Comparing the rate at which two chains fall in different scenarios

[Art Vandelay]

I don't know that much math, but this is interesting. Could you explain step 3, above?
How do you get -1? I can't see any way that (2+4+8...)-(1+2+4...) could give a finite number. But I don't know anything about infinite sets and doing arithmetic with them...
Which seems to be the point you're making. It'd just be easier to understand if you could spell out where the -1 comes from.
(If this is very complicated feel free to ignore, I just can't figure it out.)

What if I put it like this:

  0+2+4+8...
-(1+2+4+8...) 
------------
 -1+0+0+0...

The two are exactly the same, except that the second has an extra one. So the first minus the second is negative one? And yes, the point is that this is not valid logic.

 

[CaveDave]

What if I put it like this:

  0+2+4+8...
-(1+2+4+8...) 
------------
 -1+0+0+0...

The two are exactly the same, except that the second has an extra one. So the first minus the second is negative one?

To be consistant, wouldn't that be:

  0+2+4+8+16...
-(0+1+2+4+8...) 
------------
  0+1+2+4+8...

And yes, the point is that this is not valid logic

No doubt.

Cheers,
Dave

Edit: corrected stupid mistake...<kick self>

 

[Anacoluthon64]

You're probably right, but I can't pretend that I follow you.

Any piece of mathematical reasoning can be "much simplified" by handwaving and use of the phrase "immediately emerges".

All right then, here's the demo sans offending gesticulations, using your nomenclature:

r'' = r.g.D / R.D = r.g / R

The above D.E., though a standard form, is fairly demanding to solve from first principles, i.e. in the absence of a lookup table or an elephantine memory. If instead we let v = r', then r" = v.(dv/dr), as shown in my earlier post. Thus,

v.(dv/dr) = (g/R).r,

which is easy to solve through separation of variables:

Int{v.dv} = Int{(g/R).r.dr}

=> (v^2)/2 = g.(r^2)/2.R+C

With the initial condition (v, r) = (0, 0), we find C = 0, and finally, therefore,

v = Sqrt(g/R).r

Simple, no? For r = R, this gives v = Sqrt(g.R), as you maintain.

After some further rumination, it occurred to me that there's a yet simpler approach through energy considerations for the situation in which the entire mass of chain accelerates throughout the process: Using the top of the precipice as the reference level, the chain's centre of gravity is at r = 0. As the last link clears the edge, the c.o.g. is at r = R/2 (the chain is vertically oriented and extended to it's full length without kinks due to the absence of any friction). Therefore, the chain has lost E = (R.D).g.(R/2) gravitational potential energy, and has gained K = (R.D).v^2/2 kinetic energy of motion. Since, by energy conservation, K = E, solving for v in terms of R again yields v = Sqrt(g.R).

This approach can be generalised for any r between 0 and R. The c.o.g.'s position is calculated by weighting that of the vertical and horizontal pieces by their respective lengths:

A chain length of r has passed over the edge, 0 <= r <= R. The vertical piece's c.o.g. is at r/2 below the edge, while the remainder is at the reference level, i.e. where r = 0. The combined c.o.g. is therefore at

((r.D).(r/2) + ((R - r).D).0)/(R.D) = (r^2.D/2 + 0)/R.D) = r^2/2.R

below the edge. Thus and as before,

K = (R.D).g.(r^2/2.R) = D.g.r^2/2 = E = (R.D).v^2/2,

whence v = Sqrt(g/R).r

So it is, in the assumed situation, possible to obtain without resort to calculus an expression for the instantaneous chain velocity in terms of the length of chain that has already passed over the edge.

'Luthon64

 

[Anacoluthon64]

I suspect that the question posed in the OP has become contentious in this thread because the problem was not described with sufficient clarity.

In order to resolve this, it would be productive if ImaginalDisc could answer the following questions:

(1) Is the chain at the top of each precipice bunched up at the edge of the drop, or is it laid out in a straight line so that the entire chain moves as links pass over the edge?

(2) In the short drop scenario, does the chain that lands at the bottom bunch up and lie there unmoving while more chain falls, or does it continue moving by being pushed along by the falling chain?

'Luthon64

 

[hammegk]

Remembering that friction=0, at the top the coiled scenario adds angular momentum to the problem as I see it.

The hitting bottom effects at friction=0 have more variables than I know how to consider.

 

[Anacoluthon64]

... the coiled scenario adds angular momentum to the problem as I see it.

Coiled? Who ordered that?

'Luthon64

 

[Jimbo07]

r'' = r.g.D / R.D = r.g / R

v = Sqrt(g/R).r

For r = R, this gives v = Sqrt(g.R), as you maintain.

This is all great and good. For any length r, we now have the length, velocity and acceleration. For the 1000 m chain, everything checks (r'' of the last link = Rg/R = g, as expected). Also, v=r(Sqrt(g/R)) holds for both chains up to the first 10 m falling over.

What about the case where the second chain has reached 10 m and is piling up. Now, r is the constant, 10m, giving:

r''=10g/(10 + r2), where r2 is now decreasing?

 

[ImaginalDisc]

I suspect that the question posed in the OP has become contentious in this thread because the problem was not described with sufficient clarity.

In order to resolve this, it would be productive if ImaginalDisc could answer the following questions:

(1) Is the chain at the top of each precipice bunched up at the edge of the drop, or is it laid out in a straight line so that the entire chain moves as links pass over the edge?

The chain is laied out. The chain could also be on a frictoinless spool and I believe that would have the same effect.

(2) In the short drop scenario, does the chain that lands at the bottom bunch up and lie there unmoving while more chain falls, or does it continue moving by being pushed along by the falling chain?

Hmm, I tried to make it clear in the op that the chain in the short drop lies flat as it hits the ground, and the ffect of the chain piling up should be ignored.

'Luthon64[/QUOTE]

 

[Anacoluthon64]

The chain is laied out.

Yes, but does the whole chain move as it starts going over the edge or does only a small part of the chain move?

The chain could also be on a frictoinless spool and I believe that would have the same effect.

As in "rolled up on a spool"? If so, you need more information regarding the geometry of the chain and the spool.

Hmm, I tried to make it clear in the op that the chain in the short drop lies flat as it hits the ground, and the ffect of the chain piling up should be ignored.

Yes, but does it stop moving or does the chain falling behind it carry on pushing it away?

'Luthon64

 

[Jimbo07]

Yes, but does the whole chain move as it starts going over the edge or does only a small part of the chain move?

The model we've used would have to have the whole chain moving. I've assumed no slipping between the links. A better analogy might be those little old beaded chains that you'd find on lamps.

Yes, but does it stop moving or does the chain falling behind it carry on pushing it away?

'Luthon64

I read the situation to be this: since we're already 'magically' assuming no friction, let's imagine some magical scenario, where the chain that has already dropped the 10 m has no further effect on the system... maybe a big cutting machine at the bottom or something.

I'm still stuck back at the case where r1 is now 10m and r2 is decreasing, and I'm at work, so it's bugging me, but there's really nothing I can do about it...

... although this has turned out to be a more interesting problem than I had originally thought!

 

[Anacoluthon64]

This is all great and good. For any length r, we now have the length, velocity and acceleration. For the 1000 m chain, everything checks (r'' of the last link = Rg/R = g, as expected). Also, v=r(Sqrt(g/R)) holds for both chains up to the first 10 m falling over.

What about the case where the second chain has reached 10 m and is piling up. Now, r is the constant, 10m, giving:

r''=10g/(10 + r2), where r2 is now decreasing?

See also my previous posts.

Er, I don't know what you mean by "r2." I'll assume that the reduced drop length is a metres, and that the chain at the bottom comes to a standstill without being acted on any further. Then for r >= a, with r metres of chain over the edge, there are (R - r) metres at the top and a metres in the process of falling. Therefore a force of D.a.g accelerates a mass of D.(R + a - r), whence

r" = a.g/(R + a - r)

Using the v = r' substitution as before, we can solve the D.E. to obtain

v²/2 = -g.a.Ln(R + a - r) + C

for which the initial condition (v, r) = (a.Sqrt(g/R), a) holds, yielding

C = g.a.(a/2.R + Ln(R)).

After suitable substitution and rearrangement, we obtain

v = Sqrt(g.a.(a/R + 2.Ln(R/(R + a - r)))) for r > a.

The complete function for v in terms of r then consists of two parts:

v = r.Sqrt(g/R) for 0 <= r <= a
v = Sqrt(g.a.(a/R + 2.Ln(R/(R + a - r)))) for a < r <= R

For r = R, we obtain v = Sqrt(g.a.(a/R + 2.Ln(R/a))), which gives about 30.06 m/s when R = 1,000 m and a = 10 m, as I reported in post #40.

'Luthon64

 

[Anacoluthon64]

... although this has turned out to be a more interesting problem than I had originally thought!

If my assumptions, as used in the given solutions, are correct, then I think ImaginalDisc may have intended illuminating the contrast between the following two situations:

(1) An increasing force accelerating a constant mass,

versus

(2) A constant force accelerating a diminishing mass.

Hence the interest.

'Luthon64

 

[ImaginalDisc]

Yes, but does the whole chain move as it starts going over the edge or does only a small part of the chain move?


As in "rolled up on a spool"? If so, you need more information regarding the geometry of the chain and the spool.


Yes, but does it stop moving or does the chain falling behind it carry on pushing it away?

'Luthon64

Maybe I was unclear.

None of these things are important. Assume they have no influence.

 

[hammegk]

Coiled? Who ordered that?

'Luthon64

Remember the friction=0 problem. I was giving you a break over the impossibility of analytic solution to a random pile of frictionless chain with an end being pulled.

As I said, the bottom end dynamics of a frictionless falling-in-a-pile chain are already far beyond analytics.


Of course looking at it as you later do; increasing force constant mass vs constant force decreasing mass I'd think holds the correct analytic solution, and probably in the limit as friction approaches zero, equating for a constant time elapsed, maybe.

 

[Jimbo07]

See also my previous posts.

Er, I don't know what you mean by "r2."

Oh, that was just an artifact of the original setup. I had arrived at

a = g m1 / (m1 + m2) I took the lengths and Dr. A's linear densities to get

r"= g r D / (r1 + r2) D, where r1 + r2 were the constant R.

r2 is R - r, and in this case what was r1, is r, or what you're now calling a. You made the appropriate substitutions on top and bottom, giving

r" = ag/(a + (R-r))

I'll assume the rest is correct. I don't always see D.E.s well in my head, and have to resort to pencil-and-paper which I don't have the luxury of right now...

However, it's important that you've arrived at the two parts for the complete function. Ignoring my r2 nonsense, this was what I was really getting at:

The complete function for v in terms of r then consists of two parts:

v = r.Sqrt(g/R) for 0 <= r <= a
v = Sqrt(g.a.(a/R + 2.Ln(R/(R + a - r)))) for a < r <= R

For r = R, we obtain v = Sqrt(g.a.(a/R + 2.Ln(R/a))), which gives about 30.06 m/s for R = 1,000 m and a = 10 m.

'Luthon64

 

[Anacoluthon64]

Maybe I was unclear.

None of these things are important. Assume they have no influence.

The things I have asked are important to the problem. They affect the mathematical models one would use in solving the problem, unless we arbitrarily rewrite Newton's second law.

'Luthon64

 

[jj]

Remember the friction=0 problem. I was giving you a break over the impossibility of analytic solution to a random pile of frictionless chain with an end being pulled.

As I said, the bottom end dynamics of a frictionless falling-in-a-pile chain are already far beyond analytics.

I think one has to assume that the chain has zero rebound, infinite thinness, and doesn't pile up as a result, or it's just beyond...

Of course looking at it as you later do; increasing force constant mass vs constant force decreasing mass I'd think holds the correct analytic solution, and probably in the limit as friction approaches zero, equating for a constant time elapsed, maybe.

This problem will go nowhere without a lot of "ideal" assumptions.

 

[Anacoluthon64]

Remember the friction=0 problem. I was giving you a break over the impossibility of analytic solution to a random pile of frictionless chain with an end being pulled.

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.

As I said, the bottom end dynamics of a frictionless falling-in-a-pile chain are already far beyond analytics.

Again, a suitable device at 10 m below the edge that "eats" the chain at its instantaneous velocity would obviate this problem.

Of course looking at it as you later do; increasing force constant mass vs constant force decreasing mass I'd think holds the correct analytic solution, and probably in the limit as friction approaches zero, equating for a constant time elapsed, maybe.

Seemed the most sensible approach to me with what was given. However, the time aspect actually presents a considerable difficulty in that the initial condition (r, t) = (0, 0) yields the result r = 0 for all t, which is obviously not true. Some thought will be required to address it, and it is the reason I have avoided solving in terms of t, and used r instead.

'Luthon64

 

[Art Vandelay]

r'' = r.g.D / R.D = r.g / R

The above D.E., though a standard form, is fairly demanding to solve from first principles, i.e. in the absence of a lookup table or an elephantine memory.

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

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

What about the case where the second chain has reached 10 m and is piling up. Now, r is the constant, 10m, giving:

r''=10g/(10 + r2), where r2 is now decreasing?

So r''=-(r2)'', right?
Let u=10+r2
u''=(r2)''

So we need to solve the equation
u u''=-10g

r" = a.g/(R + a - r)

Using the v = r' substitution as before, we can solve the D.E. to obtain

v²/2 = -g.a.Ln(R + a - r) + C

How did you get that?

 

[hammegk]

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.

Again, a suitable device at 10 m below the edge that "eats" the chain at its instantaneous velocity would obviate this problem.

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?

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