1. The problem statement, all variables and given/known data A very thin uniform rod, 2.40 m long and of weight 135 N, has a frictionless hinge at its lower end. It starts out vertically from resting on a wall and falls, pivoting about the hinge. Just as it has rotated through an angle of 55 degrees, the acceleration of the end farthest from the hinge is what? 2. Relevant equations [tex]\tau = I[/tex][tex]\alpha[/tex] [tex]\tau[/tex] = R X F = |R| |F| sin(angle between) I for thin rod, I = (1/3)MR2 3. The attempt at a solution To find R X F I tried to find the angle between the force of gravity and the pivot. If you place both vectors at the pivot, R will be 35 degrees above the horizontal (after falling 55 degrees), and gravity will be directly down, so I believe the angle between them is 90+35 = 125 degrees? This is one thing I'm not entirely sure about. [tex]\tau[/tex] = 2.4M * 135N * sin(125) Then finding I, I = (1/3)(135N/9.8)*2.42 And then doing T/I to find the angular acceleration, (135N * sin(125)) / ( (1/3)(135N/9.8)*2.4) ~= 9.69 Finally, multiply that by the radius of 2.4M to get the linear acceleration. This didn't give me right answer though.