1. The problem statement, all variables and given/known data Consider a uniform rod of mass 12kg and length 1.0m. At its end, the rod is attached to a fixed, friction-free pivot. Initially the rod is balanced vertically above the pivot and begins to fall (from rest) as shown in the diagram. Determine, a) the angular acceleration of the rod as it passes through the horizontal at B. b) the angular speed of the rod as it passes through the vertical at C. 2. Relevant equations PE = mgh KE = 1/2Iω2 Irod = 1/3ml2 ω2 = ωo2 + 2αθ 3. The attempt at a solution a) PE = KE mgh = 1/2Iω2 (12kg)(9.81m/s2)(1.0m) = (0.5)(1/3)(12kg)(1.0m)ω2 ω = 7.668rad/s Can I use the length of the rod as its height? ω2 = ωo2 + 2αθ (7.668)2 = 2α90° (90° = 1.571rad) α = 18.7rad/s2 is that right? b) ω2 = ωo2 + 2αθ ω2 = (7.668rad/s)2 + 2(18.7rad/s2)(1.571rad) assumptions: the α will stay the same and i'm trying to calculate from point B to C so θ = 90° ω = 10.84rad/s Thanks in advance!