# Advanced Classical Mechanics/Dissipative Forces

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Dissipative forces are forces of such nature that energy is lost from a system when motion takes place. Of course energy is in general conserved but it is lost from the degrees of freedom of interest into heat (the random motion of internal degrees of freedom) or radiation (the motion of new particles created by the motion -- light usually).

The force can often be represented by $\left |\vec F \right | = a\left ( \vec r, t \right ) \left | \vec v \right |^n.$ Depending on the value of the index $n$ we have different types of dissipative forces.

## Types of Dissipative Forces

### Frictional Forces (Dry Friction)

The force of kinetic friction is supposed to be proportional to the normal force and independent of area of contact or speed. It is simply a property of the materials in contact. Specifically, we have $\left| \vec F \right | = \mu \left | F_{\rm normal} \right |$ so $n=0$.

The frictional force opposes the relative motion of two surfaces.

### Viscous Forces

Here the frictional force increases as the first power of the relative speed between the surfaces and opposes the relative motion. Viscous friction is important for wet surfaces at small relative velocties.

### High-Velocity Friction

At higher velocities, the force of friction increases as a higher power of the relative velocity. For example, $n=2$ gives a good approximation to the dissipative force experiences by objects travelling through fluids at high Reynolds number $Re = (vL)/\nu$ where $\nu$ is the viscousity of the fluid.

## Determining the Generalized Force

The Lagrangian method requires us to find the generalized forces corresponding to the different coordinates that we have used to characterize the degrees of freedom of the motion. There are two general ways to determine the generalized forces.

1. Calculate them in Cartesian coordinates and convert to the generalized coordinates
2. Use the power function

### Partial derivatives

For any force that we know in Cartesian (or any other set of coordinates for that matter), we can find the generalized force by using the definition of the generalized coordinates in terms of the coordinates in which the force is known. The definition of the generalized force is

$Q_j = \sum_i \vec F_i \cdot \frac{\partial \vec r_i}{\partial q_j}.$

To do this you have to first determine the dissipative force in Cartesian coordinates and the full transformation between the Cartesian coordinates and the generalized coordinates -- it can be pretty painful for even simple problems.

#### Damped Pendulum

Let's look at a pendulum of length $l$. We will use the angle $\theta$ and the vertical to describe the single degree of freedom. The position of the bob is given by

$x=l \sin \theta, y=-l \cos \theta\,$

and the velocities are

$\dot x= l \cos \theta \dot \theta, \dot y = l \sin \theta \dot \theta \, .$

The viscous force opposes the direction of motion and it is proportional to the velocity so we have

$F_x= - a l \cos \theta \dot \theta, F_y = - a l \sin \theta \dot \theta \, .$

We also need the partial derivatives for the transformation

$\frac{\partial x}{\partial \theta}= l \cos \theta, \frac{\partial y}{\partial \theta} = l \sin \theta \,$

to calculate the generalized force

$F_\theta = F_x \frac{\partial x}{\partial \theta} + F_y \frac{\partial y}{\partial \theta} = - a l^2 \cos^2 \theta \dot \theta - a l^2 \sin^2 \theta \dot \theta = - a l^2 \dot \theta$

### Power function

The great advance of using the potential instead of the generalized forces directly was that the definition of the partial derivative took care of all of the heavy lifting involved in going from one system of coordinates to another. It turns out that one can use an analogous quantity called the power function to do the same thing (sometimes this is called the dissipation function). Like for the potential the power function for various simple forms of forces can simply be written down and used.

In analogy with the potential we define the power function such that force on particle $i$ in the $x$-direction is

$F_{i,x} = \frac{\partial P}{\partial {\dot x}_i}$

and similarly for the other directions and particles. Of course, not all forces can be written in this way but many dissipative forces can. Let's write out the generalized force using the standard formula

$Q_j = \sum_i \vec F_i \frac{\partial \vec r_i}{\partial q_j} = \sum_i \vec F_i \frac{\partial v_i}{\partial \dot q_j} = \sum_i \frac{\partial P}{\partial \vec v_i} \frac{\partial v_i}{\partial \dot q_j} = \frac{\partial P}{\partial \dot q_j}$

where we used the equality $\partial \vec r_i/\partial q_j = \partial \vec v_i/\partial \dot q_j$ in the first step, definition of the power function in the second step, and the definition of the partial derivative in the final step.

#### Damped Pendulum

Let's look at the pendulum again. The viscous force is

$\vec F = -a \vec v$

so the power function is

$P = -\frac{1}{2} a v^2 = -\frac{1}{2} a \left ( l \dot \theta \right )^2$

and

$F_\theta = \frac{\partial P}{\partial \dot \theta} = -a l^2 \dot \theta.$

Let's combine the results for the power function with the Lagrangian for the pendulum, we have

$L = T - V = \frac{1}{2} m l^2 \dot \theta^2 + g m l \cos \theta$

and Lagrange's equations including the power function are

$\frac{d}{dt}| \frac{\partial L}{\partial \dot \theta} - \frac{\partial L}{\partial \theta} = Q_{{\rm NC},\theta} = \frac{\partial P}{\partial \dot \theta}$

which yields

$m l^2 \ddot \theta + g m l \sin \theta = -a l^2 \dot \theta.$

If we take the small angle limit of this equation we get

$\ddot \theta = - \frac{g}{l} \theta - \frac{a}{m} \dot \theta,$

the equation for a damped linear oscillator.

In this example, we have included all of the forces that we could have included in the Lagrangian, leaving only the dissipative force to be included in the power function. Of course, one could have included non-dissipative forces in the power function as well, but one must be careful not to include the same force twice, so it is generally a good idea to include as much as possible in the Lagrangian, leaving only the forces that cannot be included in the Lagrangian for the power function. This has the added advantage that one can still look at the Lagrangian for first integrals if the power function does not depend on a particular coordinate.

#### More Generally

If the dissipative force is given by

$\left | \vec F \right | = a \left | \vec v \right |^n$ or more precisely $\vec F = -a \left | \vec v \right |^{n-1} \vec v$

then the power function is given by

$P = -\frac{1}{n+1} a \left | \vec v \right |^{n+1}.$

More generally if the dissipative force points in the direction of the relative velocity it can also be written as a power function. If

$\vec F_i = \phi_i \left ( \vec r_i, \left | \vec v_i \right |, t \right ) \frac{\vec v_i}{\left | \vec v_i \right |}$

then

$P = \sum_i \int \phi_i d \left | \vec v_i \right |$

## A Useful Example with Dry Friction

It turns out that the simplest looking type of friction ('dry friction') actually can yield some useful surprises. Let's calculate the frictional force between a package and a moving conveyor belt.

File:Convevor.png
Conveyor Belt

Let's assume that the belt is moving along at a velocity $v_y$ in the $y-$direction and calculate the frictional forces that resist moving the package across the belt in the $x$ and $y$ directions.

Let's write the power function for the dry friction between the belt and the package,

$P = -\mu m g v$

where $\mu$ is the coefficient of kinetic friction between the belt and the package, $m$ is the mass of the package and the relative speed between the package and the belt is given by

$v = \sqrt{ \dot x^2 + \left ( \dot y - v_y \right )^2 } .$

We can calculate the frictional force resisting motion in the two directions

$F_x = \frac{\partial P}{\partial \dot x} = -\mu m g \frac{\dot x}{\sqrt{ \dot x^2 + \left ( \dot y - v_y \right )^2 }}, F_y = \frac{\partial P}{\partial \dot y} = -\mu m g \frac{\dot y - v_y}{\sqrt{ \dot x^2 + \left ( \dot y - v_y \right )^2 }}.$

Let's imagine that $v_y$ is much larger $\dot x$ and $\dot y$ then we have

$F_x \approx -\mu m g \frac{\dot x}{v_y}, F_y \approx \mu m g.$

The $y-$ component indicates a force that tries to drag the package along the belt. It is approximately equal to the kinetic friction that you are used to. The $x-$ component is much smaller by a factor of $\dot x/v_y$ which could be large. You can try this at the supermarket by blocking a big box of corn flakes on its side as the belt is moving. Even though the friction between the belt and the box is large (you can feel this force with the hand that impedes the motion of the box), the force to move the box across the belt is quite small.

This effect is used for more important purposes when removing a cork from a bottle. It is much easier to removing the cork while twisting it round than to pull it out directly.
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