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Polar coordinates

  1. Jun 22, 2005 #1
    A particle P describes the curve with polar equation [itex]r = a e^{\theta \sqrt{3}} \cosh 2\theta[/itex] in such a manner that the radius vector from the origin rotates with uniform angular speed [itex]\omega[/itex]. Show that the resultant acceleration of the particle at any instant makes an angle of 30 degrees in the radius vector.

    Any ideas? :yuck:
     
  2. jcsd
  3. Jun 23, 2005 #2
    Use the polar unit vectors:

    [tex]\hat{\mathbf{r}} = \cos \theta \hat{\mathbf{i}} + \sin \theta \hat{\mathbf{j}} \ \ \ \ \ \hat{\mathbf{\theta}} = -\sin \theta \hat{\mathbf{i}} + \cos \theta \hat{\mathbf{j}}[/tex]

    The nice thing about these is that you can start out with the position vector

    [tex]\mathbf{r} = r \hat{\mathbf{r}}[/tex]

    and take derivatives of it to find the velocity and acceleration vectors. You are given r as a function of theta, and [tex]\dot{\theta} = \omega[/tex], so this should allow you to express all vectors in terms of theta.
     
  4. Jun 23, 2005 #3
    I'm still not getting anywhere. Can you please show me the first couple of steps?
     
  5. Jun 23, 2005 #4
    Right. I've tried some more and here's where I ended up:

    [tex]\text{acceleration} = \langle \ddot{r} - r \dot{\theta ^2}, 2\dot{r}\dot{\theta} + r\ddot{\theta} \rangle[/tex]

    Now I've got to differentiate my expression for [itex]r[/itex] and use the fact that [itex]\dot{\theta} = \omega[/itex].

    Is that correct, or is there a better method?
     
  6. Jun 23, 2005 #5

    OlderDan

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    Looks good so far. Don't forget that [tex]\ddot{\theta} = 0 [/tex] in this problem
     
  7. Jun 23, 2005 #6
    I've obtained an expression for the magnitude of the acceleration; how do I show that the particle makes an angle 30 at the radius vector?
     
  8. Jun 23, 2005 #7
    compute dot product of acceleration and the radius vector two different ways, and set them equal.
     
  9. Jun 23, 2005 #8
    Thanks :smile:
     
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