I have been wondering about a door mechanism I have seen. It has a wire attached to the upper corner of the door and from there to the corresponding corner in the door frame, where a weight hangs from it on a "frictionless" roller. So when the door is opened the weight will rise and then be pulled down by gravity when you let go of the door, closing it.
Considering a door with width $L$ and moment of inertia $J$ what is the angle of the door $A$ as a function of time with some initial angle $A_0$?
So after doing some math I get to $A''(t)=k\cos(A(t)/2)$ where $k$ is a bunch of constants
How do I solve that?
Thanks