Question

Find the centre of mass of three particles at the vertices of an equilateral triangle .the masses of the particles are 100g,150g, and 200g respectively. each side of the equilateral triangle is 0 .5m long

Answer 1

See the diagram.

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.

Answer 2

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.