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

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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).

Mathematics

1 answer:

Svet_ta [12.7K] · Answer 1 · 2026-04-11T12:54:13+03:00

Answer:Ix = Iy = a^4 / 12

Ixy = a^4 / 24

Step-by-step explanation:

Solution:-

- Start by sketching the right-angled isosceles triangle as illustrated in the attachment.

- The density (ρ) functions multivariably based on coordinates (x and y). ρ ( x, y ) = k*(x^2 + y^2)

Here, k represents the coefficient of proportionality.

- The lamina and density are symmetrical across the line y = x. (refer to the attachment). The center of mass must be along this line.

- The coordinates of centroid ( xcm and ycm) are represented as:

- Given that xcm = ycm, these points lie along the line y = x.

- Compute the mass of the lamina (m): $x_c_m = \frac{M_y}{m} = \frac{\int \int {x*p(x,y)} \, dA }{\int \int {p(x,y)} \, dA } \\\\y_c_m = \frac{M_x}{m} = \frac{\int \int {y*p(x,y)} \, dA }{\int \int {p(x,y)} \, dA } \\\\m = mass = \int \int {p(x,y)} \, dA$

- Evaluate the Moment (My):

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

- Determine ( xcm = ycm ):

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

2a/5

- Now utilizing the relations for Ix, Iy and I: We find: 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