2. Acute $\triangle A B C$ is inscribed in $\odot O$ with radius $R$, $H$ is the orthocenter of $\triangle A B C$, and the extension of $A O$ intersects $B C$ at point $M$. If $O H \perp A O, B C=10, O A=6$, then $O M=$ $\qquad$ -
2. $\frac{11}{3}$.

The auxiliary line is shown in Figure 3.
Then $\triangle A O H \backsim \triangle O N M$, and $A H=C F=2 O N$.
Thus $\frac{O M}{O N}=\frac{A H}{A O}=\frac{C F}{A O}=\frac{2 O N}{A O}$
$\Rightarrow O M=\frac{2 O N^{2}}{A O}=\frac{2\left(O B^{2}-B N^{2}\right)}{A O}=\frac{11}{3}$.