The weight of the disk is 30 lb and its kG equals 0.6 ft. The spring has a stiffness of 2 lb/ft and an unstretched length of 1 ft. Find the velocity at the instant G moves 3 ft to the left. The disk is released from rest in the position shown and rolls without slipping.(1) Initial Position (2) Final Positiong1+ Ve1+ U1-2= T2+ V+ Ve2What equation, and what terms are zero?No change in height, so Vg is zero. Remember, rolling friction doesn’t do any work (point at bottom doesn’t displace). If you have slip, then the F=μkN DOES do work.Set up your Ve terms.Ve1 = ½ k (s1)2 = ½ (2)(5-1)2 = 16 JVe2 = ½ 2 (3)2 = 9 JT2 = ½ m (vG2)2 + ½ IG (ϖ2) 2= ½ (30/32.2) (vG2)2 + ½ (30/32.2) 0.62(ϖ2) 2What is T2?CG has a velocityTwo unknowns – how can I relate?
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This note was uploaded on 01/22/2011 for the course ME 212 taught by Professor Staff during the Spring '05 term at Cal Poly.