Free energy from gravitation using Newtonian Physic

[Kator01]

Hello pequaide,

in the Atwood-Simulation you have to switch on the friction-Button and enter 5 for mass m3. The text explaines that only if you use the friction-on-simulation that you will see the influence of mass m3.

The term friction used here is misleading. What they mean is the mass-spin-inertia-resistance of m3 which consumes part of the energy generated be free-fall of m1.

For the case m3 = 5 and m1 = 1 and friction "on" you will get an accelleration of 2.8028572.
After 1 second of free fall mass m1 has momentum P = 2.8028572 kg m/s. Circumference of m3 will deliver the angular-velocity via rot/ s = 2.8028572 / 2 x pi x r .

The spin-momentum then depends on the radius of mass m3.

In order to calculate the spin-momentum of mass m3 you have to use the relevant formulas for :

Spin-Inertia Is = m3 x r exp^2
Spin-Momentum = Is x omega ( angular-velocity )

Now the spi-inertia and the angular-velocity as well can only be calculated if you have the radius of the pulley-mass m3.

It^s not so easy to calculate this. I have done it by the assumption of r = 0.1 meter and have to look over this calculation again because I am not sure if I made a mistake.

I think that this Atwood-Example is no good for any explanations of what you found.

Regards

Kator

[pequaide]

In reference to:   www.msu.edu/user/brechtjo/physics/atwood/atwood.html

Indeed the word friction in the friction button is confusing, the word adhesion may be more appropriate. In physics we almost always view friction as something lost. From their own calculations they are not viewing this friction as something lost.

If:  m1 is 2 kg, and m2 is 3 kg, and the friction button is not selected, you get an acceleration of 1.962 m/sec/sec.

If m1 is 2 kg, and m2 is 3 kg, and m3 is zero, and the friction button is selected you get an acceleration of 1.962 m/sec/sec.

The acceleration is the same with the friction button on and the pulley moving, as it is with the button off and the pulley not moving. Their calculations assume no loss for this quantity they call friction.  And certainly we can build very high quality bearings where this is almost true. But it is incorrect to think that there needs to be a frictional loss on the surface of the pulley.

The strings on the ascending and descending sides of the pulley can be independently wrapped and fastened to the pulley; you really don?t need friction to make a pulley work. If the circumference of the pulley was coated with dry ice, you could still use the pulley by wrapping the string from both sides a few times around the pulley in the appropriate direction, and attaching the other end of the string to the pulley. You would get the same results.

Suppose we use a long steel pipe mounted horizontally on dry ice for the pulley. The acceleration of the pipe?s mass, as it spins, is roughly equal to the acceleration of the masses m1 and m2. It makes a difference what the mass of the pipe is. But if we use a larger radius pipe with the same mass; you will get the same acceleration.

I don?t think it was an oversight that the MSU Atwood site makes no reference to the radius of the pulley. It does not matter what size wheel you use. Therefore F = ma is not in reference to angular momentum changes. F = ma makes linear momentum.

I think that those that constructed the Atwood simulation assumed that the mass distribution of the pulley was like that of spokes. The spoke arrangement would allow equal mass at every distance (radius) from the point of rotation. This uniformity of mass allows the average momentum to be half the distance from the center to the circumference (1/2R). So adding 4 kg to m3 (the spokes have a mass of 6 kg instead of 2 kg for example) would be the same as adding 1 kg each to m1 and m2.


The following is an E-mail to MSU

Your MSU site on Atwood?s machine has been used as a verification of an experiment that is circulating in Europe and the U.S.

Your Atwood calculations guarantee that momentum can be placed and stored in a spinning wheel. The momentum is placed in according to the relationship of F = ma. If it is placed in according to F = ma then most assuredly it must be taken out according to the same relationship. 

Your Atwood site is unusual and useful because it makes prediction concerning the mass of the pulley. Many do not deal with the mass of the pulley and assume it to be without mass.

On your site: if mass 1 and mass 2 are not equal, the unequal portion of the mass accelerates the total mass (m1 + m2) according to the relationship of F = ma.  But the unequal mass also accelerates the pulley or wheel, as you have noted.

According to your site calculations the Atwood accelerates the pulley as if the average mass of the pulley is half way between the center axis and the circumference. In a spoke wheel this would be correct. So the mass of the pulley is accelerated twice as easily as the same mass equally distributed between m1 and m2. This is because the average mass of the spoke pulley is moving half as fast as the circumference velocity, and m1 and m2 are moving at the same velocity as the circumference.

For example: #1 let?s start with 2 kg in m1, and 3 kg in m2, and 0 kg in m3.  Then adding 1 kg to each of m1 and m2 would be equal to adding 4 kg to m3. Since the hanging masses accelerate according to F = ma then the wheel must be accelerating according to F = ma as well. 

When m1 = 2kg; m2 = 3 kg; m3 = 0 kg then acceleration equals 1.962 m/sec.

When m1 = 2kg; m2 = 3 kg; m3 = 4 kg then acceleration equals 1.4014 m/sec.

When m1 = 3kg; m2 = 4 kg; m3 = 0 kg then acceleration equals 1.4014 m/sec.


Example #2    m1 with a mass of 20 kg; and m2 with a mass of 21 kg; and with m3 zero:
Should be equal to m1 equal to zero; m2 with a mass of 1 kg and m3 with a mass of 80 kg.

When m1 = 20kg; m2 = 21 kg; m3 = 0 kg then acceleration equals .2393 m/sec/sec.

When m1 = 0kg; m2 = 1 kg; m3 = 80 kg then acceleration equals .2393 m/sec/sec.


Example #3 when force remains constant the doubling of the quantity of mass that is accelerated should reduce the acceleration to half.

When m1 = 0 kg; m2 = 1 kg; m3 = 80 kg then acceleration equals .2393 m/sec/sec this is 1 kg accelerating 41 kg. 80 kg at half the velocity (1/2 R) would be equal to 40 kilograms on the circumference.

When m1 = 0 kg; m2 = 1 kg; m3 = 162 kg then acceleration equals .2393/2 = .1196 m/sec/sec (from the MSU site). This is 1 kg accelerating 82 kg. 162 kg at half the velocity (1/2 R) would be equal to 81 kilograms on the circumference.

Twice the accelerated mass 82kg / 41kg causes half the acceleration .1196/.2393 because force remained constant ( 9.81N ).

When the cylinder and spheres machine takes the momentum back out of the wheel it can place all the motion into an object with a mass as small as the mass (difference between m1 and m2) that accelerated the wheel in the first place. This is where it gets interesting.

This MSU site confirms a source of extra momentum. The spinning wheel (pulley) has a much larger amount of momentum than the momentum developed from the freefall for the same distance of the same mass that gave the momentum to the wheel. When the momentum of the wheel is given to the mass used to accelerate the wheel, the mass can rise much higher than the distance it was dropped.

[Kator01]

Hello pequaide

I have done some research on basics :

Moment of inertia ( can bee accesses only one or two  times. Then you must register )



[url]http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html][http://www.efunda.com/math/solids/IndexSolid.cfm/url]

[url]http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html


Mechanics->Rotation->   Angular momentum
         Rotational kinetic energy
         Moments of Intertia

Here you find the basic formulas.

I found a german scientific paper about calculating the accelleration of two ( or one ) falling mass ( 1 kg ) in the atwood-example using a pulley ( full cylinder, no rim-mass ) of 5 kg. Your assumtions are not true.

See here the formula right above Paragraph 4) Ein Fallversuch (3)

http://www-ik.fzk.de/~drexlin/Mechanik0607/L8.pdf

Assuming  the radius of the pulley to be 0,1 meter the value of accelleration is in exact accordance with the atwood-simulation :

accelleration a = 2.8 m/s exp^2 and not 9,81. This means then for just one mass of 1 kg  accellerating downwards ( no other upward moving mass used )  that if you let this mass fall for 1 second the velocity will be 2.8 m/s. The distance it has fallen downwards is 1.4 m. ( s = v x t / 2 )

By this you can calculate the Input-Energy ( raise the 1 kg up to 1,4 m ) = m x g x h = 1 kg x 9,81 x 1.4 m = 13.734 Joule.

The Spin-Inertia of the pulley J = m/2 x r exp^2 = 2.5 x 0,01 = 0.025 kg m exp^2

Angular velocity of the pully at v = 2.8 m/s of the falling mass -> Omega = v/r = 2.8 m/s / 0.1 m = 28 1/s

This will give you the Spin-Momentum of the pulley  L = J x omega exp^2 = 0.025 x 28 = 0.7 kg m exp^2 / s

The falling mass m at v = 2.8 m/s has momentum of 2,8 kg m/s

Rotational energy of the pulley Wrot =  J x omega exp^2 / 2 = 0.025 x 28 exp^2 / 2 = 9,8 Joule.

Energy of the falling mass m at v = 2.8 m/s = 1/2 x m x v exp^2 = 1/2 x 1kg x 2.8 exp<^2 = 3.92 Joule

Total energy after 1 second = 9.8 + 3.9 = 13.7 Joule ( see above for Input Energy )

What you have to to is to exactly measure the velocity of the spheres and compare this to the rotational energy of the cylinder-and-spheres system before you release the balls.

As I said the atwood-machine is no good example. Spin-momentum has a different dimension ( kg m exp^2 / s ) than
translatory momentum ( kg m/s ), and I have no idea how this small 0.7 Spin-Momentum will convert to translatorry momentum, increasing energy.

There is one thing one can find out by experiment : The proposal of Alan Cresswell on his homepage here :

http://www.unifiedtheory.org.uk/

See Diagramm 2-2 ONCE MORE FOR NEWTONIAN APPLEHEADS

Regards

Kator

[pequaide]

How much linear momentum does the wheel and falling mass have after the one kilogram has fallen 1.4 meters?

I think it is about 9.8 units of linear momentum.

If these 9.8 units of linear momentum is transfer to the one kilogram that has fallen, then it will rise, 4.895 meters. This is an increase to 4.895m/1.4m = 350%.

A rim mass pulley or wheel would make it easier to calculate the momentum of the pulley. And I think you would agree that such a rim mass pulley could be constructed.

I think a ring vertically mounted on dry ice would give you a nearly perfect F = ma, if you dropped an extra mass on a string wrapped around the pulley ring.

Let the ring have a mass of 5 kg with an extra dropped mass of 1 kg. Now the dropped mass and the ring have the same velocity.  Now; do the math again, the system has more momentum than it needs to return to the top. And the cylinder and spheres transfers all the motion of the cylinder and spheres (ring and dropped mass) to the spheres (dropped mass).

[pequaide]

Wrap a string around a 5 kilogram rim mass pulley; suspend a 1 kilogram mass from the string as in a one suspended mass Atwood machine. This is one kilogram accelerating 6 kilograms. Acceleration equals 1/6 * 9.81 m/sec/sec. If you let the mass drop one meter the entire system will be moving 1.808 m/sec.

You can use the kinetic energy formula to prove that the energy of this 5 kilogram rim and suspended one kilogram mass has the same energy as one kilogram that has free fallen one meter. 

The square root of (1m * 2* 1/6 * 9.81) = 1.808m/sec,    ? * 6 kg* 1.808 m/sec * 1.808 m/sec = 9.806 joules.  This is the energy of the 6 kilograms after one kilogram has dropped one meter.

The square root of (1 * 2 * 9.81) = 4.43,    ? * 1kg * 4.43 m/sec * 4.43 m/sec = 9.81 joules. This is also the energy of a one kilogram mass that has free fallen one meter. Since energy is Force times distance I guess this should not be surprising. 

They have the same kinetic energy; but they do not have the same momentum. Momentum is Force times time so this should not be surprising either.

The time over which the force acts in the one kilogram free fall for one meter is √ (2 *d / a) = t = .4515 sec

The time over which the force acts in the one kilogram dropped one meter over a five kilogram rim mass pulley is √ (2 *d / (a/6)) = t = 1.106 sec


6 kg * 1.808 m/sec = 10.848 units of momentum; 1 kg * 4.43 m/sec = 4.43 units of momentum.  10.848/4.43 = 1.106 sec / .4515 sec

A puck circling on the end of a string, which is wrapping the string around the center pin on a frictionless plane, is not conserving angular momentum. It is conserving linear momentum.

Ballistic pendulums always conserve linear momentum not angular momentum. Even if the incoming and final motion is circular, linear momentum is still (always) the quantity conserved not angular momentum.

How can you trust angular momentum conservation when it can?t pass the simplest of test?

When you reverse the phenomenon of the ballistic pendulum and send off a small projectile from a large circling object, why wouldn?t you expect the same quantity to be conserved as that which was conserved in the forward direction?