Experimental Proof That Energy Is Not A Conserved Quantity.

[exchemist]

Your data source for your numbers?

I expect you have realised this, but just in case not (and for any other readers), this poster has no understanding whatever of the mechanics of rotary motion. Concepts such as moment of inertia and angular velocity are beyond him. Consequently he tries to treat all these problems involving rotary motion in terms of mass and tangential velocity, inevitably leading to wrong conclusions.

 

The failure to appreciate how moment of inertia works is fatal, as one has to integrate mass x square of radial distance from the axis of rotation, which for an extended solid body is often not trivial.

 

He is also mad of course, since (if he is not just a troll) he must believe that none of the NASA engineers and physicists has spotted his "insight" and the enormous potential "free energy" he is claiming to exist!

 

I had some fun a year ago, on this thread, analysing some scenarios to show why he was wrong. However as this is an idee fixe of his, there seemed little point in going on with it indefinitely, so I took his advice and put him on Ignore.     

Edited April 28, 2019 by exchemist

[OceanBreeze]

He was banned at one point, but then allowed back in.

 

I never understood why.

 

He is either a troll, or militantly ignorant (of rotational dynamics) or both.

Edited April 28, 2019 by OceanBreeze

[DelburtPhend]

Some ballistic pendulums have a cup on the end of a graphite rod. The cup can receive a steel sphere shot by a spring loaded gun. Linear momentum is conserved. The final momentum is tangent to the circle of the cup pendulum. The energy of the sphere is not conserved.

 

Both the cup and the (incoming) sphere could be bobs of pendulums with the exact same length.  In this experiment both angular momentum and linear momentum are conserved.  This is because the formula for angular momentum is: the linear velocity, times the mass, times the radius. This formula would be used for the before collision side of the equation and for the after collision side of the equation.  Or v₁ * m₁ * r₁ = v₂ * m₂ * r₂ ;   but r₁ = r₂ ; so you can divide each sides by r₁ , and you get v₁ * m₁ = v₂ * m₂  which is a true statement and is The Law of Conservation of Momentum (Linear Newtonian Momentum). Energy is not conserved.

 

However: When you increase the pendulum length of the incoming sphere (say to 4 times) to that of the cup radius (and keep v₁ the same)  r₁ and r₂ are no longer equal. They do not drop out of the v₁ * m₁ * r₁ = v₂ * m₂ * r₂ equation; and the incoming angular momentum is no longer equal to the final angular momentum, because the final angular momentum does not change. Newtonian linear momentum   v₁ * m₁ = v₂ * m₂ , remains conserved. Energy is not conserved.

 

To have angular momentum (L) work you have to apply gravitation force to accelerate the velocity (tangent linear velocity) of comets. Angular momentum conservation does not work without the application of outside force. 

 

L = ω I

 

ω is the angular speed in radians and is;   linear arc velocity / radius           Arc velocity (v) is equal to linear tangent velocity. So ω = v/r

 

The moment of inertia I is mr² for a sphere on the end of a string and for a thin wall cylinder.

   

L = ω I    then it is equal to v/r * mr²      or vmr²/r;       two of the r’s cancel for vmr = L

 

Now let’s check your formula:  The product of the velocity and mass and radius of Halley's comet at Perigee should be equal to the product of the velocity and the mass and the radius of the comet at Apogee .                v₁ * m₁ * r₁ = v₂ * m₂ * r₂

 

m₁ = m₂ so the mass drops out of the equation by dividing both side by m₁

 

v₁ = speed at perigee = 54.6 km/sec

 

v₂ = speed at apogee = .908 km/sec

 

r₁ = distance to Sun at perigee .587 AU AU = 150,000,000 Km

 

r₂ = distance to Sun a apogee 35.3 AU

 

So 54.6 km/sec *.587 AU = .908 km/sec * 35.3 AU = 32.05        Source: Alan Van Heuvelen 1986 'Little and Brown'

 

A massive amount of gravitational force is applied from apogee to perigee; and this is angular momentum conservation.

Edited April 30, 2019 by DelburtPhend

[DelburtPhend]

Mass can be a means of storing motion. The production of this motion is closely defined. F = ma: or F = m Δv/Δt: or Ft = mv

 

It has been shown that a certain force applied to a mass for a certain time gives you a certain momentum. It has also been shown that the same quantity of force and time (in the opposite direction) is needed to make it stop. This is a certain quantity in and an equal quantity out; and so motion is a perfect source of the storage of momentum. Mass in motion is not however a means of storing energy (1/2mv²).

 

If you double the force you quadruplet the energy. Ft = mv   Doubling the force doubles the velocity.   Doubling the velocity quads the energy. (1/2mv²)

 

I have an 18 inch cart wheel that has a mass of 2700 grams. The rotation of this wheel can be stopped with two 152 gram spheres wrapped around it as in the cylinder and spheres.

 

When these two spheres have (and conserve) all the momentum they have a velocity increase of ten times. 3 kg * 1 m/sec = .305 kilograms * 10 m/sec.

 

But the energy: ½ 3kg * 1 m/sec * 1 m/sec = 1.5 joules becomes 15 joules = ½ .305 kg * 10 m/sec * 10 m/sec.

 

There is no Law of Conservation of Kinetic Energy because it is not conserved. There is a Law of Conservation of Momentum. So unlimited quantities of energy can be produced from gravity.

[OceanBreeze]

Mass can be a means of storing motion. The production of this motion is closely defined. F = ma: or F = m Δv/Δt: or Ft = mv

 

It has been shown that a certain force applied to a mass for a certain time gives you a certain momentum. It has also been shown that the same quantity of force and time (in the opposite direction) is needed to make it stop. This is a certain quantity in and an equal quantity out; and so motion is a perfect source of the storage of momentum. Mass in motion is not however a means of storing energy (1/2mv²).

 

If you double the force you quadruplet the energy. Ft = mv   Doubling the force doubles the velocity.   Doubling the velocity quads the energy. (1/2mv²)

 

I have an 18 inch cart wheel that has a mass of 2700 grams. The rotation of this wheel can be stopped with two 152 gram spheres wrapped around it as in the cylinder and spheres.

 

When these two spheres have (and conserve) all the momentum they have a velocity increase of ten times. 3 kg * 1 m/sec = .305 kilograms * 10 m/sec.

 

But the energy: ½ 3kg * 1 m/sec * 1 m/sec = 1.5 joules becomes 15 joules = ½ .305 kg * 10 m/sec * 10 m/sec.

 

There is no Law of Conservation of Kinetic Energy because it is not conserved. There is a Law of Conservation of Momentum. So unlimited quantities of energy can be produced from gravity.

 

 

There is no law of conservation of Kinetic Energy because of losses! That is, the final KE is always less than the initial KE, contrary to what you are claiming.

 

You do not start with 1.5 Joules of kinetic energy and end up with 15 Joules without an external source of energy, which you do not have in the de-spin mechanism.

 

What you are claiming is crackpot nonsense and this has been explained to you multiple times.

 

Why do you persist in posting this nonsense?

 

A better question may be why this forum allows you to continue to post this nonsense. I will be taking that question up with the admins.

 

Meanwhile, I am moving this from the Strange claims forum to the Silly Claims forum.

 

If there was a trash heap forum, I would move it there instead.

[DelburtPhend]

The double despin experiment (video) restores the motion to the cylinder twice; it does not lose any motion.

 

Momentum conservation is a no loss formula.

 

Energy is a loss formula.

 

Your readers are adults: let them decide.    Plus this is worth about 5 trillion dollars.

Edited May 3, 2019 by DelburtPhend

[DelburtPhend]

When 10 N of force are applied to a one kilogram mass for 3 second you will get a one kilogram mass moving 30 m/sec. This is 30 N seconds and it produces 450 joules of energy. This is 15 joule per newton second.

 

When 10 N of force are applied to a 30 kilogram mass for 3 second you will get a 30 kilogram mass moving 1 m/sec. This is 30 N seconds and it produces 15 joules of energy. This is .5 joule per newton second.

 

When a 1 kg mass moving 30 m/sec collides with a 29 kilogram mass at rest you will get a 30 kilogram mass moving 1 m/sec. This is 450 joules of energy becoming 15 joules of energy.

 

Both paragraph 2 and paragraph 3 give you a 30 kilogram mass moving 1 m/sec; But paragraph 3 also ‘allegedly’ gives you 435 joules (103.9 calories) of heat energy. 

 

The Law of Conservation of Energy pretends that 435 joules of heat energy are given of in paragraph 3. That would be the same 10 newtons of force for the same 3 seconds but you get 103.9 calories of heat free. ½ 1 kg *30 m/sec * 30 m/sec – ½ 30 kg *1 m/sec * 1 m/sec = 435 J

 

You have 30 newton seconds ONLY producing 30 kilograms at one meter per second: and then you have 30 newton seconds producing 30 kilograms at 1 m/sec AND 435 joules of heat. Doesn't physics need to be consistent.

[OceanBreeze]

When 10 N of force are applied to a one kilogram mass for 3 second you will get a one kilogram mass moving 30 m/sec. This is 30 N seconds and it produces 450 joules of energy. This is 15 joule per newton second.

 

 

 

OK. But why are you concerned with N seconds? That is a unit of impulse and cannot be used for an energy comparison.

 

The work done to get this 1 kg mass moving at 30 m/sec is the force in N times the distance over which the force is applied, not the time over which the force is applied.

 

Distance = 1/2 a t^2. a is the acceleration = F/m = 10 N/1 kg = 10 m/s^2, time is 3 seconds

Distance = 1/2 (10) 3^2 = 45 meters, Work = Fd = 10 N x 45 m = 450 Nm = 450 Joules

As you can see, 450 Joules of work/energy in gets you 450 Joules of kinetic energy out.

 

Moving on to paragraph 2:

 

When 10 N of force are applied to a 30 kilogram mass for 3 second you will get a 30 kilogram mass moving 1 m/sec. This is 30 N seconds and it produces 15 joules of energy. This is .5 joule per newton second.

 

 

 

 

Again, you are focusing on the impulse of 30 N seconds instead of the work/energy. Now you will see why this is not a good idea (unless you are just trolling, of course)

 

Distance over which the force of 10 N is applied = 1/2 a t^2. a = F/m = 10N / 30 kg = 0.333…m/s^2

 

Distance = 1/2 (0.333…) 3^2 = 1.5 meters, Work/Energy input = Fd = 10N (1.5 m) = 15 Joules

 

As you can see, the work done = the kinetic energy out of 15 Joules and the fact that the impulse here is the same as in the first paragraph means absolutely nothing as regards to the work and kinetic energy.

 

So, on to paragraph 3 we go:

 

 

When a 1 kg mass moving 30 m/sec collides with a 29 kilogram mass at rest you will get a 30 kilogram mass moving 1 m/sec. This is 450 joules of energy becoming 15 joules of energy.

 

 

 

 

That should say 450 Joules of Kinetic energy becoming 15 Joules of Kinetic energy Plus 435 Joules of heat.

 

 

 

Both paragraph 2 and paragraph 3 give you a 30 kilogram mass moving 1 m/sec; But paragraph 3 also ‘allegedly’ gives you 435 joules (103.9 calories) of heat energy. 

 

 

 

 

There is no “allegedly” about it! The Total energy out will be equal to the Total energy in, because Total energy is conserved!

 

 

The Law of Conservation of Energy pretends that 435 joules of heat energy are given of in paragraph 3. That would be the same 10 newtons of force for the same 3 seconds but you get 103.9 calories of heat free. ½ 1 kg *30 m/sec * 30 m/sec – ½ 30 kg *1 m/sec * 1 m/sec = 435 J

 

You have 30 newton seconds ONLY producing 30 kilograms at one meter per second: and then you have 30 newton seconds producing 30 kilograms at 1 m/sec AND 435 joules of heat. Doesn't physics need to be consistent.

 

 

Your little troll game of comparing impulse, in Newton seconds to Energy in Joules is a FAIL. That is why this has been moved to SILLY Claims.

 

People have spent a lot of time and effort to educate you and you refuse to learn and in fact it is clear that you are only here to troll this forum and spread misinformation. You need to be banned for the best interest of this forum.

[DelburtPhend]

In the slow motion video of the double despin device it takes 4 frames for the black square to cross from side to side. This speed of ( .02 m * 240 frame /sec / 4 frames) 1.2 m/sec is for 596 grams. This momentum (.596 kg * 1.2 m/sec = .7152 units) is present at the start; again in the middle of the experiment; and at the end.

 

Between the start and the middle of the experiment: 132 grams has all the motion; and the cylinder has no rotation.

 

Between the middle of the experiment and the end; 132 grams has all the motion; and the cylinder has no rotation.

 

If the Law of Conservation of Momentum is true this momentum at the end (.7152 units) has to be present in the 132 grams when the cylinder is at rotation rest. The 132 grams must be moving .7152 / .132 = 5.4 meters per second.

 

The .596 kg moving 1.2 m/sec is .429 joules

 

The .132 kg moving 5.4 m/sec is 1.924 joules.

 

If The Conservation of Energy theory is correct the 132 grams must lose 52.8 % of the motion in the first stop; and 52.8% of the remaining motion in the second stop. The final motion for the cylinder after the second restart would be 47% of 47%; which is only 22.3% of the original motion. The Law of Conservation of Energy would be true if it took 18 frames for the black square to cross from side to side.   Is that what you see?

[DelburtPhend]

A 399 kilogram vertically mounted rim can be accelerated by a one kilogram mass attached to a string that is wrapped around the circumference of the rim. The one kilogram force (9.81 N /kg) will accelerate the rim in an F = ma manner. So at the end of a one meter drop of the one kilogram; the 400 kilograms is moving .22147 m/sec; and it contains 88.589 units of momentum. This momentum can be transferred to the one kilogram for an energy increase from 9.81 Joules to 3924 J. All the motion can be transferred to the one kilogram and then back to the 400 kilograms so this energy increase is a provable fact. But proponents of energy conservation refute this fact by proposing a second explanation for what causes motion. They propose that motion is caused by the application of force over distance. 

 

I am going to paraphrase Principia; where Newton gave us this advice: Never attribute more to a phenomenon than that which is necessary to explain the phenomenon. This would apply to attributing two means by which force is applied to an object.

 

I took a mass of 1860 grams and hung it on an immobile force gauge; and the force gauge read 18 N. The gauge takes a reading multiple times a second and it shows present force.  The gauge is large, a 2500 N max gauge, so it is accurate to only the nearest newton.

 

I took the mass off and put it back on several time and the gauge still read 18 newtons. I left it on a few minutes and even put it on the next day; 18N.

 

While the force is being measured the mass is not moving; it was always at rest.

 

It seems to me then; that force is applied to mass in unit periods of time. Because at no time was the mass in motion. At no time was the mass moving a distance.  How can distance be required for the application of force when there is no distance over which the object has moved? How would ‘force times distance’ work anyway; do you cut layers of flux as you move through space that adds an impulse with each layer of flux you cut? Does the 18 newtons from the time concept stop as soon as motion start; and why would the ‘time * force’ stop. Wouldn’t the force remain the same when movement starts?

 

Why was a second means, for which force is applied to an object, proposed in the first place? ‘Force times Time’ can explain all the motion of objects; and the shared motion of objects. Why was the concept of force being applied over distance ever proposed? And when was it proposed? And who proposed it? 

 

I think it was proposed (or at least supported) by Leibniz because he knew that if force was not applied over distance his mv² would be false. And as it turned out; if force is not applied over distance then the Law of Conservation of Energy is clearly false.

 

His contemporary antagonist on this and other issues was Isaac Newton.

 

I would like to show you how different these two concepts are. We will use a one kilogram mass that is already falling 10 m/sec. It takes 1.019 seconds for a falling object to accelerate to 10 m/sec. When the falling object has achieved 10 m/sec it will have fallen 5.0968 meters.  d = ½ v²/a; and d = ½ at²

 

In dropping through the next second the one kilogram mass will fall an additional 14.89 meters: In the ‘force times distance’ theory it will have (14.89 M * 9.81 N / kg) 146 units of force applied to it; because it cuts 14.68 units of distance. In the ‘force times Time’ theory it only had 9.81 newton of force applied to it. Which theory is correct? Did the one second (from 1.019 second to 2.019 second of free fall) drop give you 9.81 units of motion or 146 units of motion?

 

The truth can be found by experiment: The ballistics pendulum.

 

After the one kilogram mass has fallen for 2.019 second it will have a velocity of 19.81 m/sec and it will have fallen 20.00 meters. Let this one kilogram collides with a 19 kilogram mass. And the combined 20 kilogram will be moving .99 m/sec. This same motion can be produced by a one kilogram mass draped over a pulley that is connected, by a string, to a 19 kilogram mass on a frictionless plane. This 20 kilograms will also be moving 0.99 m/sec after the one kilogram mass drops one meter.

 

The force applied (9.81N) in the 20 meter free fall; and the force applied in the one meter drop over the pulley; are the same. The time (2.019 second) of free fall for a one kilogram and the time for the 19 kg being pulled by a one kilogram mass is the same. 

 

The 20 kilograms mass moving .99 m/sec can be stopped in 2.016 seconds by making it lift one kilogram a distance of one meter.  The 20 kilogram mass can be stopped by reversing a process by which it can be made. So this is 20 kg moving .99 m/sec after 9.81 newton of force were applied for 2.019 seconds over a distance of one meter; is the same quantity of motion as a 1 kilograms mass that has fallen for the same time period.  Newton wins: the same quantity of motion can be made with one meter or 20 meters. 

 

The distance theory loses 19 meters when the one kilogram combines with the 19 kg; only one meter is needed to stop that which took twenty meters to produce. And this is verified with the kinetic energy formula: ½ 20 kg * .99 * .99 = 9.81 J; and ½ 1kg * 19.81 m/sec * 19.81 m/sec = 196 joules. 9.807 /196,    one of twenty.  

[DelburtPhend]

I looked at the specifications on a force gauge and it said that they take a reading 1000 times each second. I also decided to play a trick on the force gauge. But we need a little background information.

 

Modern force gauges are made with a load cell that can take an accurate reading when either compressed or stretched; it can do a push and a pull. The gauge will simultaneously hold the maximum reading of both push and pull. You can also place a mass on the gauge and press the zero button and that will be the starting point for the gauge measurements.

 

Well here is the trick I played on the force gauge; I placed the 1860 gram mass on the gauge and hit the zero button. The zero reading is 18 newtons; it is tared with the base line of 18 newtons.  I also hit the button that makes it hold the maximum reading. The gauge would keep the maximum reading if I pulled down on the mass. It would keep the minimum reading if I pushed up on the mass a little. If you took the mass off; the gauge would read -18 N.  So if there is any change (+ or -) in the hanging force the gauge will read it.

 

I placed the 1860 grams on the gauge; let it come to rest; and hit the zero button; I waited a minute or so and then hit the button that reads min and max. There were zero readings (18 N is the zero reading) for both the min and max. There was no increase or decrease in the force applied to the mass for 60,000 readings. The application of force is uniform and constant over time. Time constantly applies a exact force to 18 newtons upon a mass of 1.860 kg. This is when the gauge is not moving: but the gauge can be moved up and down.

 

If you choose you can make the gauge move up and down; it is on a press. With the gauge on zero I slowly moved the gauge (with the mass still attached) up and down. When I moved slow enough the gauge would not move off of zero; 18 newtons. This was over a distance of 30 cm; for 30 cm there was no application of force from the distance times force concept. Time kept applying a force just the same as if the gauge was stopped. Distance applied no force.

 

There is no experimental justification for thinking that there are two way of applying force to a mass; Time times force explains everything in motion.

Edited May 19, 2019 by DelburtPhend

[DelburtPhend]

^{You have a 4000 kg mass moving west at .3132 m/sec. It has an inelastic collision with a 6000 kilogram mass at rest. The final velocity of the 10,000 kilograms will be .1253 m/sec west.}

 

^{A one kilogram mass moving west at 19.809 m/sec has an inelastic collision with a 6000 kilogram mass at rest. The final velocity of the 6,001 kilograms will be .0033 m/sec west.}

 

^{The 4000 kilograms can cause the 6000 kg to move .1253 m/sec; and the one kilogram can make the 6000 kg move .0033 m/sec. So which of the two has more motion; the 4000 kg or the 1 kg?}

 

^{We need to use mathematical expressions that reflect that the 4000 kg has more motion than the 1 kg. The 'newton second' or mv express that the 4000 kg has more motion.  Energy says they are the same.}

 

^{Since you can experimentally prove (youtube delburt phend yo-yo despin ) that the 4000 kilograms can give all of the motion to the one kilogram, and that the one kilogram can then give all the motion back to the 4000, you then prove that the Law of Conservation of Energy is false. Ballistic pendulums prove that only momentum is given from small objects to a larger object.}

 

^{A 20 meter stack of 200 one kilogram masses can accelerate 4000 kilograms to .3132 meters per second in a drop of .1 meter, all this momentum can be transferred to one kilogram.  This is vastly more energy than a one kilogram mass moving 19.809 meters per second. It is one kilogram moving 1252.8 m/sec. A one kilogram mass moving 19.809 m/sec will rise 20 meters and it has 19.809 units of momentum.}

 

^{A one kilogram mass can be made to be .1 meter high. 200 of these masses would be 20 meters high. You could remove one from the bottom and place it on top of the stack; the stack could then be dropped .1 meter and you would have the original configuration. It takes 19.809 units of momentum to throw a one kilogram mass up 20 meters.}

 

^{These 200 kilograms could accelerate 3800 kilograms on a frictionless plane (or a rim); and after a drop of .1 meter the whole 4000 kilograms would be moving .3132 m/sec. This is 1252.8 units of momentum. You need to only remove 19.809 units of momentum from the 1252.8 and you can return the one kilogram on the bottom to the top; and you have the original starting position of the stack. That gives you 1233 units of momentum every 2.019 seconds. This is a production of about 97 joules per second.}

 

^{Each step is backed up by common experiments; with only the re-spin of the despin being new. This would be a carbon free energy source; from momentum conservation.}

Edited May 31, 2019 by DelburtPhend

[DelburtPhend]

I was recently in a discussion with an engineering friend that has several patents and is probably wealthy. He was aghast at the idea that energy could be produced free. But it also became apparent that he had no knowledge of the history of physics.

 

 

He asked; 'do you oppose the concepts of Einstein'? I told him that the concept of energy conservation precede Einstein by several hundred year. Leibniz was a primary proponent of energy and its conservation; and Newton opposed Leibniz’s ideas his entire life. My friend knew none of the names in the history of motion (including Leibniz) except Newton and Einstein.

 

 

Leibniz and Newton both knew that only one of the concept of motion (mv or mv²) could be conserved.   

 

 

Also note Voltaire comment on Leibniz's views. “Giving inert matter some kind of internal active principle, in Voltaire's view, was like going back to a mystical representation of nature.”

 

 

People might be more reasonable if they knew this was, and remains, a  matter of legitimate debate.

 

 

https://faculty.humanities.uci.edu/bjbecker/RevoltingIdeas/leibniz.html

 

 

A practical application would be a: Hydroelectric plant with no river.

 

 

A cylinder of water 30 meters high, with a diameter of 3.57 meters, will have a mass of 300 metric tons.

 

 

This cylinder could be suspended from a vertically mounted  rim with a mass of 5700 metric tons.

 

 

Three hundred tons would be accelerating 6000 tons: for an acceleration rate of (9.81/20) .4905 m/sec².

 

 

After the 300 tons has dropped .1 meter the 6000 tons will have a velocity of .3132 m/sec.  From d = 1/2v²/a

 

 

This 1,879,255 (6,000,000 kg * .3132 m/sec) units of momentum could accelerate 77.46 metric tons to a velocity of 24.26 m/sec; by using the Dawn Mission despin event.

 

 

At 24.26 m/sec the 77.46 metric tons will rise 30 meters.

 

 

The 300 metric ton cylinder has only dropped .1 meter. Only 1 ton of the 77.46 tons is needed to refill the shallow void at the top of the cylinder. The shallow void has a volume of 1000 liters.

 

 

You now have an excess of 76 metric tons at the top of the cylinder. This is a constant output of 455,000 joules per second (.1m *300000 kg *9.81 N/m /.63855 sec * 76.46 /77.46).  

 

 

The 'Delburt Phend double yo-yo despin' proves that the cylinder and spheres conserve Newtonian momentum not energy. 

[DelburtPhend]

A quote from Craig Heile in: https://www.quora.com/Who-invented-the-formula-for-Kinetic-Energy-KE-1-2-mv-2-and-when-should-it-be-used-instead-of-the-formula-for-momentum-p-mv-and-why

 

“Kinetic energy can be converted to another form, like heat, but total energy must always be conserved. It is important to note, however, that momentum, the straight un-squared velocity times mass, is also always conserved, which inhibits the manner in which energy conservation can be seen, and this had therein made kinetic energy a difficult thing for many to recognize and accept for a very long time. Today this is all but forgotten, of course.”

 

In the cylinder and spheres experiment; If energy were conserved (when the spheres have all the motion) the momentum of the spheres would be too small to restart the cylinder.

 

I suggest that the debate should be reopened.

[DelburtPhend]

Leibniz proposed this experiment to promote his conservation of energy theory.

 

“...the case in which a body A freely suspended in the air whose speed is 2, and the mass supposed as 1, at the time hits an angle of 60 degrees B and B, the mass of each of which is 2, for, in this case the body striking A stays at rest after the hit, and the bodies B and B, whose mass is 2 and who have each received a degree of speed, have each acquired 2 of force, whichever way one looks at it. Thus body A with a speed of 2 communicated a force of 4 at one and the same time.”

 

I am going to propose that this is a thought experiment that was never conducted by Leibniz; or anyone else.

 

Apparently Leibniz's argument is that the original vector momentum of A must be maintained.

With the collision at 60° the portion on the motion that is in the original line of travel (of A) is the side of the triangle opposite the 30°. The hypotenuse is the actual speed of B and B.

 

For the original vector momentum to be maintained the 4 units of mass (B + B) must be moving at ½ unit of speed along the original line of travel (the opposite side of the 30°). This ½ unit of speed is half the motion along the hypotenuse

.

So the speed of B and B along the hypotenuse must be one.

 

That would mean that the motion in the original line of travel (opposite side) is ½; and that the motion of B and B is one (hypotenuse).

This would leave you with 4 units of mass moving at 1 unit of speed: when you started with 1 unit of mass at 2 units of speed.

This alleged experiment would double the absolute motion without the application of outside force. This is clearly outside Newtonian Physics.

I am saying this is a thought experiment and should not be given any credibility.

Direction can be changed without the application of outside force; so vector direction should not supersede magnitude. Keeping the original vector cannot be done by added momentum. The original momentum comes first.

[DelburtPhend]

https://i.ytimg.com/vi/4ovhEkSIqV0/hqdefault.jpg

 

Lets take this motion and apply it to our discussion.

 

The drop time (4.757 sec) is confirmation that this is a near perfect F = ma experiment.

 

You have ten grams accelerating 1,110 grams for one meter. The acceleration rate is 9.81 m/sec² * 10 g / 1110 g = .08837 m/sec². This gives you a final velocity of sqr (1 m * 2 * .08837 m/sec²) = .42042 m/sec.

 

This is for 1.110 kg so we have ½ 1.110 kg * .42042 m/sec * .42042 m/sec = .0981 joules.

 

This is 1.110 kg * .42042 m/sec = .46667 units of linear momentum.

 

The velocity of free fall for one meter is 4.429 m/sec so the input energy is ½ * .010 kg * 4.429 m/sec * 4.429 m/sec .0981J.

 

It will take a velocity of 4.429 m/sec for the .010 kg to return to the top of the experiment. So it will take .010 kg * 4.429 m/sec = .04429 units of momentum to reload the system.

 

Okay lets cut to the chase: If you place .46667 units of momentum into .010 kg it will be moving 46.667 m/sec.

 

If you use the cylinder and sphere event to transfer all the momentum into the 10 grams then the ten grams, is moving 46.667 m/sec,and it will rise; d = ½ v²/ 9.81 m/sec² = 111 meters. The 10 grams was only dropped 1 meter.

 

If you assume that the law of conservation of energy controls the experiment then the ten grams can not achieve a velocity over 4.429 m/sec. And it can not exceed a rise of 1 meter.

 

When you think about it this is a very practical experiment: this cylinder and spheres would be a common size of only 1.100 kilogram and the spheres would have an uncommonly small mass of only 5 grams each. This is a mass ratio of 111 to 1 'cylinder and spheres' to spheres. I have done 40 to 1 and the Dawn Mission is 400 to 1; so this is within a applicable range. The arc velocity of .42042 m/sec is only about 35% of my standard hand started rotational velocity. But it will work; I have spun the cylinder and spheres slow before.

 

I know what happened with the 40 to 1: it was whir whir. The spheres would become difficult to see. A video would show a stopping cylinder but the spheres would be a blur. It would be nice to see this done in a vacuum using high speed video. You would see a restart of the rotation of the cylinder. And I know the sphere speed would be very much higher than 4.429 m/sec.

 

A restart of the cylinder rotation would not be possible if energy were to be conserved: there would be a 90% loss of momentum. The maximum speed for the 10 grams is restricted to 4.429 m/sec; .01 kg * 4.429 m/sec divide that by the original spin momentum (1.110 kg * .42042 m/sec) and you get 9.49% remaining momentum. And only momentum can be transferred from small masses to larger masses.

 

Well I conducted the experiment; of sorts.

 

I found a cylinder and spheres in my stack of old experiments. But I was not willing to cut off the existing spheres, on the cylinder, so I just taped the 29.75 mm copper spheres to the body of the cylinder. My scale showed that this cylinder had a mass of 1380 grams. I took two 3/8 inch nuts and attached them to the cylinder on a long string; the length of the string (tether) for each nut was about 1.8 wraps around the circumference. My balancing scale said the mass was 6.9 gram for each nut. So we have 1380 grams of cylinder and 13.8 grams of nuts. This is a 101 (1380 g + 13.8 g / 13.8 g ) to 1

'cylinder and sphere' mass to spheres mass ratio. Actually it is probably close to 105 with the copper spheres being taped on the outside at a greater rotational radius (that means they are moving faster than the cylinder).

 

Well it is a little intimidating to have nuts disappear into a blur at your feet. I am convinced: this is well in excess of 4.429 m/sec.

 

Now some may point out that the Atwood's has mass that is moving up and down. I have done it with rims and it works exactly the same. F = ma

 

I in fact just used a modified Atwood's to produce an energy increase to 840%.