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Tension in the string holding a falling chain

  1. Dec 27, 2014 #1

    Nathanael

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    1. The problem statement, all variables and given/known data
    A uniform chain of length L and [constant] mass per unit length λ is suspended at one A end by an inextensible light string. The other end of the chain B is held at rest at level of end A of the chain. [See image.] Now if the end B of the chain is released under gravity, then find the tension in the string when the end B has fallen a distance y
    jGpnPiF.png

    2. Relevant equations
    F=dp/dt

    3. The attempt at a solution
    I was thinking the tension would be equal to the weight of the chain [itex]gλ(\frac{L}{2}+y)[/itex] plus the change in momentum per time of the chain that is transitioning from the right to left side (in the image).

    The momentum per time would be equal to the speed times the mass per time, and the mass per time would be the speed times the linear density, so I get:

    [itex]T=gλ(\frac{L}{2}+y)+λv^2[/itex]

    v can be written in terms of y as [itex]v=\sqrt{2gy}[/itex] so I get:

    [itex]T=gλ(\frac{L}{2}+3y)[/itex]

    But this is apparently incorrect
     
  2. jcsd
  3. Dec 27, 2014 #2

    mfb

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    If the right end went down by y, how long is the left side?
     
  4. Dec 27, 2014 #3

    Nathanael

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    L/2 + y
     
  5. Dec 27, 2014 #4

    mfb

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    If the right side is fully down (y=L) then the length of the left side is L/2+L > L? I see a problem with that assumption.
     
  6. Dec 27, 2014 #5

    Nathanael

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    Oh, wow... hahaha o0)

    The length is (L+y)/2 thanks
     
  7. Dec 27, 2014 #6

    Nathanael

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    [itex]T=\frac{gλ}{2}(L+5y)[/itex]

    But it still says this is the wrong answer :confused:
     
  8. Dec 27, 2014 #7

    mfb

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    You have to change it at two points.
     
  9. Dec 27, 2014 #8

    Nathanael

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    Yes I just realized this o0) The amount of mass transitioning per second would not be λv, it would be λv/2

    Thanks mfb :) and sorry for being careless
     
  10. Oct 26, 2016 #9
    The speed of the bottommost tip of the left string is ##\sqrt{2g(\frac{y}{2})}## .

    But what is the speed of the bottommost tip of the right string ? Is it ##\sqrt{2gy}## ?

    Thanks
     
    Last edited: Oct 26, 2016
  11. Oct 26, 2016 #10

    mfb

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    No, they should differ by a factor 2.

    This thread is 2 years old.
     
  12. Oct 26, 2016 #11
    I know :smile: . I hope this should not be a problem .

    I came across the same question and somehow worked and got the correct answer . But thinking more about the setup , wasn't sure about some things .

    You mean , the speed of the bottommost tip of the right string is ##2\sqrt{2g(\frac{y}{2})}## ? Could you explain why they should differ by a factor 2.

    Shouldn't the speed of the topmost point on right string and that of the bottommost point on right string be same ? Isn't speed of the topmost point ##\sqrt{2gy}## ?
     
  13. Oct 26, 2016 #12

    mfb

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    The length of the right side gets shorter, so if we track "the position of the currently lowest point" (identical for left and right side) and "the position of the upper end of the right side", their speeds differ (by the factor of 2 mentioned before).
     
  14. Oct 26, 2016 #13
    The thing I am not understanding is that , despite the right string being in free fall , how can the two ends move with different speeds ?
     
  15. Oct 26, 2016 #14

    mfb

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    The right string part gets shorter. This is only possible if its ends have different speed.
     
  16. Oct 26, 2016 #15
    I agree , but isn't right part of chain in free fall which means every part of the chain has same acceleration 'g' ?

    @TSny , @haruspex
     
  17. Oct 26, 2016 #16

    TSny

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    Yes.

    It's important to distinguish between the downward velocity of a point of the right side of the chain and the downward velocity of the lowest point of the string. Specifically, consider a red dot painted on one of the links on the right side. Also, consider a blue horizontal line that moves down so that it is always level with the lowest point of the chain. Note that the red dot will catch up to the blue line.

    upload_2016-10-26_20-5-17.png
     

    Attached Files:

    Last edited: Oct 26, 2016
  18. Oct 26, 2016 #17
    Thank you TSny .

    Could you please reply to post#13 and post#15 .
     
    Last edited: Oct 26, 2016
  19. Oct 26, 2016 #18
    If the two end points are moving with different speeds then clearly they cannot be having same acceleration , but then correct answer was obtained by treating the lowermost point to be in free fall .

    I am still not clear how the two ends are moving with different speeds despite having same acceleration 'g' .

    Please explain this inconsistency .
     
  20. Oct 26, 2016 #19

    TSny

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    I think you are right that all points on the right portion of the chain have the same free-fall acceleration, g. Once a link on the right side catches up to the blue line, it is abruptly brought to rest.
     
  21. Oct 26, 2016 #20

    haruspex

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    It depends what you mean by the speed of the bottom of the right hand section of chain.
    An individual link approaching the bottom moves at the same speed as the top of the section, gt, or √(2gy). The locus of "the bottom of the chain" descends at half that speed. Consequently the transference from right to left is also at the half speed.
    It's a bit like rolling contact of a wheel. The locus of the point of contact moves with the same velocity as the centre of the wheel, but the part of the wheel instantaneously at the point of contact is stationary.
     
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