Finding the centroid of a semi-circle using integration
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- Опубликовано: 2 июн 2024
- Learn how to use integration to find the centroid (i.e. center of mass) of a semi-circle. This is an application of integration that you will learn in Calculus 2 and also in statics.
The integral formulas for the centroid of a region (center of mass)
• The integral formulas ...
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The integral formulas for the centroid of a region (center of mass)
ruclips.net/video/ij77YxE9elU/видео.html
Now compute the area moment of inertia about the centroid about an axis parallel to the x axis and again about the x-axis. Compare the results using the Parallel Axis Theorem. Signed, an engineer.
**NOOOOOOOOOOOOOOOOOOOO**
*He is not a phy teacher bro*
I just used pappus' centroid theorem + other formula for volume of sphere and it worked more easily
Yup 👍
I just did this problem in my engineering textbook
and more, the rate of change of r is directly proportional to the rate of change of the y coordinate of the centroid!
Can you explain how those formulas were found? There must be a geometric understanding. Are we calculating a point where the area above the point equals the area below the point?
He showed this in another video eight days ago.
Here: The integral formulas for the centroid of a region (center of mass)
ruclips.net/video/ij77YxE9elU/видео.html
When computing y-bar, where does the 1/2 inside the integral sign come from?
Here The integral formulas for the centroid of a region (center of mass)
ruclips.net/video/ij77YxE9elU/видео.html
@@bprpcalculusbasics Makes sense. Thanks!
Why multiply by one half when calculating y centroid?
You don't do that in general. It's for this case specifically, due to the area of the semicircle having a 1/2 factor in it.
@@carultch I see now. I think a better way to show it (especially when explaining to people) would be to say we are integrating y as y goes from 0 to r (instead of -r to +r), so the 1/2 factor is not needed. It makes more sense and matches what is actually happening.
this video explains the formula of the centroids ruclips.net/video/ij77YxE9elU/видео.html
How I think of it: If you were to find the centroid of a shape composed of a finite number of rectangles you essentially take weighted average of the centroid each rectangle. It's fairly easy to find the centroid of a rectangle without any special tools.
So centroid of full shape = Σ [ (Area of rectangle k / Area of full shape) ⋅ centroid of rectangle k ]
Area of full shape is a constant so we can factor that out
centroid of full shape = Σ [ Area of rectangle k ⋅ centroid of rectangle k ] / Area of full shape --> this finite sum is the general structure both the integral expressions follow
A key idea of integral calculus is that if you take approximations of the area under the curve with rectangles, and then take the limiting process of those rectangles as the number of rectangles go to infinity you get the exact area. So if we take the weighted average of all of these rectangles we can get the centroid of the region under any curve.
for the x coordinate: the x coordinate of the centroid of any given rectangle drawn under the curve is x+width/2. The area of any rectangle is f(x) * width. Multiplying the two gets us xf(x)*width + width²/2. The width of each rectangle is already very small so width²/2 is so small we can ignore it. We can think of dx in an integral as the width of the rectangle. We can use an integral to sum up an infinite number of things so we get x̄ = ∫xf(x)dx / Area of full shape
for the y coordinate: the y coordinate of the centroid of any given rectangle is f(x)/2 (this is why there is a 1/2 in the formula for ȳ). Again, the area of any rectangle is f(x) * width. Multiplying the two gets us f(x)²/2 * width. Using simalar logic as before from this we get ȳ = ∫f(x)²/2 dx / Area of full shape
and of course if we need to we can find the area of the full shape using another integral if we need to. It wasn't needed in this case since it was just a semi-circle
The bar on top varieble represent what? (Bad england)
Average
I love mechanics
Today were the Pangellenic exams in math. I think I did good.
Huh. I wonder how this works conceptually/geometrically. I noticed you are effectively dividing the area by itself. But you alter the first area by multiplying the equation by x before integrating. The area then sorta cancels out to give you the average value if x.
But what does that multiplication before integration actually represent?
What you're doing, is taking a "weighted average" of how the area is distributed through space. You calculate the first moment of area (i.e. y*dA) contribution of each area element. You then add up (integrate) to find the total first moment of area, and divide by the area when done.
Moment in general means quantity multiplied by distance from a reference point.
Why is the centroid of a solid hemisphere not the same by symmetry?
One of the two must be wrong.
The centroid is (a,b). Obviously, a=0. b = integral by the semicircle of y / area of this semicircle = integral from 0 to r of 2sqrt(r^2-y^2)ydy / (pi r^2/2) = [y=r sin(t)] = integral from 0 to pi/2 of 2r cos(t) r sin(t) r cos(t) dt / (pi r^2 / 2) = r^3 (integral from 0 to pi/2 of 2cos^2(t) -d(cos(t)) / (pi r^2 / 2) = [u = cos(t)] = 4r/pi * integral from 0 to 1 of u^2 du = 4r/pi * (1^3/3 - 0^3/3) = 4r/(3pi). So the centroid is (0, 4r/(3pi)).
Why can we bring out ½ but not r²?
Since they are both just numbers
Because the 1/2 is being multiplied and you can pull constant factors out of an integral. The r^2 isn't a factor.
I mean, if you wanted to, after pulling the 1/2 out front, you could break the integral into 2 integrals:
2/(pi r^2) [∫r^2 dx - ∫x^2 dx], and then factor the r^2 out of the first one to get:
2/(pi r^2) [r^2 ∫dx - ∫x^2 dx]
But that doesn't really make things all that much simpler than just the single integral.
@@phiefer3 ah, so
Got it
Thanks
So, the centroid of quadrant circle is (4r/3pi , 4r/3pi).
That's right??
yes thats correct
i believe this was the method used to design the atomic bomb
Why the integrands? (xy)dx for xbar, and .5y^2dx for ybar. Seems pulled out of nowhere.
I tried this problem geometrically in my head, and ran into so trouble, so it would be cool to know where I'm coming up stupid
Edit: i loked at the video from last week, I suppose I assumed the centroid means the areas are evenly divided by a vertical and horizontal line at that point, but how could I be so foolish??? That's not how center of mass works
I don't really understand how you designed the initial integral for y. It looks quite different from the x- integral.
... all in all we meet at the same result 😉
yeah same question, ive never done this but the x symmetry seems clear enough - never known about this ybar function though, so is it just true of all centroids ?
this video explains the formula of the centroids ruclips.net/video/ij77YxE9elU/видео.html
How I think of it: If you were to find the centroid of a shape composed of a finite number of rectangles you essentially take weighted average of the centroid each rectangle. It's fairly easy to find the centroid of a rectangle without any special tools.
So centroid of full shape = Σ [ (Area of rectangle k / Area of full shape) ⋅ centroid of rectangle k ]
Area of full shape is a constant so we can factor that out
centroid of full shape = Σ [ Area of rectangle k ⋅ centroid of rectangle k ] / Area of full shape --> this finite sum is the general structure both the integral expressions follow
A key idea of integral calculus is that if you take approximations of the area under the curve with rectangles, and then take the limiting process of those rectangles as the number of rectangles go to infinity you get the exact area. So if we take the weighted average of all of these rectangles we can get the centroid of the region under any curve.
for the x coordinate: the x coordinate of the centroid of any given rectangle drawn under the curve is x+width/2. The area of any rectangle is f(x) * width. Multiplying the two gets us xf(x)*width + width²/2. The width of each rectangle is already very small so width²/2 is so small we can ignore it. We can think of dx in an integral as the width of the rectangle. We can use an integral to sum up an infinite number of things so we get x̄ = ∫xf(x)dx / Area of full shape
for the y coordinate: the y coordinate of the centroid of any given rectangle is f(x)/2 (this is why there is a 1/2 in the formula for ȳ). Again, the area of any rectangle is f(x) * width. Multiplying the two gets us f(x)²/2 * width. Using simalar logic as before from this we get ȳ = ∫f(x)²/2 dx / Area of full shape
and of course if we need to we can find the area of the full shape using another integral if we need to. It wasn't needed in this case since it was just a semi-circle
first!