Center of rotation and trajectory of a rigid body in a plane with applied *fixed* forces This is my first question so please excuse me if my format is a bit off.
Given a 2D rigid body with forces applied to it in such a way that the angle the force vector makes with the surface of the object remains constant (think of a spaceship with fixed rockets attached to it), I have problems mapping out it's trajectory and angle it has rotated respect the starting position over a certain period of time.
The net force from the rockets applied on the centre of mass of this object is easy to calculate on a local reference solitary to said object, but given the existence of external torque, it is a reference possessed by angular acceleration, which unsettles me as I do not know if this requires any extra considerations when translating that to a general reference.
Determining the centre of rotation is also difficult, as from what I have researched, it's bound to be the centrer of mass, wherever the forces I've applied are at, although this seems unintuitive to me.
 A: The problem as is stated is somehow ambiguous, but using some simplifications we can manage to get something: if we assume that forces don't change depending of the angle (i.e. there is no "correcting trayectory rocket" that acts depending of its orientation), and that the center of mass is fixed, then you can express net force F' and net torque T' with respect to the ship frame as functions only of time. Let's call $\theta$ the angle between the ship frame and the "ground" frame (some external frame at rest), then:
$ \frac{d^2 \theta}{dt^2} = \frac{T'(t)}{I_0(t)}$,
where $\ I_0$ is the moment of inertia with respect to the center of mass, posibly a function of time. you can find the angle $\theta$ by double integration and then express the force in the ground frame in terms of the force in the ship frame using the matrix rotation
$ R_\theta = \begin{bmatrix} cos\theta  & -sin\theta \\ sin\theta & cos\theta \end{bmatrix}$
Then, wrt to the ground, $ \bar{F} = R_\theta \bar{F'} $ and $\ \bar{T} = \bar{T'} $, hence
$ \frac{d^2\bar{r}}{dt^2} = \frac{R_\theta}{m} \bar{F'} $
In the simplest case, with $ \bar{F'}, \bar{T'}, I_0, m $ constant, and the ship starting at rest, you get
$ \theta(t) = \frac{T}{2I_0} t^2 $, and 
$x(t) = \frac{F_{x'}}{m} \iint cos(wt^2) dt^2 + \frac{F_{y'}}{m} \iint sin(wt^2) dt^2 $
$y(t) = -\frac{F_{x'}}{m} \iint sin(wt^2) dt^2 + \frac{F_{y'}}{m} \iint cos(wt^2) dt^2 $
for $ w = \frac{T}{2I_0} $.
Unfortunately those integrals are not doable, see this wikipedia entry. Interestingly, this kind of fuctions are used in freeway design, where forces between vehicles and the road at constant speed are also perpendicular to the trayectory(see Euler spiral).
