The formula for rotational kinetic energy is Ke = ½ I Omega².

The moment of inertia I is mr² for a point mass or a thin walled cylinder that is rotating around its central axis. Omega is angular velocity v/r: Omega squared is then v²/r². The v is arc velocity.

So the formula is ½ mr² * v²/r²: The r²s cancel each other and they drop out, leaving you with 1/2mv².

Note that this is for any size circle; a 10 cm cylinder or a ten meter tube. A large cylinder or a small cylinder will have the same energy if the arc velocity and mass are the same.

SmOky2 please list the sites that have the experiments you mentioned, I would like to take a look at their data.

I use frame by frame slow motion, I hold a paper on the screen on the left side of the black square.  I count the number of frames it takes for the right side of the black square to reach the paper. It is often four frames. This method may not be able to tell the difference between 3.9 and 4.1 frames but you can be sure that you can tell the difference between 4 frame and 12 frames. 10 m/sec and 3.16 m/sec for the isolated spheres' motion would give you 4 frame and 12 frames; respectively; for the final motion of the  cylinder.

“The actual experiments took place in physics laboratories and even on board the shuttles and spacelab”

Did they keep quiet about any data that had been collected?

One was the Dawn Mission despin event where NASA predicted about a 20 m/sec velocity for the spheres. 20 m/sec is half the speed of a major league baseball throw. But they gave no data. Newtonian physics predicts a sphere speed more like a gun: a centrifugal gun.

I think I am the only one that has left the strings attached; in order to show how much momentum the spheres have.

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

Lets take this motion and apply it to our discussion.

The drop time 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. It 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 of 4.430 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 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 done it 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. You have ((.01 kg * 4.429 m/sec
 / (1.110 kg * .42042 m/sec) = 9.4% 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 stuff. But I was not willing to cut off the existing spheres 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 closer to 111 with the copper spheres being taped on the outside at a greater rotational radius (that would change the moment of inertia); but the extra math might make for a difficult read. 

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