A rod of mass m and length L, pivoted at one of its ends, is...

The net torque on the system (rod+bullet) will be zero about hinge point. Therefore, we can apply the principle of angular momentum conservation about hinge point.

From angular momentum conservation,

$mvx = \left(\dfrac{m{L}^{2}}{3} + m{x}^{2}\right)\omega$

$⇒ \omega = \dfrac{3vx}{{L}^{2} + 3{x}^{2}} = \dfrac{3v}{\left(\dfrac{{L}^{2}}{x} + 3x\right)}$

For maximum angular velocity,

$\dfrac{d\omega}{dx} = 0$

$⇒ \dfrac{d}{dx}\left(\dfrac{{L}^{2}}{x} + 3x\right) = 0 ⇒ \dfrac{- {L}^{2}}{{x}^{2}} + 3 = 0 ⇒ x = \dfrac{L}{\sqrt{3}}$

⇒ {\omega}_\max = \dfrac{3v}{\left(\dfrac{\sqrt{3}{L}^{2}}{L} + \sqrt{3}L\right)} = \dfrac{\sqrt{3}v}{2L}