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

enter image description here

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?

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    $\begingroup$ Why do you think the force of the string is applied at the center of mass of the wheel in the second procedure? $\endgroup$ – Jahan Claes Mar 26 '18 at 19:32
  • $\begingroup$ The axle is the only thing that can apply a force to the wheels and it is attached to the center of mass of the wheel. $\endgroup$ – Gamabunto Mar 26 '18 at 19:50
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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.

  • 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:

Wheel

The torque $M$ is equal and opposite acting on the wheel such as to balance the forces on the axle.

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  • $\begingroup$ But if my system is the wheel what torque am I neglecting? $\endgroup$ – Gamabunto Mar 26 '18 at 19:52
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    $\begingroup$ The support to the weight of the wheels is on an axis offset from the center of mass. This results in a reaction moment between the axle and the wheel. $\endgroup$ – ja72 Mar 26 '18 at 20:03
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    $\begingroup$ @Gamabunto I think the fundamental problem is that if you view the center of the wheel as a single point, you've simplified the problem too much--the wheel has a finite width, and if you neglect this you neglect the torque the axle exerts on the wheel. You need to view the wheel as having width, and realize the axle doesn't exert a force on the wheel's center, but exerts forces along the entire width of the wheel in a way that causes a torque. $\endgroup$ – Jahan Claes Mar 26 '18 at 20:08
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    $\begingroup$ @Gamabunto - see FBD in the answer in order to clarify what moment I am talking about. $\endgroup$ – ja72 Mar 26 '18 at 20:12
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    $\begingroup$ Yes, there are two opposite but not equal forces acting some distance apart. The net result is a force to counteract the weight and a moment. $\endgroup$ – ja72 Mar 27 '18 at 12:26
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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.

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