Answer
(a) $\omega = 5.88~rad/s$
(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 ball and let $v_2$ be the speed of the ball after it rebounds.
$L_2=L_1$
$I\omega-mv_2r = mv_1r$
$\frac{1}{3}ML^2\omega = mr(v_1+v_2)$
$\omega = \frac{3mr(v_1+v_2)}{ML^2}$
$\omega = \frac{(3)(3.00~kg)(1.50~m)(10.0~m/s+6.00~m/s)}{(90.0~N/9.80~m/s^2)(2.00~m)^2}$
$\omega = 5.88~rad/s$
(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.
