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Mechanics rope of mass problem

  1. Oct 5, 2005 #1
    A uniform rope of mass 'i' per unit length is coiled on a table, one end s pulled straight up with constant speed Vo.

    a. find the force exerted on the end of the rope as a function of distance of the top of the rope above table (y).
    b. compare the power delivered to the rope with the rate of change of the ropes total mechanical energy.

    Part a seems to be pretty simple... F-mg = dp/dt, so F(y) = i*V^2+i*y*g.

    Now the problem is when i integrate F(y) with respect with y, I should get the total energy of the rope system. But i don't. The total energy = .5mv^2 + mgh = .5(i*y)(v^2) + .5(i*y)(y*g), which clearly isn't the integral of F. So what's going on???
  2. jcsd
  3. Oct 5, 2005 #2


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    Not sure I follow what you did, but you basically have mass that is a function of time.
  4. Oct 5, 2005 #3
    That's why I set m(t) = y(t)*density*g.
  5. Oct 5, 2005 #4


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    Pls explain where [itex]iV^2[/itex] came from in part a.

    Here's my line of thinking.

    [tex]F = \dot p = ma[/tex]

    Since v is constant, that means a = 0, meaning the net force on the rope is zero. No?
    Last edited: Oct 5, 2005
  6. Oct 5, 2005 #5
    iv^2 is the change in momentum. This is because the change in mass is I*v, and velocity is a constant V, so dp/dt = Iv^2.
    Last edited: Oct 6, 2005
  7. Oct 6, 2005 #6
    Hehe simultaneous post.

    In any case, no. The momentum is not constant. The mass of rope moving at velocity V is increasing. Which implies that F is not zero.
  8. Oct 6, 2005 #7


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    Wait a minute asdf you mean

    [tex] \frac{dp}{dt} = \frac{dm}{dt}v [/tex]


    [tex] \frac{dm}{dy} = \frac{M}{L} [/tex]


    [tex] \frac{dp}{dt} = \frac{M}{L}v^2 [/tex]
  9. Oct 6, 2005 #8
    which is precisely what i posted by saying dp/dt = i*v^2, where i = M/L (if M = mass of entire rope, and L = length of entire rope), just as you have.
  10. Oct 6, 2005 #9
    Let me just be a bit more clear. The question can be simplified.

    Forget gravity, and just say the force is pulling on the string. Then as was proven above, F = dp/dt = M/L * v^2. This makes sense, we can integrate to show that the momentum at time t is the integral of the force up to t.

    But now, if we consider the kinetic energy = .5 m v ^2, where m = y*M/L, it breaks down. The kinetic energy as a function of y should be the integral F dy from 0 to y, but this is clearly not the case, as the integral of F dy is M/L * v^2 * y, which is close to KE(y) = .5 y*M/L v^2, but the .5 term is missing.

    Why is the .5 term missing? Am a making an incredibly stupid mistake, or is something really wrong?
    Last edited: Oct 6, 2005
  11. Oct 6, 2005 #10


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    I see it now.

    [tex]F = \dot p = ma[/tex] is only true when m is constant.

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