Problem 3.12
First we sketch the rotation diagrams:
a. To find velocity \(\boldsymbol{v}_C =\boldsymbol{v}_{C\in {\cal C}/{\cal E}}\) we take the time-derivative of \(\boldsymbol{r}_{OC}= L \boldsymbol{\hat{a}}_1 + l \boldsymbol{\hat{c}}_3\) relative to \({\cal E}\) (we differentiate term by term): \[\begin{align*} \boldsymbol{v}_C &= L \dot{\psi}\boldsymbol{\hat{e}}_3 \times \boldsymbol{\hat{a}}_1 + l (\dot{\psi}\boldsymbol{\hat{e}}_3 + \dot{\theta}\boldsymbol{\hat{a}}_2 + \dot{\phi}\boldsymbol{\hat{b}}_1) \times \boldsymbol{\hat{c}}_3 \\ & = L \dot{\psi}\boldsymbol{\hat{a}}_2 + l \dot{\psi}(\cos\theta\sin\phi\boldsymbol{\hat{b}}_1 + \sin\theta\boldsymbol{\hat{c}}_2) + l \dot{\theta}\cos\phi \boldsymbol{\hat{b}}_1 - l \dot{\phi}\boldsymbol{\hat{c}}_2 \end{align*}\]
Note: We cannot find \(\boldsymbol{v}_C\) as \(\boldsymbol{\omega}_{{\cal C}/{\cal E}}\times \boldsymbol{r}_{OC}\).
b. We set \(l=0\). Since we want to set \(\boldsymbol{G}\cdot\boldsymbol{\hat{c}}_1 = \boldsymbol{G}\cdot\boldsymbol{\hat{c}}_2 =0\), we need to find vector \(\boldsymbol{G}= -g \boldsymbol{\hat{e}}_3 - \boldsymbol{a}_C\) on basis \((\boldsymbol{\hat{c}}_1, \boldsymbol{\hat{c}}_2, \boldsymbol{\hat{c}}_3)\): with \(\boldsymbol{a}_C = L \ddot{\psi}\boldsymbol{\hat{a}}_2 - L \dot{\psi}^2 \boldsymbol{\hat{a}}_1\) we find \[\begin{align*} \boldsymbol{G}&= - g \cos\theta(\cos\phi\boldsymbol{\hat{c}}_3+ \sin\phi\boldsymbol{\hat{c}}_2) + g \sin\theta\boldsymbol{\hat{c}}_1 - L \ddot{\psi}\boldsymbol{\hat{a}}_2 + L \dot{\psi}^2 \boldsymbol{\hat{a}}_1 \\ & = -g \cos\theta(\cos\phi\boldsymbol{\hat{c}}_3+ \sin\phi\boldsymbol{\hat{c}}_2) + g \sin\theta\boldsymbol{\hat{c}}_1 - L \ddot{\psi}(\cos\phi\boldsymbol{\hat{c}}_2 - \sin\phi \boldsymbol{\hat{c}}_3) + L \dot{\psi}^2 (\cos\theta\boldsymbol{\hat{c}}_1 + \sin\theta\cos\phi \boldsymbol{\hat{c}}_3 +\sin\theta\sin\phi \boldsymbol{\hat{c}}_2) \end{align*}\] This gives the components \((G_1,G_2)\) on basis \((\boldsymbol{\hat{c}}_1,\boldsymbol{\hat{c}}_2)\): \[\begin{align*} G_1 & = \boldsymbol{G}\cdot\boldsymbol{\hat{c}}_1 =g \sin\theta+ L \dot{\psi}^2 \cos\theta\\ G_2 & = \boldsymbol{G}\cdot\boldsymbol{\hat{c}}_2 =- g \cos\theta\sin\phi -L \ddot{\psi}\cos\phi + L \dot{\psi}^2 \sin\theta\sin\phi \end{align*}\] The first equation gives angle \(\theta\): \[ \boxed{\tan\theta= - \frac{L}{g} \dot{\psi}^2} \] The second equation gives angle \(\phi\): \[ \tan\phi = L \frac{\ddot{\psi}}{-g \cos\theta+ L \dot{\psi}^2 \sin\theta} = \boxed{- \frac{L\ddot{\psi}}{\sqrt{g^2 + L^2 \dot{\psi}^4}} = \tan\phi} \] (assuming \(0\leq \theta\leq \pi/2\)).