Drag versus centripetal force I'm trying to see how extensive a gaming engine's physics are.
Consider a spinning disc with a box on it which is free to spin around a vertical pole connecting it to the disc. The pole is more towards one end of the box.
How would this box align itself as the disc spins? Would it be tangential to the disc? Or would the end of the box farthest from the pivot be thrown outwards, like the rear of a car going too fast around a corner which has lost rear wheel grip?
Or is this a battle between the centripetal force not going through the center of gravity of the box, and air resistance, and thus dependent on disc speed and the box's mass?

 A: If there is drag, the final position will always be: at rest with the center of mass pointing away from the center, although slightly skewed from directly pointing away. To see this, notice that, in a reference frame rotating with the disk,  there are four forces acting on the block's center of mass: Centrifugal, internal tension, Coriolis and drag. The centrifugal is always radially out the disk and the tension in relative to the pivot, so we can imagine that the box is a pendulum where we replace centrifugal by gravity. Thus it is a pendulum with two additional forces that depend on speed. The drag, opposite to the speed, can in turn be decomposed into two forces: one that does not depend on relative speed and is proportional to $(R\omega)^2$, and another one that will depend on the relative speed relative to the pivot, will act as friction and dissipate energy. The Coriolis force will act always radially relative to the pendulum's center (either away or in) , and so will change speed direction but not speed magnitude (will not increase the pendulum's kinetic energy). Thus the initial "potential" (from the centrifugal force) energy of the pendulum will be dissipated by the relative drag and reach an equilibrium position at rest (relative to the rotating disk). This position will be with the center of mass at the "bottom", that is, radially out but skewed in a tangential direction opposite to the angular speed. The equilibrium being at the direction of the "effective" gravity: the combination of "external" forces that do not depend on speed relative to the pivot: the centrifugal force and the component of the drag that is proportional to proportional to $(R\omega)^2$.
Note: The velocity of the drag force can be split into two components, one rotating with the disk and the other relative to the disk (that is, relative to the pivot):  $v=R\omega+v_{pivot}$, so the total drag force, is proportional to $(R\omega)^2+v_{pivot}^2+2R\omega v_{pivot}$. The first term introduces a tangential force, and the other two a dissipation force. The non-dissipative term points always negative relative to the pivot, so an eventual equilibrium position must be reached.
