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

Question

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?

Answer 1

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

Answer 2

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).

Answer 3

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.)