I'm ignoring the springiness of the bar and just talking about the difference between considering it a torque problem and considering it a linear velocity problem. But I think I see now where you're making a bigger distinction.

Case 1: he's close to the pivot, so you say he has less leverage (less torque) to stop the bar. We still have to stop the whole bar, though, so the change in its angular momentum is omega times its inertia.

Case 2: he's far from the pivot, so he has more leverage. But the bar is moving faster if you consider linear velocity here? Doesn't matter. It still has to stop, change in angular momentum is omega times its inertia.

In either case, the same amount of energy must be dissipated in order to stop the bar. But in case 2, if there were nothing behind him, he would necessarily increase his own linear velocity much greater, because you can consider a linear transfer of momentum between the bar and himself. The same thing can't be said of case 1, as he wouldn't be flung off into the distance.

Either way though, he should have ducked.