Trajectory of two joined points I'm wondering how could I calculate the trajectory of two joined points with some weight (let's say they are joined in the way they are forced to be always at the same distance) if there exerts a constant force in the perpendicular angle to the line between the two points in one of the points. So as these points move and rotate, the force F moves and rotates with them.

 A: Let's assume the two masses are equal. You must first solve for the rotation in the referential of the center of mass.
$$F. \frac{L}{2}=I \frac{d \omega }{dt} $$
From which you get the angular velocity and angle:
$$ \omega (t)= \frac{F.L}{2I}t $$ and the angle: $$ \theta (t)=\frac{F.L}{4I}t^{2}$$
Now you have to calculate the trajectory of the center of mass. The force $
 \overrightarrow{F}$ is time dependent because of the rotation. $$ \overrightarrow{F} = \begin{cases}F.cos \big( \theta (t)+ \frac{ \pi }{2} \big)  \\F.sin\big( \theta (t)+ \frac{ \pi }{2} \big)\end{cases}$$
It comes:
$$ \begin{cases}2m \frac{d V_{x} }{dt} =-F.sin \big(\frac{F.L}{4I}t^{2}\big)\\2m \frac{d V_{y} }{dt} =F.cos \big(\frac{F.L}{4I}t^{2}\big)\end{cases}  $$
The system can be integrated with the help of the Fresnel integrals:$$\begin{cases}C(x)= \int_0^x cos \big( x^{2} \big) dx  \\S(x)= \int_0^x sin \big( x^{2} \big) dx  \end{cases} $$
$$\begin{cases}x_{cm}(t) = -\frac{1}{m} \sqrt{ \frac{ \pi .I.F}{2.L} } t.S \big(\sqrt{ \frac{F.L}{2 \pi .I}t } \big)- \frac{I}{m.L}cos \big(\frac{F.L}{4I}t^{2}\big)\\y_{cm}(t) = \frac{1}{m} \sqrt{ \frac{ \pi .I.F}{2.L} } t.C \big(\sqrt{ \frac{F.L}{2 \pi .I}t } \big)- \frac{I}{m.L}sin \big(\frac{F.L}{4I}t^{2}\big)\end{cases} $$
The coordinates of B are for instance: $$\begin{cases} x_{B}(t)= x_{cm}(t)+ \frac{L}{2}cos \big(\theta  \big(t\big) \big)     \\y_{B}(t)= y_{cm}(t)+ \frac{L}{2}sin \big(\theta  \big(t\big) \big)\end{cases}$$
