Centrifugal guns...

[sm0ky2]

The final energy of the .132 kg balls in the free-falling frame of reference is also about .66J.
The balls does not move ten time as fast. They simply move with constant velocity, while initially unwinding the wire then winding it back.There are no forces to accelerate the balls, hence their speed remain constant.

This is a well known physics experiment, which is best done by a machine.
the basic level college demonstration done by hand is too inconsistent to take accurate measurements.


The rotational velocity of the balls does exceed that of the initial system,
as the cylinder loses momentum to the string tension.
The system as a whole loses net energy rapidly, in air, however in a vacuum in the absence of
gravity and other effects, it transfers momentum rather efficiently and momentum depletes more slowly.


There are also desirable conditions in terms of diameters, proportionate masses, length and strength of the strings, etc. to make it work best.

[Delburt Phend]

This is a big clue smOky2. to me this is saying one formula (F = ma) works and the other formula (1/2mv²) doesn't. 

"The system as a whole loses net energy rapidly, in air, however in a vacuum in the absence of
gravity and other effects, it transfers momentum rather efficiently and momentum depletes more slowly."

If 1/2mv² 'conservation' was correct the velocity in the middle would be 3.16 m/sec.

You could not be conserving momentum with a loss of 68.4% of your momentum.

One of these formulas is not applicable to the experiment. 

[Delburt Phend]

You are correct Tinu; the final energy and the initial energy are the same. But this is because the final linear momentum (arc velocity * mass) and the initial linear momentum are the same.  The initial and final configurations are the same so everything is the same at these two points in time. But what is the linear momentum and energy in the middle when the spheres have all the motion?

The initial momentum is 1.320 units: 1.320 kg * 1m/sec.  The initial kinetic energy is .66 joules; ½ 1.320 kg *1 m/sec * 1 m/sec.

The middle linear momentum is 1.320 units: .132 kg * 10 m/sec. The middle kinetic energy is 6.6 joules; ½ .1320 kg *10 m/sec * 10 m/sec.

The final momentum is 1.320 units: 1.320 kg * 1m/sec.  The final kinetic energy is .66 joules; ½ 1.320 kg *1 m/sec * 1 m/sec.

When a 132 g projectile moving 10 m/sec collides with a 1188 gram block at rest we will get a combined mass of 1.320 kg moving 1 m/sec. This is consistent with the loss of energy between middle and final.

But what if the spheres are not left attached, when they are moving 10 m/sec. Lets direct them upward and cut them loose; at 10 m/sec they will rise 5.10 meters.      d = ½ v²/a

Initially we started with 1320 grams moving 1 m/sec. This one meter per second velocity is equal to dropping this mass .0510 meters.        square root of (d * 2 * a) = v

1320 gram at a height of 5.1 cm could lift 132 grams to 51 cm.

132 grams moving 10 m/sec is equal to 132 grams at a height of 510 cm.

Dynamic application: take a 1188 gram mounted rim and wrap a weighted string around it; let 132 grams drop .51 meters. This will accelerate the entire 1320 grams to 1 m/sec. Transfer all this momentum to the 132 grams and throw it upwards. Only one tenth of the distance up is needed to reload the system.

[tinu]

The initial momentum is 1.320 units: 1.320 kg * 1m/sec.  The initial kinetic energy is .66 joules; ½ 1.320 kg *1 m/sec * 1 m/sec.

I have an issue with that. There is also an initial rotational energy of the cylinder, given by E_cylinder=(Moment of inertia) x sqr(angular velocity).
The way you've got to .66J doesn't seem right to me. Or is it?

[Delburt Phend]

Place 1188 grams on a frictionless plane. Tie a string to the mass and drape the string over a pulley. On the suspended end of the string place 132 grams. Allow the 132 grams to drop until the total mass of 1320 grams reached a velocity of one meter per second. Upon measurement you will find that the 132 grams has dropped .50968 meters. One formula for energy is the newton/meter and this would be .132 kg * 9.81 N / kg = 1.2949 newtons; which would be multiplied by .50968 meters for .65999 joules.

Any mass moving one meter per second can rise .050968 meters. We start with 1320 grams moving 1 m/sec.  So that is 1.320 kg * 9.81 N / kg * .050968 meters = .65999 joules

Take a thin wall tube with a mass of 1188g.  Place the tube in dry ice; then accelerate it with a 132 g weighted string wrapped around its surface, it will have a near perfect F = ma acceleration. It is F = ma  for any radius tube. It would work for a 10 cm tube or a 2 meter tube.  You will have a velocity of one meter per second after the 132 grams has dropped .50968 meters.       And .132 kg * 9.81 N / kg = 1.2949 newtons; which would be multiplied by .50968 meters for .65999 joules.

But is it .65999 joules in the middle when the spheres have all the motion.  ½ * .132 kg * v * v = .65999 joule ?; solving for v we have 3.16 m/sec. Can .132 kg * 3.16 m/sec = .417 units of momentum give you 1.320 (1.320 kg * 1 m/sec) units of momentum?

The main point is that only linear momentum (arc velocity  * mass) conservation can give you a middle velocity that would be sufficient for the end velocity to equal the beginning velocity.