There is a typical undergrad problem involving an inclinated wheel travelling on a circle.

I did the quick back-of-the-~~envelope~~ travel agency paper calculation (attached). On this extremely simplified example, you basically have to equate two torques : one from the inclination of the wheel (which tends to make it fall) and one from the change of wheel rotation vector due to orbital movement (which tends to set it back upright). (I neglected precession -due to the rotating wheel inclination- as wheels on a bike are much more constrained; the front one is free but controlled by the driver and the rear one is fixed. I didn't checked if the effect is actually negligible, but it seems an honest approximation, espically for a bike.) Note that for this example, the normal and friction forces can be made irrelevant by choosing the origin of axes at the contact point, but these simplifications most probably won't be possible when you fully consider the bike.

I probably made some calculation mistakes (i make mistakes virtually all the time), but the answer for the equilibrum lean angle seems, at least, reasonable. It only depens on the whell mass and radius, its speed and the radius of the big circle it travels (i left the rotational speed versions of these, but it's the same thing).

What changes for a full bike ? Obviously the center of mass vector won't be as simple (depends on the bike mass distribution) ; the inertia moment (but i left I anyway) ; and finally the radius R is to be changed with the curvature radius of the bike's path (which is now a function of theta) ; plus some other things (like neglected friction or what the driver does with the front wheel).

But to answer your initial question, yes it is possible even for a real bike, and solving everyting numerically should be easier in this case.

edit: On the lowest line, i wrote dr/d(theta), it's of course d(u_r)/d(theta), the unit vector. But that's what i meant on the next step.