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Eliminate constants/Moment of Inertia

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the moment of inertia, Iy, about the y-axis using a vertical strip, width dx.
    Use the given equation and eliminate constants k and c, express answer in terms of number and powers of a and b.


    2. Relevant equations

    I can't figure out how to eliminate the constants k and c?

    3. The attempt at a solution

    I believe I have the integration ready:

    [itex]\int^{a}_{0}[/itex] x[itex]^{2}[/itex] (b-y)dx
    where y=k(x-a)2 + c
     

    Attached Files:

  2. jcsd
  3. Nov 19, 2012 #2

    gneill

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    Staff: Mentor

    Pick out two points on the curve where you know the coordinates. Use these values to obtain two equations in two unknowns (c and k).
     
  4. Nov 19, 2012 #3
    Ok. So far this is what I have:

    I know the point at the origin (0,0). And x=a and y=b.

    0 = k(0 - a)2 + c = ka2 + c ==> k = -[itex]\frac{c}{a^2}[/itex]

    So,

    y=(-[itex]\frac{c}{a^2}[/itex])(x-a)2 + c

    I'm still not sure how to get rid of the c?
     
  5. Nov 19, 2012 #4

    gneill

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    Start with your second known point.
     
  6. Nov 20, 2012 #5
    I see..... so x=a, y=b

    b=(-[itex]\frac{c}{a^2}[/itex])(a-a)2 + c
    b=(-[itex]\frac{c}{a^2}[/itex])(0)2 + c
    b=c

    ....I think??
     
  7. Nov 20, 2012 #6

    gneill

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    Yup. So replace c in the y(x) function and then plug in the (0,0) point to see what k is all about...
     
  8. Nov 23, 2012 #7
    Thanks! Would you also check if my integration is set up correctly?

    [itex]\int^{a}_{0}[/itex] x2 [b-([itex]\frac{-b}{a^2}[/itex])(x-a2) + b] dx

    How would I get my answer in terms of a number and powers of constants a and b (per instructions)?
     
    Last edited: Nov 23, 2012
  9. Nov 23, 2012 #8

    gneill

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    For your integration, be sure that you're squaring the right thing in the integrand; is it just "a" that is squared?

    Once you perform the integration the result will be in terms of a constant and powers of a and b.

    Note that the tacit assumption is that the 'material' that the figure is made of has a uniform density of 1 mass unit per unit area. If instead you want to assume a mass M for the object, you should calculate its density from M and the total area and incorporate this into the determination of the moment of inertia.
     
  10. Nov 23, 2012 #9
    I thought that the equation for Iy was

    A x2 dA

    Why would the "a" be the only thing squared?
     
  11. Nov 23, 2012 #10

    gneill

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    Isn't the function ##-\frac{b}{a^2}(x - a)^2 + b##, so that it's the whole term ##(x - a)## that's squared, not just the a in the term?
     
  12. Nov 23, 2012 #11
    My answer came out to be:

    [itex]\frac{3ba^3}{10}[/itex]

    I don't understand how to get a power of a constant with this problem?
     
  13. Nov 23, 2012 #12

    gneill

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    a and b are constants. So in your answer b is to the power of 1 and a is to the power of 3.

    The numerical constant 3/10 looks off to me though. Can you show more detail in your work?
     
  14. Nov 23, 2012 #13
    I've attached a scan of my work. Now that I've looked at the problem, I think I left out a "b-" for my "ydx". I think that it should be (b-y)dx. Seem right?
     

    Attached Files:

    Last edited: Nov 23, 2012
  15. Nov 23, 2012 #14

    gneill

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    Yup. b-y(x) is the length of the differential strips.
     
  16. Nov 24, 2012 #15
    After including "b-" in my problem my answer came out to be:

    [itex]\frac{8ba^3}{15}[/itex] - [itex]\frac{2ba}{3}[/itex],

    without factoring.

    Radius of Gyration: Ky = [itex]\sqrt{\frac{(\frac{8ba^3}{15} - \frac{2ba}{3})}{(b-y)dx}}[/itex]
     

    Attached Files:

  17. Nov 24, 2012 #16

    gneill

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    Your integral should look something like:

    $$\int_0^a x^2\left[ b - \left[-\frac{b}{a^2}*(x - a)^2 +b\right]\right]dx$$
    $$= \int_0^a x^2\left[\frac{b}{a^2}(x - a)^2\right]dx$$
    $$= \int_0^a b x^2 - 2\frac{b}{a}x^3 + \frac{b}{a^2}x^4 \, dx$$

    Evaluating the last form term by term should leave you with terms having b to the power of one, a to the power of three only. Try the integral again.
     
  18. Nov 24, 2012 #17
    I see what I did wrong. Instead of having (-[itex]\frac{2b}{a}[/itex]x3), I had: (-[itex]\frac{2b}{a^2}[/itex]x3).

    I now have: ba2*([itex]\frac{8ba}{15}[/itex] - [itex]\frac{2}{3}[/itex])??

    I believe I integrated correctly.
     
    Last edited: Nov 24, 2012
  19. Nov 24, 2012 #18

    gneill

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    If that's your result for the integration, it's not correct. Integrate term by term and show your work.
     
  20. Nov 25, 2012 #19
    Here is the new integration that I did. I didn't know if I should leave it as is or try to simplify it. There are no instructions that say to do so.
     

    Attached Files:

  21. Nov 25, 2012 #20

    gneill

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    In the first line of your attachment, the term with the coefficient -2b/a should be an x3 term, not x2. There's one x2 term, one x3 term, and one x4 term going into the integration.
     
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