The question as it appears in the 03/20 edition of Specs & Techs from GlobalSpec:
You have a frictionless cone that subtends an angle of 90 degrees. The cone is standing on its tip, and a very small particle moves in a circle on the inside of the cone. If it is required that the frequency of its circular motion remains at 2.2 radians/second, at what height should the particle stay?
(Update: March 27 8:32 AM) And the Answer is...
As seen by the following figure, the forces actuating on the particle are the force of gravity (mg) and the normal force from the cone pushing the particle inward (N).
The net vertical force on the particle is
From this equation we get
Now, the horizontal component of the inward horizontal force () must be equal to the centripetal force of the particle () where r is the radius of the circle traversed by the particle. This radius is . The horizontal balance equation becomes
Substituting the value of N in the above equations, we get
Using , , and
, we get

Re: Cones and Particles: Newsletter Challenge (03/20/07)