A 2 meter long bar weighing 90N hangs vertically from a frictionless pivot. The ball is hit at a point 1.5 Meters below the ceiling by a ball that weighs 3Kg and is traveling with a velocity of 10 m/s. The ball bounces back with a velocity of -6 m/s. Find the angular speed of the bar just after the collision.
L = Iw = mv x r
The Attempt at a Solution
I ended up finding the answer but I don't understand one step.
At first I set up
(3kg)*(10m/s)(1.5m) = (3kg)*(-6m/s)(1.5m) + (1/12)(9.18kg)((2^2)m)w
Solving for w gives 23.5 rad/s, the WRONG answer.
When I act like the ball's velocity ISN"T in the opposite direction as it was initially, I get the right answer, wtf!?
(3kg)*(10m/s)(1.5m) = (3kg)*(6m/s)(1.5m) + (1/12)(9.18kg)((2^2)m)w
w = 5.88m/s
Is there some weird angular momentum absolute value stuff going on here? I don't get it... Part B asks "Why is the systems angular momentum conserved but the linear momentum is not," I hope it has something to do with the observation I just noticed.