Question

Calculate the approximate moment of inertia of a copper uniform disc relative to the symmetry axis perpendicular to the plane of the disc, if its thickness is equal to $b=2.0mm$ and its radius to $R=100mm$.

Answer 1

Consider an elementary disc of thickness $dx$. Moment of inertia of this element about the z-axis, passing through its C.M.
$d{I}_z=\frac{\left(dm\right)R^2}{2}=\rho Sdx\frac{R^2}{2}$
where $\rho =$ density of the material of the plate and $S=$ are of cross section of the plate.
Thus the sought moment of inertia
$I_z=\frac{\rho S{R}^2}{2}{\int }_0^{b}dx=\frac{R^2}{2}\rho Sb$
$=\frac{\pi }{2}\rho b{R}^4(asS=\pi R^2)$
putting all the values we get, $I_z=2.8g{m}^2$