Free energy from gravitation using Newtonian Physic

[TinselKoala]

Hello TinselKoala,

I also had a difficult time to visualize this process.
For better understanding have a look at the pics way back at the pages 9, 12 and 17. Especially the three phases of the spheres-movement until they are fully swung out ( 90 degrees to tangent) are very well shown on page 17.
The only difference is that here instead of the puc steel-balls are used and a cylinder instead of the white disc.
The prime mover for the spinnig ot the complete setup is not shown there as this can be done in very different ways.

Regards

Kator01

If one picture is worth a thousand words, then one video is worth a thousand still pictures.

[pequaide]

TinselKoala quote; As seen from the definition, the derived SI units of angular momentum are newton metre seconds (N•m•s or kg•m2s-1 or joule seconds).Because of the cross product, L is a pseudovector perpendicular to both the radial vector r and the momentum vector p and it is assigned a sign by the right-hand rule."

Kepler’s formula worked without all this hocus pocus. A joule is a unit of energy. A “pseudovector” they have got to be joking. Are you sure this isn’t pseudoscience.

Take a 9 kg rim (or ring) that has a 2 meter diameter and place it vertically on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/sec². The speed of the string and the speed of the particles in the ring will be roughly equal; you could place the string in a grove in the rim to make the speeds even closer.  After one second all parts will be moving .981 m/sec. After one second everything will have moved .4905 m, the dangling mass will have moved down .4905 m and the mass in the ring will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum of the rim would be (mass * radians/sec * radius * radius) 9kg * .981 radians/sec *1m * 1m = 8.829

Take a 9 kg thin wall pipe that has a .1 meter diameter and place its length horizontally so that it rotates in a vertically plane on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/sec². The speed of the string and the speed of the particles in the pipe will be roughly equal; you could place the string in a grove in the pipe to make the speeds even closer. After one second all particles will be moving .981 m/sec. After one second everything will have moved .4905 m, the hanging mass will have moved down .4905 m and the mass in the pipe will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum would be (mass * radians/sec * radius * radius) 9kg * 19.62 radians/sec *.05m * .05m = .4414

I did this experiment with about a 40 kg wheel 1.05 m in diameter. The calculations were a little more complex because the wheel had bearings with a small steel hub and steel spokes. The majority of the mass was in the 2 inch by 1-1/4 inch steel rim. The resulting data proved that F = ma; I probably did not even keep the data because the results where what I expected.

I am sure you could find similar experiments on the internet if you knew what the experimenter called it. 

It is similar to an Atwood’s machine.

TinselKoala quote; Please don't make me construct an apparatus to prove it.

What is you construct one and it proves you are wrong; would that be worth the effort.

In the picture: I cut away the lower circle (forward of the entry hole) that allowed the fishing line to enter inside the circumference which determined tether length. This slows the reacceleration because now the line swings free and is attached closer to the point of rotation. This slowed reacceleration allows the experimenter to more easily observe the motion of the system as it comes to a stop.

I also lowered the center of mass of the sled by using rusty bars instead of the lead anchor ball. The sled has a mass of 2758g; the 456g gray puck easily stops the sled and white disk (300g rotational mass).

[TinselKoala]

TinselKoala quote; As seen from the definition, the derived SI units of angular momentum are newton metre seconds (N•m•s or kg•m2s-1 or joule seconds).Because of the cross product, L is a pseudovector perpendicular to both the radial vector r and the momentum vector p and it is assigned a sign by the right-hand rule."

Kepler’s formula worked without all this hocus pocus. A joule is a unit of energy. A “pseudovector” they have got to be joking. Are you sure this isn’t pseudoscience.

Take a 9 kg rim (or ring) that has a 2 meter diameter and place it vertically on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/sec². The speed of the string and the speed of the particles in the ring will be roughly equal; you could place the string in a grove in the rim to make the speeds even closer.  After one second all parts will be moving .981 m/sec. After one second everything will have moved .4905 m, the dangling mass will have moved down .4905 m and the mass in the ring will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum of the rim would be (mass * radians/sec * radius * radius) 9kg * .981 radians/sec *1m * 1m = 8.829

Take a 9 kg thin wall pipe that has a .1 meter diameter and place its length horizontally so that it rotates in a vertically plane on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/sec². The speed of the string and the speed of the particles in the pipe will be roughly equal; you could place the string in a grove in the pipe to make the speeds even closer. After one second all particles will be moving .981 m/sec. After one second everything will have moved .4905 m, the hanging mass will have moved down .4905 m and the mass in the pipe will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum would be (mass * radians/sec * radius * radius) 9kg * 19.62 radians/sec *.05m * .05m = .4414

I did this experiment with about a 40 kg wheel 1.05 m in diameter. The calculations were a little more complex because the wheel had bearings with a small steel hub and steel spokes. The majority of the mass was in the 2 inch by 1-1/4 inch steel rim. The resulting data proved that F = ma; I probably did not even keep the data because the results where what I expected.

I am sure you could find similar experiments on the internet if you knew what the experimenter called it. 

It is similar to an Atwood’s machine.

TinselKoala quote; Please don't make me construct an apparatus to prove it.

What is you construct one and it proves you are wrong; would that be worth the effort.

In the picture: I cut away the lower circle (forward of the entry hole) that allowed the fishing line to enter inside the circumference which determined tether length. This slows the reacceleration because now the line swings free and is attached closer to the point of rotation. This slowed reacceleration allows the experimenter to more easily observe the motion of the system as it comes to a stop.

I also lowered the center of mass of the sled by using rusty bars instead of the lead anchor ball. The sled has a mass of 2758g; the 456g gray puck easily stops the sled and white disk (300g rotational mass).

First, the quote is from Wikipedia, as I clearly state. Second, it's not hocus-pocus, it's recognized vector mechanics. It's how we landed a robot probe on Titan, for one thing. Third, note that the units of angular momentum are kilogram-meters squared-per second,  that is, (kg*m*m)/sec, which is exactly the units you have in your calculations, which would be correct, if only they were, well, correct.  F does indeed = MA, because that's how they are defined. Acceleration is the response of mass to a force. Force is what it takes to accelerate a mass. Mass responds to force by acceleration. Mass is that which resists acceleration by force. And so forth.
In your thought experiment you have made some incorrect assumptions, and without examining the details of your confirmatory "experiment" I cannot evaluate it. 
I can imagine your thought experiment as outlined above. But I still cannot figure out how the device in the pictures is supposed to move. You'll just have to show a video, for me to get it, I'm afraid.
And yes, the Joule is a unit of energy, and angular momentum is a means of energy " storage " which is why it can also be expressed in Joule seconds. It's called a "pseudovector" because, by convention, it points out the top axis of whatever is spinning or curving, according to a right-hand rule. By convention. Whereas a "real" vector looks like it points in the direction it's going. Again, by convention.

[pequaide]

Pardon me: the formula is correct. It is Kepler’s formula. You have my apologies.

\mathbf{r} is the position vector of the particle relative to the origin,

  ...is Radius

\mathbf{p} is the linear momentum of the particle, 

...is p or (m * radians/sec * Radius)

\times\, is the vector cross product.

This is where I got mixed up; I thought they were multiplying by yet another cross produce. But they are merely saying r * p, which is correct.

Thank you for sticking to your guns and not getting irritated with me.

But I will reiterate; this is a formula that works for satellites because in space radius has an effect upon (p) linear velocity. In the laboratory radius has no effect upon p.

TinselKoala quote: In your thought experiment you have made some incorrect assumptions,

Pequaide: Please be more specific.

I accelerate the sled and white disk and gray puck by tying a string to the opposite end of the sled. I drape the string over a pulley and suspend a mass from it. That brings the red ribbon up tight against the white disk and the (duck tape) flap that holds the gray puck up against the white disk. I hold one of the upright bolts until every thing is ready. After release and acceleration the suspended mass hits the floor about the same time that the ribbon releases the gray puck.  I assume the puck holds its position up against the white disk until the ribbon feeds past the flap as the system spins. Once the puck begins to swing out on the end of the fishing line (or tether) it absorbs the motion of the white disk and the sled. The puck has a mass of 456g and the system has a total mass of 3514g.

The consolidation of motion in the puck has been used by NASA, what is in question is “does the puck have the same linear Newtonian momentum as the original system”?

[pequaide]

The MSU Atwood’s site (page 18 reply 175 this thread) allows the pulley wheel to have mass without throwing off their calculations. So apparently MSU thinks that the pulley itself accelerates in an F = ma relationship.

I searched the web for a while and did not come up with many vertical wheels being accelerated by a string wrapped around the circumference with a mass on the end. Maybe I will have to repeat the experiment and keep the data this time. I never thought I would be required to prove that F = ma.

Does anyone expect the rim to accelerate under some other Law? The block on a frictionless plane, attached to a string draped over a pulley with a mass on the end, accelerates according to F = ma.  How is the rim different?

If the block on the frictionless plane is accelerated linearly and then is captured into a circular path by a string does it lose its linear quantity of motion when it enters the circle? If it does then how does it gain the linear momentum back after the string is cut?

Most texts say angular momentum is conserved in the laboratory and then give the ludicrous experiment of the ice skater as an example.