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Find the cetre of mass of three particles at the vertices of an equilateral triangle. Themasses of the particles are 100g, 150g and 200g respectively. Each side of the equilateraltriangle is 0.5m long.*(Mar-18),Ts-Mar-18 |
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Answer» Let m1 = 100 g. m2 = 150 g. m3 = 200 g. a = 0.5 m = 50 cm. Center of mass coordinates : x = [-a/2 * m1 + 0 * m2 + a/2 * m3] / (m1 + m2 + m3) = (m3 - m1) a /[2 (m1+m2+m3) ] = (200 - 100)*50 / [2 * 450] cm. = 50/9 cm y = [0 * m1 +√3/2 * a * m2 + 0 * m3 ] / (m1 + m2 + m3) =√3/2 * a * m2 /(m1 + m2 + m3) =√3/2 * 50 * 150 /450 cm = 25/√3 cm y coordinate of the centroid of the triangle: 1/3 * √3/2 * a = 25/√3 cm Thus the center of mass is 50/9 cm to the right of the centroid G of the triangle ABC. |
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