Recall that the orbital radius is defined by

$\begin{align} R \equiv \frac{hc}{2 \pi } \frac{ \sqrt{𝘑 \, }}{E} \end{align}$

where $E$ is the mechanical energy. Then if the total angular momentum quantum number $𝘑$ is zero, the radius is also zero, and the cylinder shrinks to a line that occupies no volume. If P is in slow motion then $E = m c^{2}$ and the area can be written as

$\begin{align} A \equiv \pi R^{2} = \left( \frac{ 𝘑 h }{ m } \right) \left( \frac{h}{4 \pi E} \right) \end{align}$

The WikiMechanics quark models for atoms specify two spatial rotations per atomic cycle, one turn for quarks of each phase. And each atomic cycle has a generic frequency of $\nu$. So with Planck's postulate the angular frequency of P is given by $\omega = 4 \pi \nu = 4 \pi E / h$. Using this to eliminate $E$ obtains

$\begin{align} \pi R^{2} = \frac{ 𝘑 h }{m \omega } \end{align}$

Then recall that the total angular momentum quantum number $𝘑$ is always related to the total angular momentum by

$\begin{align} \left\| \, \overline{ J} \, \right\| = \frac{ h }{ 4 \pi } \sqrt{𝘑 \, (𝘑+1)} \end{align}$

so that

$\begin{align} m R^{2} = \frac{ 2 J }{ \omega } \end{align}$

The moment of inertia $\overline{I}$ relates angular frequencies and momenta as $\overline{I} \equiv \overline{ J} / \omega$ and so for the cylinder is