Atwoods Analysis

[Trevor Lyn Whatford]

Hi Pequaide, Wubbly,

Have you guys done the impact test I suggested yet to prove the theory?

screw two bolt in the bottom of the two heavy weights to keep them balanced then add a third bolt to one of them place a peace of high density foam under the falling side of the system and record the impact, then drop the same bolt from the same height onto the foam and compare the impressions, this well tell you if Atwood's works or not.

Regards Trevor

[pequaide]

I am not arguing that the kinetic energy is different for the different mass arraignments in this Atwood’s; the kinetic energy is the same. But should it be? How can you apply the same quantity of force for different quantities of time and get the same quantity of motion.

11.56 newtons applied for .326 sec give you 3.22 joules of motion. 11.43 newtons apples for .820 seconds give you 3.189 joules of motion. If force and time are fundamental quantities then energy apparently is not.

[Wubbly]

pequaide wrote

... Angular momentum is Kepler’s and it was used for comets and other satellites that were under huge gravitational acceleration (which doesn’t occur in the lab, ...

Interesting thought. I'm not sure I agree with it though. The proverbial ice skater has angular momentum and need not be orbiting a planet to show that pulling the arms in and out affects the angular velocity and conserves angular momentum. IMO F=ma can be applied to a rotating system as well as a linear system. My physics books seems to think so too. One of these days I'm going to figure out rotational motion. Bear with me as I'm still working on it.

Further; objects moving in a straight line (as are the masses in an Atwood’s) have no radius of rotation. This means they can’t even have angular momentum. Straight line means no arc, no arc no radius.

Angular momentum can apply to objects moving in a circle, and also apply to objects moving in a straight line.

Attached is a page from a physics book showing angular momentum of a particle moving in a straight line. It looks a lot like the path of each mass in an Atwoods machine.

[Wubbly]

pequaide wrote

How can you apply the same quantity of force for different quantities of time and get the same quantity of motion.

because force is mass times acceleration. The mass changes and the acceleration changes, but the resultant quantity can stay the same. It also depends on your definition of 'quantity of motion' and whether the quantity of motion is determined by force x time or by something else.

11.56 newtons applied for .326 sec give you 3.22 joules of motion. 11.43 newtons apples for .820 seconds give you 3.189 joules of motion. If force and time are fundamental quantities then energy apparently is not.

This is from line 1 and line 6 of the spreadsheet. We must see the world from completely different perspectives because I'm not sure what you're trying to say here. Newtons is [kg m/s^2]. So Newtons x time is [kg m/s^2] x [s] which yields [kg m/s], which is the units of momentum not the units of joules [kg m^2/s^2]. So your statement that 11.56 N applied for .326 s gives you 3.22 joules of motion doesn't make sense to me. The force x time is not producing the energy. The units don't work out correctly even though they are on the same line in the spreadsheet. What was measured was energy, so the conclusion I would come to would be that force x time is not an accurate predictor of energy.

Trevor, I dismantled the apparatus months ago but still think it was an interesting experiment. No further Atwoods experiments on my end.

[pequaide]

In the formula you show for angular momentum (for objects not moving in a circle); you can set the point of origin as close or as far from the straight line motion as you choose. In other words; it is pick a number any number.

It is true that the distance between two straight lines is a constant distance (K). Whoopee; so what. So the formula is saying that linear momentum mv multiplied by any constant K is angular momentum. That would render angular momentum mathematically meaningless.

If Kepler would have proposed this formula he would have been laughed to scorn. It is amazing the professors of this day can get away such totally nonsense formulas.

[Wubbly]

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You are too funny. You can choose not to believe in an accepted equation. That's your choice. It's pick a number, any number because by definition, angular momentum is referenced to a point. You can define any point you want. If you pick a different point, you will get a different value for the angular momentum from the new point. It doesn't make the values from a different point wrong; they are just referenced from the different point. The trick is to pick the best point that is meaningful to your experiment.
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[Wubbly]

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After further study, I discovered that the linear momentum numbers should be added. The results of the experiment are still the same. In an Atwoods, for a fixed mass difference between the two sides, an increase in system mass, (resulting in an increase in linear momentum), does not equate to an increase in energy.
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[Wubbly]

Momentum is defined as Mass x Velocity.
Energy is defined as 0.5 x Mass x Velocity Squared.
They are both simply mathematical equations. If an object has Mass and Velocity, it has both Momentum and Energy. Neither can be denied. However, it should be obvious that it is mathematically impossible to say that Both are conserved quantities - because the equations are not the same, ...

In an elastic collision, mathematically, both are conserved quantities. But if you believe momentum and energy are just scalar numbers that you can manipulate, then you would be confused into thinking that both equations can not be conserved. In reality, energy is a scalar number and momentum is not a scalar (it is a vector), so in elastic collisions the equations are both conserved at the same time.
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[Wubbly]

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Here's an updated spreadsheet where the total mass goes up by powers of ten. With a large system mass, huge amounts of linear momentum are generated with no increase in energy.
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[Wubbly]

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Here's an interesting example of a Hockey Table Experiment where the large mass moves horizontally and the small mass moves down. It has a fixed input energy (9.81N of potential energy) and the total system mass goes up by powers of ten. As the system mass increases, after a 1 meter drop, the linear momentum increases, but the total energy (potential energy plus kinetic energy) stays the same.

If you compare the spreadsheet numbers from the Hockey Table Experiment to the Atwoods Experiment in the post above, you will see that the total system mass numbers are identical, the acceleration numbers are identical, the energy numbers are identical, the total momentum numbers are identical... The Hockey Table Experiment is basically the same as the Atwoods Experiment.

To think that you could take all of the momentum of the large mass in the Hockey Table Experiment, feed it back into the smaller 1 kg mass, and get the 1 kg mass to raise hundreds of times higher than it started with would display a complete misunderstanding not only of momentum, but also a complete misunderstanding of the law of conservation of momentum.

You could, however, capture the energy of the large mass, and transfer that back into the smaller mass, but you would gain no extra height above and beyond what you started with.
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[nicbordeaux]

It is the same as an atwoods assuming the "linear" friction is not higher than that of an atwoods pulley, which you have anyway.

What happened to Greendoor ? 10/1 he's been abducted by MIBs . Why they would want to abduct a guy who believes that there is more energy in a 100 kg each side atwoods with a 0.1 kg mass added to one side, compared to a 1 kg each side w/ the same 0.1 kg on one side, experiment performed over the same drop height is anybody's guess ;-)

[pequaide]

Okay; lets see, Work is: the product of a force times the distance over which the force acts.

Energy is: the ability to do work.

Lets say we have a balanced wheel with a radius of 200 cm. We can place a point mass of 200 kilograms at 2 cm and 2 kilograms at 200 cm and the wheel remains balanced. If the 2 kilograms is moving 1 m/sec and the 200 kilograms is moving .01 m/sec the wheel is balanced and moving. Neither the 2 kg or the 200 kilogram mass exerts a force upon the other mass.

A 200 kilogram Atwood's being accelerated by .1 kilograms has an acceleration of .981 N /200.1 kg = .0049 m/sec/sec. After a drop of one meter the system will be moving .099 m/sec.

A 2 kilogram Atwood's being accelerated by .1 kilograms has an acceleration of .981 N /2.1 kg = .467 m/sec/sec. After a drop of one meter the system will be moving .966 m/sec.

In the wheel above the 2 kilograms is moving 100 times faster than the 200 kilograms, so if the 200 kilograms is moving .099 m/sec then the 2 kilograms must be moving 9.9 m/sec. But it is not, it is only moving .966 m/sec so the 200 kilograms is going to apply a force to the 2 kilograms over a distance so the the 2 kilograms comes up to speed.

So which of the two masses is able to do work?

Which of the two masses has more energy?

[Wubbly]

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You give an example of a balanced wheel with two masses with different tangential velocities and different kinetic energies, but they are both travelling at the same angular velocity. Then you give an example of an unbalanced Atwoods with .981 Joules of energy. Then you give another example of an unbalanced Atwoods with .981 Joules of energy. Then you use some kind of wierd pequaidian logic where a large mass has to bring the smaller mass up to speed, even though in the first example both masses are happily travelling at the exact same angular velocity, and in the Atwoods examples they follow their appropriate equations of acceleration.

pequaide, Sometimes you make absolutely no sense. The discrepency you indicate does not exist. The Atwoods are linear systems following their equations, and the balanced wheel (with rim masses attached) is a rotational system with a different set of equations.

Did you miss the significance of the Hockey Table experiment?
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[pequaide]

The .1 kilogram dropped one meter gives you 200 kilograms moving .099 m/sec or it can give you 2 kilograms moving .966 m/sec, and yes this is the same in put energy.

But the 200 kilograms and the 2 kilograms moving at these speeds can't be placed on the same balanced wheel without one accelerating the other. And the 200 will accelerate the two. I do it all the time.

[Wubbly]

The .1 kilogram dropped one meter gives you 200 kilograms moving .099 m/sec or it can give you 2 kilograms moving .966 m/sec, and yes this is the same in put energy.

Yes, they do have the same input energy. And in this thread, the experiments transferred the input potential energy into kinetic energy, and then the kinetic energy was transferred into mechanical potential energy in a spring. The spring length was the same in the experiments (despite a huge increase in momentum). This proves the output energy for the various experiments is the same, despite a huge increase in the momentum of the heavier mass experiments.

But the 200 kilograms and the 2 kilograms moving at these speeds can't be placed on the same balanced wheel without one accelerating the other. And the 200 will accelerate the two. I do it all the time.

What are you talking about? Is this something other than the Atwoods? Two posts ago you said (200 kg @ 2 cm) balances (2 kg @ 200 cm), now you say they are not balanced. Make up your mind.