How to solve for the trajectory of the center of mass? I'm working on the physics engine component of a game engine I'm building, and I need some guidance with this particular situation.
Consider a square with mass M that is free to translate in the xy plane and free to rotate about any axis perpendicular to the page (Fig. 1)
If a linear impulse J is applied at a point above the center of mass (CM) as shown below, I know there must be some angular impulse (momentary torque) generated since there is a component of J that is perpendicular to the displacement vector from CM. I imagine this angular impulse will tend to rotate the square clockwise.
However, I can also imagine that the CM will also undergo translation since the square is not constrained. How would I go about computing the overall rotational + translational motion of this system?

 A: The motion (acceleration) of the center of mass (CM) is $\vec a_{CM} = {\vec F_{ext} \over M}$ where $\vec F_{ext}$ is the total applied external force and M is the mass of the square.  The rotation about the CM is ${d\vec L_{CM} \over dt} = \vec \tau_{ext\enspace CM}$ where $\vec L_{CM}$ is the angular momentum about the CM and $\vec \tau_{ext \enspace CM}$ is the total external torque about the CM.
For a short duration applied force $\vec F_{applied}$, the impulse $\vec P = \int_{t_1}^{t_2} \vec F_{applied}\enspace dt$ and the velocity of the CM after the impulse is $\vec v_{CM} = {\vec P \over M}$.  Similiarly, the angular velocity after the impulse is $\vec \omega = {{\vec r_c \times \vec P} \over I_c}$  where $\vec r_c$ is the vector from the CM to the point where the force is applied, and $I_C$ is the moment of inertia with respect to the CM.
If there other forces besides the force for the impulse- gravity and any constraints- these need to be also considered.
A: If $\mathbf J(x(t),y(t))$ is the external force acting on square between times $t_0, t$, then the total impulse is $\int_{t_0}^{t}\mathbf J(x(u),y(u)))du$. So we get $\int_{t_0}^{t}\mathbf J(x(u),y(u)))du=\int_{t_0}^{t}m\frac{d\mathbf v}{du}(u)du=m\mathbf v (t)-m\mathbf v (t_0)=\mathbf p(t)-\mathbf p (t_0)$, where $m$ is the mass of the square, $\mathbf v$ the velocity and $\mathbf p$ the total linear momentum. Similarly you get that $\int_{t_0}^{t}\mathbf τ du=\mathbf L (t)-\mathbf L (t_0)$, that is the total torque is the change of angular momentum from $t_0$ to $t$.
