Suppose that the function of time $t \rightarrow q \in S^{2}$ represents a rotational motion of the spherical pendulum. Since $q \in S^{2}$, it follows that the time derivative $\dot{q} \in \mathrm{T}_{q} \mathrm{~S}^{2}$ is a tangent vector of $\mathrm{S}^{2}$ at $q \in \mathrm{S}^{2}$. Thus, $\dot{q}$ is orthogonal to $q$, that is $(\dot{q} \cdot q)=0$. This implies that there is a vector-valued function of time $t \rightarrow \omega \in \mathbb{R}^{3}$ such that
$$
\dot{q}=\omega \times q
$$
or equivalently
$$
\dot{q}=S(\omega) q
$$
where
$$
S(\omega)=\left[\begin{array}{ccc}
0 & -\omega_{3} & \omega_{2} \
\omega_{3} & 0 & -\omega_{1} \
-\omega_{2} & \omega_{1} & 0
\end{array}\right]
$$
is a $3 \times 3$ skew-symmetric matrix function introduced. There is no loss of generality in requiring that $(\omega \cdot q)=0$. This observation is true since $\omega \in \mathbb{R}^{3}$ can be decomposed into the sum of a part in the direction of $q$ and a part that is orthogonal to $q$ and the former part does not contribute to the value of $\dot{q}$.
MATH4077 COURSE NOTES ：
Consider the rotational kinematics of a rigid body given by
$$
\dot{R}=R S(\omega),
$$
where $R \in \mathrm{SO}(3), \omega \in \mathbb{R}^{3}$ and
$$
S(\omega)=\left[\begin{array}{ccc}
0 & -\omega_{3} & \omega_{2} \
\omega_{3} & 0 & -\omega_{1} \
-\omega_{2} & \omega_{1} & 0
\end{array}\right] .
$$
Suppose that the angular velocity vector is given in terms of the configuration by
$$
\omega=\sum_{i=1}^{3} e_{i} \times R^{T} e_{i}
$$
where $e_{1}, e_{2}, e_{3}$ denote the standard unit basis vectors in $\mathbb{R}^{3}$. This defines a closed loop kinematics system.