physics
posted by Anonymous .
A small 5.00kg brick is released from rest 2.00m above a horizontal seesaw on a fulcrum at its center, as shown in the figure below . (radius of horizontal seesaw is 1.6m)
a.)Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant the brick is released.
b.)Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant before it strikes the seesaw.

the answer to part a) is zero. I don't know why right now though. and the answer to part b) would be 45.1 if it was a 4.50kg brick. I also don't remember how I got that answer.

For part B:
angular momentum =
m*r*sqrt(2gh), where r=1.6 and h=2
5*1.6*sqrt(2*9.8*2) 
Part 1. Velocity is 0.
(mass)(velocity)(r) = 0
Part 2:
v= (sqrt (2gh))= (sqrt(2(9.8)(2))=6.26
(mass)(velocity)(r)
(4.5)(6.26)(1.6)= 45.072=45.1