1. The problem statement, all variables and given/known data A uniform rod 1.1 m long with mass 0.7 kg is pivoted at one end, as shown in Fig. 9-46, and released from a horizontal position. Find the torque about the pivot exerted by the force of gravity as a function of the angle that the rod makes with the horizontal direction. (Use theta for θ.) http://www.webassign.net/fgt/9-43.gif 2. Relevant equations T = rFsin(theta) 3. The attempt at a solution My attempt has consisted of the following: 1st attempt: Using the basic formula, and then applying it to obtain ~7.6sin(theta), trying the negative version as well. I think realized that sin might not be valid to use due to the nature of the direction of application of the force and tried 7.6cos(90-theta), which also did not work. 2nd attempt: T = angular acceleration * moment of inertia I solved the moment of inertia to be ~.2823 (mL^2/3). I defined omega as d/(dt)[int(g)sin(theta)/r]. I then plugged in the values and came up with .2823d/(dt)[int(9.8)sin(theta)/1.1] Any help would be appreciated.