What forces are exerted on a clothespin in space? Let's say a clothespin is modeled as a simple torsion spring as follows.
Given:


*

*$p_1,\ p_2,\ p_3$: point-like objects of equal mass in 2-D space.

*All objects float in space, i.e. the center of mass will not change.

*At time $t_0$ a torsion spring is inserted at $p_1$, such that it exerts torque on $p_2$ and $p_3$, with:


*

*$\theta$: the angle of twist from its equilibrium position in radians

*$\kappa$: the spring's torsion coefficient

*$\tau = -\kappa \theta$ is the torque exerted by the spring



Question: what are the resulting forces on $p_1$, $p_2$ and $p_3$?
My answer:
Because all objects have equal mass, we can leave mass out of the equation.
F2 is a force perpendicular to $p_1,\ p_2$ of magnitude $\dfrac{\tau }{ |p_1-p_2|}$ .
By Newton’s 3rd law, F2’ is a force of equal magnitude and opposite direction as F2
Similar for F3 and F3’


If $\theta$ is the angle between the arms, displaced from the equilibrium $\theta_0$ by $\Delta \theta$ and the torque applied is $\tau =-\kappa \Delta \theta$, assuming equal masses of $m$ with initially motionless parts.
The first step is the kinematics, whereas the acceleration of 2 and 3 is related to the acceleration of 1 and the common angle. For simplification we have that 1 is not accelerating in the horizontal direction $\ddot{x}_1=0$ (as seen in figure below).

$$ \begin{aligned}  
\ddot{x}_2 &= \ddot{x}_1 - \ell \cos \left( \frac{\theta}{2} \right) \frac{ \ddot{\theta}}{2} & \ddot{x}_3 &= \ddot{x}_1 + \ell \cos \left( \frac{\theta}{2} \right) \frac{ \ddot{\theta}}{2} \\  
\ddot{y}_2 &= \ddot{y}_1 + \ell \sin \left( \frac{\theta}{2} \right) \frac{\ddot{\theta}}{2} & \ddot{y}_3 &= \ddot{y}_1 + \ell \sin \left( \frac{\theta}{2} \right) \frac{\ddot{\theta}}{2}
\end{aligned} $$
Now for the equations of motion of each part. We start with free body diagrams in order to sum up the forces on each part.

$$\begin{aligned}
-Fr_2 \sin \left( \frac{\theta}{2} \right) + Fr_3 \sin \left( \frac{\theta}{2} \right) + Fn_2 \cos \left( \frac{\theta}{2} \right) + Fn_3 \cos \left( \frac{\theta}{2} \right) & = m \ddot{x}_1 = 0 \\
-Fr_2 \cos \left( \frac{\theta}{2} \right) - Fr_3 \cos \left( \frac{\theta}{2} \right) + Fn_2 \sin \left( \frac{\theta}{2} \right) + Fn_3 \sin \left( \frac{\theta}{2} \right) &= m \ddot{y}_1 
\end{aligned} $$

The EOM are done in a direction along the arm
$$\begin{aligned}
 m \ddot{x}_2 \cos \left( \frac{\theta}{2} \right) - m \ddot{y}_2 \sin \left( \frac{\theta}{2} \right) &= -Fn_2 \\
 m \ddot{y}_2 \cos \left( \frac{\theta}{2} \right) + m \ddot{x}_2 \sin \left( \frac{\theta}{2} \right) & = Fr_2 \\
 0 & =\ell Fn_2  + \tau
\end{aligned} $$ with $\Rightarrow Fn_2 =-\frac{\tau}{\ell}$

$$\begin{aligned}
 m \ddot{x}_3 \cos \left( \frac{\theta}{2} \right) + m \ddot{y}_3 \sin \left( \frac{\theta}{2} \right)  &= -Fn_3 \\
 m \ddot{y}_3 \cos \left( \frac{\theta}{2} \right) - m \ddot{x}_3 \sin \left( \frac{\theta}{2} \right) & = Fr_3 \\
 0 & =\ell Fn_3  - \tau
\end{aligned} $$ with $\Rightarrow Fn_3 =\frac{\tau}{\ell}$
Combined the all of the above equations substituted into the kinematics are
$$ \begin{aligned} 
 -Fn_2 \cos \left(\theta \right) + Fr_2 \sin \left(\theta \right) - 2 Fn3 &= m \ell \frac{\ddot{\theta}}{2} \\ 
Fr_2 \cos \left(\theta \right) + Fn_2 \sin \left(\theta \right) + 2 Fr_3 &= 0 \\ 
 - Fn_3 \cos \left(\theta\right) - Fr_3 \sin \left(\theta \right) + 2 Fn_2 &= - m \ell \frac{\ddot{\theta}}{2} \\ 
Fr_3 \cos \left(\theta \right) - Fn_3 \sin \left(\theta \right) + 2 Fr_2 &= 0 
\end{aligned} $$
The above is solved with $$\boxed{\frac{3 \tau (\cos\theta-2)}{\ell (\sin^2\theta+3)} = m \ell \frac{\ddot{\theta}}{2}}$$
and 
$$\begin{aligned} 
  Fr_2 & = \frac{\tau \sin\theta ( 2-\cos\theta)}{\ell (\sin^2\theta+3)} \\
  Fn_2 &= -\frac{\tau}{\ell} \\
  Fr_3 &=  \frac{\tau \sin\theta ( 2-\cos\theta)}{\ell (\sin^2\theta+3)} \\
  Fn_3 &= \frac{\tau}{\ell} 
\end{aligned}$$
