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Sagot :
The equation of the line through (0, 1) and (c, 0) is
y - 0 = (0 - 1)/(c - 0) (x - c) ==> y = 1 - x/c
Let L denote the given lamina,
L = {(x, y) : 0 ≤ x ≤ c and 0 ≤ y ≤ 1 - x/c}
Then the center of mass of L is the point [tex](\bar x,\bar y)[/tex] with coordinates given by
[tex]\bar x = \dfrac{M_x}m \text{ and } \bar y = \dfrac{M_y}m[/tex]
where [tex]M_x[/tex] is the first moment of L about the x-axis, [tex]M_y[/tex] is the first moment about the y-axis, and m is the mass of L. We only care about the y-coordinate, of course.
Let ρ be the mass density of L. Then L has a mass of
[tex]\displaystyle m = \iint_L \rho \,\mathrm dA = \rho\int_0^c\int_0^{1-\frac xc}\mathrm dy\,\mathrm dx = \frac{\rho c}2[/tex]
Now we compute the first moment about the y-axis:
[tex]\displaystyle M_y = \iint_L x\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}x\,\mathrm dy\,\mathrm dx = \frac{\rho c^2}6[/tex]
Then
[tex]\bar y = \dfrac{M_y}m = \dfrac{\dfrac{\rho c^2}6}{\dfrac{\rho c}2} = \dfrac c3[/tex]
but this clearly isn't independent of c ...
Maybe the x-coordinate was intended? Because we would have had
[tex]\displaystyle M_x = \iint_L y\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}y\,\mathrm dy\,\mathrm dx = \frac{\rho c}6[/tex]
and we get
[tex]\bar x = \dfrac{M_x}m = \dfrac{\dfrac{\rho c}6}{\dfrac{\rho c}2} = \dfrac13[/tex]
The center of mass for a uniform triangular shape is on its centroid. The y-coordinate of the center of mass of the lamina is 1/3 (independent of c).
What is the center of mass for a triangular shape?
If the surface is plane triangle approximately and mass is uniformally distributed, then its center of mass will lie on the centroid of that triangle.
What is centroid of a triangle and its coordinates?
The point of intersection of a triangle's medians is its centroid (the lines joining each vertex with the midpoint of the opposite side).
If the triangle has its vertices as [tex](x_1, y_1), (x_2, y_2) , \: (x_3, y_3)[/tex], then the coordinates of the centroid of that triangle is given by:
[tex](x,y) = \left( \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{y_1 + y_2 + y_3}{3} \right)[/tex]
For this case, the triangular lamina has vertices (0, 0), (0, 1) and (c, 0)
Assuming its mass is spread regularly, the coordinates of its center of mass would be:
[tex](x,y) = \left( \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{y_1 + y_2 + y_3}{3} \right)\\\\(x,y) = \left( \dfrac{0+0+c}{3} + \dfrac{0+1+0}{3} \right) = (c/3, 1/3)[/tex]
Thus, the y-coordinate of the center of mass of the lamina is 1/3 (independent of c).
Learn more about centroid here:
https://brainly.com/question/7358842
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