A rigid rod of length L= 1 m and mass M = 2.5 kg is attached to a pivot mounted d = 0.17 m from one end. The rod can rotate in the vertical plane, and is influenced by gravity. What is the period for small oscillations of the pendulum shown? Solution Here , for the rod , MOment of inertia , I = m *L^2/12 + m * (L/2 - d)^2 I = 2.5 * (1^2/12) + 2.5 * (1/2 - 0.17)^2 I = 0.481 Kg.m^2 distance from center of mass , x = L/2 - d = 1/2 - 0.17 x = 0.33 m for the time period T = 2pi * sqrt(I/(m * g * x)) T = 2pi * sqrt(0.481/(2.5 * 9.8 * 0.33)) T = 1.53 s the time period is 1.53 s .