A rod of length 2.4 m is pivoted at its center and masses of 2.3 kg and 3.1 kg are hung from its two ends. Find the initial angular acceleration of the rod when the system is released from a starting position of the rod that is horizontal. The radius of the rod would be r=1/2(2.4).. which is 1.2. Should I use the equation I=mr^2 or I=1/12ml^2? I=mr^2 is I=(2.3)(1.2)^2=3.312 and I=(3.1)(1.2)^2=4.464. I=1/12ml^2 is I=1/12(2.3)(2.4)^2=1.104 and I=1/12(3.1)(2.4)^2=1.488. Do i even need these for this problem? Where do i go next?