1. The problem statement, all variables and given/known data Assume your arm is a solid rod, 0.740 m in length, free to rotate about your shoulder (so its moment of inertia is (1/3)ML^2). If you hold your arm out sideways, and then completely relax your muscles, what is the initial downward angular acceleration of your arm? 2. Relevant equations τ=LF=Iα F=mg I=(1/3) ML^2 3. The attempt at a solution I'm not given a specific formula to use, but I think the above formula for Torque applies. I'm not using conservation of energy because I dont have a value for the height to use for potential energy. If I plug the formulas for inertia and Fg into the torque formula: Lmg=(1/3)ML^2 α α=3g/L = (3*9.81)/0.740 = 39.77 but the answer key says the answer is 19.9 (half). What am I doing wrong here?