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Centripetal acceleration and velocity derivation

  1. Dec 20, 2009 #1
    in physics, i've been taught that centripetal acceleration is given by v^2/r, but when i try to derive it myself, i get a=2v^2/(pi*r). i said that if the time period is T, after a time of T/4, the velocity would be going in the next direction, i said west to south, and this gives the equation a=4v/T, i then said that T=2*pi*r/v, so a=2v^2/(pi*r). how do you derive the known equation? please help me!?!
  2. jcsd
  3. Dec 20, 2009 #2

    Doc Al

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    Staff: Mentor

    Several problems here. First, by taking a quarter of a circle you are only able to calculate the average acceleration during that time interval, aave = Δv/Δt. You want the instantaneous acceleration, not the average. Second, you found the average acceleration for a quarter circle incorrectly; Δv ≠ v.

    To learn how the centripetal acceleration formula is derived, read this: http://dev.physicslab.org/Document....e=CircularMotion_CentripetalAcceleration.xml"
    Last edited by a moderator: Apr 24, 2017
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