Three particles, each having a mass of 200 g , are placed at the corners of an equilateral triangle with a side length of 10 cm . Determine the moment of inertia of the system about an axis joining two of the particles.

Step-by-step solution

Let the three particles be A, B, and C, each of mass m = 200 g = 0.2 kg . Side length of the equilateral triangle, a = 10 cm = 0.1 m . Let the axis of rotation pass through particles A and B. The perpendicular distances of particles A and B from the axis are zero ( r_A = 0 , r_B = 0 ). The perpendicular distance of particle C from the axis AB is the altitude of the equilateral triangle: r_C = a 60^ = 3 2 a The moment of inertia of the system about the axis AB is given by: I = m_A r_A^2 + m_B r_B^2 + m_C r_C^2 I = m