What is the trajectory of a particle under a constant angular force always perpendicular to its position vector? It is a question of classical mechanics, related but indeed distinguished from the central force problem.

The equation of motion can be expressed as below in polar coordinates:
  $$\ddot{r}-r\dot{\theta}^2=0 \\ r\ddot{\theta}+2\dot{r}\dot{\theta}=\frac{F}{m}$$
  with initial condition of zero velocity (whatever, just for simplification):
  $$r(0)=r_0,\ \theta(0)=0,\ \dot{r}(0)=0,\ \dot{\theta}(0)=0.$$

However, I am not talent to analytically solve this equation. 
Can you help me?
Of course, the famous first integral in central force problem, the angular momentum, dose not help to solve it. I believe that the trajectory has a name, I mean some meaningful curve, but I cannot find it.
Also I am curious that can this problem map to any real physical system? It might help to solve it.
 A: This definitely corresponds to a physical system: consider a mass inside a frictionless semi-infinite tube with one end at the origin. Your situation corresponds to rotating the tube so that the normal force on the mass is constant. From this setup it's intuitively clear that this tends to force the mass away from the origin.
I've encountered this setup before, but with a constant torque, which would probably be the more realistic case. In that case it's easy to solve because the angular momentum increases linearly in time. Your case is harder. After eliminating $\theta$ I arrived at
$$3 \dot{r} \ddot{r} + r \dddot{r} = (r \ddot{r})^{1/2} \frac{F}{m}$$
from which we can observe that one solution is
$$r(t) = \frac{F}{6 \sqrt{2} m} \, t^2$$
which also implies that $\theta(t) \propto \log t$. But I don't know how to solve it for general initial conditions.
I don't have time to pursue this further, but one technique you may want to try is complexification. That is, represent the position of the particle in the $xy$ plane as a complex number $z$. Then you only have one equation, $\ddot{z} \propto i z/|z|$, which is rather simple. Hopefully somebody can finish this analysis!
