A meterstick of length $L$ is set on the very edge of a table. Almost all of the meter sticks length is off the table. The meterstick is then released from rest.
I may be wrong in my thinking but here is how it goes:
Originally I chose the very edge of the meterstick (the one on the table) as my pivot point. If you choose the center of mass as the pivot the normal force will exert less torque because the meterstick falls. My only explanation would be that the rotational inertia about the center of mass would be less than that about the edge. Would these two effects cancel out?