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Motion of a Uniform Chain falling I desperatly need help

  1. Jan 31, 2007 #1
    Motion of a Uniform Chain falling....I desperatly need help

    1. The problem statement, all variables and given/known data
    A uniform heavy chain of length "a" hangs initially with a part of length, "b", hanging over the edge of a table. The remaining part of length a-b is coiled up at the edge of the table. If the chain is released show that the speed of the chain when the last link leaves the end of the table is [2g(a^3-b^3)/(3a^2)]^1/2


    2. Relevant equations
    mg=m*dv/dt+dm/dt*v


    3. The attempt at a solution
    I have tried many times to do this problem, following the above equation, which my professor claims is correct for this problem; however, I get stuck everytime in the following situation (which my professor also claims is correct):

    From the begining

    Let m= z*dm/dx, than note that dm/dx=y. m=z*y. Than dm/dt=dz/dt*y.

    z*y*g=z*y*dv/dt + v*dz/dt*y

    Note that dv/dt= (d^2Z/dt^2)
    (note from now on that z(dot)= dz/dt, as is common in this notation)

    So dv/dt= (dz(dot)/dz)*z(dot)
    Again note that d/dz{(1/2)*(z(dot))^2} = dz(dot)/dz * z(dot)

    And then I get stuck, as I will end up in a loop of doing the same things over and over again. Worse yet, my professor told me explictly that this is what must be done for doing this problem using the equation given, and that I will (after I figure out the next step), need to use the method of frobenius to solve the problem.

    The alternative method of solving this, which he himself did mention, is to use largrange or hamaltionan mechanics (but he said we would have to figure those out on our own, as he hasn't refreshed himself on the topic).
    ------

    If I could get a hand setting this up with largrange, that would be awesome, I know that I need to some how split the problem into two to use the largrange, additionally I believe that I will need to take into account that it only has two degrees of freedom (motion on a plane); however, I can't figure out how to set up the largragian from that (I am a pretty inexperanced with the method; however, I would much rather learn to do this problem in largrange, than trying to muscule my way through the rest of the problem, when I don't quite understand how to exactly go about it).


    Thank you in advanced for anyhelp.
     
    Last edited: Jan 31, 2007
  2. jcsd
  3. Feb 1, 2007 #2

    AlephZero

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    Sorry, I can't follow your notation here.

    In m = z.dm/dx you seem to be using "m" for two different things in the same equation, and you don't say what x y and z are - so its hard to tell whether this is right or wrong.

    The equation your prof gave you is applying conservation of momentum with variable mass. In that equation, m is the mass of the chain that is off the table.

    As to alternative methods, you don't really need to use Lagrangian or Hamiltonian equations for this, but you could use conservation of energy.
     
  4. Feb 1, 2007 #3
    just a clarification, it should be m = z*dm/dx, where y=dm/dx, thus m= z*y. We have to do this because the mass falling is increasing over time.

    ------

    I have tried energy conservation, and I end up getting that v= [2g*(a-b)]^1/2, which according to the book and my professor, is wrong.
     
  5. Feb 2, 2007 #4
    Since the chain is coiled, shouldn't the term [itex]m\frac{dv}{dt}[/itex] vanish? If so then this becomes an easy seperable DE.
    My arguement goes as follows: Your main equation comes from
    [itex]F=\frac{dP}{dt}\Longrightarrow mg=m\frac{dv}{dt}+\frac{dm}{dt}v[/itex]
    [itex]F=mg[/itex] is the force pulling the chain off the table and since the chain is COILED this means that at any instant only an infinitesimal amount of mass is being jerked by this force. Consequently the m in the [itex]F=mg[/itex] term refers to all of the mass already off the table contributing to the force, while the m in the [itex]m\frac{dv}{dt}[/itex] term refers to the infinitesimal mass whose momentum is being affected by this force F. These m values are not the same.
     
    Last edited: Feb 2, 2007
  6. Feb 2, 2007 #5

    AlephZero

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    Let the chain have mass m per unit length and let x be the length off the table. The mass of the hanging part of the chain is mx.

    Force = rate of change of momentum:

    mxg = d/dt(mx x') = mxx" + mx'^2

    So the equation of motion is

    xx" + x'^2 = gx ... (1)

    Boundary conditions: at t = 0, x = b and x' = 0

    To solve (1), let y = (1/2)x'^2

    Then y' = x'x"

    x" = y' / x' = (dy/dt) / (dx/dt) = dy/dx

    So (1) becomes

    x dy/dx + 2y = gx ... (2)

    which is solvable by the standard "trick": let p = dy/dx, differentiate wrt x to get an DE connecting p and x, solve, then use (2) to eliminate p.

    Doing that gives

    gx^3 - 3yx^2 = C where C is the constant of integration

    and it's all downhill after that.
     
    Last edited: Feb 2, 2007
  7. Feb 3, 2007 #6
    Cool, thanks for the help (Your explaination was much better than when my professor was trying to explain how to approach this problem).
     
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