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Lets place a fixed coordinate system $xyz$ on the Earth, and a moving coordinate system $x'y'z'$on the surface of the Earth such that the $z'$ axis makes an angle of $\lambda$ with the equator, and an angle of $\pi/2 - \lambda$ with the $z$ axis. This takes place in the northern hemisphere. Let $\omega$ be the rotation of the Earth. Why is it then, that the components of angular velocity in the rotating frame are $$\omega_{x'} = -\omega\cos\lambda$$ $$\omega_{y'}=0$$ $$\omega_{z'} = \omega \sin \lambda$$

This makes sense of course. When $\lambda = 0$, the $x'$ component should be equal but opposite to $w$. But how is this shown mathematically?

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the sphere position vector is:

$$\mathbf R= r\,\left[ \begin {array}{c} \cos \left( \theta \right) \sin \left( \lambda \right) \\ \sin \left( \theta \right) \sin \left( \lambda \right) \\ \cos \left( \lambda \right) \end {array} \right] $$

where $\lambda$ is the latitude line and and $\theta~$ is the longitude line

The angular velocity vector is:

$$\mathbf \omega=\left[ \begin {array}{c} 0\\ 0\\ \omega\end {array} \right] $$

with:

$$\mathbf\omega=\omega_r\,\vec{\hat{e}}_r+\omega_\lambda\,\vec{\hat{e}}_\lambda+\omega_\theta\,\vec{\hat{e}}_\theta$$

and :

$$\vec{e}_r=\frac{\partial \mathbf R}{\partial r}~, \vec{\hat{e}}_r=\frac{\vec{e}_r}{\parallel\,\vec{e}_r\parallel}$$

analog $~\vec{\hat{e}}_\lambda$ and $~\vec{\hat{e}}_\theta$

\begin{align*} &\vec{\hat{e}}_r=\left[ \begin {array}{c} \cos \left( \theta \right) \sin \left( \lambda \right) \\ \sin \left( \theta \right) \sin \left( \lambda \right) \\ \cos \left( \lambda \right) \end {array} \right] \\ &\vec{\hat{e}}_\lambda=\left[ \begin {array}{c} \cos \left( \theta \right) \cos \left( \lambda \right) \\ \sin \left( \theta \right) \cos \left( \lambda \right) \\ -\sin \left( \lambda \right) \end {array} \right] \\ &\vec{\hat{e}}_\theta= \left[ \begin {array}{c} -\sin \left( \theta \right) \\ \cos \left( \theta \right) \\ 0 \end {array} \right]\\\\ & \vec{\hat{e}}_r\perp \vec{\hat{e}}_\lambda\perp \vec{\hat{e}}_\theta \end{align*}

thus:

$$\omega_r=\vec{\hat{e}}_r\cdot\,\mathbf\omega=\omega\,\cos(\lambda)\\ \omega_\lambda=\vec{\hat{e}}_\lambda\cdot\,\mathbf\omega=-\omega\sin(\lambda)\\ \omega_\theta=\vec{\hat{e}}_\theta\cdot\,\mathbf\omega=0$$

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