Question

Four particles at the corners of a square with a side length L=4.00m are connected by massless rods. The particle masses are m₁= m₄=2.00kg and m₂= m₃ = 16.0 kg. Pairs of particles with equal masses are located at opposite corners of the square. Find the moment of inertia of the system about the z-axis that passes through a corner of the square where the particle has a mass of m=16.00kg.

Answer 1

Moment of Inertia will be,

$I = m_1L^2 + m_4L^2 + m_3(L\sqrt{2})^2 = 4 \times 4^2 + 16 \times 2 \times 4^2 = 576 \ kg.m^2$