A bike wheel, of mass 1.36 kg is placed at the end of a rod 0.530 m in length, which can pivot freely about the other end. The rod is of negligble mass. The wheel is turning rapidly such that it has an angular momentum of 10.8 kgm^2/s. At what angular speed does the wheel revolve horizontally about the pivot? I used the equation L= Iw, substituting .5MR^2 for I L= .5MR^2 w 10.8=.5(1.36)(.530)^2 w Solving for w gives 56.5 rad/s which isn't right Can anyone tell me what I did wrong?