8.2 Conservative and non-conservative forces (Page 2/6)

University physics volume 1 Page 2 / 6

\frac{d F_{x}}{d y} = \frac{d F_{y}}{d x} .

You may recall that the work done by the force in [link] depended on the path. For that force,

F_{x} = (5 N/m) y and F_{y} = (10 N/m) x .

Therefore,

(d F_{x} / d y) = 5 N/m \neq (d F_{y} / d x) = 10 N/m,

which indicates it is a non-conservative force. Can you see what you could change to make it a conservative force?

A photograph of a grinding wheel being used. — A grinding wheel applies a non-conservative force, because the work done depends on how many rotations the wheel makes, so it is path-dependent.

Conservative or not?

Which of the following two-dimensional forces are conservative and which are not? Assume a and b are constants with appropriate units:

(a) $a x y^{3} \hat{i} + a y x^{3} \hat{j},$ (b) $a [(y^{2} / x) \hat{i} + 2 y ln (x / b) \hat{j}],$ (c) $\frac{a x \hat{i} + a y \hat{j}}{x^{2} + y^{2}}$

Strategy

Apply the condition stated in [link] , namely, using the derivatives of the components of each force indicated. If the derivative of the y -component of the force with respect to x is equal to the derivative of the x -component of the force with respect to y , the force is a conservative force, which means the path taken for potential energy or work calculations always yields the same results.

Solution

$\frac{d F_{x}}{d y} = \frac{d (a x y^{3})}{d y} = 3 a x y^{2}$ and $\frac{d F_{y}}{d x} = \frac{d (a y x^{3})}{d x} = 3 a y x^{2}$ , so this force is non-conservative.
$\frac{d F_{x}}{d y} = \frac{d (a y^{2} / x)}{d y} = \frac{2 a y}{x}$ and $\frac{d F_{y}}{d x} = \frac{d (2 a y ln (x / b))}{d x} = \frac{2 a y}{x},$ so this force is conservative.
$\frac{d F_{x}}{d y} = \frac{d (a x / (x^{2} + y^{2}))}{d y} = - \frac{a x (2 y)}{{(x^{2} + y^{2})}^{2}} = \frac{d F_{y}}{d x} = \frac{d (a y / (x^{2} + y^{2}))}{d x},$ again conservative.

Significance

The conditions in [link] are derivatives as functions of a single variable; in three dimensions, similar conditions exist that involve more derivatives.

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Check Your Understanding A two-dimensional, conservative force is zero on the x - and y -axes, and satisfies the condition $(d F_{x} / d y) = (d F_{y} / d x) = (4 {N/m}^{3}) x y$ . What is the magnitude of the force at the point $x = y = 1 m?$

2.83 N

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Before leaving this section, we note that non-conservative forces do not have potential energy associated with them because the energy is lost to the system and can’t be turned into useful work later. So there is always a conservative force associated with every potential energy. We have seen that potential energy is defined in relation to the work done by conservative forces. That relation, [link] , involved an integral for the work; starting with the force and displacement, you integrated to get the work and the change in potential energy. However, integration is the inverse operation of differentiation; you could equally well have started with the potential energy and taken its derivative, with respect to displacement, to get the force. The infinitesimal increment of potential energy is the dot product of the force and the infinitesimal displacement,

d U = − \vec{F} \cdot d \vec{l} = − F_{l} d l .

Here, we chose to represent the displacement in an arbitrary direction by $d \vec{l},$ so as not to be restricted to any particular coordinate direction. We also expressed the dot product in terms of the magnitude of the infinitesimal displacement and the component of the force in its direction. Both these quantities are scalars, so you can divide by dl to get

F_{l} = - \frac{d U}{d l} .

This equation gives the relation between force and the potential energy associated with it. In words, the component of a conservative force, in a particular direction, equals the negative of the derivative of the corresponding potential energy, with respect to a displacement in that direction. For one-dimensional motion, say along the x -axis, [link] give the entire vector force, $\bar{F} = F_{x} \hat{i} = - \frac{\partial U}{\partial x} \hat{i} .$

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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