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Homework Help: A rope falling off an inclined plane

  1. Jan 20, 2017 #1
    1. The problem statement, all variables and given/known data
    A rope of length L is falling off an incline. Part of the rope is still on top of the horizontal surface. There is no friction between the incline and the rope. The incline is at an angle theta above the horizontal.

    a) What is the magnitude of the acceleration of the rope, when the part of the rope on the incline is x (i.e., the part left on the horizontal surface has a length of L-x)

    b) What is the speed of the rope at this moment?

    c) What is the value of x, when the acceleration reaches maximum?

    2. Relevant equations
    ∂L/∂x - d/dt(∂L/∂v) = 0 where v = x-dot
    a = dv/dt = (dv/dx)*(dx/dt)

    3. The attempt at a solution

    So I worked out this problem but I am unsure of my result, specifically my result for the potential energy in the lagrangian.

    L = ½mv2 - U(x) = ½λLv2 + (λx)(g)(½x sin(θ))

    where λ=m/L is the linear mass density

    the Euler-Lagrange equation then gives

    1.) a = (x/L)g sin(θ)

    assuming my acceleration for # 1 is correct, I used the identity a = v*(dv/dx) to find the velocity

    a*dx = v*dv → ∫(x/L)g sin(θ)dx = ∫vdv

    2.) v = √[v20 + 1/L (x2 - x20)g sin(θ) ]

    then # 3 would be x = L
    Last edited: Jan 20, 2017
  2. jcsd
  3. Jan 20, 2017 #2


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    Why ? You take the horizontal plane as a reference, which you are allowed to do.
  4. Jan 20, 2017 #3


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    All looks right. I get the same for 2 using conservation of work. In going from x0 on the slope to x on the slope, a length x-x0 has descended from the horizontal surface to be the lowest part of the rope. Its mass centre has descended (x+x0)/2.
  5. Jan 20, 2017 #4
    The part that sort of bothers me is the "h" part in mgh for U(x). For continuous bodies I am used to having to integrate over the whole part that contributes potential energy instead of just taking the position of the center of mass.
  6. Jan 20, 2017 #5


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    If an element of mass dm starts at height yi and finishes at height yf, ΔGPE=∫Δyg.dm=g(∫yfdm-∫yidm).
    ∫yfdm is the final height of the mass centre, by definition.
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