12–2. When a train is traveling along a straight track at 2 m/s, it begins to accelerate at a = 160 v-42 m>s2, where v is in m/s. Determine its velocity v and the position 3 s after the acceleration. s SOLUTION a = dv dt dt = dv a v 3 dt = L0 3 = dv -4 L2 60v 1 (v5 - 32) 300 v = 3.925 m>s = 3.93 m>s Ans. ads = vdv ds = s L0 ds = s = 1 5 vdv = v dv a 60 1 60 L2 3.925 v5 dv 1 v6 3.925 a b` 60 6 2 = 9.98 m Ans. v 12–13. The acceleration of a rocket traveling upward is given by ap = b + c sp. Determine the rocket’s velocity when sp = sp1 and the time needed to reach this altitude. Initially, vp = 0 and sp = 0 when t = 0. b Given: 6 m c 2 s 0.02 1 sp1 2 2000 m s Solution: ap ´ µ ¶ b c sp vp vp dvp 0 ´ µ ¶ dvp dsp sp b c sp dsp 0 vp 2 b sp 2 dsp vp t vp dt µ́ µ µ ¶ sp 0 c 2 sp 2 2b sp c sp 1 2 2b sp c sp 2 dsp 2 vp1 t1 2b sp1 c sp1 µ́ µ µ ¶ sp1 0 vp1 1 2b sp c sp 2 dsp t1 322.49 m s 19.27 s Ans. Ans. 12–14. The acceleration of a rocket traveling upward is given by ap = b + c sp. Determine the time needed for the rocket to reach an altitute sp1. Initially, vp = 0 and sp = 0 when t = 0. b Given: 6 m c 2 s 0.02 1 sp1 2 100 m s Solution: ap ´ µ ¶ b c sp vp ´ µ ¶ vp dvp 0 dvp dsp sp b c sp dsp 0 vp 2 b sp 2 dsp vp t vp dt µ́ µ µ ¶ sp 0 c 2 sp 2 2b sp c sp 1 2 2b sp c sp 2 dsp 2 vp1 t1 2b sp1 c sp1 µ́ µ µ ¶ sp1 0 vp1 1 2b sp c sp 2 dsp t1 37.42 m s 5.62 s Ans. 12–26. When two cars A and B are next to one another, they are traveling in the same direction with speeds vA and vB , respectively. If B maintains its constant speed, while A begins to decelerate at aA , determine the distance d between the cars at the instant A stops. A d SOLUTION Motion of car A: v = v0 + act 0 = vA - aAt t = vA aA v2 = v20 + 2ac(s - s0) 0 = v2A + 2( - aA)(sA - 0) sA = v2A 2aA Motion of car B: sB = vBt = vB a vA vAvB b = aA aA The distance between cars A and B is sBA = |sB - sA| = ` v2A vAvB 2vAvB - v2A ` = ` ` aA 2aA 2aA B Ans. 12–28. A particle travels to the right along a straight line with a velocity v = [5>14 + s2] m>s, where s is in meters. Determine its deceleration when s = 2 m. SOLUTION v = 5 4+s v dv = a ds dv = -5 ds (4 + s)2 5 - 5 ds b = a ds a (4 + s) (4 + s)2 a = - 25 (4 + s)3 When s = 2 m a = - 0.116 m>s2 Ans. 12–34. A boy throws a ball straight up from the top of a 12-m high tower. If the ball falls past him 0.75 s later, determine the velocity at which it was thrown, the velocity of the ball when it strikes the ground, and the time of flight. SOLUTION Kinematics: When the ball passes the boy, the displacement of the ball in equal to zero. Thus, s = 0. Also, s0 = 0, v0 = v1, t = 0.75 s, and ac = - 9.81 m>s2. A+cB s = s0 + v0t + 1 2 at 2 c 0 = 0 + v110.752 + 1 1 - 9.81210.7522 2 v1 = 3.679 m>s = 3.68 m>s Ans. When the ball strikes the ground, its displacement from the roof top is s = - 12 m. Also, v0 = v1 = 3.679 m>s, t = t2, v = v2, and ac = - 9.81 m>s2. A+cB s = s0 + v0t + 1 2 at 2 c -12 = 0 + 3.679t2 + 1 1 - 9.812t22 2 4.905t22 - 3.679t2 - 12 = 0 t2 = 3.679 ; 21- 3.67922 - 414.90521 -122 214.9052 Choosing the positive root, we have t2 = 1.983 s = 1.98 s Ans. Using this result, A+cB v = v0 + act v2 = 3.679 + 1 -9.81211.9832 = - 15.8 m>s = 15.8 m>s T Ans. 12–85. A particle travels along the curve from A to B in 1 s. If it takes 3 s for it to go from A to C, determine its average velocity when it goes from B to C. y B 20 m SOLUTION A Time from B to C is 3 - 1 = 2 s vang = (rAC - rAB) 40i - (20i + 20j) ¢r = = = {10i - 10j} m>s ¢t ¢t 2 Ans. C x 12–87. Pegs A and B are restricted to move in the elliptical slots due to the motion of the slotted link. If the link moves with a constant speed of 10 m/s, determine the magnitude of the velocity and acceleration of peg A when x = 1 m. y A D C SOLUTION v Velocity: The x and y components of the peg’s velocity can be related by taking the first time derivative of the path’s equation. x2 + y2 = 1 4 1 # # (2xx) + 2yy = 0 4 1 # # xx + 2yy = 0 2 or 1 xv + 2yvy = 0 2 x (1) At x = 1 m, (1)2 + y2 = 1 4 y = 23 m 2 Here, vx = 10 m>s and x = 1. Substituting these values into Eq. (1), 23 1 (1)(10) + 2 ¢ ≤ vy = 0 2 2 vy = -2.887 m>s = 2.887 m>s T Thus, the magnitude of the peg’s velocity is v = 2vx 2 + vy 2 = 2102 + 2.8872 = 10.4 m>s Ans. Acceleration: The x and y components of the peg’s acceleration can be related by taking the second time derivative of the path’s equation. 1 # # ## ## # # (xx + xx) + 2(yy + yy) = 0 2 1 #2 ## ## # A x + xx B + 2 A y 2 + yy B = 0 2 or 1 A v 2 + xax B + 2 A vy 2 + yay B = 0 2 x Since vx is constant, ax = 0. When x = 1 m, y = (2) 23 m, vx = 10 m>s, and 2 vy = -2.887 m>s. Substituting these values into Eq. (2), 23 1 a d = 0 A 102 + 0 B + 2 c ( - 2.887)2 + 2 2 y ay = - 38.49 m>s2 = 38.49 m>s2 T Thus, the magnitude of the peg’s acceleration is a = 2ax 2 + ay 2 = 202 + ( -38.49)2 = 38.5 m>s2 Ans. B x2 4 y2 1 x 10 m/s 12–9 1. The pitching machine is adjusted so that the baseball is launched with a speed of W" 30 ms. If the ball strikes the ground at B, determine the two possible angles u" at which it was launched. vA 30 m/s A uA 1.2 m B 30 m Coordinate System: The x–y coordinate system will be set so that its origin coincides with point A. x-Motion: Here, (v")Y 30 cos u", Y" 0 and Y# 30 m. Thus, Y# Y" 30 0 U (v")YU 30 cos u"U 1 cos u" (1) y-Motion: Here, (v")Z 30 sin u", BZ H 9.81 ms2, and Z# 1.2 m. Thus, D Z# Z" 1.2 0 1 B U2 2 Z 1 ( 9.81)U2 30 sin u"U 2 (v")ZU 4.905U2 30 sin u" U 1.2 0 (2) Substituting Eq. (1) into Eq. (2) yields 4.905 A 2 1 1 I 30 sin u" A I 1.2 0 cos u" cos u" 1.2 cos2 u" 30 sin u" cos u" 4.905 0 Solving by trial and error, u" 7.19° and 80.5° Ans. 12–93. The balloon A is ascending at rate vA and is being carried horizontally by the wind at vw. If a ballast bag is dropped from the balloon when the balloon is at height h, determine the time needed for it to strike the ground. Assume that the bag was released from the balloon with the same velocity as the balloon. Also, with what speed does the bag strike the ground? Given: vA 12 km hr vw 20 km hr h 50 m g 9.81 m 2 s Solution: Thus ax 0 ay g vx vw vy g t vA sx vw t sy 1 2 g t vA t h 2 0 vx 1 2 g t vA t h 2 vw vy g t vA t 2 vA vA 2g h g v 2 2 vx vy t 3.55 s v 32.0 m s Ans. Ans. 12–95. A projectile is given a velocity v0 at an angle f above the horizontal. Determine the distance d to where it strikes the sloped ground. The acceleration due to gravity is g. y v0 φ SOLUTION + B A: d cos u = 0 + v0 (cos f) t s = s0 + v0 t + 1 2 a t 2 c d sin u = 0 + v0 (sin f)t + 1 ( - g) t2 2 Thus, d sin u = v0 sin f a d cos u 1 d cos u 2 b - ga b v0 cos f 2 v0 cos f sin u = cos u tan f - gd cos2 u 2v20 cos2 f d = (cos u tan f - sin u) d = x d s = s0 + v0 t A+cB θ 2v20 cos2 f g cos2 u v 20 sin 2f - 2 tan u cos2 f g cos u Ans. A projectile is fired from the platform at B. The 12–98. shooter fires his gun from point A at an angle of 30°. Determine the muzzle speed of the bullet if it hits the projectile at C. B C vA Coordinate System: The x–y coordinate system will be set so that its origin coincides with point A. Y$ Y" 20 0 (v")YU v" cos 30° U (1) y-Motion: Here, Z" 1.8, (v")Z v" sin 30°, and BZ H 9.81 ms2. Thus, D 1 BZU2 2 Z$ Z" (v")ZU 10 1.8 v" sin 30°(U) 1 ( 9.81)(U)2 2 Thus, 10 1.8 A 20 sin 30° I (U) 4.905(U)2 cos 30°(U) U 0.8261 s So that v" 20 28.0 ms cos 30°(0.8261) 10 m 30 1.8 m 20 m x-Motion: Here, Y" 0 and Y$ 20 m. Thus, A Ans. 12–101. It is observed that the skier leaves the ramp A at an angle uA = 25° with the horizontal. If he strikes the ground at B, determine his initial speed vA and the speed at which he strikes the ground. vA uA A 4m 3 5 4 100 m SOLUTION Coordinate System: x - y coordinate system will be set with its origin to coincide with point A as shown in Fig. a. 4 x-motion: Here, xA = 0, xB = 100 a b = 80 m and (vA)x = vA cos 25°. 5 + B A: xB = xA + (vA)xt 80 = 0 + (vA cos 25°)t 80 t = vA cos 25° (1) 3 y-motion: Here, yA = 0, yB = - [4 + 100 a b ] = - 64 m and (vA)y = vA sin 25° 5 and ay = - g = - 9.81 m>s2. A+ c B yB = yA + (vA)y t + 1 a t2 2 y - 64 = 0 + vA sin 25° t + 1 ( - 9.81)t2 2 4.905t2 - vA sin 25° t = 64 (2) Substitute Eq. (1) into (2) yieldS 4.905 ¢ ¢ 2 80 80 ≤ = vA sin 25° ¢ ≤ = 64 vA cos 25° yA cos 25° 2 80 ≤ = 20.65 vA cos 25° 80 = 4.545 vA cos 25° vA = 19.42 m>s = 19.4 m>s Substitute this result into Eq. (1), t = 80 = 4.54465 19.42 cos 25° Ans. B 12–101. continued Using this result, A+ c B (vB)y = (vA)y + ay t = 19.42 sin 25° + ( -9.81)(4.5446) = - 36.37 m>s = 36.37 m>s T And + B A: (vB)x = (vA)x = vA cos 25° = 19.42 cos 25° = 17.60 m>s : Thus, vB = 2(vB)2x + (vB)2y = 236.372 + 17.602 = 40.4 m>s Ans. 12–122. If the roller coaster starts from rest at A and its speed increases at at = (6 – 0.06s) m>s2, determine the magnitude of its acceleration when it reaches B where sB = 40 m. y y 1 x2 100 A s SOLUTION B Velocity: Using the initial condition v = 0 at s = 0, x v dv = at ds v L0 s vdv = L0 A 6 - 0.06s B ds 30 m v = a 212s - 0.06s2 b m>s Thus, (1) vB = 412 A 40 B - 0.06 A 40 B 2 = 19.60 m>s Radius of Curvature: 1 2 x 100 dy 1 = x dx 50 y = d2y dx2 = 1 50 B1 + a r = 2 dy 2 3>2 b R dx d2y dx2 B1 + a = 2 2 3>2 1 xb R 50 2 1 2 50 5 = 79.30 m x = 30 m Acceleration: # a t = v = 6 - 0.06(40) = 3.600 m>s2 an = 19.602 v2 = 4.842 m>s2 = r 79.30 The magnitude of the roller coaster’s acceleration at B is a = 2at 2 + an 2 = 23.6002 + 4.8422 = 6.03 m>s2 Ans. 12–126. y If the car passes point A with a speed of 20 m>s and begins to increase its speed at a constant rate of at = 0.5 m>s2, determine the magnitude of the car’s acceleration when s = 100 m. y B 16 m SOLUTION Velocity: The speed of the car at C is vC 2 = vA 2 + 2a t (sC - sA) vC 2 = 202 + 2(0.5)(100 - 0) vC = 22.361 m>s Radius of Curvature: y = 16 - 1 2 x 625 dy = -3.2 A 10-3 B x dx d2y dx2 = -3.2 A 10-3 B B1 + a r = 2 dy 2 3>2 b R dx d2y dx2 c1 + a -3.2 A 10-3 B x b d 2 3>2 = 2 - 3.2 A 10-3 B 4 = 312.5 m x=0 Acceleration: # a t = v = 0.5 m>s an = vC 2 22.3612 = = 1.60 m>s2 r 312.5 The magnitude of the car’s acceleration at C is a = 2a2t + a2n = 20.52 + 1.602 = 1.68 m>s2 16 Ans. 1 2 x 625 s A x 12–133. v Car B turns such that its speed is increased by (at)B = (0.5et) m>s2, where t is in seconds. If the car starts from rest when u = 0°, determine the magnitudes of its velocity and acceleration when t = 2 s. Neglect the size of the car. B SOLUTION 5m dv Velocity: The speed v in terms of time t can be obtained by applying a = . dt A dv = adt v L0 When t = 2 s, t dv = t L0 0.5e dt v = 0.5 A et - 1 B v = 0.5 A e2 - 1 B = 3.195 m>s = 3.19 m>s Ans. Acceleration: The tangential acceleration of the car at t = 2 s is at = 0.5e2 = 3.695 m>s2. To determine the normal acceleration, apply Eq. 12–20. an = v2 3.1952 = 2.041 m>s2 = r 5 The magnitude of the acceleration is a = 2a2t + a2n = 23.6952 + 2.0412 = 4.22 m>s2 Ans. u 12–139. The motorcycle is traveling at a constant speed of 60 km> h. Determine the magnitude of its acceleration when it is at point A. y y2 ⫽ 2x A x 25 m SOLUTION Radius of Curvature: y = 22x1>2 dy 1 = 22x - 1>2 dx 2 d2y dx2 = - 1 22x - 3>2 4 B1 + a r = ` dy 2 3>2 b R dx d2y ` dx2 1 2 2 3>2 B 1 + ¢ 22x - 1>2 ≤ R = 1 ` - 22x - 3>2 ` 4 4 = 364.21 m x = 25 m Acceleration: The speed of the motorcycle at a is v = ¢ 60 an = 1h km 1000 m ≤¢ ≤¢ ≤ = 16.67 m>s h 1 km 3600 s v2 16.672 = = 0.7627 m>s2 r 364.21 Since the motorcycle travels with a constant speed, at = 0. Thus, the magnitude of the motorcycle’s acceleration at A is a = 2at 2 + an2 = 202 + 0.76272 = 0.763 m>s2 Ans. 12–145. The particle travels with a constant speed v along the curve. Determine the particle’s acceleration when it is located at point x = x1. Given: mm s v 300 k 20 u 10 mm 3 x1 2 200 mm Solution: y ( x) k x y' ( x) d y ( x) dx y'' ( x) d y' ( x) dx 1 y' ( x) y'' ( x) U ( x) 2 § sin T 1 · ¨ ¸ U x1 © cos T 1 ¹ 3 T ( x) atan ( y' ( x) ) 2 a v a § 144 · mm ¨ ¸ © 288 ¹ s2 T1 T x1 a T1 322 mm 2 s 26.6 deg Ans. 12–150. The two particles A and B start at the origin O and travel in opposite directions along the circular path at constant speeds vA = 0.7 m>s and vB = 1.5 m>s, respectively. Determine the time when they collide and the magnitude of the acceleration of B just before this happens. y SOLUTION st = 2p(5) = 31.4159 m 5m sA = 0.7 t B sB = 1.5 t A vB Require 0.7 t + 1.5 t = 31.4159 aB = (1.5)2 v2B = 0.45 m>s2 = r 5 x O vA sA + sB = 31.4159 t = 14.28 s = 14.3 s 1.5 m/s Ans. Ans. 0.7 m/s 12–163. At the instant shown, the watersprinkler is rotating with an angular speed T' and an angular acceleration T''. If the nozzle lies in the vertical plane and water is flowing through it at a constant rate r', determine the magnitudes of the velocity and acceleration of a water particle as it exits the open end, r. Given: T' 2 T'' 3 rad s rad 2 s m s r' 3 r 0.2 m Solution: v a 2 r' rT' r T' 2 2 2 rT'' 2r' T' 2 m s v 3.03 a 12.63 Ans. m 2 s Ans. 12–167. If arm OA rotates counterclockwise with a constant angular velocity of u 2 rads, determine the magnitudes of the velocity and acceleration of peg P at u 30°. The peg moves in the fixed groove defined by the lemniscate, and along the slot in the arm. r2 (4 sin 2 u)m2 r O Time Derivatives: S2 4 sin 2u 2SS 8 cos 2uu 4 cos 2uu S ! ms S 2 SS S2 8 2 sin 2uu2 u 2 rads cos 2u u 4 cos 2uu 2 sin 2uu2 S2 S " ms2 S u 0 At u 30°, S u 30° 24 sin 60° 1.861 m (4 cos 60°)(2) S u 30° 2.149 ms 1.861 4 0 2 sin 60° 22 (2.149)2 S u 30° 17.37 ms2 1.861 Velocity: vu Su 1.861(2) 3.722 ms vS S 2.149 ms Thus, the magnitude of the peg’s velocity is v 2 2vS 2 2 vu 2 22.149 2 3.7222 4.30 ms Ans. Acceleration: B S S Su2 17.37 1.861 22 24.82 ms2 B u S u 2S u 0 2(2.149)(2) 8.597 ms2 Thus, the magnitude of the peg’s acceleration is B 2B 2 S2 Bu 2 2( 2 24.82)2 8.5972 26.3 ms2 Ans. u P The peg moves in the curved slot defined by the 12–168. lemniscate, and through the slot in the arm. At u 30°, the angular velocity is u 2 rads, and the angular acceleration is u 1.5 rads2. Determine the magnitudes of the velocity and acceleration of peg P at this instant. r2 (4 sin 2 u)m2 r O Time Derivatives: 2SS 8 cos 2uu 4 cos 2uu S A I ms S 2 3SS S2 4 8 3 2 sin 2uu u 2 rads cos 2uu2 4 4 cos 2uu 2 sin 2uu2 S2 S " ms2 S u 1.5 rads2 At u 30°, S u 30° 24 sin 60° 1.861 m (4 cos 60°)(2) S u 30° 2.149 ms 1.861 4 cos 60°(1.5) 2 sin 60° 22 (2.149)2 S u 30° 15.76 ms2 1.861 Velocity: vu Su 1.861(2) 3.722 ms vS S 2.149 ms Thus, the magnitude of the peg’s velocity is v 2 2BS 2 2 Bu 2 22.149 2 3.7222 4.30 ms Ans. Acceleration: B S S S u2 15.76 1.861 22 23.20 ms2 Bu Su 2S u 1.861(1.5) 2(2.149)(2) 11.39 ms2 Thus, the magnitude of the peg’s acceleration is B 2B 2 S2 Bu 2 2( 2 23.20)2 11.392 25.8 ms2 Ans. u P 12– 170. The pin follows the path described by the equation r = a + bcos T. At the instant T = T1. the angular velocity and angular acceleration are T' and T''. Determine the magnitudes of the pin’s velocity and acceleration at this instant. Neglect the size of the pin. Given: a 0.2 m b 0.15 m T1 30 deg T' 0.7 T'' 0.5 rad s rad 2 s T Solution: T1 r a b cos T v r' rT' a r'' rT' 2 2 r' b sin T T' 2 2 rT'' 2r' T' 2 r'' 2 b cos T T' b sin T T'' v 0.237 a 0.278 m s m 2 s Ans. Ans. 12–171. The slotted link is pinned # at O, and as a result of the constant angular velocity u = 3 rad>s it drives the peg P for a short distance along the spiral guide r = 10.4u2 m, where u is in radians. Determine the radial and transverse components of the velocity and acceleration of P at the instant u = p>3 rad. 0.5 m P r · u r 3 rad/s SOLUTION # u = 3 rad>s r = 0.4 u # # r = 0.4 u $ $ r = 0.4 u At u = p , 3 u O r = 0.4189 # r = 0.4(3) = 1.20 $ r = 0.4(0) = 0 # v = r = 1.20 m>s # vu = r u = 0.4189(3) = 1.26 m>s # $ ar = r - ru2 = 0 - 0.4189(3)2 = - 3.77 m>s2 $ ## au = r u + 2ru = 0 + 2(1.20)(3) = 7.20 m>s2 Ans. Ans. Ans. Ans. 0.4 u 12–172. Solve Prob. # 12–171 if the slotted link has . an angular # # acceleration u = 8 rad>s 2 when u = 3 rad>s at u = p>3 rad. SOLUTION # u = 3 rad/s u = 0.5 m r = 0.4 u # # r = 0.4 u $ $ r = 0.4 u r · u r 3 rad/s u p 3 O # u = 3 $ u = 8 r = 0.4189 # r = 1.20 $ r = 0.4(8) = 3.20 # vr = r = 1.20 m>s # vu = r u = 0.4189(3) = 1.26 m>s # $ ar = r - ru2 = 3.20 - 0.4189(3)2 = - 0.570 m>s2 $ ## au = r u + 2 ru = 0.4189(8) + 2(1.20)(3) = 10.6 m>s2 P Ans. Ans. Ans. Ans. 0.4 u 12–173. The slotted link is pinned # at O, and as a result of the constant angular velocity u = 3 rad>s it drives the peg P for a short distance along the spiral guide r = 10.4u2 m, where u is in radians. Determine the velocity and acceleration of the particle at the instant it leaves the slot in the link, i.e., when r = 0.5 m. 0.5 m P r · u SOLUTION u r = 0.4 u # # r = 0.4 u $ $ r = 0.4 u # u = 3 $ u = 0 O At r = 0.5 m, u = 0.5 = 1.25 rad 0.4 # r = 1.20 $ r = 0 # vr = r = 1.20 m>s # vu = r u = 0.5(3) = 1.50 m>s # $ ar = r - r(u)2 = 0 - 0.5(3)2 = - 4.50 m>s2 $ ## au = ru + 2ru = 0 + 2(1.20)(3) = 7.20 m>s2 r 3 rad/s Ans. Ans. Ans. Ans. 0.4u 12–179. A block moves outward along the slot in the platform with # a speed of r = 14t2 m>s, where t is in seconds. The platform rotates at a constant rate of 6 rad/s. If the block starts from rest at the center, determine the magnitudes of its velocity and acceleration when t = 1 s. · θ = 6 rad/s r θ SOLUTION # r = 4t|t = 1 = 4 # $ u = 6 u = 0 1 L0 dr = $ r = 4 1 L0 4t dt r = 2t2 D 10 = 2 m # # v = 3 A r B 2 + A ru B 2 = 2 (4)2 + [2(6)]2 = 12.6 m>s # $ $ ## a = r - ru2 2 + ru + 2 ru 2 = [4 - 2(6)2 ]2 + [0 + 2(4)(6)]2 = 83.2 m s 2 Ans. Ans. 12–180. The rod OA rotates counterclockwise with a constant angular velocity of T'. Two pin-connected slider blocks, located at B, move freely on OA and the curved rod whose shape is a limaçon described by the equation r = b(c cos(T)). Determine the speed of the slider blocks at the instant T T1. Given: rad s T' 5 b 100 mm c 2 T1 120 deg Solution: T T1 r b c cos T r' b sin T T' v 2 r' rT' 2 v 1.323 m s Ans. 12–181. The rod OA rotates counterclockwise with a constant angular velocity of T'. Two pin-connected slider blocks, located at B, move freely on OA and the curved rod whose shape is a limaçon described by the equation r = b(c cos(T)). Determine the acceleration of the slider blocks at the instant T T1. Given: rad s T' 5 b 100 mm c 2 T1 120 deg Solution: T T1 r b c cos T r' b sin T T' r'' b cos T T' a r'' rT' 2 2 2 2r' T' 2 a 8.66 m 2 s Ans. 12 –183. The slotted arm AB drives pin C through the spiral groove described by the equation R A .. If the angular velocity is constant at ., determine the radial and transverse components of velocity and acceleration of the pin. B C Time Derivatives: Since . is constant, then . 0. R A. R A. R A. 0 r . Velocity: Applying Eq. 12–25, we have 2R R A. 2. R. A.. Acceleration: Applying Eq. 12–29, we have AR R R.2 0 A..2 A..2 A. R. 2R . 0 2(A.)(.) 2A.2 Ans. Ans. Ans. Ans. A 12–184. The slotted arm AB drives pin C through the spiral groove described by the equation R (1.5 .) m, where . is in radians. If the arm starts from rest when . 60° and is driven at an angular velocity of . (4T) rads, where t is in seconds, determine the radial and transverse components of velocity and acceleration of the pin C when T 1 s. B C r Time Derivatives: Here, . 4 T and . 4 rads2. R 1.5 . R 1.5 . 1.5(4T) 6T R 1.5 . 1.5 (4) 6 ms2 . Velocity: Integrate the angular rate, ' ) 3 T D. '0 4TDT, we have . 1 1 Then, R ( (6T2 )) 8 m. At T 1 s, R 6(12) 2 2 and . 4(1) 4 rads. Applying Eq. 12–25, we have 1 (6T2 3 . )) rad. ) 4.571 m, R 6(1) 6.00 ms 2R R 6.00 ms 2. R. 4.571 (4) 18.3 ms Ans. Ans. Acceleration: Applying Eq. 12–29, we have R.2 6 4.571 42 67.1 ms2 A. R . 2R. 4.571(4) 2(6) (4) 66.3 ms2 AR R Ans. Ans. A 12 –194 . At the instant u 30°, the cam rotates with a clockwise angular velocity of u 5 rads and and angular acceleration of u 6 rads2. Determine the magnitudes of the velocity and acceleration of the follower rod AB at this instant. The surface of the cam has a shape of a limaçon defined by S (200 100 cos u) mm. r (200 100 cos u) mm u Time Derivatives: S (200 100 cos u) mm S (100 sin uu) mms S 100 sin uu cos uu2 mms2 When u 30°, S u 30° 200 100 cos 30° 286.60 mm S u 30° 100 sin 30°(5) 250 mms S u 30° 100 sin 30°(6) cos 30° 52 2465.06 mms2 Velocity: The radial component gives the rod’s velocity. vS S 250 250 mms Acceleration: The radial component gives the rod’s acceleration. BS S Su2 2465.06 286.60 52 9630 9630 mms2 Ans. Ans. A B Determine the displacement of the log if the 12 –203. truck at C pulls the cable 1.2 m to the right. B S#) L 2S" (S" 3S" S# L 3bS" bS# 0 Since bS# 3bS" bS" 1.2, then 1.2 0.4 m 0.4 m Ans. C 12–209. If the point A on the cable is moving upwards at vA, determine the speed of block B. Given: vA m s 14 Solution: L1 0 sD sA sD sE 2vD vA vE L2 0 sD sE sC sE vD vC 2vE L3 0 sC sD sC sE 2vC vD vE m s Guesses vC Given 0 2vD vA vE 0 vD vC 2vE 0 2vC vD vE § vC · ¨ ¸ ¨ vD ¸ ¨v ¸ © E¹ vB 1 vD vB 2 m s § vC · ¨ ¸ ¨ vD ¸ ¨v ¸ © E¹ Find vC vD vE vC 1 m s vE 1 m s § 2 · ¨ 10 ¸ m ¨ ¸ s © 6 ¹ Positive means down, Negative means up Ans. 12–210. The motor draws in the cable at C with a constant velocity vC. The motor draws in the cable at D with a constant acceleration of aD. If vD = 0 when t = 0, determine (a) the time needed for block A to rise a distance h, and (b) the relative velocity of block A with respect to block B when this occurs. Given: vC 4 aD 8 m s m 2 s h 3m Solution: L1 sD 2sA 0 vD 2vA 0 aD 2aA L2 sB sB sC 0 2vB vC 0 2aB aC aA aD 2 vA aA t sA h vA aA t § t2 · ¸ ©2¹ aA¨ t vB 2h aA 1 vC 2 t vAB 1.225 s vA vB Ans. vAB 2.90 m s Ans. 12–217. The crate C is being lifted by moving the roller at A downward with a constant speed of vA = 2 m>s along the guide. Determine the velocity and acceleration of the crate at the instant s = 1 m. When the roller is at B, the crate rests on the ground. Neglect the size of the pulley in the calculation. Hint: Relate the coordinates xC and xA using the problem geometry, then take the first and second time derivatives. 4m B xA xC 4m SOLUTION s xC + 2x2A + (4)2 = l # # 1 xC + (x2A + 16) - 1/2(2xA)(xA) = 0 2 1 # 2 $ # 2 $ xC - (x2A + 16) - 3/2 (2x2A)(xA ) + (x2A + 16) - 1/2 (xA) + (x2A + 16) - 1/2 (xA)(xA) = 0 2 l = 8 m , and when s = 1 m , xC = 3 m xA = 3 m # vA = xA = 2 m>s $ aA = x A = 0 Thus, vC + [(3)2 + 16] - 1/2 (3)(2) = 0 vC = -1.2 m>s = 1.2 m>s c Ans. aC - [(3)2 + 16] - 3/2 (3)2(2)2 + [(3)2 + 16] - 1/2 (2)2 + 0 = 0 aC = -0.512 m>s2 = 0.512 m>s2 c A C Ans. 12–221. At a given instant, two particles A and B are moving with a speed of v0 along the paths shown. If B is decelerating at v'B and the speed of A is increasing at v'A, determine the acceleration of A with respect to B at this instant. Given: v0 a 8 m s 1m v'A 5 m 2 s v'B 6 m 2 s Solution: 3 2 y ( x) §x· a¨ ¸ © a¹ U 1 y' ( a) y'' ( a) d y ( x) dx y' ( x) 2 d y' ( x) dx y'' ( x) 3 T U atan ( y' ( a) ) 2 aA aAB § cos T · v0 § sin T · v'A ¨ ¸ ¨ ¸ © sin T ¹ U © cos T ¹ aA aB aAB § 0.2 · m ¨ ¸ © 4.46 ¹ s2 aB aAB 7.812 m v'B § 1 · ¨ ¸ 2 © 1 ¹ 4.47 m 2 s Ans. 12–222. Two planes, A and B, are flying at the same altitude. If their velocities are vA = 600 km>h and vB = 500 km>h such that the angle between their straight-line courses is u = 75°, determine the velocity of plane B with respect to plane A. A vA B u SOLUTION vB vB = vA + vB/A 75° [500 ; ] = [600 c u ] + vB/A + ) (; 500 = - 600 cos 75° + (vB/A)x (vB/A)x = 655.29 ; (+ c) 0 = -600 sin 75° + (vB/A)y (vB/A)y = 579.56 c 1vB>A2 = 31655.2922 +1579.5622 vB/A = 875 km/h u = tan-1 a 579.56 b = 41.5° b 655.29 Ans. Ans. 12 –231. At the instant shown, cars A and B travel at *12–224. At the instant shown, cars A and B travel at speeds speeds of 70 kmh and 50 kmh, respectively. If B is of 70 mih and 50 mih, respectively. If2 B is increasing its speed increasing its speed by 1100 kmh , while A maintains a by 1100 mih2, while A maintains a constant speed, determine constant speed, determine the velocity and acceleration of the velocity and acceleration of B with respect to A. Car B B with respect to A. Car B moves along a curve having a moves along a curve having a radius of curvature of 0.7 mi. radius of curvature of 0.7 km. A VA 70 kmh Relative Velocity: v" v! v"! 30° 50 cos 30°j 70j 50 sin 30°i v"! {25.0i v"! 26.70j} kmh Thus, the magnitude of the relative velocity vB/A is 2"! 25.02 ( 26.70)2 36.6 kmh Ans. The direction of the relative velocity is the same as the direction of that for relative acceleration. Thus 1 . tan 26.70 46.9° 25.0 Ans. Relative Acceleration: Since car B is traveling along a curve, its normal 22" 502 acceleration is (A")N 3571.43 kmh2. Applying Eq. 12–35 gives + 0.7 a" a! (1100 sin 30° 3571.43 cos 30°)i a"! (1100 cos 30° a"! {3642.95i 3571.43 sin 30°)j 0 a"! 833.09j} kmh2 Thus, the magnitude of the relative velocity aB/A is A"! 3642.952 ( 833.09)2 3737 kmh2 Ans. And its direction is tan 1 833.09 12.9° 3642.95 VB 50 kmh Ans. B •12–225. 12 –232. At the instant shown, cars A and B travel at speeds of 70 kmh mih and 50 kmh, mih, respectively. respectively. If B is decreasing its speed speed by at 1400 kmh mih22,while while A A is is increasing its speed at 800 mih kmh22,, determine determine the the acceleration acceleration of B with respect to A. Car B moves along a curve having a radius of curvature of 0.7 mi. km. A VA 70 kmh Relative Acceleration: Since car B is traveling along a curve, its normal acceleration 22" 502 is (A")N 3571.43 kmh2. Applying Eq. 12–35 gives + 0.7 a" a! (3571.43 cos 30° 1400 sin 30°)i a"! ( 1400 cos 30° a"! {2392.95i 3571.43 sin 30°)j 800j a"! 3798.15j} kmh2 Thus, the magnitude of the relative acc. aB/A is A"! 2392.952 ( 3798.15)2 4489 kmh2 Ans. And its direction is tan 1 3798.15 57.8° 2392.95 Ans. VB 50 kmh 30° B 12–239. Both boats A and B leave the shore at O at the same time. If A travels at vA and B travels at vB, write a general expression to determine the velocity of A with respect to B. A B SOLUTION O Relative Velocity: vA = vB + vA>B vA j = vB sin ui + vB cos uj + vA>B vA>B = - vB sin ui + (vA - vB cos u)j Thus, the magnitude of the relative velocity vA>B is vA>B = 2( -vB sin u)2 + (vA - vB cos u)2 = 2vA2 + vB2 - 2vA vB cos u Ans. And its direction is u = tan-1 vA - vB cos u vB sin u b Ans.