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?