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Say I have this setup. The two round objects have equal mass and their centers of gravity at the same distance from the shaft. The objects only differ in that they have different surface areas (assume they are both flat-faced).

If this apparatus is spun, will it fall over?

I believe it won't. The centrifugal forces are balanced since they only depend on mass, and the aerodynamic drag will be in the direction of motion, so it won't pull the shaft in any outward direction.

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If drag is ignored, it won't fall. Note: There is no centrifugal force, only centripetal force. Without drag, both of them have the same moment of inertia, ... etc. such that they are balanced so they won't fall.

However if drag is considered obviously the object with a larger surface area will be exerted on a larger force, causing a net force + net torque, in turn causing an imbalance in the system, making the apparatus fall over.

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Actually, the parallel axis theorem implies that you need one more condition than the OP offers to insure the same moment of inertia. It could perhaps be rigged to be stable in this way, but it would have to be rigged. – dmckee Dec 9 '12 at 5:35
What conditions are needed then? – namehere Dec 9 '12 at 6:10
Dear namehere, I think that you have ignored the main observation by the OP that (at least according to the simplest formulae) the drag forces are in the direction of the motion, horizontal, so in this approximation it's plausible that the horizontal bars will stay horizontal and vertical ones will stay vertical. Is this an exact property of the system? – Luboš Motl Dec 9 '12 at 6:10
@LubošMotl I'm sorry I don't really understand your comment. Can you please clarify? And I'm pretty sure the inequivalance of the drag forces cause a net force + net torque. – namehere Dec 9 '12 at 6:14
The masses must also have the same moment of inertia about their own centers. Arranging that may not be trivial, but should be manageable for some geometries and materials. – dmckee Dec 9 '12 at 6:24

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