Find the centre of mass of 3 particles 100gms,150gms and 200gms at the vertices of an equilateral triangle
Answers
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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kvnmurty
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Rajendrapatel25Ambitious
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) ]
= (200b)*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.