We define two bases of unit vectors: \((\boldsymbol{\hat{x}},\boldsymbol{\hat{y}},\boldsymbol{\hat{z}})\) attached to referential 0, and \((\boldsymbol{\hat{i}}, \boldsymbol{\hat{j}}, \boldsymbol{\hat{k}})\) attached to plate 1. The orientation of 1 relative to 0 is defined by the two angles \(\theta\) and \(\phi\). The angle \(\theta\) defines the orientation of (right-handed) basis \((\boldsymbol{\hat{u}}, \boldsymbol{\hat{j}}, \boldsymbol{\hat{z}})\) where \(\boldsymbol{\hat{j}}\) is directed along side \(AB\). \(\boldsymbol{\hat{u}}\) is directed along \(OP\) perpendicular to \(AB\). The angle \(\phi\) defines the orientation of \((\boldsymbol{\hat{i}}, \boldsymbol{\hat{k}})\) relative to \((\boldsymbol{\hat{u}}, \boldsymbol{\hat{z}})\): \(\boldsymbol{\hat{k}}\) is directed along line \(PQ\), \(\boldsymbol{\hat{i}}\) is perpendicular to the plate.
The variables \(x, y, \phi\) can all be expressed in terms of angle \(\theta\). With the diagrams shown below, we read the relationships: \[ \boxed{ x = 2b \sin\theta, \quad y = 2b \cos\theta, \quad 2a \sin \phi = 2b \sin\theta \cos\theta } \] The last equation is found by expressing length \(OP\) first in terms of \(2a\) and \(\phi\) (in triangle OQP) then in terms of \(2b\) and \(\theta\) (in triangle OAP).
Then we find \(\boldsymbol{\omega}_{1/0} = \dot{\theta} \boldsymbol{\hat{z}}- \dot{\phi} \boldsymbol{\hat{j}}\) and \(\boldsymbol{v}_{A/{\cal E}} = \dot{x} \boldsymbol{\hat{x}}= 2b \dot{\theta} \cos\theta\boldsymbol{\hat{x}}\). Hence, \[ \{ {\cal V} _{1 / 0} \} = \left\{ \begin{array}{c} \boldsymbol{\omega}_{1 /0 } = \dot{\theta} \boldsymbol{\hat{z}}- \dot{\phi} \boldsymbol{\hat{j}}\\ \boldsymbol{v}_{A/0} = 2b \dot{\theta} \cos\theta\boldsymbol{\hat{x}} \end{array} \right\} \] In particular we can find the velocity of \(Q \in 1\): \[ \boldsymbol{v}_{Q \in 1/0} = \boldsymbol{v}_{A/0} + \boldsymbol{\omega}_{1/0} \times \boldsymbol{r}_{AQ} \] With \(\boldsymbol{r}_{AQ} = \boldsymbol{r}_{AP} + \boldsymbol{r}_{PQ}= 2b \sin^{2} \theta\boldsymbol{\hat{j}}+ 2a \boldsymbol{\hat{k}}\), we find \[ \boldsymbol{v}_{Q\in 1/0} = 2b \dot{\theta} \cos\theta\boldsymbol{\hat{x}}+ (\dot{\theta} \boldsymbol{\hat{z}}- \dot{\phi} \boldsymbol{\hat{j}})\times (2b \sin^{2} \theta\boldsymbol{\hat{j}}+ 2a \boldsymbol{\hat{k}}) = 2b \dot{\theta}\cos\theta(\cos\theta\boldsymbol{\hat{u}}-\sin\theta\boldsymbol{\hat{j}}) -2b \dot{\theta}\sin^2 \theta\boldsymbol{\hat{u}}-2a \dot{\theta}\sin\phi \boldsymbol{\hat{j}}-2a \dot{\phi}\boldsymbol{\hat{i}} \] With \(\boldsymbol{\hat{i}}= \cos\phi \boldsymbol{\hat{u}}+\sin\phi \boldsymbol{\hat{z}}\), we obtain \[ \boldsymbol{v}_{Q\in 1/0} =\Big( 2b(\cos^2\theta-\sin^2\theta) -2a \dot{\phi}\cos\phi \Big) \boldsymbol{\hat{u}} -2a \dot{\phi}\sin\phi \boldsymbol{\hat{z}} - \dot{\theta}(2a \sin\phi +2b \cos\theta\sin\theta) \boldsymbol{\hat{j}} \] After using the constraint equation \(2a \sin \phi = 2b \sin\theta \cos\theta\), the \(\boldsymbol{\hat{u}}\)-component vanishes and we obtain the final expression \[ \boxed{ \boldsymbol{v}_{Q\in 1/0} = -4b \dot{\theta}\sin \theta\cos\theta\boldsymbol{\hat{j}}- 2a \dot{\phi} \sin\phi \boldsymbol{\hat{z}} } \] which, as expected, is not equal to \[ \boldsymbol{v}_{Q/0}= \left( { d {\bf r}_{OQ} \over dt}\right)_{0}= - 2a \dot{\phi} \sin\phi \boldsymbol{\hat{z}} \] Note: we can also obtain $_{Q/0} $ as the difference \(\boldsymbol{v}_{Q/0}-\boldsymbol{v}_{Q/1}\). The velocity \(\boldsymbol{v}_{Q/1}\) is obtained from \[ \boldsymbol{v}_{Q/1}= \left( \frac{ d {\bf r}_{AQ}}{dt}\right)_{1} = \left( \frac{d}{dt}( 2b \sin^{2} \theta\boldsymbol{\hat{j}}+ 2a \boldsymbol{\hat{k}})\right)_{1} = 4b \dot{\theta}\sin \theta\cos\theta\boldsymbol{\hat{j}} \] We recover the same result.