Classically, what would be the average energy of a particle in a system of particles fine to move in the xy-plane while rotating about the $\mathit{i}$-axis?

The equipartition theorem says that any quadratic term in the expression for the total energy E of a particle contributes to the average energy $\overline{\mathbf{E}}$ by an amount$\stackrel{}{\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}$, where${\mathbf{k}}_{\mathbf{B}}$is the Boltzmann constant, and T is temperature.

For a particle undergoing translational motion in the x y-plane and rotational motion about the z- axis, the total energy writes:

$E=\frac{1}{2}m{v}_x^2+\frac{1}{2}m{v}_y^2+\frac{1}{2}I{\omega }^2$

where$V_x$ and$V_y$ are the$x$ and$y$ components of the velocity$\overrightarrow{v},I$ is the moment of inertial, and$\omega$ is the angular speed.

From the above equation for$E$ , there are three quadratic terms for which$E$ depends. Thus, use the equipartition theorem, each of these three degrees of freedom has an average energy of $\frac{1}{2}{k}_{B}T$. The average energy of the particle is thus:

$\begin{array}{c}\overline{\mathrm{E}}=\frac{1}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}+\frac{1}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}+\frac{1}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\\ =\frac{3}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\end{array}$