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Motion of heavy chain

  1. Apr 11, 2007 #1
    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. jcsd
  3. Apr 11, 2007 #2

    Päällikkö

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    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.
     
  4. Apr 11, 2007 #3
    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.
     
  5. Apr 11, 2007 #4
    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.
     
  6. Apr 11, 2007 #5

    Päällikkö

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

    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
  7. Apr 11, 2007 #6
    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?
     
  8. Apr 11, 2007 #7
    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
     
  9. Apr 11, 2007 #8
    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?
     
  10. Apr 11, 2007 #9
    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.
     
  11. Apr 12, 2007 #10

    Päällikkö

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    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 :smile:.
     
    Last edited: Apr 12, 2007
  12. Apr 13, 2007 #11
    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?
     
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