"Assume that the spokes are all attached (hinged) at the very center of the bicycle's axle. Take a wild guess at how many degrees of rotation will be required to realign the weight's plane with the axle's plane"
Depends on moment of inertia, angular velocity, and friction in the flapping hinge. If there is no friction in the flapping direction (ie not hinge, no air resistance, etc.) then the weight will never realign.
The weight acts like a pendulum. The restoring force (force that brings pendulum to it's median position) comes from the rotational motion. In a frictionless environment, pendulums would continue to swing forever.
"Assume that the hub now has a very large diameter, which is 90% of the 'wheels' diameter, and that spokes are all attached (hinged) to the rim of this hub. Take a wild guess at how many degrees of rotation will be required to realign the weight's plane with the axle's plane."
Same answer, it depends on.....
In A, the natural frequency of the weight would be the same as the angular frequency of the wheel. Thus, the weight would go through one full cycle as the wheel turns once.
In B, the natural frequency of the weight would be higher than that of the wheel, so it would go through more than one cycle for one turn of the wheel. To get an idea of how much of a difference it makes, if we neglect the weight of the spoke and simplify the system, the weight will do 3.16 cycles for one turn of the wheel.
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I had to convince myself that this was accurate, so I built a teetering pencil, spun it on my cordless scredriver. I saw the exact same motion that I expected, and for a sudden change of the axle by about 30 degrees, it took about 7 revolutions for the teetering pencil to return to perpendicular to the axle.
Pictures available by email if interested.