# Torque produced by conservative forces vs Torque produced by non-conservative forces

One clear observation is that we could write torque in the following way for conservative forces:

$$\vec{\tau} = \sum_{i=1}^n \vec{r}_i \times (-\vec{\nabla} U(\vec{r}_i))$$

Where $$U$$ is the potential function defined over space and $$\vec{r}_i$$ is the position vector from the origin to the point where force is acting.

My question: Physically speaking, what would be the difference in rotational effects shown by torque produced by each kind of force? Further what would be the nature of torque produced by a mixture of conservative and non-conservative forces causing a moment?

• If you have nonconservative forces, by definition you can't write ${\vec F} = {\vec \nabla}U$, because there is no well-defined potential function. Commented Dec 26, 2020 at 15:58

At each moment, the torque depends on the direction of, but not the nature of the force.

It is not much different, except you cannot write the expression for the torque like you did for a conservative force: by definition, non-conservative forces cannot be expressed as the gradient of a potential. Other than that, once a force is present, it is straightforward to calculate its moment, or the torque: $$\mathbf{\tau} = \mathbf{r} \times \mathbf{F}.$$

Example: The torque due to friction on a cylinder on an inclined plane enables it to roll down the plane (as opposed to simply translating).

If $$U$$ is a continuous potential-energy density, then

$$\vec \tau = \int _ D \vec r \times - \vec \nabla U ( \vec r ) dV = \oint _ S - U ( \vec r ) \vec r \times \vec n d\sigma$$

where $$S$$ is the surface of the volume $$D$$ and $$\vec n$$ is the unit vector normal to the surface area $$d\sigma$$.

Consequently, the total torque on a finite body due to a conservative (i.e. irrotational) force density is zero. (Just choose $$S$$ so that it encompasses the body.)