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Angular Momentum of two masses

  1. Apr 8, 2010 #1
    1. The problem statement, all variables and given/known data
    A light, rigid rod l = 9.44 m in length rotates
    in the xy plane about a pivot through the
    rod’s center. Two particles of masses m1 =
    9.1 kg and m2 = 2.5 kg are connected to its
    ends.


    Determine the angular momentum of the
    system about the origin at the instant the
    speed of each particle is v = 2.1 m/s.
    Answer in units of kgm2/s.

    2. Relevant equations
    Lz=Iw
    I = mr2
    I = (1/12)ML2
    w = v/r

    3. The attempt at a solution
    I used the above equations:
    Lz=Iw
    I = mr2
    I = (1/12)ML2
    w = v/r
    I = 1/12(9.1kg)((9.44)2)=67.578
    I = 1/12(2.5kg)((9.44)2)=18.57
    sum of I = 86.1431
    Used w = v/r = ((2.1m/s))/(9.44/2)=0.444915
    Lz=Iw
    (86.1431)(0.444915)=38.32639
    I got it wrong, please help, what did I do wrong? Did I even approach the right way? Thanks.
     
  2. jcsd
  3. Apr 8, 2010 #2
    To find the correct value of I, you would need the moment of inertia about the centre of mass?
     
  4. Apr 9, 2010 #3
    For the MOI about the centre of mass, take take the radius as 4.72m and the centre x=0.
    Hence,

    centre of mass, [tex]x_{cm}= \frac{2.5\times -4.72 +9.1\times 4.72}{2.5+9.1}
    =2.68m[/tex]

    MOI, [tex]I_{cm}=\frac{1}{12}(M_{system})(x_{cm})^{2}=\frac{1}{12}(11.6)(2.68)^{2}=6.94[/tex]

    [tex]\omega=\frac{v}{r}
    =\frac{2.1}{4.72}
    = 0.44 rad/s[/tex]

    Therefore, [tex]L=I\omega=6.94\times 0.44=3.05 kgm^{2}/s[/tex]
     
  5. Apr 9, 2010 #4

    I tried that, but it is not right either. ;(
     
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