1. The problem statement, all variables and given/known data A uniform thin rod with mass 4.59 kilograms pivots about an axis through its center and perpendicular to its length. Two small bodies with a mass of 0.543 kilograms, are attached to the ends of the rod. What is the length of the rod such that the moment of inertia of the three-body system with respect to the described axis is 0.941 kg·m^2? 2. Relevant equations Length of Rod = L (unknown) For the rod, MoI = 1/12ML For the two small masses, MoI of each is: MoI = MR^2 MoI-rod + MoI-mass + MoI-mass = ƩI 3. The attempt at a solution Moment of Inertia of Rod: 1/12(4.59 kg )L Moment of Inertia of Mass: (0.543 kg) * (L/2)^2 For moment of inertia of small mass, I picked the radius from the axis of rotation to be L/2 since the axis of rotation is at the rod's center and perpendicular to its length. I thought the radius from that point to the small mass would be L/2. So the setup is: 1/12(4.59)L + (0.543)(L/2)^2 + (0.543)(L/2)^2 = 0.941 When I solve for L, I get L = 1.28609 m, which is apparently incorrect. I do not know where I could have screwed up on this problem.