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Conservation of Energy Question

  1. Nov 16, 2004 #1

    ek

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    Okay guys and gals, I need some help on this one. It has me pretty stumped.

    A circus trapeze consists of a bar suspended by two parallel ropes, each of length L, allowing performers to swing in a vertical circular arc. Suppose a performer with mass m holds the bar and steps off an elevated platform, starting from rest with the ropes at an angle xi with respect to the vertical. Suppose the size of the performer's body is small compared to the length L, that she does not pump the trapeze to swing higher, and that air resistance is negligible. Show that when the ropes make an angle x with the vertical, the performer must exert a force:

    mg(3cosx - 2cosxi)

    Any help would be much appreciated.

    Diagram:
     

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  3. Nov 16, 2004 #2
    Use an energy argument to calculate the velocity of the acrobat at the desired angle. Consider that the acrobat must be in circular motion and account for any forces necessary to maintain this circular motion (think: centripetal). Then account for the force of gravity.

    --J
     
  4. Nov 16, 2004 #3
    Where does the performer exert the force? Onto the rope? Is it the tension of the rope?
     
  5. Nov 16, 2004 #4

    ek

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    Your guess is as good as mine. That is the exact question I posted, there's no more information.

    Where the hell do the 3 and the 2 come from? And how does the initial x angle get factored in somewhere along the line?
     
  6. Nov 16, 2004 #5

    Doc Al

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    Staff: Mentor

    Start by applying conservation of energy to find the KE of the girl at angle x. (The initial and final angles allow you to find the change in height and thus the change in gravitational PE.)

    Then consider the forces acting on the girl. (Note: Whatever force the bar pulls on the girl, the girl pulls on the bar.) And the fact that she's centripetally accelerated.
     
  7. Nov 16, 2004 #6
    I have tried this method already. In fact, I tried the other method using integration which also yields the same answer but not the same as the answer given. Can you try to show us your whole working?
     
  8. Nov 16, 2004 #7

    Doc Al

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    Staff: Mentor

    I'd like to give ek a chance to solve it.

    But if you like, post your work and I'll give my comments. (Start by finding the KE.)
     
  9. Nov 16, 2004 #8
    Ok, I mixed up with + and - sign in my calculation by using method of KE. My other method is as follow:
    [tex]\tau=mgL\sin{\Theta}[/tex]
    [tex]\alpha=\frac{g\sin{\Theta}}{L}[/tex]
    [tex]\int \omega d\omega=\frac{g}{L} \int \sin{\Theta} d\Theta[/tex]
    After integration, I got [tex]\omega^2=\frac{2g}{L}(\cos{\xi}-\cos{x})[/tex]
    Then I find the tension which is
    [tex]mL\omega^2+mg\cos{x}[/tex]
    However I don't get the answer by using this method. I know if I put a -sign for the [tex]\alpha[/tex], I can get the answer. I suspect the reason for putting a - sign there is the same as the - sign in SHM formula that the direction of the acceleration is always opposing the displament. Is it? Please help me to clear my doubt. Thanks
     
  10. Nov 16, 2004 #9

    ek

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    Does the solution have to involve integration?
     
  11. Nov 16, 2004 #10
    No need if you use the KE method. It's just an alternative method from me, I just think that it might be possible if the - sign can be explained.
     
  12. Nov 16, 2004 #11

    Doc Al

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    Doing it your way, I get [tex]\omega^2=\frac{2g}{L}(\cos{x}-\cos{x_i})[/tex]. (Check your limits of integration: you reversed initial and final angles.)

    Should work out fine now.

    Here's how I did the problem, using conservation of energy: [itex]\Delta {KE} = - \Delta {PE}[/itex]

    [tex]1/2 m v^2 = mgL(cos{x} - cos{x_i})[/tex]
    Thus:
    [tex]m v^2/L = 2mg(cos{x} - cos{x_i})[/tex]
    Then, using F = ma for radial forces:
    [tex]T - mgcos{x} = m v^2/L = 2mg(cos{x} - cos{x_i})[/tex]
    Rearrange to show T.
     
  13. Nov 16, 2004 #12
    what is the reason for reversing the angle? At [tex]\Theta=\xi[/tex], the [tex]\omega=0[/tex] and when the [tex]\Theta=x[/tex], its [tex]\omega=\omega[/tex]. Shouldn't them be parallel to each other? Every integration follows this law, right?
     
  14. Nov 16, 2004 #13

    ek

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    Thanks Doc. I didn't implement the tension/gravity of the girl correctly. Thanks for the help. Thanks Cartoonkid too.
     
  15. Nov 16, 2004 #14
    No prob, ek. In fact, the answer for this question itself is not totally correct. It didn't make itself clear. If it only wants the tension of the rope on one side, then the tension should be half of the answer given. I also have to thank you because I can revise my physics and maths from this question. Thank you Doc AI too. However I still need your explaination for the integration part.
     
    Last edited: Nov 16, 2004
  16. Nov 17, 2004 #15

    Doc Al

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    Staff: Mentor

    integrating torque

    I think I know where the problem is: you are finding the work done by the torque, so you had better make sure that your [itex]d\theta[/itex] is positive, which means that [itex]\theta[/itex] ranges from [itex]x[/itex] to [itex]x_i[/itex].

    Thus you are calculating
    [tex]\int_{x}^{x_i} sin\theta d\theta = \left[- cos\theta \right]_{x}^{x_i} = cos(x) - cos(x_i)[/tex]

    If that doesn't feel right, you can always keep the angle range the same, but then integrate [itex]-sin\theta d\theta[/itex] from [itex]x_i[/itex] to [itex]x[/itex].
     
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