2. As shown in Figure 3, $A_{0} A_{1}$ is the diameter of a semicircle, $A_{0} A_{1}=2$, $A_{2}, A_{3}, \cdots, A_{k}$, $A_{k+1}, \cdots$, are points on the semicircle, $\angle A_{0} A_{1} A_{2}=1^{\circ}, \angle A_{1} A_{2} A_{3}=2^{\circ}$, $\angle A_{2} A_{3} A_{4}=3^{\circ}, \cdots, \angle A_{k-1} A_{k} A_{k+1}=k^{\circ}(k$ is a positive integer). If the length of the chord $A_{k} A_{k+1}$ is less than 1, then the minimum value of $k$ is $\qquad$.
2. 11 .

As shown in Figure 6, connect
$$
A_{k} O, A_{k+1} O, A_{k} A_{1} \text {, }
$$

then
$$
\begin{array}{l}
\angle O A_{k} A_{k+1}= \\
\angle A_{1} A_{k} A_{k+1}+\angle A_{1} A_{k} O \\
=\angle A_{1} A_{k} A_{k+1}+\angle A_{k} A_{1} O=\left(\frac{k(k+1)}{2}\right)^{\circ} .
\end{array}
$$

Since $A_{k} A_{k+1}60^{\circ}$, that is
$$
\begin{array}{l}
\frac{k(k+1)}{2}>60, \\
\left(k+\frac{1+\sqrt{481}}{2}\right)\left(k-\frac{-1+\sqrt{481}}{2}\right)>0 .
\end{array}
$$

Solving gives $k>\frac{\sqrt{481}-1}{2}$.
Also, $10<\frac{\sqrt{481}-1}{2}<11, k$ is a positive integer, so the minimum value of $k$ is 11.