= Solution
Let $s(t)$ be the <length> of the free straight segment from the current contact point to the <mass>, and let $\theta$ measure the contact point's <rotation> from first contact. For an inextensible taut rope, winding through angle $\theta$ uses <length> $R\theta$:
$$
s=L-R\theta.
$$
The point <mass> follows an <involute> of the pole's circular cross-section. Its instantaneous <velocity> is perpendicular to the free rope. To verify the geometry explicitly, use planar polar <unit vectors> at the contact point:
$$
\mathbf x=R\mathbf e_r(\theta)+s\mathbf e_\theta(\theta),
\qquad \frac{d\mathbf x}{d\theta}
=(R+s')\mathbf e_\theta-s\mathbf e_r=-s\mathbf e_r.
$$
The rope's <tension> therefore does no <work>, and the <speed> stays at its initial value $v$. The geometry gives $v=s\dot\theta$. Consequently
$$
\frac{ds}{dt}=-R\dot\theta=-\frac{Rv}{s},
\qquad s^2=L^2-2Rvt.
$$
The distance from the pole's axis is $\sqrt{R^2+s^2}$, so a point <mass> first reaches the pole when $s=0$. The <winding time around a circular pole> is thus
$$
\boxed{t_{\rm hit}=\frac{L^2}{2Rv}.}
$$
This has units of <time>. The model assumes a light inextensible rope, no <energy> loss at contact and a point <mass>. The required <tension> $mv^2/s$ diverges at the ideal final contact, so finite strength and size matter very close to impact.
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