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Atwood Machine with a twist - String with Mass

  1. Mar 24, 2009 #1
    Hello all. I've read the forums a few times for help but this is my first time posting, so please bear with me if I make a few mistakes.

    I'm studying for my final undergrad assessment, so I thought a good way to study would be to make "advanced" versions of problems I'm already familiar with. Here's one I came up with but I'm not quite sure I'm on the right track exactly.

    1. The problem statement, all variables and given/known data
    A rope, of mass [tex]M[/tex] and length [tex]L[/tex] (and it's uniform density [tex]\lambda[/tex], for simplicity) is suspended over an idealized pulley (no masses attached to the rope for now). It's released from rest at time [tex]t = 0[/tex] with the length of one side (The right side, for the heck of it) being length [tex]y[/tex]0. I want to find the equation of motion of the string.


    2. Relevant equations
    I like using Energy methods, so I'll go with that here:
    [tex]E = T + V[/tex]
    [tex]T = 1/2 m \dot{y}^2[/tex]
    [tex]V = m g y_{cm}[/tex]


    3. The attempt at a solution
    Find the total Kinetic Energy of the system:
    [tex]T = T_{right} + T_{left} = 1/2y\lambda\dot{y}^2 + 1/2(L-y)\lambda\dot{y}^2 = 1/2M\dot{y}^2[/tex]
    And the total potential:
    [tex]V = V_{right} + V_{left} = -\frac{\lambda g}{2} y^2 - \frac{\lambda g}{2}(L-y)^2 = -\frac{\lambda g}{2}(L^2 - 2Ly + 2y^2)[/tex]

    So the total energy is constant and has the value
    [tex]E = T + V = 1/2M\dot{y}^2 - \frac{\lambda g}{2}(L^2 - 2Ly + 2y^2)[/tex]
    So lets plug in our initial conditions to find the equation of motion at any time!
    At [tex]t=0, y=y_0, \dot{y}=0[/tex]:
    [tex]-\frac{\lambda g}{2}(L^2 - 2Ly_0 + 2y_0^2) = 1/2M\dot{y}^2 - \frac{\lambda g}{2}(L^2 - 2Ly + 2y^2)[/tex]
    Which simplifies to:
    [tex]2g(Ly_0-y_0^2) = L\dot{y}^2 + 2g(Ly-y^2)[/tex]

    This is a differential equation of the form [tex]\dot{y}^2 + \alpha y^2 + \beta y = C[/tex].

    I have no idea how to solve this final differential equation. Is it possible that if you choose some clever variable to substitue in you can turn it into an easily solvable linear equation? Or did I mess up somewhere in my Energy equations?

    Thanks in advance!
     
  2. jcsd
  3. Mar 24, 2009 #2

    turin

    User Avatar
    Homework Helper

    First of all, good luck and I hope that you have fun with this. I'm unfamiliar with your terminology. Are you just now entering grad school, or just now entering college?

    Good. That is always my first way to attempt a problem, as well. I still remember the day that I realized, under the guidance a faculty member, that energy is an extremely powerful tool to solve physics problems.

    While I don't have time to check the details of your calculation here (it seems correct), you may consider using the Hamiltonian method. You start with the Lagrangian, L=T-V, and then obtain the Hamiltonian, H=T+V, which is basically E=T+V fancied up a bit. The disadvantage is that you get twice as many variables. The advantage is that you get differential equations that are typically much more approachable.

    I don't have time right now to check if you messed up somewhere, but I will say that the way that you approached the problem typically does lead to a scary looking differential equation. Assuming that you have obtained the correct equations, then you could always take the square root. Then you will end up with a first order ordinary differential equation, and "everyone" knows how to solve those. ;) (in principle, you can solve a first order ordinary differential equation by "brute force".)
     
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