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October 19, 2009 PHY2048 Discussion-Fall ‘09 Quiz 7 Name: UFID: In the figure below, two particles, each with mass m = 0.60 kg are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 5.00 cm and mass M = 0.80 kg. The combination rotate around the rotation axis with angular speed ω = 0.50 rad/s. a) What is the rotational inertia of the combination about the rotation axis? The rotational inertia of the two particles is Ip = md2+m(2d)2 = 5md2 = 5×0.6×0.052 = 0.0075 kg m2 The two connecting rods can be regarded as a rod with length 2d and mass 2M Thus, Ir = (1/3)(2M)(2d)2 = (8/3)Md2 = (8/3)×0.8×0.052 = 0.00533 kg m2 Adding them together, the total rotational inertia is I = Ip+Ir = 0.0075+0.00533 = 0.0128 kg m2 b) What is the kinetic energy of the combination? The kinetic energy is K = (1/2)Iω2 = (1/2)×0.0128×0.52 = 1.60×10-3 J