Answer
(a) $L=6.39$ m
(b) $I_{O}=53.2$ kg*m$^2$
Work Step by Step
$y=\frac{(6)(1.5)+(1.3)(2)(0.65)}{6+(1.3)(2)+(L)(2)}=0.5$
$L=6.39$ m
$I_{O}=\Sigma I_{G}+md^2=((\pi) (0.2)^2(0.25)+(6)(1)^2)+(\frac{1}{12})(1.3)^3(2)+(2)(1.3)(0.15)^2+(\frac{1}{12})(2)(6.39)(6.39)^2+(2)(6.39)(0.5)^2)$
$=53.2$ kg*m$^2$