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Math Help - Center of mass

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    MHF Contributor arbolis's Avatar
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    Center of mass

    A lattice is on the portion of the disk x^2+y^2 \leq 1 in the first quadrant. Find its center of mass if the density at any point is proportional to its distance with the x axis.
    My attempt : I first calculated the total mass of the lattice which gave me \frac{k}{3}, where k is the constant of proportionality.
    Then I found out that the center of mass is located at \left( \frac{3}{8},\frac{3}{16} \right).

    I just want to know if this is the correct result because I find it quite strange.
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    I agree with your total mass and the x coordinate of the c of m. How did you arrive at the y coord.?
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Danny View Post
    I agree with your total mass and the x coordinate of the c of m. How did you arrive at the y coord.?
    I've made many errors and now I'm not able to read my notes. Is y=\frac{3\pi}{16}?
    y=\frac{k \int_0^{\pi/2} \int_0^{1} \rho ^3 \cos ^2 (\theta) d\rho d\theta}{\frac{k}{3}}=\frac{3\pi}{16}.
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    Quote Originally Posted by arbolis View Post
    I've made many errors and now I'm not able to read my notes. Is y=\frac{3\pi}{16}?
    y=\frac{k \int_0^{\pi/2} \int_0^{1} \rho ^3 \cos ^2 (\theta) d\rho d\theta}{\frac{k}{3}}=\frac{3\pi}{16}.
    Yes, that's what I obtained. Side note - most people use r and not \rho for the radius in 2D problems.
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Danny View Post
    Yes, that's what I obtained. Side note - most people use r and not \rho for the radius in 2D problems.
    You're right. I'm the odd one.
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    Two things I notice immediately are that the figure and density function are both symmetric in x and y. The x and y coordinates of the center of mass must be the same.
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    Quote Originally Posted by arbolis View Post
    A lattice is on the portion of the disk x^2+y^2 \leq 1 in the first quadrant. Find its center of mass if the density at any point is proportional to its distance with the \color{red}{x} axis.
    My attempt : I first calculated the total mass of the lattice which gave me \frac{k}{3}, where k is the constant of proportionality.
    Then I found out that the center of mass is located at \left( \frac{3}{8},\frac{3}{16} \right).

    I just want to know if this is the correct result because I find it quite strange.
    Quote Originally Posted by HallsofIvy View Post
    Two things I notice immediately are that the figure and density function are both symmetric in x and y. The x and y coordinates of the center of mass must be the same.
    I beg to differ.
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