Question

The uniform solid block in right figure has mass 0.2 kg and edge lengths a = 5.0 cm, b = 8.0 cm, and c = 2.0 cm. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces.

Answer 1

The moment of inertia at the midddle of center= 1/12 M( b^2 + a^2)= 0.2/ 12 ( 0.0025 + 0.0064) = 1.483 x 10^-4

Distance between the center and axis at corner = 1/2 sqroot ( a^2+b^2)

Using the paralllel axis theorem,

I = 1/12 M ( b^2 +a^2) + M ( a^2+b^2) / 4 = 1.483 x 10^-4 + 4.45 x 10^-4= 5.933 x 10^-4