Answer
$I_A=7.67Kg\cdot m^2$
Work Step by Step
The required moment of inertia can be determined as follows:
We know that
$I_G=m_1r_1^2+\frac{m_2l_2^2}{12}$
We plug in the known values to obtain:
$I_G=10(0.5)^2+\frac{4(2)(1)^2}{12}$
$\implies I_G=3.1667Kg\cdot m^2$
Now $I_A=I_G+(m_1+m_2)d^2$
We plug in the known values to obtain:
$I_A=3.1667+(10+4(2))(0.5)^2$
This simplifies to:
$I_A=7.67Kg\cdot m^2$
