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Eli
Eli
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yourthe pseudo forces are:

$$\vec F_p=\underbrace{m\,\left(\vec \Omega\times (\vec \Omega\times \vec R)\right)}_{\text{Centrifugal force}}+ \underbrace{2\,m\,(\vec \Omega\times {\dot{\vec{R}}})}_{\text{Coriolis force}}$$

with:

$$\vec R=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}\qquad,{\dot{\vec{R}}}=\begin{bmatrix} v_x \\ v_y \\ v_z \\ \end{bmatrix}\\ \vec \Omega=\Omega\,\begin{bmatrix} 0 \\ \cos(\lambda) \\ \sin(\lambda) \\ \end{bmatrix}$$

$\Rightarrow$

$$\vec F_p=\left[ \begin {array}{c} m{\Omega}^{2}x-2\,m \left( \cos \left( \lambda \right) v_{{z}}-\sin \left( \lambda \right) v_{{y}} \right) \Omega\\ -m\sin \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) { \Omega}^{2}-2\,m\sin \left( \lambda \right) v_{{x}}\Omega \\ m\cos \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) {\Omega}^{2}+2 \,m\cos \left( \lambda \right) v_{{x}}\Omega\end {array} \right] $$

with $~\Omega=7.27\,10^{-5}~[\frac 1s]$ and $~\lambda=\frac{\pi}{2},~\vec\Omega=[0~,0~,\Omega]^T$

$$F_p=\left[ \begin {array}{c} 5.2 \,10^{-9}\,mx+ 1.4 \,10^{-4}\, mv_{{y}}\\ 5.2 \,10^{-9}\,my- 1.4 \,10^{-4}\,mv_{{x}}\\ 0.0\end {array} \right] $$

hence the pseudo force components (unit [N]) are very small and can neglected

enter image description here

your pseudo forces are:

$$\vec F_p=\underbrace{m\,\left(\vec \Omega\times (\vec \Omega\times \vec R)\right)}_{\text{Centrifugal force}}+ \underbrace{2\,m\,(\vec \Omega\times {\dot{\vec{R}}})}_{\text{Coriolis force}}$$

with:

$$\vec R=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}\qquad,{\dot{\vec{R}}}=\begin{bmatrix} v_x \\ v_y \\ v_z \\ \end{bmatrix}\\ \vec \Omega=\Omega\,\begin{bmatrix} 0 \\ \cos(\lambda) \\ \sin(\lambda) \\ \end{bmatrix}$$

$\Rightarrow$

$$\vec F_p=\left[ \begin {array}{c} m{\Omega}^{2}x-2\,m \left( \cos \left( \lambda \right) v_{{z}}-\sin \left( \lambda \right) v_{{y}} \right) \Omega\\ -m\sin \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) { \Omega}^{2}-2\,m\sin \left( \lambda \right) v_{{x}}\Omega \\ m\cos \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) {\Omega}^{2}+2 \,m\cos \left( \lambda \right) v_{{x}}\Omega\end {array} \right] $$

with $~\Omega=7.27\,10^{-5}~[\frac 1s]$ and $~\lambda=\frac{\pi}{2},~\vec\Omega=[0~,0~,\Omega]^T$

$$F_p=\left[ \begin {array}{c} 5.2 \,10^{-9}\,mx+ 1.4 \,10^{-4}\, mv_{{y}}\\ 5.2 \,10^{-9}\,my- 1.4 \,10^{-4}\,mv_{{x}}\\ 0.0\end {array} \right] $$

hence the pseudo force components (unit [N]) are very small and can neglected

enter image description here

the pseudo forces are:

$$\vec F_p=\underbrace{m\,\left(\vec \Omega\times (\vec \Omega\times \vec R)\right)}_{\text{Centrifugal force}}+ \underbrace{2\,m\,(\vec \Omega\times {\dot{\vec{R}}})}_{\text{Coriolis force}}$$

with:

$$\vec R=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}\qquad,{\dot{\vec{R}}}=\begin{bmatrix} v_x \\ v_y \\ v_z \\ \end{bmatrix}\\ \vec \Omega=\Omega\,\begin{bmatrix} 0 \\ \cos(\lambda) \\ \sin(\lambda) \\ \end{bmatrix}$$

$\Rightarrow$

$$\vec F_p=\left[ \begin {array}{c} m{\Omega}^{2}x-2\,m \left( \cos \left( \lambda \right) v_{{z}}-\sin \left( \lambda \right) v_{{y}} \right) \Omega\\ -m\sin \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) { \Omega}^{2}-2\,m\sin \left( \lambda \right) v_{{x}}\Omega \\ m\cos \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) {\Omega}^{2}+2 \,m\cos \left( \lambda \right) v_{{x}}\Omega\end {array} \right] $$

with $~\Omega=7.27\,10^{-5}~[\frac 1s]$ and $~\lambda=\frac{\pi}{2},~\vec\Omega=[0~,0~,\Omega]^T$

$$F_p=\left[ \begin {array}{c} 5.2 \,10^{-9}\,mx+ 1.4 \,10^{-4}\, mv_{{y}}\\ 5.2 \,10^{-9}\,my- 1.4 \,10^{-4}\,mv_{{x}}\\ 0.0\end {array} \right] $$

hence the pseudo force components (unit [N]) are very small and can neglected

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Eli
Eli
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enter image description here

your pseudo forces are:

$$\vec F_p=m\,\left(\vec \Omega\times (\vec \Omega\times \vec R)\right)+ 2\,m\,(\vec \Omega\times {\dot{\vec{R}}})$$$$\vec F_p=\underbrace{m\,\left(\vec \Omega\times (\vec \Omega\times \vec R)\right)}_{\text{Centrifugal force}}+ \underbrace{2\,m\,(\vec \Omega\times {\dot{\vec{R}}})}_{\text{Coriolis force}}$$

with:

$$\vec R=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}\qquad,{\dot{\vec{R}}}=\begin{bmatrix} v_x \\ v_y \\ v_z \\ \end{bmatrix}\\ \vec \Omega=\Omega\,\begin{bmatrix} 0 \\ \cos(\lambda) \\ \sin(\lambda) \\ \end{bmatrix}$$

$\Rightarrow$

$$\vec F_p=\left[ \begin {array}{c} m{\Omega}^{2}x-2\,m \left( \cos \left( \lambda \right) v_{{z}}-\sin \left( \lambda \right) v_{{y}} \right) \Omega\\ -m\sin \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) { \Omega}^{2}-2\,m\sin \left( \lambda \right) v_{{x}}\Omega \\ m\cos \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) {\Omega}^{2}+2 \,m\cos \left( \lambda \right) v_{{x}}\Omega\end {array} \right] $$

with $~\Omega=7.27\,10^{-5}~[\frac 1s]$ and $~\lambda=\frac{\pi}{2},~\vec\Omega=[0~,0~,\Omega]^T$

$$F_p=\left[ \begin {array}{c} 0.000000005285290000\,mx+ 0.0001454000000\, mv_{{y}}\\ 0.000000005285290000\,my- 0.0001454000000\,mv_{{x}}\\ 0.0\end {array} \right] $$$$F_p=\left[ \begin {array}{c} 5.2 \,10^{-9}\,mx+ 1.4 \,10^{-4}\, mv_{{y}}\\ 5.2 \,10^{-9}\,my- 1.4 \,10^{-4}\,mv_{{x}}\\ 0.0\end {array} \right] $$

hence the pseudo force components (unit [N]) are very small and can neglected

enter image description here

your pseudo forces are:

$$\vec F_p=m\,\left(\vec \Omega\times (\vec \Omega\times \vec R)\right)+ 2\,m\,(\vec \Omega\times {\dot{\vec{R}}})$$

with:

$$\vec R=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}\qquad,{\dot{\vec{R}}}=\begin{bmatrix} v_x \\ v_y \\ v_z \\ \end{bmatrix}\\ \vec \Omega=\Omega\,\begin{bmatrix} 0 \\ \cos(\lambda) \\ \sin(\lambda) \\ \end{bmatrix}$$

$\Rightarrow$

$$\vec F_p=\left[ \begin {array}{c} m{\Omega}^{2}x-2\,m \left( \cos \left( \lambda \right) v_{{z}}-\sin \left( \lambda \right) v_{{y}} \right) \Omega\\ -m\sin \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) { \Omega}^{2}-2\,m\sin \left( \lambda \right) v_{{x}}\Omega \\ m\cos \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) {\Omega}^{2}+2 \,m\cos \left( \lambda \right) v_{{x}}\Omega\end {array} \right] $$

with $~\Omega=7.27\,10^{-5}~[\frac 1s]$ and $~\lambda=\frac{\pi}{2},~\vec\Omega=[0~,0~,\Omega]^T$

$$F_p=\left[ \begin {array}{c} 0.000000005285290000\,mx+ 0.0001454000000\, mv_{{y}}\\ 0.000000005285290000\,my- 0.0001454000000\,mv_{{x}}\\ 0.0\end {array} \right] $$

hence the pseudo force components (unit [N]) are very small and can neglected

enter image description here

your pseudo forces are:

$$\vec F_p=\underbrace{m\,\left(\vec \Omega\times (\vec \Omega\times \vec R)\right)}_{\text{Centrifugal force}}+ \underbrace{2\,m\,(\vec \Omega\times {\dot{\vec{R}}})}_{\text{Coriolis force}}$$

with:

$$\vec R=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}\qquad,{\dot{\vec{R}}}=\begin{bmatrix} v_x \\ v_y \\ v_z \\ \end{bmatrix}\\ \vec \Omega=\Omega\,\begin{bmatrix} 0 \\ \cos(\lambda) \\ \sin(\lambda) \\ \end{bmatrix}$$

$\Rightarrow$

$$\vec F_p=\left[ \begin {array}{c} m{\Omega}^{2}x-2\,m \left( \cos \left( \lambda \right) v_{{z}}-\sin \left( \lambda \right) v_{{y}} \right) \Omega\\ -m\sin \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) { \Omega}^{2}-2\,m\sin \left( \lambda \right) v_{{x}}\Omega \\ m\cos \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) {\Omega}^{2}+2 \,m\cos \left( \lambda \right) v_{{x}}\Omega\end {array} \right] $$

with $~\Omega=7.27\,10^{-5}~[\frac 1s]$ and $~\lambda=\frac{\pi}{2},~\vec\Omega=[0~,0~,\Omega]^T$

$$F_p=\left[ \begin {array}{c} 5.2 \,10^{-9}\,mx+ 1.4 \,10^{-4}\, mv_{{y}}\\ 5.2 \,10^{-9}\,my- 1.4 \,10^{-4}\,mv_{{x}}\\ 0.0\end {array} \right] $$

hence the pseudo force components (unit [N]) are very small and can neglected

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Eli
Eli
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enter image description here

your pseudo forces are:

$$\vec F_p=m\,\left(\vec \Omega\times (\vec \Omega\times \vec R)\right)+ 2\,m\,(\vec \Omega\times {\dot{\vec{R}}})$$

with:

$$\vec R=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}\qquad,{\dot{\vec{R}}}=\begin{bmatrix} v_x \\ v_y \\ v_z \\ \end{bmatrix}\\ \vec \Omega=\Omega\,\begin{bmatrix} 0 \\ \cos(\lambda) \\ \sin(\lambda) \\ \end{bmatrix}$$

$\Rightarrow$

$$\vec F_p=\left[ \begin {array}{c} m{\Omega}^{2}x-2\,m \left( \cos \left( \lambda \right) v_{{z}}-\sin \left( \lambda \right) v_{{y}} \right) \Omega\\ -m\sin \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) { \Omega}^{2}-2\,m\sin \left( \lambda \right) v_{{x}}\Omega \\ m\cos \left( \lambda \right) \left( \cos \left( \lambda \right) z-\sin \left( \lambda \right) y \right) {\Omega}^{2}+2 \,m\cos \left( \lambda \right) v_{{x}}\Omega\end {array} \right] $$

with $~\Omega=7.27\,10^{-5}~[\frac 1s]$ and $~\lambda=\frac{\pi}{2},~\vec\Omega=[0~,0~,\Omega]^T$

$$F_p=\left[ \begin {array}{c} 0.000000005285290000\,mx+ 0.0001454000000\, mv_{{y}}\\ 0.000000005285290000\,my- 0.0001454000000\,mv_{{x}}\\ 0.0\end {array} \right] $$

hence the pseudo force components (unit [N]) are very small and can neglected