Four particles of masses 1 kg , 2 kg , 3 kg , and 4 kg are placed at the corners A , B , C , and D respectively of a square of side 10 m . The moment of inertia of the system about an axis perpendicular to the plane of the square and passing through the center of mass of the system is ________ kg m ^2 .

Step-by-step solution

Let the corner A be at the origin (0,0) . The coordinates of the corners are: A(0, 0) with m_A = 1 kg B(10, 0) with m_B = 2 kg C(10, 10) with m_C = 3 kg D(0, 10) with m_D = 4 kg Total mass of the system, M = 1 + 2 + 3 + 4 = 10 kg . The coordinates of the center of mass (X_ cm , Y_ cm ) are: X_ cm = m_A x_A + m_B x_B + m_C x_C + m_D x_D M X_ cm = 1(0) + 2(10) + 3(10) + 4(0) 10 = 50 10 = 5 m Y_ cm = m_A y_A + m_B y_B + m_C y_C + m_D y_D M Y_ cm = 1(0) + 2(0) + 3(10) + 4(10) 10 = 70 10 = 7 m The center of mass is at (