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Math Help - Compute the moment of inertia

  1. #1
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    Compute the moment of inertia

    Hi,

    Can anybody help with this question please:

    Compute the moment of inertia of the object in the attached figure (a thin disk missing a portion, where the angle is θ is 45) with respect to the centre point, P.

    Note: 1/(cosx)^4 .dx = [sinx / 3(cosx)^3] + [2tanx/3]. The integral can be split into two integrals, one over a sector of disk one over a triangle.
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  2. #2
    MHF Contributor chisigma's Avatar
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    The general formula for the moment of inertia is…

    I_{z}= \int_{V} \rho \cdot r^{2} \cdot dv (1)


    The object you have proposed is a sort of cilinder that i suppose homogeneous and of height H. In this case the (1), developped in cilindrical coordinates, becomes…

    I_{z}= H \cdot \rho \cdot (\int_{\frac{\pi}{4}}^{\frac{7 \pi}{4}} \int r^{3} \cdot d \phi \cdot dr + 2 \cdot \int_{0}^{\frac{\pi}{4}} \int_{0}^{R \cdot \cos \phi} r^{3} \cdot d \phi \cdot dr) (2)

    The first integral in (2) gives…

    H \cdot \rho \cdot \int_{\frac{\pi}{4}}^{\frac{7 \pi}{4}} |\frac {r^{4}}{4}|_{0}^{R} \cdot d\phi= \frac{3\cdot \pi}{8} \cdot R^{4} H \cdot \rho

    For the second integral in (2) you have to keep in mind that…

    \int \cos^{4} \phi \cdot d \phi = \frac{1}{32} \cdot (12 \cdot \phi + 8 \cdot \sin 2 \phi + \sin 4 \phi) + c

    … so that…

    H \cdot \rho \cdot \int_{0}^{\frac{\pi}{4}}|\frac {r^{4}}{4}|_{0}^{R \cdot \cos \phi} \cdot d \phi= \frac{1}{2} \cdot H \cdot \rho \cdot R^{4} \cdot \int_{0}^{\frac{\pi}{4}} \cos^{4} \phi \cdot d\phi= \frac{1}{2} \cdot H \cdot \rho \cdot R^{4} \cdot (\frac{3 \cdot \pi}{32} + \frac{1}{4})

    The requested moment of inertia so is…

     I_{z}= H \cdot \rho \cdot R^{4} \cdot (\frac{27 \cdot \pi}{64} + \frac{1}{8})= H \cdot \rho \cdot R^{4} \cdot 1,4503594 \dots

    In case of a ‘perfect cilinder’ the moment of inertia would be…

     I_{z}= H \cdot \rho \cdot R^{4} \cdot \frac{\pi}{2}= H \cdot \rho \cdot R^{4} \cdot 1,5707963\dots

    … and that is not a surprise…

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor chisigma's Avatar
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    My previous message did contain errors ... i apologize for that and reply the message with the new solution... i hope without errors ...

    The general formula for the moment of inertia is…

    I_{z}= \int_{V} \rho \cdot r^{2} \cdot dv (1)


    The object you have proposed is a sort of cilinder that i suppose homogeneous and of height H. In this case the (1), developped in cilindrical coordinates, becomes…

    I_{z}= H \cdot \rho \cdot (\int_{\frac{\pi}{4}}^{\frac{7 \pi}{4}} \int r^{3} \cdot d \phi \cdot dr + 2 \cdot \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac {R}{\sqrt {2} \cdot \cos \phi}} r^{3} \cdot d \phi \cdot dr) (2)

    The first integral in (2) gives…

    H \cdot \rho \cdot \int_{\frac{\pi}{4}}^{\frac{7 \pi}{4}} |\frac {r^{4}}{4}|_{0}^{R} \cdot d\phi= \frac{3\cdot \pi}{8} \cdot R^{4} H \cdot \rho

    For the second integral in (2) you have to keep in mind that…

    \int \cos^{-4} \phi \cdot d \phi = \frac{1}{3} \cdot (sec^{2} \phi +2) \cdot tan \phi + c

    … so that…

    H \cdot \rho \cdot \int_{0}^{\frac{\pi}{4}}|\frac {r^{4}}{4}|_{0}^{\frac {R}{\sqrt {2} \cdot \cos \phi}} \cdot d \phi= \frac{1}{4} \cdot H \cdot \rho \cdot R^{4} \cdot \int_{0}^{\frac{\pi}{4}} \cos^{-4} \phi \cdot d\phi= \frac{1}{3} \cdot H \cdot \rho \cdot R^{4}

    The requested moment of inertia so is…

     I_{z}= H \cdot \rho \cdot R^{4} \cdot (\frac{3\cdot \pi}{8} + \frac{1}{3})= H \cdot \rho \cdot R^{4} \cdot 1,5114305 \dots


    In case of a ‘perfect cilinder’ the moment of inertia would be…

     I_{z}= H \cdot \rho \cdot R^{4} \cdot \frac{\pi}{2}= H \cdot \rho \cdot R^{4} \cdot 1,5707963\dots

    … and that is not a surprise…

    Kind regards

    \chi \sigma
    Last edited by chisigma; March 15th 2009 at 12:08 PM.
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