Monday
March 27, 2017

Post a New Question

Posted by on .

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.

  • physics - ,

    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.

  • physics - ,

    For part B:

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

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

  • physics - ,

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question