Answer
a) $\omega=\sqrt{\frac{3g}{L}}$
b) $v_t=\sqrt{3gL}$
Work Step by Step
(a) We know that
$K_i+U_i=K_f+U_f$
$\implies 0+Mg(\frac{L}{2})=\frac{1}{2}I\omega^2+0$
This simplifies to:
$\omega=\sqrt{\frac{MgL}{I}}$
$\omega=\sqrt{\frac{MgL}{\frac{1}{3}ML^2}}$
$\omega=\sqrt{\frac{3g}{L}}$
(b) We know that
$v_t=r\omega$
$\implies v_t=L\sqrt{\frac{3g}{L}}$ $as \omega=\sqrt{\frac{3g}{L}}$ from part (a)
$v_t=\sqrt{3gL}$
