Question

2-43. In order to determine the mass moment of inertia of a wheel, it is supported on a knife-edge, as shown and permitted to oscillate with small angular motion. The period is measured as 1.14 sec. Determine the moment of inertia of the wheel about its mass-center axis at C. The wheel has a mass of 34 kg 냐 寸 Problem 2-43

Answer 1

Time period of oscillator = 1.14 sec therefore T = 2 A squareroot m/k m = 34 kg k = radius of gyration = squareroot I/A 1.14 = 2 pi squareroot 34/ k k = 1032.83 m A = pi/4 times (0.4^2 - 0.3^2) = 0.055 m^2 I_D = k^2 A I_D = 58617.24 m^4 therefore I_D = I_C + delta times d^2 I_D = I_C + A times d^2 = I_D - 0.055 times 0.15^2 I_C = 58617.24 - 0.055 times 0.15^2 I_C = 58617.239 m^2