A rod of length l forming an angle $\mathrm{\theta }$ with the horizontal strikes a frictionless floor at A with its centre of mass velocity ${\mathrm{v}}_0$ and no angular velocity. Assuming that the impact at A is perfectly elastic. Find the angular velocity of the rod immediately after the impact.

Let v= linear velocity of rod after impact (upwards),

$\mathrm{\omega }=$ angular velocity of rod

and J= linear impulse at A during impact

Then,

$$\begin{gathered} \mathrm{J}=\mathrm{\Delta P}={\mathrm{P}}_{\mathrm{f}}-{\mathrm{P}}_1 \\ \mathrm{J}=\mathrm{mv}-\left(-{\mathrm{mv}}_0\right) \\ \mathrm{J}=\mathrm{m}\left(\mathrm{v}+{\mathrm{v}}_0\right) \\ \therefore \end{gathered}$$

Angular impulse $=\mathrm{\Delta L}$

$\therefore \mathrm{J}\left(\frac{\mathrm{l}}{2}\mathrm{cos\theta }\right)=\mathrm{I\omega }=\frac{{\mathrm{ml}}^2}{12}\mathrm{\omega }$

Angular impulse $=\mathrm{\Delta L}$

$\therefore \mathrm{J}\left(\frac{l}{2}\mathrm{cos\theta }\right)=\mathrm{I\omega }=\frac{{\mathrm{ml}}^2}{12}\mathrm{\omega }$

Collision is elastic (e=1)

$\therefore$ Relative speed of approach = Relative speed of separation at point of impact

${\mathrm{v}}_0=\mathrm{v}+\frac{l}{2}\mathrm{\omega cos\theta }$

Solving above equations, we get

$\mathrm{\omega }=\frac{12{\mathrm{v}}_0\mathrm{cos\theta }}{\mathrm{l}\left(1+3{\cos }^2\mathrm{\theta }\right)}$