Find the center of mass for the lamina of uniform density rh

Find the center of mass for the lamina of uniform density rho occupying the region between the graph f(x) = x and g(x) = -2 for 0 lessthanorequalto x lessthanorequalto 2.

Solution

the figure formed by the lines in the question forms a trapezium

with coordinates of vertices as follows (0,0) (0 , -2) ( 2 , -2) (2 , 2)

this trapezium contains a triangle with vertex (0,0) (2,0) (2,2)

and a rectangle with vertex (0,0) (0, -2) ( 2 ,-2) (2 ,0)

as the density is uniform mass of the triangle is half of rectangle

let the mass of triangle be M mass of rectangle is 2M

centre of mass of the triangle = centroid of the triangle = (0+2+2)/3 , (0+0+2)/3= 4/3 , 2/3

centre of mass of the rectangle = point of intersection of diagonals = (1 , -1)

centre of mass of the figure =

X com = (M x 4/3 + 2M x 1) / 3 M = 10/9

Y com = (Mx 2/3 + 2M x -1 ) / 3M = - 4/9

CENTRE OF MASS = (10/9 , -4/9) ANS