion is partitioned; and inelastic collisions can not conserve both energy and momentum.

 

Exchemist quote: “Right. But that is NOT saying that 2.12 "units of momentum" (in fact kg m/sec) gives rise to 4.5 kg m/sec. I explicitly pointed out that you cannot add momenta ignoring their direction. The figure of 4.5 kg m/sec is your figure, not mine and, as I told you, it is wrong. You can't do that.”

 

The 4.5 kg m/sec comes from an experiment. The experiment ends (with the same four frames) with the same 4.5 kg m/sec motion as it started. You are saying that in the middle of that experiment it can drop to 2.12 kg m/sec; I am saying that it can not. It must be 4.5 kg m/sec through out.

 

Atwood added 'up' and 'down' motion and he was correct in doing so.

 

Exchemist quote: “You are also making another, even more fundamental, wrong assumption, namely that you can choose to convert all the momentum from one body to a second, in an interaction. Whether this takes place or not depends on energy considerations as well: you do not have freedom to specify this.”

 

You can choose what experiment you are to conduct; some experiments convert all the motion and some don't. If the cylinder is stopped rotating then all the rotational motion is now in the spheres.

 

The slow motion version of the cylinder and spheres experiment takes 7 seconds; talk you administrators into giving me 7 second space and I will post the experiment in question. It has the 4.5 kg  m/sec three times: and in between are two stops of the cylinder.

Certainly the partitioning of momentum depends on the system - the experiment you run. The point I am making is that you will find, if you measure properly, that partitioning of momentum is always such as to conserve energy as well as momentum. You have put forward several scenarios in which you assume total transfer of momentum from one body to another. The purpose of my last, very simple, scenario is to show you that this assumption is in general incorrect.

 

You complain that I am "avoiding" inelastic collisions, but that is because it is only in elastic collisions that all the initial energy remains in mechanical form (k.e. and Fd work), enabling us to do the maths and illustrate my point. We can discuss inelastic collisions certainly, but then some of the energy is converted to heat and so it becomes impossible to calculate how much will remain in mechanical form, which means that no quantitative analysis of the system can be done.

 

I am definitely avoiding anything involving rotary motion, because early in this thread you made some remarks that suggest you don't understand angular momentum at all. I think you need to walk before you can run and the first thing you need to understand is that in linear motion both energy and momentum are always conserved, and it is that simultaneous conservation of BOTH that determines the unique outcome of a given experiment.  

 

Do you agree with my analysis of the 1kg and 2kg balls, i.e. the 1kg ball will rebound at 1/3m/sec while the 2kg ball acquires a speed of 2/3m/sec.?   Or do you think something different will happen and if so, what?

[OceanBreeze]

Yes I don't know why I bother really, but it seems to me possibly useful for some readers to take this right back to basics, in order to clarify it.

 

It is always the same with free energy nuts: devise a system that exceeds their capacity to analyse, analyse it wrongly and, hey presto, the whole of physics for 200 years is wrong, up to and including Noether's Theorem. Brilliant, Nobel Prizes all round!   :)

 

I told him the exact same things in my first two posts in this thread:

 

 

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.

 

 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.

 

 

And in Post 49 I gave him the solution for the Dawn satellite:

 

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

 

The information just bounces off him, like an elastic collision!

He isn’t stupid; he can do the correct math if he wants to. He doesn’t want to!

This has all the aspects of a troll and I predict you will get nowhere with him, but I do admire your patience!