1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integrals to Solve Area and Center of Mass of a Cut Circle

  1. Jan 16, 2016 #1
    1. The problem statement, all variables and given/known data
    I am after finding the centroid of the remaining area (hatched) when a circle is cut by a line. I made a diagram in CAD that demonstrates the problem.

    lw8F3pr.jpg

    The idea is that, starting from the bottom of the circle, a cut is taken leaving a remaining shape whose area and centroid can be calculated. [itex]dc[/itex] denotes the height from the bottom of the circle to the center of mass and [itex]c[/itex] is the height from the bottom of the circle where the cut line is.

    I believe the problem I am having is to do with the equation of the circle being written in polar coordinates and the equation of the line is in cartesian coordinates.



    2. Relevant equations
    The mass of a lamina is per the following definition:

    [tex]M = ρ\int_{a}^{b} f(x)-g(x) dx[/tex]

    Where I am taking:
    $$f(x) = \sqrt{r^2-x^2}$$ and,
    $$g(x) = c$$

    I am ignoring the density [itex]ρ[/itex] in the mass equation above because it will cancel out when I divide using moments to solve for the center of mass:

    [tex]M_x = ρ\int_{a}^{b} \frac{1}{2}([f(x)]^2-[g(x)]^2) dx[/tex]

    Thus the center of mass of a lamina is:

    [tex]\bar{x}=\frac{M_x}{M}=\frac{\int_{a}^{b} \frac{1}{2}([f(x)]^2-[g(x)]^2) dx}{\int_{a}^{b} f(x)-g(x) dx}=\frac{1}{A}\int_{a}^{b} \frac{1}{2}([f(x)]^2-[g(x)]^2) dx[/tex]


    3. The attempt at a solution
    My first step is to use the same coordinate system for both the line at distance [itex]c[/itex] and the circle.

    The equation of the circle is thus:
    [tex](x-h)^2+(y-k)^2=r^2[/tex]
    rewritten in terms of y:
    [tex]f(x) = \sqrt{r^2-(x-h)^2}+k[/tex]
    Where [itex](h,k)[/itex] represents in the cartesian [itex](x,y)[/itex] coordinates that define the center of the circle. Now is makes more sense that I also use:
    [tex]g(x) = c[/tex]
    Next step is solving for the bounds on the above intergrals [itex]a,b[/itex].
    [tex]x = \sqrt{r^2-(c-k)^2}+h[/tex]
    Thus:
    [tex] b = +\sqrt{r^2-(c-k)^2}+h[/tex]
    [tex] a = -\sqrt{r^2-(c-k)^2}+h[/tex]

    Mass integral with constant density:
    [tex]M = ρ\int_{-\sqrt{r^2-(c-k)^2}+h}^{+\sqrt{r^2-(c-k)^2}+h} (\sqrt{r^2-(x-h)^2}+k-c) dx[/tex]
    I want to hold here for some further advice because the resulting expression is quite large. Does anyone have any comments on my working so far? Is there an easier approach? Perhaps remaining in polar coordinates?
     
  2. jcsd
  3. Jan 16, 2016 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The way you've set your integrals up, you're omitting the left and right ends of the circle. I suggest you integrate with respect to ##y##.

    I'm not sure how you got ##M_x = \frac \rho 2 \int (f^2-g^2)\,dx##.
     
  4. Jan 17, 2016 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    (1) Polar coordinates are the worst way to go.
    (2) Why do you choose to integrate with respect to x? Since you have a 2D region, you could equally well integrate with respect to y.
     
  5. Jan 17, 2016 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Instead of frying your brain trying to write integrals to calculate the area and moment of the shaded part, why not try calculating the area and moment of the circular segment which is not shaded? You can write a couple of integral expressions for the area and moment of this segment which are much simpler than trying to do it for the rest of the circle. You can always subtract the area and moment of the segment from the whole circle, and then find an expression for locating the centroid of the shaded area using algebra.

    You should also take advantage of symmetry to simplify calculations as well. Look at this figure:


    706-quarter-circle-differential-strip.gif

    By using the setup from the figure above, you can write a couple of integral expressions for the area and moment of the segment and evaluate them without having to lobotomize yourself in the process.
     
  6. Jan 18, 2016 #5
    Thank you all for your input.

    SteamKing - your approach was great thank you for your help.

    Final solution:

    [tex]C_y = \frac{\pi r^3-\frac{r^2\left(r+\frac{4rsin^3(\frac{\theta}{2})}{3(\theta-sin\theta)}\right)\left(\theta-sin\theta\right)}{2}}{\pi r^2-\frac{r^2\left(\theta-sin\theta\right)}{2}}[/tex]

    Fairly certain this only works between 0 and r which is precisely what I was after.

    Thanks again.
     
  7. Jan 18, 2016 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You should check your results against these values for the centroid:

    http://www.efunda.com/math/areas/circularsegment.cfm

    If you set θ = π radians in your expression for Cy, you should get ##C_y=r+\frac{4r}{3π}##, which is the centroid of a semi-circle, measured as indicated by the distance dc in the diagram in the OP.
     
  8. Jan 18, 2016 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Alternately, if you had integrated wrt y instead of x you would have gotten ##\text{Area} = A = \int_{-(r-c)}^r 2 \sqrt{r^2 - y^2} \, dy ##
    and the y-component of the centroid involves the integral ##\int_{-(r-c)}^r 2 y \sqrt{r^2 - y^2} \, dy##, which is particularly elementary. (Of course, the x-component of the centroid is zero, by symmetry.)
     
  9. Jan 18, 2016 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Also, if you did it this way, it's essentially the same calculation for both the portion of the circle cut off and the part you were actually interested in. The only difference would have been the limits. Steamking's suggestion had occurred to me, but I realized it wasn't actually all that helpful for this particular problem.
     
  10. Jan 18, 2016 #9

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    This result doesn't look right to me. It's way too complicated. And what is ##\theta##? How come ##c## doesn't appear in your answer?
     
  11. Jan 18, 2016 #10
    I simplified the problem definition by using a set of expressions that already solved for the circle segment rather than deriving anything myself. [itex]\theta[/itex] is a function of the cut location [itex]c[/itex].

    CircularSegmentGen00.gif

    I subtracted moments between a full circle and the shape above to leave me with the "remaining" geometry. The same applies for the area. Dividing the moments by area leaves me the aforementioned expression. SteamKing's approach is correct.

    vela - good thing I made sure it was right posted the solution!
     
  12. Jan 19, 2016 #11

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    If you use this figure as a guide to calculating the moments of the full circle and the circular segment, you can make this calculation simpler by computing moments about the center of the full circle.

    In this case, the moment of the full circle about the center is zero, since the center point lies on all axes of symmetry for the circle.

    The moment of the circular segment is Ms = As ⋅ Cy, where As is the area of the segment.

    The centroid of the remaining area will be displaced a distance ##\bar y## below the center of the full circle,

    $$\bar y = \frac{0 - A_s ⋅ C_y}{πr^2 - A_s}$$

    If you want to refer ##\bar y## from the center of the full circle to say the point on the +y axis where the full circle crosses (0, y), then

    $$\bar y' = r - \bar y$$,

    recognizing that ##\bar y## as formulated above will be negative and lie on the negative y-axis somewhere.
    vela is right. The result given in Post #5 is way too complicated, and it doesn't look right.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Integrals to Solve Area and Center of Mass of a Cut Circle
Loading...