Centre of mass of the given system of particles will be at

Step-by-step solution

Let the square ABCD have side length a . We place the origin at the center O of the square. The coordinates of the vertices are A(-a/2, a/2) , B(a/2, a/2) , C(a/2, -a/2) , and D(-a/2, -a/2) . The masses at the vertices are m_A = 2m , m_B = 4m , m_C = 2m , and m_D = 2m . The coordinates of the center of mass (X_ cm , Y_ cm ) are given by: X_ cm = m_A x_A + m_B x_B + m_C x_C + m_D x_D m_A + m_B + m_C + m_D = 2m(-a/2) + 4m(a/2) + 2m(a/2) + 2m(-a/2) 2m + 4m + 2m + 2m = -ma + 2ma + ma - ma 10m = ma 10m = a 10 Y_ cm = m_