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Rope through a hole

  1. Aug 29, 2013 #1
    1. The problem statement, all variables and given/known data
    A rope of mass M and length ##l## lies on a friction less table, with a short portion, ##l_0## hanging through a hole. Initially the rope is at rest.

    a. Find a general solution for x(t), the length of rope through the hole.

    (Ans: ##x=Ae^{\gamma t}+Be^{-\gamma t}##, where ##\gamma^2=g/l##)

    b. Evaluate the constants A and B so that the initial conditions are satisfied.


    2. Relevant equations



    3. The attempt at a solution
    The forces acting on the rope are weight and tension (T) due to the part of rope on the table. If x is the length of rope hanging, l-x is the length of rope on the table. Let ##\lambda## be the mass per unit length of rope.
    Newton's second law for hanging part,
    $$\lambda xg-T=\lambda xa$$
    Newton's second law for rope on table,
    $$T=\lambda (l-x)a$$
    From the two equations,
    $$a=\frac{gx}{l+2x}$$
    I can substitute a=d^2x/dt^2 but Wolfram Alpha gives no solution for this. :confused:

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Aug 29, 2013 #2
    Hi Pranav-Arora. Your formulation is correct, but check your algebra. It should be la=xg.
     
  4. Aug 29, 2013 #3
    Oh yes, sorry about that. Thanks a lot! :smile:

    At t=0, ##x(0)=l_0##, x'(0)=0
    ##x(0)=A+B=l_0##

    Since ##x'(t)=A\gamma e^{\gamma t}-B\gamma e^{-\gamma t}\Rightarrow x'(0)=0=A-B##
    Solving the two equations, ##A=B=l_0/2##.
    Hence,
    $$x(t)=\frac{1}{2}\left(l_0e^{\gamma t}+l_0e^{-\gamma t}\right)$$
    Looks good?
     
  5. Aug 29, 2013 #4
    It can be solved this way too, by Newton's second law :

    [tex]F=Ma=\rho g x A[/tex]

    [tex]M\frac{d^2x}{dt^2}=\rho g x A[/tex]

    [tex]M\frac{d^2x}{dt^2}=\frac{M}{l^3} g x l^2[/tex]

    [tex]\frac{d^2x}{dt^2}=\frac{gx}{l} [/tex]

    and the solution of this DE is

    [tex]x(t)=x=A\cdot exp(\sqrt{\frac{g}{l}}t)+B\cdot exp(-\sqrt{\frac{g}{l}}t)[/tex]

    same as yours...
     
    Last edited: Aug 29, 2013
  6. Aug 29, 2013 #5
    @janhaa: What are ##\rho## and ##A##? :confused:
     
  7. Aug 29, 2013 #6
    [tex]\rho[/tex] is density
    and
    [tex]A: area = l^2[/tex]
     
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