A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter 10 and altitude 12, and the axes of the cylinder and cone coincide. Find the radius of the cylinder. Express your answer as a common fraction.
Let the cylinder have radius $r$ and height $2r$. Since $\triangle APQ$ is similar to $\triangle AOB$, we have $$\frac{12-2r}{r} = \frac{12}{5}, \text{ so } r = \boxed{\frac{30}{11}}.$$[asy]
draw((0,2)..(-6,0)--(6,0)..cycle);
draw((0,-2)..(-6,0)--(6,0)..cycle);
draw((0,1)..(-3,0)--(3,0)..cycle);
draw((0,-1)..(-3,0)--(3,0)..cycle);
fill((-6,0.01)--(-6,-0.01)--(6,-0.01)--(6,0.01)--cycle,white);
draw((0,14)--(0,0)--(6,0),dashed);
draw((0,8)..(-3,7)--(3,7)..cycle);
draw((0,6)..(-3,7)--(3,7)..cycle);
fill((-3,7.01)--(-3,6.99)--(3,6.99)--(3,7.01)--cycle,white);
draw((0,7)--(3,7),dashed);
draw((-6,0)--(0,14)--(6,0));
draw((-3,0)--(-3,7));
draw((3,0)--(3,7));
label("{\tiny O}",(0,0),W);
label("{\tiny B}",(6,0),E);
label("{\tiny P}",(0,7),W);
label("{\tiny Q}",(3,7),E);
label("{\tiny A}",(0,14),E);
draw((0,-2.5)--(6,-2.5),Arrows);
draw((-6.5,0)--(-6.5,14),Arrows);
label("{\tiny 5}",(3,-2.5),S);
label("{\tiny 12}",(-6.5,7),W);
draw((10,0)--(15,0)--(10,12)--cycle);
draw((10,6)--(12.5,6));
draw((15.5,0)--(15.5,12),Arrows);
label("{\tiny O}",(10,0),W);
label("{\tiny P}",(10,6),W);
label("{\tiny A}",(10,12),W);
label("{\tiny 2r}",(10,3),W);
label("{\tiny 12-2r}",(10,9),W);
label("{\tiny B}",(15,0),E);
label("{\tiny Q}",(12.5,6),E);
label("{\tiny 12}",(15.5,6),E);
label("{\tiny 5}",(12.5,0),S);
[/asy]