Comparing the rate at which two chains fall in different scenarios

[Anacoluthon64]

That's basic first year DE. It hardly takes an "elephantine" memory to remember how to do it.

As I remarked, it's a bugger to solve from first principles - "solve" here indicates a general solution as opposed to a particular one. However, I'm impressed that you find it so easy.

Your noncalculus solution, however, is an elegant alternative to some truly ugly arguments.

Thank you, I'm pleased that you approve.

How did you get that?

It's basic first-year calculus, in this case the chain rule of differentiation, as I indicated in a prior post:

r' = v = (dr/dt)
=> r" = v' = (dv/dt) = (dv/dr).(dr/dt) = v.(dv/dr)

Substituting, r" = v.(dv/dr) = a.g/(R + a - r)

Separating the variables v & r and integrating yields the result you query.

'Luthon64

 

[Anacoluthon64]

Given those added conditions, it seems to me that at any time t while chain remains to fall the velocity of the link going over the edge will be the same for that case and for the free-fall case. Or am I missing something?

Your assessment agrees with mine. In the full-length drop scenario, only the feeding arrangement at the top would be required, while the reduced drop would require both arrangements. In both these cases it is obvious that at each instant the total mass of chain subject to gravitational attraction is equal to the total mass of chain being accelerated. Therefore the accelerations would be equal and have a value of g, i.e. free fall.

With friction=0 the chain needs to be stretched out and eaten / not eaten for the difference to occur.

Agreed. It is these differences that make the problem interesting.

'Luthon64

 

[69dodge]

Your assessment agrees with mine. In the full-length drop scenario, only the feeding arrangement at the top would be required, while the reduced drop would require both arrangements. In both these cases it is obvious that at each instant the total mass of chain subject to gravitational attraction is equal to the total mass of chain being accelerated. Therefore the accelerations would be equal and have a value of g, i.e. free fall.

Not sure what sort of feeding arrangement at the top you have in mind. But if there's nothing on the ledge that's powered---rather, the chain is just bunched up at the edge, so that each bit doesn't move at all until it gets pulled off the ledge and starts falling---then the situation differs from free-fall because, at each instant, gravity, in addition to continuing to accelerate the part of the chain that's already falling, has to instantaneously accelerate, from rest to the speed of the falling part, the bit that's getting pulled over the edge and is just starting to fall.

 

[Anacoluthon64]

Not sure what sort of feeding arrangement at the top you have in mind.
...snip...

Reviewing a very few of the previous posts before despatching your above epistle would have rendered it wholly unnecessary to pose such a comment. I invite you now please to have a look.

'Luthon64

 

[69dodge]

Reviewing a very few of the previous posts before despatching your above epistle would have rendered it wholly unnecessary to pose such a comment. I invite you now please to have a look.

I've read the whole thread. I still wasn't sure what you had in mind.

Sorry if I came across as overly argumentative. I didn't intend to.

 

[Anacoluthon64]

I've read the whole thread. I still wasn't sure what you had in mind.

Sorry if I came across as overly argumentative. I didn't intend to.

No problem - see post #78.

'Luthon64

(Edit: URL fix - sorry...)

 

[69dodge]

Well, if we're going to assume frictionlessness, there's no reason that, in such an ideal world, the chain can't be piled up right at the precipice's edge in such a manner that the horizontal force required to "free" each successive link from the pile is arbitrarily close to zero, and that each link clears precisely at the edge (i.e. not above it). This might be accomplished through a suitably sensitive chain feeding device that continuously monitors the instantaneous chain velocity and adjusts its feed rate accordingly.

I guess this is the part that you're referring to?

I had read it; I'm just not sure what it means. The first sentence talks only about piling the chain up right at the edge. Why is a feeding device needed for that?

I guess the real question is, what do you mean by "'free' each successive link from the pile"? Do you mean, "move the link an infinitesimal horizontal distance over the edge, so that it's no longer supported by the ledge"? Or do you mean, "get the link over the edge and, additionally, accelerate it up to the speed of the part of the chain that's already falling"? The former requires only an infinitesimal force acting for an infinitesimal time anyway, so there's no need for a feeding device to satisfy your condition of a "horizontal force [that's] arbitrarily close to zero". But the latter requires some work to be done on the link that's going over the edge, so if you don't want gravity to have to do it, you'd need a powered feeding device that does it instead.

Agree? Disagree? Am I making any sense at all here?

(In any case, I've learned a new word: epistle. )

 

[Anacoluthon64]

I guess this is the part that you're referring to?

I had read it; I'm just not sure what it means. The first sentence talks only about piling the chain up right at the edge. Why is a feeding device needed for that?

I guess the real question is, what do you mean by "'free' each successive link from the pile"? Do you mean, "move the link an infinitesimal horizontal distance over the edge, so that it's no longer supported by the ledge"? Or do you mean, "get the link over the edge and, additionally, accelerate it up to the speed of the part of the chain that's already falling"? The former requires only an infinitesimal force acting for an infinitesimal time anyway, so there's no need for a feeding device to satisfy your condition of a "horizontal force [that's] arbitrarily close to zero". But the latter requires some work to be done on the link that's going over the edge, so if you don't want gravity to have to do it, you'd need a powered feeding device that does it instead.

Hmm, yes, you've got a point there - after re-reading it, I see that it is quite ambiguous. I had used the word "free" in the second, wider, sense you note.

The case where successive chain links are jerked one-by-one from rest over the edge requires more data about the chain than are given. One would need to bring in considerations of momentum transfer between chain links. Energy considerations on their own would not suffice as there is energy dissipated (read "effectively lost") on each momentum transfer. I think an attempt to solve this case could become inordinately ugly quite quickly.

Agree? Disagree? Am I making any sense at all here?

(In any case, I've learned a new word: epistle. )

Agree, ferociously good sense, and glad to edify you, in that order .

 

[ImaginalDisc]

I very much appreciate the work you've all done in regards to this problem, but I am having trouble following the math. Is this problem best addressed by differential calculus? I'm afraid I only understand basic caclulus.

 

[Anacoluthon64]

I very much appreciate the work you've all done in regards to this problem ...

Speaking for myself, only a pleasure. It was interesting and fun.

... , but I am having trouble following the math. Is this problem best addressed by differential calculus? I'm afraid I only understand basic caclulus.

You could address it through the energy considerations I gave in an earlier post, at least one part of it. Alternatively, you could address it by considering one-link-at-a-time and their mutual interactions with one other. The calculus approach is an idealisation that results from the limiting case where each link is assumed to be infinitesimal in size.

'Luthon64

 

[kevin]

I'm not terribly good at math, so I need help with a comparision of two different kinematic systems, please. It's been puzzling me.

There are two chains at the very edges of two different precipices. Each chain is 1000 meters long. Chain A is just over a 1000 meter drop. Chain B is just over a 10 meter drop. The entire system is completely frictionless, and the chains are assumed not to stack, but to just lie on the ground like limp noodles.

Here is where the kinematics kicks in:

The first link of each chain is nudged over their respective precipices.

Does each chain slip over the edge at the same rate, or does the chain slipping over the shorter drop slip over the edge at a lower rate?

I haven't read all the posts in the thread, and my calculus has been unused for many years so quite a few of the ones I've read I'd have to puzzle through but it seems to me this is just simple physics.

Take the 1000 meter cliff:
a) any link, once gone over the edge, has to be travelling at the same speed as the first link in the chain (unless the chain isn't taut, but I don't see why that would be the case.) Same as if you had dropped a rod - the back end of the rod travels the same speed as the front end.

b) the speed of the first link, after falling 1000 meters would be (starting at 0):
1000 = 1/2 (9.8) t^2
t = 14.3 seconds
v = 1000/14.3 = 70 meters/sec

so the last link on the 1000 meter chain is travelling 70 meters/sec because the first link, just before it hits the ground, is travelling that speed.

For the 10 meter cliff no link falls further than 10 meters. The chain, at that point, is lax while the remainder is taut. So the calculation is:
10 = 1/2 (9.8) t^2
t = 1.43 seconds
v = 10/1.43 = 7 meters/second

 

[kevin]

I haven't read all the posts in the thread, and my calculus has been unused for many years so quite a few of the ones I've read I'd have to puzzle through but it seems to me this is just simple physics.

now i'm not so confident in my second answer. The first still looks ok to me. but for the second I've really calculated the speed of the link at the 10th meter of chain. If we suppose there is some sort of disintegration ray at meter 10 that zaps the first link, link 2 would then accelerate for the distance of 1 link, since when link 1 is disintegrated link 2 has the same speed as link 1 then link 2 would be disintegrated at a higher velocity.

I'm starting to lean towards the last 2 links would be at the same speed.

 

[Anacoluthon64]

b) the speed of the first link, after falling 1000 meters would be (starting at 0):
1000 = 1/2 (9.8) t^2
t = 14.3 seconds
v = 1000/14.3 = 70 meters/sec

You have calculated the average velocity of the chain free-falling its entire length over the edge. It seems to me that a more realistic conception is having the weight of chain over the edge accelerate the entire chain, in which case a rather different answer is obtained.

For the 10 meter cliff no link falls further than 10 meters. The chain, at that point, is lax while the remainder is taut. So the calculation is:
10 = 1/2 (9.8) t^2
t = 1.43 seconds
v = 10/1.43 = 7 meters/second

You have calculated the average velocity of the first 10 metres of chain free-falling over the edge. A similar conception as described before again yields a very different result.

'Luthon64

 

[kevin]

You have calculated the average velocity of the chain free-falling its entire length over the edge. It seems to me that a more realistic conception is having the weight of chain over the edge accelerate the entire chain, in which case a rather different answer is obtained.

No, I calculated the speed of the first link after falling 1000 meters.

I used the equations for objects under constant acceleration:
http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html#c2

The weight of a free falling body does not affect the speed or acceleration, so I have no idea what you mean by "weight of chain over the edge accelerate the entire chain". The weight of the chain over the edge can have no affect on the first link. and as long as the chain is taut the back link can never travel faster than the first link.

I would say the weight still on the cliff could slow the first link down, except that a frictionless environment was called for, so that won't happen.

 

[Anacoluthon64]

No, I calculated the speed of the first link after falling 1000 meters.

Clearly you don't understand what you have calculated, so permit me the liberty of enlightening you: First, you assume that the chain accelerates at a constant rate of g m/s², i.e. free fall, through a distance of 1,000 metres. You calculate (from s = ¹/₂.a.t²) that this process takes 14.3 seconds. You then take the distance travelled, i.e. 1,000 metres, and divide it by 14.3 seconds, yielding 70 m/s. But this is the constant velocity (i.e. without acceleration) that the chain would have had to travel to cover 1,000 metres in 14.3 seconds. Ergo, it is the average velocity, as I said. Also, you haven't provided any justification for why the chain accelerates at g m/s², which would only be the case if you threw the whole lot of chain over the edge all in one go.

I used the equations for objects under constant acceleration:
http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html#c2

Uncritically, I'm afraid.

The weight of a free falling body does not affect the speed or acceleration, so I have no idea what you mean by "weight of chain over the edge accelerate the entire chain". The weight of the chain over the edge can have no affect on the first link. and as long as the chain is taut the back link can never travel faster than the first link.

But that's the whole point: if there's chain over the edge and chain left at the top, the chain over the edge cannot accelerate at g m/s² since the chain at the top is being accelerated horizontally by the pull of the part that is over the edge. It may help to think of the chain as laid out horizontally on top so that its entire length is perpendicular to the edge of the cliff. If what you're saying were correct, then a 10-ton mass at the top being pulled horizontally via a thin cord connected to a 10-gramme weight over the edge would accelerate at the same rate, i.e. g m/s², as the reverse situation where a 10-ton weight over the edge pulls a 10-gramme weight horizontally at the top. That would be in total violation of Newton's Second Law of motion.

I would say the weight still on the cliff could slow the first link down, except that a frictionless environment was called for, so that won't happen.

The assumption of a frictionless environment does not remove the inertia of a mass. All it does is remove from consideration energy losses resulting from friction.

'Luthon64

 

[69dodge]

The weight of the chain over the edge can have no affect on the first link.

Why not? They're attached. In order for the first link to move, it has to drag the rest of the chain along with it. That will slow it down.

I would say the weight still on the cliff could slow the first link down, except that a frictionless environment was called for, so that won't happen.

Inertia and friction are two different things. Even though there's no friction, the chain still has inertia, including the part of it that hasn't fallen off the edge yet.

ETA: Oops. I was beaten to it, I see.

 

[Anacoluthon64]

Why not? They're attached. In order for the first link to move, it has to drag the rest of the chain along with it. That will slow it down.

Inertia and friction are two different things. Even though there's no friction, the chain still has inertia, including the part of it that hasn't fallen off the edge yet.

Thanks for making the same points independently.

'Luthon64

 

[Jimbo07]

A similar conception as described before again yields a very different result.

Or Kevin, speaking as one who made a mistake early in this thread, it might be helpful to go back and read it...

 

[rehn]

I can't see the problem here.
The chain with the low drop will go slower. Only common sense is needed.
( Well.. I know common sense might lead astray sometimes but in this
case I think it is valid )
Just assume that the short drop is approaching zero m.
I can't image how it would be possible that the speed would be constant
regardless of drop and then suddenly go to zero when the drop goes to zero.....