# hw30 - 4 5-1 22 ω C F ˆ j = v P C 3 5 96 ω C F Solving...

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TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 30 (3.147) 3.147 1.2 m 0.3 m 0.6 m 0.4 m C B R Q 3 4 O v P / F = ω R × r R = 1 ˆ k × 0 . 3 ˆ j = - 0 . 3 ˆ i m / s v P / F = v P / C + v Q / F + ω C / F × r QP = v P / C + ω C / F × r OQ + ω C / F × r QP = v P / C + ω C / F × ( r OQ + r QP ) = v P / C + ω C / F × ± - 3 5 0 . 4 ˆ i + 4 5 0 . 4 ˆ j + 1 . 2 ˆ i + 0 . 9 ˆ j ² = v P / C + ω C / F ˆ k × ( 0 . 96 ˆ i + 1 . 22 ˆ j ) = v P / C - 1 . 22 ω C / F ˆ i + 0 . 96 ω C / F ˆ j = v P / C ± - 4 5 ˆ i + 3 5 ˆ j ² - 1 . 22 ω C / F ˆ i + 0 . 96 ω C / F ˆ j 1

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Equating the ˆ i and ˆ j components separately, ˆ i : - 0 . 3 = v P / C
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Unformatted text preview: 4 5-1 . 22 ω C / F ˆ j : = v P / C 3 5 + . 96 ω C / F Solving, we get v P / C =-. 192 m / s ω C / F = . 12 rad / s ω C / D = . 12 ˆ k rad / s 2...
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## This note was uploaded on 05/24/2010 for the course TM 212 taught by Professor Richard during the Spring '10 term at University of Illinois, Urbana Champaign.

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hw30 - 4 5-1 22 ω C F ˆ j = v P C 3 5 96 ω C F Solving...

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