Why does the heavier end of a falling object orient toward the ground?

[TheExibo]

If an object that has more mass at one end of it was dropped, how come the heavier end would be facing closer to the ground than the lighter end? For instance, a toliet plunger is dropped, and the heavier, rubber end of it positions itself at the bottom during the fall. What is this called?

 

[Bandersnatch]

Air resistance. The lighter end has a lower mass/area ratio than the heavier end, so it is affected by air drag more than the heavier end. In vacuum the plunger would fall without rotating at all, as the force of gravity accelerates all parts at the same rate and there is no friction to introduce imbalance.

 

[TheExibo]

Air resistance. The lighter end has a lower mass/area ratio than the heavier end, so it is affected by air drag more than the heavier end. In vacuum the plunger would fall without rotating at all, as the force of gravity accelerates all parts at the same rate and there is no friction to introduce imbalance.

That makes sense, thanks!

 

[TheExibo]

Air resistance. The lighter end has a lower mass/area ratio than the heavier end, so it is affected by air drag more than the heavier end. In vacuum the plunger would fall without rotating at all, as the force of gravity accelerates all parts at the same rate and there is no friction to introduce imbalance.

Also, is there a way to explain it using forces? For example, would it be correct to say that the force of friction from the air molecules accelerated the lighter half of the object back up to be re-positioned behind the heavier half? Or maybe the collision of the object with the air molecules decreased the momentum of the lighter half of the object, while the heavier object had more momentum to spare? So the lighter half had lost more velocity than the heavier half?

 

[Bandersnatch]

These are all viable ways of describing the situation, providing you choose the correct reference frame.
Air resistance is a force, so calling it that is defninitely correct. Similarly, decreasing (chaning in general) momentum over time is the definition of Force: $F=\frac{dp}{dt}$ (as per Newton's second law of motion; if the mass is constant, it reduces to the familiar $F=ma$).

To be able to say that air friction (drag) accelerates something upwards, you'd need to choose a correct reference frame. In the FoR of the ground, the net force is always down. It's just that drag reduces the net force down differently for different bits of the object.
You could choose the reference frame of a falling object "A" with negligible air resistance - it would fall purely under the force of gravity. In this FoR (that is, with respect to the object A), drag does accelerate the plunger upwards, with different bits going up at different rates.
Unless by "up" you just mean the drag component acting "in the direction against gravity", which is correct.

Note two things:
1.the situation is similar to having two equal-size balls of different masses falling next to each other. Say, one is a ping-pong ball, the other is an iron-cast copy. It's easier to intuitively imagine how the two behave when falling. If you connect the balls with a very thin string, you have yourself an object that is behaving similarly to the plunger.
2.the heavier end needen't necessarily be the one less affected by air resistance. If the light end is very aerodynamic (e.g., a vertical, thin plank of wood) while the other is very non-aerodynamic (like a horizontal, thin sheet of rubber), then drag would affect the heavier end more than the light one. This is similar to two people falling from the sky: a parachutist and somebody without a parachute. Person+parachute is always heavier than just person, but one will connect with the ground at much more agreable speed than the other nevertheless.