= m. 30 m. The angle that the tangent at B makes with the x axis is f = tan-1

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1 1 11. When the roller coaster is at B, it has a speed of 5 m>s, which is increasing at at = 3 m>s. Determine the magnitude of the acceleration of the roller coaster at this instant and the direction angle it makes with the x axis. y y 1 x 100 A s B Radius of Curvature: x 1 y = x 100 dy 1 = x dx = dx 1 B1 + a r = dy 3> b R dx dy dx Acceleration: # a t = v = 3 m>s an = B1 + a = 3> 1 xb R 5 = m th an Th sa eir d i is w le co s p or w of a urs rov k is ill de ny es a ided pro st pa nd s te ro rt o c y of as lel ted th t se y e his s fo by s r in te wo ing the Uni gr rk s u te ity ( tu s d of inc de e o Sta th lud nt f in te e in lea s s c w g r tru o or o ni c p k n ng to yri an th. rs gh d e W Dis in t l is t a no orl sem eac ws t p d W in hi er id ati ng m e on itt W o ed e r. b) dy 30 m 1 x = 30 m vb 5 = m>s = r The magnitude of the roller coaster s acceleration is a = at + an = = 8.43 m>s The angle that the tangent at B makes with the x axis is f = tan-1 dy 1 = tan-1 c A 30 B d = dx x = 30 m As shown in Fig. a, an is always directed towards the center of curvature of the path. Here, a = tan-1 a an b = Thus, the angle u that the roller coaster s acceleration makes b = tan-1 a at 3 with the x axis is u = a - f = 38. b 013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

2 1 13. The speedboat travels at a constant speed of 15 m> s while making a turn on a circular curve from A to B. If it takes 45 s to make the turn, determine the magnitude of the boat s acceleration during the turn. A B r Acceleration: During the turn, the boat travels s = vt = 15(45) = 675 m. Thus, the radius of the circular path is r = s Since the boat has a constant speed, p = 675 p m. a t = 0. Thus, a = a n = v r = 15 a 675 p b = 1.05 m>s This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. 013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

3 *13 5. A girl, having a mass of 15 kg, sits motionless relative to the surface of a horizontal platform at a distance of r = 5mfrom the platform s center. If the angular motion of the platform is slowly increased so that the girl s tangential component of acceleration can be neglected, determine the maximum speed which the girl will have before she begins to slip off the platform. The coefficient of static friction between the girl and the platform is m = 0.. Equation of Motion: Since the girl is on the verge of slipping, F f = m s N = 0.N. Applying Eq. 13 8, we have F b = 0; N - 15(9.81) = 0 N = N 5m z F n = ma n ; 0.(147.15) = 15a v 5 b v = 3.13 m>s This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. 013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

4 *13 7. The ball has a mass of 30 kg and a speed v = 4m>s at the instant it is at its lowest point, u = 0. Determine the tension in the cord and the rate at which the ball s speed is decreasing at the instant u = 0. Neglect the size of the ball. u 4m +a F n = ma n ; +Q F t = ma t ; T - 30(9.81) cos u = 30a v 4 b -30(9.81) sin u = 30a t a t = sin u a t ds = vdvsince ds = 4 du, then u v sin u (4 du) = vdv L L 0 4 u 9.81(4) cos u = 1 0 (v) - 1 (4) 39.4(cos u - 1) + 8 = 1 v At u = 0 v = m>s a t = m>s = 3.36 m>s T = 361 N b This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. 013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

5 * The 5-lb collar slides on the smooth rod, so that when it is at A it has a speed of 10 ft> s. If the spring to which it is attached has an unstretched length of 3 ft and a stiffness of k = 10 lb>ft, determine the normal force on the collar and the acceleration of the collar at this instant. y A 10 ft/s y 8 1 x y = 8-1 x - dy dx = tan u = x = u = x = d y dx = -1 O ft x r = B1 + a dy dx b d y dx 3 R y = 8-1 () = 6 OA = () + (6) = F s = kx = 10( ) = lb tan f = 6 ; f = = (1 + (-) ) 3-1 +b F n = ma n ; 5 cos N cos 45.0 = a 5 3. ba(10) b N = 4.4 lb = ft +R F t = ma t ; 5 sin sin 45.0 = a 5 3. ba t This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. a t = 180. ft>s a n = v r = (10) = ft>s a = (180.) + (8.9443) a = 180 ft>s 013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by

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