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Thread: Moment of inertia

  1. #1
    MHF Contributor arbolis's Avatar
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    Moment of inertia

    I'd like to know where I went wrong :
    A lattice with constant density $\displaystyle \rho (x,y)= \rho$ is on the region under the curve $\displaystyle y= \sin x$ from $\displaystyle x=0$ to $\displaystyle x=\pi$. Find the moments of inertia $\displaystyle I_x$ and $\displaystyle I_y$.
    My attempt : $\displaystyle I_x$ :
    I realize that the region is an upper semi circle with radius $\displaystyle r=1$.
    $\displaystyle I_x=\iint _D y^2 \rho (x,y)dA=\int_0^{\pi} \int_0^{1} (\rho \sin (\theta))^2 \rho ^2 d \rho d\theta=\frac{\pi}{10}$, which mean that $\displaystyle I_x$ does not depend on the mass of the lattice!
    I'm guessing that my error comes from the first step, that is $\displaystyle \iint _D y^2 \rho (x,y)dA=\int_0^{\pi} \int_0^{1} (\rho \sin (\theta))^2 \rho ^2 d \rho d\theta$. Should I replace $\displaystyle y$ by $\displaystyle \sin x$, so $\displaystyle y=\sin (\cos (\theta))$?
    If so, then I don't understand why I can't take $\displaystyle y$ as $\displaystyle \rho \sin (\theta)$...
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  2. #2
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    Quote Originally Posted by arbolis View Post
    I'd like to know where I went wrong :
    A lattice with constant density $\displaystyle \rho (x,y)= \rho$ is on the region under the curve $\displaystyle y= \sin x$ from $\displaystyle x=0$ to $\displaystyle x=\pi$. Find the moments of inertia $\displaystyle I_x$ and $\displaystyle I_y$.
    My attempt : $\displaystyle I_x$ :
    I realize that the region is an upper semi circle with radius $\displaystyle r=1$.
    $\displaystyle I_x=\iint _D y^2 \rho (x,y)dA=\int_0^{\pi} \int_0^{1} (\rho \sin (\theta))^2 \rho ^2 d \rho d\theta=\frac{\pi}{10}$, which mean that $\displaystyle I_x$ does not depend on the mass of the lattice!
    I'm guessing that my error comes from the first step, that is $\displaystyle \iint _D y^2 \rho (x,y)dA=\int_0^{\pi} \int_0^{1} (\rho \sin (\theta))^2 \rho ^2 d \rho d\theta$. Should I replace $\displaystyle y$ by $\displaystyle \sin x$, so $\displaystyle y=\sin (\cos (\theta))$?
    If so, then I don't understand why I can't take $\displaystyle y$ as $\displaystyle \rho \sin (\theta)$...
    You've got to be careful. You are using $\displaystyle \rho $ for two different things here. A density and a changing radius.
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