Free energy from gravitation using Newtonian Physic

[Kator01]

Hello pequaide,

I repost here what we have found some time ago concernig the Atwood-Simulation :

https://www.msu.edu/user/brechtjo/physics/atwood/atwood.html

Now I used 1kg for m1, 0 kg for m2 and 9 kg for m3 ( pulley ) which returns ( friction must be set, meaning pulley-mass influence is taken into consideration ) a accelleration of 1.7836363 m/ sec exp2 which is double the
value you have calculated in your above post.
I have to do the math again according to the correct formulas I had found in the net way back then.

There is a simple answer to tinselkoala concerning this angular-momentum-thingy which seems to get stuck im many peoples head :

Once the drive-wheel and the white-wheel has stopped there is no angular-momentum any more. This moment in time is a transition-point and the only thing one can use then is the velocity of the rim-mass which when suddenly brought to zero will mean a negative-accellation ( value is the velocity) which will fetch the Force = M x ( - a )


Regards

Kator01

[pequaide]

Kator01: a acceleration of 1.7836363 m/ sec exp2 which is double the value you have calculated in your above post.

That is correct: MSU assumes that the pulley wheel does not have all the mass in the rim (which without doubt is true) but is something more like a solid disk. They assume that the average rotational mass is half way from the center of rotation to the circumference. This would be true if the pulley were only spokes with a very light circumference mass.

[pequaide]

The MSU Atwood’s site accepts the acceleration of the pulley wheel as an F = ma relationship. If they thought there was a significantly different type of momentum in the pulley they would have set up a different category for the motion to fall into, but they didn’t. If they thought there was a significantly different type of momentum in the pulley they would have probably mentioned the radius of the pulley for that would be necessary to evaluate the output angular momentum, but they didn’t. If they thought there was a significantly different type of momentum in the pulley they would have had another window for angular momentum when the friction was turned on, and the quantity in the window would change as the mass of the pulley was changed, but they didn’t. All of their motion falls into one category: Linear Newtonian momentum.

If you tripled the radius of the pulley (but kept the mass distribution of the pulley wheel the same) what effect would there be on the analysis?

I think there would be no effect. But with triple the radius angular momentum would be significantly different.   

Take a 10 kg rim (or ring) that has a 2 meter diameter that has been placed vertically on dry ice. After accelerating the rim to 1 m/sec connect its spinning circumference to a string that is wrapped around the circumference of a 10 kg thin wall pipe, at rest, that has a .1 meter diameter. Place the pipe on its length horizontally so that it rotates in a vertically plane on dry ice. What kind of motion will the large rim share with the pipe, and what will be the final circumference velocity of the larger rim. 

[pequaide]

Pictured: Four masses (brass bushings) were used to accelerate the wheel and a flag (hexagonal wrench) past two photo gates held at a uniform distance. This should be enough information to determine if F = ma. Do you agree?

I took four consecutive readings that were within 2/10,000th of a second from each other, amazing isn’t it.

[pequaide]

F = ma is a mathematical relationship between force, mass, and acceleration. To prove that the statement is true you must show that a proportional change in force causes a proportional change acceleration. For example if force increases by 20% then acceleration increases by 20%, etc.

The experiment; four different descending masses were hung from the ribbon, The force they produce is proportional to their mass, this mass is only a small portion of the accelerated mass. The four descending masses were a brass bushing or a combination of several brass bushings. Bushing 1 was 37.4g, bushing two was 26.7g, three was 26.5g, and the fourth was 73.0g

   Descending mass 1 was 37.4g, Descending mass 2 was 64.1g (37.4 + 26.7), descending mass 3 was 90.6g (37.4 + 26.7 + 26.5g), descending mass 4 was163.6g (37.4 + 26.7 + 26.5g + 73.0g),   

The time interval for each force to make the flag cross the same distance between the photo gates (from the same starting point) was; F1 .6957 sec, F2 .5336 sec, F3 .4556 sec, and F4 .3420 sec.

All forces worked over the same distance but they did not take the same amount of time to do it. To find acceleration from time we can us the formula d = ½ a * t * t because all the distance are the same the only variable are time and acceleration.  Rearranging the formula to solve for a we get 2 * d / t * t = a, again distances are all the same so the proportional  acceleration of F1 is 1 / .6957 *.6957 = 2.078, F2 is 1 / .5336 * .5336 = 3.512, F3 is 1 / .4556 *.4556 = 4.818, F4 is 1 / .3420 * .3420 = 8.55.

The proportion of force to the proportion of acceleration is close to as follows.
F1   37.4g      2.078

F2   64.1g     3.512

F3   90.6 g    4.818

F4   163.6g     8.55

But there is one more correction to make. The fourth force F4 is accelerating a larger mass than the first force. This is in a proportion to its total mass and the total mass accelerated by the smaller force F1, that proportion is the rotational mass (inertia of the wheel, 2600g) of the wheel plus each mass that caused the force. F4 accelerates the larger mass slower in a proportion of 2637.4g / 2763.6g = .954; this means that F4 would have accelerated the smaller mass 2637.4 faster giving it a higher acceleration by 1 / .954 = 1.048 * 8.55 = 8.96.

The proportion of force to the proportion of acceleration is as follows.
F1   37.4g      2.078

F2   64.1g     3.512 corrected for the greater mass being accelerated 2664.1g / 2637.4g to 3.55

F3   90.6 g    4.818 corrected for the greater mass being accelerated 2690.6g / 2637.4 to 4.92

F4   163.6g     8.55 corrected for the greater mass being accelerated 2763.6 / 2637.4 to 8.96


So does a proportional change in force (which equals a proportional change in descending mass) equal a proportional change in acceleration? Or does: 37.4g / 163.6g = 2.078 / 8.96


37.4g / 163.6g = .2286,    2.078 / 8.96 = .2319    .2319 / .2286 = 1.0145

37.4g / 64.1g = .5834,   2.078/3.55 = .5853    .5853/.5834 = 1.0033

37.4g / 90.6g = .4128,   2.078/4.92 = .4223     .4223/.4128 = 1.023

These are within two percent of a perfect F = ma relationship.  So yes; F = ma is true for tangent forces working on the circumference of a circle. This is true even though the wheel is not a rim mass wheel, and this is Linear Newtonian Momentum.