# Trajectory of particle thrown from the center of rotating frame of reference

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?

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 $$O'$$.