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Angular momentum in projectile motion

  1. Apr 19, 2010 #1
    1. The problem statement, all variables and given/known data

    A rock of mass 55.0 g is thrown with initial horizontal velocity v_x = + 32.00 m/s off a building from a height of 29.00 m. Take UP to be the +y direction. You will need to calculate the angular momentum of the rock about the line along the edge of the roof as a function of time.

    Find the angular momentum of the rock about the line along the edge of the roof at 0.28 s.

    2. Relevant equations

    L = r*m*v

    3. The attempt at a solution

    to find the distance traveled in .28 seconds

    d = 32m/s(.28) + .5(9.8m/s^2)(.28^2)
    d = 9.344 m

    then L = r*m*v

    L = 9.344(.055kg)(32m/s)

    I know this is wrong but i am not sure how else to go about this problem :(

    thank you for any help
  2. jcsd
  3. Apr 20, 2010 #2


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    Homework Helper
    Gold Member

    The angular momentum is a vector product:

    [tex]\vec L = m \vec r \times \vec v [/tex]

    For motions in the xy plane, L has only z component and it is

    [tex] L = m ( x v_y - y v_x ) [/tex].

    Determine the x, y coordinates of the projectile as function of time with respect to the edge as origin, and the x and y components of its velocity too, and substitute in the above expression for L.

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