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Massive cable and massive pulley

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data
    A cable of length L and mass density λ is rolled on a pulley of mass m and radius r, with its tip hanging. The cable starts to unravel and fall due to gravity; the system starts from rest. Assume the cable has negligible thickness and rolls without slipping on the pulley, whereas the pulley spins without friction.
    Solve the equation of motion for the cable, denoting its unraveled length as x(t).

    2. Relevant equations
    τ = F⋅r = dL/dt
    L = Iω
    Moment of inertia for a solid disk: I=1/2mr2
    Moment of inertia for a ring: I=MR2

    3. The attempt at a solution
    The torque is given by τ=λgxr
    The angular momentum is given by L=(Ipulley+Icable
    where Ipulley=1/2mr2 and Icable=λ(L-x)r2
    therefore L=(1/2mr2 + λ(L-x)r2
    ⇒ L=1/2mr2ω + λLr2ω - λxr2ω
    Due to non-slipping condition ω=x/r and ω=x⋅⋅/r
    So after differentiating L we get dL/dt=(1/2mr+λLr)x⋅⋅ - λrx⋅x - λrxx⋅⋅

    And this is where I'm stuck - I have no idea how to solve the differential equation after equating dL/dt with the torque, due to the dx/dt squared (sorry, couldn't figure out how to make the latex work, tried the manual but the previews didn't appear to work).
    I guess there supposed to be a more elegant expression so to make the differential equation more reasonable, but I can't seem to find it.

    Thanks everyone!
     
  2. jcsd
  3. Mar 1, 2015 #2

    TSny

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    Hi and welcome to PF!

    You have a good approach. Does the hanging part of the cable contribute to the angular momentum of the system?
     
  4. Mar 1, 2015 #3
    Hi, thanks!
    And yes, it applies a torque τ=λgxr.
     
  5. Mar 1, 2015 #4

    TSny

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    OK, you already had the torque expression correct. You just want to make sure that you have the total angular momentum expressed correctly.
     
  6. Mar 2, 2015 #5
    Yes! Got it!
    I've neglected the hanging part of the cable in the expression for the angular momentum.
    The correction gives L=1/2mr2ω + λLr2ω - λxr2ω + rMcablev
    L=1/2mr2ω + λLr2ω - λxr2ω + rλxx
    So after substituting ω=x/r the last two terms cancel and after differentiating we get a much nicer equation!
    (1/2mr+λLr)a-λgrx=0

    And from here on it's just a matter of plugging in the exponent and solving...

    Am I correct or have I missed anything else?

    Thank you very much, and also thanks for not giving it away entirely!
    Yoni
     
  7. Mar 2, 2015 #6

    TSny

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    Looks good. Nice work!
     
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