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Homework Help: Center of mass, moment of inertia for x = y^2 and y = x - 2 with density d = 3x

  1. Apr 13, 2005 #1
    Here is the problem:

    A region is defined as being bounded by the parabola [tex]x = y^2[/tex] and the line [tex]y = x - 2[/tex].

    The density of this region is [tex]\delta = 3x[/tex].

    a) Find the center of mass.

    b) Find the moment of inertia about the y-axis.

    c) Find the radius of gyration about the y-axis.

    Here is what I have:

    [tex]M = \int_{-1}^{2}\int_{y^2}^{y + 2}\;3x\;dx\;dy = \frac{108}{5}[/tex]

    [tex]M_{x} = \int_{-1}^{2}\int_{y^2}^{y + 2} 3xy\;dx\;dy = \frac{135}{8}[/tex]

    [tex]M_{y} = \int_{-1}^{2}\int_{y^2}^{y + 2} 3x^2\;dx\;dy = \frac{1269}{28}[/tex]

    [tex]\bar{x} = \frac{\frac{1269}{28}}{\frac{108}{5}} = \frac{235}{112}\;\;and\;\;\bar{y} = \frac{\frac{135}{8}}{\frac{108}{5}} = \frac{25}{32}[/tex]

    [tex]I_{y} = \int_{-1}^{2}\int_{y^2}^{y + 2} x^2 \left(3x\right)\;dx\;dy = 110.7[/tex]

    [tex]R_{y} = \sqrt{\frac{110.7}{\frac{108}{5}}} \approx 2.26[/tex]

    Does this look correct?
     

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    Last edited: Apr 13, 2005
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  3. Apr 13, 2005 #2

    HallsofIvy

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    I get what you do for M but 153/8 rather than 135/8 forMx. I get completely different answer for the others- but I may have tried to do them too fast.
     
  4. Apr 13, 2005 #3

    dextercioby

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    I dunno what Halls did,but all your calculations are perfect.I double checked them with Maple...


    Daniel.
     
    Last edited: Apr 13, 2005
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