On a frictionless table, a glob of clay of mass 0.40 kg strikes a bar of mass 1.74 kg perpendicularly at a point 0.15 m from the center of the bar and sticks to it. If the bar is 1.14 m long and the clay is moving at 6.6 m/s before striking the bar, what is the final speed of the center of mass? I got this part by using conservation of momentum. The answer was 1.23m/s . At what angular speed does the bar/clay system rotate about its center of mass after the impact? I tried using conservation of angular momentum. [tex] L_o = L_f [/tex] [tex] I \omega= I \omega [/tex] [tex] \omega _ initial = v/r= 6.6/.15= 44 rad/s [/tex] [tex] I= (1/2)ML^2 = (1/2)(1.74)(1.14)^2 = 1.13 kgm^2 [/tex] so [tex] L_o= 1.13* 44 [/tex] = 49.72 [tex]L_f= (1/2)(1.74+ .40)(1.14^2) * \omega = .452 *\omega [/tex] Solving for omega gave me 109.9 rad/s.. which doesn't really make sense. Can someone help me? Thanks in advance.