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Bead on wire tilted at angle theta.

  1. Jul 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider a bead of mass m constrained to slide without friction along a rigid wire that rotates about the vertical at a fixed angle [itex]\theta[/itex] with constant angular velocity [itex]\omega[/itex]. Write down the Lagrangian in terms of z as the general coordinate. Find the equation of motion of the bead, and determine whether there are positions of equilibrium. If there are equilibrium positions, are they stable?

    2. Relevant equations
    [itex]z=rcos\theta[/itex]

    [itex]U=mgz[/itex]

    3. The attempt at a solution
    [itex]\dot{r}=\frac{\dot{z}}{cos\theta}[/itex]

    [itex]T=\frac{1}{2}m(\dot{r}^{2}+r^{2}\dot{\theta}^{2})[/itex]

    [itex]L=\frac{2}{3}m(\frac{\dot{z}^{2}}{cos^{2}\theta}+(z\omega)^{2}/(cos^{2}\theta}))-mgz[/itex]

    [itex]m\ddot{z}=2z\omega^{2}-mgcos^{2}\theta[/itex]

    [itex]\ddot{z}=\frac{2z\omega^{2}}{m}-gcos^{2}\theta[/itex]

    Just wondering if my equation of motion is correct. Thanks for the help
     
    Last edited: Jul 9, 2012
  2. jcsd
  3. Jul 9, 2012 #2

    TSny

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    Note that [itex]\theta[/itex] is a fixed angle. So, [itex]\dot{θ}[/itex] = 0.

    You will need a kinetic energy term that takes into account the azimuthal motion ("swinging around motion") around the axis of rotation at the angular speed ω. So, see if you can construct a kinetic energy term involving ω, z, and [itex]\theta[/itex]. Of course, you will still have the kinetic energy term [itex]\frac{1}{2}[/itex]m[itex]\dot{r}[/itex]2.
     
  4. Jul 9, 2012 #3

    TSny

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    Note that the distance of the particle from the axis of rotation is rsinθ = ztanθ.
     
  5. Jul 11, 2012 #4
    TSny,

    Using your advice I get:
    [itex]L=\frac{1}{2}m(\frac{\dot{z}^{2}}{cos^{2}\theta}+(z^{2}\omega^{2})/(cos^{2}\theta))-mgz[/itex]

    [itex]\ddot{z}=\frac{2z\omega^{2}}{m}-gcos^{2}\theta[/itex]

    Does this look ok?

    Thanks.
     
  6. Jul 11, 2012 #5

    TSny

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    The kinetic energy associated with the rotation about the z axis would be (1/2)m(ωd)2 where d is the radius of the circular motion about the z-axis. If the bead is a distance r out along the wire, then check that d = r sinθ. d can then be expressed in terms of z by using r = z/cosθ.

    So, verify that the second term of the Lagrangian would have a tan2θ in place of the 1/cos2θ.

    You should find that the mass m cancels out in the equation of motion for [itex]\ddot{z}[/itex]. Also, you have a factor of 2 in the first term on the right hand side of the equation for [itex]\ddot{z}[/itex] that should not be there.
     
  7. Jul 12, 2012 #6
    Awesome that made it perfectly clear and I got the correct answer. Thanks so much for the help!
     
  8. Jul 12, 2012 #7
    Also to find any equilibrium positions I looked at what would make [itex]\ddot{z}=0[/itex].
    I got only 1 equilibrium position of [itex]z=\frac{g}{tan^{2}\theta\omega^{2}}[/itex].
    I also concluded that this is stable since [itex]\omega[/itex] and [itex]\theta[/itex] are constant.
     
  9. Jul 12, 2012 #8

    TSny

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    Your equilibrium value for z looks correct to me.

    However, I believe it's an unstable equilibrium. One way to see this is to go back to your result for [itex]\ddot{z}[/itex].

    [itex]\ddot{z}[/itex] = sin2θ ω2 z - g cos2θ

    Imagine placing the bead on the rod at the equilibrium position where [itex]\ddot{z}[/itex] = 0. Then displace the bead a little bit farther out on the rod. This will make z greater than the equilibrium value of z and an inspection of the expression for [itex]\ddot{z}[/itex] shows that [itex]\ddot{z}[/itex] will become positive. So, if you let the bead go from this position, which way would it begin to move?

    Similarly, analyze what would happen if you displaced the bead a small amount in the opposite direction from the equilibrium position.
     
    Last edited: Jul 12, 2012
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