= Solution
The initial <angular momentum> is $J=mlv_0/2$, and the initial total <energy> is
$$
\mathcal H_i=\frac12mv_0^2-\frac12Mgl.
$$
When the hanging <mass> reaches the table, $r=l$. Conservation of <angular momentum> fixes the tabletop tangential <speed> there to $J/(ml)=v_0/2$. At the limiting case of just reaching, radial <velocity> is zero. Equating the two <energies>,
$$
\frac12mv_0^2-\frac12Mgl=\frac12m\left(\frac{v_0}{2}\right)^2,
$$
yields
$$
\boxed{v_{0,\rm just}^{\,2}=\frac{4Mgl}{3m}.}
$$
More generally, the boundary radial <kinetic energy> is
$$
\frac12(M+m)\dot r(l)^2=\frac38mv_0^2-\frac12Mgl,
$$
so reaching requires $v_0^2\ge4Mgl/(3m)$, with equality for just reaching and strict inequality for nonzero outward radial <speed> at contact. This condition is also sufficient: $U_J$ is strictly convex, and the allowed interval connecting the initial radius to $l$ has no intervening higher <effective potential> barrier. At the threshold the initial radial <acceleration> is outward and $r=l$ is the outer <turning point>.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2019-resit-effective-radial-potential.png]
{title=Effective radial potential at the threshold for the hanging mass to reach the table}
{description=The horizontal energy level intersects the effective potential at r/l equal to one half and one. The interval between them is accessible. The curve is shifted by the irrelevant constant Mgl and normalized by Mgl; the physical region ends at r/l equal to one.}
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