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).

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Answer from: Izzyfizzy

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

Find the moments of inertia Ix, Iy, I0 for a lamina in the shape of an isosceles right triangle with

Answer from: Quest

mr. flanders’ class sold 47 doughnuts and mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations:

$\left\{\begin{array}{l}x+y=85\\1.35x+1.08y=104.49\end{array}\right.\rightarrow \left\{\begin{array}{l}x=85-y\\1.35(85-y)+1.08y=104.49\end{array}\right.$

$114.75-1.35y+1.08y=104.49,\\ \\114.75-104.49=1.35y-1.08y,\\ \\10.26=0.27y,\\ \\y=38,\\ \\x=85-38=47.$

Answer from: Quest

step-by-step explanation:

to find if any function is an inverse of the other, put the first equation in as x in the next one. then simplify and solve. if at the end you get the statement y = x, then it is an inverse. if you get anything else, it is not.

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