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Stable Equilibrium II

  1. Dec 17, 2005 #1
    say I'm givin the potential energy fxn

    V(r) = Eo [ (a/r)^3 - (a/r) ]

    Eo a constant

    After taking the derivatives to find the equilibrium pts and checking for stability we get

    r = sqrt(3)*a and this pt is stable

    r = -sqrt(3)*a and this pt is unstable

    Now let's pose the question

    b) What is the max kinetic energy that the particle can have and still remain bounded (ie. remain within a region of finite spatial extent)

    Tmax = Vmax - Vmin right?

    I am still all little confused about the concept from "stable eq I" Only because that question didn't have 2 equilibrium pts.
     
  2. jcsd
  3. Dec 17, 2005 #2
    I got Tmax = Vmax - Vmin = Eo[ 2/(3sqrt(3)) - ( -2(3sqrt(3)))
    Tmax = 4Eo/(3sqrt(3))

    Is this correct?
     
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