Question

Determine the centroid of the shaded area shown in figure 2. Determine the moment of inertia about y-axis of the shaded area shown in figure 2

Answer 1

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

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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 ...

  $\displaystyle A=\int_0^{156.25}{dA}=\int_0^{156.25}{(y_1-y_2)}\,dx$

The y-coordinate of the centroid is ...

  $\displaystyle \overline{y}=\dfrac{S_x}{A}=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}}\,dA=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}(y_1-y_2)}\,dx=137.5$

Similarly, the x-coordinate is ...

  $\displaystyle \overline{x}=\dfrac{S_y}{A}=\dfrac{1}{A}\int_0^{156.25}{x}\,dA=\dfrac{1}{A}\int_0^{156.25}{x(y_1-y_2)}\,dx=81.25$

That is, centroid coordinates are (x, y) = (81.25, 137.5) mm.

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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 $\overline{x}$, 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.