sometimes, i will confuse on changing frame, especially between rotating frame and lab frame.

the notation plays an important role. for a rotating vector. $V_S' = R_n ( \omega t ) V_S$

where $V_S$ is the static frame vector. $R_n ( \theta )$ is the rotation operator, with right -hand rotation is positive $\theta$ around direction $n$. for clearance, when we think on a rotating, $\omega$ is always positive. $R_n ( \theta ) \Rightarrow \begin {bmatrix} \theta > 0 & \rightarrow & + \\ \theta <0 & \rightarrow & - \end {bmatrix}$

however, if we regard the vector does not rotate but is the frame rotating, which is equivalent as rotating backward ( left-hand or negative ). the expression is the same but with different notation. $V_R = R_n ( \omega_R t ) V_S$ $V_R$ is the rotating frame vector, $\omega_R$ is the rate of rotating of the rotating frame respect to the static frame. for the rotating frame is rotating forward ( positive) $V_R = R_n ( - \omega_R t ) V_S$

if both the vector is rotating an the rotating frame is rotating, in same direction, with different rate. $V_R' = R_n ( - \omega_R t ) V_S' = R_n ( - \omega_R t ) R_n( \omega t ) V_S$

and combine the rotation operator. $V_R' = R_n ( (\omega - \omega_R ) t ) V_S$

which make perfect sense.

the principle can apply to any frame transformation.

__________________  Examples _____________________

a rotation on positive $\theta$ direction for the vector. $V_S = ( cos (\alpha ) , sin(\alpha ) )$ $V_S' = ( cos( \theta + \alpha ) , sin(\theta + \alpha) ) = V_R$ $(A_x)_S = (1,0)_S$ $(A_x)_R = ( cos ( \theta ) , sin ( \theta ) )_R$ $( A_x)_R = ( 1, 0 )_R = ( cos ( \theta ) , sin ( - \theta ) )_S$

if we add the frame reference on as a subscript, things will be much clear. since the same vector can have different coordinate expression. so, we better mark the coordinate with reference system.