https://www.youtube.com/watch?v=boLk57cKNao

https://www.youtube.com/watch?v=KgT70vSIUgA

It is actually impossible that Newton can be wrong in this energy source:

I have one inch spheres that have a mass of 152 grams, they are tungsten spheres and should weight a little more. But I will do some math with what is real. (152 grams)

If I accelerate one of these spheres to 20 m/sec it will rise 20.387 meters. This 20.387 m is 802.6 inches.

This means I can stack 802 of the 152 grams spheres on top of each other. This is 121.9 kilograms. And I can drop the entire stack one inch (2.54 cm).

When the stack is dropped 2.54 cm it will have a final velocity of: the square root of (.0254 m * 2 * 9.81 m/sec/sec) = .7059 m/sec.  This is a momentum of 121.9 kg * .7059 m/sec = 86.054 units

The sphere needs only .152 g * 20 m/sec = 3.04 of these 86 units to travel back up to the top and reconfigure the pre-drop arrangement.

The 121.9 kilograms is equal to a rim of the same mass; and moving at the same speed of .7059 m/sec around the arch of the circle.
121.9 kg * .7059 m/sec = 86

Or it could be a 60.95 kilogram rim moving 1.4118 m/sec around the arch of the circle.

Or 30.475 kg moving 2.836 m/sec.

Or 15.2375 kg moving 5.647 m/sec.

Or 7.615 kg moving 11.29 m/sec.

Or 3.809 kg moving 22.588 m/sec. = 86.046 units;   this 22.588 m/sec will send 3.8 kilograms up 26.0 meters.

When this 22.588 m/sec for 3.8 kilograms is given to .152 kilograms it will send it to the next county.
     
You would hear a whir and you would never see the sphere again; if you were lucky enough not to get hit.

https://www.youtube.com/watch?v=dHxdmKmAfl8

It takes (19 frames) to go from 1.2 m/sec of rotation of the cylinder and spheres to the first stop of the cylinder. It takes the same amount of time (19 frames) to go from the last stop, of the cylinder's rotation, to the last full return of the cylinder's rotational motion. This motion appears to be the same 1.2 m/sec (from counting the frames needed to cross the black square). The 19 frames confirms that the rotational velocity is indeed the same. The accelerations are of the same magnitude (19 frames) therefore linear Newtonian momentum is conserved.

It would be moving about a fourth that fast (83 frames) if energy was conserved and the alleged heat was lost. It would take 83 frame to go from the last stop to the last restart instead of 19 frames.  The entire experiment is only 75 frames. 

A mass moving on the end of a string can wrap around a stationary post. The string will become shorter and the radius will be reduced; but the linear Newtonian momentum will remain the same.

In this (two 86 gram bolts) cylinder and sphere experiment the total mass of a spinning object is reduced from 1448g (cylinder and spheres) to 304g (spheres); but the linear Newtonian momentum remains the same.

In this (two 86 gram bolts taped to the cylinder) cylinder and sphere experiment the energy increases from (.5 * 1.448 kg * 1.2 m/sec 1.2 m/sec) = 1.0425 joules to (.5 * .304 kg * 5.716 m/sec * 5.716 m/sec) = 4.966 joules: but the linear Newtonian momentum remains the same.

This is an unlimited source of free energy. Because you can transform a 400 kilogram rim moving 1 m/sec into a 1 kilogram rim moving 400 m/sec; for an increase from 200 J to 80,000 J.

I made a 10 (1296g) to 1 (132 g) cylinder and spheres and it follows the same pattern as the 4.5 to 1 and others cylinder and spheres. It takes 25 frames to stop the cylinder's spin and it takes 25 frames to fully restart the spin of the cylinder. There are three frames needed to cross the black square at the release; and it takes three frames to cross from one side of the black square to the other side after 50 frames (after a full stop and a full restart).

If energy were conserved when the cylinder was stopped it would only have one third of the linear Newtonian momentum needed to return the cylinder to full rotation. That means it would take 75 frames to return the less than one third of the motion. It takes 25.

If energy were conserved when the cylinder was stopped it would only have one third of the linear Newtonian momentum needed to return the cylinder to full rotation. That means it would take 9.5 frames to cross the black square from side to side after the restart. It takes 3.

The Linear Newtonian momentum formula (mv) would be satisfied with a velocity increase of 10. This is an energy increase to 1000%.

The kinetic energy formula (½ * m * v * v) would be satisfied with a velocity increase of the square root of ten: 3.16.  Ballistic pendulum experiments prove that only Linear Newtonian Momentum is conserved as the small mass spheres collide with the cylinder; kinetic energy is never conserved.

The cylinder and spheres event keeps its Linear Newtonian Momentum.

This means that a 400 kilogram rim moving 1 m/sec would throw off weighted strings of 40 kilograms moving 10 m/sec. This 40 kilograms could throw off weighted strings of 4 kilograms moving 100 m/sec. This 4 kilograms could throw off 1 kilogram moving 400 m/sec.   Now we have ½ * 1 kg * 400 m/sec * 400 m/sec = 80,000 joule and you started with 200 joules.

Is the only way to capture this excess energy by sending the projectile upward?
And then using it as a potential energy?