2. Given that $A B C D-A_{1} B_{1} C_{1} D_{1}$ is a unit cube, $O_{1}$ is the center of the base $A_{1} B_{1} C_{1} D_{1}$. $M$ is the midpoint of edge $B B_{1}$. Then the volume of the tetrahedron $O_{1} A D M$ is $\qquad$ .
2. $\frac{1}{8}$.

As shown in Figure 4, it is easy to know that $A C \perp$ plane $D_{1} B_{1} B D$. Let $O$ be the center of plane $A B C D$, then $A O \perp$ plane $D O, M$. Therefore,
$$
\begin{aligned}
& V_{4-\omega_{1}} M I \\
= & \frac{1}{3} S_{\Delta X_{1} M} \cdot A O \\
= & \frac{1}{3}\left(1 \times \sqrt{2}-2 \times \frac{1}{2} \times 1 \times \frac{\sqrt{2}}{2}-\frac{1}{2} \times \frac{1}{2} \times \frac{\sqrt{2}}{2}\right) \times \frac{\sqrt{2}}{2} \\
= & \frac{1}{3} \times \frac{1}{2}\left(2-1-\frac{1}{4}\right)=\frac{1}{8} .
\end{aligned}
$$