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Mathematics & Science Learning Center
Computer Laboratory


Applications of Differential Equations

The Simple Pendulum

(continued from last page...)

A Real Pendulum—Facing Up to Friction

Finally, then, what about a real pendulum? We've been leaving out friction. Now it's time we considered it. First we need to decide how this might be done realistically.

It seems reasonable to assume that the force of friction is directly proportional to the angular velocity of the pendulum. (The faster it goes, the more the friction.) So, if we want to incorporate friction into the model, we need to introduce a new term to the mix of forces—a term directly proportional to velocity.

Now let's see...angular velocity is given by the derivative of angular position—in this case y'. So a term that is directly proportional to angular velocity would be c y'.

Back in the introduction to this lab we derived the following formulas for force:

F = m L y'',

which was found by analyzing the angular acceleration, and:

F = -m g sin(y),

which came about when we analyzed the gravitational acceleration. It is this second force into which we must incorporate our frictional force—after all, friction, like gravity, acts against the motion of the pendulum as it attempts to move in the positive (counter-clockwise) direction away from equilibrium. This makes the last equation become:

F = - c y' - m g sin(y).

As we did in the introduction, we equate the two formulations of force to get:

m L y'' = - c y' - m g sin(y),

or, moving all the terms to the left-hand side:

m L y'' + c y' + m g sin(y) = 0.

This could use a little more work. We could divide through by m L, which would lead us to:

y'' + (c/(m L)) y' + (g/L) sin(y) = 0.

But that's a bit too ugly. Hmmm! Hang on a second. c was just an unknown constant of proportionality, so we could just rename the constant c/(m L) as a new constant, say k. This makes our equation become:

y'' + k y' + (g/L) sin(y) = 0,

which is where I'm going to stop messing with it. Finally, then, if we bring back our old initial conditions we have a new, friction aware initial value problem:

Friction Aware Initial Value Problem

y'' + k y' + (g/L) sin(y) = 0
y(0) = yo
y'(0) = vo

Let's go on and look at an actual example using this new initial value problem...

Compass If you're lost, impatient, want an overview of this laboratory assignment, or maybe even all three, you can click on the compass button on the left to go to the table of contents for this laboratory assignment.

ODE Laboratories: A Sabbatical Project by Christopher A. Barker

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