Let the orientation of a coordinate frame $\{b\}$ w.r.t. a static coordinate frame $\{a\}$ be expressed by a rotation matrix $R_{ab}\in SO(3)$ whose columns represent the coordinates of the unitary axis of $\{b\}$ w.r.t. $\{a\}$. Furthermore, let $R_{ab}$ depend on time $t$ i.e. we have $R_{ab}(t)$.
It is well known that the angular velocity, $\omega_{b}\in\mathbb{R}^3$, of the frame $\{b\}$ w.r.t. the frame $\{a\}$ can be obtained by:
$$ [\omega_b]_{\times} = \dot{R}_{ab}{R_{ab}}^T \label{1}\tag{1}$$ Where $[\cdot]_\times$ represents the anti-symmetric matrix related to $\omega_b$.
However I've not been able to read anywhere a similar expression for the angular velocity $\dot{\omega}_b$. I believe it can be directly obtained by differentiating eq. \ref{1} w.r.t. time i.e.:
$$ [\dot{\omega}_b]_\times = \ddot{R}_{ab}{R_{ab}}^T + \dot{R}_{ab}{\dot{R}_{ab}}^T \label{2}\tag{2}$$ In this way $[\dot{\omega}_b]_\times$ is also an anti-symmetric matrix, something that contradicts the answer in this post. Why $\dot{\omega}_b$ should not be computed as eq. \ref{2} ?