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Energy problem - Chain on low friction surface.

  1. Dec 12, 2006 #1
    1. The problem statement, all variables and given/known data
    Consider a chain of total mass M = 5kg that is coiled into a tight ball on a low friction floor(like an ideal ice rink). You pull on a link at one end with a constant horizontal force F=20 N, just until it reaches its full length L = 3 m. What is the final velocity of the chain? (Hint: consider the point particle system. Make reasonable assumptions, like treating the coiled ball as if it has negligible size.)

    2. Relevant equations
    Im pretty sure this involved the Energy Principle so delta_E = Wext + Q

    the enlongated chain is 3 meters, with the center of mass having moved 1.5 m

    I wish I had more to go on, but I just need a bump in the right direction.

    3. The attempt at a solution

    ah, well thats kindof in section 2. although its only a feable attempt to solve it, its as far as I can get right now.
  2. jcsd
  3. Dec 13, 2006 #2


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    If the chain were stretched out, energy principles would work fine, but how are you going to find Q in this problem? (Although the equation you wrote is not quite what you would need here.) I think you will do better looking at the change in momentum of the system in some small amount of time Δt. There are two contributions to this change. One contribution is the gradual change in velocity of the mass that is already moving. The other is the abrupt change in velocity of the next little piece of the chain of mass Δm from zero to the current velocity of the moving part of the chain.

    You could look at the motion of the center of mass, but although it is relatively easy to find the position of the center of mass as a function of the postion of the end of the chain, it is not quite so easy to find it as a function of time. There is the derivative chain rule that could help if you choose this approach.
  4. Dec 13, 2006 #3
    Thank you for your reply. Here we are to consider the ball of chain as a point mass.

    W = F *d (here) so W = 20(1.5) = 30 J

    Im guessing that the Energy when enlongated is nearly all kinetic so
    30 = 1/2 m v^2 = (.5)(5)(v^2)

    v^2 = 12

    v ~ 3.46 m/s

    Which is the correct answer. Thanks for the help!
  5. Dec 13, 2006 #4


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    It is not a point mass if part of the chain starts to move before all of it moves. If you were treating it as a point mass, all parts of the chain would have the same speed at all times.

    How do you justify W = 20N(1.5m)? The point where you apply the force moves 3m, not 1.5m. The work you would have to do while applying this force is twice what you calculated. Where did the energy go?
  6. Dec 13, 2006 #5
    This problem was discussed in lecture and I think was a poor problem. The way I did it was how it was intended. But you are correct in the apparent loss of energy which was due to friction?!? Thats what the prof said.... then he discussed how the microscopic model of friction where "peaks" slide across each other, resulting in a shorter distance. Thats what I got from the explanation but it doesnt all make sense to me.
  7. Dec 13, 2006 #6


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    There was no friction included in the problem. It was done as if the floor were perfectly frictionless. What you have in this problem can be considered to be a series of inelastic collisions between mass in motion and mass at rest. Each time a bit of mass dm starts to move, the momentum change of the system is exactly equivalent to an inelastic collision between that bit of mass (dm) initially at rest and the part of the chain already in motion while accelerataing under the action of the applied force. Energy is not conserved in this process. A lot of the work done gets converted into thermal energy of the chain.

    A justification for your calculation follows:

    x is the position of the end of the chain where F is applied, with x = 0 at the coiled end. When an external force is applied to a system of particles, the CM of the system accelerates obeying Newton's second law.

    [tex] x_{CM} = \frac{1}{M}\frac{x}{2}\frac{{Mx}}{L} = \frac{{x^2 }}{{2L}} [/tex]

    but that is not explicitly needed to do the problem

    [tex] F_{ext} = Ma_{CM} = M\frac{{dv_{CM} }}{{dt}} = M\frac{{dv_{CM} }}{{dx_{CM} }}\frac{{dx_{CM} }}{{dt}} = Mv_{CM} \frac{{dv_{CM} }}{{dx_{CM} }} [/tex]

    where the chain rule has been used so the variables can be separated as follows

    [tex] F_{ext} dx_{CM} = Mv_{CM} dv_{CM} [/tex]

    [tex] F_{ext} \int_0^{L/2} {dx_{CM} } = M\int_0^V {v_{CM} dv_{CM} } [/tex]

    [tex] F_{ext} \frac{L}{2} = M\frac{{V^2 }}{2} [/tex]

    [tex] F_{ext} L = MV^2 [/tex]

    The work done is twice the final kinetic energy of the chain.
    Last edited: Dec 13, 2006
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