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The angular velocity vector of a rigid object rotating about the z-axis is given by $\vec \omega = \omega \hat z$. At any point in the rotating object, the linear velocity vector is given by $\vec v = \vec \omega \times \vec r$, where $\vec r$ is the position vector to that point.

a) Assuming that $\omega$ is constant, evaluate $\vec v$ and $\vec \nabla \times \vec v$ in cylindrical coordinates.

b) Evaluate $\vec v$ in spherical coordinates.

c) Evaluate the curl of $\vec v$ in spherical coordinates and show that the resulting expression is equivalent to that given for $\vec \nabla \times \vec v$ in part a.

So for part a.) I get the following, \begin{align} \vec v = \vec \omega \times \vec r & = \begin{vmatrix} \hat \rho & \hat \phi & \hat z \\ 0 & 0 & \omega \\ \rho & 0 & z \\ \end{vmatrix} \\ & = \rho \omega \hat \phi \end{align}

Then \begin{align} \vec \nabla \times \vec v &=\frac{1}{\rho} \begin{vmatrix} \hat \rho & \rho\hat \phi & \hat z \\ \frac{\partial}{\partial \rho} & \frac{\partial}{\partial \phi} & \frac{\partial}{\partial z} \\ 0 & \rho^2 \omega & 0 \\ \end{vmatrix}\\ &= 2\omega \hat z \end{align}

for part b.) I use the relations $x = r\cos\theta \sin\phi$, $y = r \sin\theta \sin\phi$, and $ z = r \cos\theta$ to change $\vec \omega$ to spherical form

$\vec \omega = \omega \hat r$, the position vector in spherical form is

$\vec r = r \hat r$

so the velocity in spherical form is

\begin{align} \vec v = \omega \times \vec r & = \begin{vmatrix} \hat r & \hat \theta & \hat \phi \\ \omega & 0 & 0 \\ r & 0 & 0 \\ \end{vmatrix} \\ &= 0 \end{align}

I don't know what I'm doing wrong.

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    The mistake is $\vec \omega=\omega \hat r$. That says whatever point you look at, the object is rotating around an axis through that point. Rotation is still around the same axis as before, as detailed in Dr. MV's answer. – 2017-02-04

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In spherical coordinates, $(r,\theta,\phi)$,the axial unit vector $\hat z$ is

$$\hat z=\hat r \cos(\theta)-\hat \theta \sin(\theta)$$

Then,

$$\begin{align} \vec v&=\vec \omega\times \vec r\\\\ &=\omega \hat \omega\times \vec r\\\\ &=\omega (\hat r \cos(\theta)-\hat \theta \sin(\theta))\times \vec r\\\\ &=\hat \phi\omega r \sin(\theta) \end{align}$$

Finally, we have

$$\begin{align} \nabla \times \vec v&= \omega r \nabla \times (\hat \phi r\sin(\theta))\\\\ &=\hat r\omega r\left(\frac{1}{r\sin(\theta)}\frac{\partial }{\partial \theta}(r\sin^2(\theta))\right)+\hat \theta \omega r\left(\frac1r \frac{\partial (r^2\sin(\theta))}{\partial r}\right)\\\\ &=\hat r 2\omega r\cos(\theta)+ \hat \theta 2\omega r\sin(\theta)\\\\ &=2\omega r\underbrace{\color{blue}{\left(\hat r \cos(\theta)+ \hat \theta \sin(\theta)\right)}}_{\color{blue}{=\hat z}}\\\\ \end{align}$$

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    Ah ok I see. The part I was confused about was the changing of $\vec \omega$ from cylindrical to spherical coordinates. I see how substituting for the unit vector $\hat z$ rectifies the problem, but do you mind explaining why my original approach didn't work. I used the relations that take you from cartesian to spherical, since there $\vec \omega$ only had a z component, and I got that $\theta = \phi = 0$ and $r = \omega$. So shouldn't the coordinates in spherical be ($r$, $\theta$ , $\phi$ )? – 2017-02-04
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    No, I meant that since I got that $\theta = \phi = 0$ and $r = \omega$ then since the spherical coordinates are $(r , \theta , \phi)$ $\to$ $(\omega , 0 , 0)$ that means that $\vec \omega = \omega \hat r + 0 \hat \theta + 0 \hat \phi $ which is obviously wrong, but that's what I thought. – 2017-02-04
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    $\omega$ has both $r$ and $\theta$ components. – 2017-02-04
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    I didn't post the formula for the curl in spherical coordinates, but the one I posted for cylindrical which is $\nabla$ X $\vec a$ = $$ \frac{1}{\rho} \begin{vmatrix} \hat \rho & \rho \hat \phi & \hat z \\ \frac{\partial}{\partial \rho} & \frac{\partial}{\partial \phi} & \frac{\partial}{\partial z} \\ a_\rho & \rho a_\phi & a_z \\ \end{vmatrix} $$ is directly from my book. – 2017-02-04
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    Yes, I missed the scale factors. But why are you setting $\theta=\phi=0$? $\vec \omega =\hat z\omega$. So, we write $\hat z=\hat r \cos(\theta)-\theta \sin(\theta)$. And then $\omega$, which is a constant, requires no coordinate transformation. – 2017-02-04
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    Never mind, I just realized why I was getting the wrong answer. Thanks for your time, you've been a big help! – 2017-02-04
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    You're welcome. My pleasure. -Mark – 2017-02-04
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    And feel free to up vote and best vote as you see fit. – 2017-02-04
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    I tried to up vote but I don't have enough reputation for it to show up. Not sure what best vote is though. – 2017-02-04