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Find the radius of curvature of a particle

  1. Aug 17, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle is moving on a path parameterized as such:
    $$x(t)=a\sinωt \quad y(t)=b\cosωt$$
    Find the radius of curvature ρ as a function of time. Give your answer in Cartesian coordinates.
    2. Relevant equations

    $$\frac{1} {Radius~of~curvature}=|\frac{de_t}{ds}| $$, where et is the unit tangent vector and s the path length.
    3. The attempt at a solution
    $$\frac{de_t}{ds}=\frac{de_t}{dt}\frac{dt}{ds}=\frac{1}{v}\frac{de_t}{dt}$$, where v is the speed.
    $$\frac{de_t}{dt}=\frac{d}{dt}\frac{\vec v}{v}$$ so$$|\frac{de_t}{ds}|=\frac{1}{v^2}|{\vec a}|$$
    Evaluating the derivative gives me $$\frac{\sqrt{a^2ω^4sin^2ωt+b^2ω^4cos^2ωt}}{a^2ω^2cos^2ωt+b^2ω^2sin^2ωt}$$ which is wrong. At which step did I make a mistake?
     
  2. jcsd
  3. Aug 17, 2015 #2

    SteamKing

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    I'm not sure where you got your definition of the radius of curvature from, but this article contains an alternate formulation:

    http://mathworld.wolfram.com/RadiusofCurvature.html

    The osculating circle is given by these coordinates (parametrically):

    https://en.wikipedia.org/wiki/Osculating_circle
     
  4. Aug 17, 2015 #3
    Thanks; that formulation is much easier to work with.
    As for the definition I used, it was illustrated using a circle. The full equation would be:
    $$|\frac{de_Θ}{ds}|=\frac{1}{R}e_r$$
    , where $$e_Θ \quad and \quad e_r$$ are the unit tangent vector and unit normal vector respectively.
    What I gleaned from watching online tutorials is that the rate of change of the direction of motion w.r.t path length gives the curvature.
     
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