re: energy producing experiments

[Delburt Phend]

What is the acceleration of a one kg rim mass wheel; that has a one meter diameter and is accelerated by 100 kg suspended by a string from a 1 cm shaft?

[Delburt Phend]

A massless one meter beam balance has 4 kg on one end and 1 kg on the other. It is rotating about its center of mass at 6.19 radians/sec. What is the arc speed of the one kg?

The mass relationship is 4 kg to 1 kg so when the beam is rotating about it center of mass: the 4 kg will have a radius of 20 cm and the 1 kg will have a radius of 80 cm.

At 6.19 radians per second the 4 kg is moving .2 m * 6.19 rad/sec = 1.238 m/sec around the arc of the circle.

At 6.19 radians per second the 1 kg is moving .8 m * 6.19 rad/sec = 4.952 m/sec around the arc of the circle.

At 4.952 m/sec the 1 kg can rise 1.25 m.

One kilogram in a modified Atwood's will accelerate a 4 kg block to 1.98 m/sec after the 1 kg has dropped 1 meter.

Okay lets go through this again: You drop a one kilograms mass to accelerate a 4 kilogram block in a modified Atwood's.

This gives you a final velocity of: the square root of (1 m * 2 * 9.81 m/sec² * 1/5) = 1.9809 m/sec

And this is 9.9045 units of momentum. Because this is 5 kg moving 1.9809 m/sec

A balanced beam that has 4 kg on one end and 1 kg on the other end; and is moving 6.19 rad/sec also has 9.9045 units of momentum. 

The one kilogram on the beam has half the momentum (4.95 units) and it can rise 1.25 m.

It was dropped 1 m.

If the one kilogram had all the momentum it would rise 5 m.

Physics with George 5：Rotation about the center of mass of a system like the Moon-Earth - YouTube

[sm0ky2]

2 things dilbert


1 - to get your 9 units you had to spend 10 getting the 5kg up to 60rpm


2 - to transfer all of the momentum into the 1kg mass:
     you have to stop the 4kg mass.....

[kolbacict]

will anytime any picture here ?

[Delburt Phend]

smOky2 quote: 1 - to get your 9 units you had to spend 10 getting the 5kg up to 60rpm

No: it only cost 4.429 units of momentum. If you drop a one kilogram mass one meter it will have a velocity of 4.429 m/sec. To throw a one kilogram mass up one meter it will cost you 4.429 units of momentum. The one kilogram was dropped one meter.

smOky2 quote: 2 - to transfer all of the momentum into the 1kg mass:
     you have to stop the 4kg mass.....

This is true. We can place this rotating beam in a horizontal plane. The 4 kilograms will have lost 7.9236 units of momentum: but it is in the same plane as that plane where it started; and it is at rest which is also how it started.

Now we have one kilogram that adds those 7.92 units of momentum to 1.98 and we have one kilogram moving (7.92 = 1.98) 9.9 m/sec and it can rise   d = ½ v²/a;  5 meters.      It was dropped one meter.

It seems possible that we can place a 4 kg block on one end of a horizontal beam and a 1 kg mass on the other end of a horizontal beam. Even though both masses are moving; I think this is doable.

As for pictures;  "Delburt Phend youtube " shows rotating objects giving all their motion to smaller mass spheres.

And the 'MIT Atwood's machine' has an input momentum of .0443 units and an output of .4624 units of momentum.    .01 kg * 4.429 m/sec    1.11 kg * .41666 m/sec

If this MIT Atwood's was a modified Atwood's you would have 1.100 kg moving .4166 m/sec (on a plane) and one meter away there would also be .01 kg moving .4166 m/sec. If these two masses were placed on the ends of a balanced beam; and the beam were rotating about its center of mass, then the 1.100 kg would be moving .2102 m/sec and the .01 kg would be moving 23.135 m/sec. At 23.135 m/sec the .01 kg will rise 27.255 m.

The .01 kg will rise 27.255 m and it was dropped 1 meter.

So one way to make energy is to combine a modified Atwood's with a  rotating balanced beam.