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Center of mass

  1. Jun 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Given a homogenous rectangular surface, sides of length a and b=4*a, with a circular hole in x=a and y=a/2, find the center of mass.

    jf7uj9.jpg

    2. Relevant equations

    R=1/M*Ʃmi*ri , M= total mass, r= position vector

    area of a circle = pi*r2 , r being the radius

    3. The attempt at a solution

    I determined the center of mass of the surface and circle and tried to subtract one from another, however the result did not agree with the solutions.

    Acircle=pi*r2 = pi*a2/4
    Arectangle= a*4a = 4a2
    Center of mass of the rectangle alone:
    Rr=(2a,a/2)
    Center of mass of the circle alone:
    Rc=(a,a/2)

    R=[Rr*Arectangle - Rc*Acircle] / Total area ⇔
    ⇔R=a*(32-pi)/(16+pi)

    The correct center of mass is (2,05,a/2)

    Thanks!

    D.
     
  2. jcsd
  3. Jun 15, 2012 #2

    vela

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    You need to subtract the areas in the denominator because you're considering the circle to have negative mass. Right now, your calculation is for the solid slab plus a positive-mass circle located at (-a, -a/2).

    Also, to get the answer you cited, I think you need a different radius for the circle. Is the radius of the circle really equal to a/2? If so, you won't get that answer.
     
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