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Lagrangian of a system

  1. Feb 16, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    A bead of mass m slides on a long straight wire which makes an angle alpha with, and rotates with constant angular velocity omega about, the upward vertical. Gravity acts vertically downard.
    a)Choose an appropriate generalized coordinate and find the Lagrangian.
    b)Write down the explicit Lagrange's equation of motion.


    2. Relevant equations
    L=T-V.


    3. The attempt at a solution
    I'm a bit confused. They mean only 1 generalized coordinate so this mean the system has only 1 degree of freedom. Ok.
    The center of my coordinate system is the point where the wire and the vertical meet. I call x the distance from this point to the mass. This means [itex]V=mgx \cos \alpha[/itex].
    I'm having a hard time in finding the velocity of the mass. Not only it has a circular motion with tangential speed [itex]\omega x \sin \alpha[/itex] but can also move along the wire with velocity... [itex]\dot x[/itex]? Adding up these 2 speeds and squaring them in order to get the kinetic energy would not match the given answer.
    I'd like some help to get the kinetic energy of this particle/mass. Thank you.
     
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  3. Feb 16, 2012 #2

    vela

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    Are you squaring then adding or adding then squaring?
     
  4. Feb 17, 2012 #3

    fluidistic

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    Adding then squaring. Basically I say that [itex]v=\dot x +x \omega \sin \theta[/itex] which is addition of tangential speed + speed along the wire. Wait a minute, since these 2 velocities aren't making "a right angle" it means that the total velocity isn't the sum as I thought. The total velocity would be worth the square root of their sum, meaning that the kinetic energy is [itex]\frac{m}{2}(\dot x+\omega x \sin \theta )[/itex].
     
  5. Feb 17, 2012 #4

    I like Serena

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    The tangential speed and the speed along the wire do make a right angle with each other.
     
  6. Feb 17, 2012 #5

    fluidistic

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    Whoops right, I didn't see this. I played with my fingers, now I see it. I reported this on my draft. I have made a right triangle. One side is [itex]\dot x[/itex], another one is [itex]\omega x \sin \alpha[/itex] and the hypotenuse is [itex]\sqrt {\dot x^2 + \omega ^2 x^2 \sin ^2 \alpha }[/itex]. This is the speed of the particle. I hope this is right now... thanks for pointing this out!
     
  7. Feb 17, 2012 #6

    I like Serena

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    Yes, it's right now. ;)
     
  8. Feb 17, 2012 #7

    fluidistic

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    a)[itex]L=\frac{m}{2} (\dot x ^2+\omega ^2 x ^2 \sin ^2 \alpha )-mgx \cos \alpha[/itex].
    b)[itex]\ddot x -x\omega ^2 \sin ^2 \alpha+g\cos \alpha =0[/itex]. Apparently this result is correct.
    I think I'm interested in solving this ODE.
    I first solve the homogeneous DE, this gave me [itex]x_h(t)=Ae^{\omega x \sin \alpha}+Be^{-\omega x \sin \alpha}[/itex].
    A particular solution is [itex]x_p(x)=c_3[/itex]. Plugging and chugging this into the DE gives [itex]c_3=\frac{g\cos \alpha }{\omega ^2 \sin ^2 \alpha}[/itex].
    So the general solution to the ODE is [itex]x(t)=Ae^{\omega x \sin \alpha}+Be^{-\omega x \sin \alpha}+\frac{g\cos \alpha }{\omega ^2 \sin ^2 \alpha}[/itex].

    Edit: Hmm I probably made an error. It looks like there's a term that blows up if omega is worth 0... cannot be right...
     
  9. Feb 17, 2012 #8

    vela

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    You want t in the exponents, not x. The ##\omega=0## case needs to be handled separately by setting it to 0 in the original DE and then solving it. You can't set it to 0 at the end because you assumed it wasn't 0 while solving the problem.
     
  10. Feb 17, 2012 #9

    fluidistic

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    Oh I see. I've made the exact same mistake in a previous exercise... I will try to be more careful in future.
     
  11. Feb 17, 2012 #10

    fluidistic

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    If [itex]\omega =0[/itex] or in fact [itex]-\omega ^2 \sin ^2 \alpha =0[/itex] (so alpha could be worth 0), I reach that [itex]x(t)=\frac{-g\cos ( \alpha ) t^2}{2}+v_0 t+x_0[/itex]. Basically the equation of motion of a free fall with gravity worth [itex]-g\cos ( \alpha )[/itex] instead of [itex]-g[/itex]. I can consider alpha to be worth 0 and I reach a common free fall equation which makes sense.
     
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