Question

Time Rmm m the weel with a 300 mm radius has a mass of 51 kg and a radius of gyration of 300 mm. load mis 158 kg and all masses are suspended by the cable and the spring, the stiffness of the spring is 2522 Nm of the system is released from rest with the spring is unstretched initially determine the velocity of O after it has dropped 50 mm

Answer 1

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Given that = 0.3 m R = 300 mm mw = 51 kg. Ky = radius of Gyration 158 kg Ky 300 mm = 0.3 m m = 155 kg 2522 Nim. spring potential energy love) { K 12 x will become double of locd fall. n = 2x so mm loc mm colm i Dve = į (2822) (0.12 = 1261 J - © Goavitational P.E lovg) =-(mwtm) gh --(54+158) (9.41)(0.05) Ang =-102051 J KE of wheel (OT) - 4 m, v2 + £ 12 m = mwim Iemk2 Ź (81+ 258) v2+ ] ( S1) v2 416-3)? (03)2 3 DT = 130 V2 Consider the balance ecuation. work energy OT + ove + Dug = 0 130 V2 + 12.61 102.51 = 0 V = 0.83 m/s