The question shows a horizontal uniform rod with a pivot point on the left end and a weight tied to the right end. The rod and weight have mass 7 kg each. The rod is 35m long It asks what the inital torque of the system on the pivot point will be when dropped with gravity being 9.8 m/s^2 What I tried to do was solve for the torque of the weight and the rod then add the two results for the total torque on the pivot point. For the weight since it is perpendicular force then T = r * F so I did 35m * -9.8 m/s^2 * 7 kg = -2401 kg (m/s)^2 For the rod I tried to use the theorem that involved parallel axis from the center of mass in the book. The book also gave several Formulas for common objects for inertia of center of mass. Since I didn't know how to integrate r^2 with respect to dm (only in calc II)this is what I did. I = I (com) + mh^2 (h is the distance from the axis perpendicular com) so.. I = 1/12 ML^2 (from the book pg. 227 for rod) + 1/4 M L^2 (since h is L/2) so I = 1/3 ML^2 = 1/3 7 kg (35m)^2 = 2858 kg*m^2 for the inertia Therefore Torque equals alpha * I so I had T = 2858 * (-9.8 m/s^2) / 35m = -800.3 so I had -800.3 kg*m^2/s^2 - 2401 kg*m^2/s^2 = -3201 kg*m^2/s^2 I tried this and several other ideas but to no avail.. Can you please help me. What mistake am I making?