The Dynamics of a Heavy Chain on a Smooth Peg

[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?

 

[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.

 

[sauravbhaumik]

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.

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.

 

[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.

 

[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).

 

[sauravbhaumik]

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).

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?

 

[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

 

[sauravbhaumik]

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

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?

 

[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.

 

[Päällikkö]

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?

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 .

 

[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?