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

  • #1
3
0

Homework Statement


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.



Homework 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]


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?
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,813
1,390
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##.
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement


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.



Homework 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]


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?
(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.
 
  • #4
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668

Homework Statement


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.



Homework 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]


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?
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.
 
  • #5
3
0
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.
 
  • #6
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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.
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.
 
  • #7
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
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.
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.)
 
  • #8
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,813
1,390
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.)
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.
 
  • #9
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,813
1,390
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.
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?
 
  • #10
3
0
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!
 
  • #11
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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!
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.
This result doesn't look right to me. It's way too complicated.
vela is right. The result given in Post #5 is way too complicated, and it doesn't look right.
 

Related Threads on Integrals to Solve Area and Center of Mass of a Cut Circle

  • Last Post
2
Replies
26
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
24
Views
4K
Replies
6
Views
11K
Replies
2
Views
5K
  • Last Post
Replies
4
Views
987
Replies
2
Views
11K
Replies
7
Views
16K
Replies
8
Views
517
  • Last Post
Replies
3
Views
3K
Top