Polar coordinates

  • Thread starter devious_
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  • #1
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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:
 

Answers and Replies

  • #2
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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.
 
  • #3
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I'm still not getting anywhere. Can you please show me the first couple of steps?
 
  • #4
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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?
 
  • #5
OlderDan
Science Advisor
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devious_ said:
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?
Looks good so far. Don't forget that [tex]\ddot{\theta} = 0 [/tex] in this problem
 
  • #6
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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?
 
  • #7
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compute dot product of acceleration and the radius vector two different ways, and set them equal.
 
  • #8
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Thanks :smile:
 

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