Velocity in spherical coordinates

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Why the velocity in spherical coordinates equal to ## v^2 = v \dot{} v = \dot{r}^2 + \dot{r}^2\dot{\theta}^2##

maybe

## v^2 = v \dot{} v = (\hat{ \theta } \dot{ \theta } r +\hat{r} \dot{r} + \hat{ \phi } \dot{\phi } r \sin{ \theta}) \dot{} (\hat{ \theta } \dot{ \theta } r +\hat{r} \dot{r} + \hat{ \phi } \dot{\phi } r \sin{ \theta}) = \dot{\theta}^2 r^2 + \dot{r}^2 + r^2 \dot{\phi}^2 \sin^2{\theta} ##

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  • #2
Ibix
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The movement is in a plane, so they've picked coordinates such that the plane is the equatorial plane of the coordinates - hence ##\dot{\phi}=0##.

Incidentally, it should be ##r^2{\dot{\theta}}^2## in your first expression. I presume that's just a typo since it's correct in the text and your next expression.
 
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  • #3
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The question about why the velocity in spherical coordinates takes the form it does has been answered, so any further discussion of this problem should be in a new thread in the homework forums.

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