Assume the rod to be of mass $m$, uniform and completely rigid. The floor is also perfectly rigid and there's no friction between floor and rod, which prevents any sideways motion on or after impact.

The collision is therefore perfectly elastic.

The translational kinetic energy is completely converted to rotational kinetic energy, for rotation about the CoG:

$$\frac12 mv^2=\frac12 I\omega^2$$ where $v$ is the translational velocity of the rod at impact. $$\frac12 mv^2=\frac12 \frac{1}{12}mL^2 \omega^2$$ $$v^2=\frac{1}{12}L^2\omega^2$$ $$\omega=\sqrt{12}\frac{v}{L}$$

Assume the rod to be of mass $m$, uniform and completely rigid. The floor is also perfectly rigid and there's no friction between floor and rod, which prevents any sideways motion on or after impact.

The collision is therefore perfectly elastic.

The translational kinetic energy is completely converted to rotational kinetic energy, for rotation about the CoG:

$$\frac12 mv^2=\frac12 I\omega^2$$ where $v$ is the translational velocity of the rod at impact. $$\frac12 mv^2=\frac12 \frac{1}{12}mL^2 \omega^2$$ $$v^2=\frac{1}{12}L^2\omega^2$$ $$\omega=2\sqrt{3}\frac{v}{L}$$