Statement of the Problem
In frame $F'$ with co-ordinates $(t',x',y',z')$, a straight rod rotates in the $x',y'$ plane with angular velocity $\omega ′$ about one of its ends. The fixed end is located at the spatial origin in $x'$ which is $x'=y'=z'=0$. At time $t′ = 0$ the rod lies along the positive $x′$ axis. Frame $F′$ moves with constant velocity $v$ in the $x$-direction with respect to frame $F$. The origins of the two frames coincide at time $t′ = t = 0$ (i.e. the points $(t,x,y,z)=0$ and $(t′,x′,y′,z′)=0$ are the same). Also, there is no spatial rotation between the two frames. Find an equation which gives the shape of the rod in frame $F$ at time $t = 0$. In other words, find an equation for $y$ as a function of $x$.
It's clear to me that the rod won't be straight in $F$. Points closer to the origin have a lower velocity in $F'$, so those points move with different velocities than points further from the origin, in a nonlinear way.
However, I'm not sure how to encode this mathematically within the framework of special relativity and thus ultimately get the equation that describes the shape of the rod in $F$. I considered taking a small element $dl'$ of the rod, and modifying its shape according to length contraction in the direction of the relative velocity between the frames, but that doesn't seem to be leading anywhere. It's also possible that one could use the relativistic velocity addition formula to get the velocity of some point on the rod, but that doesn't seem to lead to the necessary function, either.
Any help is greatly appreciated.