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Time in polar coordinates

  1. Apr 18, 2013 #1
    1. The problem statement, all variables and given/known data

    I have a path defined in polar coordiantes defined as r=a*cos2(θ). I also have the velocity along this path as a function of θ. I want to find the time take to move between two given angles on the path.


    2. The attempt at a solution

    I know that this problem will involve some kind of integration but what is the general method behind solving this kind of problem.
     
  2. jcsd
  3. Apr 18, 2013 #2
    $$ \vec{v} = \frac {d\vec{R}} {dt} = \frac {d\vec{R}} {d\theta} \frac {d\theta} {dt } $$
     
  4. Apr 18, 2013 #3
    The rules are that you need to show some attempt to solve the problem before anyone can help you. So show us what you have tried so far.
     
  5. Apr 18, 2013 #4
    Well my idea is that since I have the velocity of the particle in the direction of the path, then would the time just be ∫(a*cos2(θ)/v(θ))dθ integrated between the two angles I want ( f(v) is the velocity along the path as a function of theta). The main problem is see with this is that r dθ is for a circular loop but can I assume this because dθ is very small.
     
  6. Apr 18, 2013 #5
    This isn't quite right, but you are on the right track. Let me help you. If the particle trajectory is arbitrary, what is the equation for a differential length ds along the trajectory in terms of r, dr, and dθ? What is the equation for the tangential velocity v in terms of ds and dt?
     
  7. Apr 21, 2013 #6
    Differential length is √(r2+(dr/dθ)2).

    Do you mean the tangential component of the velocity? For that, I get √(ds2-dr2). (Using v=r dθ and ds2=(r dθ)2+dr2).

    Or do you mean the tangent to the curve at that point, if ds is a straight line at that point, then v=ds/dt. This looks better, so would time just be ∫ds/v= ∫√(r2+(dr/dθ)2)/v dθ , where v is the velocity along the curve?
     
  8. Apr 21, 2013 #7
    √(r2+(dr/dθ)2) isn't quite right. The differential distance ds along the trajectory is this multiplied by dθ. [tex]ds=\sqrt{(dr)^2+(rd\theta)^2}[/tex]

    So, [tex]dt=\frac{\sqrt{(\frac{dr}{d\theta})^2+r^2}}{v(\theta)}d\theta[/tex]
     
  9. Apr 21, 2013 #8
    That makes sense now. Thank you for your help.
     
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