energy producing experiments

[Omnibus]

@Fletcher,

Read very carefully what I write and try to understand it without imagining things. What I've shown is different from what you've understood.

[Omnibus]

Wubbly F = ma; v = at or a = v/t, substituting v/t for a in F = ma we get F = mv /t. Solving for mv we get Ft = mv. For a certain quantity of force applied for a certain quantity of time there is a certain quantity of linear Newtonian momentum produced.

The Atwood’s is a machine that was used to prove Ft = mv.

You could use a 100 kg block on dry ice and accelerate it with a one kilogram mass on a string draped over a pulley, and you would prove Ft = mv.

You could use a rim mass wheel and you will prove Ft =mv.

I know, because I have done all these experiments.

You could place all the mass of the MSU Atwood’s in the pulley and you would still prove Ft = mv. (the radius of gyration is .5)

The rule or Law (Ft = mv) will not be tricked or violated by the improper use of vectors or the improper use of angular momentum. It (Ft = mv) is a cornerstone concept in physics.

So, then, what's your point? There's no excess momentum and there's no basis to suppose there's a perpetuum mobile effect.

[Omnibus]

Wubbly I don't like to burst your bubble but anyone who understood what was happening so far acknowledged that vectors are meaningless.

In these experiments it's not the direction of the bodies that matters but their quantity of momentum/inertia/motion. In your textbook this quantity alone is meaningless not because it makes sense but because up till now it has had no use as a quantity. The direction was always needed in collision analysis. Not to mention that same textbook will tell you energy cannot be created.

What we are doing is using LOGICAL newtonian physics for practical solutions. Do you think I or the others would care less that the vector sum is 0? I already have shown how conservation of angular momentum can be debunked using conservation of linear QUANTITY momentum. Later on I will post the equation for the radius increase in time. Not because I have anything to prove for the skeptics but because I want to help pequaide out to have some mathematical model he can compare his results to.

Try to understand what @Wubbly wrote if you really care about helping @pequaide get out of his confusion. The above makes no sense.

[greendoor]

@greendoor,

It is at this point that the energy calculations will show the massive increase in energy.

No, they will not. Read carefully what @Wubbly wrote and try to understand it.

I have read what Wubbly wrote, and I understand it to be wrong.

Do you mean his incorrect vector analysis? I've already explained that the lever performs a 180 degree translation of force that negates his incorrect analysis ... maybe I didn't make that clear enough.

I also proposed eliminating the Atwood machine and replacing it with a single horizonal moving mass, as an alternative energy storage.

You anti guys seem freaked by a scary new idea and are clutching at straws to bag it with wrongly applied maths without picturing the proposed system in your mind.

My crude experiments appear to be showing momentum increases as I increase the ratio between the small and the heavy masses. Pequaid has been far more accurate in his tests - but I hope to improve my methods very soon.

What are you afraid of? Like I said earlier - if you have a vested interest in keeping this idea supressed, you would be best to stay silent. The more you try to dismiss this idea with pathetic bad science, the more I will try to validate it with logical reasoning and experiments.

By all means - give me sound reasons to doubt it. So far, i'm not impressed with the poor arguments given.

[Omnibus]

@greendoor,

What vector analysis? You may start here:

M1 = 1000 kg
v1 = +0.313 m/s

M2 = 1001 kg
v2 = -0.313 m/s

P1 = M1 * V1 = 1000 kg * +0.313 m/s = 313 kg-m/s

P2 = M2 * V2 = 1001 kg * -0.313 m/s = -313.313 kg-m/s

System momentum = P1 + P2 = 313 kg-m/s + (-313.313 kg-m/s) = -0.313 kg-m/s

I get a system momentum of -0.313 kg-m/s (not 626 kg-m/s)


Note that a 1 kg mass travelling at -0.313 m/s has a momentum of -0.313 kg-m/s
and the mass difference between M1 and M2 is 1 kg

@Wubbly's energy arguments are also correct but they are redundant. The above suffices to see there's no excess momentum and all is quite trivial.

[greendoor]

Yup - that's the incorrect vector analysis I thought you were refering to. Can't you see how wrong this? Really??

Can you picture the simple machine being described? There is no way that this vector based math is appropriate for the system in question.

Did you understand what I meant about the lever applying a 180 degree transformation? That solves this crazy +/- vector rubbish. Factor that in, by multiplying by -1, and you see that it fixes part of Wubbly's misunderstanding.

Another way to look at it: consider the two heavy balanced masses comprising the Atwood system. Pick one of them - let's say the Descending mass - and chose that as our frame of reference. At the same time - the opposite mass is Ascending. But is the momentum of that Ascending (negative) vector summing with the Descending (positive) vector? NO - of course not. The lever is transforming the momentum of the Ascending mass, by 180 degrees around the fulcrum, and adding this momentum to the Descending mass in the Descending direction.

Please try harder - this is embarassing.

[Omnibus]

The misunderstanding is yours not @Wubbly's. His is not a vector analysis but a correct application of the simple concepts involving velocities, forces and momentum. No "level applying 180 degree transformation" or any other dreamed up adjustments can get you out of abiding by these simple considerations.

[Omnibus]

Of course, for the system observed the two momenta should be summed up, if you don't want to invent some kind of fairy tale mechanics having nothing to do with reality. Each weight has its own momentum and the system as a whole will have a resultant momentum equal to the sum of these two momenta. What isn't clear?

[greendoor]

I will also add that even if the momentum of one of the masses was not available - there can be so much surplus momentum that we could afford to lose half anyway. In this example we have 44 times the momentum available compared to free fall. I think we could make do with a mere 22 times surplus ..

But in reality - all the momentum can be used. This is why flywheels are a very efficient energy storage system. By your fault reasoning, all flywheels of any mass and any speed would have zero momentum.

Yes - i'm aware that this is normally fudged by reverting to angular momentum mathematical gymnastics to disguise the truth. Which is why i'm sticking to basicly linear machines to avoid you diving for cover in that direction.

If the two 1000 kg masses were tethered and moving together along a horizontal track, we wouldn't be having this silly argument. It's quite obvious that the momentum of each mass would sum together and be available upon impacting into another system.

The momentum of these two masses do not sum to zero just because they are moving in different directions. You are just being silly and not thinking about the real application.

Also note that if you have been following pequaides suggestion - the small weight is removed from the Atwoods prior to taking all the momentum out of the system. Having achieved maxium velocity from it's descent, it has servd the purpose and now needs to be returned. So at the point where we want to extract the total momentum, we are back to a perfectly balanced system.

Do try to keep up.

[greendoor]

Sorry - bit of cross posting there. Omnibus - you seemed to do a 180 yourself. Are you saying Wubbly is wrong, or right? From your last post, you seem to agree that they should be summed, which is not what Wubbly was saying ...

Anyway - experiments will reveal the truth. Never fear.

[broli]

Greendoor don't feed the fire it will grow and consume you. These same people believe a ballistic pendulum loses 99% of the bullet energy while conserving momentum completely.

[pequaide]

I had the same thought; is Omnibus agreeing with Wubbly or not? But I assume he means to sum a negative and a positive and that would make the momentum near zero. So I prepared this post.

After the red and blue masses are moving .44 m/sec drop the extra 1 kg mass from the one sphere. Attach a horizontal string to the top of the Atwood’s and let the string wrap around the pulley of the Atwood’s. Attach this string to a 10 kilogram mass horizontally mounted on a dry ice frictionless plan. Let the Atwood’s pull the string tight to accelerate the 10 kilogram mass; you will get a conservation of linear Newtonian momentum. The Atwood’s will have 44 units of momentum and it will share that momentum with the 10 kilogram block. If it had only .44 units, as Wubbly and Omnibus propose, it would have almost nothing to share and the 10 kilogram block would be given also no motion.

I think I did this experiment too; somewhere along the way. Wubbly and Omnibus; you are just not going to trick F = ma.

[Michael]

I have heard about an physics teacher who would set up a very large pendulum, and then pull it up to a students face, and then let it swing down and back up again. The idea is for the student to be brave enough not to flinch - knowing for certain that the pendulum will never return to the exact same spot, and therefore can't hurt him/her.

I propose a similar experiment to prove/disprove this theory - which I shall call "Pequaide A".

Set up a very large balanced see-saw. Have 1000 kg masses on each end. Let's imagine the see-saw can fall, say, 2 meters. Lie with your head underneath one of the 1000 kg weights. It's as light as a feather, because it is perfectly balanced by the opposite 1000 kg weight.

OK so far?

Now - for the experiment control, let's take a box of Cornflakes. Let's drop the box of Cornflakes from 2 meters, right onto your face. I think most of us are tough enough to withstand a full frontal assault from a box of Cornflakes dropped flatside onto our face.

OK?

Now - place your face underneath the 1000 kg mass at one end of the see-saw. Raise the see-saw so this mass is now 2 meters above your face. Place the box of cornflakes on this see-saw.

You will see the 1000 kg weight slowly - very slowly - start to accelerate. It gets faster and faster - you can see it accelerating ...

Do you pull you face away, and run away like a chicken?

Or do you let the 1000 kg weight smack you in the face?

Remember - it's fully balanced, and effectively weightless ...

But it has momentum - a lot of momentum by my calculations ...

Who do you trust? Michael, Broli & Fletcher? Would you leave your face underneath it?

Personally - I wouldn't. I value my head too much.

I would greendoor, in fact I will. Forget the seesaw, all you have is a flywheel. Think you can get more power from a flywheel than in? I'll make you a bet okay. A thousand bucks. You build the thing and I'll place my face under the area you mentioned. Heck I'll place any body part under it ( and you know what I mean ). If it does more damage to me than the box of corn flakes dropping the same distance the thousand bucks is yours. If not it's mine.


I'll do this one better. I'll make this really simple. A 10 pound weight is going to free fall for the time period of 1 second. Any weight falling 1 second has a velocity of roughly 33 feet a second, rounded off. Using these figures the energy the weight will have at the end of the travel is 33 x 33 ( velocity squared ) x 10 ( the weight ). Since we are not applying any other systems to these equations we don't need to half the mass like it is usually done. So using the units in that formula the total energy is 10890.
33 x 33 x 10 = 10890.
A second weight that weighs half the first, 5 pounds, will have 5445 units of energy after its 1 second free fall journey.
33 x 33 x 5 = 5445.
This is how much energy is needed to return the weight to the top.

Now, tie a string to one weight, loop it over the ideal pulley ( no weight, no friction ) and tie it to the other weight. Have the larger weight at the top and the smaller weight at the bottom. Since it takes 5445 units of energy to return the smaller weight to the top this figure becomes subtracted from the ( driving ) larger weight. 10890 - 5445 = 5445.
5445 is how much energy the larger weight will have at the end of the journey. The velocity will be; 5445 divided by 10 and then find the square root.
5445 / 10= 544.5, then,
the square root of 544.5 = 23.33452..........
23.33452.......... is the final velocity of both the larger weight and the smaller weight.
Now, you guys are saying there's two weights traveling at 23.33452.......... added together this is much more than the 33 feet a second velocity. Your thinking there are two usable units of momentum because both weights are traveling at that velocity, but your forgeting only one of the weights is doing the work, the larger one, while the smaller one is being pulled up against gravity. The smaller weight does not have energy, it's only taking it from the larger one.
So, don't believe it? Cut the string while it is happening and see how much energy you'll get from the smaller weight.

[broli]

It's not difficult folks. When a mass increases it radius in a rotating system, the radial speed increases due to the centrifugal force. But as this happens the tangential speed goes down in order for momentum to be conserved.

The only way momentum can change is due to a force. This is what I have showed in my last drawing. Whenever the force increases the radial speed an equal force decreases the tangential speed.

This theory can be expanded to a system where only one mass is increasing its radius while the other is not.

The exact same logic applies. One radial force increases the radial speed and one tangential force decreases the tangential speed.

Now care should be taken. Because even though the forces are equal the acceleration is not. One force is working on m1 radially the other is working on m1+m2 tangentially. So obviously the radial acceleration is bigger than the tangential one.

What does this mean? it means that the tangential force will slow down m1+m2 at a lower rate than the radial velocity of m1 is increasing. When the tangential velocity is completely slowed down centrifugal force vanishes and we are left with m1 moving with the initially started momentum.

[Omnibus]

@pequaide and @greendoor,

Focus your attention on this @Michael’s sentence:

Your thinking there are two usable units of momentum because both weights are traveling at that velocity, but your forgeting only one of the weights is doing the work, the larger one, while the smaller one is being pulled up against gravity.