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Finding the center of mass of laminae

  1. May 3, 2012 #1
    Calculus 3 project – any and all help is appreciated.
    We've just gone over center of mass with double integrals, so it's a bit peculiar to see this project feature only one integral. I went over that in calc 2 – and as a result, know how to calculate it that way.

    However (as you will see in the formula below), I don't think he wants me to calculate it that way. I have no idea how to proceed and the book is of absolutely no help.

    Anything you could do to help me out would be fantastic.

    And here's the problem:

    1. The problem statement, all variables and given/known data
    There are three type of laminae we are trying to find:
    1) A triangle with sides 3", 4", and 5" respectively.
    2) A semicircle
    3) A "horseshoe", i.e. a half-ring, whose inner and outer edges are composed of circular arcs of radii 3" and 5" respectively.

    Definition 1:
    The first moment of a planar mass distribution about a line l is the integral over the region of the (area) density σ(P) times the distance of P from the line. For example, the first moment about the line x=k is
    [itex]M_k=∫_R\left(x-k\right) σ\left(P\right) dA[/itex]

    Definition 2:
    In order for a line to be a balancing line, the first moment about this line must be zero.

    2. Relevant equations
    See the M_k above
    I found, I believe, the equations for the laminae:
    For the triangle,
    [itex]y=4-\frac{4}{3}x[/itex]
    For the semicircle,
    [itex]x^2+y^2=1[/itex]
    And the horseshoe,
    [itex]x^2+y^2=25[/itex]
    [itex]x^2+y^2=9[/itex]

    3. The attempt at a solution
    Welp, I'm lost. As I said, I can figure it out easily using the calculus 2 method of [itex]x_{cm}=\frac{M_y}{M}[/itex], but I'm not sure the way he wants it done.

    What's the density formula? Is it just the equations I listed above? What's this (x-k) nonsense?
     
  2. jcsd
  3. May 3, 2012 #2
    *cricket*
    Anyone?
     
  4. May 3, 2012 #3
    The [itex]M_k[/itex] gives, roughly speaking, the potential torque on the lamina if one were to try to balance it on the line [itex]x=k[/itex]. In order for the lamina to be balanced on that line, this "tourque" would need to be zero.

    If you set [itex]M_k=0[/itex] and solve for [itex]k[/itex], I think you'll see what the "[itex]x-k[/itex] nonsense" is for.
     
  5. May 3, 2012 #4

    HallsofIvy

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    Wow, you waited a whole hour before bumping? Sorry, I was eating supper!
     
  6. May 3, 2012 #5

    HallsofIvy

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    You are not given any "density" function so this is a purely mathematical problem, not physics. You are really looking for the "centroid", not the "center of mass". That is the same as assuming the "density" is a constant.

    This is a single lamina. "laminae" is the plural.

    Well, you mean [itex]x_{cm}=\frac{M_x}{M}[/itex], and [itex]y_{cm}=\frac{M_y}{M}[/itex]

    x- k is, just your quote above says, the horizontal distance from point (x, y) to the line x= k.
    So this is a right triangle. Set up a coordinate system so the right angle is at the origin, one vertex is to (3, 0), and another at (0, 4). The equation of the hypotenuse is y= -(4/3)x+ 4. The area of the triangle is, of course (1/2)(3)(4)= 6 so the x-coordinate of the centroid is [itex](1/6)\int_{x=0}^3\int_{y= 0}^{-(4/3)x+ 4} x dydx[/itex] and the y coordinate is [itex](1/6)\int_{x=0}^3\int_{y= 0}^{-(4/3)x+ 4} x dydx[/itex].

    The area of a semicircle of radius R (you don't give a radius) is [itex]\pi R^2/2[/itex]. It's probably best to use polar coordinates to do the integrals:
    [tex](2/\pi R^2)\int_{r= 0}^R\int_{\theta= 0}^{\pi} x d\thetadr= (2/\pi R^2)\int_{r= 0}^R\int_{\theta= 0}^{\pi} rcos(\theta) d\thetadr[/tex]
    (It might occur to you that, by symmetry, the x coordinate of the centroid is 0)
    and
    [tex](2/\pi R^2)\int_{r= 0}^R\int_{\theta= 0}^{\pi} y d\thetadr= (2/\pi R^2)\int_{r= 0}^R\int_{\theta= 0}^{\pi} rsin(\theta) d\thetadr[/tex]

    So this is the area between two semi-circles. Its area is [itex](pi(5)^2/2- \pi(3)^2/2= 8\pi[/itex].The coordinates of the centroid are
    [tex](1/8\pi)\int_{r=3}^5\int_{\theta= 0}^\pi r cos(\theta)d\theta dr[/tex]
    (Again, by symmetry, this should be 0.)
    and
    [tex](1/8\pi)\int_{r=3}^5\int_{\theta= 0}^\pi r sin(\theta)d\theta dr[/tex]
     
    Last edited: May 3, 2012
  7. May 3, 2012 #6
    Ah, you both are absolutely fantastic! Thank you.
     
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