= Solution
Initially $r_i=l/2$, $\dot r_i=0$ and $\dot\theta_i=v_0/r_i$, so the conserved <angular momentum> is $J=mlv_0/2$. For $v_0\ne0$, the <effective potential> contains the <centrifugal barrier>
$$
U_J(r)\sim\frac{J^2}{2mr^2}\longrightarrow+\infty\quad(r\to0^+).
$$
The finite conserved <energy> therefore forbids the hole from being reached:
$$
\boxed{v_0\ne0\ \Longrightarrow\ r(t)>0\ \text{throughout the ideal motion}.}
$$
The exceptional case $v_0=0$ has no <centrifugal barrier>. Then $J=0$ and $\ddot r=-Mg/(M+m)$; starting at rest at $l/2$, the tabletop <mass> reaches the hole after $t=\sqrt{(M+m)l/(Mg)}$. The model ends at that event. Thus the usual nonzero initial tangential <speed> assumption matters.
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