Question

A. Find the centroid of each of the following systems of masses.

1. Equal masses of 100 kg at (0, 3), (2, 2) and (2, -1).

2. Masses of 200 lb, 500 lb and 1000 lb at (3, 2), (4, 0) and (1, 5).

B. Find the centroid of each of the areas bounded by the following curves.

1. y = 6 - 2x ; x = 0 ; y = 0

2. Y = X³, Y = 4X

Answer 1

$A.1\newline \text{Let (x,y) be the centroid.}\newline x=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}\newline y=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3} \newline \text{Given, mass is 100 kg at each point.} \newline x=\frac{100×2+100×2+100×0}{100+100+100}=\frac{4}{3}\newline y=\frac{100×2+100×(-1)+100×3}{100+100+100}=\frac{4}{3}\newline \text{Thus, centroid is} (\frac{4}{3}, \frac{4}{3}).\newline A.2\newline \text{Given, mass is 200 lb, 500lb and 1000lb at (3, 2), (4, 0) and (1, 5) respectively.}\newline x=\frac{200×3+500×4+1000×1}{200+500+1000}=\frac{3600}{1700}=\frac{36}{17}\newline y=\frac{200×2+500×0+1000×5}{200+500+1000}=\frac{5400}{1700}=\frac{54}{17}\newline \text{Thus, centroid is} (\frac{36}{17}, \frac{54}{17}).\newline B.1.\newline area(A)=\int_0^3y(x)dx=\int_0^3(6-2x)dx=(6x-x^2)_0^3=9\newline \text{Let (X, Y) be the centroid.}\newline X=\frac{\int_0^3xy(x)dx}{A}=\frac{1}{9}\int_0^3(6x-2x^2)dx=\frac{1}{9} (3x^2-\frac{2x^3}{3})_0^3=1\newline Y=\frac{\frac{1}{2}\int_0^3y^2(x)dx}{A}\newline =\frac{1}{9}.\frac{1}{2}\int_0^3(6-2x)^2dx\newline =\frac{1}{18}\int_0^3(36-24x+4x^2)dx\newline =\frac{1}{18}(36x-12x^2+\frac{4x^3}{3})_0^3\newline =2\newline \text{Thus, centroid is} (1, 2).\newline B.2.\newline area(A)=\int_0^2(4x-x^3)dx+\int_{-2}^0(x^3-4x)dx \newline =(2x^2-\frac{x^4}{4})_0^2+(\frac{x^4}{4}-2x^2)_{-2}^0\newline =8\newline \text{Let} ( \alpha, \beta)\text{be the centroid.}\newline \alpha=\frac{\int_0^2x(4x-x^3)dx+\int_{-2}^0x(x^3-4x)dx}{A}\newline =\frac{\int_0^2(4x^2-x^4)dx+\int_{-2}^0(x^4-4x^2)dx}{A}\newline =\frac{1}{8}((\frac{4x^3}{3}-\frac{x^5}{5})_0^2+(\frac{x^5}{5}-\frac{4x^3}{3})_{-2}^0)\newline =0\newline \beta=\frac{\frac{1}{2}( \int_0^2(4x-x^3)^2dx+\int_{-2}^0(x^3-4x)^2dx)}{A}\newline =\frac{\int_0^2(16x^2-8x^4+x^6)dx+\int_{-2}^0(16x^2-8x^4+x^6)}{16}\newline =\frac{(\frac{8x^3}{3}-\frac{8x^5}{5}+\frac{x^7}{7})_{0}^2+(\frac{8x^3}{3}-\frac{8x^5}{5}+\frac{x^7}{7})_{-2}^0}{16}\newline =\frac{128}{105}\newline \text{Thus, centroid is} (0, \frac{128}{105}).$