68&71 - TAM 212 Spring, 2004 1.68 Given: The bar of...

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TAM 212 Spring, 2004 1.68 Given: The bar of length 2L leans up against a vertical wall. It makes an angle q with the horizontal. Required: The velocity and acceleration of the center of the bar as a function of q and its derivatives. (Essentially, we are solving for the velocity and acceleration of C, assuming we might know q and its derivatives, as functions of time.) Solution: We first need to express j i r OC OC OC y x + = , where O is some fixed point (origin). Let’s chose the origin to be where the floor meets the wall. If you can’t picture the expression for the position for C immediately, it may be simpler to first think of point B where, B is where the bar touches the floor. ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 j i i j i j i r r r q q q sin cos 0 cos 2 L L L y x y x BC BC OB OB BC OB OC + - + + = + + + = + = ( 29 ( 29 j i r q q sin cos L L OC + = , so q q sin ; cos L y L x OC OC = = 1.27, 1.28 give us: j i v OC OC OC y x + = and j i a OC OC OC y x + = So we simply take derivatives (w.r.t. time) of
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This note was uploaded on 04/14/2010 for the course TAM 212 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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