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Homework Help: Center of Mass problem!

  1. Jan 30, 2005 #1
    A uniform square plate L = 5.9 m on a side as a square section d = 2.5 m cut out of one side.

    What is the center of mass of the remaining plate? Use a coordinate system with origin (0, 0) at the center of the plate.

    The answer must be given in components.

    X = ?
    Y = ?. From the picture it is obvious the Y component is 0, so Y = 0 is correct. (and the auto-grade says correct also, so Y = 0 ).

    However, exactly how to find the X component I can't figure out.

    I tried 2-3 different ways of doing it, and none have worked, does anyone know exactly how to solve this problem for the X component? I tried a few things like splitting it up and using weighted averages but it is not working or I have the wrong equation.

    Thanks! I am attaching a picture of the problem.

    Attached Files:

  2. jcsd
  3. Jan 30, 2005 #2
    This should work. Can you show your calculations ?

  4. Jan 30, 2005 #3
    (Assume it is constant 1kg)

    (5.9m(1kg) ) / (1kg) = 5.9m <- from the Y axis, for center 5.9/2 = 2.95m
    This goes for top and bottom sections.

    (5.9m-2.5m)(1kg) / (1kg) <- 3.4m <- from the Y ais, for center 5.9/2 = 1.7m

    1.7m(1kg) + 2.95m(1kg) + 2.95(1kg) / (3kg) = 2.53m

    Since the origin (0,0) is in center of Square, center of x = 2.95m, 2.53 - 2.95 = -.417 m from the origin.

    I dont think this is right, this is my work tho
  5. Jan 30, 2005 #4


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    First of all, the question is not complete. Are we given that the plate is cut out of the center of one side, so that the figure has a line of symmetry parallel to a side of the large square ?

    Failing that condition, we have to take a general case, letting the distance between the top edge of the cut piece and one edge of the main plate be y, and then working out the center of mass in terms of y.

    Divide the figure into 3 rectangles and find the center of mass of each in the coordinate system specified. See attached figure.

    Then figure out the mass of each rectangle. In this case, you can use area as a proxy for mass, since the thickness and density are uniform.

    Finally use the formula [tex]\Sigma m_i(x_i,y_i) = M(X,Y)[/tex] to derive an expression for the overall center of mass.

    This is a rather tedious problem if you have to assume the general case.

    Attached Files:

  6. Jan 30, 2005 #5


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    Decompose the remaining plate into 3 rectangles and use the axis simmetry to find the result...

    Last edited: Jan 30, 2005
  7. Jan 30, 2005 #6
    I am absolutely sure about Y = 0. Besides that, only things known are D and L
  8. Jan 30, 2005 #7


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    A far more elegant solution than decomposing into 3 rectangles is to take the big plate before cutting as one mass with center of mass at the origin. The second mass is a negative mass of the cut portion with the center of mass given in terms of the variable (depending on position). The final mass M is the difference between the two. This cuts down on a lot of calculation and yields the correct answer much more smartly.
  9. Jan 30, 2005 #8


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    Using my last method, here is the short and elegant solution.

    Let the cut portion be taken out of the right edge such that the distance between the top edge of the cut portion and the top edge of the main plate is y.

    M is the mass of the whole plate before cutting, m is the mass of the cut portion. Coordinates represented in square brackets to avoid confusion. [X, Y] is the position of the center of mass of the final figure.

    Then :

    [tex]M[0,0] - m[1.7,1.7 - y] = (M-m)[X,Y][/tex]

    [tex](-(2.5)^2)[1.7, 1.7 - y] = (5.9^2 - 2.5^2)[X,Y][/tex]

    [tex][X,Y] = \frac{6.25}{28.56}[-1.7, (y - 1.7)][/tex]

    Attached Files:

    Last edited: Jan 30, 2005
  10. Jan 30, 2005 #9
    So the answer to the X component would be -1.7? or 6.25/28.56 * -1.7?
  11. Jan 30, 2005 #10


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    The latter. The factor should be multiplied by each coordinate.
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