Conservation of angular momentum in a free rod When a collision is elastic and no external torque acts on a system, angular momentum is conserved
I found this example and checked the results:
A ball (m = 1 Kg , v = p =+22 m/s, Lm = +11, Ke = 242 J) hits the tip of a rod (M = 10Kg , length = 1m, $I = 10*1^2/12$ = 5/6 ) in an elastic collision.
If the rod is pivoted, the ball bounces back with v, p = -11.846 m/s , L = -5.923,  (Ke = 70.16) and the rod rotates with $\omega$ = 20.3 , L is conserved : Lr = (20.3 *5 /6) = 16.923 and Ke = 70.166 + 171.834 = 242 J
the rod translates with v = 3.3846m/s , p = 33.846 , (Ke = 57.2J) and rotates about its CoM with $\omega$ = 16.58 (Ke = 114.556).
If the rod is not fixed to a pivot, in order to conserve linear momentum the rod must translate with p = (11.846 + 22) = 33.846 (v = 3.3846, Ke = 57.28) and the energy of the rotating rod becomes Ke = 114.5 and $\omega = \sqrt(2E/I)$ = 16.58
angular momentum  L was +11
after the impact we have
Lm = -5.923
Lr = 16.58 ( $\omega * I$ )  5/6 = 13.82
13.82 - 5.92 = +7.9
It seems that angular momentum is not conserved. Is there a case in which also L is conserved? 
In that case the bouncing speed must necessarily be different from 11.8 m/s, if that case exists, can you explain why the bouncing speed is different, whereas the masses are the same?
 A: Yes, angular momentum is conserved if you do the problem correctly.
If you assume the ball bounces back along exactly the same path it followed before the collision, there are three degrees of freedom: the velocity of the ball, the velocity of the rod, and the rotational rate of the rod. There are three constraints: conservation of linear momentum in the direction of motion of the ball, conservation of angular momentum, and conservation of kinetic energy. You seem to understand this.
From there, it's unclear what your approach was. How did you get the particular numbers you cite? There are infinitely-many ways to conserve linear momentum and kinetic energy. Conserving just those two applies two constraints to a problem with three degrees of freedom, so there is an entire one-dimensional manifold of solutions. IE you could give the ball any velocity you like (up to a maximum), then choose the rod's velocity and rotation rate to fit the two constraints. If you simply pick one of these solutions at random, then it is very unlikely to conserve angular momentum. You must use all three constraints to solve the problem.
You should do this an confirm that the correct figures are
$$v_{ball} = -\frac{66}{7} m/s$$ 
$$v_{rod} = \frac{22}{7} m/s$$ 
$$\omega = \frac{132}{7}s^{-1}$$
