 Solution

Introduce ${\mathbf{e}}_i=(\cos {\theta }_i,\sin {\theta }_i)$ and ${\mathbf{e}}_i^{\perp }=(-\sin {\theta }_i,\cos {\theta }_i)$. The joint and tip positions aresoUsing the point drags ${\mathbf{F}}_A=-\zeta {\dot{\mathbf{r}}}_A$, ${\mathbf{F}}_B=-\zeta {\dot{\mathbf{r}}}_B$, and the follower force $\mathbf{\Gamma }=-\Gamma {\mathbf{e}}_2$ in the principle of virtual work gives the independent coefficients of $\delta {\theta }_1$ and $\delta {\theta }_2$:With scaled time $s=kt/(\zeta {\ell }^2)$ and dimensionless follower load $\Sigma =\Gamma \ell /k$, these are exactly the stated equations with primes denoting $d/ds$.

If the two links are constrained to remain collinear, their admissible virtual rotations satisfy $\delta {\theta }_1=\delta {\theta }_2$. Adding the two generalized equations and putting ${\theta }_1={\theta }_2=\theta$ givesThe drag factor five is the sum of the squared lever arms $1^2+2^2$. The follower force lies along the straight filament and has no moment, so it cannot affect this rigid rotational relaxation.

For unrestricted perturbations, linearization about the straight state givesFor modes proportional to $e^{\lambda s}$,and thereforeThe roots are negative and real below $\Sigma =2$, coalesce at $-1$, and then form a complex-conjugate pair. For $2<\Sigma <4$ they trace the unit circle from $-1$ to $+1$; they cross the imaginary axis at $\lambda =\pm i$ whenwhich is a Hopf bifurcation. Above $\Sigma =4$ they separate along the positive real axis.

Viscous drag and elastic spring forces alone have a symmetric positive mobility and a symmetric potential Hessian, so an overdamped gradient system has only real decay rates. The follower force is nonconservative: its linearized generalized-force matrix is nonsymmetric and cannot be derived from a potential. This broken variational structure permits complex eigenvalues and hence an oscillatory instability even though inertia is absent.