Discussion
Consider S' a rotating frame rotating with constant angular velocity \(\omega\) with respect to stationary reference frame S (S and S' are having a common origin). If \(\bar{A}\) is any vector represented in S' frame, then \(\frac{d^{2}\bar{A}}{dt^{2}}\) in S frame is given by :
Answer: A
The correct expression is the one with \(+2\bar{\omega}\times\frac{d\bar{A}}{dt}\big|_{S'}\) and \(+\bar{\omega}\times\bar{\omega}\times\bar{A}\).
For any vector, the rates of change in the two frames are related by \(\frac{d}{dt}\big|_{S} = \frac{d}{dt}\big|_{S'} + \bar{\omega}\times\). Applying this operator twice:
\[ \begin{aligned} \ddot{\bar{A}}_{S} &= \Big(\tfrac{d}{dt}\Big|_{S'} + \bar{\omega}\times\Big)(\dot{\bar{A}}' + \bar{\omega}\times\bar{A}) \\ &= \ddot{\bar{A}}' + \dot{\bar{\omega}}\times\bar{A} + 2\bar{\omega}\times\dot{\bar{A}}' \\ &\quad + \bar{\omega}\times(\bar{\omega}\times\bar{A}) \end{aligned} \]
where primes mean rates seen in \(S'\). The cross term appears twice, once from each application, which gives the factor 2 of the Coriolis term; the last term is the centripetal term. Both carry a plus sign here. The minus signs in the other options are what appear when the equation is solved for the acceleration in \(S'\), where the Coriolis and centrifugal terms move to the other side.
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