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)

Question

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.

Answer 1

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.

Answer 2

For part B:

angular momentum =
m*r*sqrt(2gh), where r=1.6 and h=2

5*1.6*sqrt(2*9.8*2)

Answer 3

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