# 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)?

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Hi - edited your math to be more readable (open your question for editing to see the syntax). –  Kyle Jul 10 '13 at 19:41

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 $\tau_B$.

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Granted, if you consider the connection between the disk and the rod to be a non-trivial one, the problem is more difficult and becomes ill-posed. I was assuming that the rod is connected to the disk right at the rim using a point-like connection of super-glue. It seems like the natural thing to do with these problems... –  Jonas Jul 10 '13 at 21:00
@Jonas You can't have a "point-like connection". The stress would be infinite. –  Mark Eichenlaub Jul 11 '13 at 0:51
You can't have point-like anything in reality, yet it that doesn't stop us from considering point-like car, trains, atoms, ... . The force due to the hand on the rod is also really a pressure, since hands are not point-like... –  Jonas Jul 11 '13 at 5:46
Also, all of this doesn't really matter. The answer to the question will be $\tau_B=0$ since the ensemble of disk and rod is stationary just as the net torque around point $A$ is zero. This is independent of how you arrive at it -- by considering complicated pressures on finite areas or by modelling point-like forces. –  Jonas Jul 11 '13 at 5:51
While it is true that a point is a model, sometimes it is a good model, and sometimes it is a bad model. This this case, the force between the disk and the rod being a point is a bad model. If there are only two forces on the rod (one from the hand and one from the disk) it is impossible to have zero torque and zero net force on the rod. Therefore the rod would have to accelerate or have angular acceleration. By assumption is has neither. One must have at least three forces for the rod to remain at rest. Therefore the point force is a bad model. –  Mark Eichenlaub Jul 11 '13 at 6:53

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!)

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$$\tau_B = F L \\ \tau_A = F (L+r)$$
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.