06_InstSolManual_PDF_Part3 - Solve(a The tension in the...

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(b) gives gives and 6.5. Set Up: The person moves in a circle of radius The acceleration of the person is directed horizontally to the left in the figure in the problem. The time for one revolution is the period Solve: (a) The free-body diagram is given in Figure 6.5. is the force applied to the seat by the rod. Figure 6.5 (b) gives and gives Combining these two equations gives Then the period is (c) The net force is proportional to m so in the mass divides out and the angle for a given rate of rotation is independent of the mass of the passengers. Reflect: The person moves in a horizontal circle so the acceleration is horizontal. The net inward force required for circular motion is produced by a component of the force exerted on the seat by the rod. 6.6. Set Up: The free-body diagram for the seat plus rider is shown in Figure 6.6. is the angle the cable makes with the horizontal. Let be horizontal, toward the center of the circular path. Figure 6.6
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Unformatted text preview: Solve: (a) The tension in the cable must have a vertical component equal to the weight of the seat and passengers, no matter what the rotation rate is. It is not possible for the cable to be precisely horizontal, because then its tension would have no vertical component. (b) so and As the cable approaches horizontal, and T S ` . f S T 5 w sin f . T sin f 5 w T y 5 w y x T x w T y T f a rad 1 x f g F S 5 m a S T 5 2 p R v 5 2 p 1 5.50 m 2 5.58 m / s 5 6.19 s. v 5 " Rg tan u 5 "1 5.50 m 21 9.80 m / s 2 2 tan 30.0° 5 5.58 m / s. F sin 30.0° 5 m v 2 R . g F x 5 ma x F 5 mg cos 30.0° . F cos 30.0° 5 mg g F y 5 ma y F y F cos30° F sin30° 30° mg x a rad F S T 5 2 p R v . a rad 5 v 2 / R , R 5 3.00 m 1 1 5.00 m 2 sin 30.0° 5 5.50 m. m s 5 v 2 gR 5 1 25.0 m / s 2 2 1 9.80 m / s 2 21 220 m 2 5 0.290 m s mg 5 m v 2 R m s n 5 m v 2 R . g F x 5 ma x n 5 mg . g F y 5 ma y Circular Motion and Gravitation 6-3...
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