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Deriving Centripetal Acceleration

  1. Apr 13, 2009 #1
    Hi. I'm trying to derive a = (v^2)/r from the position of a particle moving in a circle. Since the equation for a particle moving in a circle is x(t) = r(isin(wt) + jcos(wt)) where w is constant. I took two derivatives to find it's acceleration. I got a(t) = -rw^2(icos(wt)+jsin(wt)). Now I'm stuck. How the heck and I getting rid of those trig functions? Thanks.

    -Dan
     
  2. jcsd
  3. Apr 14, 2009 #2

    atyy

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    Find the magnitude of the vector a. You will get i.icos^2+j.jsin^2. Then i.i is unit vector dotted with itself equals 1. Then sin^2+cos^2 is some trig identity.
     
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