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Math Help - Center of mass

  1. #1
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    Center of mass

    What is the center of mass of the region bounded by y=\sqrt{3}|x| and the line y=3?

    A) (0,0)
    B) (0,1)
    c) (0,2)
    d) (0,3)

    Ok, I know the x-coordinate of the center of mass is 0 because this is an even function. The y-coordinate of the center of mass can be found by this correct? \frac{\int\int_{R}xdydx}{\int\int_{R}dydx}?

    If so I first must find the bounds. I set \sqrt{3}|x|=3 and get that the two graphs intersect at +- \frac{3}{\sqrt{3}} or \sqrt{3}

    So if this correct then the y-coordinate of the center of mass is \frac{\int_{-\sqrt{3}}^{\sqrt{3}}\int_{\sqrt{3}|x|}^{3}xdydx}{\  int_{-\sqrt{3}}^{\sqrt{3}}\int_{\sqrt{3}|x|}^{3}dydx}

    I hope someone can check me up to here. Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Jameson
    What is the center of mass of the region bounded by y=\sqrt{3}|x| and the line y=3?

    A) (0,0)
    B) (0,1)
    c) (0,2)
    d) (0,3)

    Ok, I know the x-coordinate of the center of mass is 0 because this is an even function. The y-coordinate of the center of mass can be found by this correct? \frac{\int\int_{R}xdydx}{\int\int_{R}dydx}?

    If so I first must find the bounds. I set \sqrt{3}|x|=3 and get that the two graphs intersect at +- \frac{3}{\sqrt{3}} or \sqrt{3}

    So if this correct then the y-coordinate of the center of mass is \frac{\int_{-\sqrt{3}}^{\sqrt{3}}\int_{\sqrt{3}|x|}^{3}xdydx}{\  int_{-\sqrt{3}}^{\sqrt{3}}\int_{\sqrt{3}|x|}^{3}dydx}

    I hope someone can check me up to here. Thanks.
    You seem to be making a bit of a meal out of this one.

    The region is a triangle symmetric about the y-axis, with a vertex at (0,0)
    and the two other vertices on the line y=3.

    So the centre of mass is on the axis of symmetry x=0, also it cannot be at
    either (0,0) or (0,3), as these are a vertex and on a side respectively.

    Also it cannot be at (0,1), as this leaves much more area above than below
    the line y=1.

    So we conclude that the centre of mass must be at (0,2) if it is at any of the
    given candidate answers.

    RonL
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  3. #3
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    Good reasoning. I should have also added that on these tests there is always an option E. NOTA (None of the above).
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Jameson
    Good reasoning. I should have also added that on these tests there is always an option E. NOTA (None of the above).
    Actually a simple dissection of the triangle into 9 congruent triangles all similar to
    the big triangle will show that the centre of mass divides the axis of symmetry
    in the ratio of 2:1, again showing that (c) is the answer.

    RonL
    Last edited by CaptainBlack; March 5th 2006 at 02:04 PM.
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  5. #5
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    This is an elementary way to do this problem but it works here. No need to use the formula for finding a lamina just use a compass and straightedge. But that I mean the centroid of this isoseles triangle can easiliy be found by algebra.
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  6. #6
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    Quote Originally Posted by CaptainBlack
    Actualy a simple disection of the triangle into congruent triangles all similar to
    the big triangle will show that the centre of mass divides the axis of symetry
    in the ratio of 2:1, again showing that (c) is the answer.

    RonL
    Sorry, CaptainBlack you posted that before me while I was posting.
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  7. #7
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    Thanks guys. Am I correct in my setup though? Usually these problems don't make regular polygons where geometry can be used like this one.
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Jameson
    What is the center of mass of the region bounded by y=\sqrt{3}|x| and the line y=3?

    A) (0,0)
    B) (0,1)
    c) (0,2)
    d) (0,3)

    Ok, I know the x-coordinate of the center of mass is 0 because this is an even function. The y-coordinate of the center of mass can be found by this correct? \frac{\int\int_{R}xdydx}{\int\int_{R}dydx}?
    This is correct.

    [quote]
    If so I first must find the bounds. I set \sqrt{3}|x|=3 and get that the two graphs intersect at +- \frac{3}{\sqrt{3}} or \sqrt{3}

    So if this correct then the y-coordinate of the center of mass is \frac{\int_{-\sqrt{3}}^{\sqrt{3}}\int_{\sqrt{3}|x|}^{3}xdydx}{\  int_{-\sqrt{3}}^{\sqrt{3}}\int_{\sqrt{3}|x|}^{3}dydx}[QUOTE]

    The limits of integration also look OK.

    RonL
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by Jameson
    Good reasoning. I should have also added that on these tests there is always an option E. NOTA (None of the above).
    Another reason for choosing (c) now that I think of it is that the centre
    of mass of a triangle is the point of intersection of the medians, and that
    the medians divide one another into segments in the ratio 2:1.

    RonL
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