4. Figure 4 is a schematic diagram of a clock hanging on a wall. $O$ is the center of rotation of its second hand, $M$ is the other end of the second hand, $O M$ $=10 \mathrm{~cm}, l$ is a vertical line passing through point $O$. There is an ant $P$ crawling on the second hand $O M$, the distance from ant $P$ to point $O$ is always equal to the distance from $M$ to $l$. In 1 minute, the distance the ant $P$ is carried by the second hand $O M$ (not the distance the ant crawls on the second hand) is $\qquad$ $\mathrm{cm}$.
$4.20 \pi$.
As shown in Figure 7, with point $O$ as the center and $10 \, \text{cm}$ as the radius, draw $\odot O$. Draw $MN \perp l$ at point $N$, and draw the perpendicular from $O$ to $l$ intersecting $\odot O$ at points $Q_{1}$ and $Q_{2}$. Connect $PQ_{1}$. Then,
$$
\begin{array}{l}
MN \parallel OQ_{1}, \\
\angle M = \angle MOQ_{1}.
\end{array}
$$

Since $OM = OQ_{1}$ and $MN = OP$, we have $\triangle OMN \cong \triangle Q_{1}OP$.
Therefore, $\angle OPQ_{1} = \angle ONM = 90^{\circ}$.
Thus, point $P$ lies on the circle with $OQ_{1}$ as its diameter.
Similarly, point $P$ lies on the circle with $OQ_{2}$ as its diameter.
Therefore, the path traced by ant $P$ in 1 minute as it is carried by the second hand $OM$ is the two circles with diameters $OQ_{1}$ and $OQ_{2}$. The distance traveled is
$$
2 \times 10 \times \pi = 20 \pi.
$$