Find the center of mass of a lamina in the shape of an isosceles right triangle with equal sides of length if the density at any point is proportional to the square of the distance from the vertex opposite the hypotenuse. Assume the vertex opposite the hypotenuse is located at , and that the sides are along the positive axes.
E)None of these

Question

Find the center of mass of a lamina in the shape of an isosceles right triangle with equal sides of length  if the density at any point is proportional to the square of the distance from the vertex opposite the hypotenuse. Assume the vertex opposite the hypotenuse is located at  , and that the sides are along the positive axes.
E)None of these

Quizplus · Accepted Answer

Since the sides of the right triangle are along the positive axes, we can use the coordinates of the vertices as the measurements for the center of mass. Let the vertex opposite the hypotenuse be (0,0), and let the other two vertices be (11,0) and (0,11).

We want to find the center of mass (x, y), where x is the coordinate along the x-axis and y is the coordinate along the y-axis.

Since the density at any point is proportional to the square of the distance from the vertex opposite the hypotenuse, we can use the distance formula to find the density function:

d(x, y) = k((x-11)^2 + (y-11)^2)

We know that the total mass of the lamina is equal to its area multiplied by its density. The area of the isosceles right triangle is (1/2)(11)^2 = 121/2. Thus, the mass of the lamina is:

m = (121/2)k

Next, we use the formulas for the center of mass:

x = (1/m)∬R x dA
y = (1/m)∬R y dA

where R is the region corresponding to the triangle. We can split R into two integrals based on the boundaries of x and y:

x = (1/m)∫0^11 ∫0^x x dA + (1/m)∫0^11 ∫x^11 x dA
y = (1/m)∫0^11 ∫0^x y dA + (1/m)∫0^11 ∫x^11 y dA

Evaluating the integrals and simplifying, we get:

x = 11/3
y = 11/3

Therefore, the center of mass is (11/3, 11/3).

Find the Center of Mass of a Lamina in the Shape