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Math Help - Basic moment of inertia question

  1. #1
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    Basic moment of inertia question

    26-5-7

    Find moment of inertia with respect to x axis.
    x = -1, x = 1, y = 0, y = 1

    Attempt:
    Ix = 2pi(k) (int) (1)y^3 dy
    boundaries 1 and 0
    = 2pi(k)1/4(y^4)
    = (2pi(k))/4 = Ix

    m = 2pi(k) (int) xy dy
    = 2pi(k) (int) (1)y dy
    = 2pi(k)1/2(y^2)
    = pi(k)(y^2) = pi(k)

    R^2 (subscript x) = [(2pi(k))/4]/[(pi(k)]
    K = 1 (from appearances of book example questions)
    R^2 (subscript x) = 0.5
    R (subscript x) = sqrt(0.5)

    the book answer is 2/3k though. What did I do wrong? thanks
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  2. #2
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    Re: Basic moment of inertia question

    Find the moment of inertial of what with respect to the x-axis? The square bounded by x= -1, x= 1, y= 0, and y= 1? And what is "k"? The density?
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  3. #3
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    Re: Basic moment of inertia question

    Density is K pretty sure. K or k. So it has been used in the examples as such, and the example questions calculate K as 1. However the answer includes K for this question, so apparently they didn't just plug 1 into it. You could assume easily enough that this question considers the area of x = -1, x = 1, y = 0 and y= 1 as the area of interest.
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  4. #4
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    Re: Basic moment of inertia question

    So your problem looks like this?


    Its not a square but anyway,
    Consider a tiny strip of material with width dy a distance y away from the x axis


    Inertia is given by I= mr^2

    For this strip of material the mass is dm and the distance from the axis 'r' is y. Let the inertia for a tiny strip be I' while the inertia for the whole body is I
    I'= y^2dm

    As the length of the strip is 2 (from -1 to 1) the area of this strip dA is given by dA= 2dy

    The mass per unit area, the density k is equal to \frac{dm}{dA}
    So dm= k(dA)

    From your expression of area
    dm= k(2dy)

    Use the expression of dm in the formula for the inertia of the piece

    I'= y^2k(2dy)

    Sum up the inertia of all the tiny strips along the material from y=0 to y=1

    \int I'= \int_0^1 y^2k(2dy)
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