Pivot point on a Uniform Rod and Acceleration I'm stuck on this question, which seems like it should be fairly simple: A uniform rod of length 1.15 m is attached to a frictionless pivot at one end. It is released from rest from an angle theta = 21.0° above the horizontal. Find the magnitude of the initial acceleration of the rod's CM. theres a graphic of it here: http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype20/prob28_1012rod.gif ok so what I've done so far is: Rmg(sinθ) = Iα Rmg(sinθ) = (1/3)mL^2α getting the expression I = (1/3)mL^2 from my text book for a uniform rod with a pivot at its end. working that out I get: (0.575)(9.8)(sin21) = (1/3)(1.15)^2 α α = 4.58 rad/s then converting α to a with: a = Rα a = (.575)(4.58) a = 2.63 m/s^2 but apparently this isn't right. can anyone help me out?