hw18 - 3 2 v A + 15 3 v A = 30 f t / sec 1 The velocity of...

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TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 18 (3.16, 3.18) 3.16 v B = v A + ω β 2 × r AB 13 µ 5 13 ˆ i + 12 13 ˆ i = v A ˆ i + ω β 2 ˆ k × (20) ( cos60 ˆ j + sin60 ˆ j ) = v A ˆ i + ω β 2 ˆ k × (20) ˆ 1 2 ˆ j + 3 2 ˆ j ! 5 ˆ i + 12 ˆ j = v A ˆ i + ω β 2 (20) ˆ 1 2 ˆ j - 3 2 ˆ i ! Equating the ˆ i and ˆ j coefFcients separately, we get ˆ j : 12 = ω β 2 (20) 1 2 ω β 2 = 1 . 2 ˆ k rad / sec ˆ i : 5 = v A - ω β 2 (20) 3 2 v A = 5 + 10 3(1 . 2) = 25 . 8 v A = 25 . 8 ˆ i f t / sec 3.18 v B = v A + ω × r AB v B ˆ i = v A ˆ 1 2 ˆ i - 3 2 ˆ j ! + 1 . 5 ˆ k × (20) ˆ 3 2 ˆ i - 1 2 ˆ j ! = v A ˆ 1 2 ˆ i - 3 2 ˆ j ! + 1 . 5(20) ˆ 3 2 ˆ j + 1 2 ˆ i ! Equating the ˆ j coefFcients, we get 0 = -
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Unformatted text preview: 3 2 v A + 15 3 v A = 30 f t / sec 1 The velocity of the center is given by v C = v A + r AB = 30 1 2 i- 3 2 j ! + 1 . 5 k (10) 3 2 i + 1 2 j ! = 15 i-15 3 j + 7 . 5 3 j + 7 . 5 i = 22 . 5 i-12 . 99 j f t / sec 2...
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hw18 - 3 2 v A + 15 3 v A = 30 f t / sec 1 The velocity of...

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