Graviman,
A piston starts at rest and ends at rest, net momentum change is zero,
we can't ascertain anything about the work done during the cycle from that.
What assumptions did you make when translating to 1D? Jiff has given us no indication of how the torque varies with angular position. The torque gives acceleration (and tangential force), which gives us velocity (and centrifugal force), which gives us position (and then torque).
I have used a simulation (stepping at 0.001s) where the torque is constant at 1 positive 0-π and negative π-2π. Using mass, radius (and therefore I) as 1, simplifies the numbers. Start the thing off with a slight clockwise velocity and round and round it goes. The result is a net force not surprisingly to me) in the direction π.
That is the extent to which Jiff has defined the system.
As a thought exercise,
Imagine the masses in Jiff's machine are rockets aligned fore and aft tangentially. On the accelerating side the rocket would have to burn pointing aftward to accelerate, and on the decelerating would have to burn pointing forward to decelerate. Outside of the frame of reference, the rocket has rotated and so the rocket thrust has a cyclical lateral component that varies in sign, and a longitudinal component that is also cyclical but always positive. Lateral effects are taken out by having contra-rotating pairs. In this instance I am sure all would agree that Jiff has created an extremely complex and inefficient way of producing linear force.
What we were presented at the beginning is a very simple concept.
As I stated several posts ago, the difficulty lies in the creation of the torque (or the changing of angular momentum), and managing its reaction on the body of the thing that is creating it. If the model is built using conventional electric motors, and started up in free space, I imagine it will begin to rotate on an axis 90° to the axis of rotation. Not unlike gyroscopic precession. As the rotation of the motor body creates an equal but opposite effect at the other end of the driveshaft. I've not done the maths, but this may go someway to explain the effect of the unit on the string, as the torque changes the equilibrium between the CG and the attachment point.
In ballistic calculations, I have never included the momentum change of the Earth. Within my frame of reference the behaviour of the bullet appears none Newtonian.
I have said it before, the clever bit is not what Jiff has drawn but what he hasn't, i.e. the means of generating the torque that is none Newtonian within our frame of reference.