Mathematics, 07.04.2020 21:15 nails4life324
Find the moments of inertia Ix, Iy, I0 for a lamina in the shape of an isosceles right triangle with equal sides of length a if the density at any point is proportional to the square of the distance from the vertex opposite the hypotenuse. (Assume that the coefficient of proportionality is k, and that the lamina lies in the region bounded by x = 0, y = 0, and y = a − x).
Answers: 1
![m = mass = \int \int {p(x,y)} \, dA = \int\limits^0_a \int\limits_0 {k*(x^2 + y^2)} \, dy.dx \\\\m = k\int\limits^a_0 {(x^2y + \frac{y^3}{3}) } \, dx|\limits^a^-^x_0 = k\int\limits^a_0 {(\frac{-4x^3}{3} + 2ax^2-ax^2+\frac{a^3}{3}) } \, dx\\\\m = k* [ \frac{-x^4}{3} + \frac{2ax^3}{3} - \frac{a^2x^2}{2} +\frac{a^3x}{3}] | \limits^a_0 \\\\m = k* [ \frac{-a^4}{3} + \frac{2a^4}{3} - \frac{a^4}{2} +\frac{a^4}{3}] = \frac{ka^4}{6}](/tpl/images/0587/3203/252fb.png)
![M_y = \int \int {x*p(x,y)} \, dA = \int\limits^0_a \int\limits_0 {k*x*(x^2 + y^2)} \, dy.dx \\\\M_y = k\int\limits^a_0 x*{(x^2y + \frac{y^3}{3}) } \, dx|\limits^a^-^x_0 = k\int\limits^a_0 x*{(\frac{-4x^3}{3} + 2ax^2-ax^2+\frac{a^3}{3}) } \, dx\\\\M_y = k* [ \frac{-4x^5}{15} + \frac{ax^4}{2} - \frac{a^2x^3}{3} +\frac{a^3x^2}{6}] | \limits^a_0 \\\\M_y = k* [ \frac{-4a^5}{15} + \frac{a^5}{2} - \frac{a^5}{3} +\frac{a^5}{6}] = \frac{ka^5}{15}](/tpl/images/0587/3203/6da26.png)
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Find the moments of inertia Ix, Iy, I0 for a lamina in the shape of an isosceles right triangle with...
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