If the position of the c.g. is $\vec{r}_C$ and the location of the force application $\vec{r}_A$ then the Euler-Newton equations of motion for rigid body are:
$$ \vec{F} = m\,\vec{a}_C \\ (\vec{r}_A-\vec{r}_C)\times \vec{F} = I_C \vec{\alpha} + \vec{\omega}\times I_C \vec{\omega} $$
with c.g. velocity $\vec{v}_C = \dot{\vec{r}_C}$, c.g. acceleration $\vec{a}_C = \ddot{\vec{r}_C}$, $I_C$ the moment of inertia tensor about the c.g.
In 2D when $(x,y)$ is the location of the c.g. Point C this becomes
$$ \begin{vmatrix} F_x \\ F_y \\ 0 \end{vmatrix} = m \begin{vmatrix} \ddot{x} \\ \ddot{y} \\ 0 \end{vmatrix} \\ \begin{vmatrix} c_x \cos\theta \\ c_y \sin\theta \\ 0 \end{vmatrix} \times \begin{vmatrix} F_x \\ F_y \\ 0 \end{vmatrix} = \begin{vmatrix} I_x & &\\& I_y & \\ & & I_z \end{vmatrix} \begin{vmatrix} 0 \\ 0 \\ \ddot{\theta} \end{vmatrix} + \begin{vmatrix} 0 \\ 0 \\ 0 \end{vmatrix} $$
where $(c_x,c_y)$ is the position of point A from the c.g. when the body orientation is $\theta=0$ (initially).
By component then the equations are $$ \ddot{x} = F_x/m \\ \ddot{y} = F_y/m \\ \ddot{\theta} = \frac{-c_y \sin\theta F_x + c_x \cos\theta F_y}{I} $$
If the force is rotating with the body, and initially located at $(cx,0)$ pointing in the +y direction then
$$ \ddot{\theta} = \frac{c_x F_y}{I} $$