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Lagrange - A rope sliding off a table - quite difficult?

  1. Dec 17, 2008 #1
    Hi people, here's my problem:
    A uniform, flexible rope of length D, mass M, hangs off a frictionless table-top of height greater than D. The length of the section of rope hanging off is A. Gravity accelerates the part of the rope that is hanging off, so the length of the hanging part increases - A is a function of time t.

    This system is entirely characterised by a single dimensionless coordinate, f(t) = A(t)/D, which is the fraction of the rope that is hanging over, which also is a function of time. Only consider times so that A(t) </= D, or in other words, f(t) </= 1.

    I have to find the kinetic and potential energy of the rope as a function of df/dt and a function of f respectively. For the potential energy V I have to normalise V so that it is zero for f=0, ie. zero when the whole length of the rope is on the table.

    As a hint for the potential I am told to consider just the centre of mass of the part hanging off the edge.

    Here's my solution attempt:

    a) Kinetic energy T = (1/2)mv2.

    here, the velocity of the rope is the rate of change of A with time: v = dA/dt = d/dt[Df(t)]
    and the mass is M here, so for my kinetic energy, i said:

    T = (1/2)MD2(df/dt)2.

    I don't think there's anything wrong with what I've done here but please point it out if I've made a mistake.

    b) The gravitational field can be considered uniform in this situation, so potential energy is given by V = mgh.

    The only part of the rope taking part in the gravitational acceleration is the part hanging off the end, and since the mass is uniformly spread throughout, then the fraction of mass M involved here is the same as the fraction of the length, ie.

    m = Mf(t).

    The centre of mass of the hanging part is half way up it, which is A(t)/2 = Df(t)/2.

    this corresponds to a negative h.

    So, I said V = Mf(t)g.(-Df(t)/2) = -(1/2)MgD(f(t))2

    however, when i put this into the Lagrangian, L = T - V and then differentiate L by f(t) and by d/dt[f(t)] = f(dot)

    and I put this into the lagrangian equation of motion, ie.

    d/dt dL/df(dot) = dL/df(t)

    i find that the left hand site does not equal the right hand side, so i must be making a mistake somewhere.

    I can't see what I am doing wrong though. Can anybody help me out please, I'd really appreciate it.

  2. jcsd
  3. Dec 17, 2008 #2
    Your values for T and U look good. The equation of motion is given by

    [tex] \frac{\partial L}{\partial f}=\frac{d}{dt}\frac{\partial L}{\partial \left(df/dt\right)}} [/tex]

    The LHS and RHS do not need to be equal. By using the Lagrangian, you obtain a differential equation for the equation of motion. You should find something of the form:

    [tex] \frac{d^{2}f}{dt^{2}} = Cf [/tex]

    where C is a constant.
  4. Dec 17, 2008 #3
    thanks again buffordboy, you keep popping up just when i need you :biggrin:
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