Law of conservation of energy wrong?

[Lakshya]

Law of conservation of energy wrong??

a problem...pls check..
imagine there's a horizontal planeA bove which the effects of gravity are not felt. any object above the plane just floats and gravity has no influence on it.
now take a chain(for e.g:made of small steel ring) of length hundred metres above the planeA and arrange it so that its in a heap just above the planeA. but take care the chain is not entangled and should be easy to stretch along its length.

now pull the lower end of the thread below the planeA and the chain runs down like a single thread and hits the ground and as links pull down on the ones they are linked to , the whole length of the chain will run like a thread and hit the ground in some time. now take planeA is 20metres above the ground surface. the terminal velocity of links when they hit ground will be around 14 metres .
now consider hundred metres of the chain weighs 100 units. the potential energy of the chain just above the planeA will be m*g*h =100 *9.8*20
=19600 units.

but kinetic energy recovered would be 1/2 * m * v*v = 0.5 * 100* 14 *14 = 9800 units. so in this method only half the energy is recovered hence violating the law of conservation of energy.

SIMPLY PUT, there's DIFFERANCE IN ENERGY RECOVERED WHEN U TAKE 100 METRES OF CHAIN TO HEIGHT OF 10 METRES AND DROP IT AS WHOLE TOGETHER THAN WHEN YOU ARRANGE IT SO THAT IT RUNS DOWN LIKE A THREAD , EACH LINK PULLING DOWN ON THE OTHER LINK IT IS CONNECTED TO...


now what will the terminal velocity be when the planeA is above 40 metres...?
i guess its around 19.6m/s instead of 28m/s...

ur thoughts pls correct me if I am wrong...

 

[nealh149]

One of your first premises is that there is not gravitational field, but then you use gravitational potential energy to calculate energy.

 

[cyruskoras]

a problem...pls check..
imagine there's a horizontal planeA bove which the effects of gravity are not felt. any object above the plane just floats and gravity has no influence on it.
now take a chain(for e.g:made of small steel ring) of length hundred metres above the planeA and arrange it so that its in a heap just above the planeA. but take care the chain is not entangled and should be easy to stretch along its length.

now pull the lower end of the thread below the planeA and the chain runs down like a single thread and hits the ground and as links pull down on the ones they are linked to , the whole length of the chain will run like a thread and hit the ground in some time. now take planeA is 20metres above the ground surface. the terminal velocity of links when they hit ground will be around 14 metres .
now consider hundred metres of the chain weighs 100 units. the potential energy of the chain just above the planeA will be m*g*h =100 *9.8*20
=19600 units.

but kinetic energy recovered would be 1/2 * m * v*v = 0.5 * 100* 14 *14 = 9800 units. so in this method only half the energy is recovered hence violating the law of conservation of energy.

SIMPLY PUT, there's DIFFERANCE IN ENERGY RECOVERED WHEN U TAKE 100 METRES OF CHAIN TO HEIGHT OF 10 METRES AND DROP IT AS WHOLE TOGETHER THAN WHEN YOU ARRANGE IT SO THAT IT RUNS DOWN LIKE A THREAD , EACH LINK PULLING DOWN ON THE OTHER LINK IT IS CONNECTED TO...


now what will the terminal velocity be when the planeA is above 40 metres...?
i guess its around 19.6m/s instead of 28m/s...

ur thoughts pls correct me if I am wrong...

Here when you consider plane A you said it has no gravity , but you have used g = 9.8 I don't know how is it possible?

 

[Lakshya]

Here when you consider plane A you said it has no gravity , but you have used g = 9.8 I don't know how is it possible?

I said that on the plane no effect of gravity can be felt. This can be achieved by accelerating it upward with an acceleration of 9.8 m/s. Then it will have no effect of Earth's gravity.

 

[Doc Al]

imagine there's a horizontal planeA bove which the effects of gravity are not felt. any object above the plane just floats and gravity has no influence on it. now take a chain(for e.g:made of small steel ring) of length hundred metres above the planeA and arrange it so that its in a heap just above the planeA. but take care the chain is not entangled and should be easy to stretch along its length.

The stuff about "gravity has no influence" sounds mysterious, but I assume you just mean that the chain is resting on a frictionless horizontal plane.

now pull the lower end of the thread below the planeA and the chain runs down like a single thread and hits the ground and as links pull down on the ones they are linked to , the whole length of the chain will run like a thread and hit the ground in some time. now take planeA is 20metres above the ground surface. the terminal velocity of links when they hit ground will be around 14 metres .

Please show your calculation. What do you mean by terminal velocity? The speed of the chain when the first link hits the floor? When the last link hits the floor?

Realize that when the chain runs down like a thread, it loses energy as it hits the floor inelastically--kinetic energy is transformed into thermal energy. Could that be what you are concerned with?

 

[AlephZero]

... arrange it so that its in a heap just above the planeA. but take care the chain is not entangled and should be easy to stretch along its length.

No, that's logically impossible.

Part of the chain is moving at a finite speed and the other part is at rest. However you try to arrange it, when the next link starts moving it has to accelerate "instantaneously" from rest, to match the speed of the moving part.

It is impossible for moving part to accelerate at 9.8m/s, because part of its KE will used to accelerate the stationary part to the same speed. It will only accelerate at 9.8m/s when all the chain has left the table.

This has nothing to do with imagining a "real" chain with finite sized links - the same applies to an idealized flexible string.

 

[Idjot]

I said that on the plane no effect of gravity can be felt. This can be achieved by accelerating it upward with an acceleration of 9.8 m/s. Then it will have no effect of Earth's gravity.

Accelerating upward at 9.8 m/s would not cancel out the g's. It would double them. Free fall is the way to cancel g's.

This hypothetical experiment would be better conducted in a vacuum with the chain hanging from its top link. Then the chain would fall without bunching up and you wouldn't need the confusing anti-gravity plane to find the suspected difference from the bunched up chain. Of course there will be no difference other than that which might be measured if the bunched up chain is released at the wrong altitude required to compensate for the incremental altitudes of the links on the hanging chain. To make it easier - how about a twenty pound ball compared to twenty, one pound balls released one after the other? Same thing, right?

 

[Lakshya]

No, that's logically impossible.

Part of the chain is moving at a finite speed and the other part is at rest. However you try to arrange it, when the next link starts moving it has to accelerate "instantaneously" from rest, to match the speed of the moving part.

It is impossible for moving part to accelerate at 9.8m/s, because part of its KE will used to accelerate the stationary part to the same speed. It will only accelerate at 9.8m/s when all the chain has left the table.

This has nothing to do with imagining a "real" chain with finite sized links - the same applies to an idealized flexible string.

Please clear yourself. If you will perform the experiment mentioned by me you will get the law of conservation of energy wrong.

 

[Doc Al]

If you will perform the experiment mentioned by me you will get the law of conservation of energy wrong.

Please answer the questions I asked in post #5. (Show how you calculated terminal velocity.)

 

[txshender]

PE is not constant

Supposing that the distance to fall is 20 meters (it matters not that this chain is 100 meters long or 200 meters long the moment it passes through the "opening" from the plane) and that all of the chain drops together...

x=.5*a*t^2
-20=.5*9.8*t^2
t~= 2.02031 s

v(final)=a*t + v(initial)
v=9.8*2.02031 + 0
v~=19.8 m/s

KE=1/2 * m * v^2
KE=.5*100*19.8^2
KE=19602 ~= 19600 J which is the same as PE=mgh=100*9.8*20

However, this is done ideally and is only a free fall problem...

If I am not mistaken... F=ma... When just one link of that chain passes through the "hole," there will only be the force being exerted on that one chain by gravity, and that force will be transferred to the entire mass... Assuming one link of the chain weighed 1 kg, and the whole chain weighed 100 kg... 1*9.8= 100 * a... That means that the whole chain will come out of the plane eventually, and it will do so in a jerking accelerating (I don't know how else to describe non-uniform acceleration) manner until all chains are out of the plane... Another point is that energy will be used accelerating the other chains out of the other plane, so I do not believe it should be expected that total KE should equal PE (which is what you used in your first calculation, and the concept of using it here is wrong as I will explain later)...

Besides, what is your definition of KE in this case? Since each part of the chain will hit the ground with a different terminal velocity, is it correct to calculate KE as if it were one mass like you did? And what is the terminal velocity that you used? Did you use the last ring? Because if so, it would be expected to be somewhere near (not exactly, but close to) 19.8 m/s at the moment before it touches the ground and experiences an impulse.

I am sure somebody could model this with an equation... But it would be... very messy because you have to deal with the non-uniform acceleration (1/6 * j * t^3?), the mass outside of the plane is the only one experiencing force, and that force is applied to the entire mass of the object, which causes the acceleration at any moment to be completely governed by how much chain has already been let out... Run-ons for the lose...

But my final words are that potential energy is CONSTANTLY CHANGING and you will not be able to calculate it using m*g*h. Potential energy is merely defined as E=F*d with F being the weight force... The weight of that entire chain is constantly changing until the last chain leaves the plane...

Edit: Apologies for the lack of focus throughout this post

 

[Idjot]

To the OP:

I'm wondering if there is any possible scenario where you might prove this theory. Otherwise the calculations you're doing are entirely dependent upon the links of the chain magically entering Earth's gravity as if through a doorway one by one. Don't get me wrong. I have nothing against hypothetical props, even impossible props, if the theory itself can at least be proven somehow in a possible way.

But I reiterate:
The "working" theory here seems to be wholly dependent on an impossible prop. Please correct me if I'm wrong - anyone.

 

[DaveC426913]

"...imagine there's a horizontal planeA bove which the effects of gravity are not felt. any object above the plane just floats and gravity has no influence on it..."

The rest of the entry is merely gilding the lily.


Let me see if I can summarize:

"Imagine we can nullify the effects of gravity at will. Under these circumstances, we can violate the law of conservation of energy."


I would say YEP! With your imaginative premise, there's no reason why CoE couldn't be violated. But why bother with a gravity nullifier? Why not ghosts or tooth fairies?

 

[Gokul43201]

Actually, Dave, there's no reason that the OP's construction should result in energy violation. There's nothing wrong with requiring potentials that vary differently than 1/r. Even a discontinuous potential is fair game, and will not lead to non-conservation.

If you wish, replace the gravitational field with an electrostatic field created by a giant, parallel plate capacitor (field is uniform everywhere inside and zero everywhere outside) and make the chain be a charged insulator.

What is missing in this thread, however, is the calculation performed by the OP in determining the "terminal velocity". Unless we see this calculation, we can't say where it's wrong.

 

[jhat21]

I don't know how you got 14m/s for the velocity instead of 20m/s, but here is the calculation. it looks like a straightforward high school physics problem.

h = 1/2 g t^2 + v0 t
20m = 1/2 10m/s^2 t^2 + 0

t = sqrt( 2h/g )
t = sqrt( 2*20/10) = sqrt(4) = 2

vf= g t
vf= 10m/s^2 * 2 = 20m/s

1/2 m v^2 = mgh
1/2 v^2 = gh
v^2 = 2gh
so
20^2 = 2 * 10 * 20 = 400

mass is irrelevant and
your initial velocity is zero

i don't see any problem with energy conservation here, not today

 

[jhat21]

If you calculate for each link on the chain, you will get the same equality.

Now let's assume that the link is horizontally placed, so that mgh is the same for each link.
It appears that the potential energy is the same for all links, and it also appears that the later links will hit the ground with more velocity than the first as it will accelerate over a longer time. This seems like kinetic energy does not equal potential energy.
But this is wrong.
Because we cannot express the potential energy as mgh anymore! The potential energy is due to the force of gravity acting upon the link, F, over a distance, d, it is equal to work the integral of F*d. Therefore we do need to consider the length of chain that the link must traverse horizontally also! since it is being acted upon with the same force of gravity over this distance as the link that is falling.
You would then write mg(h+L) as the potential energy, where L is the horizontal distance of the link from the drop point.

 

[Lakshya]

If you calculate for each link on the chain, you will get the same equality.

Now let's assume that the link is horizontally placed, so that mgh is the same for each link.
It appears that the potential energy is the same for all links, and it also appears that the later links will hit the ground with more velocity than the first as it will accelerate over a longer time. This seems like kinetic energy does not equal potential energy.
But this is wrong.
Because we cannot express the potential energy as mgh anymore! The potential energy is due to the force of gravity acting upon the link, F, over a distance, d, it is equal to work the integral of F*d. Therefore we do need to consider the length of chain that the link must traverse horizontally also! since it is being acted upon with the same force of gravity over this distance as the link that is falling.
You would then write mg(h+L) as the potential energy, where L is the horizontal distance of the link from the drop point.

And why is it necessary to take the horizontal distance.

 

[russ_watters]

The explanation was clear about that: w=fd

 

[Doc Al]

Until Lakshya clarifies the scenario he has in mind--by explicitly showing his calculations--this thread is rather pointless.

As long as the stretched out chain doesn't hit the ground--thus introducing dissipative forces--its speed as it slides off the (frictionless) surface can be found at any point using conservation of energy. The chain loses PE and gains KE; it's as simple as that.

Having the chain stretched out, as opposed to being bunched up, simplifies things greatly. The chain moves as a unit with a single speed.

 

[Gokul43201]

Because we cannot express the potential energy as mgh anymore! The potential energy is due to the force of gravity acting upon the link, F, over a distance, d, it is equal to work the integral of F*d. Therefore we do need to consider the length of chain that the link must traverse horizontally also! since it is being acted upon with the same force of gravity over this distance as the link that is falling.
You would then write mg(h+L) as the potential energy, where L is the horizontal distance of the link from the drop point.

There are many things incorrect here and in the previous paragraph.

U = \vec{F} \cdot \vec{d} = Fd cos \theta

The horizontal distance, L, is traversed perpendicular to the gravitational force F. So the change in gravitational PE is 0 over that section.

Lakshya has had ample opportunity to show his/her calculations, and hasn't done so. I recommend the thread be locked.

 

[Lakshya]

Oh I made a mistake

I have found a mistake in my calculation when a scientist named Milo Wolff told me about that I ahve calculated the velocity wrong. It is (2gh)^1/2 and the value comes out 19.79 but still the energy is not conserved. The potential energy comes out to be 19600 units and kinetic energy comes out to be 19582 units.

 

[rcgldr]

Did you take into account that the end of the chain will be accelerating once it clears the edge of the plane?

 

[Doc Al]

I have found a mistake in my calculation when a scientist named Milo Wolff told me about that I ahve calculated the velocity wrong. It is (2gh)^1/2 and the value comes out 19.79

That's the speed of an object dropped from rest after falling 20 m. What does this have to do with your chain scenario? (A chain sliding off of a surface is hardly an object in free fall.)

but still the energy is not conserved. The potential energy comes out to be 19600 units and kinetic energy comes out to be 19582 units.

How did you calculate those values of PE and KE? What makes you think your calculations are accurate to 5 significant figures?

Last chance: Define your scenario clearly and show your calculations. Otherwise you're just wasting everyone's time.

 

[FedEx]

I said that on the plane no effect of gravity can be felt. This can be achieved by accelerating it upward with an acceleration of 9.8 m/s. Then it will have no effect of Earth's gravity.

But if you acclerate the system by 9.8m/s2 upwards then we will have to consider a noninertila frame and we will have to apply pseudo force in the downward direction. So the acceleration would than be equal to g+pseudo=19.6 downwards. And if you acclerate it by 9.8 in the downward direction than the psedo will act upwards. And in this case the total acceleration would be g-9.8=o. which will cause weightlessness.

 

[HallsofIvy]

Please clear yourself. If you will perform the experiment mentioned by me you will get the law of conservation of energy wrong.

"Clear yourself"? If you are just going to ignore what people say, why did you post this in the first place. Have you forgotten that your "experiment", assuming a plane above which gravity is not "felt", is purely imaginary and can't be performed?

Actually, you don't need the change in gravity part. Just having the chain lying on a plane is sufficient. This experiment has been done and showed that there was no error in the conservation of energy.

In any case, what people are telling you is correct. You can't have part of the chain accelerating downward and the rest not moving. As soon as part of the chain starts falling it will pull the chain still on the plane after it. You are not taking into account the kinetic energy of that part of the chain.

 

[Lakshya]

That's the speed of an object dropped from rest after falling 20 m. What does this have to do with your chain scenario? (A chain sliding off of a surface is hardly an object in free fall.)

How did you calculate those values of PE and KE? What makes you think your calculations are accurate to 5 significant figures?

Last chance: Define your scenario clearly and show your calculations. Otherwise you're just wasting everyone's time.

No no I was just calculating terminal velocity with (2gh)^1/2. The figure comes out to be 19.79 and if we calculate KE i.e. 1/2mv^2 = 19582 and PE = mgh = 19600. Now is that clear to you or you need some other explanation.

 

[Doc Al]

No no I was just calculating terminal velocity with (2gh)^1/2. The figure comes out to be 19.79 and if we calculate KE i.e. 1/2mv^2 = 19582 and PE = mgh = 19600. Now is that clear to you or you need some other explanation.

All you've demonstrated here is roundoff error in your calculations. Since you calculated terminal velocity as v = (2gh)^1/2, plug that directly into 1/2mv^2 and you'll get KE = mgh exactly.

What does this have to do with a chain sliding off of a surface?

 

[vibgyor]

hope this thread can be revived. act'y i posted this question in a group a few months agoand I am surprised to see it here.

yes it would be enough to describe the scenario as bunchedup(not stretched along) chain or lace lyin on frictionless plane at height, say 40 metres. now one end of the chain is made to slid over the edge(or go over a fictionless cylindrical bar just above the edgge of the plane) and hit the ground due to the pull of gravity. air resistance is zero.
now for the eqn mgh =1/2 m v^2 to hold good , every link has to hit the ground at exactly 28 .28 metres per second ( g = 10 m/s^2) and remain the same except the last 40 metres of chain...

 

[vibgyor]

should i repost this as a new thread to continue discussing this ..?

 

[Lakshya]

hope this thread can be revived. act'y i posted this question in a group a few months agoand I am surprised to see it here.

yes it would be enough to describe the scenario as bunchedup(not stretched along) chain or lace lyin on frictionless plane at height, say 40 metres. now one end of the chain is made to slid over the edge(or go over a fictionless cylindrical bar just above the edgge of the plane) and hit the ground due to the pull of gravity. air resistance is zero.
now for the eqn mgh =1/2 m v^2 to hold good , every link has to hit the ground at exactly 28 .28 metres per second ( g = 10 m/s^2) and remain the same except the last 40 metres of chain...

Look Vibgyor, if it's really made by you then see I have corrected our velocity taken. You had taken 14m/s and acually it's 19.79m/s. It's no use to start a new thread because this thread is already updated and everybody can see it.

 

[Dale]

There is no problem with the OP's hypothetical "gravitational" field. As Gokul mentioned we can make conservative electric fields with similar configurations. Nor is there a problem with the conservation of energy, but there are a few problems anyway.

The most important problem from a conservation of energy standpoint is that your chain is longer than your field. Thus you never have a situation where the chain's potential energy is entirely converted to kinetic energy. In other words, a given link of chain has a certain KE the instant before it strikes the ground and no PE, but the instant after it strikes the ground it has no KE and no PE. The energy has been lost from the chain in the plastic collision with the earth.

If you really want a "clean" conservation of energy problem you should make the field at least as high as the chain is long. That way you will have a situation where the PE is merely exchanged for KE. Stop the analysis as soon as the first link hits the ground.

Now, if you make the field exactly as high as the chain then it is easy to see that the KE when the chain is fully stretched out is exactly 1/2 of the initial PE since the center of mass of the chain has fallen 1/2 of the distance.

This problem should NOT be approached using any of the constant acceleration equations that I have seen throughout this thread. The acceleration is not constant so none of those equations are applicable. It should be approached entirely via conservation of energy principles. Otherwise you will need to do a much more detailed and difficult derivation of the equations of motion.