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\noindent Solution 35: \bigskip

\noindent $\dot{\vec{\phi}}\ =\ \dot\phi\,\hat{Z}\,$. 
\bigskip

\noindent $\dot{\vec{\theta}}$ is along the node axis ON. Therefore,
$\dot{\vec{\theta}}\ =\ \dot{\theta}\,\cos\phi\,\hat{X}+\dot{\theta}
\,\sin\phi\,\hat{Y}\,$.
\bigskip

\noindent $\dot{\vec{\psi}}\ =\ \dot{\psi}\,\hat{x}_3$ and $\hat{x}_3\ 
=\ \cos\theta\,\hat{Z} + \sin\theta\,\hat{\rho}$, where $\hat\rho$ is 
in the $X-Y$ plane, $\hat\rho\ =\ \sin\phi\,\hat{X} - \cos\phi\,\hat{Y}$.
Therefore, $\dot{\vec{\psi}}\ =\ \dot{\psi}\,\cos\theta\,\hat{Z} +
\dot{\psi}\,\sin\phi\,\hat{X} - \dot{\psi}\,\cos\phi\,\hat{Y}\,$.
\bigskip

\noindent Collecting all terms ($\Omega_X=\hat{X}\cdot\vec\Omega$ and
so on):
\begin{eqnarray} \nonumber
  \Omega_X &=& \dot{\theta}\,\cos\phi + \dot{\psi}\,\sin\phi\,,
  \\ \nonumber
  \Omega_Y &=& \dot{\theta}\,\sin\phi - \dot{\psi}\,\cos\phi\,,
  \\ \nonumber
  \Omega_Z &=& \dot\phi + \dot\psi\,\cos\theta\,.
\end{eqnarray}


 
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