How to take in to account torques applied at different points Suppose we have a rigid body with known moment of inertia through some axis ($J$) and that there are multiple torques being applied in different points in that body.
I know that for a rigid body, the equation 
$$J \space \ddot{\theta}=\sum\tau_i$$
holds. However, how am I supposed to sum the torques if they aren't applied at the same point?
 A: Any force not applied through the center of mass of an object will impart both linear motion and torque. You can consider the force applied at the center of mass for the computation of the linear motion, and you can then consider the torque as being the moment applied by the force at the center of mass. So for both linear and rotational motion, you end up considering "everything as seen from the c.o.m.".
In general, you need to take the vector sum of all torques - torque as a vector is perpendicular to both the force vector, and the arm. Or if you like, for a force $\vec{F}$ at distance $\vec{r}$ from the c.o.m., the torque vector $\vec{\Gamma}$ is (note: formula was edited. Originally sign was wrong. Thanks to Physicist137 for pointing it out)
$$\vec\Gamma = \vec{r}\times\vec{F}$$
that is, the cross product of distance and force. You sum this expression over each individual force on the object and will get the net torque.
Incidentally, if you took a disk and tried to apply a torque "off center" (for example by sticking a rod in the disk at some off center location and trying to just rotate the disk about the rod) you might intuitively think that you are applying an off center torque. However, if the rod is to remain in place, you actually need to apply a linear force as well (because the center of mass tries to move when you spin your rod, a net force needs to be applied to cause that motion so the desired center of rotation remains stationary). So "spinning about a point other than the center of mass" cannot be done with just a torque... All of this is of course in complete agreement with everything I said above.
A: You can sum them, without taking their respective positions into account. 
For example: Lets say you have a 2D body, with two forces applied to them, forming a couple. When the reference point is exactly the center position between the points where the forces are applied, you experience in the reference point
$ \tau = Fa $
with $F$ as one force and $a$ as the distance between the applied forces. Let's take another reference position, located by distance $b$ from the center of the applied forces. The torque in this new reference point is
$ \tau = F (b + 1/2 a) - F (b - 1/2 a) $
The sign of the second term on right hand side is negative, because it causes a negative torque. The terms with $b$ cancel out, leaving the first equation.
So the torque at a point can be moved to any point of a rigid body.
Therefore, you can simply add any torques acting on different points to get a total torque.
