Now, for chuckles and grins a direct counterpoint to your angle-based counterpoint: in the (non-applicable) framework of SR, the rod is parallel to the car floor at all times in the car frame. The car floor is parallel to the platform in both the car and the platform frame. Therefore, the rod is parallel, at all times, to the platform.
This is wrong.
Here's the correct analysis (for a rod moving inertially to hit the floor of the train in flat spacetime).
I'm renaming the frames a little, since I'm starting from the rod frame and progressing to the platform frame.
(It's not actually necessary to begin with the rod frame, since the worldines of the rod are perfectly well defined in the train frame, but this is more complete.)
$$S = $$ rest frame of the rod
$$S' = $$ rest frame of the train
$$S'' = $$ rest frame of the platform
$$v = (0, u)$$ is the velocity of $$S'$$ relative to $$S$$
$$v' = (V, 0)$$ is the velocity of $$S''$$ relative to $$S'$$
$$\gamma = 1/\sqrt{1-u^2/c^2}$$
$$\gamma' = 1/\sqrt{1-V^2/c^2}$$
The rod has two ends, A and B. In the rod frame, we place them at (x,y) = (0,0) and (1,0) respectively.
In $$S$$:
$$\begin{align}
A &= (x_A, \ y_A) \\
&= (0, \ 0)
\end{align}$$
$$\begin{align}
B &= (x_B, \ y_B) \\
&= (1, \ 0)
\end{align}$$
The angle of the rod with the x-axis at time t is:
$$\begin{align}
\tan(\theta) &= \frac{y_B-y_A}{x_B-x_A} \\
&= 0
\end{align}$$
In $$S'$$:
$$\begin{align}
A' &= (x_A'(t'), \ y_A'(t')) \\
&= (x_A, \ \frac{y_A}{\gamma} - ut') \\
&= (0, \ -ut')
\end{align}$$
$$\begin{align}
B' &= (x_B'(t'), \ y_B'(t')) \\
&= (x_B, \ \frac{y_B}{\gamma} - ut') \\
&= (1, \ -ut')
\end{align}$$
The angle of the rod with the x-axis at time t' is:
$$\begin{align}
\tan(\theta') &= \frac{y_B'(t') - y_A'(t')}{x_B'(t')-x_A'(t')} \\
&= 0
\end{align}$$
In $$S''$$:
$$\begin{align}
A'' &= \left(x_A''(t''), \ y_A''(t'') \right)\\
&= \left(\frac{x_A'}{\gamma'} - Vt'', \ -ut' \right) \\
&= \left(-Vt'', \ -u\gamma'(t'' + \frac{Vx_A''}{c^2}) \right ) \\
&= \left(-Vt'', \ \frac{-ut''}{\gamma'} \right)
\end{align}$$
$$\begin{align}
B'' &= \left(x_B''(t''), \ y_B''(t'') \right) \\
&= \left(\frac{x_B'}{\gamma'} - Vt'', \ -ut' \right) \\
&= \left(\frac{1}{\gamma'} - Vt'', \ -u\gamma'(t'' + \frac{Vx_B''}{c^2}) \right) \\
&= \left(\frac{1}{\gamma'} - Vt'', \ \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \right)
\end{align}$$
The angle of the rod with the x-axis at time t'' is:
$$\begin{align}
\tan(\theta'') &= \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} \\
&= \frac{uV\gamma' / \gamma c^2}{1 / \gamma'} \\
&= \frac{uV\gamma'^2}{\gamma c^2}
\end{align}$$
This answer coincides with the one based on Thomas precession
Your Thomas rotation answer is wrong, as demonstrated in (edit) [post=3062610]post 157[/post].
