1
$\begingroup$

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.

$\endgroup$

2 Answers 2

4
$\begingroup$

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 $I\alpha$.

$\endgroup$
3
  • $\begingroup$ But what happens if the force is non conservative? $\endgroup$
    – WHOOP
    Jun 17, 2021 at 14:18
  • $\begingroup$ "Short answer is yes" yes to what? Did he ask any question? I can't find one. $\endgroup$ Dec 28, 2021 at 8:38
  • $\begingroup$ "so can we say that relation between torque about an axis and potential energy to be ....?" $\endgroup$ Jan 20, 2022 at 17:26
1
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.