Confusion have vector-nature of velocity in rotating frame

Question

Consider an inertial frame of reference, $s$, (henceforth coordinates in this system will be small letters), and a frame $S$ rotating coaxially ($z$-axis) with respect to this frame (all coordinates in big letters), with constant angular velocity $\omega$.

Now consider a particle with position $\mathbf{x}$ in $s$, so that its coordinate in $S$ can be written as $\mathbf{X} = T(\theta)\mathbf{x}$, where $T(\theta)$ is the appropriate rotation/transformation matrix for angle $\theta$. (Note: Consider motion in $s$ in only plane $xy$, and $z$ axis is common, so motion in $S$ is in plane $XY$).

So $\mathbf{\dot{X}} = T(\theta) \mathbf{\dot{x}} + \omega\dfrac{dT}{d\theta}\mathbf{x} $.

But any vector in $s$ must satisfy that its coordinates will have to transform by the $T(\theta)$ transformation rule (to qualify as a vector).

By that logic, velocity in $S$, $\mathbf{V}$ must be such that $\mathbf{V} = T(\theta)\mathbf{\dot{x}}$, as $\mathbf{\dot{x}}$ is velocity is $s$. This must imply that $\mathbf{\dot{X}}$ is not the velocity in the $S$ frame, as it has an extra term as shown above.

What is going wrong here?

Answer 1

If you write $$ \boldsymbol{X}=T(\theta)\boldsymbol{x} $$ then indeed $$ \dot{\boldsymbol{X}}=\dot{T}(\theta)\boldsymbol{x}+T(\theta)\dot{\boldsymbol{x}}\tag{1} $$ Thus, $\dot{\boldsymbol{X}}$ has two parts: one is due to the motion of the particle in the rotating frame - that's $\dot{\boldsymbol{x}}$ - and the other is due to the velocity of the rotating frame w/r to the inertial frame - that's the $\dot{T}(\theta)\boldsymbol{x}$ part.

Answer 2

If the position vector of the particle with respect to $\:s\:$⁽¹⁾ is $\:\mathbf{x}\:$ then its velocity $\:\mathbf{V}\:$ with respect to $\:S\:$ is the resultant of two velocities : \begin{equation} \mathbf{V}=[\mathbf{v}]_{S}+\mathbf{W} \tag{01} \end{equation} where $\:[\mathbf{v}]_{S}\:$ is its velocity with respect to $\:s\:$ but expressed in $\:S-$coordinates and $\:\mathbf{W}\:$ its velocity due to the motion (here rotation) of system $\:s\:$ as a whole with respect to $\:S\:$.

Now \begin{equation} \mathbf{v}=\mathbf{\dot{x}} \tag{02} \end{equation} and there is no objection, as you claim, that its expression in $\:S-$coordinates is \begin{equation} [\mathbf{v}]_{S}=T(\theta)\mathbf{\dot{x}}=T(\theta)\mathbf{v} \tag{03} \end{equation} The extra term we have is $\:\mathbf{W}\:$ \begin{equation} \mathbf{W}=\dfrac{\mathrm{d}T}{\mathrm{d}t}\mathbf{x}=\dot{T}\mathbf{x} \tag{04} \end{equation} and if we express $\:\mathbf{x}\:$ in $\:S-$coordinates \begin{equation} \mathbf{x}=T^{-1}(\theta)\mathbf{X}=T^{\boldsymbol{\top}}(\theta)\mathbf{X} \tag{05} \end{equation} then \begin{equation} \mathbf{W}=\dot{T}T^{\boldsymbol{\top}}\mathbf{X} \tag{06} \end{equation} But \begin{equation} TT^{\boldsymbol{\top}}=I=T^{\boldsymbol{\top}}T \quad \Longrightarrow \quad\dot{T}T^{\boldsymbol{\top}}=-\left(\dot{T}T^{\boldsymbol{\top}}\right)^{\boldsymbol{\top}} \tag{07} \end{equation} that is the $\:3\times 3\:$ matrix $\:\dot{T}T^{\boldsymbol{\top}}\:$ is antisymmetric, so it could be expressed as \begin{equation} \dot{T}T^{\boldsymbol{\top}} \equiv \begin{bmatrix} 0&-\Omega_3&+\Omega_2\\ &&\\ +\Omega_3&0&-\Omega_1\\ &&\\ -\Omega_2&+\Omega_1&0 \end{bmatrix} \:=\:\boldsymbol{\Omega}\:\boldsymbol{\times} \tag{08} \end{equation} and finally \begin{equation} \mathbf{W}=\boldsymbol{\Omega}\boldsymbol{\times}\mathbf{X} \tag{09} \end{equation}

---

^{(1) since we talk about kinematics and not kinetics there is no need for any one of the systems to be inertial.}