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Beads on a thread - What stops the acceleration?

  1. Nov 9, 2015 #1
    I would like to ask for your help with understanding a few things connected to the following problem:

    1. The problem statement, all variables and given/known data

    There is an (infinitly) long thread, on which small beads can move without friction. The beads with mass m are lined up on the thread with a constant distance d between them. Whe start to push one bead with the constant force F and continue to do so all the time. The velocity will eventually reach a final, constant value. What is this value of the collisions between the beads are completely inelastic?

    2. Relevant equations
    F * Δt = Δp
    v * Δt = Δs

    3. The solution
    When the first bead is accelerated, it will collide into the next one. The beads stuck together and collide in the third one, then the fourth one and so on. If the final velocity is v, the number of new beads attached to the moving "chain" in a time Δt is: n = s/d = v*Δt/d
    Therefore the change in mass of the moving "chain" is: Δm = m*n = m*v*Δt/d
    Because the velocity is now constant, the change in momentum is only due to the change in mass:
    F*Δt = Δp = v*Δm = v2*m*Δt/d
    Thus: v = (F*d/m)^(1/2)

    Now, there is nothing wrong with the solution. The thing I don't understant is exactly why a constant final velocity is reached. I know that when the total mass approaches infinity, the acceleration will become zero (a=F/m), but is there anything else which causes the acceleration to stop? I feel like the condiction of infinit number of beads is not directly included in the solution. I do therefore wonder: Is there any outer force which counteracts the pushing force F?

    I have also thught about an energy aspect. In this case, the work done by F on a distance d should equal the kinetic energy of a new bead, shouldn't it? However, this yields: v = (2F*d/m)^(1/2) , which is one factor √2 to much compared to the answer obtained in the solution above. Why?

    Thank you very much!

    Attached Files:

  2. jcsd
  3. Nov 9, 2015 #2


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    Nope. The beads that are moving already are not accelerated further. every ##\Delta p## involves only one extra bead !

    Last question: energy is lost because the collisions are inelastic.
  4. Nov 9, 2015 #3
    What an interesting and totally unnatural problem to solve!
    Remember that work is force times distance. The distance between collisions is fixed, so the average work between each collision is fixed. If you assume that the velocity does asymptote to a value (averaged over the collision period), then the average work is the average kinetic energy of a bead.

    Edit: Oops, this is wrong.
    Last edited: Nov 9, 2015
  5. Nov 9, 2015 #4
    How can we be sure that the movig beads do not accelerate futher?

    Of course! Thank you so much! This was a very stupid question from me.
  6. Nov 9, 2015 #5
    Thank you so much for your help Sir! The only thing I wonder: Why can we assume that the velocity asymptot to a value?
  7. Nov 9, 2015 #6


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    As BvU pointed out, collisions are inelastic here. You cannot assume work conservation.
  8. Nov 9, 2015 #7


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    You don't need to assume it. It will drop out of the equations.
  9. Nov 11, 2015 #8
    From which equation Sir? This is the thing I can't really see.
  10. Nov 11, 2015 #9


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    We could treat this as a smooth process, but it might be clearer if we treat it as discrete.
    Suppose we have just picked up the Nth bead and they're all moving at speed vN. What will the speed be after picking up the next bead?
  11. Nov 11, 2015 #10
  12. Nov 12, 2015 #11
    Well, if we suppose that a work W = F*d*k is done between the beads, where k is the efficiency (0<k<1), energy conservation gives:
    Fdk+vN2Nm/2 = vN+12(N+1)m/2
    which yields:
    vN+1 = ((2Fdk/(N+1)m)+(vN2N/m))^(1/2)
  13. Nov 12, 2015 #12


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    You are told the collisions are completely inelastic. What is conserved?
  14. Nov 12, 2015 #13
    F dt = M dV + V dM
    Consider a glob of putty thrown at a "fixed" wall and sticking to the wall..
    Then the impulse on the wall is F dt = V dM because the wall does not move.
    Then you get the term V^2 = F d / M where M = L m / d since dM / dt = m V / d.
    So the accumulated beads behave as a fixed wall which is being moved by a force F.
    Also, this holds at all times after t = 0, and the impulse as each bead is added continually corresponds to the force F.
    So one could accumulate an unlimited mass of beads with the application of force F at speed V.
  15. Nov 12, 2015 #14
    Momentum is conserved! Which means that: ## V_N *Nm = V_{N+1} *(N+1)m##
    so ## V_{N+1} = \frac{V_N *N}{N+1}##
    What should be done next? :)
  16. Nov 12, 2015 #15
    Thank you for your reply! I like the example with the wall, but exactly how do I relate it to the bead problem? I can't see this clearly because the are not still after a collision, but go on moving.
    I do also have some problems following your mathematics, would you please mind explaining it a bit more? :)
    I am also slightly unsure about F dt = M dV + V dM , because my textbook says d(mv)/(dt) ≠ dm/dt + dv/dt and that this can be motivated with "symmetry" (although this is not done).
  17. Nov 12, 2015 #16
    The accumulated string of beads (wall) is moving at constant speed so the problem takes dV / dt = 0.
    The beads evidently are taken to be so close together so as to simulate a constant accumulation of mass.
    The applied force to the wall then just equals the impulse of the beads striking the wall so the net force on the wall is zero.
    So the accumulated mass of beads moves at constant speed from t = zero to any indefinite time.
    I'm not familiar with the example to which your textbook refers in which dP/dt would not equal M dV/dt + V dM/dt.
  18. Nov 12, 2015 #17
    I know it is assumed in the problem, but how does one motivate that the speed will become constant?

    What physical consequences are there of a constant accumulation of mass? (will mass not accumulation constantly even if the speed is not constant?) In there is an impuls on the wall, how can the force be zero?
  19. Nov 12, 2015 #18


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    No, you've forgotten about the force. The velocity will increase before it strikes the next bead.

    Now, there is a much easier way, but I went this route because you asked how we know it will converge to a finite speed.
    If you assume a steady state is reached, you can work in the reference frame of the moving mass. The force F applied one side is to be balanced by the steady stream of impacts from beads on the other side. You can just write down the solution.
    Last edited: Nov 12, 2015
  20. Nov 12, 2015 #19


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    That equation is popular but not really valid, and many rail against it.
    Logically, the dm/dt term is saying that mass is magically created or destroyed. In reality, it is coming from or going to some other system, so may bring with it or take with it some momentum. The equation can work, but only if the added mass had no initial momentum (or departing mass takes none with it).
  21. Nov 14, 2015 #20
    so should the equation be:
    ## (V_N + a* \Delta t) Nm = V_{N+1} *(N+1)m ##
    ? where ##a = F/m## and ##\Delta t = \sqrt 2d/a ## thus:
    ## (V_N + \sqrt \frac{2Fd}{m}) Nm = V_{N+1} *(N+1)m ## ?

    If it is assumed that a steady state is reached (but I do still not see exactly how we can assume this), then the impuls of the new bead must be balanced by the force pushing from the other side. In the frame of reference of the moving mass, the mass is moving towards it with velocity v. Thus:
    ## F * \Delta t = \Delta p = mv ##
    The time is the time between two collisons, thus ## v = \sqrt \frac{Fd}{m} ## , which is the correct answer.
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