Linear impulse $$\int \mathrm{Fdt}=\Delta$$ momentum

$$\begin{aligned} & =\mathrm{m}\left(\mathrm{V}_{\mathrm{cm}}-0\right) \\ & \mathrm{P}=\mathrm{m}\left(\omega \mathrm{r}_{\mathrm{cm}}\right) \\ & =\mathrm{m} \omega\left(\mathrm{L}+\frac{\mathrm{L}}{2}\right) \\ & \mathrm{P}=\mathrm{m} \omega\left(\frac{3 \mathrm{~L}}{2}\right) \quad \text{... (i)} \end{aligned}$$

Angular impulse $$\int \tau \mathrm{dt}=\Delta$$ angular momentum

$$\int \mathrm{r} \times \mathrm{Fdt}=\Delta \mathrm{L}$$

$$\mathrm{r} \times \int \mathrm{Fdt}=\mathrm{I}(\omega-0)$$, and I is moment of inertia about axis of rotation.

$$\begin{aligned} \left(\mathrm{L}+\frac{\mathrm{L}}{2}\right. & +\mathrm{x}) \times \mathrm{P}=\left(\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^2\right) \omega \\ & =\left(\frac{\mathrm{mL}^2}{12}+\mathrm{m}\left(\mathrm{L}+\frac{\mathrm{L}}{2}\right)^2\right) \omega \end{aligned}$$

$$\begin{aligned} \left(\frac{3 L}{2}+x\right) P & =m L^2\left(\frac{1}{12}+\left(\frac{3}{2}\right)^2\right) \omega \\ \left(\frac{3 L}{2}+x\right) P & =m L^2\left(\frac{7}{3}\right) \omega \quad \text{... (ii)} \end{aligned}$$

Divide eq.-(i) & (ii)

$$\begin{aligned} & \left(\frac{3 L}{2}+x\right)=\frac{L\left(\frac{7}{3}\right)}{\left(\frac{3}{2}\right)} \\ & \frac{3 L}{2}+x=L\left(\frac{14}{9}\right) \\ & x=\frac{L}{18} \end{aligned}$$