This is the common problem of a charged particle moving in a static electric and magnetic field. Say $\textbf{E}=(E_x,0,0)$ and $\textbf{B}=(0,0,B_z)$.
In the inertial frame of reference, the equation of motion is (1): \begin{equation} \frac{d \textbf{v} }{dt} = -\frac{q \textbf{B} }{m}\times \textbf{v} + \frac{q}{m}\textbf{E} \end{equation}
We can find equations for $v_x$ an $v_y$ and see that the resulting motion is a circular orbit with a constant drift velocity $v_d=\frac{E_x}{B_z}$.
Surely I should get the same answer if I solve the problem in a rotating frame of reference?
I know that (2): $$ \frac{d \textbf{v} }{dt} \vert_{Inertial} = \frac{d \textbf{v} }{dt} \vert_{Rotational} + \boldsymbol{\omega}\times\textbf{v};$$
If I use Eq. (1) as the LHS of Eq. (2), and choose $ \boldsymbol{\omega}=-\frac{q \textbf{B} }{m}$, then I get (3):
$$ \frac{d \textbf{v} }{dt} \vert_{Rotational} = \frac{q}{m}\textbf{E};$$
How do I obtain a constant drift velocity (as mentioned before) from this? Have I used any formula incorrectly? Does the electric field $\textbf{E}=(E_x,0,0)$ change form in the rotating frame?