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Motion of a bead on a rotating linear rod

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data
    The center of a long frictionless rod, pivoted at the origin, is forced to rotate at a constant angular velocity Ω in the horizontal xy-plane. Write down the equation of motion for a bead threaded on the rod, using the coordinates x and y where x is measured along the rod and y perpendicular to it. Solve for x(t). What is the role of the centrifugal and coriolis force?

    2. Relevant equations
    Newton's Second Law in a rotating frame?


    3. The attempt at a solution
    Since the bead's fixed to move along the wire, I've eliminated the equation for the motion along the y-axis. The bead's position along the x-axis varies with time, and based on the coriolis effect, the bead should slide out away from the origin.

    I'm not sure how to deal with the math from here though. Any tips on how to get started would be greatly appreciated.
     
  2. jcsd
  3. Nov 28, 2011 #2

    BruceW

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    There are several different ways to do it. What have you been taught so far? Have a go using the method you are familiar with.
     
  4. Nov 28, 2011 #3
    Right now we're working on nonlinear mechanics, but we've covered Lagrangian mechanics. I'm not sure how I should start this problem using either method. I know that the rotation will cause the bead to slide out, but I'm not sure how to model it...
     
  5. Nov 28, 2011 #4

    BruceW

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    To start with, you know it is 2d motion of an object (the bead), so you can first write down the general equation of motion for an object in 2d.

    Edit: Also, the question is talking about polar coordinates, so that's the coordinate system you should use.
     
  6. Nov 28, 2011 #5
    The question itself is asking for the equation of motion in cartesian coordinates. My question is, how can I incorporate the rotation into my answer?
     
  7. Nov 28, 2011 #6

    BruceW

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    How do you usually incorporate rotational motion? I can't really tell you how I would do it, because I think that would be giving too much help.
     
  8. Nov 28, 2011 #7
    Ohh, I see. Thanks for the help!
     
  9. Nov 28, 2011 #8

    BruceW

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    If you don't help me by giving it a go, I can't help you.
     
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