re: energy producing experiments

[Kator01]

also I noticed recurring discussions whether formulas "moment of inertia" or "momentum" should be used . We dont need any further discussion on this as moment of inertia in the cylinder & the spheres experiment is not involved for any further calculations as the spheres - after getting all the momentum -  are freely rotating and do not create any torque via a spoke to the axis.


Its just simple as that. I have built a simple setup with a small flywheel from an old record player ( eighties) and can confirm the full-stop of the wheel if
the small mass is out at 90 degress. I used just one small mass. It will take me a longer time to build the complete setup as I need the assistance of a mechanic  which has his workshop 25 mile away



Mike_G

[Delburt Phend]

At about 2:36 they talk about going supersonics before full extension; so a larger portion of the whip (in addition to the feathered end) must be moving at about the same speed.

Backyard scientist also says they went supersonic with rubber bands and a string. I am still working on that one.

[Delburt Phend]

Laws of levers proves you can make energy in the lab.

Let's apply a force to the lever arm at the same radius as the accelerated mass. Let's use numbers to keep this clear. The accelerated mass is on a lever arm of .1 m radius; and the force is applied on a lever arm of .1 m radius.

The mass being accelerated is more difficult to accelerate, or more easily accelerated, in accordance with the Laws of Levers.    Fr

If the two lever arms are the same .1 m /.1 m them the mechanical advantage of the accelerated mass is 1. In this case you have a direct F = ma situation. If you apply 4 newtons of force to a 1 kg mass it will be moving 4 m/sec after one second. The energy is 8 joules

If you move the mass to .2 m radius; then the accelerated mass has a mechanical advantage of .2 m / .1 m = 2. To keep the torque the same at the .1 m position of the applied force you will have to reduce the mass at .2 m to .5 kg. The torque at the applied force location (.1 m) is the same so you will still get an F = ma acceleration: except now you will get .5 kg moving 8 m/sec. 

If you move the mass to 4 m radius; then the accelerated mass has a mechanical advantage of 4 m / .1 m = 40. To keep the torque the same at the .1 m position of the applied force you will have to reduce the mass at 4 m to .025 kg. The torque at the applied force location (.1 m) is the same so you will still get an F = ma acceleration: except now you will get .025 kg moving 160 m/sec. The energy is 320 joules

Remember the second Law is "The change of momentum is proportional to the applied Force." Note that all these rotating masses have the same momentum.

[Kator01]

Delburt


are you talking about a seesaw setup ? Mass on one side...force beeing applied on the other ?


Mike

[Delburt Phend]

Consider a seesaw with 40 kg on one side and 1 kg on the other side. To be balanced the one kilogram must be 40 times further from the point of rotation than the radius of rotation of the 40 kilograms.

An extra force could be applied at the same radius as the 40 kg.

First: let us focus on the radius and rotation of the 40 kg. Let's put the radius at 2 cm. So we have 40 kg at 2 cm on one side. We could also put 40 kg at 2 cm on the outer side of this balanced beam. There would be no rotation because we have no unbalanced force.

When we put an extra 40 kg at 2 cm on one side we will get acceleration. We have (40 kg + 40 kg+ 40 kg) 120 kg accelerated by 40 kg. This is 1/3 standard acceleration; 9.81 m/sec/sec / 3 = 3.27 m/sec/sec. We know this to be true from Atwood's machine experiments.

Now let us switch out a 40 kg mass on both sides for a 1 kg mass at (2 cm * 40 = 80 cm) 80 cm. The two 1 kg masses at 80 cm are the same rotational resistance as the two 40 kg masses at 2 cm, because they are at forty times the radial distance.

Because the resistance remained the same the acceleration will also remain the same; 3.27 m/sec/sec for the remaining extra 40 kg at 2 cm. But the two one kilogram masses are moving 40 times faster than the 40 kg drive mass. At forty times faster they have forty times the energy. ½ * 40 kg *(2.557 m/sec)² = 130.76 joules     ½ * 1 kg * (102.28 m/sec)² = 5230.6 J

So let's change this into a practical arrangement. The 2 cm radius could be a 4 cm shaft that is attached to a 160 cm beam. On the end of each side of the beam are 1 kg masses. The shaft can be accelerated by a string that is wrapped around the shaft; the string is suspending a 40 kg mass from its end. This would be 3.27 m/sec/sec acceleration at the surface of the shaft.

After a one meter drop of the 40 kg; the 40 kg will be moving (1 m = ½ v²/ 3.27 m/sec/sec) 2.557 m/sec. But the two 1 kg masses will be moving 102.28 m/sec. Their kinetic energy would be KE = ½ * 1 kg * (102.28 m/sec)² = 5230.6 Joules.

The input energy is 40 kg at 1 meter of height which is (40 kg * 9.81 N/kg) * 1 m = 394.4 joules. This is the force produced by the drive mass of 40 kg that was dropped 1 meter.

The output energy is 5230.6 joules * 2 + (½ *40 kg * 2.557 m/sec * 2.557 m/sec) = 10,591.96 Joules.

There were 394.4 joules put in: and 10,591.96 joules came out.