The angular momentum of a body can be decomposed into two vectors: (a) angular momentum
about the center of mass, more descriptively the spin angular momentum and (b) angular momentum
of the center of mass, more descriptively the orbital angular momentum. Both vectors require a torque to change. A torque ##\vec \tau=\vec r \times \vec F## will have different effects depending on ##\vec r## which is the position vector from the origin to the point of application of the force.
Consider a fixed force ##\vec F## applied on a body. The question is, "what is the position vector ##\vec r~##? The answer is, "it depends on where you choose the origin." If you choose it at the center of mass of the object, its spin angular momentum will change in a direction perpendicular to the plane defined by the force and the position vector; its orbital angular momentum will not change. This is the situation described by
@Baluncore in post #2. If you choose the origin about any point other than the center of mass, then both spin and orbital angular momentum will change in a way that depends on your choice.
Specifically, for a mass falling from rest, if you choose the origin directly above the center of mass, the angular momentum will not change about that origin which means, of course, that the torque will be zero. However, if you choose the origin at perpendicular distance ##R## from the straight line trajectory, the magnitude of the torque will be ##\tau = mgR## and the orbital angular momentum about the origin will vary in time according to ##L=mgtR##. If you tack on the mechanism suggested by
@Baluncore, there will also be a change in the spin angular momentum.
