= Solution
The contact radius is $r_1$, so <rolling without slipping> gives $v=r_1\omega$. With $I=Mr_2^2/2$, the energy at the minimum of the stated parabolic motion model is
$$
Mgh_0=\frac12Mv_{\max}^2+\frac12\frac{Mr_2^2}{2}\frac{v_{\max}^2}{r_1^2}.
$$
Consequently
$$
\boxed{v_{\max}=\frac{2\sqrt{gh_0}}{\sqrt{(r_2/r_1)^2+2}}=0.148\,\mathrm{m\,s^{-1}}.}
$$
The corresponding spin <angular frequency> is \b[$\omega_{\max}=v_{\max}/r_1=14.8\,\mathrm{rad\,s^{-1}}$]. It is the rod radius, not the disc radius, that relates translation and rotation; the discs determine the larger rotational inertia.
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