# 1&7&14 - Equating(2 with(3 cm cm cm s cm j i j i 1...

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3.1, 3.7, 3.12, 3.14 Spring, 2004 3.1 Given: Pictures with geometry shown Find: Which objects are in simple plane motion Solution: a, c, d, and e are all in plane motion. A good indication is that the rotational axis – i.e. the point of zero velocity for each plane of rotation – is not moving. 3.7 Given: Orientations and sizes of the objects shown, where bar 3 has angular velocity 3 rad/s k . Given the orientation, all angular velocities are in the ± k direction. Find: The angular velocities of the other two bars: 1 ω v , 2 ω v Solution: Given the orientation, all angular velocities are in the ± k direction, j r cm OA 10 = , i r cm AB 25 = , () cm CB j i r 15 8 + = On bar 3, the “2-point” formula gives us s cm OA O A / 10 3 0 3 j k r v v × + = × ω + = v , so i s cm v A / 30 = (1) On bar 2, the “2-point” formula gives us () cm s cm cm s cm AB A B j i i k i r v v 2 2 2 25 / 30 25 / 30 ω = × ω + = × ω + = v (2) On bar 1, the “2-point” formula gives us () cm cm cm CB C B i j j i k r v v 1 1 1 1 15 8 15 8 0 ω ω = + × ω + = × ω + = v (3)
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Unformatted text preview: Equating (2) with (3) cm cm cm s cm j i j i 1 1 2 8 15 25 / 30 ω − ω − = ω − − looking just at the x-components gives us k s rad s rad cm s cm / 2 / 2 15 / 30 1 1 1 = ω ⇒ = ω ⇒ ω − = − v looking just at the y-components gives us k s rad s rad s rad x cm cm / 64 . / 64 . / 2 8 25 8 25 2 2 2 1 2 = ω ⇒ = ω ⇒ = ω ⇒ ω − = ω − v 3.12 Done in class 3.14 Given: The wheel has an angular velocity of -2 rad/s k , and the v c = -0.3m/s i . Find: v A Solution: ( ) ( ) ( ) cm m CA j i r 13 / 5 13 / 12 48 . + = The “2-point” formula gives ( ) ( ) ( ) ( ) s m AC C A / 13 / 5 13 / 12 48 . 2 3 . j i k i r v v + × − + − = × ω + = v i.e., ( ) ( ) ( ) [ ] ( ) s m s m A / 886 . 069 . / 13 / 5 13 / 12 96 . 3 . j i i j i v − = + − + − = A B O C k i j k i j...
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## This note was uploaded on 04/18/2008 for the course TAM 212 taught by Professor Keane during the Spring '08 term at University of Illinois at Urbana–Champaign.

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