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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


    3. The attempt at a solution





    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


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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


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

    Using your advice I get:


    Does this look ok?

  6. Jul 11, 2012 #5


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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


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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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