06_InstSolManual_PDF_Part5

06_InstSolManual_PDF_Part5 - passenger has mass Solve: (a)...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
6.9. Set Up: At the safe speed no friction force is required. Consider a car of mass m . The free-body diagram for the car is given in Figure 6.9. Use coordinates where is toward the center of the horizontal turn. Each turn is one- quarter of a circle so the radius R is given by is Figure 6.9 Solve: gives so gives so and Reflect: Race cars travel much faster than this. The track must exert a large friction force on the cars to maintain their circular motion. 6.10. Set Up: The ball has acceleration directed toward the center of the circular path. When the ball is at the bottom of the swing, its acceleration is upward. Take upward, in the direction of the acceleration. The bowling ball has mass Solve: (a) upward. (b) The free-body diagram is given in Figure 6.10. gives Figure 6.10 6.11. Set Up: The period is and The apparent weight of a person is the normal force exerted on him by the seat he is sitting on. His acceleration is directed toward the center of the circle. The
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: passenger has mass Solve: (a) Note that (b) The free-body diagram for the person at the top of his path is given in Figure 6.11a. The acceleration is down-ward, 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. 912 r 0.25 mi 5 1 4 1 2 p R 2 . 1 x Circular Motion and Gravitation 6-5...
View Full Document

Ask a homework question - tutors are online