I have been trying to find ways to solve: $$J\frac{d²\theta(t)}{dt²}-K_m cos(\theta(t))=-\tau_f$$ With the initial conditions $$\theta(t=0)=0$$ $$\frac{d\theta}{dt}(t=0)=0$$ Without success. Is that this complicated?
Which should lead to theta(t) = a step to pi/2 with damped oscillations.
$$J\frac{d²\theta(t)}{dt²}-K_m \cos(\theta(t))=-\tau_f$$ $$2J\frac{d²\theta(t)}{dt²} \frac{d\theta(t)}{dt} -2K_m \cos(\theta(t))\frac{d\theta(t)}{dt}=-2\tau_f\frac{d\theta(t)}{dt}$$
$$J\left(\frac{d\theta(t)}{dt} \right)^2 -2K_m \sin(\theta(t)) =-2\tau_f \theta(t) +c_1$$
Condition : $t=0\quad \theta=0\quad\frac{d\theta(t)}{dt}=0 \quad\to\quad c_1=0$
$$\frac{d\theta}{dt}= \pm \sqrt{J\left(2K_m \sin(\theta) -2\tau_f \theta \right) }$$
$$t=\pm \int_{x=0}^{x=\theta} \frac {dx}{\sqrt{J\left(2K_m \sin(x) -2\tau_f \: x \right) }}$$
There is no closed form for this integral (this isn't an elliptic integral)