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Speed of a rope sliding off a peg?

  1. Mar 9, 2014 #1
    A flexible rope of mass m and length L = L1 + L2 hangs over a frictionless peg. What is the speed of the rope when it just slides off the peg? (it is reasonable to assume that the rope is uniform and that the masses of each section 1 and 2 are concentrated at the midpoint)

    In the picture L1 is shorter than L2. I set the change in kinetic energy equal to the negative change in gravitational potential energy, with the final height (of the center of mass of the rope) being L/2 and the initial height being L2/2 + L1:

    (1/2)mv^2 = mg(L/2 - L2/2 - L1)
    The masses cancel and I got that v = √(gL1)

    I'm just asking about this because setting the heights right in terms of L, L1 and L2 is slightly confusing, and the question seems to ask for an actual speed but I think you can only find it in terms of the length.
     
  2. jcsd
  3. Mar 9, 2014 #2

    SammyS

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    You are correct to conclude that the velocity will depend on the length -- or rather more precisely, depend on the two lengths, L1 & L2.

    What is the mass of length L1 relative to the total mass, m ?

    What is the mass of length L2 relative to the total mass, m ?
     
  4. Mar 9, 2014 #3
    m1 = m(L1/L) and m2 = m(L2/L)
     
  5. Mar 9, 2014 #4

    SammyS

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    Thus the masses don't cancel.
     
  6. Mar 9, 2014 #5
    When I use the conservation of mechanical energy though with the relative heights: h2 = L/2 and h1 = L2/2 + L1 and thus a change in height of L/2 - (L2/2 + L1), and then substitute in L = L1 + L2, the L2's cancel and I'm just left with the velocity in terms of L1. Is this incorrect?
     
  7. Mar 9, 2014 #6

    SammyS

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    Δ(m1gh1) = 0 , assuming this is the change in P.E. for length 1, because, Δh for the section of rope is zero.

    However, for the change in P.E. for length 2, it is true that Δh = L1 , but the mass of the portion
    is not m.


    What is the mass involved when finding the final overall Kinetic Energy ?
     
  8. Mar 9, 2014 #7
    I'm a bit confused. Can't you relate the change in potential energy for the entire rope to the change in kinetic energy for the entire rope? So the mass cancels on both side and mass is no longer a factor? As long as the initial and final height for the change in potential energy is the location of the center of mass of the rope.
     
  9. Mar 9, 2014 #8

    SammyS

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    Show what you get for all those quantities.


    ... but as I see it now, no, the mass doesn't cancel.
     
  10. Mar 9, 2014 #9
    Yes you can .

    But the change in potential energy is with respect to the Center of Mass of the system.First calculate the COM of the rope at t=0 .Then at t =t when the rope leaves the peg .The difference in the heights Δh is what you are going to use in the change in potential energy mgΔh .

    I think you are not calculating the COM properly .For calculating the masses of the lengths ,work with the mass density λ=m/L .The masses on the two sides are unequal.
     
    Last edited: Mar 9, 2014
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