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Rotating frames (non-homework)

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data

    This is an example problem from the book. However, I have absolutely no clue how they solved it. I cannot follow their logic. If someone can just solve it and explain it for me lucidly step by step, I would really appreciate it. Here it is:

    "A bicycle travels with constant speed around a track of radius [tex]\rho[/tex]. What is the acceleration of the highest point on one of its wheels? Let [tex]V_{0}[/tex] denote the speed of the bicycle and [tex]b[/tex] the radius of the wheel."

    What is confusing me is that in a very similar example preceding this one (the only difference being the wheel following a straight path, rather than curved one) they picked a different coordinate system in which a' and v' turned out to be zero, where as in the example I just posted above, this is not the case. I cannot see the logical leap they make.

    So, it seems I'm having difficulty understanding how to establish the coordinate systems and how I would derive the important variables which proceed from it.

    Thanks for your help.



    2. Relevant equations

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    3. The attempt at a solution

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  2. jcsd
  3. Nov 22, 2009 #2

    tiny-tim

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    Hi Void123! :smile:

    (have a rho: ρ :wink:)

    Any frame of reference (coordinate system) will do …

    just choose whichever is more convenient.

    I suspect that, in the straight case, they decided that a linearly moving frame was easy, and so used it, but in the circular case, a rotating frame is more complicated (you need to introduce a fictional "centrifugal force"), so they stayed with the simpler stationary frame. :wink:
     
  4. Nov 22, 2009 #3
    Thanks for your advice. But my problem is more in depth than that.

    For instance, for an observer in a frame translating along with the wheel, located at the axle, the x'y'z' frame is described as this:

    r' = b (in the jth direction)
    v' = [tex]v_{0}[/tex] (in the ith direction)
    a' = -b * (omega)^2 (in the jth direction)

    (b = radius)

    r = [tex]v_{0}[/tex]t (ith direction) + 2b(jth direction)
    v = 2[tex]v_{0}[/tex](ith direction)
    a = -([tex]v_{0}[/tex]/b)^2 (jth direction)

    How did they derive a', v, and a? I'm missing something here.

    Also, why did they decide that r' is going to be in the jth direction, rather than the ith?
     
  5. Nov 23, 2009 #4

    tiny-tim

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    Hi Void123!± :smile:

    (just got up :zzz: …)
    (try using the X2 tag just above the Reply box :wink:)

    r' v' and a' are the position velocity and acceleration of the top of the wheel, relative to a frame whose (moving) origin is always at the centre of the wheel,

    while r v and a are the position velocity and acceleration of the top of the wheel, relative to a frame whose (stationary) origin is the position of the bottom of the wheel at time t = 0.

    The centre is b higher than the bottom, so r' = bj while r = 2bj + v0ti.

    The top is going twice as fast as the centre (and incidentally the bottom is stationary), so v' = 2v = 2v0i.

    And the top's acceleration is the same in both frames: it equals the acceleration of the centre plus the relative acceleration, which is 0 - bω2j, = v02j/b (your formula for a is wrong).
     
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