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Homework Help: Angular Momentum of a system of particles relative to another point

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Two particles move in opposite directions along a straight line. Particle 1 of mass m1 = 23 kg at x1 = 16 m moves with a speed v1 = 34 m/s (to the right), while the particle 2 of mass m2 = 64 kg at x2 = −25 m moves with a speed v2 = −43 m/s (to the left). Given: Counter-clockwise is the positive angular direction.

    1)What is the total angular momentum of the system about the z-axis relative to point A along y axis if d1 = 13 m? Answer in units of kgm2/s.

    2) What is the total angular momentum of the system about the z-axis relative to point B along y axis if d2 = 22 m? Answer in units of kgm2/s.

    m1 = 23
    Position1 = {16 i, 0 j, 0 k}
    v1 = {34 i, 0 j, 0 k}

    m2 = 64
    Pos2 = {-25 i , 0 j, 0 k}
    v2 = {-43 i, 0 j, 0 k}

    d1 = {0 i, -13 j, 0 k}
    d2 = {0 i, 22 j, 0 k}



    2. Relevant equations

    L = Position [tex]\times[/tex] m v

    3. The attempt at a solution

    I got number 1 correct, and number 2 should be the same. however just changing d1 to d2 does not give the correct answer.


    1)
    {16, 0 , 0} + {0, -13, 0} = {16, -13, 0}
    {-25 , 0, 0}+ {0, -13, 0} = {-25, -13, 0}

    {16, -13, 0} [tex]\times[/tex] (23 {34 , 0 , 0 }) +
    {-25, -13, 0} [tex]\times[/tex] (64 {-43 , 0 , 0 }) = {0, 0, -25610} = 25610 k

    2) should be the same but I get {0, 0, 43340} which is incorrect
     

    Attached Files:

  2. jcsd
  3. Feb 14, 2009 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi DarkerStorm! Welcome to PF! :smile:

    Looks ok to me (opposite sign to 1), of course). :confused:

    (btw, you could have omitted the {16, 0 , 0} and {-25, 0 , 0} :wink:)

    hmm … you could try 43440, in case someone hit the wrong key. :redface:
     
  4. Feb 17, 2009 #3
    Thanks the problem was that number 2 have to be negative.
     
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