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Falling Chain (3.11)

  1. Mar 5, 2014 #1

    Radarithm

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

    A chain with length ℓ is held stretched out on a frictionless horizontal table, with a length y0 hanging down through a hole in the table. The chain is released. As a function of time, find the length that hangs down through the hole (don't bother with t after the chain loses contact with the table). Also, find the speed of the chain right when it loses contact with the table.

    2. Relevant equations

    [tex]F_y=m\ddot{y}[/tex]
    [tex]y_0\ge y\ge l[/tex]

    3. The attempt at a solution

    Conceptually, more of the chain must be through the hole so that y-naught is greater than y (initially). To cause an acceleration to the left (or towards the hole; however you picture the scenario), the tension T must be equal to mg (correct?). What I'm having trouble with is setting up an appropriate differential equation; should I just do the old: [tex]\ddot{y}=-g[/tex] because this would (should) hold true if I claim that the tension T is equal to mg. This means that something fundamental is getting past me, I just don't know what. Also this doesn't have to do with the question, but is K&K harder than the book I already have (Morin)?
     
    Last edited: Mar 5, 2014
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  3. Mar 5, 2014 #2

    BvU

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    The m in ##\ddot y## is different from the m in ## mg##. Make a drawing.
     
  4. Mar 5, 2014 #3

    Radarithm

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    http://postimg.org/image/vprzi0y6z/
    The FBDiagram on the left is for a piece of the chain that is present on top of the table, whereas the one on the right is for a piece of the chain that is falling through the hole in the middle.
     
  5. Mar 5, 2014 #4

    BvU

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    frictionless, so T isn't needed.
    My point is that mg is not constant because that m = y x m/L
     
  6. Mar 5, 2014 #5

    Radarithm

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    Would you mind proving how m = y x m/L? Is that a cross product?
     
  7. Mar 5, 2014 #6

    BvU

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    This is one of these totally artificial exercises that confuse bright students. Just imagine what is needed to make the chain take a 90 degree turn with 0 radius at the rim of the hole. Yuch.

    Never mind. Questions about your FBDs: On the left I see d M g. What is that ? T is the accelerating tension I suppose.
    If there is an FBD for a (the ?) piece on top of the table, why is there no FBD for a (the) piece hanging vertically ?

    On the right I see rather nothing. T is gone ?
     
  8. Mar 5, 2014 #7

    BvU

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    No. If the whole chain has length L, mass m, then a piece with length y has mass ## y\ m / L ##
     
  9. Mar 5, 2014 #8

    Radarithm

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    This is probably incorrect (I'm sure it is) but here I go:
    [tex]dm\frac{dv}{dt}=-dm_2g[/tex]
    [tex]dmdv=-dm_2gdt[/tex]
    edit: Nevermind the diff eqs, that was before I saw your replies.
    Can you at least give me a hint as to how I can relate the length to all of this? I can't seem to set up a diff eq.
     
  10. Mar 5, 2014 #9

    Radarithm

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    In the left FBD, I now know that:
    [tex]gdm=\frac{ym}{dL}[/tex]
    Forgot to include the tension in the FBD on the right, and it is the one for the vertical piece.
     
  11. Mar 5, 2014 #10

    Radarithm

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    Ok so because:
    [tex]m\frac{d^2y}{dt^2}=\frac{ym}{dL}g-T[/tex]
    for the vertical piece of the chain, then:
    [tex]T=m\left(\frac{d^2y}{dt^2}-\frac{gy}{dL}\right)[/tex]

    Correct?
    I don't know how to relate the diff eqs for the vertical and horizontal pieces of chain in order to find L(t).

    I do know that the limits of integration (for the integral that I have to calculate) are [itex]y_0[/itex] and [itex]l[/itex]
     
  12. Mar 5, 2014 #11

    Radarithm

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    Ok so if what I got for T is correct then:
    [tex]m\frac{d^2y}{dt^2}-m\frac{gy}{dL}=m\frac{d^2x}{dt^2}[/tex]
    [tex]\frac{d^2y}{dt^2}-\frac{gy}{dL}=\frac{d^2x}{dt^2}[/tex]

    I guess the only problem I have now is with setting up the diff eq. If I am right, does the [itex]dL[/itex] have to be on the left side with everything else on the right? (there are multiple dt's; I don't suppose that this is problematic, assuming that I can group them up together if they are similar).
     
  13. Mar 5, 2014 #12

    BvU

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    This is going at a posts per few minutes. I can't follow. What is d in dL and what is x ?
    You can't write [itex]gdm=\frac{ym}{dL}[/itex] if you take a differential in the lhs, you also need a d in the numerator on the rhs; it can't end up in the denominator.

    We know the chain remains stretched. So vertical piece, length y(t) means horizontal piece length L - y(t).

    Vertical: F = y(t) m/L g - T
    Horizontal: F = T

    Someone else will have to take over; I have to go to a training.
     
  14. Mar 5, 2014 #13

    BvU

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    You may need to go back to F = dp/dt = dm/dt v + m dv/dt if you write a differential equation for only one of the sections (because for a section dm/dt is not 0). Either that , or only write the differential equation for the entire chain.
     
  15. Mar 5, 2014 #14
    Well, the accelerations for x and y must be equal -- otherwise the chain would stretch out or it would create extra slack. How about something like this?
    $$\frac{d^{2}y}{dt^{2}} = \frac{g \cdot \lambda y}{M}$$ Where [itex]\lambda[/itex] is the linear mass density -- [itex]\frac{M}{L}[/itex], which reduces to: $$\frac{d^{2}y}{dt^{2}} = \frac{gy}{L}$$

    This may be totally off though -- I'll think more about it.
     
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