Acceleration in the rotating frame

Question

I understand that it is possible to express a vector $\textbf{A}$ in two different frames such that $\textbf{A}_{I} = A_{x, I}\hat{\textbf{i}}_{I} + A_{y, I}\hat{\textbf{j}}_{I} + A_{z, I}\hat{\textbf{k}}_{I}$ or $\textbf{A}_{R} = A_{x, R}\hat{\textbf{i}}_{R} + A_{y, R}\hat{\textbf{j}}_{R} + A_{z, R}\hat{\textbf{k}}_{R}$. Then differentiating both in terms of the inertial frame $I$ and applying product rule to the rotating frame gives: $\frac{d \textbf{A}_{I}}{dt}\big|_{I} = \frac{d \textbf{A}_{R}}{dt}\big|_{I} = \big(\frac{dA_{x, R}}{dt} \hat{\textbf{i}}_{R} + \frac{dA_{y, R}}{dt} \hat{\textbf{j}}_{R} + \frac{dA_{z, R}}{dt} \hat{\textbf{k}}_{R}\big) + \big( A_{x, R}\frac{d\hat{\textbf{i}}_{R}}{dt} + A_{y, R}\frac{d\hat{\textbf{j}}_{R}}{dt} + A_{z, R}\frac{d\hat{\textbf{k}}_{R}}{dt} \big) = \frac{d\textbf{A}_{R}}{dt}\big|_{R} + \mathbf{\omega}\times \textbf{A}_{R}$.

If this is then applied to a velocity vector such that $\frac{d\textbf{v}_{I}}{dt}\big|_{I} = \frac{d\textbf{v}_{R}}{dt}\big|_{R} + \mathbf{\omega}\times \textbf{v}_{R}$, how is the usual form for acceleration in the inertial frame derived from this ? (also in proofs I have read online most say that $\textbf{A}_{I} = \textbf{A}_{R}$ but why not $\textbf{A}_{I} = R \textbf{A}_{R}$ where R is the rotation between the two bases?)

Answer 1

This is a common confusion in non-inertial dynamics. $A_R$ and $A_I$ are the same vectors, just expressed in a different basis, so $A_R = A_I$. Now for your other question:

$\frac{d\textbf{v}_{R}}{dt}\big|_{I}= \frac{d^2\textbf{x}_{R}}{dt^2}\big|_{I}= \frac{d}{dt}\left(\frac{d\textbf{x}_{R}}{dt}\big|_{I}\right)\big|_{I}=\left(\frac{d}{dt}\big|_{R} + \omega \times\right)\left(\frac{d\textbf{x}_{R}}{dt}\big|_{R} + \mathbf{\omega}\times \textbf{x}_{R}\right) =\frac{d^2\textbf{x}_{R}}{dt^2}\big|_{R} + \omega \times \frac{d\textbf{x}_{R}}{dt}\big|_{R} + \omega \times (\omega \times x_R) + \frac{d}{dt}\left(\omega \times x_R\right)\big|_{R} = \frac{d^2\textbf{x}_{R}}{dt^2}\big|_{R} + 2\omega \times \frac{d\textbf{x}_{R}}{dt}\big|_{R} + \omega \times (\omega \times x_R) + \frac{d\omega}{dt}\big|_{R} \times x_R $

If the origin of the non-intertial frame is also moving with respect to the fixed frame, then we get:

$\frac{d\textbf{v}_{R}}{dt}\big|_{I}= \frac{d^2R}{dt^2}\big|_{I} +\frac{d^2\textbf{x}_{R}}{dt^2}\big|_{R} + 2\omega \times \frac{d\textbf{x}_{R}}{dt}\big|_{R} + \omega \times (\omega \times x_R) + \frac{d\omega}{dt}\big|_{R} \times x_R $,

where $\frac{d^2R}{dt^2}\big|_{I}$ is the acceleration of the origin of the non-intertial frame with respect to the fixed frame.