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Ans: 1.0733 rad/s CCW Checkpoints: lo 31.6487 kg-mA2, lo'-34.8987 kg-mA2 The structure is made of thin metal rods with mass per unit length p-10 kg/m. The8 short rods inside the ring are symmetrically arranged. The pendulum has an initial angular velocityo-0.1 rad/s CW. A 5-kg magnet flying at 20 m/s hits the structure at the top of the ring and becomes attached to it. Determine the angular velocity of the structure right after impact. R 0.4m Wo-o.ra 4m 0.2n 5, 0.2M

Answer 1

Given data: - rho = 10 kg/m Critical Angular Velocity W_0 = 0.1 rad/s weight = 5kg we have to find angular velocity of the structure right of the impact V = 20 m/s The distance OB from Pythagoras theorem is OB = squareroot (0.4)^2 + (0.6)^2 OB = squareroot 0.16 + 0.36 OB = squareroot 0.52 OB = 0.72 M Velocity of point B is V = 0.B.W_0 = 0.72 times 0.1 V = 0.072 m /s After impact velocity of combined system is OC on conserving the momentum is 'y' direction taking positive Mass of the ring M = 2 times (10)(0.4) = 2 times 3.14 times 10 times 0.4 = 25.12 kg m = 5 kg m 20 cos 30 degree + M 0.072 sin 30 = (M + m) Vc sin 30 degree V_c = m.20 cos 30 degree/(M + m) + M/(m + M) 0.072 = 5 times 20 times 1.7/(25.12 + 5) + 25.12/(25.12 + 5) times 0.072 = 5.44 + 0.057 V_c = 5.49 m/s O.B w' =