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Centripetal acceleration

  1. Jun 17, 2009 #1
    What happens to the period when you cut the radius in half for a centripetal acceleration problem?


    a=V^2/r
    T=2pi*r/v


    I need some background on this question. I believe the answer is the period is decreased by a factor of 2. I am just slightly confused.
     
  2. jcsd
  3. Jun 17, 2009 #2
    Well, that would depend. Are you keeping the acceleration on the body the same?
    If so, then the problem isn't quite as trivial:
    ainitial = v²/rinitial
    afinal = u²/rfinal
    Tinitial = 2πr/v
    Tfinal = 2π(½r)/u

    Since we want the acceleration before we cut the radius in half to be the same as after, all that remains to find u and Tfinal is a simple equation.
     
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