Unformatted text preview: j Combining with the equation for v 3 developed above, we have 3 Ω R = w 2 ω B − 5 4 Ω W R = ⇒ 3 Ω = 2 ω B − 5 4 Ω Solving for ω B , there follows ω B = 17 8 Ω (b) Using the two equations developed for v 2 in Part (a), 2 ω S R = w ω B − 5 4 Ω W R = ⇒ 2 ω S = ω B − 5 4 Ω We found in Part (a) that ω B = 17 8 Ω . Substituting into the equation above and solving for ω S yields ω S = 7 16 Ω...
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This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.
 Spring '06
 Shiflett

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