Fictitious force

Remarks

The formula comes from differentiating $x = o + R X$ twice: with $\dot R = R\, [\Omega]_\times$ one gets $\ddot x = \ddot o + R\big(\ddot X + 2\,\Omega \times \dot X + \dot\Omega \times X + \Omega \times (\Omega \times X)\big)$, and $m\ddot x$ is the force in the inertial frame. The four terms are the translational force $-m R^T \ddot o$, the Euler force $-m\,\dot\Omega \times X$, the Coriolis force $-2m\,\Omega \times \dot X$ and the centrifugal force $-m\,\Omega \times (\Omega \times X)$. Example: in a car moving in a straight line with constant acceleration vector $A$, $R$ is constant and $\ddot o = A$, so a passenger feels $F_{\mathrm{fict}} = -m R^T A$, a push backward. Non-example: the gravitational force on the passenger is not fictitious, because it is exerted by another body and is present in every frame. For a frame moving at constant velocity without rotation, $F_{\mathrm{fict}} = 0$, in agreement with Galilean invariance.