Why is the propeller torque, which is cancelled by drag, considered in calculation of net torque about the rotation axis?

Question

I am calculating the net torques and forces about the yaw axis of a quadrotor. If $I$ is the moment of inertia about the rotation axis, and $\omega$ is the angular velocity of the propeller, then the torque is given by:

$$ \tau = I \frac{d \omega}{dt} $$

This represents the net force $F$ acting on the propeller's edge at a radius $r$, giving it a net torque of $\tau = r \times F$.

From the force perspective, there is the driving force of the propeller, and the drag force of the air. For a constant angular velocity, the two forces/torques cancel each other out. Therefore, according to the formula, a constant speed propeller will contribute no torque to the body.

And yet, torque calculations in literature (see here, page 4) will express the torque of a propeller as about the yaw axis as:

$$ \tau = I \frac{d\omega}{dt} + b\cdot\omega^2 $$

Where the latter term is the drag force and $b$ is the drag coefficient. But hasn't the drag force cancelled out, leaving the residual $d\omega/dt$? Why is it being counted again? For example, when calculating net forces about the body, I don't count the weight twice like this ($v$ is velocity, $g$ is gravitational acceleration, $m$ is mass):

$$ F = m \frac{dv}{dt} + m\cdot g $$

So my question is, why is there a net yaw torque about a quadrotor body, when the propeller is spinning at constant speed?

Answer 1

It's important to distinguish torques on the propeller vs those on the body of the aircraft. The propeller is experiencing 0 net torque, while the aircraft is experiencing non-zero net torque. For a constant speed propeller, the drag torque and the driving torque on the propeller cancel. But in order to apply the driving torque, the motor must exert an equal and opposite torque on the aircraft. This torque remains uncancelled, leaving a net torque on the aircraft.