Do spinning object near a spinning black hole obey the equivalence principle? According to the equivalence principle the path of an object should not depend on it's composition.
But on the other hand a spinning object (e.g. an electron) moving past a rotating Kerr black hole will experience an additional attractive/repulsive force according to the relative spin of the two objects. This is the spin-connection term.
Thus by "composition" one could say if an object contained an object spinning one way or another, then this would have a different gravitational attraction towards a Kerr black hole. Alternatively a beam of photons, neutrinos and anti-neutrinos would follow different paths near a Kerr black hole.
How can one reconcile this with the equivalence principle and even the Galilean/Newtonian principle that gravity should not depend on the composition of an object? Should that not also refer to the angular momentum of it's constituent parts?
 A: Well, I'll take a shot at a classical explanation.  Imagine a large spinning star, and a small spinning object passing close by.
The principle you mention is really the "weak" equivalence principle, and it says that any free particle at a given place and time, with a given velocity, will follow the same path.
In general relativity, Einstein made that principle even "stronger" by saying that those paths which free particles follow are in fact geodesics of the space-time manifold.  They are the closest thing to straight lines in space-time - that is, gravitational motion is actually equivalent to inertial motion.
However, the notion of a free particle is basically a mathematical idealization.  To be a "particle", it has to exist at a single point with a single velocity, so that it can follow a single geodesic.  And, to be "free", it can not be feeling any force such as electromagnetism; such a force will divert it from its geodesic, as described for example by the relativistic electromagnetic equation of motion.
A real object can be thought of as a distribution of point particles.  However, they are not free; they are bound to each other.  They are all diverting each other from the geodesics they would naturally follow, so as to hold the object together.
In a weak gravitational field, and when all the points of the object have pretty similar velocities, their geodesics do not diverge from each other much anyway, so this effect is pretty small.
But for an object spinning very fast, the points on it all have very different velocities.  One is trying to pass the star one way, while the other is trying to move past it in the reverse direction.
Now, if the star itself is also spinning, then the motion of its matter affects the geodesics it creates, per the Einstein field equation.  For example, the geodesic corresponding to the "forward-moving" object point may curve a lot over time, while the one along the "reverse-moving" direction isn't as curved.
Since the two points have to stay together, they end up following a path in between the two geodesics - in fact, the relativistic version of Newton's third law (action-reaction) basically says that the binding force between them diverts them equally from their respective geodesics.  But, if the rotation of the star made one of those geodesics more curved than the other, then the path they follow will be curved, relative to the path they would have followed if they were not spinning.  That's the spin-coupling force you mention.
