Calculating the torque at a point when a motor is stopped? So, I'm trying to solve for the torque $\tau_A$ of a motor. I have attached a strong stick to the motor, like so:

I apply a force $F$ on the stick which stops the motor. The distance from the outside edge of the cylinder to the end of the stick is $L$. The torque for the motor is $\tau_A=F(L+r)$, $r$ is the radius.
My friends believe that the torque at point $B$ is $\tau_B=-FL$, but I believe even though the motor is not moving (due to the force), it still applies a torque at point $B$. It would be less than $\tau_A$ since it doesn't push around the point uniformly, but it should be $\tau_B=-FL+\tau_Ac$ ($c$ is a constant). Using their method, they got that $\tau_A=0$, which I believe happened because in calculating the torque at point $B$, they make $\tau_A=0$. 
Who is right? How do I calculate how much $\tau_A$ is applied about point $B$ (assuming I'm correct)?
 A: Well, nothing is starting to move, right? So there's no net torque at all. Neither around point $A$, not around point $B$.
Let me clarify:
Around point $A$, there is the torque of the motor, $\tau_A$, and the torque due to the force at the end of the stick, $-(r+L)F$. So, the net torque is indeed zero (as is should). There is also a force attacking at point $A$ of magnitude $F$ which is pointing up. This is required to cancel the translatory motion that would otherwise be due to the force on the stick. Since it is attacking at $A$, it will not exert a torque around that point.
Also, one can model the inherent torque of the motor by a force of magnitude $F'$ attacking in $B$ and pointing upwards. To get the same torque, we shall require $rF'=\tau_A$ leading to $F'=(1+L/r)F$. Again, we need to cancel that force by one of the same magnitude attacking in $A$ in the downward direction. Note, that the particular direction and position of this force are arbitrary as long as they oppose each other and lead to the same torque on $A$. However, the above choice is most practical.
Now that we know all the forces, we can consider the torque on $B$. Your friends are right in saying that a torque of $-LF$ acts due to the force on the stick. Yet, there are also the forces on $A$, $F$ (up) and $F'$ (down) leading to an additional torque of $+r(F'-F)$. The force $F'$ that attacks in $B$ is now irrelevant for the torque on $B$. Hence, the net torque on $B$ is
$\tau_B=-LF+r(F'-F)=-LF+LF=0$.
(Again, the torque around $A$ is also zero. It is misleading to call $\tau_A$ the torque around $A$ as it is counteracted by the force on the stick!)

A: The first part of your analysis is right. Your hand exerts a torque 
$$\tau_{hand} = F(L+r)$$
about the center of the disk. The disk is stationary, so it must have zero net torque on it, so
$$\tau_A = -F(L+r)$$
As long as the system you want to analyze is the disk and rod together, that's the end of the story.
Evidently, your friend wants to analyze the system of just the disk, with force from the rod on the disk considered as an external force. You can do that, but the story is a little more complicated if you do. Although the net force from the stick on the disk is $F$, the force takes the form of a pressure distributed over the area over which the rod and disk are in contact. This pressure creates a force acting in different directions at different points over that contact area. The net torque from this force on the disk about the disk's center of mass is still $-F(L+r)$, which comes from integrating the moment of the pressure over the area of contact. You still have the same equation for $\tau_A$. Anyone whose ultimate conclusion was $\tau_A = 0$ was incorrect.
I find the part about $\tau_B$ confusing because it is not clear what body the torque is acting on and what forces are being considered as making up $\tau_B$.
A: $$ \tau_B = F L \\ \tau_A = F (L+r) $$
You just have to separate the objects to calculate the above from statics. What torque on B would keep the stick immobile? What torque on A would keep the rotor and stick immobile?
The torque on B is supplied by the joint, and on A by the motor. If the stick was pinned on the rotor then $\tau_B=0$ and $\tau_A = F r$. It is the disallowing of relative rotation between the parts that transfers the torque from the stick to the rotor. 
A: (ANSWER ASSUMPTION: In this answer, I have assume that the stick is not hinged to the rotor, but actually properly fixed. Also, I've interpreted $\tau_B$ as an internal torque/moment in the rotor-stick system because, unless there is some external effect that is causing a torque to be applied, like someone twisting with their fingers at B, there is nothing to cause an external $\tau_B$ to exist. Feel free to correct me if I have misunderstood your question)
So, to analyse this problem, you have to have a pretty good grasp on exactly what system it is you are analysing. You could look at the whole rotor and stick thing a few ways: you could consider the rotor and stick as two rigid bodies, with equal and opposite reaction forces and moments at the point of attachment, or you could consider it as one whole rigid body, and that the moment you are looking for is an internal force (i.e. if you find the net force/moment a rigid body, internal forces are not involved in the analysis. This is to do with how internal forces and moments have equal and opposite pairs that cancel with each other from an external analysis, like finding net force. Newton's 3rd Law is to blame for this cancelling.) A problem seems to arise in your analysis as you appear to be treating the system as a single rigid body, and you are treating $\tau_B$ as an external force, when it is an internal force for the rotor+stick body.
For the sake of analysis, let's treat the stick and rotor as individual bodies with equal and opposite reaction forces/moments at the point of attachment. The value of the reaction moment will be equal to $\tau_B$. Here is a diagram depicting the two separate bodies:

If you were to consider the rotor and stick as one rigid body for analysis, then the moments and force with purple arrows would be the internal forces in that system. This makes sense as these forces and moments occur in equal and opposite pairs that cancel for external analysis.
Now, all we have to do is to apply the equations of equilibrium for each rigid body.
Taking moments about, say, the left end of the stick:
$$-\tau_B + FL = 0$$
$$\therefore \tau_B = FL$$
Whether the value is positive or negative depends on the sign convention useful for inter-body reaction forces/internal forces or moments.
Just to check that the calculation is consistent, take moments about the centre of the rotor:
$$-\tau_A + \tau_B + Fr = 0$$
$$\therefore \tau_B = \tau_A - Fr$$
Substitute the result of $\tau_A$ you derived earlier and you get the same result. The best point I can make for this issue is to be careful and aware of what system you are analysing.
