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, THERES 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, THERES 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 accleration 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.

 

[arildno]

End conclusion is:
If gravity, as the sole force, remains a conservative force, then energy conservation is inevitable, whatever experiment you try to set up.
If you get anything else, you haven't calculated it correctly.

This is not an unproven dogma, but a fairly trivial mathematical result.

 

[vibgyor]

posted by dalespam
"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".

in this case when height is taken as 40 metres and chain length 40 m , the first link should hit the floor with velocity 20m/s.


but when the chain is much longer every link has to hit the floor with a velocity of 28.28 m/s. the force of gravity acting on 40 metres of chain that is vertically running between the floor and the plane as to be balanced at some point and at this point there will be no further acceleration.rgt.? now gravity acting on (g =10m/^2) 40 metres of chain balances out when 28.2 metres chain hit the floor per sec. (since v = 28.2)

 

[Dale]

I am not convinced that 28.2 m/s is correct. That would be correct if you broke the links and dropped them individually. In other words it would apply if the links were non-interacting. However, what makes the problem interesting is the fact that the links are interacting. You need to treat the chain as a whole, not just analyze an individual link.

The problem needs to be analyzed in more depth. First, you need to consider the 3 "pools" of energy I mentioned above. Second, you need to consider 3 portions of chain (chain coiled on top, chain falling, chain that hit the ground). Third, you need to consider 3 stages of time, no chain on the ground, chain on ground and on top, no chain on top.

 

[russ_watters]

This is the classic crackpot conundrum: devise a scenario that is complicated enough that you can't analyze it correctly, then assert that the result you got proves a scientific principle/theory wrong. Sorry, but the reality is that all you did was prove you made a mistake. Conservation of energy is not wrong.

You guys need to do two things:

1. Listen to people who know what they are talking about. That is why you are here, isn't it?
2. Simplify your scenarios so that you can properly learn the scientific principles you are investigating.

 

[DaveC426913]

I'll expound upon Russ' statement.

If this setup were to violate CoE, then that means some more granular part of it violates CoE. That part could be isolated and studied simply and independently. Come back when you've found the isolated component that seems to violate CoE, and we can study that. Because unless you can isolate it, you can't rule out the possibility that you've simply made a mistake in your analysis.

Really, your question is more properly "where are we failing in our analysis of this setup?"

 

[vibgyor]

i revived this closed thread yesterday becoz whe i chanced upon tthis i thought i could take the thread starter's place and post my doubts. hope the experimental set up is clear ad iwant you to tell me if the exp is perfomed ehat would the velocity of links could be and how they all add up so that LoC holds true. i have expressed my doubts in the previous post .

 

[Doc Al]

No, your scenario is not clearly defined. Is the chained stretched out on the table (as in the OP's problem) or bunched up at the edge? That makes a big difference! (In the first case all parts of the chain will move at the same speed--at least until the chain hits the floor. But not so for the second case--here only the hanging part is moving. Mechanical energy is not conserved in that second case--some of it is converted to thermal energy as each chain link is jerked from rest to the speed of the falling chain.)

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...

As has been pointed out several times already, this calculation treats each link as if it were in free fall, which is not true regardless of which scenario you are talking about.

Here's what you should do if you would like to continue this discussion: Define your scenario clearly and present your calculations. Your initial attempt was shown to be incorrect, so it's up to you to try again.

 

[vibgyor]

i described the scenario as a bunched up chain or lace at the edge in my opening post.

"
Mechanical energy is not conserved in that second case--some of it is converted to thermal energy as each chain link is jerked from rest to the speed of the falling . ""

i thought about that too and in that case I am not sure how much of mechanical
energy gets converted to thermal energy.

here, the force of gravity acting on 40 metres of chain or lace that is vertically running between the floor and the plane as to be balanced at some point and at this point there will be no further acceleration.rgt.? now gravity acting on (g =10m/^2) 40 metres of chain balances out when x metres chain hit the floor per sec. (and v = xm/s). and how to arrive at the value of x which is the velocity(can i call it terminal..?) of the chain .

 

[Dale]

Are you certain that there even is a "terminal" or "steady-state" velocity? You certainly have not shown that yet. And the one calculation that you have shown is clearly wrong since the equation doesn't apply for your situation.

Since this is your problem you can certainly make your chain into a "perfectly flexible rope" and "neglect friction and heat". However, even doing all of that will not make the problem very easy to solve. You will have to do some calculus, perhaps even some differential equations, to handle the KE lost due to the collision with the earth.

If I get bored sometime I may try to work this problem just for kicks, but don't hold your breath.

 

[Dale]

I had to babysit tonite, so I got bored. So here is my problem statement:

A length L of perfectly flexible rope is coiled at the edge of a cliff of height H (H<L). One end of the rope begins to fall off the edge under gravity, with the rest of the rope smoothly uncoiling as each coil is pulled down in turn. Determine the speed of the rope as it falls from the time that the leading end of the rope hits the ground to the time that the trailing end of the rope leaves the cliff. Neglect friction.

First, I parameterized the system with a parameter y which represents the amount of rope that has already hit the ground. The parameter y ranges from 0 to L-H. I then wrote the conservation of energy expression in terms of y:

g H L \rho =\frac{1}{2} H \rho v(y)^2+g H \left(-\frac{H}{2}+L-y\right)<br /> \rho +\frac{1}{2} \rho \int_0^y v(\psi )^2 \, d\psi

Where \rho is the linear density of the rope, and v is the velocity of the rope as a function of y. The left hand side is the initial potential energy before the rope begins to fall. The first term on the right hand side is the kinetic energy in the falling rope, the second term is the gravitational potential energy, the third term is the (plastic) collision energy as the rope hits the ground.

Setting y=0 and solving for v(0) we obtain:
v(0)=\sqrt{g H}

Differentiating both sides wrt y we obtain the following differential equation:
0=\frac{1}{2} \rho v(y)^2+H \rho v&#039;(y) v(y)-g H \rho

Solving this differential equation with the above initial condition we obtain:
v(y)=\sqrt{2 g H-e^{-\frac{y}{H}} g H}

Thus there is no constant velocity although the velocity exponentially approaches:
v(\infty )=\sqrt{2} \sqrt{g H}=\sqrt{2} v(0)

Bottom line, there is no problem with conservation of energy here, only a problem with misapplying a formula and making several assumptions about the results without doing the math. Conservation of energy has been proved in general, as mentioned by arildno and others, but it is not terribly difficult to come up with a complicated scenario with so many variables and details that even the best "problem solver" might work it incorrectly. Whenever you find that some complicated scenario seems to violate some known physical law you can be sure that "the devil is in the details". There is always some ambiguity in the problem set up or some error in the work or assumptions. If you really want to demonstrate a violation of a physical law look for a very simple and clear example where the violation cannot possibly be due to any other factor.

 

[vector3]

I can't believe I read this whole thread...

I'm not leaving however, without posting something...

I once heard that if you put a scale in an elevator and stood on it while the elevator was free falling the scale would read zero. I'm suggesting that the weight of the chain above and the Plane A equals the weight of chain in free fall below the plane regardless of how much chain is above or below plane A. In other words, the chain does not weight anything above the plane or below the plane... until it hits the ground...?

Personally I believe links enter the gravitational field with the current velocity that the first link is at. I assume the chain will not break if the links are quickly accelerated and that there is a slight tension between the links in order to pull a new link into the field.

Any given link would then have it's normal sum of potential energy plus it's kinetic energy. I guess I don't see the significance that plane A has on the problem? Except to make it more clear that that mgh + 1/2mv^2 is the total energy.

The problem starts with no gravitational field. If no g then then no mGh. Also no initial 1/2 m v^2.

 

[stewartcs]

I can't believe I read this whole thread either.

I can't believe this thread hasn't been moved https://www.physicsforums.com/forumdisplay.php?f=5" yet either.

 

[sadhu]

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.

what struck me was the word "heap of links"

i think that the chain was kept on something like a table

in respect to it the statement i have quoted is correct only if we assume that no internal frictional forces are acting within the chain.

the very first link will pull the next and it willdo the same to next ..

hence at any instant the force is conveyed to all links and they start moving
but assume that somehow only some links start moving ..
then there must be some force preventing others to move ..
the force must also act on links just starting to move and must do some negtive work..

when this negative work is added to net K.E
i am sure it equalise the net potential energy

 

[sadhu]

I had to babysit tonite, so I got bored. So here is my problem statement:

A length L of perfectly flexible rope is coiled at the edge of a cliff of height H (H<L). One end of the rope begins to fall off the edge under gravity, with the rest of the rope smoothly uncoiling as each coil is pulled down in turn. Determine the speed of the rope as it falls from the time that the leading end of the rope hits the ground to the time that the trailing end of the rope leaves the cliff. Neglect friction.

First, I parameterized the system with a parameter y which represents the amount of rope that has already hit the ground. The parameter y ranges from 0 to L-H. I then wrote the conservation of energy expression in terms of y:

g H L \rho =\frac{1}{2} H \rho v(y)^2+g H \left(-\frac{H}{2}+L-y\right)<br /> \rho +\frac{1}{2} \rho \int_0^y v(\psi )^2 \, d\psi

Where \rho is the linear density of the rope, and v is the velocity of the rope as a function of y. The left hand side is the initial potential energy before the rope begins to fall. The first term on the right hand side is the kinetic energy in the falling rope, the second term is the gravitational potential energy, the third term is the (plastic) collision energy as the rope hits the ground.

Setting y=0 and solving for v(0) we obtain:
v(0)=\sqrt{g H}

Differentiating both sides wrt y we obtain the following differential equation:
0=\frac{1}{2} \rho v(y)^2+H \rho v&#039;(y) v(y)-g H \rho

Solving this differential equation with the above initial condition we obtain:
v(y)=\sqrt{2 g H-e^{-\frac{y}{H}} g H}

Thus there is no constant velocity although the velocity exponentially approaches:
v(\infty )=\sqrt{2} \sqrt{g H}=\sqrt{2} v(0)

Bottom line, there is no problem with conservation of energy here, only a problem with misapplying a formula and making several assumptions about the results without doing the math. Conservation of energy has been proved in general, as mentioned by arildno and others, but it is not terribly difficult to come up with a complicated scenario with so many variables and details that even the best "problem solver" might work it incorrectly. Whenever you find that some complicated scenario seems to violate some known physical law you can be sure that "the devil is in the details". There is always some ambiguity in the problem set up or some error in the work or assumptions. If you really want to demonstrate a violation of a physical law look for a very simple and clear example where the violation cannot possibly be due to any other factor.

well i can't understand your assumption.

to neglect friction and to uncoil the pile one by one

what i only figured out that you prooved that velocity cannot remain same , by taking law of con. of energy as true

 

[Dale]

well i can't understand your assumption.

to neglect friction and to uncoil the pile one by one

Sorry about that. I didn't know how to express the "smoothly uncoiling" assumption well. What I wanted to describe was that I was assuming that no energy was lost in uncoiling the rope, and that the velocity of all of the rope at the top of the cliff was zero (i.e. all the KE was in the falling rope).

 

[sadhu]

well yesterday i thought to derive a mathematical formulation to include frictional forces
wthin the chain so as to complete the proof

what i thought was this...

if we only consider friction forces then the force must reach up to that link where net friction due to the links before it equalise the net force due to gravity,but then i thought what when suppose the orientation of pile disturbes and friciton decreases
then also (according yo my assumption)then the force must reach up to that link where net friction due to the links before it equalise the net force due to gravity,but how come all links come to same velocity from rest instantaniously
more over whatever the velocity the rope was having in one orientation that will remain same for all orie. that's impossible

that factor which i left was "bends"

imagine a rope coiled in circle,now you began to uncoil it from outside,let you apply a constant force and let the rope do reduce it radius due to friction,

at any instant the force acting on the coiled rope depend upon the angle of force with respect to that uncoiling end,the tangetial stress ,etc
then only we can suppose that one end of rope can move while other do not..

this is just one case but imagine a rope spread randomly thus the above factors will act randomly and hence the accelration of rope will also be random ,but less then g.

if you consider a straight rope it has no bends hence the rope will move only if net friction is more that gravity at that instant and acceration keep on increasing
but whole rope move at once...

logically the terminal velocity is never reached ..if i am correct

well don't expect me to come up with a mathematical model to proove it

 

[Doc Al]

modeling a falling chain

Sorry about that. I didn't know how to express the "smoothly uncoiling" assumption well. What I wanted to describe was that I was assuming that no energy was lost in uncoiling the rope, and that the velocity of all of the rope at the top of the cliff was zero (i.e. all the KE was in the falling rope).

There are two simplified ways to model this problem of a chain falling over an edge:
(1) Have the chain layed out in a line (not coiled) so that the entire chain has one speed.
(2) Have the chain coiled so that only the "falling" section of chain has a velocity and the coiled portion has speed zero.

Using model #1, mechanical energy is conserved as the chain falls. (Ignoring the collision of chain with the ground.)

But with model #2, as each segment of chain goes over the edge it is jerked from a speed of zero to the speed of the falling segment. I don't see anyway of doing this without losing some mechanical energy to thermal energy--a point I made in post #37. (Each segment undergoes an inelastic collision.) So you cannot assume conservation of mechanical energy.

 

[Dale]

But with model #2, as each segment of chain goes over the edge it is jerked from a speed of zero to the speed of the falling segment. I don't see anyway of doing this without losing some mechanical energy to thermal energy--a point I made in post #37. (Each segment undergoes an inelastic collision.) So you cannot assume conservation of mechanical energy.

Of course you cannot physically uncoil a rope without losing some mechanical energy to thermal energy. Just as you cannot physically push a box up an incline without losing some mechanical energy to thermal energy, or a roller coaster cannot go without losing some thermal energy. We solve these kinds of problems all the time anyway. Just because it is physically impossible doesn't mean that you cannot get some useful insight from the solution.

 

[Himanshu]

But with model #2, as each segment of chain goes over the edge it is jerked from a speed of zero to the speed of the falling segment. I don't see anyway of doing this without losing some mechanical energy to thermal energy--a point I made in post #37. (Each segment undergoes an inelastic collision.) So you cannot assume conservation of mechanical energy.

Can't we assume the collision to be elastic?

 

[Doc Al]

the uncoiling rope

Of course you cannot physically uncoil a rope without losing some mechanical energy to thermal energy. Just as you cannot physically push a box up an incline without losing some mechanical energy to thermal energy, or a roller coaster cannot go without losing some thermal energy. We solve these kinds of problems all the time anyway. Just because it is physically impossible doesn't mean that you cannot get some useful insight from the solution.

It's a different kind of loss than simply loss due to friction. (I'm perfect OK with assuming a frictionless rope! ) And I think you'll get more useful--and more interesting--insight by trying to figure out the energy loss, rather than just by assuming energy conservation.

Can't we assume the collision to be elastic?

Well, no. Especially since the original theme of this thread was to question whether mechanical energy was conserved.

Let's crank it out. Let the amount of rope hanging over the edge be y; let the mass per unit length be \lambda. Applying Newton's 2nd law (in momentum form, since the hanging mass changes):

F = dp/dt = d(mv)/dt = (dm/dt)v + m(dv/dt)

\lambda y g = \lambda \dot{y}^2 + \lambda y\ddot{y}

The solution to this is:
y = \frac{gt^2}{6}

Thus:
\dot{y} = \frac{gt}{3}

After a length L of rope has slid off the table, the KE will be:
(1/2)mv^2 = \frac{1}{3}\lambda g L^2

But the decrease in PE is:
mg(L/2) = \frac{1}{2}\lambda g L^2

Note that the decrease in PE is greater than the resulting KE; the difference is the mechanical energy "lost". (Unless I made an error somewhere. )