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59 - B(0s v B(0s)t ½ a B t 2(substituting our results we...

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TAM 212 Spring, 2004 1.59 Given: v A (t=0) = 10 m/s; a A = -2 m/s 2 (where positive is to the right) Configuration of blocks and pulleys in Fig. 1.59 Required: The total distance traveled from t=0s to t=8s Solution: Label diagram to help identify constraints. We let L = length of string; r 1,2,3 be the radii of the three pulleys as labeled. There are only two variables(x A , x B ), easily identified especially when care is taken in labeling the diagram Our constraints are the configuration of the diagram and that L is constant and equals: L = x A + ½ π r 1 + x B + r 2 + π r 2 + r 2 + x B + K + π r 3 + K + x B = x A + 3x B + ½ π r 1 + 2r 2 + π r 2 + 2K + π r 3 (1) Because of the constraints, when we take the first and second derivatives, we get 0 = v A + 3v B , and (2) 0 = a A + 3a B . Since a A is constant and given above, a B , is also constant, and a B = 2/3 m/s 2 Given v A (t=0) = 10 m/s and substituting into (2) v B (t=0) = (-10/3) m/s Now, taking into account a B is constant, we can use the formula derived in class: x B (t) = x
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Unformatted text preview: B (0s) + v B (0s)t + ½ a B t 2 (substituting our results we get:) x B (t=8s) – x B (t=0s) = (-10/3) m/s * (8s) + ½ (2/3 m/s 2 )(8s) 2 = (-16/3) m/s or 5.3 m down. This is the magnitude of the displacement. We need to check if this is the total distance traveled as well. To do this, since the pulley is decelerating, we need to find if/when the pulley turns around, that is, when v B (t) = 0 m/s. Differentiating x B (t) = x B (0s) + v B (0s)t + ½ a B t 2 we get v B (t) = v B (0s) + a B t This = 0 when t = -v B (0s)/a B or t = 10/3 m/s / (2/3 m/s 2 ) = 5s Just as we feared, the pulley has gone down and is on its way up before 8s has passed. The displacement before turning around is x B (t=5s) = (-10/3) m/s * (5s) + ½ (2/3 m/s 2 )(5s) 2 = -25/3 m From the calculations above, 3s later it has traveled 25/3 – 16/3 m or 9/3 m. Therefore, the total distance traveled is 25/3 + 9/3 m = 34/3 m = 11.3 m...
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