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Legrangian Eqn. of Motion

  1. Sep 1, 2007 #1
    Two blocks, each of mass M, are connected by an extensionless, uniform string of length l. One block is placed on a smooth horizontal surface, and the other block hangs over the side, the string passing over a frictionless pulley. Describe the motion of the system when the string has a mass m.

    By Hamilton's principle
    [tex]L = T - U[/tex]

    the kinetic energies will be
    [tex]T = 1/2 m \dot{x}^2 + 1/2 m \dot{y}^2[/tex]

    and if the potential is defined to be zero at the horizontal, the potential will be
    [tex]U = -Mgy + U_{string}[/tex]

    This is the part I need a quick help on. The x block has a zero potential because it stays along the horizontal where the zero potential is defined, and the hanging block will have a potential of -Mgy, and I know that the mass of the string contributing to the potential will increase until finally it reaches as the string moves down. So I was thinking that

    [tex]U_{string} = -\frac{m}{t}*g*y[/tex]

    That gives the mass per unit time for a given length y, which would also be

    [tex]U_{string} = -m g \dot{y}[/tex]

    But units don't work out correctly unless I divide U_string by t, which would create a discontinuity and not make any sense. I don't know why I am having so much trouble with such a simple prospect.
     
  2. jcsd
  3. Sep 2, 2007 #2
    Why did you use (m/t)? You should find the center of mass of the part of the string which hangs below the table. If the total mass is m, then the hanging part is
    [tex]m'=m\frac{y}{l}[/tex]
    and mass center is in the middle (y/2). This information should give you the potential energy of the string.
     
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