Hi i have a question with this problem A thin, uniform, metal bar, 2.00 m long and weighing 90.0 N, is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.50 m below the ceiling by a small 3.00-kg ball, initially traveling horizontally at 10.0 m/s. The ball rebounds in the opposite direction with a speed of 6.00 m/s. Find the angular speed of the bar just after the collision. I understand that angular momentum is conserved. what i have done is found the moment of inertia of the bar and the ball I_bar=(1/3)ml^2 I_ball=m1.5^2 i then added the 2 together to get the total moment of inertia I_bar+I_ball=I i then know that the angular momentum is equal to I(w)_init=I(w)_f therefore w=(mvl)/I which means angular velocity is equal to the massof the ball times the velocity (im am not sure which velocity i should use though, should i use both ?)times the distance from the pivot point. all this over I is this methodology correct?