Langevin equations in translational and rotational direction I want to describe the following system.


*

*A bead is connected with a tether.

*There is a force $\vec{F}_{up}=F_{up}\hat{y}$ that acts on the bead.

*The tether acts with a force on the bead, this force $F_{spring}$ is only dependent on the length/extension of the spring.

*A Brownian force acts in the rotational and translational directions. Just consider these as a random variable in x-y-theta direction.


Now I want to formulate the Langevin equation for the rotational motion. 
I know that if there is no angular dependence, $\theta=0$ for all time, the Langevin equation are:
$\gamma_{y}\frac{dy}{dt}+F_{y}^{tether}(x,y)+\frac{dF_{y}^{tether}(x,y)}{dy}\hat{y}=F_{y}^{therm}+F_{mag}-mg$
$\gamma_{x}\frac{dx}{dt}+F_{x}^{tether}(x,y)+\frac{dF_{y}^{tether}(x,y)}{dx}\hat{x}=F_{x}^{therm}$
What I know:
The $\gamma_{theta}$ and $F_{them}^{\theta}$ are known from literature.
Should I construct a construct a Langevin equation in the same way, such as
$\gamma_{\theta} \frac{d\theta}{dt}+\frac{d\tau_{tether}}{d\theta}\theta+\tau_{tether}=F^{tether}_{\theta}$ ?
More specifically, should it include the term $\frac{d\tau_{tether}}{d\theta}\theta+\tau_{tether}$, which is completely analogous to the translational equations?

Trivia: In the end I want to extend this into 3 dimensions and solve the system numerically, using Langevin dynamics. But I'm stuck on how to describe the force.
EDIT:
I've edited the question into something more specific.
 A: All the forces acting on the bead will accelerate the center of mass as though the force is acting there - so the mass will accelerate with a magnitude and direction given by the vector sum of $$\frac{\vec{F_{up}} + \vec{F_{spring}}}{m}$$ .
The bead will further undergo a rotation. The torque is entirely due to the spring - but we are missing the angle of the spring to the vertical. Let's call that angle $\alpha$. Then the component of the force of the spring causing rotation is $F_{spring}\cos(\pi - \alpha - \theta - \pi / 2)=F_{spring}\sin(\alpha + \theta)$. 
Diagram:

From this it follows that the angle $\beta$ can be calculated from $x$, $y$ and $\theta$ (as I defined them in the drawing - I assume this is how you define them as well... not quite clear from your diagram).
Now you can see that 
$$\tan\alpha = \frac{x + r\sin\theta}{y - r\cos\theta}$$
After which you can calculate $\beta$ from
$$\beta = \pi/2 - (\alpha + \theta)$$
And finally, the torque on the bead is
$$\Gamma = F_{spring} r \cos\beta$$
