# Finding the curl of velocity in spherical coordinates

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1. Feb 4, 2017

### John004

1. The problem statement, all variables and given/known data
The angular velocity vector of a rigid object rotating about the z-axis is given by
ω = ω z-hat. At any point in the rotating object, the linear velocity vector is given by v = ω X r, where r is the position vector to that point.

a.) Assuming that ω is constant, evaluate v and X v in cylindrical coordinates.

b.) Evaluate v in spherical coordinates.

c.) Evaluate the curl of v in spherical coordinates and show that the resulting expression is equivalent to that given for X v in part a.
2. Relevant equations
The expressions for the curl in cylindrical and spherical coordinates. Since I don't know how to put the determinant here ill just leave them out.

For spherical

x = r sinθ cosΦ

y = r sinθ sinΦ

z = r cos θ

3. The attempt at a solution
So I worked out part a correctly (I think) which is in the attachment, but I'm stuck on part b.

b.) So for this part I have to convert ω to spherical coordinates. Since ω only lies along the z-axis, that means that Φ and θ are equal to zero, so

ω = ω r-hat

and the position vector in spherical polars is

R =
r r-hat

so that means that when I cross ω and R I get zero, I don't know what I'm missing.

#### Attached Files:

• ###### methods 3.jpg
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Last edited: Feb 4, 2017
2. Feb 4, 2017

### TSny

part (a) looks good.
Go to an arbitrary point in space and try to write ω in terms of the unit vectors $\hat{r}$, $\hat{\theta}$, and $\hat{\phi}$ at that point.

OK

3. Feb 4, 2017

### John004

Is there some resource you can point me to so I can learn how to type out symbols and equations neatly like you just did?

I can't really picture it in the way you're asking me too. What if I substitute z-hat = r-hat cosθ - θ-hat sinθ?

4. Feb 4, 2017

### TSny

https://www.physicsforums.com/help/latexhelp/
You can learn a lot by just examining how others have used Latex in their posts. That's how I picked it up. I still have a lot to learn.

Yes, that's it.