First Derivation.
Starting from:
$$ \frac{d \mathbf r}{dt} = \frac{d}{dt} \left [ \sum_i r_i \mathbf e_i \right ] = \sum_i\frac{d r_i}{dt} \mathbf e_i + \sum_i r_i\frac{d \mathbf e_i}{dt} = $$ $$ = \sum_i\frac{d r_i}{dt} \mathbf e_i + \mathbf \Omega \times \sum_i r_i \mathbf e_i $$
$$ \left ( \frac{d\mathbf r}{dt} \right )_S = \left ( \frac{d\mathbf r}{dt} \right )_{S'} + \mathbf \Omega\times \mathbf r $$ Where $(d\mathbf r / dt)_S$, denotes the rate of change of $\mathbf r$, as measured from the $S$ frame. Same with $(d\mathbf r / dt)_{S'}$ but in the $S'$ frame. For knowing the second derivative of $\mathbf r$, we have: $$ \frac{d ^2\mathbf r}{dt ^2} = \ddot{\mathbf r}' + \dot{\mathbf{\Omega}} \times \mathbf r' + 2 \mathbf \Omega \times \dot{\mathbf r}' + \mathbf \Omega \times (\mathbf \Omega \times \mathbf r') $$ to simplify I've defined $\mathbf r$ and $\mathbf r'$ as the vector seen in the $S$ frame and $S'$ frame, respectively. I've took this derivation from the Book "Mechanics - R.Taylor" pg. 339.
Second Derivation:
Starting from: $$ \mathbf r' = \mathcal R \mathbf r $$ where $\mathcal R$ is a rotation matrix and knowing that: $$ \hat \Omega = \dot{\mathcal R}^T \mathcal R $$ $$ \hat \Omega \mathbf r = \mathbf \Omega \times \mathbf r $$ performing a second order derivation on both sides we arrive to: $$ \frac{d ^2\mathbf r}{dt ^2} = \mathcal R^T \left ( \ddot{\mathbf r}' + \dot{\mathbf{\Omega'}} \times \mathbf r' + 2 \mathbf \Omega' \times \dot{\mathbf r}' + \mathbf \Omega' \times (\mathbf \Omega' \times \mathbf r') \right ) $$
I found this derivation in our professor notes. This equations are almost identical (excluding the $\mathcal R^T$ factor and the primed $\mathbf \Omega$ on the second one), but shouldn't it be the same? I cannot see the relation (or difference) between assuming that $\mathbf r' = \mathcal R \mathbf r$, or relating both frames of reference via their derivatives? Is there any relation, or a way to go from one equation to another.
