ch11-p012 - seek the velocity v p at x = 13 m. If we obtain...

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12. (a) Let the turning point be designated P . By energy conservation, the mechanical energy at x = 7.0 m is equal to the mechanical energy at P . Thus, with Eq. 11-5, we have 75 J = 1 2 mv p 2 + 1 2 I com ω p 2 + U p Using item ( f ) of Table 10-2 and Eq. 11-2 (which means, if this is to be a turning point, that p = v p = 0), we find U p = 75 J. On the graph, this seems to correspond to x = 2.0 m, and we conclude that there is a turning point (and this is it). The ball, therefore, does not reach the origin. (b) We note that there is no point (on the graph, to the right of x = 7.0 m ) which is shown “higher” than 75 J, so we suspect that there is no turning point in this direction, and we
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Unformatted text preview: seek the velocity v p at x = 13 m. If we obtain a real, nonzero answer, then our suspicion is correct (that it does reach this point P at x = 13 m). By energy conservation, the mechanical energy at x = 7.0 m is equal to the mechanical energy at P . Therefore, 75 J = 1 2 mv p 2 + 1 2 I com p 2 + U p Again, using item ( f ) of Table 11-2, Eq. 11-2 (less trivially this time) and U p = 60 J (from the graph), as well as the numerical data given in the problem, we find v p = 7.3 m/s....
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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