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Rotational inertia of particle system

  1. Oct 1, 2007 #1
    The masses and coordinates of four particles are indicated in the following table.
    40 g x = 1.0 cm y = -4.0 cm
    25 g x = 2.0 cm y = 4.0 cm
    30 g x = -3.0 cm y = 5.0 cm
    35 g x = 4.0 cm y = 4.0 cm

    (a) What is the rotational inertia of this collection about the x axis?
    g·cm2

    (b) What is the rotational inertia of this collection about the y axis?
    g·cm2






    3. The attempt at a solution

    I honestly don't really know where to start with this problem. I know it's not a continuous body, so I wouldn't solve it through integration, but there's not really anything in the chapter that discusses multiple points in the xy plane.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 1, 2007 #2

    Chi Meson

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    Rotation inertia is simply the resistance to rotational (or "angular" acceleration). Every object will have a rotational inertia with respect to any point in the universe. The simple equation for rotational inertia is I=mr^2.

    Careful: when x=1 cm, that means it is 1 cm from the y-axis.

    For a group of individual points, you just sum all the individual inertias. The next step will be integration for solid objects.
     
  4. Oct 1, 2007 #3
    - That's what I was thinking about doing, but my problem was that I wasn't sure of where to take my origin at so I could get my r. Would I just take the point (0,0) as the origin and get the r from that?
     
  5. Oct 1, 2007 #4

    Chi Meson

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    No, it's simpler. The system of points will be rotated about the x-axis first, so the "r's" are the straight line distances to the axes.

    If you were to rotate the system about the z-axis, then you would find the distance to the origin
     
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