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

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?

• If you push against the air, the air still pushes against you just as if you were pushing off the ground when running. Feb 28 at 1:05