4. A frustum of a cone has an upper base radius of $1 \mathrm{~cm}$, a lower base radius of $2 \mathrm{~cm}$, and a generatrix $A B$ of $4 \mathrm{~cm}$. A certain ant starts from point $A$ on the circumference of the lower base, and crawls around the lateral surface of the frustum (i.e., it must intersect every generatrix) to point $B$ on the circumference of the upper base. The shortest path the ant can take is ( ).
(A) $4 \sqrt{3}+\frac{2 \pi}{3}$
(B) $4 \sqrt{3}+\frac{4 \pi}{3}$
(C) $2 \sqrt{3}+\frac{2 \pi}{3}$
(D) $2 \sqrt{3}+\frac{4 \pi}{3}$
4. A.

Let the radii of the upper and lower bases of the frustum be $r^{\prime}, r$, and the central angle of the sector ring in the lateral development of the frustum be $\theta$. It is easy to get
$$
\begin{aligned}
\theta & =\frac{r-r^{\prime}}{A B} \cdot 2 \pi \\
& =\frac{2-1}{4} \cdot 2 \pi \\
& =\frac{\pi}{2} \\
O B= & \frac{r^{\prime}}{r-r^{\prime}} \cdot A B=\frac{1}{2-1} \cdot 4=4 .
\end{aligned}
$$

The required shortest distance should be the sum of the length of the tangent $A T$ and the arc $\overparen{T B^{\prime}}$.
$$
A T=\sqrt{O A^{2}-O T^{2}}=\sqrt{8^{2}-4^{2}}=4 \sqrt{3} .
$$

Since $O A=2 O T$, we have $\angle A O T=\frac{\pi}{3}$, thus $\angle B^{\prime} O T=$ $\frac{\pi}{6}$. Therefore, the arc $\overparen{T B^{\prime}}=4 \cdot \frac{\pi}{6}=\frac{2 \pi}{3}$. So, the shortest distance is $4 \sqrt{3}+\frac{2}{3} \pi$.