A relativistic meter stick and a thin disk I have a question like the "pole in the barn" special relativity "paradox", but I'm not sure what to make of this:

Question
A meter stick lies along
the $x$-axis and approaches the origin, moving along its length, with velocity $v_x$. A very
thin plate, parallel to the $xz$ plane in the laboratory, moves upward in the $y$ direction with
velocity $v_y$. 
The plate has a circular hole with diameter $1$ m centered on the y-axis. The
center of the meter stick arrives at the origin at the same time as the center of the hole
in the plate. 
In the laboratory frame, the meter stick is Lorentz contracted, so it's in the
hole in the plate, and plate and stick continue on their paths without a collision. In the
rest frame of the meter stick, however, it is the plate, and the hole in it, that is contracted
along the $x$-axis. 
This would seem to predict a collision, contradicting the requirement
that physical predictions must be invariant. Which of these two predictions are correct?

My thinking is that simultaneity will play a role here, but I'm not sure how to write down the Lorentz contraction of the disk from the meter stick's point of view. It will be contracted in both the $x$ and the $y$ direction by a factor of $\gamma_y = {1\over \sqrt{1 - v_y^2}}$ but will this cause the disk to seem "tilted" from the stick's point of view? A hint would be great thanks.  
 A: Lab frame: $L$, Rod frame: $R$.
Allow me to use slightly different notation. Please ignore the $z$-coordinate as it is not relevant.
We denote the rod's velocity in $L$ by $(u,0)$ , the circular hole's velocity in $L$ by $(0,u_y)$ and velocity of hole in $R$ by $(v_x,v_y)$. $\gamma = \sqrt{1-(u^2/c^2)}$.
In $L$, the two events corresponding to the two endpoints of the effectively 2-D hole coinciding with the $x$-axis are given as $I_1 = (1/2, 0, 0)$ and $I_2 = (-1/2, 0, 0)$.
Using the Lorentz transformation these events are represented as 
\begin{align*}I_1'= (\gamma/2,0,-\gamma u/2c^2) \\ I_2' = (-\gamma/2,0,\gamma u/2c^2)\end{align*} 
in $R$. Since $\gamma > 1$, we conclude that whenever an end of the hole crosses the $x$-axis (note here that the times are different), in $R$, it is on the correct/expected side of the rod's closest end. In my opinion, this suffices as the resolution to the main problem of the "collision".
As noted by the OP, the hole will be tilted w.r.t. the $x$-axis in $R$, unlike $L$. 
Using the velocity transformation on the centre of the hole:
\begin{align*}
v_x &= \frac{u_x-u}{1-(u_xu/c^2)} = -u \\
v_y &= \frac{u_y / \gamma}{1-(u_xu/c^2)} = u_y / \gamma
\end{align*}
We may simply translate 
\begin{equation} I_1' \rightarrow (I_3')_{(x,y)} = (I_1')_{(x,y)} + v\cdot ((I_2')_t-(I_1')_t) \quad , \quad  (I_3')_t = (I_2')_t \end{equation}
 to know where the front end of the hole was at $(I_2')_t$.
After solving that we obtain 
\begin{equation} I_3' = (\gamma \cdot(1/2 - u^2/c^2), u_y u /c^2,\gamma u/c^2)\end{equation}
We now observe that 
\begin{equation}
\frac{(I_3')_y-(I_2')_y}{(I_3')_x-(I_2')_x} = \frac{u_yu/c^2}{1/\gamma}> 0
\end{equation}
which gives us the required tilt. The entire motion of the hole in $R$ can now be understood by considering the velocity found out earlier.
