SOLUTION v = v 0 + a c t. v = 8 + 6(0.5) = 11 rad>s. v = rv; v A = 2(11) = 22 ft>s. Ans. a t = ra; (a A ) t = 2(6) = 12.0 ft>s 2. Ans.

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Download "SOLUTION v = v 0 + a c t. v = 8 + 6(0.5) = 11 rad>s. v = rv; v A = 2(11) = 22 ft>s. Ans. a t = ra; (a A ) t = 2(6) = 12.0 ft>s 2. Ans."

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1 16 3. The disk is originally rotating at v 0 = 8 rad>s. If it is subjected to a constant angular acceleration of a = 6 rad>s 2, determine the magnitudes of the velocity and the n and t components of acceleration of point at the instant t = 0.5 s. 1.5 ft V 0 2ft 8 rad/s v = v 0 + a c t v = 8 + 6(0.5) = 11 rad>s v = rv; v = 2(11) = 22 ft>s a t = ra; (a ) t = 2(6) = 12.0 ft>s 2 a n = v 2 r; (a ) n = (11) 2 (2) = 242 ft>s 2

2 The vertical-axis windmill consists of two blades that have a parabolic shape. If the blades are originally at rest and begin to turn with a constant angular acceleration of a c = 0.5 rad>s 2, determine the magnitude of the velocity and acceleration of points and on the blade when t=4 s. a c 0.5 rad/s 2 ngular Motion: The angular velocity of the blade at t = 4s can be obtained by applying Eq v = v 0 + a c t = (4) = 2.00 rad>s Motion of and : The magnitude of the velocity of points and on the blade can be determined using Eq ft 20 ft v = vr = 2.00(20) = 40.0 ft>s v = vr = 2.00(10) = 20.0 ft>s The tangential and normal components of the acceleration of points and can be determined using Eqs and respectively. (a t ) = ar = 0.5(20) = 10.0 ft>s 2 (a n ) = v 2 r = (20) = 80.0 ft>s 2 (a t ) = ar = 0.5(10) = 5.00 ft>s 2 (a n ) = v 2 r = (10) = 40.0 ft>s 2 The magnitude of the acceleration of points and are (a) = 2(a t ) 2 + (a n ) 2 = = 80.6 ft>s 2 (a) = 2(a t ) 2 + (a n ) 2 = = 40.3 ft>s 2

3 The disk starts from rest and is given an angular acceleration a = (2t 2 ) rad>s 2, where t is in seconds. Determine the angular velocity of the disk and its angular displacement when t = 4 s. 0.4 m P When t = 4 s, a = dv dt v = 2 3 t3 2 0 v = 2 3 t3 = 2 t 2 v t dv = 2 t 2 dt L0 L 0 t v = 2 3 (4)3 = 42.7 rad>s u t 2 du = L0 L 0 3 t3 dt u = 1 6 t4 When t = 4 s, u = 1 6 (4)4 = 42.7 rad

4 The vacuum cleaner s armature shaft S rotates with an angular acceleration of a = 4v 3>4 rad>s 2, where v is in rad>s. Determine the brush s angular velocity when t = 4 s, starting from v 0 = 1 rad>s, at u = 0. The radii of the shaft and the brush are 0.25 in. and 1 in., respectively. Neglect the thickness of the drive belt. Motion of the Shaft: The angular velocity of the shaft can be determined from L dt = L t v s dt = L0 L 1 t 2 t v s 0 = v S 1>42 1 1>4 t = v S 1 dv S a S dv S 4v S 3>4 S S v S = (t+1) 4 When t = 4s v s = 5 4 = 625 rad>s Motion of the eater rush: Since the brush is connected to the shaft by a non-slip belt, then v r = v s r s v = r s v r s = a 0.25 b(625) = 156 rad>s 1

5 The driving belt is twisted so that pulley rotates in the opposite direction to that of drive wheel. If the angular displacement of is u = (5t t 2 ) rad, where t is in seconds, determine the angular velocity and angular acceleration of when t = 3 s. 200 mm v 125 mm v Motion of Wheel : The angular velocity and angular acceleration of wheel can be determined from v = du dt = 15t t rad>s and a = dv dt = 30t + 20 rad>s When t = 3 s, v = (3) = 195 rad>s a = 30(3) + 20 = 110 rad>s Motion of Wheel : Since wheels and are connected by a nonslip belt, then v r = v r v = a r bv r = a 200 b(195) = 312 rad>s 125 a r = a r a = a r ba r = a 200 b(110) = 176 rad>s2 125

6 The rod assembly is supported by ball-and-socket joints at and. t the instant shown it is rotating about the y axis with an angular velocity v = 5 rad>s and has an angular acceleration a = 8 rad>s 2. Determine the magnitudes of the velocity and acceleration of point at this instant. Solve the problem using artesian vectors and Eqs and v = v * r x 0.4 m z 0.4 m 0.3 m V y v = 5j * (-0.4i + 0.3k) = {1.5i + 2k} m>s v = = 2.50 m>s a = a * r - v 2 r = 8j * (-0.4i + 0.3k)-5 2 (-0.4i + 0.3k) = {12.4i - 4.3k} m>s 2 a = (-4.3) 2 = 13.1 m>s 2

7 Rotation of the robotic arm occurs due to linear movement of the hydraulic cylinders and. If this motion causes the gear at D to rotate clockwise at 5 rad>s, determine the magnitude of velocity and acceleration of the part held by the grips of the arm. 4ft 45 2ft Motion of Part : Since the shaft that turns the robot s arm is attached to gear D, then the angular velocity of the robot s arm v R = v D = 5.00 rad>s. The distance of part from the rotating shaft is r = 4 cos sin 45 = ft. The magnitude of the velocity of part can be determined using Eq D 3ft v = v R r = 5.00(4.243) = 21.2 ft>s The tangential and normal components of the acceleration of part can be determined using Eqs and respectively. a t = ar = 0 a n = v 2 R r = (4.243) = ft>s 2 The magnitude of the acceleration of point is a = 2a 2 t + a 2 n = = 106 ft>s 2

8 The mechanism shown is known as a Nuremberg scissors.if the hook at moves with a constant velocity of v, determine the velocity and acceleration of collar as a function of u. The collar slides freely along the vertical guide. u u L/2 L/2 v x = 3L sin u v = x # = 3L cos u u # y = L cos u y # = -L sin u u # y # v = - L sin u u # 3L cos u u # y # = (v tan u)>3 T y $ = v 3 (sec2 u u # ) = v 3 a 1 cos 2 u ba v 3L cos u b y $ = v 2 9L cos 3 u T

9 The bridge girder G of a bascule bridge is raised and lowered using the drive mechanism shown. If the hydraulic cylinder shortens at a constant rate of 0.15 m>s, determine the angular velocity of the bridge girder at the instant u = 60. G u Position oordinates:pplying the law of cosines to the geometry shown in Fig. a, s 2 = (3)(5) cos 180 -u s 2 = cos 180 -u However, cos 180 -u = - cos u. Thus, s 2 = cos u Time Derivatives: Taking the time derivative, 2ss # = sin uu # ss # = - 15sin uu # (1) When u = 60, s = cos 60 = 7m. lso, s # = m>s since s # is directed towards the negative sense of s. Thus, Eq. (1) gives = -15sin 60 u # 5m 3m v = u # = rad>s

10 ar rotates uniformly about the fixed pin with a constant angular velocity V. Determine the velocity and acceleration of block, at the instant u = 60. V L L u L cos u + L cos f = L cos u + cos f = 1 L sin u u # ] + sinf f # = 0 cos u(u # ) 2 + sin uu $ + sinff $ + cos f (f # ) 2 = 0 (1) (2) When u = 60, f = 60, thus, u # = -f # = v (from Eq. (1)) u $ = 0 f $ = v 2 (from Eq.(2)) lso, s = L sin f - L sin u v = L cos ff # - L cos uu # a = -L sin f (f # ) 2 + L cos f (f $ ) - L cos u(u $ ) + L sin u(u # ) 2 t u = 60, f = 60 s = 0 v = L(cos 60 )(-v) - L cos 60 (v) = -Lv = Lvc a = -L sin 60 (-v) 2 + L cos 60 (-1.155v 2 ) L sin 60 (v) 2 a = Lv 2 = Lv 2 c

11 If the wedge moves to the left with a constant velocity v, determine the angular velocity of the rod as a function of u. L v u f Position oordinates:pplying the law of sines to the geometry shown in Fig. a, x sin(f - u) = However, sin180 - f = sinf. Therefore, L sin180 - f L sin(f - u) x = sin180 - f x = L sin (f - u) sin f Time Derivative:Taking the time derivative, x # = L cos (f - u)(-u# ) sin f v = x # L cos (f - u)u # = - sin f (1) Since point is on the wedge, its velocity is v = -v. The negative sign indicates that v is directed towards the negative sense of x. Thus, Eq. (1) gives u # = v sin f L cos (f - u)

12 The Geneva wheel provides intermittent rotary motion v for continuous motion v D = 2 rad>s of disk D. y choosing d = mm, the wheel has zero angular velocity at the instant pin enters or leaves one of the four slots. Determine the magnitude of the angular velocity V of the Geneva wheel at any angle u for which pin is in contact with the slot. 100 mm v D u d 100 u 2 mm f v f D 100 mm tan f = 0.1 sin u cos u = sin u 22 - cos u sec 2 ff # = 22 - cos u(cos uu# ) - sin u(sin uuª ) 22 - cos u 2 = 22 cos u cos u 2 u# (1) From the geometry: r 2 = (0.1 sin u) 2 + [ cos u] 2 = cos u sec 2 f = r cos u = = 2 [ cos u] [ cosu] 2 (3-222 cos u) 22 - cos u 2 From Eq. (1) cos u 22 cos u - 1 fª= cos u 22 - cos u u# 2 f # = 22 cos u cos u u# Here f = v and u # = v D = 2 rad>s 22 cos u - 1 v = 2a cos u b

13 The shaper mechanism is designed to give a slow cutting stroke and a quick return to a blade attached to the slider at. Determine the velocity of the slider block at the instant u = 60, if link is rotating at 4 rad>s. 125 mm 45 v = v + v * r > -v i = -4(0.3) sin 30 i + 4(0.3) cos 30 j + vk * ( cos 45 i sin 45 j) v 4rad/s u 300 mm -v = v 0 = v Solving, v = 6.79 rad> s v = 1.64 m>s

14 If the angular velocity of link is v = 3 rad>s, determine the velocity of the block at and the angular velocity of the connecting link at the instant u = 45 and f = 30. θ =45 3ft v = v + v > 2ft ω = 3 rad/s φ =30 v ; R = 6 S + Dv (3) T 30 c 45 b ( : + ) -v = 6 sin 30 - v (3) cos 45 (+ c) 0 = -6 cos 30 + v (3) sin 45 v = 2.45 rad>s v = 2.20 ft>s ; d lso, v = v + v * r > -v i = (6 sin 30 i - 6 cos 30 j) + (v k) * (3 cos 45 i + 3 sin 45 j) a: + b -v = v (+ c) 0 = v v = 2.45 rad s v = 2.20 ft s ; d

15 Determine the velocity of point on the rim of the gear at the instant shown. 4ft/s ft O 1.50 ft 3ft/s General Plane Motion: pplying the relative velocity equation to points and and referring to the kinematic diagram of the gear shown in Fig. a, v = v + v * r > 3i = -4i + -vk * 2.25j 3i = 2.25v - 4i Equating the i components yields 3 = 2.25v - 4 v = rad>s (1) (2) For points and, v = v + v * r > v x i + v y j = -4i k * i j v x i + v y j = i j Equating the i and j components yields v x = ft>s v y = ft>s Thus, the magnitude of v is v = 2v x 2 + v y 2 = = 5.16 ft>s and its direction is u = tan - 1 v y S = tan v x = 39.8

16 If the slider block is moving at v = 3 m>s, determine the angular velocity of and the crank at the instant shown. 0.5 m 60 1 m Rotation bout a Fixed xis: Referring to Fig. a, 45 v 3 m/s v = v * r = (-v k) * (0.5 cos 60 i sin 60 j) = v i v j General Plane Motion: pplying the relative velocity equation and referring to the kinematic diagram of link shown in Fig. b, v = v + v * r > v i v j = -3j + (-v k) * (-1 cos 45 i + 1 sin 45 j) v i v j = v i + (0.7071v - 3)j Equating the i and j components yields, Solving, v = v -0.25v = v - 3 v = 2.69 rad>s v = 4.39 rad>s

17 If crank rotates with a constant angular velocity of v = 6 rad>s, determine the angular velocity of rod and the velocity of the slider block at the instant shown.the rod is in a horizontal position. 0.3 m 30 v 6 rad/s 0.5 m 60 Rotation bout a Fixed xis: Referring to Fig. a, v = v * r = (6k) * (0.3 cos 30 i sin 30 j) = [-0.9i j] General Plane Motion: pplying the relative velocity equation to the kinematic diagram of link shown in Fig. b, v = v + v * r > (-0.9i j) = (-v cos 60 i - v sin 60 j) + (-v k) * (-0.5 i) -0.9i j = -0.5v i + (0.5v v )j Equating the i and j components yields -0.9 = -0.5v = 0.5v v (1) (2) Solving Eqs. (1) and (2) yields v = 1.80 m>s v = 6.24 rad>s

18 * If the slider block is moving downward at v = 4 m>s, determine the velocity of block at the instant shown. 250 mm mm E 400 mm 30 D 300 mm v 4 m/s General Plane Motion: pplying the relative velocity equation by referring to the kinematic diagram of link shown in Fig. a, v = v + V * r > v i = -4j + (-v k) * c -0.55a 4 5 b i a 3 5 bj d v i = 0.33v i + (0.44v - 4) j Equating j component, 0 = 0.44v - 4 v = rad>sb Using the result of, v v D = v + V * r D> = -4j + (-9.091k) * c -0.3a 4 5 bi + 0.3a 3 5 bj d = {1.636i j} m>s Using the result of to consider the motion of link DE,Fig. b, v D v = v D + V D * r >D v i = (1.636i j) + (-v D k) * (-0.4 cos 30 i sin 30 j) v i = ( v D )i + (0.3464v D )j Equating j and i components, 0 = v D v D = rad>sb v = (5.249) = m>s :

19 The oil pumping unit consists of a walking beam, connecting rod, and crank D. If the crank rotates at a constant rate of 6 rad>s, determine the speed of the rod hanger H at the instant shown. Hint: Point follows a circular path about point E and therefore the velocity of is not vertical. 9ft 1.5 ft E 9ft 9ft 3ft 6 rad/s 10 ft D H Kinematic Diagram: From the geometry, u = tan - 1 a 1.5 and 9 b = r E = = ft. Since crank D and beam E are rotating about fixed points D and E, then v and v are always directed perpendicular to crank D and beam E, respectively. The magnitude of v and v are y = v D r D = 6(3) = 18.0 ft>s and y = v E r E = 9.124v E. t the instant shown, v is directed vertically while v is directed with an angle with the vertical. Instantaneous enter: The instantaneous center of zero velocity of link at the instant shown is located at the intersection point of extended lines drawn perpendicular from v and v. From the geometry r >I = 10 = ft sin r >I = 10 = 60.0 ft tan The angular velocity of link is given by v = y = 18.0 = rad>s r >I 60.0 Thus, the angular velocity of beam E is given by y = v r >I 9.124v E = 0.300(60.83) v E = 2.00 rad>s The speed of rod hanger H is given by y H = v E r E = 2.00(9) = 18.0 ft>s

20 If the collar at is moving downward to the left at v = 8m>s, determine the angular velocity of link at the instant shown. 350 mm sin 75 = v = r I- sin 45 = r I- = m r I- = m r I- sin 60 8 = rad>s v = (0.2562) = m>s 45 V mm v = = 13.1 rad>s

21 The wheel is rigidly attached to gear, which is in mesh with gear racks D and E. If D has a velocity of v D = 6 ft>s to the right and the wheel rolls on track without slipping, determine the velocity of gear rack E. D 1.5 ft 0.75 ft v E O v D 6ft/s E General Plane Motion: Since the wheel rolls without slipping on track, the I is located there, Fig. a. Here, r D>I = 2.25 ft r E>I = 0.75 ft Thus, the angular velocity of the gear can be determined from Then, v = v D = 6 = rad>s r D>I 2.25 v E = vr E>I = 2.667(0.75) = 2ft>s ;

22 If rod is rotating with an angular velocity v = 3 rad>s, determine the angular velocity of rod at the instant shown. 3ft 4ft 2ft v 3 rad/s D Kinematic Diagram: From the geometry, u = sin sin 60-2 sin 45 a b = Since links and D is rotating about fixed points and D, then v and v are always directed perpendicular to links and D, respectively. The magnitude of v and v are y = v r = 3(2) = 6.00 ft>s and y = v D r D = 4v D. t the instant shown, v is directed at an angle of 45 while v is directed at 30 Instantaneous enter: The instantaneous center of zero velocity of link at the instant shown is located at the intersection point of extended lines drawn perpendicular from v and v. Using law of sines, we have r >I sin = 3 sin 75 r >I = ft r >I sin = 3 sin 75 r >I = ft The angular velocity of link is given by v = y r >I = = rad>s = 1.98 rad>s

23 t a given instant the top end of the bar has the velocity and acceleration shown. Determine the acceleration of the bottom and the bar s angular acceleration at this instant. v 5ft/s a 7ft/s 2 10 ft v = 5 5 = 1.00 rad>s 60 a = a + a > a = a(10) : T h 30 a 30 ( : + ) a = 0-10 sin 30 + a(10) cos 30 (+ c) 0 = cos 30 + a(10) sin 30 a = rad>s 2 = rad>s 2 b a = ft>s 2 = 7.88 ft>s 2 ; lso: a = a - v 2 r > + a * r > a i = - 7j - (1) 2 (10 cos 60 i - sin 60 j) + (ak) * (10 cos 60 i - 10 sin 60 j) : + a = - 10 cos 60 + a(10 sin 60 ) (+ c) 0 = sin 60 + a(10 cos 60 ) a = rad>s 2 = rad>s 2 b a = ft>s 2 = 7.88 ft>s 2 ;

24 t a given instant, the slider block has the velocity and deceleration shown. Determine the acceleration of block and the angular acceleration of the link at this instant. 300 mm v = v r >I = 1.5 = 7.07 rad>s 0.3 cos 45 v = 1.5 m/s 45 a =16m/s 2 a = a + a * r > - v 2 r > -a j = 16i + (ak) * (0.3 cos 45 i sin 45 j) - (7.07) 2 (0.3 cos 45 i sin 45 j) a : + b 0 = 16 - a(0.3) sin 45 - (7.07) 2 (0.3) cos 45 (+T) a = 0 - a(0.3) cos 45 + (7.07) 2 (0.3) sin 45 Solving: a > = 25 4 rad>s 2 d a = 5.21 m s 2 T

25 The hydraulic cylinder is extending with a velocity of v and an acceleration of a = 1.5 ft>s 2 = 3 ft>s. Determine the angular acceleration of links and at the instant shown. 3 ft v 3 ft/s a 1.5 ft/s 2 D 1.5 ft 45 ngular Velocity: Since link rotates about a fixed axis,fig. a, then The location of the I for link is indicated in Fig. b. From the geometry of this figure, Then v = v r = v (1.5) r >I = 3 tan 45 = 3 ft v = v r >I = 3 3 = 1 rad>s r >I = 3 = ft cos 45 and v = v r >I v (1.5) = (1)(4.243) v = rad>s cceleration and ngular cceleration: Since crank rotates about a fixed axis, Fig. c, then a = a * r - v 2 r = (a k) * (-1.5 cos 45 i sin 45 j) (-1.5 cos 45 i sin 45 j) = ( a )i - ( a )j Using this result and applying the relative acceleration equation by referring to Fig. d, a = a + a * r > -v 2 r > ( a = -1.5i + (a k) * (-3i) )i - ( a )j (-3i) ( a )i - ( a )j = 1.5i - 3a j Equating the i and j components, yields a = 1.5 -( a ) = -3a Solving Eqs. (1) and (2), a = 6.59 rad>s 2 a = 5.16 rad>s 2 (1) (2)

26 The disk is moving to the left such that it has an angular acceleration a = 8 rad>s 2 and angular velocity v = 3 rad>s at the instant shown. If it does not slip at, determine the acceleration of point m v 3 rad/s a 8 rad/s 2 D 45 a = 0.5(8) = 4m>s 2 a = a + a > a = c 4 d + D(3) 2 (0.5)T + D (0.5)(8) T ; a 30 f 30 : + (a ) x = cos sin 30 = m>s 2 + c (a ) y = sin 30-4 cos 30 = m>s 2 a = 2(1.897) 2 + (-1.214) 2 = 2.25 m>s 2 u = tan - 1 a b = 32.6 c lso, a = a + a * r > - v 2 r > (a ) x i + (a ) y j = -4i + (8k) * (-0.5 cos 30 i sin 30 j) - (3) 2 (-0.5 cos 30 i sin 30 j) : + (a ) x = (0.5 sin 30 ) + (3) 2 (0.5 cos 30 ) = m>s 2 + c (a ) y = 0-8(0.5 cos 30 ) + (3) 2 (0.5 sin 30 ) = m>s 2 u = tan - 1 a b = 32.6 c a = 2(1.897) 2 + (-1.214) 2 = 2.25 m>s 2

27 cord is wrapped around the inner spool of the gear. If it is pulled with a constant velocity v, determine the velocities and accelerations of points and. The gear rolls on the fixed gear rack. 2r G r v Velocity analysis: v = v r v = vr >I = v (4r) = 4v : r v = v r >I = v r 2(2r)2 + (2r) 2 = 222v a45 cceleration equation: From Example 16 3, Since a G = 0, a = 0 r >G = 2r j r >G = -2r i a = a G + a * r >G - v 2 r >G = a v 2(2rj) r b 2v 2 = - r j a = 2v2 r T a = a G + a * r >G - v 2 r >G = a v r b 2(-2ri) = 2v 2 r i a = 2v2 r :

28 The wheel rolls without slipping such that at the instant shown it has an angular velocity V and angular acceleration. Determine the velocity and acceleration of point on the rod at this instant. 2a a O V, v = v + v / (Pin) ;+ v = 1 22 Qv22aR + 2av a 1 2 b + c O = Qv22aR + 2av a 22 2 b v = v 23 v = 1.58 va a = a O + a /O (Pin) (a ) x + (a ) y = aa + a(a) + v 2 a ; T ; T : (a ) x = aa - v 2 a (a ) y = aa a = a + a / (Pin) a = aa - v 2 a + 2a(a )a 1 2 b - 2a a v 23 b O = -aa + 2aa a 2 23 b + 2aa v 23 b 2a 1 2 b a =0.577a v 2 a = 1.58aa v 2 a

29 Pulley rotates with the angular velocity and angular acceleration shown. Determine the angular acceleration of pulley at the instant shown. 50 mm v 40 rad/s a 5 rad/s 2 ngular Velocity: Since pulley rotates about a fixed axis, v = v r = 40(0.05) = 2m>s c 50 mm 125 mm The location of the I is indicated in Fig. a. Thus, v = v r >I = 2 = rad>s E cceleration and ngular cceleration: For pulley, (a ) t = a r = 5(0.05) = 0.25 m>s 2 c Using this result and applying the relative acceleration equation to points and D by referring to Fig. b, a D = a + a * r D> - v 2 r D> (a D ) n i = (a ) n i j + (-a k) * (0.175i) (0.175i) (a D ) n i = [(a ) n ]i + ( a )j Equating the j components, 0 = a a = 1.43 rad>s 2

30 all moves along the slot from to with a speed of 3ft>s, which is increasing at 1.5 ft>s 2, both measured relative to the circular plate. t this same instant the plate rotates with the angular velocity and angular deceleration shown. Determine the velocity and acceleration of the ball at this instant. v 6 rad/s a 1.5 rad/s 2 z Reference Frames: The xyz rotating reference frame is attached to the plate and coincides with the fixed reference frame XYZ at the instant considered, Fig. a. Thus, 2ft 45 the motion of the xyz frame with respect to the XYZ frame is v O = a O = 0 v = [6k] rad>s v # 2ft = a = [-1.5k] rad>s 2 x y For the motion of ball with respect to the xyz frame, (v rel ) xyz = (-3 sin 45 i - 3 cos 45 j) ft>s = [-2.121i j] ft>s (a rel ) xyz = (-1.5 sin 45 i cos 45 j) ft>s 2 = [-1.061i j] ft>s 2 From the geometry shown in Fig. b, r >O = 2 cos 45 = ft. Thus, r >O = ( sin 45 i cos 45 j)ft = [-1i + 1j] ft Velocity:pplying the relative velocity equation, v = v O + v * r >O + (v rel ) xyz = 0 + (6k) * (-1i + 1j) + (-2.121i j) = [-8.12i j] ft>s cceleration: pplying the relative acceleration equation, we have a = a O + v # * r >O + v * (v * r >O ) + 2v * (v rel ) xyz + (a rel ) xyz = 0 + (1.5k) * (-1i + 1j) + (6k) * [(6k) * (-1i + 1j)] + 2(6k) * (-2.121i j) + (-1.061i j) = [61.9i j]ft>s 2

31 lock, which is attached to a cord, moves along the slot of a horizontal forked rod. t the instant shown, the cord is pulled down through the hole at O with an acceleration of 4m>s 2 and its velocity is 2m>s. Determine the acceleration of the block at this instant. The rod rotates about O with a constant angular velocity v = 4 rad>s. V y O 100 mm x Motion of moving reference. v O = 0 a O = 0 Æ=4k Æ # = 0 Motion of with respect to moving reference. r >O = 0.1i v >O = -2i a >O = -4i Thus, a = a O +Æ # * r >O +Æ*(Æ *r >O ) + 2Æ *(v >O ) xyz + (a >O ) xyz = (4k) * (4k * 0.1i) + 2(4k * (-2i)) - 4i a = {-5.60i - 16j}m>s 2

32 The quick-return mechanism consists of a crank, slider block, and slotted link D. If the crank has the angular motion shown, determine the angular motion of the slotted link at this instant. 100 mm D v 3 rad/s a 9 rad/s 2 30 v = 3(0.1) = 0.3 m>s (a ) t = 9(0.1) = 0.9 m>s mm (a ) n = (3) 2 (0.1) = 0.9 m>s 2 v = v +Æ*r > + (v > ) xyz 0.3 cos 60 i + 0.3sin 60 j = 0 + (v D k) * (0.3i) + v > i v > = 0.15 m>s v D, a D v D = rad>s d a = a +Æ # * r > +Æ*(Æ *r > ) + 2Æ *(v > ) xyz + (a > ) xyz 0.9 cos 60 i cos 30 i sin 60 j sin 30 j = 0 + (a D k) * (0.3i) +(0.866k) * (0.866k * 0.3i) + 2(0.866k * 0.15i) + a > i i j = 0.3a D j i j + a > i a > = m>s 2 a D = 3.23 rad>s 2 d

33 t the instant shown, the robotic arm is rotating counter clockwise at v = 5 rad>s and has an angular acceleration a = 2 rad>s 2. Simultaneously, the grip is rotating counterclockwise at v = 6 rad>s and a = 2 rad>s2, both measured relative to a fixed reference. Determine the velocity and acceleration of the object heldat the grip. y 300 mm 125 mm ω,α x ω,α v = v +Æ*r > + (v > ) xyz a = a +Æ # * r > +Æ*(Æ *r > ) + 2Æ *(v > ) xyz + (a > ) xyz (1) (2) Motion of moving reference Motion of with respect to moving reference r > = {0.125 cos 15 i sin 15 j}m Æ={6k} rad>s Æ # = {2k} rad>s 2 (v > ) xyz = 0 (a > ) xyz = 0 Motion of : v = v * r > = (5k) * (0.3 cos 30 i sin 30 j) = {-0.75i j}m>s a = a * r > -v 2 r > = (2k) * (0.3 cos 30 i sin 30 j) - (5) 2 (0.3 cos 30 i sin 30 j) = { i j}m>s 2 Substitute the data into Eqs. (1) and (2) yields: v = (-0.75i j) + (6k) * (0.125 cos 15 i sin 15 j) + 0 = {-0.944i j}m>s a = ( i j) + (2k) * (0.125 cos 15 i sin 15 j) + (6k) * [(6k) * (0.125 cos 15 i sin 15 j)] = {-11.2i j}m s 2

34 If the slotted arm rotates about the pin with a constant angular velocity of v = 10 rad>s, determine the angular velocity of link D at the instant shown. 600 mm D Reference Frame: The xyz rotating reference frame is attached to link and coincides with the XYZ fixed reference frame at the instant considered, Fig. a.thus, the motion of the xyz frame with respect to the XYZ frame is v 10 rad/s mm v = 0 v = [10k] rad>s v # = 0 For the motion of point D relative to the xyz frame, we have r D> = [0.6i] m (v rel ) xyz = (v rel ) xyz i Since link D rotates about a fixed axis, can be determined from v D = v D * r D = (v D k) * (0.45 cos 15 i sin 15 j) = v D i v D j Velocity: pplying the relative velocity equation, we have v D v D = v + v * r D> + (v rel ) xyz v D i v D j = 0 + (10k) * (0.6i) + (v rel ) xyz i v D i v D j = (v rel ) xyz i + 6j Equating the i and j components Solving, v D = (v rel ) xyz v D = 6 v D = rad>s = 13.8 rad>s (v rel ) xyz = m>s

35 If the piston is moving with a velocity of v = 3 m>s and acceleration of a = 1.5 m>s 2, determine the angular velocity and angular acceleration of the slotted link at the instant shown. Link slides freely along its slot on the fixed peg. Reference Frame: The xyz reference frame centered at rotates with link and coincides with the XYZ fixed reference frame at the instant considered, Fig. a.thus, the motion of the xyz frame with respect to the XYZ frame is v 3 m/s a 1.5 m/s m 30 v = a = 0 v = -v k a = -a k The motion of point with respect to the xyz frame is r > = [-0.5i] m (v rel ) xyz = (v rel ) xyz i (a rel ) xyz = (a rel ) xyz i The motion of point with respect to the XYZ frame is v = 3 cos 30 i + 3 sin 30 j = [2.598i + 1.5j] m>s a = 1.5 cos 30 i sin 30 j = [1.299i j] m>s Velocity: pplying the relative velocity equation, v = v + v * r > + (v rel ) xyz 2.598i + 1.5j = 0 + (-v k) * (-0.5i) + (v rel ) xyz i 2.598i + 1.5j = (v rel ) xyz i + 0.5v j Equating the i and j components, (v rel ) xyz = m>s 0.5v = 1.5 v = 3 rad>s cceleration: pplying the relative acceleration equation, a = a + v # * r > + v * (v * r > ) + 2v * (v rel ) xyz + (a rel ) xyz 1.299i j = 0 + (-a k) * (-0.5i) + (-3k) * [(-3k) * (-0.5i)] + 2(-3k) * (2.598i) + (a rel ) xyz i 1.299i j = (a rel ) xyz Di + (0.5a )j Equating the j components, 0.75 = 0.5a a = rad>s 2 = 32.7 rad>s 2

36 The two-link mechanism serves to amplify angular motion. Link has a pin at which is confined to move within the slot of link D. If at the instant shown, (input) has an angular velocity of v = 2.5 rad>s, determine the angular velocity of D (output) at this instant. D mm r sin 120 = 0.15 m sin 45 v 2.5 rad/s r = m v = 0 a = 0 Æ=-v D k Æ # = -a D k r > = {-0.15 i} m (v > ) xyz = (y > ) xyz i (a > ) xyz = (a > ) xyz i v = v * r > = (-2.5k) * ( cos 15 i sin 15 j) = {0.1189i j} m>s v = v +Æ*r > + (v > ) xyz i j = 0 + (-v D k) * (-0.15i) + (v > ) xyz i i j = (v > ) xyz i v D j Solving: (v > ) xyz = m>s v D = 2.96 rad>s b

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