EX23.1 Rolling Wheel and Spring-Aud

# EX23.1 Rolling Wheel and Spring-Aud - EX23.1 Rolling Wheel...

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The weight of the disk is 30 lb and its k G 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 Position g1 + V e1 + U 1-2 = T 2 + V + V e2 What 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= μ k N DOES do work. Set up your Ve terms. V e1 = ½ k (s 1 ) 2 = ½ (2)(5-1) 2 = 16 J V e2 = ½ 2 (3) 2 = 9 J T 2 = ½ m (v G2 ) 2 + ½ I G ( ϖ 2 ) 2 = ½ (30/32.2) (v G2 ) 2 + ½ (30/32.2) 0.6 2 ( ϖ 2 ) 2 What is T 2 ? CG has a velocity Two 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.

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