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Moment of inertia and integration

  1. Jan 31, 2009 #1
    1. The problem statement, all variables and given/known data

    A uniform rectangular plate has mass m and sides a and b.

    a) Show by integration that the moment of inertia of the plate about an axis is perpendicular to the plate and passes through one corner is (1/3)*m*(a^2+b^2)

    2. Relevant equations

    I= integral( r^2, dm)

    3. The attempt at a solution

    do I account this problem as 3 dimensions or 2 dimensions ??

    What should be r in this case since I have a and b
     
  2. jcsd
  3. Jan 31, 2009 #2

    LowlyPion

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    You could approach it in one of two ways.

    1)find the moment for a rectangular plate a x b about the center and then use the parallel axis theorem to rotate it about the corner on a line through the center.

    2)set up an integral along the diagonal from the opposite corner and relate Δ mass as little arcs that integrate over that distance.

    Seems like 1) would be an easier approach.

    Since it is uniform in thickness, then just treat it as a 2 dimensional object.
     
  4. Jan 31, 2009 #3
    I can use the 1st approach for part 2 which they ask for moment of inertia about the axis going through the center o of mass.

    If I use the 2nd approach, I will integrate (r^2,dm) from 0 to b (let say I choose b as the limit).

    how can I represent r^2 as ? Is that a^2 ?
     
  5. Jan 31, 2009 #4

    LowlyPion

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    Then do the second part first then.

    To find the moment of a rectangle isn't it simply something like:

    I = ∑m*r² = ρ*∫∫r²*dm = ρ*∫∫(x² +y²)*dx*dy

    where ρ = m/(h*w) and the limits of your double integration are |½h-½h and |½w-½w

    Then apply the || axis theorem along the diagonal at 1/2 the length of the diagonal away?
     
  6. Jan 31, 2009 #5
    Thanks. A dumb question. How do I solve double integration ??
     
  7. Jan 31, 2009 #6

    LowlyPion

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    Choose one variable - either one - to integrate keeping the other as a constant. Then you are left with a single integral that you can integrate treating the first variable as a constant.
     
  8. Jan 31, 2009 #7
    Consider any plane shape and scale it up by two, keeping the mass
    per unit area the same. Each "element" has quadrupled in mass and
    doubled its distance from the C of M. It follows that the moment of
    inertia about an axis through the C of M normal to the plane has gone up
    by a factor of 16.

    For the a x b rectangle, let I be the moment of inertia about the C of M
    and J that about a corner.

    The || axis theorem gives J = I + m(a^2 + b^2)/4

    Now consider the double sized rectangle formed from placing 4 of the
    originals together. We deduce that

    16I = 4J = 4I + m(a^2 + b^2)

    David
     
  9. Feb 2, 2009 #8
    Somehow I cannot get the answer of (1/3)m(a^ 2+b^2).

    I use the double integration technique and get

    Icom=(1/12)*m*a^2*b^2

    Then when I substitute in I= Icom+ m*h^2

    I use h=a/2

    I cannot get the answer. Did I do wrong somewhere ??
     
  10. Feb 2, 2009 #9
    h is the distance from centre to corner, not a/2
     
  11. Feb 2, 2009 #10
    Can I assume either it can be (1/2)a since stack the small dm along side a ??

    Is not the center of mass is 1/2 from each corner ?
     
  12. Feb 2, 2009 #11

    LowlyPion

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    Are you sure you didn't get 1/12*(a2 + b2) ?

    Because armed with that then and the || axis you can say the moment from the corner is Imiddle + m*(a2 + b2)/4

    The hypotenuse2 being (a/2)2 + (b/2)2 from the center.
     
  13. Feb 2, 2009 #12
    I am sure I did not get (1/12)*m*(x^2+y^2).

    I got x^2*y^2 .

    So I got out from the integration as (x^3/3)*(y^3/3)* m/(x*y).

    So plug in the limit I got (x^3/12)*(y^3/12)*m/(x*y)

    Then I cancel x and y and got x^2*y^2.

    How did you get the "+" ??
     
  14. Feb 2, 2009 #13

    LowlyPion

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    For :
    I = ∑m*r² = ρ*∫∫r²*dm = ρ*∫∫(x² +y²)*dx*dy

    I'd recheck your calculation, being sure to recognize ρ*h*w as being the mass of the rectangular object.

    There is also the approach suggested by the elegant thinking in this post by davieddy:
    https://www.physicsforums.com/showpost.php?p=2057709&postcount=7
    ...that gets the same result.
     
  15. Feb 2, 2009 #14
    Thank you:)
    I felt justified in presenting that solution because the question specifically
    asked for integration.

    By the sound of it, the original poster could use the practice with integrating.
    Maybe Pythagoras needs some brushing up too :)

    I = ∑m*r² = ∫r²dm = m/(ab)*∫(∫(x² +y²)dx)dy

    integral limits 0 to a and 0 to b

    David
     
    Last edited: Feb 2, 2009
  16. Feb 2, 2009 #15

    LowlyPion

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    I enjoyed it anyway because of the less traditional approach of deducing it from the || axis.
     
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