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Gravity powered devices => Gravity powered devices => Topic started by: iacob alex on August 16, 2019, 11:15:34 AM
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.....at : https://strangengines.wordpress.com/category/centrifugal-guns/
Centrifuges,centrifugal pumps,centrifugal guns,centrifugal casting...trebuchet...merry-go-round...so many applications of the curvilinear motion...
Why not a method of applying the "classic" long-arm short-arm concept of our forum...for a possible gravity ( and centrifugal...) powered device ???
Al_ex
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maybe something like this (see attached) ...? Applied on a gravitational device like:
https://www.youtube.com/watch?v=EumMNeAVD-8 (https://www.youtube.com/watch?v=EumMNeAVD-8)
or
https://www.youtube.com/watch?v=YTLh_vYBpd4 (https://www.youtube.com/watch?v=YTLh_vYBpd4)
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This looks like the work of some indian/paki trailer trash.
Beware of their scams.
The problem is not fossil fuels, the problem is that they are too many indian/paki trailer trash in this world for solar, wind and nuclear to be viable.
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Hi Cipbranea !
Sorry , but your comment ( YouTube movies...bouncing frames) has no connection with up here subject...you know..."Two wrongs do not make a right".Again,sorry!
Al_ex
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Hi Alex,
I think the animated gif attached is not working, but it shows the long arm-short arm principle (as a mechanical trebuchet) you mention in your post. It may have an mechanical advantage on a bouncing frame setup (more momentum on the right side, less effort to rise it back on the left side). Maybe the sketch attached will clarify my idea.
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Hi Cipbranea !
Regarding your proposal : make a search on the front page of this forum ( or Wikipedia...) with "Roberval balance " to meet people thinking alike you regarding this particular four bar linkage.
In my opinion , the only result is the internal mechanical stress...to release this ...it's a different problem...
Al_ex
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Directing me to Robertval balance proves me you didn't get my point, but because my poor drawing skills and language barrier I'll not going any further on this.
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Hi Cipbranea !
Sometimes , the disappointment tracks the step of hope.
My suggestion (...it's a different problem...): try to replace the fixed mass with a sliding mass-spring system...so to release the inner mechanical stress (destresser...) into a possible centrifugal push...but play four synchro-counter-rotating arrangements (four bar/parallelogram linkage )...to "harness" the four centrifugal "horses", as a possible variable one-way tractive vector?!
Al_ex
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You finally got it :) . And this: https://www.youtube.com/watch?v=gxsmCX5zTRw (https://www.youtube.com/watch?v=gxsmCX5zTRw) is the device I tried to explain (and the connection with the topic).
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https://jpreporter.net/v/%D0%B8%D1%81%D0%BF%D0%BE%D0%BB%D1%8C%D0%B7%D0%BE%D0%B2%D0%B0%D0%BD%D0%B8%D0%B5-%D1%86%D0%B5%D0%BD%D1%82%D1%80%D0%BE%D0%B1%D0%B5%D0%B6%D0%BD%D1%8B%D1%85-%D1%81%D0%B8%D0%BB-%D0%B2-%D0%B3%D0%B5%D0%BD%D0%B5%D1%80%D0%B0%D1%82%D0%BE%D1%80%D0%B0%D1%85-%D1%8D%D0%BD%D0%B5%D1%80%D0%B3%D0%B8%D0%B8-0CC8kHds0Ys.html
That design does not work. Bobby Amarasingam ttried that and his attitude made it clear that it did not work.
Cannot work, time to find a new hobby.
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https://www.youtube.com/watch?v=YaUmzekdxTQ
https://www.youtube.com/watch?v=boLk57cKNao
I don't call them centrifugal devices but they made energy.
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https://www.youtube.com/watch?v=YaUmzekdxTQ
https://www.youtube.com/watch?v=boLk57cKNao
I don't call them centrifugal devices but they made energy.
I farted today, it made energy.
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https://www.youtube.com/watch?v=w-7d66JscI8
In the 1.320 kg model the original energy is about .66 joules.
The final energy of the .132 kg moving ten times as fast is 6.6 joules.
That is a production of 5.94 joules.
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https://www.youtube.com/watch?v=w-7d66JscI8
In the 1.320 kg model the original energy is about .66 joules.
The final energy of the .132 kg moving ten times as fast is 6.6 joules.
That is a production of 5.94 joules.
How to redirect the weight upward in order to harvest a potential energy instead of the kinetic?
May be with some kind of the 45 degrees inclined plates?
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.....as a spiral ( trajectory ) accelerator , at :
https://youtu.be/LH8G7atCFSQ (https://youtu.be/LH8G7atCFSQ)
It can develop a suggestion to play the "classical" variable arm(short arm-long arm ) , hoping for a continuous gravity unbalance on to the same side of the fulcrum...
Al_ex
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If it hits a funnel like target, it should go up and then overboard
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No mater what the length of the acceleration the steel sphere can not be moving at a velocity greater than the speed of the stick that is pushing it. With a radius of 25 cm the circumference is 1.57 meters. If the drill is up to 1200 rpm then the spheres are moving; 1200 rpm / 60 sec * 1.57 m = 31.4m/sec.
But when you transfer the momentum of a rim mass wheel to the spheres, by using a weighted string, then the spheres can be moving much much faster.
If the sphere on the end of the string has 1/40th the mass of the rim then it will be moving 31.4 m/sec * 40 = 1256.6 m/sec when released.
I think you could use the weighted string concept to achieve speeds to put things in orbit; but you have to get them past the atmosphere.
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A one kilogram mass that is on the end of a string draped over a 99 kilogram rim will produce 44.29 units of momentum after the one kilogram has dropped only one meter.
A one kilogram mass with 44.29 units of momentum will rise 100 meters.
Transfer the momentum cause by an unbalancing mass to the mass that caused the imbalance and you create very large amounts of energy.
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https://www.asc.ohio-state.edu/durkin.2/treb2.htm
This is a 50 to one mass ratio between the accelerating mass (brick) and the thrown mass (golf ball). The brick is stopped even though it is still being accelerated by gravity.
This could be an Atwood's with 1250 grams on one side and 1300 grams (1250 + 50) on the other side; that is accelerated down for .1 meters. Only the 50 grams (golf ball) is lowered .1 meters. The center of the 2500 gram (balanced) mass remains in the same position. At the end of a drop of .1 meter (for the 50 grams golf ball) the 2550 grams would be moving .19614m/sec.
That would be 2.550 kg *.19614 m/sec =.5001 units of Newtonian momentum.
A 50 gram golf ball with .5001 units of momentum, would have to be moving 10.003 m/sec.
A golf ball moving 10.003 m/sec will rise 5.1 meters; it was only dropped .1 meters.
The 2.550 kg Atwood's moving .19614 m/sec would have .04905 joules of energy. ½ mv²
A 50 gram golf ball moving 10.003 m/sec will have 2.50155 joules.
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http://garydgordon.com/Despin/ this is interesting and it goes back in time (RCA). But do you see a slow down when the smaller disks open up and have all the motion. There is an air resistance slowdown but if the disks where left attached they would have restarted the larger disk.
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http://en.m.wikipedia.org/wiki/Centrifugal_gun (http://https://en.m.wikipedia.org/wiki/Centrifugal_gun)
Hmm
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https://www.youtube.com/watch?v=w-7d66JscI8 (https://www.youtube.com/watch?v=w-7d66JscI8)
In the 1.320 kg model the original energy is about .66 joules.
The final energy of the .132 kg moving ten times as fast is 6.6 joules.
That is a production of 5.94 joules.
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.
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The spheres are only one tenth the origin quantity of rotational motion. The total mass is 1320 grams and the spheres only have a mass of 132 grams.
The spheres have all the rotational motion when the cylinder is stopped.
Ballistic pendulums prove that only linear Newtonian momentum can be given from a small mass to a large mass. Energy is not conserved in the motion.
The spheres are smaller masses and they have all the motion and they return all the rotational motion back to the cylinder.
The only thing that can be happening is that the spheres receive all the linear Newtonian momentum and then they give it back again.
I use slow motion and go frame by frame; and all the rotational motion is returned to the cylinder after the cylinder has been stopped. Only Newton's momentum can do this. The spheres are moving 10 times faster when they have all the motion.
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There are forces acting within the closed system. (132 g / 1320 g)
The original motion of the 1188 (1188g + 132g = 1320g) gram cylinder is about one meter per second. In a short period of time (around .1 second) the cylinder's rotation has stopped. This stop requires the application of 11.88 newtons of force; applied for .1 seconds. This force is applied by the tension in the string; and the force is applied equally to the cylinder and to the spheres. (Third Law)
When you apply 11.88 newtons of forces to a .132 kg mass for .1 seconds you will get a velocity change of 9 m/sec. F = ma and V = at
Like the cylinder: the original velocity of the spheres was one meter per second; when the additional 9 m/sec is added we have 10 m/sec.
After the cylinder has stopped: and the strings and spheres are left attached the 11.88 newtons kicks in again and it reverses the process. The cylinder is accelerated up to one m/sec and the spheres slow back down to one meter per second. This 11.88 newtons is of course the average force. And even to my surprise; the spheres do not crash into the cylinder. They simply fly in formation just above its surface.
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There are forces acting within the closed system. (132 g / 1320 g)
The original motion of the 1188 (1188g + 132g = 1320g) gram cylinder is about one meter per second. In a short period of time (around .1 second) the cylinder's rotation has stopped. This stop requires the application of 11.88 newtons of force; applied for .1 seconds. This force is applied by the tension in the string; and the force is applied equally to the cylinder and to the spheres. (Third Law)
When you apply 11.88 newtons of forces to a .132 kg mass for .1 seconds you will get a velocity change of 9 m/sec. F = ma and V = at
Like the cylinder: the original velocity of the spheres was one meter per second; when the additional 9 m/sec is added we have 10 m/sec.
After the cylinder has stopped: and the strings and spheres are left attached the 11.88 newtons kicks in again and it reverses the process. The cylinder is accelerated up to one m/sec and the spheres slow back down to one meter per second. This 11.88 newtons is of course the average force. And even to my surprise; the spheres do not crash into the cylinder. They simply fly in formation just above its surface.
Indeed, you're right. My bad.Still, the energy is being transferred from the cylinder to the balls.The movie ends to soon to tell but so far I can not see that more energy is transferred back from the balls to the cylinder.Maybe you can shed some light on this issue?Best regards!
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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.
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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.
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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.
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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?
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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.
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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.
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SmOky2 please list the sites that have the experiments you mentioned, I would like to take a look at their data.
I have seen a few fan sites on the internet but none that perform the actual experiment
or even attempt to accurately measure anything beyond the frame rate of their digital camera.
The actual experiments took place in physics laboratories and even onboard the shuttles and spacelab
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“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.
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If web site translations, of an experiment, are an indication the interest; then the winner is the double despin. https://www.youtube.com/watch?v=YaUmzekdxTQ
Here the ratio of spheres' mass to total 'cylinder and spheres' mass is smaller (132 g / 596 g) but the cylinder stops and restarts twice. For energy conservation to be true it would take 18 frames (at the end of the experiment) for the black square to cross from one side to the other.
In the slow motion video of the double despin device it takes 4 frames for the black square (.02m) to cross from side to side. This speed of ( .02 m * 240 frame /sec / 4 frames) = 1.2 m/sec is for 596 grams. This momentum (.596 kg * 1.2 m/sec = .7152 units) is present at the start; again in the middle of the experiment; and at the end.
Between the start and the middle of the experiment: 132 grams has all the motion; and the cylinder has no rotation.
Between the middle of the experiment and the end; 132 grams has all the motion; and the cylinder has no rotation.
If the Law of Conservation of Momentum is true this momentum at the end (.7152 units) has to be present in the 132 grams when the cylinder is at rotation rest. The 132 grams must be moving .7152 / .132 = 5.4 meters per second.
The .596 kg moving 1.2 m/sec is .429 joules
The .132 kg moving 5.4 m/sec is 1.924 joules.
If The Conservation of Energy theory is correct the 132 grams must lose 52.8 % of the motion in the first stop; and 52.8% of the remaining motion in the second stop. The final motion for the cylinder after the second restart would be 47% of 47%; only 22.3% of the original motion. The Law of Conservation of Energy would be true if it took 18 frames for the black square to cross from side to side. Is that what you see?
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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