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Answer: Usually, a FBD is used along with Newton's 2nd, and then the small angle approximation is used twice (once for siintheta in the weight component and once for tantheta to convert linear to rotational). Eventually you get the usual second-order Diffyq for angular displacement.

Question: Some of my students instead used the centripetal force requirement, setting the normal force (which is equal to the perp. component of the particle's weight) equal to the usual centripetal force equation. Using small angles, the costheta term in the normal force becomes 1, and the problem reduces to the familiar form.

Do you all feel that this is also a valid way to solve the problem? It seems OK to me; it involves Newton's 2nd, as well as the small angle approx., and I can't find anything wrong with it. But I thought I'd see what you all thought.