jmc

Let $C$ and $D$ be the centers of the larger and smaller semicircles, respectively, and let $r$ be the radius of the smaller semicircle. We have $QD=QB-DB=14-r$ and $QC=7$, so we can apply the Pythagorean theorem to triangle $QCD$ to obtain $$(14-r{)}^2+7^2=(7+r{)}^2.$$ After squaring both binomials and subtracting $7^2+r^2$ from both sides simplifies to $196-28r=14r$. Adding $28r$ to both sides and dividing by 42, we find $r=\boxed{\frac{14}{3}}$ inches.