Angular momentum paradox I want to analyze the following system, where the mass of the axle is negligible and the Wheel is spinning.

In particular I need to calculate the amount of torque about the point where the string meets the axle.
Unfortunately I get two different answers depending on the procedure:
First (and correct) way
Taking into account the whole system, wheel and axle, there are two forces. The weight, with point of application the center of mass of the wheel; and a force with equal magnitude but opposite orientation, with point of application the place where the string meets the axle. Therefore there is a net torque.
Second (and wrong) way
Taking into account only the wheel, there are the same two forces, but both are applied to the center of mass of the Wheel. Therefore there is no torque.
Question
There must be some amount of torque because otherwise the wheel wouldn't experience precession, and it does. So why is my second approach wrong?
 A: Essentially you are asking why is neglecting the net torque on a system wrong, and the obvious answer it is because you fail to account for the change in angular momentum properly without torque.


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*Linear equations of motion relate net force to the motion of the center of mass. $$ \boldsymbol{F} = m \boldsymbol{a}_C $$

*Rotational equations of motion relate net torque to the motion about the center of mass. $$ \boldsymbol{T}_C = \mathtt{I}_C \dot{\boldsymbol{\omega}} + \boldsymbol{\omega} \times \mathtt{I}_C \boldsymbol{\omega} $$


By neglecting the net torque you are not accounting for the gyroscopic forces correctly.
Look at a free body diagram of the separated parts from the side:

The torque $M$ is equal and opposite acting on the wheel such as to balance the forces on the axle.
A: If you take the wheel as the system, you will realize that unless the wheel has a finite width, you cannot give it a torque. The torque come from the point of force from the axle to the wheel being distributed along the inside of the cylinder where the axle fits in.
Indeed the thinner you make the cylinder, the more will be the force on the contact points. In the limit of the infinitesimally thin wheel you implicitly have in mind the force will go to infinity. Of course in reality the metal will give much before that.
