Question

s) Use integration with respect to x (only) to find the center of mass of the lamina with uniform density and with the given shape shown. (see picture). The area of the lamina can be determined using simple geometry without the use of calculus. The lamina is bound by the equations y-4,x-0, y 2x. 4ys4 (2,4) a) Area 0 b)X

Answer 1

In triangle OAB, area of the square = (4 times 2)m^2 = 8 m^2 rightarrow area of triangle = 1/2 times 8 A = 4 m^2 x = integral x dm/integral dm (Because mass, m = rho A rightarrow dm = rho dA) = integral x rho dA/integral rho dA = integral integral x dx dy/integral integral dx dy (Because dA = dx dy) (integral dA = total area = 4) = 1/4 times integral^4_0 integral^y/2_0 x dx dy (Because y is going from 0 to 4 x is going from 0 to y/2) rightarrow 1/4 times integral^4_0 1/2 y^2/4 dy x = 1/32 times [y^3/3]^4_0 = 1/3 times 32 times 64 rightarrow x = 2/3 = 0.6667 y = integral y dm/integral dm = integral y rho dA/integral rho dA = integral y dA/integral dA.