# Acceleration in the rotating frame

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?)

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.

• Thank you very much for your reply! What I am still struggling however is how to reconcile (assuming the origins are shared and fixed, and $\omega$ is constant): $\frac{dv_{I}}{dt}\big|_{I} = \frac{dv_{R}}{dt}\big|_{I}= \frac{dv_{R}}{dt}\big|_{R} + \omega \times v_{R}$ using $A_{I} = v_{I} = v_{R} = A_{R}$ with the form above where, instead the transformation is applied twice as a second derivative to give instead $\frac{dv_{I}}{dt}\big|_{I} = frac{dv_{R}}{dt}\big|_{I}= \frac{dv_{R}}{dt}\big|_{R} + 2\omega\times v_{R} + \omega \times \omega \times x_{R}$? May 5, 2019 at 14:15
• I think you are assuming that $v_R= \frac{dx_R}{dt}\big|_R$, which is not the case. May 5, 2019 at 21:53
• Oh, I thought $\textbf{v}_{R} = v_{x, R}\textbf{i}_{R} + v_{y, R}\textbf{j}_{R} + v_{z, R}\textbf{k}_{R}$ and $\frac{d\textbf{x}_{R}}{dt}\big|_{R} = \frac{dx_{x, R}}{dt}\big|_{R}\textbf{i}_{R} + \frac{dx_{y, R}}{dt}\big|_{R}\textbf{j}_{R} + \frac{dx_{z, R}}{dt}\big|_{R} \textbf{k}_{R}$ (and then product rule also differentiates the basis vectors but these are fixed in the rotating frame giving zero). What is the form of $v_{R}$ in component form? Is $\frac{dx_{x, R}}{dt}\big|_{R} = v_{x, R}$ not the case? May 5, 2019 at 22:31
• No, $v_{x,R} = \frac{dx_{x, R}}{dt}\big|_{R} + \left(\omega \times x_R\right)_x$ May 5, 2019 at 22:40
• So $\textbf{v}_{R} = \frac{d\textbf{x}_{R}}{dt}\big|_{I} = \frac{d\textbf{x}_{R}}{dt}\big|_{R} + \omega \times \textbf_{x}_{R}$? What does $v_{R}$ represents physically and compared to $v_{I}$? Is velocity vector as seen by a observer fixed in the rotating frame $\frac{d\textbf{x}_{R}}{dt}\big|_{R}$? (Thank you very much for the help, this has always confused me) May 5, 2019 at 22:52