One end of a straight uniform $$1 \; m$$ long bar is pivoted on horizontal table. It is released from rest when it makes an angle $$30°$$ from the horizontal (see figure). Its angular speed when it hits the table is given as $$\sqrt{n}$$ rad $$s^{-1}$$, where $$n$$ is an integer. The value of $$n$$ is ___________.

Moment of inertia of a uniform bar about an end pivot: $$I = \frac{1}{3}ML^2$$

Conservation of mechanical energy: $$E_i = E_f \implies U_i + K_i = U_f + K_f$$

Taking the horizontal table surface as the reference potential level:

$$Mg\left(\frac{L}{2}\sin 30^\circ\right) + 0 = 0 + \frac{1}{2}I\omega^2$$

$$Mg\frac{L}{4} = \frac{1}{2}\left(\frac{1}{3}ML^2\right)\omega^2 \implies \frac{g}{4} = \frac{1}{6}L\omega^2$$

$$\frac{10}{4} = \frac{1}{6}(1)\omega^2 \implies \omega^2 = \frac{60}{4} = 15$$

$$\omega = \sqrt{15}$$