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Chain falling through hole

  1. Nov 22, 2013 #1
    1. The problem statement, all variables and given/known data
    One end of the chain falls through a hole in its support and pulls the remaining links after it in a steady flow. If the links which are initially at rest, acquire the velocity of the chain suddenly and without frictional resistance or interference from the support or from adjacent links. Choose the INCORRECT statement. (when x=0, then v=0) (length of chain is L and ##\rho## is the mass per unit length of the chain).

    A) the velocity v of the chain as a function of x is ##\sqrt{\frac{2gx}{3}}##.

    B) the acceleration of a of the falling chain as a function of x is ##\frac{g}{3}##.

    C) the energy Q lost from the system as the last link leaves the platform is ##\frac{\rho gL^2}{6}##.

    D) tension at the middle point of the falling chain is ##\frac{\rho gx}{3}##.

    2. Relevant equations

    3. The attempt at a solution
    I am completely lost on this one. I could only write the following (from Newton's second law on falling part of chain):

    $$\rho gx-T=\rho xa$$
    where T is the tension acting on the part just getting out of the support. How do I write more equations?

    Any help is appreciated. Thanks!

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  3. Nov 22, 2013 #2
    I would use conservation of energy here.
  4. Nov 22, 2013 #3
    At what instants do I use conservation of energy? When the x length is out and the other when a dx part comes out, right?
  5. Nov 22, 2013 #4
    With x out, consider the potential/kinetic energy balance.
  6. Nov 22, 2013 #5
    What does "potential/kinetic energy balance" mean? :confused:

    If you ask me to write down the energy when x length is out, this is what I get:

    $$-\rho gx \frac{x}{2}+\frac{1}{2}\rho xv^2$$
    where U=0 is at the support and v is the velocity of falling chain.
  7. Nov 22, 2013 #6
    Very well, but why did you not write the entire equation for conservation of energy? Do you think it is not conserved here?
  8. Nov 22, 2013 #7
    I did not understand what you meant by "potential/kinetic energy balance". I agree that energy is conserved.

    Initial energy of system is zero. Equating it with what I wrote above, I get


    which is wrong. The answer states D as the answer which means other options are correct.
  9. Nov 22, 2013 #8
    Hmm. It seems that with that assumption, more than one statement is incorrect. Yet I do not immediately see why energy would not be conserved here.
  10. Nov 22, 2013 #9
    I have the solutions but I don't want to look at them before I reach the answers. Should I open the solution booklet and post the solution if you find something wrong here?
  11. Nov 22, 2013 #10
    I think it is because links are supposed to acquire the velocity of the chain suddenly. That means there is an impulse accelerating them down, which does non-zero work.

    Then conservation of energy is not usable here.
  12. Nov 22, 2013 #11
    So how should I proceed now?
  13. Nov 22, 2013 #12
    You could try treating that as a variable mass system, where mass is acquired, plus everything is accelerated by gravity.
  14. Nov 22, 2013 #13
    At any instant, momentum of system is ##\rho xv##.
    $$\Rightarrow \rho gx=\rho \frac{d(xv)}{dt}$$
    $$\Rightarrow gx=v\frac{dx}{dt}+x\frac{dv}{dt}$$
    I can substitute ##dx/dt=v##, but then how do I solve the D.E, I have three variables. :confused:
  15. Nov 22, 2013 #14

    D H

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    The question asks for things such as velocity v of the chain as a function of x. Time is not mentioned. This suggests you want to find velocity and acceleration as functions of x rather than as functions of t.
  16. Nov 22, 2013 #15
    I am confused, can you please give a few more hints?

    The only thing I can think of is this:

    $$gx\,\, dt=d(xv)$$
    I am thinking of integrating the above and replacing ##\int_0^t xdt## with v. Does that sound good?

    I am confused at one more point. I wrote
    $$\rho gx=\rho\frac{d(xv)}{dt}$$

    Shouldn't that be
    $$\rho gx-T=\rho \frac{d(xv)}{dt}$$
  17. Nov 22, 2013 #16

    D H

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    You need to get rid of the dependence on time. You're making it worse!

    Hint: By the chain rule, ##\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}##.
  18. Nov 22, 2013 #17
    Ah yes, I forgot about that. Can I blame that on my bad health? :redface:

    So we have the equation,
    $$gx=v^2+xv\frac{dv}{dx} \Rightarrow gx=v\left(v+x\frac{dv}{dx}\right)=v\frac{d(xv)}{dx}$$
    $$\Rightarrow gxdx=vd(vx)$$
    Multiplying by x on both sides,
    $$\Rightarrow gx^2dx=vxd(xv) \Rightarrow \int_0^xgx^2dx=\int_0^{xv} d(xv)$$

    Does this look correct?

    Can you please clear my doubt I posted in my previous reply?
  19. Nov 22, 2013 #18


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    Yes. That looks good to me (except for a typo in the last integral where you left out part of the integrand)

    ##T## represents the tension at what point?
  20. Nov 22, 2013 #19


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    Yes, that looks fine.
    This is very similar to https://www.physicsforums.com/showthread.php?t=669505, which you posted last Feb. It caused a schism between those who insisted work would be conserved, and those (including me) who declared it wasn't.
    When you're given the form of the answer (##v=\sqrt{\frac{2gx}{3}}##), you can use shortcuts. Clearly that answer means x = Atn, some A and n. So just plug that into your ODE and see what A and n satisfy it.
  21. Nov 22, 2013 #20

    D H

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    No, it shouldn't.

    The reason is that tension is an internal force. By Newton's third law, the net internal force acting on an object is zero. It's only the external forces that count with regard to Newton's second law.
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