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Math Help - Statics Problem: Finding Center of Mass

  1. #1
    Member Em Yeu Anh's Avatar
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    Angry Finding a Center of Mass




    The object above has a uniform cross-section. It is made of two materials that have different densities. Part A's density is p_o(kg/m^3) and for part B (p_o/2)(kg/m^3) Find the coordinates (x,y) of the center of mass for this object. You may consider a part of the object that is one meter long in the z direction.

    I don't even know where to begin, the workbook I have does not seem to have any similar examples nor my lecture notes.
    Last edited by Em Yeu Anh; November 24th 2009 at 08:40 PM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    It is convenient to perform two separate integrations for the areas A and B. First we compute the 'whole mass'...

    M= \int \int_{A} \rho_{A}\cdot dx\cdot dy + \int \int_{B} \rho_{B}\cdot dx\cdot dy=

     = \int_{0}^{.1} \int _{.2}^{.7} dx\cdot dy + \frac{1}{2}\cdot \int_{.1}^{.5} \int _{.2}^{.3} dx\cdot dy + \frac{1}{2}\cdot \int_{.5}^{.6} \int _{0}^{.3} dx\cdot dy= .085 (1)

    ... and then we compute x_{m} and y_{m} ...

    x_{m}= \frac{1}{M} \cdot \{ \int \int_{A} \rho_{A}\cdot x \cdot dx\cdot dy + \int \int_{B} \rho_{B}\cdot x\cdot dx\cdot dy \}=

     =\frac{1}{.085}\cdot \{ \int_{0}^{.1} \int _{.2}^{.7} x\cdot dx\cdot dy + \frac{1}{2}\cdot \int_{.1}^{.5} \int _{.2}^{.3} x\cdot dx\cdot dy + \frac{1}{2}\cdot \int_{.5}^{.6} \int _{0}^{.3} x\cdot dx\cdot dy \}=

     =\frac{1}{.085}\cdot \{ \int_{0}^{.1} .5\cdot x\cdot dx + \frac{1}{2}\cdot \int_{.1}^{.5} .1\cdot x\cdot dx + \frac{1}{2}\cdot \int_{.5}^{.6} .3\cdot x\cdot dx \}=

    =\frac{1}{.085}\cdot \{ .5\cdot |\frac{x^{2}}{2}|_{0}^{.1} + .05\cdot |\frac{x^{2}}{2}|_{.1}^{.5} + .15\cdot |\frac{x^{2}}{2}|_{.5}^{.6} \} = .1970588... (2)


    y_{m}= \frac{1}{M} \cdot \{ \int \int_{A} \rho_{A}\cdot y \cdot dx\cdot dy + \int \int_{B} \rho_{B}\cdot y\cdot dx\cdot dy \}=

     =\frac{1}{.085}\cdot \{ \int_{0}^{.1} \int _{.2}^{.7} y\cdot dx\cdot dy + \frac{1}{2}\cdot \int_{.1}^{.5} \int _{.2}^{.3} y\cdot dx\cdot dy + \frac{1}{2}\cdot \int_{.5}^{.6} \int _{0}^{.3} y\cdot dx\cdot dy \}=


     =\frac{1}{.085}\cdot \{ \int_{.2}^{.7} .1\cdot y\cdot dy + \frac{1}{2}\cdot \int_{.2}^{.3} .4\cdot y\cdot dy + \frac{1}{2}\cdot \int_{0}^{.3} .1\cdot y\cdot dy \}=

    =\frac{1}{.085}\cdot \{ .1\cdot |\frac{y^{2}}{2}|_{.2}^{.7} + .2\cdot |\frac{y^{2}}{2}|_{.2}^{.3} + .05\cdot |\frac{y^{2}}{2}|_{.0}^{.3} \} = .39235294... (3)

    The results are shown in the figure...



    Kind regards

    \chi \sigma
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