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Acceleration in arbitrary trajectory (differential geometry)

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that for any trajectory r(t) the acceleration can be written as:

    [itex]\mathbf{a}(t)=\frac{dv}{dt}\hat{T}(t)+\frac{v^2}{\rho}\hat{N}(t)[/itex]

    where v is the speed, T is a unit vector tangential to r and N is a unit vector perpendicular to T, at time t. rho is the radius of curvature.

    2. The attempt at a solution

    [itex]\mathbf{a}=\frac{d\mathbf{v}}{dt}=\frac{d}{dt}(v\hat{T})[/itex]

    [itex]=\frac{dv}{dt}\hat{T}+v\frac{d\hat{T}}{dt}[/itex]

    Since we can write

    [itex]\frac{1}{\rho}\hat{N}=\frac{d\hat{T}}{dt}[/itex]

    This gives

    [itex]\mathbf{a}=\frac{dv}{dt}\hat{T}+\frac{v}{\rho}\hat{N}[/itex]

    I can't see where to get the other 'v' from, in the second term.
     
  2. jcsd
  3. Jan 21, 2014 #2

    LCKurtz

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    ##v\frac{d\hat{T}}{dt}= v \frac{d\hat T}{ds}\frac{ds}{dt}= v^2\frac{d\hat T}{ds}##. But the curvature ##\kappa## is ##\kappa = \left|\frac{d\hat T}{ds}\right|## so ##
    v^2\frac{d\hat T}{ds} = v^2\kappa \hat N(t) = \frac{v^2}{\rho}\hat N(t)##.
     
  4. Jan 21, 2014 #3
    Got it, thanks.

    Missed that it is the derivative of [itex]\hat{T}[/itex] with respect to arc length that measures curvature.
     
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