1. The problem statement, all variables and given/known data A uniform bar AB of length L is freely hinged at one end A and released from a horizontal position. a) Find the initial angular momentum. Ans: 3g/2L b)Find the angular velocity when the bar is vertical. Ans: (3g/L)^(1/2) 2. Relevant equations momentum of inertial of a rod through end =(ML^2)/3 3. The attempt at a solution I have no problem with part a, I did it like this Let N be the moment, I be the inertia of moment, a be the angular acceleration. N = Ia mg(L/2) = ML^2 /3 a a= 3g/2L But I have problem with part b, I tried to do it with energy conservation Kr + Kt = PE Iw^2 /2 + mw^2(L^2)/2 = mgL/2 1/3 L w^2 + w^2L = g I would be correct if I neglect the linear KE, but I don't think it should be neglected, is my method of doing part b wrong?