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So we have a rotating platform with two frames o reference: the one which is static, $O:\{x,y,z\}$, and the one wich is rotating along the platform, $O':\{x',y',z'\}\ (z\equiv z')$. The platform is spinning at $\vec{\omega}=\omega\hat{k}$, the particle is initially at $(0,0,z_o)$ and has initial velocity $\vec{v}_o=v_o\hat{i}$. Therefore, the trajectory of the particle measured in $O$ would be $\vec{r}= v_ot\hat{i}+(z_o-\frac{1}{2}gt^2)\hat{k}$. Now, the problem comes when you measure it in $O'$. The question states it'd describe the Archimedean spiral, $r'=(v_o/\omega)\theta$, in the $x'-y'$ plane, meaning $x'=v_ot\cos{\omega t}$ and $y'=v_ot\sin{\omega t}$, and that $z'=z=z_o-\frac{1}{2}gt^2$. That's one way, but I wanted to try it by using fictitious forces (Coriolis and centrifugal ones). After some calculations, I got that $\vec{a}'=-\omega^2v_ot\hat{i}-2\omega v_o\hat{j}-g\hat{k}$. Integrating these expressions twice, we should get $\vec{r}'=(v_ot-\frac{1}{6}\omega^2v_ot^3)\hat{i}-\omega v_ot^2\hat{j}+(z_o-\frac{1}{2}gt^2)\hat{k}$. When I animate both trajectories, they only coincide at the very start, but they aren't equal. Did I do this right?

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I just noticed that $\hat{i}$ and $\hat{j}$ are: $$\begin{pmatrix} \hat{i}\\ \hat{j} \end{pmatrix} =\begin{pmatrix} \cos(\omega t) & \sin(\omega t)\\ -\sin(\omega t) & \cos(\omega t) \end{pmatrix} \begin{pmatrix} \hat{i}'\\ \hat{j}' \end{pmatrix} $$ So when I develop these in $\vec{a}'$, I get the actual correct acceleration expressed in the unit vectors of $O'$.

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