Answer
$I_{x^{\prime}}=7.19g\cdot m^2$
Work Step by Step
We can determine the required moment of inertia as follows:
$I_x^{\prime}=I_{crank}+I_{plate}$ [eq(1)]
We know that
$I_{crank}=\frac{1}{2}m_cr^2+(\frac{1}{2}m_cr^2+m_cd^2)$
As $m_c=\rho V_c=7.85\times 10^3\pi r^2 t=7.851\times 10^3\times \pi (0.01)^2\times 0.05=0.1233Kg$
$\implies I_{crank}= \frac{1}{2}\times 0.1233(0.01)^2+[\frac{1}{2}(0.1233)(0.01)^2+0.1233(0.12)^2]$
$\implies I_{crank}=1.7878\times 10^{-3}Kg\cdot m^2$
Similarly $I_{plate}=\frac{1}{12}m_p(l^2+t^2)+m_pd^2$
As $m_p=\rho V_p=7.85\times 10^3\times 0.02\times 0.18\times 0.02=0.8478Kg$
$\implies I_{plate}= \frac{1}{12}\times 0.8478(0.18)^2+(0.03)^2+(0.8478)(0.06)^2=5.4047\times 10^{-3}Kg\cdot m^2$
Now, we plug in the known values in eq(1) to obtain:
$I_{x^{\prime}}=1.7878\times 10^{-3}+5.4047\times 10^{-3}=7.19\times 10^{-3}Kg\cdot m^2=7.19g\cdot m^2$
