3. The attempt at a solution Hello, I'm having trouble with energy in rotational motion. The question asks: Consider a uniform rod of mass 12 kg and length 1.0 m. 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. Ok. So, here's my logic. The answer is wrong, I'm aware, but I'd like to know why. Einitial=Efinal Einitial = mg(L)/2 =58.8J Efinal = (1/2)(I)(w^2) (I can use h=0 as a reference point in the middle of the motion, right?) =(1/6)mL^2 w^2 w = 5.42rad/s Ok, so correct me if I'm wrong but this is the velocity as soon as the rod is horizontal. Let's continue -- using this equation: Where a = angular acceleration d = θ (in rad) Didn't the rod rotate 90 degrees (vertical to horizontal)? -- if so, its radians are 1.57 w^2 = wo^2 + 2ad (5.42 ) /(2(1.57)) = 9.3rad/s^2 The answer is something like 14.7rad/s^2... I feel like something's wrong with the theta.