smc

Let $A$ be the center of the circle with radius $5$, and $B$ be the center of the circle with radius $4$. Let $\overline{CD}$ be the common internal tangent of circle $A$ and circle $B$. Extend $\overline{BD}$ past $D$ to point $E$ such that $\overline{BE}\perp \overline{AE}$. Since $\overline{AC}\perp \overline{CD}$ and $\overline{BD}\perp \overline{CD}$, $ACDE$ is a rectangle. Therefore, $AC=DE$ and $CD=AE$. Since the centers of the two circles are $41$ inches apart, $AB=41$. Also, $BE=4+5=9$. Using the Pythagorean Theorem on right triangle $ABE$, $CD=AE=\sqrt{41^2-9^2}=\sqrt{1600}=40$. The length of the common internal tangent is $\boxed{40\text{ inches}}$