How to calculate the change in angle due to the righting moment of a ship? I trying to calculate how the forces on a boat (initially out of equilibrium) work to right it. The cross section of the boat is a convex polygon and constant throughout its length. I can find the center of mass, $G$, and hence the weight of the boat, and for a given water level I can find the center of buoyancy, $B$, and hence the buoyant force due to the displaced water (the boat has a uniform density less than that of water).

My question is: am I correct to take the torque around $G$ to be $\boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F}_b$ and then calculate the instantaneous change in angular velocity about $G$ as
$$
\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = \frac{\boldsymbol{\tau}}{I}
$$
where $I$ is the moment of inertia of the boat about $G$? Here I take it that the component of $\boldsymbol{F}_b$ perpendicular to $r$ is responsible for the torque; leaving a parallel component: what happens to this component of $\boldsymbol{F}_b$ parallel to $r$? Do I add it to $\boldsymbol{F}_g$ to give the change in linear momentum of the boat?
 A: There are two relationships to consider.  (1) The equation of motion for the center of mass (CM), and (2) the equation for rotation about the CM.
For motion of the CM, $\vec F_b - \vec F_g = m\vec a$ where $\vec a$ is the acceleration of the CM taking upward as positive, and $m$ is the total mass of the boat.  Assuming the boat can only rotate about an axis into the page, as you say you take the CM as the origin and $\vec r \times \vec F_b = \vec \tau =  I d \vec \omega/dt$ for rotation around an axis through the CM ($\vec \omega$ is into the page through the CM).  As you say, the component of $\vec F_b$  perpendicular to $\vec r$ contributes to the magnitude of the torque.  (For the CM moving, you should always take the torque about the CM , as shown in texts on mechanics, such as Mechanics by Symon and Classical Mechanics by Goldstein.)
As the equation of motion for the CM indicates, you add the vectors $\vec F_b$ and $-\vec F_g$ to evaluate the motion for the CM.  Both these vectors are only in the vertical direction, so the magnitude of the acceleration upward of the CM is $a = {{F_b - F_g} \over m}$.
All of $F_b$ contributes to counter $F_g$, not just the component of $\vec F_b$ parallel to $\vec r$.
