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Polar coords of motion

  1. Oct 14, 2006 #1
    Question goes something like:
    A cylindrical particle is placed in a certain metre long, frictionless tube, where it can slide back and forth. If the particle is not positioned in the centre of the tube, then as the tube is turning at a constant angular velocity of U, determine the acceleration of the particle along the tube.

    I understand from my textbook that the equation to use in such case should be
    a_r = r "doubledot" (second derivative w.r.t time) - r x U^2
    where r = the distance between the particle to the centre of the tube.

    But I do not know how to work out r "doubledot".

    Any help is much appreciated.
  2. jcsd
  3. Oct 14, 2006 #2
    you need a function that represents the distance r so you can differentiate it twice to get the acceleration (which is r double dot)
  4. Oct 14, 2006 #3
    But I wasn't provided with any sort of function in relation to r. It's just a particle sliding inside a tube. ><
  5. Oct 14, 2006 #4
    ah ok sorry i didnt see the angular velocity bit.


    Tangental velocity, v, is equal to angular velocity, w, times radius, r. Ie v/r = w. So if centripetal acceleration, a, is (v^2)/r you should be able to work out a in terms of r and/or w
  6. Oct 14, 2006 #5
    Oh orite.

    So what do I do with the formula

    a = r double dot - r*w^2?

    can I just ignore it and use a = r*w^2 instead?
  7. Oct 16, 2006 #6
    its the same thing :) r double dot is simply the 2nd time derivative of r, which being a spacial value, is acceleration of that value.
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