Answer
$I_x=0.0334Kg\cdot m^2$
Work Step by Step
The required moment of inertia can be determined as follows:
We know that
$m_p=\rho V_p=7.85\times 10^3[(0.03)(0.18)(0.02)]=0.848Kg$
and $m_c=\rho V_c=7.85\times 10^3 [(0.05)\pi (0.01)^2]=0.123Kg$
Now the moment of inertia of the crank is given as
$I_{xc}=\frac{1}{2}m_Cr_C^2+m_Cd^2$
$\implies I_{xc}=\frac{1}{2}(0.123)(0.01)^2+(0.123)(0.06)^2=4.4895\times 10^{-4}Kg\cdot m^2$
and the moment of inertia of the plate is given as
$I_{xp}=\frac{1}{12}m_pt_p^2+m_pl^2$
$\implies I_{xp}=\frac{1}{12}(0.848)(0.03)^2+0.848\times (0.18)^2=0.0275Kg\cdot m^2$
Now we can find the moment of inertia of the overhung as
$I_x=I_{xp}+2I_{xc}$
We plug in the known values to obtain:
$I_x=0.0275+2\times (4.4895\times 10^{-4})$
This simplifies to:
$I_x=0.0334Kg\cdot m^2$
