# Energy of Falling chain

Can someone explain this solution for the motion of a falling chain?

My Question is based on the above mentioned question on PSE. Suppose we have a chain attached on one end, while the other end is let go from the same height.

Now suppose we take a part of the wire, just above the lower bend, but on the right arm. Now once we let the chain fall, this part of the wire gains some velocity, so there is an increase in K.E. But once it goes on to the left arm, it comes to rest. SO the K.E turns back to 0. But by that time, it has dropped in height, so has lost potential.

So, where is this potential energy going?

If it is lost due to inelastic collisions with the neighbouring part of the wire, how can we still apply energy conservation as done in the above linked problem. Near the bend, the rope looses velocity almost instantly,so an infinite impulse must be required which would produce almost infinite heat. But, instead they have just averaged the value to (0+v)/2

I read one paper, on bunjee jumping physics, where they have used energy conservation too. What point am I missing over here?

The energy conservation analysis in this problem is, at best, an approximation. But there are some reasons why this approximation holds good.

First, to model the kind of chain the question is talking about, imagine a line of point masses connected to each other by massless, rigid line segments (or sticks). These sticks can freely rotate about the joints giving the chain flexibility. But since all the links are rigid, the chain is incompressible. Its length is fixed.

Now, if you drop such a chain in the manner given in the question, the kinetic energy of the links that come to rest after fall is not dissipated. It gets transferred to the rest of the chain below it which 'can' move. So, every link that comes to rest, transfers its share of kinetic energy to the links below it. Hence, the total energy is conserved.

You can see this more clearly in following manner. What happens when you let a rigid pendulum oscillate? Ideally, its energy does not dissipates. Now, add another rigid pendulum to the lower edge of the first. What we get is a compound pendulum. The energy again does not dissipates in this system, but keeps on being transferred from one pendulum to the other.

The chain in the question is nothing but a very large set of rigid compound pendulums.

Now, the reason I said that this analysis is only an approximation is- the chains that are assumed to come to complete rest in the question would in reality oscillate along horizontal direction about their fixed position like a plucked string in guitar, hence retaining a small part of their kinetic energy which they were supposed to transfer down. So, this residual part of the total energy will become a good source of error.

The potential energy is what gets converted to Kinetic Energy as the chain falls, As it gets abruptly to rest, as ja72 said in his answer, it flips, and folds itself raising to a height of $x=0$, and falls back again and raises again, this process continues, but in reality some of the energy is lost due to inelastic conditions, so it wont sustain its flipping motion.

You can apply the Energy conservation as we have assumed it to be perfectly elastic, and since we have assumed it to be elastic no heat is lost too.

If the end moves at speed $v$ the center of mass moves with speed $\frac{v}{2}$

• I do not get what you mean by flips and folds itself. Can you provide a diagram to show how the string behaves. – Apiastos Dec 16 '15 at 7:49
• You could try the experiment yourself, take an example of bungee jumping, notice how it flips back after reaching the lowest point – Oswald Dec 16 '15 at 8:11
• Still it doesnt resolve my question. For a tiny part of the string near the bend, while the string starts falling for the first time, it looses K.E as it passes to the left arm and it looses P.E too. Overall there is a net decrease in P.E without increase in K.E. – Apiastos Dec 17 '15 at 7:05
• @Avik it doesn't loose its P.E, even in the tiny part of the string P.E is what gets converted to K.E while it falls, when it can no more fall in the vertical direction it flips or moves in an arc to the left – Oswald Dec 17 '15 at 10:23
• But, then there should be a flipping motion as soon as the chain is dropped, which is not obderved. – Apiastos Dec 18 '15 at 19:51

Perhaps to correctly model this you need to either make the whole chain elastic, or to put a spring at location A holding the chain up. Once you do that, it becomes obvious what's going on.

• But,adding a spring is not essential, and I want to know what happens to the energy in this case. – Apiastos Dec 16 '15 at 7:57

Considering the chain to be elastic.

As the chain falls down, the lower end of chain has K.E and it comes to rest and the K.E gets converted to potential energy as the chain[left part] will get stretched. this will continue till whole chain falls and after that the restoring force will again start pulling the chain and it will again attain its original position. Thus this process will go on continuing. so actually you will find that the center of mass of the chain is in simple harmonic motion.

Coming to your question, P.E is not lost. First, all of the P.E is due to its height and after falling the chain still has P.E but now it is due to its stretching.

• Was the question on kind of elastic chain? "Uniform mass density" doesnt make speculations about stretching, tension and potential energy any relevant. – jaromrax Feb 21 '17 at 15:25
• The falling chain will extend the left part and produce strain in it that it was i meant stretching and hence store P.E. – ATHARVA Feb 21 '17 at 16:15

The energy will appear as one of the following option,

1. The rope will swing side by side, or

2. Something will stop the rope from swinging (maybe a hand) so the energy will be transferred to this object.