Rotational behavior of objects in zero-g with forces I would like to know how this object would rotate in free space out of curiosity.
Below is a diagram of the object. It is of uniform density and has a center of mass in the center of the object, denoted by the green cross.
On each long end of the bar, on opposing sides, are thrusters. They are placed at exactly the same distance from the ends and from the center of mass. They provide the same amount of force. They are labeled "A" and "B" in the diagram, and the red arrows show which direction the thruster fires.
Edit: Assume that the thrusters are ideal and do not lose fuel/mass over time.

If both thrusters are fired simultaneously, I believe the bar will spin about its center of mass, and there will be no translation.
My questions are about what happens when you fire only one of the thrusters.
Question 1
What happens if you fire only thruster B?
I think the bar will rotate, but not about it's center of mass, and not it's endpoints.
I think it will also translate, but not sure how.
Question 2
What equation(s) govern the motion of the body in this scenario? 
I am almost certain that center of mass is a variable in this, because if you put a thruster directly in line with the center of mass, you would have no rotation at all, only translation.
Question 3
If left to spin to high velocities, will the object be "locked in" on it's translation vector?
Because it is spinning so fast, the thrust wouldn't have a significant net force in any particular direction, so the object would forever travel at some particular velocity and heading.
Question 4
What happens if you fire thruster B at 100% force, and thruster B at 50% force? 
This question may already be answered by one of the other questions above.
Edit 2:
I wanted to see what this motion looked like, so I coded something up in an attempt to model it. I only implemented angular and translational acceleration/velocity from the equations Physics Llama provided.
 A: By way of analogy, think of what happens when you blow up a balloon and let it go. It spins around, goes this way and that. A balloon rarely goes straight, without spinning. The thrust from a balloon rarely goes through the center of mass. It rotates and translates. Because the thrust vector itself turns with the rotating balloon, the translation is not along a straight line path.
Equations of motion
The equations of motion separate into translational and rotational parts by looking at things from the perspective of the center of mass and by ignoring motion of the center of mass within the vehicle. The equations of motion don't decouple of the fuel tanks are not at the center of mass (which they never are). I'll ignore this.
In this case, the center of mass translational motion is given by Newton's second law, $\vec F = m\vec a$, where $\vec F$ is the net force acting on the object, total thrust plus all other external forces. (Note: Not $F=\dot p$. That is valid for constant mass objects, but then again, so is $\vec F=m\vec a$.)
The rotational behavior is more complex. The rotational analog of $\vec F=m\vec a$ is $\vec T = I\vec \alpha$ in simple, freshman-level problems. That is not true in general, and it is not true for your object. Your object is boxy; it does not have a spherically symmetric mass distribution. This means that the inertia tensor is time varying from the perspective of an inertial frame. You don't want to go there. Much better is to look at things from the perspective of a frame fixed with respect to the rotating object. This is a non-inertial frame, so fictitious torques will arise.
For any vector quantity $\vec q$, the time derivative of the vector from the perspective of a co-moving inertial frame and this body-fixed frame are related via
$$\left(\frac {d\vec q}{dt}\right)_\text{inertial} =
\left(\frac {d\vec q}{dt}\right)_\text{body} + \vec \omega \times \vec q$$
This is sometimes called the dynamical transport theorem (e.g., see section 2.2 of http://ocw.mit.edu/courses/mechanical-engineering/2-003sc-engineering-dynamics-fall-2011/newton2019s-laws-vectors-and-reference-frames/MIT2_003SCF11Kinematic.pdf).
Applying this to angular momentum $\vec L = \mathbf I\, \vec \omega$ yields $$\dot{\vec L}_{\text{inertial}} = \dot{\vec L}_{\text{body}} + \vec \omega \times (\mathbf I\, \vec \omega )$$
The left-hand side is the external torque, including that from the thrusters. Once again ignoring mass depletion, the inertia tensor is constant in the body frame, yielding
$$\dot {\vec \omega} = \mathbf I^{-1}\left(\vec \tau_{\text{ext}} - \vec \omega \times (\mathbf I\, \vec \omega ) \right)$$
That second term on the right hand side results in some bizarre motion: "Hence the jabberwockian sounding statement: the polhode rolls without slipping on the herpolhode lying in the invariant plane." (Goldstein, Classical Mechanics).
Things get messier yet when mass depletion and motion of the center of mass within the body is brought into the picture.
Question 1
You've just turned the vehicle into a kid's balloon. Not good.
Ideally, thrust will be straight through the center of mass, resulting in pure translational motion, or the net thrust will be zero (e.g., thrusters A and B firing in unison), resulting in pure rotational motion. That ideal never happens. Thrusters designed to produce translational acceleration inevitably produce a bit of rotational acceleration, thrusters designed to be used in pairs as a force couple inevitably produce a bit of translational acceleration. A real vehicle has to watch out for these undesired forces and torques and control for them.
Question 2
The simplified equations of motion are above.
Question 3
That's the subject of scifi movies. And vehicles with failed-on thrusters. Real thrusters are designed to fail off.
By way of analogy, what happens when you spin a CD too fast? The answer is that they tear themselves apart. Every once in a while that does happen. The CD disintegrates and the little chunks tear the daylights out of the CD drive. Space vehicles aren't made to be spun up to a high angular velocity. They would disintegrate, just like that CD, but at a much lower angular velocity.
Question 4
This is still a kid's balloon, just not as bad as your question #1.
A: Apparently it's better to ask questions only one-at-a-time on this site, but I understand why you did it like that; these are all related. Now, these are some really interesting questions, hopefully I can help you out (there's some interesting subtleties to consider due to the thrusters). 
Answer 1 
If you fire only B, then it will rotate, and it will translate- you are correct. It will rotate about its center of mass, and will accelerate in the direction the force is applied (upwards, in this case). However, it will only accelerate upwards for a bit; then the object will have rotated, so now it will accelerate upwards and a little to the left; what we have is a force that is changing direction continuously. Another thing to consider: if it relies on thrusters, then the object loses mass as the thrusters are turned on. If it only turns on one thruster, maybe it will lose mass on that side and the center of mass will shift? Not only that, but the force applied by the thrusters will change in magnitude as well as direction of the mass of the object is changing! That's something important to consider; maybe it would be simpler to just push it on side B or side A instead. ]
Hopefully now you have a clear-ish picture of what should happen. If we simplify the situation a little so we use a constant-magnitude force instead of thrusters, then there will be a constant torque applied and a force that changes direction. This force will accelerate the center of mass, first upwards, then up and right, then right, then right and downwards, etc.
Answer 2
The equations of motion governing this are Newton's second law and the definition of torque. If the thrusters (or better yet, the constant magnitude force you will apply) apply a force $\mathbf{F}$ on the object, then the acceleration of its CM will be
$$
\mathbf{a}=\frac{\mathbf{F}}{m}
$$
And the rotation of the body about the center of mass will be described by:
$$
\alpha=\frac{\mathbf{\tau}}{I}=\frac{\mathbf{r}\times\mathbf{F}}{I}
$$
Where $\alpha$ is its angular acceleration, $I$ is its moment of inertia, and $\mathbf{r}$ is the vector which goes from the center of mass to the place where the force is being applied. So it does matter, for the rotation, if you place the thruster far away from the center of mass or right on it- your intuition was correct on that point. 
Answer 3
This question I didn't understand so well. Are you saying that if it spins very quickly, then it'll be as if there's no net force accelerating the CM anywhere, since the force vector will be changing direction so fast? That's not quite true- it will move fairly erratically, changing direction quite quickly and translating a bit first upwards, then up and right, etc. 
If this wasn't your question, please comment so I can edit this section.
Answer 4
Yep, you're right- this is sort of a similar situation to part 1. There's one difference, though: the torque will be 1.5 times greater, because you'll essentially have a rotation done by two different sources- one which is the same as before, and one which is 50% of the one we had before. The net force on the center of mass, though, will be 0.5 times the previous net force. So there will be more rotation and less translation in this case. 
Hopefully this helped, I really enjoyed your questions- I'm reviewing rotational dynamics right now so it was a nice way to see if I'm getting everything!
Appendix to Answer 1
A little side-note to question 1: I've seen a sort of similar problem before where an initial impulse was applied onto one side of the mass for a very short amount of time. Here, what would happen is just a force which accelerates the CM upwards, and a torque which sets the mass spinning. This was a much simpler situation to visualize and to concretely solve than the one you proposed, so maybe you might want to try looking at that in detail. 
