Unformatted text preview: passenger has mass Solve: (a) Note that (b) The freebody diagram for the person at the top of his path is given in Figure 6.11a. The acceleration is downward, so take downward. gives mg 2 n 5 ma rad . g F y 5 ma y 1 y a rad 5 v 2 R 5 1 5.24 m / s 2 2 50.0 m 5 0.549 m / s 2 . v 5 2 p R T 5 2 p 1 50 m 2 60.0 s 5 5.24 m / s. m 5 w / g 5 90.0 kg. a rad 5 v 2 / R , T 5 2 p R v . T 5 60.0 s T 5 m 1 g 1 a rad 2 5 1 7.27 kg 21 9.80 m / s 2 1 4.64 m / s 2 2 5 105 N mg y T x a rad T 2 mg 5 ma rad . g F y 5 ma y a rad 5 v 2 R 5 1 4.20 m / s 2 2 3.80 m 5 4.64 m / s, m 5 w / g 5 7.27 kg. 1 y a rad 5 v 2 / R , v 5 " Rg tan f 5 "1 256 m 21 9.80 m / s 2 2 tan 9.2° 5 20.0 m / s 5 44.7 mph 1 mg cos f 2 sin f 5 m v 2 R n sin f 5 m v 2 R n x 5 ma rad g F x 5 ma x n 5 mg cos f n y 5 w g F y 5 ma y R 5 4 1 0.25 mi 2 / 2 p 5 0.1592 mi 5 256 m y w n y n x n x f f a rad 9.2°. 9°12 r 0.25 mi 5 1 4 1 2 p R 2 . 1 x Circular Motion and Gravitation 65...
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 Spring '07
 Shoberg
 Circular Motion, Force, Friction, Mass, freebody diagram, Arad

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