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Centripetal Acceleration derivation help

  1. Nov 14, 2009 #1
    w=2π /T
    v=rw=2πr/T
    a=rw2=r(4π2r2/T)

    This I know, but why cant I just differentiate v with respect to t to get a? (But the answer is wrong) Can anyone tell me why?

    Thanks.
     
  2. jcsd
  3. Nov 14, 2009 #2
    Also, [tex]a_{c}=\frac{v^{2}}{R}[/tex] in case you didn't have this equation at your disposal.

    Are you trying to differentiate v=rw=2πr/T with respect to time?
     
  4. Nov 14, 2009 #3

    Delphi51

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    Error in the third line - should be a = a=rw²=r(2π /T)² = 4π²r/T²
    Differentiating v = 2πr/T does no good because this formula is for the (constant) magnitude of velocity, so the dv/dt = 0. In fact the direction of the velocity is continuously changing so a vector derivative must be done to get the acceleration that way. It would be something like this:
    v = 2πr/T[cos(ωt),sin(ωt)]
    dv/dt = 2πr/T[-ω*sin(ωt), ω*cos(ωt)]
    = 2πr/T*ω[-sin(ωt), cos(ωt)] and since ω = v/r and v = 2πr/T this is
    = v²/r[-sin(ωt), cos(ωt)]
    showing that the magnitude of acceleration is good old v²/r and that its direction rotates with time.
     
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