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Uniform Circular motion

  1. Apr 10, 2009 #1
    This problem has been discussed before, but one of the parts were not discussed.

    1. The problem statement, all variables and given/known data
    Your calculation is actually a derivation of the centripetal acceleration. To see this, express the acceleration of the particle in terms of its position r(t).

    r(t) = [tex]Rcos(\omega t)\textbf{i} + Rsin(\omega t)\textbf{j}[/tex]

    2. Relevant equations

    [tex]v = \omega R [/tex]
    [tex]a = \frac{v^2}{R} [/tex]
    [tex]a = \omega v [/tex]

    3. The attempt at a solution

    [tex]v = \frac{d}{dt}r(t) \times \omega[/tex]

    btw, this is a Mastering engineering question, so i submitted and it says that:
    The correct answer does not depend on the variables and functions: d, d, r.

    is there another way to display this or am i missing something?

    thx,
     
  2. jcsd
  3. Apr 10, 2009 #2

    rl.bhat

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    Acceleration = V^2/R = w^2*R. In a uniform circular motion w and R are constant.
     
  4. Apr 10, 2009 #3
    omg....too stressed out about finals...i think i got it =_=

    please correct me if i'm wrong (again)...

    a(t) = [tex] -\omega^{2}r(t)[/tex]
     
  5. Apr 10, 2009 #4

    rl.bhat

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    If you express in the vector form, it is correct.
     
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