I've attempted this problem several times and keep getting 14 rads as the answer but its not an available answer choice... any help would be greatly appreciated. The problem: A uniform rod (mass = 2.0 kg, length = 0.6 m) is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60 degrees below the horizontal? Available answers: 15, 12, 18, 29, and 23 rads/s^2 Relevant equations: The moment of inertia for this rod is (1/3)mL^2 according to wikipedia. 3. The attempt at a solution Here's how I keep coming to 14 rads: Calculated moment of inertia to be 0.24 using the above equation. Calculated the angular acceleration using sin(60)*9.8 to get 8.49 m/s^2. Divided this by circumference of the circle the rod would make (8.49/3.77=2.25). Multiplied 2.25 * 2∏ which gives me an answer of 14 rads/s^2. As I typed this up I realized I made no use of the rods moment of inertia which I realize I probably need to use in order to solve the problem. I'm just not sure how to implement it. Again any help would be greatly appreciated! Thanks!