Motion of heavy chain

1. Apr 11, 2007

sauravbhaumik

Would anyone kindly discuss the motion of a heavy chain ?
A typical problem I am stuck with is as follows:
A uniform chain of length 2a is hung over a smooth peg, with the lengths (a+b) and (a-b) of two sides. Motion is allowed to ensue - show that the time the chain would take to leave the peg is .....

Kindly discuss the motion and hint at the problem too. Isn't it a kind of varying mass problem? How should I treat the problem - the motion of the end points, that of the CGs of each side?

2. Apr 11, 2007

Päällikkö

Indeed it is a kind of varying mass problem. As always, start by drawing a free body diagram and identify all (both, assuming negligible friction) the forces acting on the chain.

You should aim for an expression for x(t), where x is the displacement of one of the end points (you should get a differential equation). Start with the chain's acceleration at t = 0, and think how b relates to x.

3. Apr 11, 2007

sauravbhaumik

Yes I thought of that case, too. But the problem is, I am not sure about the forces that would work at the end points - the tension of the chain may not remain equal at both sides' end points, and I am not sure how the tension varies. So I'd be able to immediately see the eqn of motion of the end points, should you kindly illuminate a bit more on the principles.

4. Apr 11, 2007

denverdoc

I'm not sure that tension is explicitly required in this problem tho it is clearly operative. If the two sides were equal a=0. At the time it slips over peg, a=g. For all other times it is some fraction of g determined by the relative masses on each side.

5. Apr 11, 2007

Päällikkö

For starters, a bit unrelated, but by peg you mean something like a hook, right?

Well I wouldn't really worry about tension all that much. The net acceleration is caused by gravity. The net force on the chain is
F = (m1 + m2)a = m1g - m2g = (m1 - m2)g,
where m1 is the a+b part and m2 the a-b part. Now m1 and m2 are functions of b, which in turn is a function of x (or you could just use b if you want to, x is a bit unnecessary).

Last edited: Apr 11, 2007
6. Apr 11, 2007

sauravbhaumik

OK, but it doesn't hint at the eqn of motion. Let the linear density be d. So, the net force on the chain is {(a+b) - (a-b)}d.g = 2bdg.
But how can I have the eqn of motion? It seems that there might be some A(b) such that A.b'' = 2bdg, where b'' = d2b/dt2; but what is A, then? Is A = (a+b)d? Is A = 2a.d?

7. Apr 11, 2007

denverdoc

darn I threw the soln away while house cleaning this afternoon. Let me recollect:
I called total mass M and chose to use y(t) as the position of the longer length. I took similar strategy as you and divided M by 2a to represent mass/length and came up with something like

Md"y/dt"=g*M/2a(-2b+y(t)) where limits are 0<=y<=b

8. Apr 11, 2007

sauravbhaumik

Oh I think you meeant to write
.....=g*M/2a(-2a+2y)

Let me calculate:
y'' = (g/a).(y-a)
That gives general soln:

y-a = A.exp(kt) + B.exp(-kt), where k2 = g/a

Boundary values:
b = A + B ----------------> (1)
y' = k{ A.exp(kt) - B.exp(-kt)}

So,
0 = A - B ------------------>(2)

We get, then A = B = b/2

So, y = a + (b/2)(exp(kt) + exp(-kt))

But the chain would leave the peg when y = 2a, i.e.,

2a/b = x + 1/x, where x = exp(kt)

or, (xb)^2 - 2a.(xb) + a^2 = a^2 - b^2
or, (xb - a)^2 = a^2 -b^2

There is a problem! Taking sqrt of both sides, I've got to select only one value from right side -

Ok, if I take the +ve sign,

x = (a + sqrt(a2 - b2))/b

And t = sqrt(a/g). {(a + sqrt(a2 - b2))/b}

But that is same with the expression required in the problem.
Can you explain why the -ve sign was not taken?

9. Apr 11, 2007

denverdoc

I thought it was b and I had a ln soln, I'm off to play some online backgammon, I'll look at this again in the am. If others want to help in the meantime, feel free.

10. Apr 12, 2007

Päällikkö

Although not essential anymore, I was thinking something like:

F = (m1 + m2)a = m1g - m2g = (m1 - m2)g,
where:
M = m1 + m2 (constant)
m1 = M(a+b)/(2a), m2 = M(a-b)/(2a)
So
b'' = ((a+b)/(2a) - (a-b)/(2a))g = (g/a)b

Changing the variables (and fitting the boundary conditions to the new ones) you should get the one you've analyzed:
y'' = (g/a)(y-a)

Ok, to your current problem, assuming everything you've done is correct, except did you forget ln from the expression for t?

As both solutions are mathematically possible, there must be a physical reason to ignore the minus sign:
t = sqrt(a/g) ln{(a +- sqrt(a2 - b2))/b} goes negative when
0 < a +- sqrt(a2 - b2) < b,
which never happens with the plus sign, as:
sqrt(a2 - b2) < b - a < 0, and as sqrt(x) >= 0, it isn't possible.
.
With the minus sign, t goes negative when:
sqrt(a2 - b2) > a - b
eg. a = 5, b = 3:
sqrt(52 - 32) = 4 > 2 = 5 - 3, so it might happen.

I suppose this might seem a bit farfetched, but it seems to work. Let's wait and see if someone comes up with a better idea .

Last edited: Apr 12, 2007
11. Apr 13, 2007

sauravbhaumik

To Päällikkö:
Oh sorry I forgot ln while putting the expression for t.
Thanks, however, for your explanation for ignoring the minus sign.

Another request: would you kindly discuss the motion of a heavy chain hung from a smooth pulley/peg/hook in general? I have doubts about the procedure I have come up with at last (as shown above) to solve the problem : why the eqn of motion is as though we considered the whole chain as a particle without taking into account the nitty-gritty of varying mass problem? Why didn't we consider the motions of the CG's of the two parts (two sides of the chain)? Would you kindly illuminate?