1. The problem statement, all variables and given/known data On a frictionless table, a glob of clay of mass 0.72 kg strikes a bar of mass 1.34 kg perpendicularly at a point 0.23 m from the center of the bar and sticks to it. If the bar is 1.22 m long and the clay is moving at 8.3 m/s before striking the bar, what is the final speed of the center of mass? At what angular speed does the bar/clay system rotate about its center of mass after the impact? 2. The attempt at a solution Here is what I did Vcm = sum(mi*vi) / total mass in this case just one body has velocity: Mclay = .72 kg Mbr = 1.34 Kg Mtotal = 2.06 kg Vclay = 8.3 m/s Vcm = 2.901 m/s which is correct, I got the first part, but here is the second part that I got wrong now calculate the angular speed: angular speed - omega omega = v /r v - is the linear velocity on the trajectory of the body (tangential velocity) r - is the distance between the body which rotate and the center of rotation in your case the centre of rotation is the center of mass and r is the distance of the clay to the centre of mass D = 1.22 m d = 0.23 m the centre of the bar related to one end is D/2 the position of the clay related to the same end is d+D/2 Xcm = [Mclay *(d+D/2)+Mbr*(D/2) ]/Mtotal Xcm = .6904 m position of the center of mass the angular velocit of the clay: omega clay = Vclay / D1 D1=(d+D/2)-Xcm = .1496 m omega clay = 55.477 rad/s now about the bar: the center of the bar is situated related to the center of mass at: Dbar = Xcm - D/2 = .0804 m If you assume that the bar when it rotate has the same tangential velocity v = 8.3 m/s omega bar = 8.3 / 0.0804 = 103.249 rad/s the question is asking about At what angular speed does the bar/clay system rotate about its center of mass after the impact? and so I add up both the omega of the bar and the clay and got 158.726 rad/s but it's wrong Can anyone tell me where my mistake is?