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Centre of gravity, moment of inertia and radius of gyration

  1. May 30, 2017 #1
    1. The problem statement, all variables and given/known data
    media%2F576%2F5762c133-50e7-48d5-a518-44b7cfa007df%2FphpzTtjj8.png

    2. Relevant equations

    Centre of gravity: X=m1x1-m2x2/m1-m2

    MOI rectangle: 1/3ml^2

    MOI triangle: 1/18md^2

    Radius of gyration: Ixx=mk^2

    3. The attempt at a solution

    Mass of body 1: b*l*p = 0.8*1*10=8kg

    Mass of body 2: 1/2b*h*p = 1/2(0.4)*0.6*10=1.2kg


    1.1

    X=m1x1-m2x2/m1-m2

    = 8*(-400)-1.2*(-1000/3)/8-1.2

    = -411.7mm



    1.2

    Ixx= 1/3ml^2 - 1/18md^2

    = 1/3*8*1^2 - 1/18*1.2*0.2^2

    =2.664m^4



    1.3

    Ixx=mk^2

    2.664=(8-1.2)*k^2

    k=625.91mm
     
  2. jcsd
  3. May 30, 2017 #2

    haruspex

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    How do you get 400, why -400, and how do you get 1000/3?
     
  4. May 30, 2017 #3
    That should be 500, from the centre of the rectangle to the XX axis is 500mm.
    I believe the centre of the triangle should be 2/3*600 = 400mm
    Then added to the 200mm from the start of the triangle to the XX axis should give 400+200=600mm.
    I put a negative in front of the 400 and 1000/3 as they are "away" from the XX axis.
    I don't know why I put 1000/3, I think I was dividing the length of the rectangle in 3 because of the triangle for some reason.

    X=m1x1-m2x2/m1-m2

    = 8*(-500)-1.2*(-600)/8-1.2

    = -482.35mm
     
  5. May 30, 2017 #4

    haruspex

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    Right.
    But they are "away" from that axis in the positive y direction.
     
  6. May 30, 2017 #5
    Meaning they should be positive?
    If they were below the X axis then they would be negative? In other words if the XX axis were at the top of the rectangle.
     
  7. May 30, 2017 #6

    haruspex

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    Yes.
     
  8. May 30, 2017 #7
    Okay so then

    X=m1x1-m2x2/m1-m2

    = 8*(500)-1.2*(600)/8-1.2

    = 482.35mm

    Is this correct?

    What about my answers for 1.2 and 1.3?
     
  9. May 30, 2017 #8

    haruspex

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    The term for the triangle is wrong. What axis would be right for that expression?
     
  10. May 30, 2017 #9
    The formulae given for a rectangular triangle:

    about aa axis
    I=1/18md^2
    about bb axis (through G)
    I=1/18mh^2

    because we want the inertia about xx axis I used =1/18md^2 where d=distance to xx axis=200mm
     
  11. May 30, 2017 #10
  12. May 30, 2017 #11

    haruspex

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    I don't know what you mean by aa axis here, but that expression is for the moment of inertia about an axis through the centroid and parallel to one side. d is the width orthogonal to the axis. So here, d=0.6 for an axis through the centroid and parallel to the x axis.
    To get from that axis to the xx axis itself you need to use the parallel axis theorem.
     
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