Question

The uniform solid block in Fig. 11-36 has a mass of M = 5.05 kg and edge lengths a = 10.0 cm, b = 31.0 cm, and c = 2.65 cm. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces.

Answer 1

let me try:
\[dI=r^2dm\],
define mass density:
\[\rho=M/(abc)\]
then
\[dm=\rho dV=\rho dxdydz\]
and r around z axis
\[r=\sqrt{x^2+y^2}\]
put things together:
\[I=\int_0^c\int_0^b\int_0^a(\sqrt{x^2+y^2})^2\rho dxdydz\]
\[=\rho ((x^3/3)yz+(y^3/)3xz)=M/abc(a^3bc-b^3ac)/3=M(a^2-b^2)/3\]

Answer 2

I am not sure about the answer!