1. The problem statement, all variables and given/known data Consider a thin 14 m rod pivoted at one end. A uniform density spherical object (whose mass is 9 kg and radius is 3.2 m) is attached to the free end of the rod and the moment of inertia of the rod about an end is Irod =1/3mL^2 and the moment of inertia of the sphere about its center of mass is Isphere =2/5mr^2 . What is the angular acceleration of the rod immediately after it is released from its initial position of 35◦ from the vertical? The acceleration of gravity g = 9.8 m/s^2 If needed a graphical representation of the question can be found at the following link: http://i1221.photobucket.com/albums/dd470/jokerg908/435.jpg 2. Relevant equations τ = Iα I = (1/3)MR2 τ = FRsin(Θ) 3. The attempt at a solution Finding the position of the center of mass (x meters from pivot): (x-7) = (14+3.2-x) x=12.1 meters Using xcm = 12.1 m, we can solve for α: I = (1/3)(18 kg)(12.1 m)2 ≈ 878.46 kg·m2 τ = (18 kg)(9.8 m/s2)(12.1 m)sin(35°) ≈ 1224.264489 N·m α = τ/I ≈ 1.39361 rad/s2 This answer is taken as incorrect by my online program. Any suggestions or assumptions that I have made in my calculations would be helpful. Thanks. One thing that I was worried about was that this method didn't use the moments of inertia that I was given.