energy producing experiments

[ME]

NASA speculated that the masses would have 5% of the initial momentum; but how can 5% return 100%?

I don't know what NASA did, but Angular Momentum can explain things a lot better than Linear Momentum.

At least we know: there's initially something in/with the cylinder S[cylinder,0] and in/with the sphere's S[spheres,0]. Then at time (T) there's nothing in/with the cylinder and all in/with the spheres S[spheres,T]=S[cylinder,0]+S[spheres,0]. Finally all is returned except the tether being wound the other way. This indicates (almost by definition) that energy is conserved for such situation.

As all situations/conditions are rotating around a center point, I don't see the use of Linear momentum at all as there isn't any.

You could consider a Linear-momentum-part but only if you consider it in relation to its center (because all is attached), let's say: a centripetal-momentum-part.
It needs to be as flat and linear as the surface of a sphere in relation to its center. Hence, angular-stuff.

Anyway, I'm interested if you find it closer to 3-times or 9-times.

[pequaide]

The circumference is a line and the distance traveled around it is equal to the straight line of travel if the tether is cut. That is a direct equality.

It is 9 any way you try to measure it.

[pequaide]

The 9.00 to 1.00 model is operational and the motion is returned after the cylinder is stopped.

Even though the Dawn Mission yo-yo despin was much more massive than the 9 to 1; the principle, of a 100% spin return, would remain the same.  If the 3 kilogram masses were allowed to rewrap around the satellite; the final satellite spin rate would equal the initial spin rate. NASA predicted that 95% of the momentum would be missing; then why not let the three kilograms rewrap, if only 5% of the motion remains. They had an 8% return of motion before they cut the tether loose; and they were happy with that. Remember; ballistic pendulums prove that only momentum is conserved when a small mass gives its motion to a large mass.

The Dawn Mission had a 400 to 1 mass ratio; so when the small mass has all the motion it had to be moving 400 m/sec for momentum conservation; or only 20 m/sec for energy conservation. Where does the one kilogram come up with the other 380 units of momentum, for the complete return of motion.

[ME]

The circumference is a line and the distance traveled around it is equal to the straight line of travel if the tether is cut. That is a direct equality.

True, but it's not about cutting the tether, it's about rotation and a change in Moment of Inertia.
Also true, the momentary linear momentum (in case the tethers do break) can be calculated by v=ωr.
When the cylinder would be a solid disk with twice the weight of the hollow cylinder your results would be the same (equal MoI), but the ratio of the involved massed alone would be different.
The transfer in momentum is the angular kind as the cylinder stops rotating (with the right tether-length) while the spheres rotate at a larger radius, and where the MoI of the spheres equals the initial MoI of the spheres and cylinder combined. The transfer of motion is caused by the spheres flying outwards by centrifugal effects, causing negative torque by tether-tension (centripetal at an angle) on the cylinder.
The center of mass of this system remains at the rotational axis of the cylinder.

NASA:
As far as I have read some papers, the discrepancy was due to the remaining motion of liquid fuel. It would stick more to the outside of some unknown-shaped container with effects depending on the amount of fuel, friction on its walls and some residual motion it has (perhaps vortices included) when the tether got released. The amount is hard to predict as such depends on usage and thus depends on precise timing which must obviously obey some other (perhaps unrelated) control-systems.
Probably it slowed down enough to control things by steering-thrusters.

It's possible to return all motion even when the cylinder starts to rotate the wrong way before fully extended. The action is symmetric where the tether acts like a pendulum.

[pequaide]

So if you have one ninth the mass at three times the radius it only has to move three times as fast to conserve angular momentum? Same 3 as energy.

The problem is; it does not work; ballistic pendulums conserve linear momentum.

To return the motion you need nine units of linear Newtonian motion not nine unit of angular motion.

In the world of experiments it is only Linear Newtonian motion that is conserved.

You can't have it both ways it is one or the other; and Newton always wins.

[telecom]

The ballistic pendulum is a classic example of a dissipative collision in which conservation of momentum can be used for analysis, but conservation of energy during the collision cannot be invoked because the energy goes into inaccessible forms such as internal energy. ????????
http://hyperphysics.phy-astr.gsu.edu/hbase/balpen.html

[ME]

You can't have it both ways it is one or the other; and Newton always wins.

Yes you can, because that angular stuff is just a mathematical construction of linear motion. It's not a completely different concept.

Some object going in circles is actually a lever. Attached to a cylinder it acts on all points of the cylinder at once (as the cylinder is a solid object).
Moment of Inertia is simply the average of all the leverages towards all the points. That's why the MoI-factor of a ring is 1, and the MoI-factor of a solid disc is 1/2. Precalculated stuff.

Rotational kinetic Energy is E[rk] = ½ I·ω²
We know (by mathematics and the nature of the angle in radian): ω=v/r
For a ring the I=1·m·r²
We can replace and get: E[k] = ½ (m·r²)·(v/r)² = ½ m·v²
Luckily this coincides, but you'll get different results when you have a different MoI. (And the yo-yo is about the change in MoI not about ballistics: it does not have any impact anywhere... even on return it should gently 'land' on the cylinder)

We know kinetic energy is Newtonian, by mathematical equivalences:
To show a bit how we should 'know':
From height [h] and acceleration [a=G], the kinematics will be: h=½ a·t²
We can determine [t] to be √(2·h/a)
The velocity after dropping height [h] will be: v=a·t = √(2·h·a)
When v= √(2·h·a), then v²=2·h·a, or ½·v² = a·h
This is all mathematics, in this case derivatives of kinematics: [h] is position, [v] is change of position over time, [a] is change of velocity over time.
We can multiply by [m] to get E[k]=E[p] : ½·m·v² = m·a·h
The explanation what this actually should represents is physics...

So you're right to assume Newtonian applies, but it's still not a linear process.

When something goes 3 times as fast (3v) at 3 times the radius (3r) then
ω = v/r = (3v)/(3r) ... the angular speed remains the same.

[telecom]

We can determine [t] to be √(2·h/a)
The velocity after dropping height [h] will be: v=a·t = √(2·h·a)
When v= √(2·h·a), then v²=2·h·a, or ½·v² = a·h
This is all mathematics, in this case derivatives of kinematics: [h] is position, [v] is change of position over time, [a] is change of velocity over time.
We can multiply by [m] to get E[k]=E[p] : ½·m·v² = m·a·h
The explanation what this actually should represents is physics...

The problem is that you are applying a specific case of dropping an object,
where conservation of energy works, to other cases, which involve interaction of bodies, where it doesn't work,

[ME]

Is it? It is the most generic thing there is.
Whenever a force is applied (F=m*a) over some distance (s) we have spend some energy (E = F*s = m*a*s)
As said, it's a mathematical equality for whichever kind of distance or acceleration (could be centripetal for instance).

By leaving the tethers and spheres attached after full extension the second half of this experiment becomes a ballistic pendulum experiment.

With a yo-yo spin/despin there are no colliding bodies: no ballistics; but only a transfer of motion: perhaps a pendulum.

[telecom]

Is it? It is the most generic thing there is.
Whenever a force is applied (F=m*a) over some distance (s) we have spend some energy (E = F*s = m*a*s)
As said, it's a mathematical equality for whichever kind of distance or acceleration (could be centripetal for instance).

Except it doesn't work when two bodies interact, like with the ballistic pendulum.
Energy is an elusive concept, for example, after you lifted a weight and holding it steady for 10 minutes, you are spending energy during this time,
even though you don't lift it any higher.

[pequaide]

KE is Leibniz not Newton. Newton opposed the concept of KE. Attributing KE to Newton is a rewrite of history. Newton was mv; Leibniz was (1/2)mv². They were contemporaries but they were not friends.


Let’s not hide the Angular Momentum theory behind ‘radians per second’ and ‘moments of inertia’; let’s do an experiment that confirms (or not) your Angular Momentum Conservation theory.

We start with an 8 kilogram cylinder at rest, with a radius of R.

We place two tethers across a diameter of this cylinder that extent a length of 2R from the surface of the cylinder. The tethers are fully extended and they have spheres on the end that have a mass of .5 kilograms each. 

The spheres are moving 3 meters per second around the arc of the 2R circle. But you include the radius of the cylinder so you have a 3R length for your angular momentum formula.

So with a real velocity of 3 m/sec times a radius of three you have an angular momentum of 9. Is this correct?

Your theory predicts that when the spheres wrap around the cylinder they will share their motion with the cylinder; and with an angular momentum of 9 the (nine kilograms at one R) system should be moving 1 meter per second.

The angular momentum conservation theory predicts that you can use 3 units of linear momentum to produce 9. This is not going to happen.  There are several faulty formulas in physics but Newton’s mv is not one of them.

It is a matter of which of the three formulas are true and which are false.  You can’t have it both ways.

The final motion of the 9 to 1 cylinder and spheres is equal to the initial motion; so the initial energy and final energy are the same, the initial angular momentum and final angular momentum are the same; the initial linear momentum and final linear momentum are the same.  But what is the same in the middle of the experiment when the spheres have all the motion?

Answer: I have a cart wheel that is a ballpark 9 to 1. It is designed to release the tungsten spheres.  The spheres have a mass of 152 grams each.  The big trick is to find the spheres after the throw. It is dangerous; I wear a hard hat and eye protection; along with my winter coat.

At 3 meters per second such precautions would not have been observed by the thrower; because the throw could only rise .46 meters.  The throw would barely reach the throwers face. 

At 9 m/sec the throw is on the roof: and warrants the above precautions. Video tapes; rise, strobe lights photography; it is 9 m/sec.

That formula for motion which is the same in the middle of the experiment is linear Newtonian momentum.

[ME]

What was your initial speed-up velocity?

[pequaide]

For this comparison of the three formulas I have been using 1 m/sec for mathematical simplicity. But 1 m/sec is also in the range of what I throw. With the cart wheel I throw as slow as I can so that I can find the tungsten spheres. One sphere goes down into the grass and the other hopefully goes down range a bit; so that you don't get thumped. When I was throwing a bag of BBs I did not mind throwing over the trees; but spheres are not so safe.

[ME]

To be honest, I don't see how it possible, but if it is 1:9 then it's 1:9.
Could you share the video?
I hope you protect you camera as well (plexiglass?)

[pequaide]

I can send you a flash drive with a video on it; if you give me an address.