Answer:
- centroid: (x, y) = (81.25 mm, 137.5 mm)
- I = 8719.31 mm^2 for unit mass
Explanation:
Finding the desired measures requires we know a differential of area. That, in turn, requires we have a way to describe a differential of area. Here, we choose to use a vertical slice, which requires we know the area boundaries as a function of x.
The upper boundary is a line with a slope of 125/156.25 = 0.8, and a y-intercept of 125. That is, ...
y1 = 0.8x +125
The lower boundary is given in terms of y, but we can solve for y to find ...
100x = y^2
y2 = 10√x
Then our differential of area is ...
dA = (y1 -y2)dx
__
The centroid is found by computing the first moment about the x- and y-axes, and dividing those values by the area of the figure.
The area will be ...
[tex]\displaystyle A=\int_0^{156.25}{dA}=\int_0^{156.25}{(y_1-y_2)}\,dx[/tex]
The y-coordinate of the centroid is ...
[tex]\displaystyle \overline{y}=\dfrac{S_x}{A}=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}}\,dA=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}(y_1-y_2)}\,dx=137.5[/tex]
Similarly, the x-coordinate is ...
[tex]\displaystyle \overline{x}=\dfrac{S_y}{A}=\dfrac{1}{A}\int_0^{156.25}{x}\,dA=\dfrac{1}{A}\int_0^{156.25}{x(y_1-y_2)}\,dx=81.25[/tex]
That is, centroid coordinates are (x, y) = (81.25, 137.5) mm.
__
The moment of inertia is the second moment of the area. If we normalize by the "mass" (area), then the integral looks a lot like the one for [tex]\overline{x}[/tex], but multiplies dA by x^2 instead of x.
The attachment shows that value to be ...
I ≈ 8719.31 mm^2 (normalized by area)
The area is 16276.0416667 mm^2, if you want to "un-normalize" the moment of inertia.
