In Lanczos Variational Principles of Mechanics there's a subsection for apparent forces in a rotating reference system. It states that
If the origin O of the reference system is kept fixed, the radius vectors R and R' in the absolute system S and in the moving system S' are the same: R = R'. Nevertheless, the velocities and accelerations measured in both systems differ from each other because of the fact that rates of change observed in the two systems are different. If a certain vector B is constant in S' it rotates with the system and thus, if observed in S, undergoes in the time dt an infinitesimal change $$dB = (\Omega \times B) dt$$ Therefore $$\frac{dB}{dt} = (\Omega \times B)$$ and $$\frac{d'B}{dt}=0 $$ where $\frac{d'}{dt}$ refers to the operation of observing the rate of change of a quantity in the moving system S'. If the vector B is not constant in S', so that $\frac{d'B}{dt}\ne0$, then we find, by making use of the principle of infinitesimal processes, that $$\frac{dB}{dt}=\frac{d'B}{dt}+\Omega \times B$$ applying this to the radius vector $R=R'$ we get $$\frac{dR}{dt}=\frac{d'R'}{dt}+\Omega \times R'$$ which gives $$v = v' + \Omega \times R'$$ Differentiating then gives $$\frac{dv}{dt} = \frac{dv'}{dt} + \dot\Omega \times R' + \Omega \times \frac{dR'}{dt}$$
It is stated that this can be expressed in terms of quantities only belonging to S' as $$\frac{dv}{dt} = \frac{d'v'}{dt} + 2\Omega \times v' + \Omega \times (\Omega \times R') + \Omega \times R'$$ by using $$\frac{dB}{dt}=\frac{d'B}{dt}+\Omega \times B$$
But I'm failing to see which quantity doesn't belong to S' and how they are getting the final equation. Any help would be greatly appreciated!
