Atwoods Analysis

[Wubbly]

In a rotating system, torque = moment of inertia x angular acceleration. t = I * a (It's the equivalent of F=ma in the linear world.)

You can have a baseline experiment where a torque produces an angular acceleration to a mass based on its moment of inertia.
If you double the torque and your angular acceleration goes down by half, your moment of inertia had to have gone up by a factor of 4.
If you triple the torque and your angular acceleration is only one third, your moment of inertia had to have gone up by a factor of 9.
If the quadruple the torque and your angular acceleration is only one fourth, your moment of inertia had to have gone up by a factor of 16.

[pequaide]

Wiki says torque is radius times force, N * m. You make the world far to difficult.

Lets start with 200 kilograms at 2 cm on both sides. Are we balanced? Is the torque equal on both sides? If a top is made of it will it spin nicely?

Now lets go to 200 kilograms at 2 cm on one side and 199 kg at cm 2.01 cm (9.81N/kg * m) on the other side. Are we balanced? Is the torque equal on both sides? If a top is made of it will it spin nicely?

Now lets go to 200 kilograms at 2 cm on one side and 190 kg at cm 2.105 cm on the other side. Are we balanced? Is the torque equal on both sides? If a top is made of it will it spin nicely?

Now lets go to 200 kilograms at 2 cm on one side and 10 kg at cm 40 cm on the other side. Are we balanced? Is the torque equal on both sides? If a top is made of it will it spin nicely?

Now lets go to 200 kilograms at 2 cm on one side and 2 kg at cm 200 cm on the other side. Are we balanced? Is the torque equal on both sides? If a top is made of it will it spin nicely?

All these arrangements (2 kg at 200 cm, 190kg at 2.105cm, 10 kg at 40 cm) behave the same as 200 kg at 2 cm (9.81N/kg * m). The moments of inertia are all different but they all behave he same.

[Wubbly]

I understand your examples about the torque balancing. They were done in the vertical plane. One definition of Torque is force x radius or force x distance at which the force is acting, maybe throw a sine or cosine in there to make things perpendicular.

You mentioned they have different moments of inertia. I'm trying to explain moment of inertia. If you wikipedia moment_of_inertia, you see that torque is also defined as moment of inertia x angular acceleration (for point masses). Or if you take torque and divide it by angular acceleration, you get your moment of inertia.

In your linear world, F=ma.
If you double the force, but only get half the acceleration, then your mass quantity had to have increased by a factor of 4.
If you triple the force but only get one third the acceleration, then your mass quantity had to have increased by a factor of 9.

In the rotational world, torque = moment_of_inertia x angular_acceleration.
The torque is like your force,
the moment_of_inertia is like your mass,
and your angular_acceleration is like your acceleration.

So in the mass/rod experiment, you can calculate your torque (force acting at a distance [Newton meters]), you can calculate your angular acceleration, then by dividing torque by angular acceleration, you can calculate your moment of inertia. In the mass/rod experiment what you find is that if your 1 kg mass is moved to twice the radius, it acts like it's 4 times harder to rotate. You also find that if you move your 1 kg mass to three times the radius, it acts like it's 9 times harder to rotate.

[Wubbly]

.
The torque in Your examples balanced in the vertical plane. If you place them in the horizontal plane and try to rotate them (change their angular acceleration), the moment of inertia makes it harder to rotate one of the masses than the other (even though they balance in the vertical plane).

For example, your (200 kg @ .02 m) vs your (2 kg @ 2 m). The 200 kg mass has a moment of inertia of 0.08 kg m^2. The 2 kg mass has a moment of inertia of 8 kg m^2. It is 100 times harder to change the angular acceleration of the 2 kg mass than the 200 kg mass, even though their torques balance out in the vertical plane. If you do change their angular acceleration at the same rate, the 2 kg mass will have 100 times the energy of the 200 kg mass. That's what "2 vs 200.png" insert was trying to show.

If you were to take all of the "motion" of the 200 kg mass moving at 0.5 radians/second, and give it to the 2 kg mass (spin it up to 0.5 radians/second), you might look at linear tangential momentum, say they are equal and conclude that you just created 100 times more energy than you started with. But what you ignored was the fact that the 2 kg mass is 100 times harder to accelerate up to 0.5 radians/second. Looking at just linear tangential momentum would lead you to draw an incorrect conclusion about energy gain. If you transferred all of the "motion" of the 200 kg mass to the 2 kg mass, you would find that linear tangential momentum takes a huge drop.

[Fletcher]

I'll post this again for those following Wubbly's arguments.

Read it closely.

[pequaide]

I did some math so I might set it out.

You can indeed set both sides of the lever arm being discussed into two 200 cm radii arms with 2 kilograms on the end of each 200 cm side. In my calculation this would be equal to two 2cm lever arms on both sides with 200 kilograms on each side. So if the Atwood's with 1300 kg of extra mass is at 2 cm radius the acceleration would be 1300 / 1700 * 9.81 m/sec/sec. And those calculations were discussed.

Now lets do the calculation assuming that Wubbly is correct and moment of inertia is the determining factor and energy is conserved instead of momentum.

The initial energy is 1300 kilograms dropped one meter. These were the extra kilograms on the one side of the Atwood’s with zero mass on the other side. The 4 (2 each side of the wheel) kilograms at the end of 200 cm are balanced and they do not drop. 1300 kilograms dropped one meter is 9.81 N/kg *1300 kg * 1 m = 12,753 joules. This means that the 4 kilograms can not be moving over 78.6 m/sec: ½ * 4 kg * 78.6 m/sec * 78.6 m/sec = 12,356 joules with the 1300 kilogram moving .786 m/sec; ½ 1300 kg * .786 m/sec * .786 m/sec = 401.56 joules.

For a final velocity (at 2 cm) of .786 m/sec, after a one meter drop, the acceleration would be .308898 m/sec/sec. The accelerating mass over the accelerated mass would be proportional to .309 / 9.81. The mass doing the accelerating is 1300 kilograms; so 1300 / x = .309 / 9.81 and x or the accelerated mass would be equal to 41,271. 8 kilograms. The 1300 kilograms is part of that total accelerated mass of 41,271.8 kilograms. So the four kilograms at the end of 200 cm is acting as if it was 40,000 kilograms at 2 cm. This is consistent with Wubbly's concept that the difficulty of rotating 2 kilograms at 200 cm is 100 times greater than rotating the 200 kilograms at 2 cm. It is consistent but is it logical and more importantly is it experiential provable.

Wubbly you are predicting that 4 kilogram, by simple putting it on the end of a two meter arm, can be made to act like a semi tractor with a fully loaded trailer attached. And you have no experimental evidence.

I have done a multitude of equal or similar experiments and the results are dead on with the 400 kilogram expectations. My expectations are based upon F = ma and momentum conservation. So I am totally convinced that you are wrong, and this is your best argument for not being able to make energy from gravity.

Most of my experiments fall within a 5% error range, when you are expecting 400, the difference between 400 and 40,000 could not be missed.

[Tarsier79]

Peq, if you have a weight on the end of a massless rod of a certain lenght(feel free to use your above example), and you calculate the two different speeds of the two masses and the KE of each mass, it will tell you how much effort needs to be put in to accelerate both masses. You can calculate how much KE a mass has linearly moving at that same speed, and it will give you the same figure as the method used for calculating rotational energy.

Both methods confirm eachother, so do you disagree with the method for calculating linear KE? * note the rotational energy formula contains inertia which is mr^2.

Rotational inertia is how hard one is to accelerate compared to the other with applied torque, IE force x radius. The lighter mass at a distance will have greater KE when moving at the same rpm, and therefore takes more force to accelerate.

[pequaide]

Yes I agree with the linear formula for KE but you are assuming that KE is a conserved quantity and it is not. I have done the experiments and I know. Why don't you gentlemen sight some experiments for your theories.

Ya: force times radius is correct, but not force times radius times radius. A torque wrench works why not read one. it will show you that 2 at 200 is the same as 200 at 2.

[path_finder]

excerpt from this Web site: http://physics.kenyon.edu/EarlyApparatu ... dulum.html

[pequaide]

I looked up the experiment I did that proves your 40,000 kilogram theory is false.

It is in “energy producing experiments, pg 8, Wed Oct, 26 2011.

118.2 grams at the 17.82 diameter (8.91 radius) accelerates two Atwood's equally.

One Atwood's is 1180 grams (590 grams on each side) at the 17.82 inch diameter.

The other Atwood's was 1747 grams (873.5 grams on each side) at the 12.05 inch diameter.

1180 g * 17.82 in. = 21027 this is light one like 2 kg at 200 cm

1747 g * 12.05 in. = 21051 and this is the heavy one like 200 kg at 2 cm, less extreme of course

The times for the same distance dropped were, .0630 and .0627 and are within experimental error.

This is less than 1% error from the 400 kilogram theory. coincidence no doubt to be this close.

This would be exactly like doing the experiment with Atwood's of 400 kilograms at 4cm, and 4 kilograms at 400 cm. The same force at the same radius over the same drop distance (for time) accelerates both Atwood's at the same rotational rate. One Atwood's is just as easily moved as the other.

And of course the energy produced is different.

[Tarsier79]

Peq, I already did those experiments with my atwoods pulley, and filmed them in the MRR Experiment thread. The results of that test were IMO fairly conclusive and disagree with what you are trying to say.

[pequaide]

I have repeated the experiments on different machines and I trust my photo gate timers over your experiments. I usually can time events within 3 ten thousandths, pretty hard to miss the different between 400 and 40,000. I have done a whole lot more and I have a whole lot better equipment.

[Wubbly]

Wubbly you are predicting that 4 kilogram, by simple putting it on the end of a two meter arm, can be made to act like a semi tractor with a fully loaded trailer attached. And you have no experimental evidence.

You don't need to attach the fully loaded trailer. However, if you did, you might get a little more momentum out of it.

...and this is your best argument for not being able to make energy from gravity.

Actually, it's an excellent explanation as to why your linear tangential momentum conservation is incorrect.

pequaide, People have looked into your beliefs and done experiments and have found nothing. We have no faith in your experimental abilities.

[pequaide]

I do.

And the trebists are doing things you say are impossible.

[daxwc]

Really, seems to me one hell of a lot of energy goes in.

Peq can you give us a site that measures exactly how total much energy goes in?