Atwoods Analysis

[Wubbly]

An Atwoods machine consists of 2 masses suspended by a pully.
By varying the masses m1 and m2, and by varying the mass difference between m1 and m2, you get varying accelerations.

wikipedia reference: http://en.wikipedia.org/wiki/Atwood_machine
website reference: http://hyperphysics.phy-astr.gsu.edu/hbase/Atwd.html
website where you can play with mass numbers and view the acceleration of the masses. https://www.msu.edu/~brechtjo/physics/a ... twood.html

Formulas for an Atwoods machine:

Acceleration of an Atwoods: a = g*(m2 -m1)/(m2+m1) => from atwoods website
Force in Atwoods: F=(m2-m1)g => from atwoods website
Velocity: v = sqrt(2*a*d) => standard physics formula
Time: t = sqrt(2*d/a) => standard physics formula
Momentum: p = mass * velocity (note: mass = m1 + m1)
Momentum: p = force * time
Kinetic Energy: KE = 0.5 * m * v^2 (note mass = m1 + m2)

Put these formulas into a spreadsheet to analyze.

Analyze the Atwoods at a given height (e.g. 1 m), since height is a limiting factor, but you supposedly have unlimited amounts of time.

The spreadsheet would add 1 kg to each side of the atwoods for each new row, so the difference between the two masses would always be constant.

Graph momentum as a function mass, and graph KE as a function of mass.

What would be the results?

[Wubbly]

Attached is a spreadsheet analyzing an Atwoods at a specific drop, and with a constant difference between the two masses.

Observations for the above conditions:

As you increase the system mass (but keep the difference in masses constant):
the acceleration decreases,
the velocity decreases,
the time increases,
the momentum increases,
the Kinetic Energy remains constant.

The equation for force in an Atwoods is the difference in the masses times g. If the mass difference on both sides of the Atwoods remains constant between iterations, the force in the system remains constant.

Note: Momentum was calculated twice, (once as mass x velocity and once as force x time). Both values were the same.

The main observation from this spreadsheet is that for a given drop D, if the mass difference between both sides of the Atwoods stays the same, as the system mass increases, the momentum increases, but the kinetic energy does not change.

[Wubbly]

What if you built an experiment that measured what it took to stop the system at a given height. Add the same amount of mass to each side of the Atwoods and repeat the experiment.

As you kept adding mass to each side, would your measurement show an increasing value, or a constant value?

If your measurement showed an increase with system mass increase, then momentum would be showing up in your measurement.

If your measurement stayed relatively constant with system mass increase, then energy would be showing up in your measurement.

[Fletcher]

Thankyou for taking the time wubbly - there ought to be some interesting comments from some members to show the great big hole in the argument they can drive a bus thru ! - that is, if one exists in physical reality.

[pequaide]

The Atwood’s can be stopped with the same amount of force applied for the same amount of time as it took to cause the motion. It must of course be applied to the opposite side; or in the opposite direction. If you switch the overbalanced mass to the other side it should rise as far as it was dropped.

I have been doing more massive motion transfers; 13 units of mass to one unit of mass. That would be an Atwood’s with 6 units of mass on each side and one unit of overbalanced mass. This would be an acceleration of .7546 m/sec², and after a one meter drop a velocity of 1.228 m/sec, and a momentum of 15.97 units. After the overbalanced mass receives all the motion of the Atwood’s it could rise 13 meters. The overbalanced mass was only dropped one meter.

We are throwing masses now and frankly it is a little on the dangerous side.

I don’t know what Fletcher is talking about. But here is your real hole. How can you double the time (time) over which a real quantity (force) acts and get the same results (Kinetic energy)? Shouldn’t doubling the time mean double the quantity (momentum)? In Newton’s world it does. But then maybe Fletcher doesn’t think Newton’s world is physical reality.

[Wubbly]

The purpose of the spreadsheet is to show multiple systems with identical kinetic energies, but increasing amounts of momentum.

The second website reference states the force in an Atwoods is (m2-m1)g.
So if the mass difference between each experiment is the same, the force will be the same.
If you look at line 25 of the spreadsheet, system mass = 1 kg, and the time to drop is .452 seconds.
If you look at line 31 of the spreadsheet, system mass = 13 kg, and the time to drop is 1.628 seconds.
The time increased by 260%, the force stayed constant, and the momentum increased from 4.43 to 15.97, (a 260% increase), but the KE remained constant (a 0% increase).

So if I stopped the atwoods with a spring, are you saying that it would be 260% harder to stop the system described in line 31 than to stop the system described in line 25?

Are you saying that momentum would be the measure of the work required to stop the system, rather than the kinetic energy being a measure of the amount of work required to stop the system?

I believe kinetic energy represents the energy in the system, and it would take the same amount of effort to stop each system in the spreadsheet since they all contain equal amounts of kinetic energy.

How can you double the time (time) over which a real quantity (force) acts and get the same results (Kinetic energy)?

Force x Time has the units of momentum [kg m/s], not the units of Energy [J] or [kg m^2/s^2]. Momentum is not energy. Energy is not momentum.

Shouldn’t doubling the time mean double the quantity (momentum)?

Yes, the momentum will increase.

[pequaide]

Wubbly quote: But this is exactly what the formulas are predicting.

Formula not formulas. mv conservation does not predict 1/2mv² conservation they are two different things.

Wubbly quote: So if I stopped the atwoods with a spring, are you saying that it would be 260% harder to stop the system described in line 31 than to stop the system described in line 25?

I don’t use springs but; Yes: how could it be more obvious? You apply the same force for 3.6 times as long. Do you expect the same results?

One kilogram with 15.97 units of momentum can rise 13 meters, and it takes 1.62799 seconds. Same time and force as in line 31. You don’t have to lift the balanced portion (12 kg) of the Atwood’s, just the unbalanced portion.

[Wubbly]

This is interesting. I look at a set of numbers and see something completely obvious to me. You look at the identical set of numbers and see something completely different, yet obvious to you.

We apparently have different 'beliefs', yet the laws of physics don't care what either of us believe.

[nicbordeaux]

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[pequaide]

That is correct. But I went to the next level and determined which one of the two formulas is correct.

I don’t believe that if mv is conserved that 1/2mv² is also conserved, I think that is total nonsense. Obviously they are not mathematically equal and kinetic energy always needs imaginary friends (heat) to be conserved.

Just yesterday I stopped the spinning of a 1,858 gram steel disk with only 47 grams. I wrapped a single string around the disk several times with the 47 grams attached to its end. I spun the disk and released the 47 grams. The disk stopped spinning as the string unwrapped. Now the 47 grams mass has only half of the motion because the 47 g mass and the disk are rotating around their common center of mass. The disk is no longer spinning but it is still moving. You could totally stop the motion of the disk by using two strings with two 47 gram masses. I used 1,245 grams for the disk calculations because of the mass distribution; I think it is called the radius of gyration. This gives us a total mass of 1,245g + 47g + 47g = 1,339g; I called it 13 to 1 for margin of error. I have a ten to one (constructed of pipe) hanging in the out building, and another ten to one (constructed of pipe) in the indoor lab, so I am fairly certain I am using the correct calculations.

The point is: that the Atwood’s can throw all of its motion into the overbalanced mass and all the events are compliant with Newtonian Physics. This is the world of what works.

[pequaide]

My last post is responding to Wubbly; nicbordeaux posted in between.

[Wubbly]

pequaide, thankyou for your responses but we are doing two completely different experiments. I am just trying to understand the Atwoods and measure what it takes to stop an Atwoods. You have gone beyond that and I belive you use an Atwoods machine as an input into your secondary experiment which I believe you call "cylinder and spheres".

The Atwood’s can be stopped with the same amount of force applied for the same amount of time as it took to cause the motion.

If I stop the Atwoods machine with a spring, a spring has a force [kg m/s^2] x distance [m] relationship, which produces something with units of [kg m^2/s^2] or [J].
A spring stores mechanical potential energy which has units of [kg m^2/s^2] or [J].

A force [kg m/s^2] x time [s] relationship produces something with the units of [kg m/s], which has the same units of momentum.
Doesn't it bother you in the least that the units of your force x time relationship doesn't have the correct units of energy?

One kilogram with 15.97 units of momentum can rise 13 meters ...

I totally disagree with this statement. I see you are in the momentum camp. To raise one kilogram requires work and energy expenditure, not momentum expenditure. If I raise one kilogram I am increasing its potential energy.
Again, doesn't it bother you that momentum doesn't have the correct units of energy?

[pequaide]

In order for one kilogram to have 15.97 units of momentum it must be moving 15.97 m/sec. It is the meters per second that is used to determine the rise. Rise (d) is d = ½ v²/a; if v = 15.97 m/sec and (a) is 9.81m/sec² then rise = 13.00 meters. The d = ½ v²/9.81 formula is from d = ½ at² because v = at or t = v/a and t² =v²/a².

Wubbly quote: Doesn't it bother you in the least that the units of your force x time relationship doesn't have the correct units of energy?

Pequaide: Maybe it does: maybe your definition is wrong.

I just wrapped a 4000gram center bearing wheel with a string with 66 grams on the end, the wheel nearly stopped or did stop as the 66 grams unwound. The original spinning motion of the wheel was slow but the motion of the 66 gram sphere was violent. I see nothing to suggest that Newton’s Three Laws of Motion do not cover this event.

[nicbordeaux]

Correct Pequaide, nothing in Newtonian physics or logic contradicts that. But I'd be extremely surprised if all the energy was being transferred. Quite a bit is lost to friction ? You want to measure maximum "impact" on the small weight (call it velocity or anything you want, no passa nada) ? : try use 10 kg breaking string dacron (no elasticity, very little knot reduction in breaking strain). If it busts and the weight sails into orbit, maybe you need 15. if it doesn't break, use 5...

You have life insurance ? You need a helmet ? Window breakage insurance ? Gimme a call, I can arrange it cheap :)

[Ben]

Nick wrote,
You have life insurance ? You need a helmet ? Window breakage insurance ? Gimme a call, I can arrange it cheap :)

I'm sorry about the way I handled my interaction with Nick, but some people are So annoying. And is this the right venue to sell insurance? (It's ok, he won't see this because I'm on his ignore list.)