Free energy from gravitation using Newtonian Physic

[pequaide]

The velocities are roughly equal (actually the spheres are going slightly faster because their center of mass is further from the point of rotation) at release, but they immediately become unequal. In an eighth rotation (1/10 sec) or so the horizontal motion of the cylinder is gone. The momentum of the cylinder has disappeared, and there is scant little friction in the system. So where did the momentum go?

The momentum had to be transferred to the spheres, but their mass is constant.

The spheres? velocity had to change.

When the string is touching the end of the slit the radius of rotation of the spheres is the length of the string to the slit, when the cylinder is stopped. I often make that length equal to the radius of the cylinder. The initial and final radii are equal.

Look at the strobe light pictures. How much does the horizontal distance between the pictures of the sphere increase? Before release the sphere?s pictures were about a half overlapping each other. Now they are how much apart?

It all makes sense; the momentum has been transferred to the spheres.

If the cylinder has a mass three times greater than the spheres, the spheres at max are going four times as fast.   ? * 4 kg * 1 m/sec * 1 m/sec is less than ? * 1 kg * 4 m/sec * 4 m/sec (but the momentum is the same).  The energy of the system quadrupled.

What you can not see in the pictures is that there is a hole in the center of the black plastic strip in the middle of the cylinder. The strings going up to the top of the cylinder are the ends of the strings that come from the spheres. After being fed through the hole the strings are secured with the small screws. This is a quick and efficient way to change a broken string. In the test the string comes away from the cylinder with nearly no friction, and while in the slit the strings don?t even touch the slit. But when the bottom of the cylinder strikes the bed or padded floor the interaction is violent.

[Homer S.]

@Kator

So you state a differential gear will produce additonal energy? - Very interesting! I always thought a gear will transfer a part of energy into heat by friction loss?

...and by the way I'm an mechanical engineer as well (Konstruktion allgemeiner Maschinenbau). 

@pequaide

The velocities are roughly equal (actually the spheres are going slightly faster because their center of mass is further from the point of rotation) at release, but they immediately become unequal.

Please can you explain why a mass will going faster if its center of mass is further from the point of rotation? That is not true! The velocity of a rotating mass is always constant independent from its rotation radius. Only the angular rate (or rotary speed) will change according the value of the rotation radius! I believe that fact is the main reason for your misunderstandings.

In an eighth rotation (1/10 sec) or so the horizontal motion of the cylinder is gone. The momentum of the cylinder has disappeared, and there is scant little friction in the system. So where did the momentum go?

The momentum was used to stretch the strings and to transfer the spheres from translation direction into rotation direction.

The spheres? velocity had to change.

Yes the velocity is smaller than directly after releasing from the cylinder because the spheres were slowed down by the strings fixed at the cylinder.

When the string is touching the end of the slit the radius of rotation of the spheres is the length of the string to the slit, when the cylinder is stopped. I often make that length equal to the radius of the cylinder. The initial and final radii are equal.

Look at the strobe light pictures. How much does the horizontal distance between the pictures of the sphere increase? Before release the sphere?s pictures were about a half overlapping each other. Now they are how much apart?

It all makes sense; the momentum has been transferred to the spheres.

You cannot definitely assume that the velocity of the spheres has increased, the cylinder's velocity can slow down as well.

If the cylinder has a mass three times greater than the spheres, the spheres at max are going four times as fast.   ? * 4 kg * 1 m/sec * 1 m/sec is less than ? * 1 kg * 4 m/sec * 4 m/sec (but the momentum is the same).  The energy of the system quadrupled.

Nope, that's only your assumption founded in misunderstood physics...

This is a quick and efficient way to change a broken string.

Very interesting! Guess why a string may brake? And of course you think there is no need of energy to break a string made of steel???

Regards,
Homer

[pequaide]

The strings are made of Berkley Fireline, I think this is a fluorocarbon line that has a very low stretch modulus. It is very strong but after the cylinder hits the bed the line drags across the outer edge of the slit. You can see the line slowly being cut by the slit after a few dozen tests. The string won?t be cut until over one hundred tests so I am not too bothered by it.

Homer; You cannot definitely assume that the velocity of the spheres has increased, the cylinder's velocity can slow down as well.

Pequaide; The strobe light photo shows the sphere pictures becoming further apart, what has that got to do with what the cylinder is doing. As the sphere pictures separate the cylinder is shown stopping.

Homer quote; Yes the velocity is smaller than directly after releasing from the cylinder because the spheres were slowed down by the strings fixed at the cylinder.

Pequaide; How can both the cylinder and the spheres slow down, I have pictures where the cylinder is stopped and the spheres are a blur of motion.

Homer quote: The velocity of a rotating mass is always constant independent from its rotation radius.

Pequaide; This is true only if there is no unbalanced force being applied to the internal parts of the system. The cylinder shows that there is unbalanced force by stopping. The force also affects the spheres; it will speed them up There is no outside unbalanced force being applied to the system but the internal parts of the system experience unbalanced force.

[Homer S.]

I'll give up because that's wasted time and time is running too fast. Make your trials and setups and become happy!

Bye
Homer

[pequaide]

Iacob alex has a link to an Atwood simulation in his Pulsatory gravitational avalanche post. The simulation shows that the momentum of the pulley has to be incorporated in the F = ma equation to determine acceleration. They do not however state the mass distribution of the pulley: which is necessary to determine if their calculations are correct. Their calculations give the average mass, in motion, of the pulley to be at half the velocity of the circumference.

If the pulley was a ring mounted on dry ice, the calculations would be simple.  The entire mass m1 + m2 + m3 is accelerated by the difference between m1 and m2.      F = ma;     (m2 ? m1) * 9.81 / (m1 + m2 + m3) = a.  This is true because m3 (the rim shaped pulley) is moving at the same velocity as m1 and m2. If m3 is not a ring the calculations becomes more complex, but the moving mass m3 still has to be incorporated in the F = ma equation.

This simulation has value because m3 can be used as a cylinder and spheres experiment.

Make the pulley (mass 3) massive (9 kg), and place nearly all the mass of the pulley in the rim. From their calculations I think they are assuming a more or less uniform (from center to rim as in a solid disk) distribution of mass.

Place the pulley in a horizontal plane. Drape the string to mass 2 over a frictionless pulley. 

Make mass 1 equal to zero and mass 2 equal to 1 kg. With m1 zero there can be no more friction in the string, so you will have to wrap and tie the string to the horizontal pulley. 

This will give you 1 kg accelerating 10 kilograms and an acceleration of .981 m/sec.

When the ten kilograms achieves a velocity of 1 m/sec; mass 2 has dropped .5096 meters.  d = ? v?/a

Transfer the nine units of momentum of the horizontally mounted rim (pulley) into one of those nine kilograms as in the cylinder and spheres device and the one kilogram will rise (as in a pendulum) 4.128 meters. This is an energy increase to 810%

Mechanically arrange the kilogram at 4.128 meters to lift mass 2 back to where it started and you still have 3.618 meters left over.

I was reviewing my video tapes and I came across the ten to one model; the134 grams of spheres stopped 1258 grams of cylinder. In that model the slit was extended 3/16 in. from the sphere?s seat hole in the cylinder; the string is wrapped from one seated sphere nearly half way around the cylinder and entered the cylinder through the slit extended from the seat of the other sphere. The string length to the sphere?s center of mass of this model is about 2.8 times the cylinder?s radius.

Let?s say the mass of the cylinder with spheres (1392 g) was moving 1 m/sec around the circumference of the circle before release. That would mean that it would take one newton 1.392 seconds to bring the cylinder and seated spheres to a stop. F = ma, a = ∆v/∆t, 1N = 1.392 kg *1 m/sec / time; time = 1.392 kg * 1 m/sec / 1 newton.

After the spheres have all the motion it will still take one newton 1.392 seconds to bring the spheres to a stop. F = m * ∆v/∆t; ∆v = F * ∆t/m; ∆v = 1 newton * 1.392 sec / .134 kg = 10.38 m/sec. The spheres will have to be moving 10.38 m/sec for the one newton to stop them in 1.392 seconds

So Newtonian Physics predicts that the energy change in the system will be; initial energy ? * 1.392 kg * 1 m/sec * 1 m/sec, to final KE, ? * .134 kg * 10.38 m/sec * 10.38 m/sec; which is an increase to 7.22 / .696 = 1038%.

I have scavenged this model to make another model that has slits in the normal positions about half way between the seats, this will change the stopping capacity of the original model.