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I'm solving the following problem from the book "Analytical Mechanics for Relativity and Quantum Mechanics" by Oliver Davis Johns:

Let $R(t)=R_a(t)R_b(t)$ be a rotation matrix. Show that the angular velocity associated to the product is \begin{gather*} \vec{w}=\vec{w_a}+R_a(t)\vec{w_b}(t) \end{gather*}

So, my attempt, which seems correct, is: \begin{gather*} \vec{x}(t)=R(t)\vec{x}(0)\tag{1} \\ R(t)=R_a(t)R_b(t)\tag{2} \\ \dot{R}(t)=\dot{R_a}(t)R_b(t)+R_a(t)\dot{R_b}(t)\tag{3} \\ \dot{\vec{x}}(t)=\dot{R}(t)R^{t}(t)\vec{x}(t)\tag{4} \end{gather*} Using some matrices properties, we have
\begin{gather*} \dot{R}(t)R^{t}(t)=(\dot{R_a}(t)R_b(t)+R_a(t)\dot{R}_b(t))(R_b^{t}(t)R_a^{t}(t)) \tag{5} \\ =\dot{R_a}(t)IR_a(t)+R_a(t)\dot{R}_b(t)R_b(t)R_a^{t}(t)\tag{6} \\ =\dot{R_a}(t)R_a(t)+T(t) \\ where \\ T(t)=R_a(t)\dot{R}_b(t)R_b(t)R_a^{t}(t) \end{gather*}

On equation 6, how do I manipulate the term $T(t)$ so when mulplitying it by $x(t)$ I get $\vec{w_b}(t)$?

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You are correct that $$\dot{R}(t)R(t)^T = \dot{R}_a(t)R_a(t)^T + R_a(t)\Big(\dot{R}_b(t)R_b(t)\Big)R_a(t)^T$$ Then, define the skew-symmetric matrices $\omega(t) = \dot{R}(t)R(t)^T,\,\,\, \omega_a(t) = \dot{R}_a(t)R_a(t)^T$ and $\omega_b(t) = \dot{R}_b(t)R_b(t)^T$. Then your expression can be written as $$\omega(t) = \omega_a(t) + R_a(t) \,\omega_b(t) \,R_a(t)^T$$ Then, by a theorem from the Lie group theory of the orthogonal group of three space, there are unique vectors $\vec{w}(t), \,\,\, \vec{w}_a(t)$ and $\vec{w}_b(t)$ such that for any three dimensional vector $\vec{x}$ \begin{align} &\omega(t)\, \vec{x} = \vec{w}(t) \times \vec{x}\\ &\omega_a(t)\, \vec{x} = \vec{w}_a(t) \times \vec{x}\\ &\omega_b(t)\, \vec{x} = \vec{w}_b(t) \times \vec{x} \end{align} Moreover, \begin{align} \Big(R_a(t)\, \omega_b(t) \, R_a(t)^T\Big) \, \vec{x} &= R_a(t)\Big( \,\omega_b(t)\big(\, R_a(t)^T \, \vec{x}\,\big)\,\Big) = R_a(t)\Big( \,\vec{w}_b(t)\times \big(\, R_a(t)^T \, \vec{x}\,\big)\,\Big)\\ &= \Big( R_a(t) \,\vec{w}_b(t)\times \big(\,R_a(t) R_a(t)^T \, \vec{x}\,\big)\,\Big) \\ &=\Big( R_a(t) \,\vec{w}_b(t)\Big) \times \vec{x} \end{align} That is why for any vector $\vec{x}$ $$\omega(t)\, \vec{x} = \omega_a(t)\, \vec{x} + \Big(R_a(t) \,\omega_b(t) \,R_a(t)^T\Big) \vec{x}$$ can be written as $$\vec{w}(t)\times \vec{x} = \vec{w}_a(t) \times \vec{x} + \Big( R_a(t) \,\vec{w}_b(t)\Big) \times \vec{x}$$ which implies that $$\vec{w}(t) = \vec{w}_a(t) + R_a(t) \,\vec{w}_b(t)$$

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