#### Answer

(a) $\omega = 1.71~rad/s$
(b) During the collision, the pivot exerts an external force, which causes the linear momentum to change. However, the force from the pivot exerts zero torque. Therefore, angular momentum is conserved.

#### Work Step by Step

(a) We can use conservation of angular momentum to solve this question. Let $v_1$ be the initial speed of the raven and let $v_2$ be the speed of the raven after it rebounds.
$L_2=L_1$
$I\omega-mv_2r = mv_1r$
$\frac{1}{3}ML^2\omega = mr(v_1+v_2)$
$\omega = \frac{3mr(v_1+v_2)}{ML^2}$
$\omega = \frac{(3)(1.1~kg)(0.75~m)(5.0~m/s+2.0~m/s)}{(4.5~kg)(1.5~m)^2}$
$\omega = 1.71~rad/s$
(b) During the collision, the pivot exerts an external force, which causes the linear momentum to change. However, the force from the pivot exerts zero torque. Therefore, angular momentum is conserved.