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Whirling Rope - Find the Tension at Position r

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data
    A uniform rope of mass M and length L is pivoted at on point and whirls with uniform velocity ω
    What is the tension at the rope at r from the pivot point? Neglect gravity.
    2. Relevant equations



    3. The attempt at a solution
    Centripetal force at r is given by [itex]-mr \omega^{2}\hat{r}[/itex]

    I don't know where to go from here, some pointers might be nice.
     
  2. jcsd
  3. Feb 11, 2013 #2
    The tension in the rope supplies the centripetal force for the part of the rope beyond r.
     
  4. Feb 11, 2013 #3
    Yes, but that doesn't really help me. I have no idea how to set this up mathematically.

    What's the first step? tension is equal to centripetal acceleration here. centripetal force is equal to -mrω^2

    We need to find m, at point r. If you divide the rope into infinitely many pieces, m will equal 0 for point r, so that's not the right way.
     
  5. Feb 11, 2013 #4

    haruspex

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    It means you have to add up (i.e. integrate) all the forces for all the dr elements of the rope beyond r.
     
  6. Feb 12, 2013 #5
    μ = M/L - mass per unit length

    ∴ m = μ (L-r)
     
  7. Feb 12, 2013 #6

    haruspex

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    How is that useful?
     
  8. Feb 12, 2013 #7
    The mass is the only unknown quantity.
     
  9. Feb 12, 2013 #8

    haruspex

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    The rope as a whole is mass M. You've given the formula for the mass beyond a given radius, but how are you proposing to turn that into an expression for the tension at a given radius?
     
  10. Feb 12, 2013 #9
    The tension in the rope supplies the centripetal force for the part of the rope beyond r.
     
  11. Feb 12, 2013 #10

    haruspex

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    Sure, but it's not quite trivial to get that from the mass of that portion. Seems to me that it's easier to do the right integral in the first place.
     
  12. Feb 13, 2013 #11
    So, I believe I've solved it, not sure WHY it's correct.

    I did the following:
    [tex]m_{r}=\frac{L}{r}m_{t}[/tex]
    [tex]v_{c}=r\dot{\theta}\hat{\theta}=L\dot{\theta}= \omega[/tex]

    Now we have the following function for centripetal force on r: [tex]-\frac{L}{r}m(r\frac{\omega}{L})^{2}[/tex]

    Because the centripetal force is a function of r, we integrate with respect to r and get: [tex]-m\frac{(\omega)^{2}r^{2}}{2L}[/tex]

    My book got the same answer, but through a way that I don't get. Is this approach bad, or incorrect?

    And why exactly is the centripetal force at r not equal to: [tex]-\frac{L}{r}m(r\frac{\omega}{L})^{2}[/tex]
     
  13. Feb 13, 2013 #12

    haruspex

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    I can't follow that logic because I don't know what all your variables mean, but you ended up with the right expression for the centripetal force required for an element dr at radius r (except that you left out the dr itself).
    That cannot be right since it reaches max magnitude at the far end of the rope. You've left out the constant of integration.
    Why do you think it should be that? (You've left out a distance factor somewhere; dimensionally that would be MT-2).
     
  14. Feb 13, 2013 #13
    Pull out all the constants and integrate.

    [tex]-m\frac{\omega^{2}}{L}\int r dr [/tex]

    = [tex]-m\frac{(\omega)^{2}r^{2}}{2L}[/tex]

    From what I understand, the final function should be the centripetal force of all pieces of the rope up to, and including, r.

    [tex]-\frac{L}{r}m(r\frac{\omega}{L})^{2}[/tex] - I thought this might be the centripetal force for r, because it has the same form as the general equation for centripetal force in polar coordinates, which is (if motion is only in the [itex]\hat{\theta}[/itex] direction: [tex]\vec{F}_{c}=m(r(\frac{d \theta}{dt})^{2}\hat{\theta})[/tex]
     
  15. Feb 13, 2013 #14

    haruspex

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    What is the range of that integral?
    No, from r and beyond.
    It cannot be because it is dimensionally wrong. Where does 1/r in front come from?
     
  16. Feb 13, 2013 #15
    Sorry, the code was acting funny, the range is [r,o], so the constants cancel, and we're left with:

    [tex]-m\frac{(\omega)^{2}r^{2}}{2L}[/tex], is that not so?

    And why do you say it's the force from r and beyond? We're only adding up the pieces to r.

    The (1/r) came from rearranging the equation for mass, as a function of radius.

    [itex]\frac{r}{L}M=m[/itex], I made M in terms of small m (big M is the total mass, small m is the mass from 0 to r).
     
  17. Feb 14, 2013 #16

    haruspex

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    I assume you are measuring r from the fixed end of the rope, i.e. the centre of rotation. What contributes to the tension at that point - the rope at radius < r or the rope at radius > r?
     
  18. Feb 14, 2013 #17
    At the point r, all the force is from r<, ah, I think I get it.

    The question is asking the force AT r that is from the REST of the rope?
     
  19. Feb 14, 2013 #18

    haruspex

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    Exactly.
     
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