- #1

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Any ideas? :yuck:

- Thread starter devious_
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- #1

- 310

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Any ideas? :yuck:

- #2

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[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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[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

Homework Helper

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Looks good so far. Don't forget that [tex]\ddot{\theta} = 0 [/tex] in this problemdevious_ said:

[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

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- #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

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