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2. The weight of the disk is 30 lb and its radius of gyration equals 0.6 ft. The spring has a stiffness of 2 lb/ft and an uns

2. The weight of the disk is 30 lb and its radius of gyration equals 0.6 ft. The spring has a stiffness of 2 lb/ft and an unstretched length of 1 ft. Find the velocity of the center G at the instant G moves 3 ft to the left. Assume that the disk is released from rest in the position shown and rolls without slipping. k=2 lb/ft

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Solution

uft { h 5th Spring stretched in position 1, 5, - (5-1) = 4ft Spring stretched in position a, se = (k-1) = 3ft Kinetic Energy

At state 2, Let velocity of CG at state a angular velocity be w be VG and To - Im Call + I I GW² (9)+1(190)60-c*wse - 2-45 (

1,536 (2) 4) เฮ 9J V2 - 14 (2) (3) = 11 Titu, = Tatva 04 16 = 6-7u v + V6 3.61 (tk

Therefore velocity of center G is 3.03 ft/s.

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