I know, this may seem as a bit of a weird question. As we know that the relation between force and potential energy i.e: $$F = -\frac{\text{d}U}{\text{d}x}$$ so can we say that relation between torque about an axis and potential energy to be: $$\tau = -\frac{\text{d}U}{\text{d}\theta}$$
Any help would be appreciated.
Short answer is yes, as long as your system is conservative.
Longer answer is that the Euler-Lagrange equations tell you what the equations of motion really are for these systems, so you setup the Lagrangian $L(r, \theta,\dot{r},\dot{\theta}) = T(\dot{r},\dot{\theta}) - U(r,\theta)$, where $T$ is the kinetic energy and $U$ is potential energy. Then calculate:
${d \over dt} {\partial L \over \partial \dot{\theta}}={\partial L \over \partial \theta}$.
In most cases, ${\partial L \over \partial \theta}$ is exactly $-dU\over d\theta$,and the left hand side turns out to be $I\alpha$.
In my opinion, the best definition of torque is the work per unit angle which can be done by a force which is acting in a manner that tends to cause a rotation. If motion occurs then τ = +dU/dθ where U is the total energy of the rotating system (not just the potential energy). Your expression refers to the torque coming from an external source, such as a deformed spring which is losing potential energy as it causes a rotation.
