A thin, uniform rod is bent into a square of side length "a". If the total mass is M, find the moment of inertia about an axis through the center and perpendicular to the plane of the square. Use the parallel-axis theorem. According to Parallel Axis theorem: I = I(cm) + Md^2 The distance across the diagonal of the square (corner to corner) is 1.4142...a (= to sqrt(2a^2)). The distance bisecting the square through the center of mass is a/2. I = m(a/2)^2 + m(0.7071...a)^2 I = (1/4)ma^2 + (1/2)ma^2 I = (3/4)ma^2 This answer isn't right. I think I need to integrate since every point on the square is a different distance from the rotational axis through the center of mass. But there is not other axis specified in the problem, so I'm not sure as to which parallel axis I need to derive an expression for. Any ideas?