The uniform solid block in Fig. 10-37 has mass .172 kg and edge lengths a = 0.035 m, b = .084 m, and c = .014 m. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces.
I = Icom + M(h^2)
I = 1/12M(a^2+b^2)
The Attempt at a Solution
With the pythagorean theorem, I found the distance it has shifted which was [1/2(a^2+b^2)]^1/2 = h so,
I = 1/12M(a^2+b^2) + M(h^2) substituting h with the previous discovery giving:
I = 1/12M(a^2+b^2) + M[1/2(a^2+b^2)] I still come out with an incorrect answer.
I'm wondering if the c should be included in the problem or not, but when I do input it in the Icom, it still does not work. I must be missing something important
The answer should be fitting with 4.7 x 10^-4 kg.m^2