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Find the Moments of Inertia question

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data

    The four masses shown in the figure below are connected by massless, rigid rods.
    http://i241.photobucket.com/albums/ff4/alg5045/p13-17.gif
    a.) Find the coordinates of the center of mass if Ma=100g and Mb=Mc=Md=230g.
    b.) Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page.
    c.) Find the moment of inertia about a diagonal axis that passes through masses B and D.

    2. Relevant equations

    I = mA(rA^2) + mB(rB^2) + etc...

    3. The attempt at a solution

    I solved part a by taking the x coordinates times the masses and then dividing by all of the masses. I did the same thing for the y coordinates. For both the x and the y coordinates I got .058m. For parts b and c I tried using the above equation, but it didn't work. I assumed for part b that it would be zero because mass a is at the origin, but that was wrong.
     
  2. jcsd
  3. Oct 14, 2007 #2

    Doc Al

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    Staff: Mentor

    Only the contribution due to mass a would be zero. Show exactly what you did for b and c.
     
  4. Oct 14, 2007 #3
    For part b I just assumed it would be zero because it's at the origin, but that was wrong. For part c I calculated I = (.230kg)(.07m)^2 + (.230kg)(.07m)^2. I got .07 m by using the pythagorean theorem since I knew the other two sides of the triangle to be .1m.
     
  5. Oct 14, 2007 #4

    Doc Al

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    But only mass A is at the origin. What about the others?
    But A and C have different masses. Also: Be a bit more precise in your calculation of the side length.
     
  6. Oct 14, 2007 #5
    I don't understand what the other masses have to do with anything for one question about mass A and the other question about masses B and D.
     
  7. Oct 14, 2007 #6

    Doc Al

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    You are misunderstanding the question. In both questions they are asking for the moment of inertia of the entire assembly of all four masses. The only difference is where the axis is, which changes the moment of inertia.
     
  8. Oct 14, 2007 #7
    Oh, I see. Let me try to figure it out.
     
  9. Oct 14, 2007 #8
    For part b, would it I = (.100)(0) + (.230)(.05^2) + (.230)(.07071^2) + (.230)(.05^2)?
     
  10. Oct 14, 2007 #9

    Doc Al

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    Double check the distances from the axis.
     
  11. Oct 14, 2007 #10
    The axis is at A. So B and D would remain the same distances, and C would be 14.142cm.
     
  12. Oct 14, 2007 #11

    Doc Al

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    That's better.
     
  13. Oct 14, 2007 #12
    For part b, I got .0092 kgm^2. For part c I recognized that B and D would be zero, so I = (.1)(.07071^2) + (.230)(.07072^2) = 1.65 x 10-3. Both were right. Thanks so much!!
     
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