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

https://docs.google.com/present/embed?id=dct7rkzd_1145djq6h4g3&size=m

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.

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