etotheipi
Charles Link said:The calculation is done simply at the moment of impact, and it is assumed the bat pivots at a point near where the hands are. The speed of the bat is in the same direction all along the bat, and is ## v=\omega r ##. It is the same thing as the door and the doorstop.
It is still insufficient. Consider a point on the bat which is a distance ##\xi## from the pivot. The angular momentum about such a point is of magnitude$$L = M \omega \left( \xi - \frac{\mathscr{L}}{2} \right)^2 + \frac{1}{12} M \mathscr{L}^2 \omega$$Note that this is strictly > 0. Hence, there is no point on the bat about which there is zero angular momentum.
Further, it does not matter that the calculation is only performed within an interval ##[t_c - \epsilon, t_c + \epsilon]## around the collision. If the coordinate system has its origin connected rigidly to the bat then you must still account for an inertial impulsive Dirac delta force through the centre of mass.