leafy said:
#2 also stated the force is always orthogona. If torque = Frsin(theta) and theta always zero, then torque is always zero?
Orthogonal to the velocity vector, yes. Which means that speed is unchanged. ##\vec{F} \cdot \vec{v} = 0## so no power associated with the force. But that has little to do with torque.
Orthogonal to the radius vector, no. Which means that torque is not maximized. No big deal.
Parallel to the radius vector, not during the transfer. Which means that torque (##\vec{F} \times \vec{r}##) about the reference axis is non-zero. That is a very big deal.
Parallel to the radius vector while the object is on either the outer or the inner circle, yes. Which means that torque (##\vec{F} \times \vec{r}##) is zero during those portions of the trajectory and angular momentum about the reference axis is conserved during those phases.
Parallel to a radius vector toward the center of curvature of the transfer arc during the transfer, yes. Which means that torque
about some other reference axis is zero during the transfer and that angular momentum about that other reference axis is temporarily conserved. Not a good argument for conservation of angular momentum about our chosen reference axis.
One can try to cobble up a half-baked argument that angular momentum is conserved about one axis and then another axis and then again about the original axis. But if you keep that argument in the oven long enough to fully bake it, you'll realize that each time you changed reference axis, you needed to add a term to angular momentum to compensate for the cross product of the linear momentum of the system and the displacement of the axis. You did not add that term. So that argument falls flat.