Here is what I did initially: v = 3.77m/1.6s mgh = (1/2)mv^2 + (1/2)*I*omega^2 omega = v/R (20kg)(9.8)(3.77m) = (1/2)(20kg)(2.36 m/s)^2 + (1/2)I((2.36)^2/(0.60)^2) I = 88.3 ... which is more than I = MR^2 = 25.2 ... So this is entirely wrong, because I know it should be less, since the mass is more evenly distributed. Second approach, I believe the drops 3.77 m in 1.6 s is still due to gravity, is not constant velocity. There must also be tension then since it is less than gravitational acceleration. mg - T = ma y=(1/2)a*t^2 (3.77m) = (1/2)a(1.6)^2 a= 2.95 m/s^2 (20 kg)(9.8) - T = (20kg)(2.95) T = 137N torque = F*R Only T does something in relation to the rotation of the axis? So: torque = T*R = I * alpha alpha = a/R = 2.95m/s^2 / 0.60m = 4.92 rad/s^2 137N * 0.60m = I * (4.92 rad/s^2) I = 16.6 N s Well I got the less inertia part now. But was that the right way to do it?