ch08-p017 - 17. We use Eq. 8-18, representing the...

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h ) is the lowest point of the swing; it is also regarded as the “final” position in our calculations. (a) Careful examination of the figure leads to the trigonometric relation h = L L cos θ when the angle is measured from vertical as shown. Thus, the gravitational potential energy is U = mgL (1 – cos 0 ) at the position shown in Fig. 8-34 (the initial position). Thus, we have KU KU mv mgL mv ff 00 0 2 0 2 1 2 1 1 2 0 += + +−= + cos bg which leads to 22 21 (1 cos ) 2 (1 cos ) 2 (8.00 m/s) 2(9.80 m/s )(1.25 m)(1 cos40 ) 8.35 m/s. vm v m g L v g L m θθ ªº =+ = + «» ¬¼ ° = (b) We look for the initial speed required to barely reach the horizontal position — described by v h = 0 and = 90° (or = –90°, if one prefers, but since cos(– φ ) = cos , the sign of the angle is not a concern). mv mgL mgL hh 0 2 0 1 2 10 +=+ + cos which yields 2 2 cos 2(9.80 m/s )(1.25 m)cos40 4.33 m/s. vg L == ° = (c) For the cord to remain straight, then the centripetal force (at the top) must be (at least) equal to gravitational force: mv r mg mv mgL t t 2 2 = ¡ = where we recognize that
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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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ch08-p017 - 17. We use Eq. 8-18, representing the...

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