Chain Hangs Over a Pulley and Starts Moving

[Gundyam]

Homework Statement

A chain hangs over a pulley. Part of it rests on a table, and another part rests on the floor. When released, the chain begins to move and soon reaches a certain constant speed v. Can we find the height h of the table?

I think this question need some tricks. I've tried some general ways to do it but it can't.

Relevant Equations

v, g is known
h = ?

Homework Statement:
A chain hangs over a pulley. Part of it rests on a table, and another part rests on the floor. When released, the chain begins to move and soon reaches a certain constant speed v. Can we find the height h of the table?

I think this question need some tricks. I've tried some general ways to do it but it can't.
Homework Equations:
v, g is known
h = ?

 

[BvU]

Hello Gundyam, !

Homework Equations: v, h = ?
Can we find the height h of the table?

Not without equations. So you have to find a few more ...
What is needed for a constant velocity ?

Please read the guidelines. We can't help without a posted attempt from your side ...

 

[Gundyam]

Hello Gundyam, !Not without equations. So you have to find a few more ...
What is needed for a constant velocity ?

Please read the guidelines. We can't help without a posted attempt from your side ...

What equations? There's only v is known in this problem. I think we can also use some constants like g

 

[haruspex]

What equations? There's only v is known in this problem. I think we can also use some constants like g

Think about the tension just above the table. Call it T.
Consider a period dt and the consequence of the tension. What momentum equation can you write?

 

[jbriggs444]

What equations? There's only v is known in this problem. I think we can also use some constants like g

You could consider tracking either energy or momentum. Look at where energy is injected into the system and where it is drained out. Or do the same for momentum.

 

[haruspex]

You could consider tracking either energy

Only after satisfying oneself mechanical energy is conserved during pick up, but it won't be.

 

[hutchphd]

Only after satisfying oneself mechanical energy is conserved during pick up, but it won't be.

But in steady state it will be constant.

 

[haruspex]

But in steady state it will be constant.

You miss the point. Work is being done by the net descent of the chain. Where is that work going? Is it all going into acceleration of chain from the table?

 

[hutchphd]

In steady state the power into the moving chain segment is zero. That was my point and one way to solution.

 

[jbriggs444]

Only after satisfying oneself mechanical energy is conserved during pick up, but it won't be.

It does not need to be as long as one can quantify the rate at which it is gained or lost at the pick-up interface. However, I think I take your point. It is not a simple plus or minus on a balance sheet. The energy lost by the chain in accelerating the next link will not match the energy gained by the link (*). Momentum is simpler in that respect.

(*) Intuition suggests an efficiency of exactly 50%.

 

[haruspex]

It does not need to be as long as one can quantify the rate at which it is gained or lost at the pick-up interface.

True, but isn't the calculation sequence backwards? Use momentum to find the work done on the chain, then find how much is lost.

 

[haruspex]

In steady state the power into the moving chain segment is zero. That was my point and one way to solution.

It will not give the solution since work is not conserved during the pick up.

 

[hutchphd]

Sorry but I don't know what "work is conserved" means. The power input by gravity is equal to the rate change in KE of the newly accelerated picked-up chain.

 

[hutchphd]

I get the same result using either power or force. Force is perhaps easier. ...

 

[jbriggs444]

Sorry but I don't know what "work is conserved" means. The power input by gravity is equal to the rate change in KE of the newly accelerated picked-up chain.

As @haruspex points out, it is not. Energy is lost to the inelastic collision of the chain with each new link that it picks up.

[In practice, there are complexities with the force analysis as well. Torque and angular momentum considerations can lead to a "push-off" from the table. That effect is Googleable]

 

[hutchphd]

As @haruspex points out, it is not. Energy is lost to the inelastic collision of the chain with each new link that it picks up.

[In practice, there are complexities with the force analysis as well. Torque and angular momentum considerations can lead to a "push-off" from the table. That effect is Googleable]

I am constrained to not supply the result. We shall put it in abeyance

 

[jbriggs444]

I am constrained to not supply the result. We shall put it in abeyance

I've worked it both ways. There is an extra factor of two in one of the results that I obtained.

 

[Gundyam]

I've got my answer. In my answer the constant is 2, What's yours?

 

[hutchphd]

You need to be careful about the potential energy of the chain from floor to table. The center of mass is only h/2 higher than its final resting place...does that do it?

 

[Gundyam]

Actually I've got some different constants. Can you tell me the exact constant for the final answer?

 

[hutchphd]

You tell me your answer first...

 

[jbriggs444]

You need to be careful about the potential energy of the chain from floor to table. The center of mass is only h/2 higher than its final resting place...does that do it?

Oh, that's where you have your factor of two error.

 

[Gundyam]

I've got my answer. In my answer the constant is 2, What's yours?

I've told it

 

[hutchphd]

I've told it

OK.. Either energy or force gives me v² =gh. I've lost track of the consensus.

 

[jbriggs444]

OK.. Either energy or force gives me v² =gh. I've lost track of the consensus.

Right. Because the energy balance is likely incorrect. The position of the center of gravity of the falling chain is irrelevant.

 

[hutchphd]

But if I use Newton's Law n I get, in steady state,

ρhg= (ρv)*v​

where ρ is the linear mass density of the rope.. So where is this factor of two?...you've got me mystified.

 

[jbriggs444]

But if I use Newton's Law n I get, in steady state,

ρhg= (ρv)*v​

where ρ is the linear mass density of the rope.. So where is this factor of two?...you've got me mystified.

That analysis is correct. It is the energy analysis which accidentally gets the right answer because the two factor of two mistakes cancel out.

 

[hutchphd]

It is the energy analysis which accidentally gets the right answer because the two factor of two mistakes cancel out.

Since you haven't seen my analysis, this statement seems a trifle presumptuous.
You may be correct but please tell what are the two offsetting mistakes I made? Am I missing something here?

 

[jbriggs444]

Since you haven't seen my analysis, this statement seems a trifle presumptuous.
You may be correct but please tell what are the two offsetting mistakes I made? Am I missing something here?

You mentioned the center of gravity of the section of the chain between table and floor. You claim to have ignored the energy lost to the inelastic collision between chain and the new links.

 

[hutchphd]

You mentioned the center of gravity of the section of the chain between table and floor. You claim to have ignored the energy lost to the inelastic collision between chain and the new links.

I never "claimed" anything of the sort. What are you talking about??
And what are the "two offsetting mistakes" I made?

1. ?
2. ?

 

[haruspex]

I never "claimed" anything of the sort. What are you talking about??
And what are the "two offsetting mistakes" I made?

1. ?
2. ?

In post #26 you appear to have used, directly or indirectly, conservation of momentum and got the right result.
In post #19 you used conservation of energy and should have got the wrong result, but you erroneously took the overall descent of each link of the chain to be h/2 instead of h. That exactly canceled the error resulting from assuming energy to be conserved (because exactly half is lost).

 

[haruspex]

As @haruspex points out, it is not. Energy is lost to the inelastic collision of the chain with each new link that it picks up.

[In practice, there are complexities with the force analysis as well. Torque and angular momentum considerations can lead to a "push-off" from the table. That effect is Googleable]

Fwiw, I have never believed the kick-off-the-ground explanation for the chain fountain. It would explain a reduced loss of energy in the pick-up, but that should only make the whole movement faster. It reduces the tension needed to accelerate the added links.

To explain the mid-air arc we need an increased tension in the top of the ascent. This is where the moment of inertia of the links could do the trick. As each link rounds the bend it tends to lift the ascending side.

 

[hutchphd]

In post #26 you appear to have used, directly or indirectly, conservation of momentum and got the right result.
In post #19 you used conservation of energy and should have got the wrong result, but you erroneously took the overall descent of each link of the chain to be h/2 instead of h. That exactly canceled the error resulting from assuming energy to be conserved (because exactly half is lost).

I must plead guilty to vigorous (and unseemly) hand-waving. Thanks for calling me out.
This is a much more interesting and perplexing (to me) problem than I initially understood
Is the result for a continuous (presumably pliable but massive) rope salient? The requirement of "links" seems odd.

 

[haruspex]

I must plead guilty to vigorous (and unseemly) hand-waving. Thanks for calling me out.
This is a much more interesting and perplexing (to me) problem than I initially understood
Is the result for a continuous (presumably pliable but massive) rope salient? The requirement of "links" seems odd.

Yes, it should apply to a rope as well. I have seen it posed using string.

It's a tricky problem because the posing of it treats the chain/rope as infinitely thin and starting in a pile of zero size, whereas in each real scenario one has to think what is going on at the detailed level.
For a rope the length is constant, so, no matter how thin, it must start as a stack of horizontal sections. When lifted, these will acquire some horizontal velocity, and this will not contribute to overcoming gravity. If no KE were lost it would mean the rope acquires a transverse vibration. This may seem like a detail that could be somehow avoided by a different model, but every model I can think of - rope, chain, telescope - has a way of getting rid of KE.
The chain model, as in the chain fountain, salvages some of the KE by means of rotational inertia, as discussed in the chain fountain link.