Suppose we have a gyroscope connected to a rod, undergoing precession parallel to the ground:
I'm trying to figure out how the torque on the rod agrees with the motion. Suppose the rod has a mass $m$ and the wheel has a mass $M$, with $m \ll M$. I'm assuming the wheel is connected to the rod with some frictionless bearing, so the rod does not spin around its axis (y-axis in figure). Since the rod+wheel system does not fall, the pivot applies an upwards force of $(m+M)g$, and since the wheel system does not fall, the rod applies an upwards $Mg$ force on the wheel, so the wheel applies a downwards $Mg$ force on the rod (I ignore the radial force here). Lastly there's a downwards gravitational force of $mg$ applied on the rod's center of mass, so overall we have 3 forces on the rod.
My problem is that if we consider the rod on its own, and take the pivot to be the origin, we get a net torque acting on the rod, even though its angular momentum does not change. The problem persists even if I allow the rod to droop a little so that the angular momentum does change in time and the horizontal force from the wheel cancels out some of the torque. In this case, if the wheel spins fast enough, the rod will still spin too slowly for the change of angular momentum to account for all the torque.
My only explanation is that we have to assume the rod has finite width, so that the pivot does actually apply a torque on it. Is this the case or am I missing something?
