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Polar Coordinates

  1. Aug 8, 2016 #1
    1. The question
    The position of a particle is given by r(t) = acos(wt) i + bsin(wt) j. Assume a and b are both positive and a > b. The plane polar coordinates of a particle at a time t equal to 1/8 of the time period T will be given by _

    2. Relevant equations
    r(t) = acos(wt) i + bsin(wt) j.

    3. The attempt at a solution
    at t = T/8 the value of wt=π/4.
    therefore in Cartesian coordinates the vector is r= a/√2 + b/√2.
    so in polar coordinates this transforms to r = √(a2 + b2)/2 and tanθ = b/a. My answer does not match with the given one. Also if we were to write the polar equation of the curve how would that follow ?
     
  2. jcsd
  3. Aug 8, 2016 #2

    BvU

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    Hi,
    Strange. Don't see anything wrong. In what respect does your answer not match with the given one ?
     
  4. Aug 8, 2016 #3
    The angle does not match. tanθ = b/a√2 is given
    Also this is one of the part of the question, and I still can't understand how to wite the polar equation for this curve
     
  5. Aug 8, 2016 #4

    BvU

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    Still strange. Where would the ##\sqrt 2## come from ?
    Your ##\tan\theta## matches the expression here (##\tan t=1##).
    And the answer for the polar equation is right in the next paragraph :rolleyes:
     
  6. Aug 8, 2016 #5
    You have $$r\cos \theta = a \cos \omega t$$and $$r\sin \theta = b \sin \omega t$$
    Just solve for r and ##\theta##
     
  7. Aug 8, 2016 #6
    I calculated the polar equation to be r = √(a2cos2(wt) + b2sin2(wt)) r^
    and then I differentiated it to get the velocity as v = ((b2 - a2)sin(wt)cos(wt)w)/√(a2cos2(wt) + b2sin2(wt)) r^ + √(a2cos2(wt) + b2sin2(wt))w θ^
     
  8. Aug 8, 2016 #7
    Is this wrong ?
     
  9. Aug 8, 2016 #8
    It looks correct. For aesthetic purposes, I would go to the double angle formulas:
    $$2\sin \omega t\cos \omega t=\sin2\omega t$$
    $$\cos^2 \omega t=\frac{1+\cos 2\omega t}{2}$$
    $$\sin^2 \omega t=\frac{1-\cos 2\omega t}{2}$$
     
  10. Aug 8, 2016 #9

    BvU

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    Are we sure the exercise composer wants r(t) and v(t) instead of ##r(\theta)## when asking for the polar form ?
     
  11. Aug 8, 2016 #10
    Oops. I don't confirm your coefficient of ##\hat{\theta}##. I get ##\frac{\omega ab}{r}##.
     
  12. Aug 8, 2016 #11
    that is the problem I have been having, I can't figure out as to why I am not getting the correct angular component
     
  13. Aug 8, 2016 #12
    The angular component is ##rd\theta/dt##, not ##r\omega##. In this problem ##d\theta/dt## is not the same thing as what they call ##\omega##. You need to be able to show that ##\frac{d\theta}{dt}=\frac{\omega ab}{r^2}##.
     
  14. Aug 8, 2016 #13
    $$\tan \theta = \frac{b}{a}\tan \omega t$$
    $$\sec^2 \theta \frac{d\theta}{dt}=\frac{b\omega}{a}\sec^2\omega t$$$$r\cos \theta=a\cos \omega t$$

    Combine 2nd and 3rd equations to eliminate ##\sec \theta##.
     
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