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## Homework Statement

Consider a pendulum with a spring as in the following diagram:

Please note the 'rotated' coordinates.

The bob has a mass [itex]m[/itex]. The spring has a spring constant [itex]k[/itex] and an unextended length [itex]\ell[/itex]. We can

*not*ignore air friction.

Assume the initial velocity and horizontal displacement are parallel, so that the motion of the pendulum is in the plane. It is not a 'spherical' pendulum.

I need to show that the equations of motion satisfy the following:

[tex]m\left( r'' - r (\theta ')^2 \right) + cr' + k(r-\ell) = mg \cos \theta[/tex]

[tex]m \left( r \theta '' + 2 r' \theta ' \right) + cr \theta ' = - mg \sin \theta[/tex]

## Homework Equations

The forces on the bob are the force of gravity, the spring force and the air friction:

[tex]\mathbf{F}_g = mg \mathbf{\hat{i}}[/tex]

[tex]\mathbf{F}_s = -k \left(r - \ell \right) \mathbf{\hat{e_r}}[/tex]

[tex]\mathbf{F}_f = -c \mathbf{r}'[/tex]

(where c is a positive friction coefficient)

## The Attempt at a Solution

I can use the Lagrangian to find the equations of motion, but I don't have a clue what to do with the friction force..??

Using the Lagrangian, the kinetic energy T is:

[tex]T = \frac{1}{2} m v^2 = \frac{1}{2} m (x' ^2 + y'^2) = \frac{1}{2} m \left[ (r')^2 + r^2 (\theta ')^2\right][/tex]

The potential V is the sum of the gravitational potential V_g, the spring potential V_s, and...??? Frictional potential? I don't think I can write the frictional force as a potential, can I?

Anyway:

[tex]V = -mgr \cos \theta + \frac{1}{2}k(r-\ell)^2 + ...?[/tex]

[tex]L = T - V = \frac{1}{2} m \left[ (r')^2 + r^2 (\theta ')^2\right] +mgr \cos \theta - \frac{1}{2}k(r-\ell)^2[/tex]

For r, we have:

[tex]\frac{\partial}{\partial t}\left( \frac{\partial L}{\partial r'} \right) = \frac{\partial L}{\partial r}[/tex]

From this I get:

[tex]m\left( r'' - r (\theta ')^2 \right) + k(r-\ell) = mg \cos \theta[/tex]

For theta, we have:

[tex]\frac{\partial}{\partial t}\left( \frac{\partial L}{\partial \theta '} \right) = \frac{\partial L}{\partial \theta}[/tex]

And I get:

[tex]m \left( r \theta '' + 2 r' \theta ' \right) = - mg \sin \theta[/tex]

So as expected, I get the correct equations, without the frictional terms...

How can I calculate the frictional terms? Can I use the Lagrangian, or do I have to use normal force analysis? It doesn't make sense in my mind to use both the Lagrangian

*and*force analysis... ?

Thanks!