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Finding a centriod (center of mass)

  1. Oct 4, 2009 #1
    Studying for a test and I can't really grasp a full example problem. I would also like help on a couple further example problems that are not done for me.

    1. The problem statement, all variables and given/known data

    Find the exact coordinates of the centriod:

    a. y=4-x^2, y=0
    b. 3x+2y=6, y=0, x=0
    c. y=e^x, y=0, x=0, x=1
    d. y=1/x, y=0, x=1, x=2

    2. Relevant equations

    equation 1: xbar=(1/A) of integral from a to b of xf(x) dx
    equation 2: ybar=(1/A) of integral from a to b of (1/2)* [f(x)]^2 dx

    A= area
    f(x)=the function

    3. The attempt at a solution

    a.) I have this entire problem worked out for me. I understand why xbar=0, but not how to solve for ybar. In my notes it goes from the line ybar=equation 2 (with f(x) filled in)
    to A= integral of -2 to 2 of (4-x^2)dx
    to 2 * integral from 0 to 2 of (4-x^2)dx
    I don't know why the limits changed or where the 2 came from.

    b-d
    Could someone run down how to do these. Don't complete the problem or anything but I'm not sure about:

    *What to do with the second conditions. (y=1, x=0, y=0 etc.
    *To find x intercepts for integration do I just solve for x?

    It looks to me like you follow these steps:

    1. Find out if either xbar or ybar are zero (odd). Which saves time.
    2. Find area by integrating the function. <<Not clear to me how exactly to do this.
    3. Plug A into given equation
    4. Integrate
    1. The problem statement, all variables and given/known data

    thanks
     
  2. jcsd
  3. Oct 4, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    first think about the shape, in this case an negtive paraboloid bounded by the x axis(y=0)

    so xbar is zero as the shape is symmetric about the y axis

    to compute ybar use the follwing steps, first compute the whole area by either an intergal over x or y, as we are trying to find y bar, use an integral over y (though its a little trickier to set-up)

    now for a given y, is what is the area, dA(y), of a sliver of width dy? rearranging for x

    [tex] x = \sqrt{4-y} [/tex]
    then counting the contribution on each side of teh y axis gives:
    [tex] dA(y) = 2.x.dy = 2.sqrt(4-y).dy [/tex]
    to find the area integrate from the x axis to the turning point
    [tex] A = \int dA(y) = \int_0^4 2.sqrt(4-y).dy [/tex]
    note you could set this up in termes of x and it would be a bit easier, though we need it again later

    to work out the centre of mass now sum up the contribution of each area element using an integral
    [tex] \bar{y}.A = \int y.dA(y) = \int_0^4 2.y.sqrt(4-y).dy [/tex]

    the the average centre of mass is given by
    [tex] \bar{y} = \frac{\int y.dA(y)}{\int dA(y)} = \frac{\bar{y}.A }{A} [/tex]
     
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