4. As shown in Figure 1, in $\triangle A B C$ with perimeter $l$,
$$
\begin{array}{l}
\angle A B C=\alpha, \\
\angle A C B=2 \alpha .
\end{array}
$$

A circle $\odot O$ with radius $\frac{l}{2008}$ rolls without slipping along the sides of $\triangle A B C$ in a counterclockwise direction until it returns to its original position. The length of the path traced by the center $O$ is ( ).
(A) $\left(1+\frac{\pi}{502}\right) l$
(B) $\left(1+\frac{\pi}{1004}\right) l$
(C) depends on $l$ and $\alpha$
(D) depends on $\alpha$, but not on $l$
4. B.

As shown in Figure 6, the path taken by the center $O$ is $O \rightarrow B_{1} \rightarrow$
$$
\begin{array}{l}
B_{2} \rightarrow C_{1} \rightarrow C_{2} \rightarrow \\
A_{1} \rightarrow A_{2} \rightarrow O .
\end{array}
$$

Let $\angle A_{1} A A_{2}$
$$
=x^{\circ}, \angle B_{1} B B_{2}=y^{\circ}, \angle C_{1} C C_{2}=z^{\circ} \text {. }
$$

Then $x^{\circ}+y^{\circ}+z^{\circ}$
$$
\begin{aligned}
= & 180^{\circ}-\angle B A C+180^{\circ}-\angle A B C+ \\
& 180^{\circ}-\angle A C B=360^{\circ} .
\end{aligned}
$$

Therefore, the length of the path taken by the center $O$ is
$$
\begin{array}{l}
A_{2} B_{1}+B_{2} C_{1}+C_{2} A_{1}+\overparen{A_{1} A_{2}}+\overparen{B_{1} B_{2}}+\overparen{C_{1} C_{2}} \\
=A B+B C+C A+ \\
\quad x \pi \frac{l}{2008}+\frac{y \pi \frac{l}{2008}}{180}+\frac{z \pi \frac{l}{2008}}{180} \\
=l+\frac{(x+y+z) \pi}{180} \cdot \frac{l}{2008}=l+\frac{\pi l}{1004} .
\end{array}
$$