= Solution
Choose the cylinder's <velocity> to be $v\hat{\mathbf x}$, and the ball's laboratory <velocity> to be $u(\cos\theta,\sin\theta)$. By <Galilean velocity addition>, the ball's <velocity> relative to the cylinder is
$$
\mathbf u'=\bigl(u\cos\theta-v,\ u\sin\theta\bigr).
$$
For a suitably centred entry without hitting a wall, the cylinder's axis must be parallel to this relative <velocity>. Its oriented <angle> $\alpha$ to the cylinder's direction of travel is therefore
$$
\boxed{\alpha=\operatorname{atan2}(u\sin\theta,u\cos\theta-v).}
$$
Equivalently, $\tan\alpha=u\sin\theta/(u\cos\theta-v)$, with the quadrant chosen from the two components. For $v/u\ll1$,
$$
\boxed{\alpha=\theta+\frac vu\sin\theta+O((v/u)^2).}
$$
Thus the extra tilt relative to the ball's original direction is approximately $(v/u)\sin\theta$. The use of $\operatorname{atan2}$ also covers nearly transverse incidence, where taking an unqualified inverse tangent could select the wrong axis direction.
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