Answer
$I = 6.16\times 10^{-5}~kg~m^2$
Work Step by Step
We can find the mass of the small disk:
$m = \rho~V$
$m = (1400~kg/m^3)(\pi)(0.0200~m)^2(5.00\times 10^{-3}~m)$
$m = 8.7965\times 10^{-3}~kg$
We can find the mass of the large disk:
$M = \rho~V$
$M = (1400~kg/m^3)(\pi)(0.0400~m)^2(5.00\times 10^{-3}~m)$
$M = 3.5186\times 10^{-2}~kg$
We can find the rotational inertia of the two-disk assembly:
$I = \frac{1}{2}mr^2+m(R+r)^2+\frac{1}{2}MR^2$
$I = \frac{1}{2}(8.7965\times 10^{-3}~kg)(0.0200~m)^2+(8.7965\times 10^{-3}~kg)(0.0400~m+0.0200~m)^2+\frac{1}{2}(3.5186\times 10^{-2}~kg)(0.0400~m)^2$
$I = 6.16\times 10^{-5}~kg~m^2$
