Answer:
Ix = Iy = a^4 / 12
Ixy = a^4 / 24
Step-by-step explanation:
Solution:-
- Sketch the right angled isosceles triangle as shown in the attachment.
- The density (ρ) is a multivariable function of both coordinates (x and y).
ρ ( x, y ) = k*(x^2 + y^2)
Where, k: coefficient of proportionality.
- The lamina and density are symmetrical across the line y = x. (see attachment). The center of mass must lie on this line.
- The coordinates of centroid ( xcm and ycm) are given by:
[tex]x_c_m = \frac{M_y}{m} = \frac{\int \int {x*p(x,y)} \, dA }{\int \int {p(x,y)} \, dA } \\\\y_c_m = \frac{M_x}{m} = \frac{\int \int {y*p(x,y)} \, dA }{\int \int {p(x,y)} \, dA } \\\\m = mass = \int \int {p(x,y)} \, dA[/tex]
- The coordinates of xcm = ycm, they lie on line y = x.
- Calculate the mass of lamina (m):
[tex]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}[/tex]
- Calculate the Moment (My):
[tex]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}[/tex]
- Calculate ( xcm = ycm ):
xcm = ycm = ( ka^5 /15 ) / ( ka^4/6) = 2a/5
- Now using the relations for Ix, Iy and I: We have:
Ix = bh^3 / 12
Iy = hb^3 / 12
Where, h = b = a .... (Right angle isosceles)
Ix = Iy = a^4 / 12
Ixy = b^2h^2 / 24
Ixy = a^4 / 24
