Question

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).

Answer 1

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:

                            $x_c_m = \frac{M_y}{m} = \frac{\int \int {x*p(x,y)} \, dA }{\int \int {p(x,y)} \, dA } \\\\y_c_m = \frac{M_x}{m} = \frac{\int \int {y*p(x,y)} \, dA }{\int \int {p(x,y)} \, dA } \\\\m = mass = \int \int {p(x,y)} \, dA$

- The coordinates of xcm = ycm, they lie on line y = x.

- Calculate the mass of lamina (m):

                            $m = mass = \int \int {p(x,y)} \, dA = \int\limits^0_a \int\limits_0 {k*(x^2 + y^2)} \, dy.dx \\\\m = k\int\limits^a_0 {(x^2y + \frac{y^3}{3}) } \, dx|\limits^a^-^x_0 = k\int\limits^a_0 {(\frac{-4x^3}{3} + 2ax^2-ax^2+\frac{a^3}{3}) } \, dx\\\\m = k* [ \frac{-x^4}{3} + \frac{2ax^3}{3} - \frac{a^2x^2}{2} +\frac{a^3x}{3}] | \limits^a_0 \\\\m = k* [ \frac{-a^4}{3} + \frac{2a^4}{3} - \frac{a^4}{2} +\frac{a^4}{3}] = \frac{ka^4}{6}$

- Calculate the Moment (My):

                     $M_y = \int \int {x*p(x,y)} \, dA = \int\limits^0_a \int\limits_0 {k*x*(x^2 + y^2)} \, dy.dx \\\\M_y = k\int\limits^a_0 x*{(x^2y + \frac{y^3}{3}) } \, dx|\limits^a^-^x_0 = k\int\limits^a_0 x*{(\frac{-4x^3}{3} + 2ax^2-ax^2+\frac{a^3}{3}) } \, dx\\\\M_y = k* [ \frac{-4x^5}{15} + \frac{ax^4}{2} - \frac{a^2x^3}{3} +\frac{a^3x^2}{6}] | \limits^a_0 \\\\M_y = k* [ \frac{-4a^5}{15} + \frac{a^5}{2} - \frac{a^5}{3} +\frac{a^5}{6}] = \frac{ka^5}{15}$

- 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