**particle in cone**

A particle slides at constant speed in a horizontal circular path on the frictionless inner surface of an inverted right circular cone. How is the particle’s kinetic energy related to its potential energy?

__Solution by __**Ilkka Mäkinen**

The Virial Theorem tells us that if a particle is moving in a limited part of space with a finite velocity, and if its potential energy is a homogeneous function of its position, there is a relation between the mean value of its kinetic energy <K> and the mean value of its potential energy <U>,

2<T> = k<U>,

where k is the degree of homogeneity of the potential energy function.

In this problem, the ball is moving in a limited part of space, since it’s inside the cone The ball’s potential energy is U = mgh, i.e. a first-degree homogeneous function of the position. So, by the Virial Theorem, we now have 2<T> = <U> for the mean values of kinetic and potential energy. Since the ball is moving in a horizontal circle, its potential energy is always the same. Therefore, by conservation of energy, the ball’s kinetic energy is also the same in every point of its path. Thus the potential and kinetic energies are always equal to their mean values, so 2T = U, i.e. the kinetic energy is half the potential energy, for every point in the path of the ball.