energy producing experiments

[pequaide]

I have a real fundamental problem with this (the WM2D post). Your saying that it is 10 times as hard to rotate a one kilogram object at one meter as it is to rotate a 10 kg object at .1 meters.

If you had 10 kg at one meter then that would make sense because your lever length has increased the rotational difficulty by 10. But we have reduced the mass by ten and that should make it come up even. After all it is balanced.

Suppose I want to lift a 10 kilogram rock, I put the fulcrum at .1 meter from the rock and on the other side of the fulcrum I have a one meter lever arm. I will have to place only one kilogram (plus .001 newton to over come inertia) on the end of the lever arm to get the rock to move.

What is the meaning of the word same or equal? The rock side equals the lever arm side. 10 at .1 equals 1 at 10.

If I have such a rod handing from the sealing with 1 at 10 and 10 at 1, I could walk up to the rod and use my finder to rotate (accelerate) the rod. I would equally space my fingers from the center of mass and put equal amounts of force on each side and the rod would smoothly rotate. And as mentioned before it would not wobble.

This may well be what the books say but I will do the experiment and find out.

I am constantly impressed by the accuracy of photo gates. Just for your interest here are two sets of 4 data points taken from the double pulley Atwood’s. They are the first four reading taken from the 12 and 17.59 inch drive positions for a 93.3 gram mass. .0926, .0924, .0924, and .0924 sec: .0756, .0755, .0755, .0756 sec.

The data works out to give a 95% of Ft = mv, not impressive I guess, but it is closer than moment of inertia would have been.

[jim_mich]

I have a real fundamental problem with this (the WM2D post). Your saying that it is 10 times as hard to rotate a one kilogram object at one meter as it is to rotate a 10 kg object at .1 meters.

Yes and no.

Both weights end up moving (almost) the same velocity. (the difference is due to the weight of the bar.) The 10 kg weight at 0.1 meter must rotate 10 times as fast for it to move at the same velocity as the 1 kg weight at 1 meter.

Don't confuse RPM with weight velocity.

Note that the W2D example has about a 20% difference between the two scenarios. This is due to the weight of the lever bars.

[broli]

Jim, I think you are unnecessarily clouding the subject. A useful contribution would be the simple suggested experiment.

[jim_mich]

Broli, I simply did the experiment using WM2d. I already new what the outcome would be.

I'll crawl back into my hole now.

[greendoor]

Then there's a split in the road here. Some group want to use height while the other want to use velocities (me). Imo both are equally good if performed correctly. To call one simpler over the other is not correct, it depends on your inventory.

Sorry if that's my fault.

I just think that even if you post experimental evidence of high velocity readings, the nay sayers will quote Angular Momentum and WM2D and CF and Radius of Gyration, blah, blah, woof, woof, etc, etc ...

If we see a mass ascending a greater height than it fell, then we cut through all that armchair expert stuff.

If both are just as easy, why not go for height.

[Fletcher]

Broli - I've attempted to make two Atwoods Analogue devices - one uses a rack a pinion friction gear & the other a sliding slotted arrangement - I don't know how to make a proper rack & pinion in wm2d nor use the flexbeam to act like a pulley.

I've used 1 kg flywheel masses @ 1 m radius v's 10 kg flywheel masses @ 0.1 m radius - there are no other losses - the flywheel circle itself is virtually massless.

How do these times, velocities & Ke's fit in with your expectations ?

Feel free to alter the sim to show what you want.

They didn't fall at the same rate as I'd expect because of Inertia but you may have a different take on it ?

EDIT: although the leverage on the flywheel is balanced between the two examples for the set with the weights at the greater radius the driver has to 'work harder' contrary to what I think Pequaide was saying ? - therefore there is a timing difference in the drive weight arriving in sim world at least.

[Fletcher]

Whilst I'm at it, here is what I was talking about the other day about geometry stops you gaining Energy to get above start height.

Here I have created a simple flywheel & a pivoted drive mass which also acts as the lifted mass - this mass is 1 kg - the pivot is 180 degrees from the mass on the opposite rim & I used a rod as a pendulum rod [massless] - the green flywheel masses are 5 kgs each - because the driver mass is 1 kg it makes things easier to see - I've shown the velocity & acceleration vectors only - since f = m x a & m = 1 kg there is no need to show the force vectors [they are the same as the green acceleration vectors] - the center flywheel pivot has a one-way pin bearing on it to stop a small amount of back movement otherwise but that doesn't materially change things.

N.B. the acceleration vectors [& by default the Total Force vectors] are made up of acceleration due to gravity plus Cf's - there are no other losses such as air frictions etc.

If you follow thru the sim or drawing you will see the drive mass accelerate the flywheel CW - at 3 o'cl the pendulum bob [drive & lifted mass] is released - the next phase shows the pendulum bob falling & accelerating gaining velocity & Ke - at the same time the momentum of the flywheel is moving the pivot point downwards & across to the left then starts upwards - in this upwards phase it accelerates the bob - the next phase shows the Cf's [Cp's] then retarding the acceleration & velocities.

It is actually the Cf's which stop the flywheel in its tracks & this is a geometrical relationship of forces, vectors & symmetry - in order for the lifted mass to get higher than it started it must have a reasonable vertical component - to get the vertical component Cf's keep pulling the mass inwards & vertical - it appears the extra Ke & momentum is used up in the rod & connections strain - this is why the lifted weight can not get higher than it started from & give a gain in Pe.

Take a closer look at the Ke graph's & the Momentum graphs - both are interesting - the Momentum one shows that whilst there is a build up of flywheel momentum the driver/lifted does not inherit all the momentum as you might believe by noting the flywheel is stopped - this means IMO that Cf's took energy & momentum from the system so that CoE remains valid, at least in sim world :7) - & there appears to be not anywhere near complete transfer of momentum as thought.

So unless the Atwood device can somehow give an energy gain to launch a yo-yo de-spin device to gain height over start height then in sim world it looks very much like no energy is created to do useful work, IMO !

Feel free to rework & present the sims as required - you might cut the mass free at max velocity & bounce it upwards to see if it gets higher for instance ?

[pequaide]

Place 10 kilograms on the end of a 1.1 meter graphite rod, place 1 kilogram on the other end of the rod. Hold the rod horizontally and move the position of a fulcrum until the rod balances on the fulcrum. This fulcrum position is the point where the force needed to rotate one side is equal to the force needed to rotate the other side.

I will buy some materials to build a few arrangements of this experiment so that the photo gates will tell you the same thing.

[Fletcher]

This fulcrum position is the point where the force needed to rotate one side is equal to the force needed to rotate the other side. ???

ok ??? .. here is a sim & pic of what I think you are saying.

The pic is taken after the bottom left & right sets have completed 360 degrees of rotation - the middle set is further advanced that the offset top, set as seen by the rpms.

N.B.1. left sets driven by torque tool at -1 N-m - right sets driven by torque motor set to torque at -1 N-m for comparison.

N.B.2. rods have mass of 0.000001 kg's [virtually massless].

The important bit is that the same rotational force is applied to all sets - whilst neutrally dynamically balanced their inertia characteristics are quite different - this means that 'more effort' is required top to bottom to overcome inertia to achieve the same rpm.

Lets see how this stacks against your experiments Pequaide ?

[greendoor]

Take a closer look at the Ke graph's & the Momentum graphs - both are interesting - the Momentum one shows that whilst there is a build up of flywheel momentum the driver/lifted does not inherit all the momentum as you might believe by noting the flywheel is stopped - this means IMO that Cf's took energy & momentum from the system so that CoE remains valid, at least in sim world :7) - & there appears to be not anywhere near complete transfer of momentum as thought.

Very interesting, thanks Fletcher. I hope that this is just the sim 'writing off' what it can't account for ... as the fundamental maths is based on the impossibility of what we are trying to achieve.

If the flywheel stops, then all the momentum has gone from it - and we understand momentum to be a conserved quantity. So the question is whether it has been wasted as some other form of energy: heat, light, sound, etc. I would fully expect this to depend on the elasticity of the materials involved. I think Newton's balls suggests that using highly elastic materials means that the losses during momentum transfer can be very small indeed.

You say "it appears the extra Ke & momentum is used up in the rod & connections strain". I don't believe that strain 'uses up' force - AFAIK it merely stores it, admittedly with losses depending on elasticity.

I would like to think that if most of the momentum is transfered to the smaller mass, then we should see more motion/energy than COE predicts. Wishful thinking yes, but why admit defeat when the alternative isn't making much sense either (unexplained losses).

If all that momentum is going to go somewhere, is it a big deal whether it gets converted into motion, or heat or sound etc? I don't think we should assume that it must necessarily go to waste.

Regarding "this means IMO that Cf's took energy & momentum from the system so that CoE remains valid": The basic concept of creating momentum and transforming it to higher energy doesn't require rotation. The simplest implementation is linear momentum, using an Atwoods pully or a train track. That would take CF out of the equation? The rotary models are just more practical versions to make, but the basic principles are based on linear momentum, as I see it.

[greendoor]

The important bit is that the same rotational force is applied to all sets - whilst neutrally dynamically balanced their inertia characteristics are quite different - this means that 'more effort' is required top to bottom to overcome inertia to achieve the same rpm.

Do we need to maintain the same rpm?

If we are comparing this to linear momentum, P=MV. If we just want to use a flywheel as a means of storing momentum, then Mass and the peripheral speed of the radius of gyration are what I would consider the parameters that need to be in proportion. AFAIK, the actual rpm is spurious to this.

Edit: spurious isn't the right word - perhaps subservient.

[broli]

Fletcher I don't know what you're asking. Peq has said he saw a violation of moment of inertia in his experiments. This leads to a violation in CoE.

Instead of using the m*r*r he uses m*r. That is to say if m*r is kept the same by equally changing the mass and radius then the "difficulty to rotate a wheel" is kept the same.

Theory: 10kg at 1m is equivalent with 0.1kg at 10m
peq exp: 10kg at 1m is equivalent with 1.0kg at 10m

If the latter holds then it's another energy gaining concept.

@peq: If you build it, make sure the mass of the bar is relatively small compared to the weights.

[Fletcher]

Broli .. here is what Pequaide wrote.

Place 10 kilograms on the end of a 1.1 meter graphite rod, place 1 kilogram on the other end of the rod. Hold the rod horizontally and move the position of a fulcrum until the rod balances on the fulcrum.

This fulcrum position is the point where the force needed to rotate one side is equal to the force needed to rotate the other side. [Torque's are EQUAL ?]

I will buy some materials to build a few arrangements of this experiment so that the photo gates will tell you the same thing.

Fletcher I don't know what you're asking.

Peq has said he saw a violation of moment of inertia in his experiments. This leads to a violation in CoE.

Instead of using the m*r*r he uses m*r. That is to say if m*r is kept the same by equally changing the mass and radius then the "difficulty to rotate a wheel" is kept the same.

Theory: 10kg at 1m is equivalent with 0.1kg at 10m

peq exp: 10kg at 1m is equivalent with 1.0kg at 10m

If the latter holds then it's another energy gaining concept.

I definitely don't understand then Broli - IMO you start out with dynamically balanced systems to have Equal Torque each side of the pivot/fulcrum.

Then you can vary the position of the masses & the positions, & even split the masses one side of the fulcrum to test the m*r^2 v's m*r concept, if that's what you say he is saying.

Then you can test against a control & see what combinations require more or less torque [turning force] to maintain a rotation rate [rpm] - if you are looking at another situation then you add up the velocities & Ke's each side of the fulcrum for the same torque & see if there is Energy creation ? - at least that's how I might approach it.

See pic & sim below of dynamically balanced arrangements - the lower one with 0.1 kg's at 10 m & 9 kg's at 1 m [instead of 1 kg at 10 m] accelerates faster for the same torque - that's because of inertia calculated according to I = mr^2.




Greendoor .. the graphs show that although the flywheel was stopped about half the momentum went into the small mass - the momentum didn't disappear - it just got removed from the frame of reference IMO - the Cf's create strain in the pendulum rod & pivot & axle bearings - the Cf's would have moved the earth beneath it though so small you wouldn't materially notice it.

P.S. Greendoor .. the increase in Ke of the pendulum bob as it swings might be explained by a corresponding loss in Pe [Potential Energy of Position/Height] of the bob & a decrease in the Ke of the flywheel & other forms of Energy ?! - the theory is Total Energy is conserved in its various forms ! - definitely looks like Ke increases for a time according to the graph, but can this translate into useful work ? i.e. lift a mass higher than it started from ?



Well, I think we have a difference of opinion about what the sims predict & what is hoped to be achieved by the theories, as we knew.

When the experimental results are in to confirm or deny the theories then I can attempt to duplicate the experimental setups in WM2D to the best of my ability, as you can Broli.

Then we just may have the job of explaining the divergence of results & why !

[pequaide]

The computer program WM2D is programmed from the rules in your books (such as using moment of inertia). If the books are wrong the program is wrong.

I intend to make the equal lever arms out of graphite arrows I will get the materials ordered, but mean will look at this.

Take a ten kilogram mass and place an eye-bolt and multiple pulleys at the top. Above it on the ceiling place an eye-bolt and multiple pulleys. In this why a cord can be tied to the 10 kg mass or ties to the ceiling. The cord can go through the pulley or pulleys at the top and/or the pulley or pulleys on the 10 kilogram mass.

Thread the cord through a top pulley and then tie the cord to the 10 kg mass; you will have to place a 10 kilogram mass on (what we will call) the right side to balance the 10 kilogram mass on the left side. This pulley arrangement will give you no mechanical advantage.

If you then place an extra 20 kilogram mass on the left side, the system will accelerate as if 20 kilograms is accelerating 40 kilograms; for an acceleration of (20/40 * 9.81) 4.905 m/sec². After a drop of 1 meter, for the 20 kilogram mass, you will have a final velocity of 3.312 m/sec. This is (40 * 3.132) 125.3 units of momentum and (.5 * 40 * 3.132 * 3.132) 196.2 joules of energy. Note that the center of mass of the two 10 kilogram masses remains stationary.

Thread the cord through a top pulley and then tread it through a pulley on the 10 kg mass then tie it to the ceiling; you will have to place a 5 kilogram mass on the right side to balance the 10 kilogram mass on the left side. This will give you a mechanical advantage of 2.

If you then place an extra 20 kilogram mass on the left side the system will accelerate as if 20 kilograms is accelerating 40 kilograms; for an acceleration of (20/40 * 9.81) 4.905 m/sec². After a drop of 1 meter, for the 20 kilogram mass, you will have a final velocity of 3.312 m/sec for the 20 kilograms and the 10 kilograms but the 5 kilograms is moving 6.624 m/sec. This is (30 * 3.132 + 5 * 6.264) 125.3 units of momentum and (.5 * 30 * 3.132 * 3.132 + .5 * 5 *6.264 * 6.264) 245.24 joules of energy. Note that the center of mass of the 10 kilogram and the 5 kilogram masses remains stationary.

Thread the cord so as to make a mechanical advantage of 5 then you will have to place 2 kilograms on the right side to balance the 10 kilograms on the left.

If you then place an extra 20 kilogram mass on the left side the system will accelerate as if 20 kilograms is accelerating 40 kilograms; for an acceleration of (20/40 * 9.81) 4.905 m/sec². After a drop of 1 meter, for the 20 kilogram mass, you will have a final velocity of 3.312 m/sec for the 20 kilograms and the 10 kilograms but the 2 kilograms is moving 15.66 m/sec. This is (30 * 3.132 + 2 * 15.66) 125.3 units of momentum and (.5 * 30 * 3.132 * 3.132 + .5 * 2 *15.66 * 15.66) 392.37 joules of energy. Note that the center of mass of the 10 kilogram and the 2 kilogram mass remains stationary.

Thread the cord so as to make a mechanical advantage of 10 then you will have to place 1 kilogram on the right side to balance the 10 kilograms on the left.

If you then place an extra 20 kilogram mass on the left side the system will accelerate as if 20 kilograms is accelerating 40 kilogram; for an acceleration of (20/40 * 9.81) 4.905 m/sec². After a drop of 1 meter, for the 20 kilogram mass, you will have a final velocity of 3.312 m/sec for the 20 kilograms and the 10 kilograms but the 1 kilogram is moving 31.32 m/sec. This is (30 * 3.132 + 1 * 31.32) 125.3 units of momentum and (.5 * 30 * 3.132 * 3.132 + .5 * 2 *31.32 * 31.32) 637.6 joules of energy. Note that the center of mass of the 10 kilogram and the 1 kilogram mass remains stationary.

The final velocity of free fall for one meter is 4.429 m/sec: so the initial energy in all of the situations was (.5 * 20 kg * 4.429 m/sec * 4.429 m/sec) 196.2 joules; and the experiments produced 196.2, 245.24, 392.37, and 637.6 joules of energy.

[Fletcher]

I'm going away for few days Pequaide - perhaps in the interim you'd take the time to draw your setup so that I can duplicate & test it in WM on my return - it should take less time than it took for you write the last post.

Yes, WM conforms to the known laws of physics.

If you have ANY complete device that performs different to the program in any significant way then it will be worth investigating in depth - but that means identifying all lengths, dimensions, masses accurately, then having a start & finish cycle & a means of determining a conclusion about what you are investigating.

If you are talking about Joules of Energy [the capacity to do work] being the benchmark then whilst math calcs are a start the acid test is watching those predicted Joules doing an equivalent amount of work to confirm the math, knowing how much Energy you put into the system to start it off if other than gravity.