= Solution
Use the <rotating reference frame> equilibrium model, in which the <centrifugal potential> <energy> of a particle of <mass> $m_i$ is $U_i(r)=-m_i\omega^2r^2/2$. The <Boltzmann distribution> then gives
$$
\frac{n_i(r)}{n_i(a)}=
\exp\left[\frac{m_i\omega^2(r^2-a^2)}{2k_BT}\right].
$$
Define the concentration ratio as $c=n_1/n_2$. At the axis,
$$
\boxed{c_{\rm axis}=c_{\rm wall}
\exp\left[\frac{(m_2-m_1)\omega^2a^2}{2k_BT}\right].}
$$
Thus the axis favours the lighter species. The expression assumes a dilute classical mixture at one <temperature>, equilibrium in the <rotating reference frame>, and sampling sufficiently close to $r=0$.
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