Answer
a). Moment of inertia about an axis not passing through the center of mass is given by $I=I_{CM}+Md^{2}$, the term $Md^{2}$ is always greater than zero. Hence we can say that moment of inertia about center of mass is the minimum.
b). Hence, moment of inertia about the center of rod is $8 kg.m^{2}$ and moment of inertia about the center of mass of the system is $7.5 kg.m^{2}$.
Work Step by Step
a). Moment of inertia about an axis not passing through the center of mass is given by $I=I_{CM}+Md^{2}$, the term $Md^{2}$ is always greater than zero. Hence we can say that moment of inertia about center of mass is the minimum.
b). $x_{CM}=\frac{3.0+5.2}{3+5}=1.25m$
Now, $I_{center of rod}=3\times1^{2}+5\times1^{2}=8 kg.m^{2}$
$I_{center of mass}=3\times1.25^{2}+5\times0.75^{2}=7.5 kg.m^{2}$
Hence, moment of inertia about the center of rod is $8 kg.m^{2}$ and moment of inertia about the center of mass of the system is $7.5 kg.m^{2}$.
