On slide 6/7 and in homework 4, problem 2, is q labeled correctly to be zero in the "up" position?
(I thought one of the TAs said it should be zero in the down position).

Tim Chung, 29 Oct 02

This was understandably a little confusing. The q in the picture is correct for the linearization of the dynamics equations. Here's why:

Define an angle, f, which is defined to be zero pointing down. We will want to linearize about f = p + q, where q is a small angle away from the vertical (as defined in the picture). In doing this linearization, we thus say that cos(f) = -1, sin(f) = -q, and all second-order terms vanish.

This is how and why the q is defined the way it is in these pictures.