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Physical pendulum problem

  1. Dec 13, 2008 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    A physical pendulum is made of a cord of length S with no mass and a bar attached to the cord, of length L and mass m.
    We apart it from the vertical with an angle of [tex]\theta[/tex] °. We then release it.
    1)Find the center of mass' velocity of the pendulum when it reaches the vertical position.
    2)If the cord break up just after passing by the vertical position, find the angular velocity of the bar.
    2. The attempt at a solution
    I don't really know how to approach the problem (I'll try to do alone part b). I thought about conservation of energy but I have too many unknowns ([tex]v_{cm_{\text{initial}}}[/tex], [tex]\omega_{\text{final}}[/tex] and [tex]\omega_{\text{final}}[/tex]).
    By the way I've calculated the moment of inertia of the pendulum as being worth [tex]m\left ( \frac{L^2}{3}+SL+S^2 \right )[/tex] because I'm sure I'll have to use it.
    Should I write down the solution of the differential equation of motion of the pendulum?
    [tex]\frac{d^2 \theta}{dt^2}=\omega ^2 \theta[/tex] where [tex]\omega=\sqrt{\frac{mg(S+L/2)}{I}}[/tex] is the diff. eq. of the motion and the solution is [tex]\theta (t)= \theta _A \cos (\omega t +\phi)[/tex]...
    A little help is appreciated.
     
  2. jcsd
  3. Dec 13, 2008 #2
    Can you be a bit more descriptive about the pendulum set-up (maybe even draw a picture in Paint?). I'm a bit confused about the whole thing. Thanks!
     
  4. Dec 13, 2008 #3

    LowlyPion

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    I think you know more than that. Evaluating dθ/dt at θ = 0 with the vertical, then you can determine your horizontal velocity can't you?
     
  5. Dec 14, 2008 #4

    fluidistic

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    Here's the picture.
     

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  6. Dec 14, 2008 #5

    fluidistic

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    I think I can but I'm confused. Despite that the motion is harmonic like in the case of a mass oscillating with a spring, there's a difference. The equation of motion is not the position in function of time but an angle in function of time. Furthermore if [tex]\theta[/tex] is greater than say 15° then the solution of the equation of motion ([tex]\theta (t)= \theta _A \cos (\omega t +\phi)[/tex]) is getting quite different from the real solution.
    What I've done so far thanks to you : [tex]\Omega (t)=-\omega \theta _A \sin (\omega t + \phi)[/tex]. I found that [tex]\phi[/tex] is worth 0 so [tex]\Omega (t)=-\omega \theta _A \sin (\omega t)[/tex]. I found that the angle (and so the position) is worth 0 when [tex]t=\frac{\pi}{2\omega}[/tex].
    So [tex]\Omega (\frac{\pi}{2\omega})=-\omega \theta _A \sin (\frac{\pi }{2})=-\omega \theta _A[/tex]. What is exactly [tex]\Omega (t)[/tex]? The variation of an angle with respect to time. Isn't it like angular velocity? Or as the motion is harmonic, it should be the velocity with respect to time... I'm not sure in the case of the pendulum. Can you help me?
     
  7. Dec 14, 2008 #6

    LowlyPion

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    At the 0° angle you are surely as close to a small angle as you will get and dθ should most clearly relate to dx at that point shouldn't it by the same small angle approximation you would use to solve the differential equation?
     
  8. Dec 14, 2008 #7

    fluidistic

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    If I understand well in this case [tex]\frac{d\theta (t)}{dt} \approx v(t)[/tex].
    So [tex]v[/tex] when [tex]x=0[/tex] is worth [tex]-\omega \theta _A[/tex]. With simple trigonometry I can find [tex]\theta _A[/tex] and I'm done. Was it that simple?
     
  9. Dec 14, 2008 #8

    LowlyPion

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    Let me be careful, I think I may be misleading you. The equation (the solution of your differential equation) is describing the motion of displacement and at the instant of cutting the cord aren't you are by the statement of the problem converting circular motion to linear? The dθ/dt of your solution gives you a way to determine the magnitude of the rate of displacement at that distance from the center of rotation.

    So I think this dθ/dt is the other ω and that is v/r isn't it? Or in this case S + L/2 ?

    My comment about the small angle approximation was muddy upon reflection insofar as I was thinking it was in relation to velocity and not as it would have been used from the first term of the Taylor series for solving in the first place.
     
  10. Dec 14, 2008 #9

    fluidistic

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    Thank you LowlyPion, I think I understand.
    These [tex]\omega[/tex]s a confusing. :smile:
     
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