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Mechanics problem

  1. May 16, 2005 #1

    z0r

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    I am having difficulty understanding how to solve this mechanics problem...

    A rope rests in a bundle on a table with a small hole in it. One end of the rope slides through the hole and gravity steadily pulls on it until the total length of the rope slides through the hole. The length of the rope hanging out of the hole can be [tex] x [/tex]. What is the velocity of the rope when it has all slid through the hole? Ignore all friction.

    Let's say the rope has mass [tex] M [/tex] and length [tex] L [/tex]. Then we say the linear density is [tex] \mu = \frac{M}{L} [/tex]. The force on the rope is then [tex] F = \mu x g [/tex], where [tex] g [/tex] is just gravity. Then I need to figure out the momentum, take the time derivative, and set it equal to the force. Then we should have a simple differential equation to solve and integrate from zero to the length [tex] L [/tex] to find the final velocity. Simple enough, but I don't know how the momentum is defined here...

    Both the mass and velocity of the length of rope hanging out of the hole is changing, so then would we not say that momentum [tex] P = \dot{m}x + m\dot{x} [/tex], with [tex] m = \mu x [/tex]? That yields something that doesn't look correct. Nor does just [tex] P = \mu x \dot{x} [/tex].

    Thanks in advance for any help.
     
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  3. May 16, 2005 #2

    OlderDan

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    Since gravity is the only force doing any work in this problem, it can be approached from the point of view of energy conservation. If you are required to do it another way, then looking at it in terms of force resulting in a change of momentum is possible. Energy conservation is easier. How much has the potential energy changed when the rope has just slid through the hole?
     
  4. May 16, 2005 #3

    Doc Al

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    I agree... that's not correct.
    Actually, that looks right to me. It's just mass times speed.

    Note: I don't think you can use conservation of energy, since the pile of rope gets jerked from rest as it moves through the hole. Some of the energy transforms to heat.
     
  5. May 16, 2005 #4

    OlderDan

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    "Rope" problems of this sort always need some simplification. When they say the rope "rests in a bundle" they are telling you in a roundabout way to ignore any motion of the rope above the table. Assume that the only part of the rope that is moving is the part below the table. That part has a mass, and it has a velocity common to all parts of the rope below the table, both of which are changing with time as you have observed. The momentum is all in the vertical direction, and it is the product of the moving mass times the velocity at any time.

    [tex] P = \mu x \dot{x} [/tex]

    is the momentum
     
  6. May 16, 2005 #5

    OlderDan

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    I am having second thoughts about the energy. I'm not yet convinced, but I am leaning your way at the moment.

    If the rope were stretched out on the table, then energy conservation would be appropriate, but then all of the rope would have the same speed all the time, and that is clearly not the case in this problem.
     
  7. May 16, 2005 #6

    Doc Al

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    Right... the stretched out rope problem is different. No jerking... energy is conserved.
     
  8. May 16, 2005 #7

    z0r

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    Very well. Let us take the time derivative of the momentum [tex] P = \mu x \dot{x} [/tex] and set it equal to the force [tex] F = \mu x g [/tex].

    We get:

    [tex] \mu \dot{x}^2 + \mu x \ddot{x} = \mu x g [/tex]. Right off the bat this makes no sense to me. When the rope has completely left the table, the acceleration will be [tex] g [/tex], meaning the velocity term [tex] \dot{x}^2 [/tex] will be zero?

    Secondly, the resulting differential equation doesn't look nice. I know we can use [tex] \ddot{x} = \dot{x} \frac{d \dot{x}}{dx} [/tex]. I think I'm making it more difficult than it needs to be...
     
  9. May 16, 2005 #8

    Doc Al

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    That equation only applies while there is rope remaining on the table. It doesn't apply once the rope leaves contact. Note that the term [tex] \mu \dot{x}^2[/tex] comes from the changing mass of the hanging portion of the rope. Once the rope leaves the table, the mass no longer changes and the term does become zero. (Of course, that does not mean that the velocity becomes zero.)
     
  10. May 16, 2005 #9

    z0r

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    Hmm... so the differential equation to solve is:

    [tex] \dot{x}^2 + x \ddot{x} = xg [/tex] or [tex] xg = \dot{x}^2 + x\dot{x} \frac{d\dot{x}}{dx}[/tex].

    It doesn't appear we can easily isolate the two variables. The goal would be to get an expression for [tex] x [/tex] and one for [tex] \dot{x} [/tex] and integrate the [tex] x [/tex] from L to 0 to obtain an expression for the final velocity. I don't know of a mathematical technique to do that here...

    Again, thanks in advance.
     
  11. May 17, 2005 #10

    Doc Al

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    Assume that x is some function of t. See if you can guess the form it must have.
     
  12. May 19, 2005 #11

    z0r

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    I went to discuss the problem with my professor and he told me that indeed the form [tex] xg = \dot{x}^2 + x\dot{x} \frac{d\dot{x}}{dx}[/tex] is correct, so thanks to those who set me straight.

    The method he used for solving it is assuming that the velocity is some function of x...

    [tex]\dot{x} = v = a x^n [/tex]

    The constant in front and the exponent n are to be determined. Then one only need plug and chug, and set coefficients of the x terms equal to g and eliminate the others, and one obtains the correct expression for the final velocity.
     
  13. May 19, 2005 #12

    Doc Al

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    Cool. You can also assume that x is some function of t (which is how I was thinking); if you try [tex]x = a t^n [/tex] you will easily find a and n that satifies your equation.
     
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