Solutions to Chapter 3 Exercise Problems

Transcription

1 Solutions to hapter Exercise Problems Problem.1 In the figure below, points and have the same horizontal coordinate, and ω = 0 rad/s. Draw and dimension the velocity polygon. Identify the sliding velocity between the block and the slide, and find the angular velocity of link. 4 ω, 4 45 = 1 in = in r =.8 in Position nalysis: Draw the linkage to scale. in = 1 in = in 45.0 Ov Velocity Polygon 0 in/sec b, b4 b Velocity nalysis: 1 v = 1 v / = 1 ω r / 1 v = 1 ω r / = 0(.084) = 66.5 in / sec 1 v4 = 1 v = 1 v + 1 v 4 / (1) - -

2 1 v = 1 v / = 1 ω r / Now, 1 v = 66.5in / sec in the direction of r / 1 v = 1 ω r / ( to r / ) 1 v4 / is on the line of Solve Eq. (1) graphically with a velocity polygon. From the polygon, 1 v4 / = 15.6in / sec lso, 1 ω = 1 v / = =.94 rad / sec r / From the directions given in the position and velocity polygons 1 ω =.94 rad / sec W - 4 -

3 Problem. If ω = 10 rad/s W, find the velocity of point. E 110 = 1.5" DE =.5" D = 4.0" = 1.6" Α 18 4 Β 1ω 45 D Position nalysis Draw the linkage to scale

4 Velocity nalysis v = v + v = 0+ ω r v = ω r = 10(1.5) = 15 in/s (1) / / / v = v + v = v + v E E/ E/ v = v + v E E4 E/ E4 v = v + v = 0 + ω r E4 D4 E4/ D4 4 E/ D Now, ( to r ) 1 v = 15 in/s / v = ω r ( to r ) 1 1 E/ E/ E/ v = ω r ( to r ) 1 1 E4/ D4 4 E/ D E/ D 1 and ω = ω4, need to get ω to find v. Define the point F where F DF in position polygon. v = v + v F F/ v = v + v F F4 F/ F4 v = v + v F4 F F4/ F v = v 1 1 F4 F4/ D4 Solve Eq. (1) graphically with a velocity polygon

5 fter finding point f, construct the velocity image to find the point b a line a line to through the point a to F through the point f fine the point b From the polygon, 1 v =9.4 in/s - 7 -

6 Problem. If ω = 100 rad/s W, find v 4. D = 1.8" D = 0.75" E = 0.7" F = 0.45" FG = 1.75" = 1.0" D = 1.65" G 15 4 E F 4 D 95 Position nalysis Draw the linkage to scale

7 Velocity nalysis v = v + v = 0 + ω r (1) G G/ G/ v = v + v = v + ω r G G/ 4 G/ v = v + v G G G/ G v = v = v + v = 0 + ω r D4 4/ D4 4 / D Now, v = ω r = 100(.44) = 44 in/s ( to r ) 1 1 G G/ G/ ω = ω 1 1 v = ω r = 100(.65) = 65 in/s ( to r ) 1 1 G/ G/ G/ 1 v is on the line of EG G/ G v = ω r ( to r ) 1 1 4/ D4 4 / D / D Solve Eq. (1) graphically with a velocity polygon

8 To find the point b 4 use velocity polygon v / D = in / s

9 Problem.4 If ω = 50 rad/s W, find v D 4. = D D =.06" D 4 Position nalysis Draw the linkage to scale

10 Velocity nalysis v = v + v = 0 + ω r (1) / / v = v + v = v + ω r D / D D4 / D v = v + v / v = v 1 1 D4 D Now, v = ω r = 50(.9) = in/s ( to r ) 1 1 / / ω = ω 1 1 v = ω r = 50(.06) = 15 in/s ( to r ) 1 1 / D / D / D 1 v is on the line of / v = v (// to r ) 1 1 D4 D D/ - 4 -

11 Solve Eq. (1) graphically with a velocity polygon. From the velocity polygon 1 v D4 = in / s - 4 -

12 Problem.5 Determine the velocity and acceleration of point on link. in Y ω 4 = 1 rad/s α 4 = X Position nalysis Draw the mechanism to scale. Velocity nalysis r / = / cos0þ=.4641 The velocity of is given by and `1v / 1 = 4 v / ω4 r/ (1) 1ω 4 r / = =.464 ( to r/ in the direction indicated by the cross product) The direction for the velocity 1 v /1 must be vertical. Equation (1) can be solved for the unknowns. From the polygon, 1v / 1 = in / sec 4v/1 =.0148 in / sec cceleration nalysis The acceleration of is given by

13 1 a / 1 = 4 a / α4 r/ + 1 ω4 ( 1 ω4 r/ )+ 1 ω4 4 v / 1 () 4 b' b' Velocity Scale in/sec cceleration Scale in/sec O v o' The individual vectors are: 1ω 4 (1ω 4 r / ) = (1).464 =.464 in/sec (opposite to r/ ) 1α4 r/ = 0 r/ = 0 1 ω4 4 v / 1 = (1).0148 = in / sec ( to 4 v / 1 in the direction up and to left) 1α4 r/ (vector to r/ ) 4 a / 1 (vector along the slot) Equation () has only two unknowns and can be solved. From the polygon, the acceleration of is in/sec upward

14 Problem.6 If 1 ω = 100 rad/s W, find 1 ω 6. E (.0", 6.0") 6 = 1." = 6.0" D =.0" D = 4.0" F =.0" ω Y 5 F 4 D X Position nalysis Draw the linkage to scale

15 Velocity nalysis v = v / = ω r / v = ω r / = 100(1.) = 10 in/s (1) v = v + v = v + ω r / / v = v + v = 0 + ω r D4 4/ D4 4 / D v = v + v F6 F5 F6/ F5 v = v + v = 0 + ω r F6 E6 F6/ E6 6 F/ E

16 Now, 1 v = 10 in/s r / ( to ) v = ω r ( to r ) 1 1 / / / v = ω r ( to r ) 1 1 4/ D4 4 / D / D 1 v ( // to r ) F6/ F5 E/ F v = ω r ( to r ) 1 1 F6/ E6 6 F/ E F/ E Solve Eq. (1) graphically with a velocity polygon

17 To find the point f construct the velocity image by : F = bc : bf 1 1 and vf = v F5 From the polygon, 1 v / = 5. / sec F6 E in 6 and 1 v F6/ E6 5. ω 6 = = = 5.54 rad / sec W r F/ E

18 Problem.7 If ω = 50 rad/s W, find the velocity of point G E 5 = 1.16" = 0.70" D = 1.45" DE = 1.16" D = 1.0" DF = 1.0 EG =.0" ω D 6 F G Position nalysis Draw the linkage to scale

19 Velocity nalysis v = v / = ω r / v = ω r / = 50(1.16) = 58 in/s (1) v = v + v = v + ω r / / v = v = v + v = 0 + ω r D4 4/ D4 4 / D v = v + v = 0 + ω r E4 D4 E4/ D4 4 E/ D v = v + v = v + ω r G5 E5 G5/ E5 E4 5 G/ E v = v + v = 0 + v F5 F6 F5/ F6 F5/ F6 Now, ( to r ) 1 v = 58 in/s / v = ω r ( to r ) 1 1 / / /

20 v = ω r ( to r ) 1 1 4/ D4 4 / D / D v = ω r ( to r ) 1 1 G5/ E5 5 G/ E G/ E 1 v ( // to r ) F5/ F6 G/ E Solve Eq. (1) graphically with a velocity polygon. From the polygon, - 5 -

21 1 v 4 D = in 4 / 46 / sec and 1 v4/ D4 46 ω 4 = = = 1.7 rad / sec W r / D To find the point g construct the velocity image by EF : EG = ef : eg 5 From the polygon, 1 v G5 =.5 in / sec - 5 -

22 Problem.8 If ω = 5 rad/s W, find ω 6. E 5 D 6 4 F = 1.0" =.0" D = 4.0" DE = 1.5" EF = 1.5" F = 4.1" 50 ω Position nalysis Draw the linkage to scale

23 Velocity nalysis v = v / = ω r / v = ω r / = 5(1) = 5 in/s (1) v = v + v / v = v + v / 4 Now, 1 v = 5 in/s r / ( to ) v = ω r ( to r ) 1 1 4/ 4 4 / / Solve Eq. (1) graphically with a velocity polygon. r / = 1. in From the polygon, 1 v 4 = in 4 / 4.5 / sec and 1 v 4/ ω 4 = = =.8 rad / sec W r 1. 1 / With the value of ω 4 v = v + v / = ω4 r / v = ω4 r / =.8(.88) = 1.1 in/s D4 4 4 D4 D D4 D v = v 1 1 D5 D4 v = v + v D5 E5 D5/ E5 v = v 1 1 E6 E5 v = v + v = 0 + v E6 F6 E6/ F6 E6/ F6-55 -

24 1 v ( to r ) D5/ E5 D/ E 1 v ( to r ) E6/ F6 E/ F Solve graphically with a velocity polygon. From the polygon, 1 v E6 F = in 6 / 15.4 / sec and 1 ve 6/ F ω 6 = = = rad / sec W r E/ F

25 Problem.9 In the mechanism below, ω = 10 rad/s. Write the velocity equations and determine the following: v D 4, ω 4, v F 6, ω 6. Y G (-5.6,.65) D 4 E F = 1.0 in D = 4. in E =.0 in DE = 4.0 in GF =.5 in =.4 in 100 X Position nalysis Draw the linkage to scale. First locate the pivots, E, and G. Next draw link and locate. Then locate point D and point. Draw the line F as shown, and finally locate point F. Velocity nalysis: Now, 1 v = 1 v = 1 v / 1vD4 = 1 vd = 1 vd4 /E4 = 1 v + 1 vd/ (1) 1vF5 = 1 vf6 = 1 vf6/g6 = 1 vf + 1 vf5/f () 1vD4 /E4 = 1 ω4 rd4 /E4 1 vd4 /E4 = 1 ω4 rd4/e4 ( to rd4 /E4) 1 v / = 1 ω r / 1 v / = 1 ω r / =10 1=10 in / sec ( to r / ) 1vD/ = 1 ω rd/ 1 vd/ = 1 ω rd/ ( to rd/)

26 Solve Eq. (1) graphically with a velocity polygon. From the polygon, D 4 E 6 5 F f f 4 d 4 Velocity Scale 10 in/sec b o f 5 1vD4 =17.04 in / sec or 1ω4 = 1 vd4 /E4 = = 4.59 rad / sec W r D4 /E

27 lso, using velocity image 1vF4 = in / sec Now, 1vF =.7 in / sec(using velocity polygon) 1vF6/G6 = 1 ω6 rf6/g6 1 vf6/g6 = 1 ω6 rf6/g6 ( to rf6/g6) 1vF6/G6 along the slot Solve Eq. () graphically with a velocity polygon. From the polygon, 1vF6/G6 = 6.171in / sec or 1ω6 = 1 vf6/g6 = = 7.81 rad / sec W r F6 /G 6.5 Problem.10 If the velocity of point on link is 10 in/s as shown, find the velocity of point on link 5. DE =.5" D = 0.75" = 1.75" E = 1.5" GF = 1.5" v D Y " 0.5" G E 6 F (.15", 1.9") X Position nalysis Draw the linkage to scale

28 Velocity nalysis v = v + v = v + ω r (1) / / v = v = v = ω r / E4 4 / E v = v + v G5 G4 G5/ G4 v = v = v = ω r G6 G5 G6/ F6 6 G/ F v = v + v = v + ω r G5 5/ G5 G6 5 / G

29 Now, 1 v = 10 in/s r / D ( to ) v = ω r ( to r ) 1 1 / / / v = ω r ( to r ) 1 1 4/ E4 4 / E / E v = ω r ( to r ) 1 1 G5/ F6 6 G/ F G/ F 5 v ( // to r ) G5/ G4 / E Solve Eq. (1) graphically with a velocity polygon

30 From the polygon, 1 v 4 E = in 4 / 8.1 / sec and 1 v 4/ E4 8.1 ω 4 = = = 5.4 rad / sec r / E W With 1 ω = 1 ω

31 v = ω r = 5.4(0.6) = in/s ( to r ) 1 1 5/ G5 5 / G / G To find the point g 4 construct the velocity image by E : EG = be: eg 4 From the polygon, 1 v 5 = in / sec Problem.11 In the clamping device shown, links and 4 are an air cylinder. If the opening rate of the air cylinder is 5 cm/s and the opening acceleration of the cylinder is cm/s, find the angular velocity and acceleration of link, and the linear velocity and acceleration of point D on Link. D s s = 5 cm = 17 cm = 0 cm D = 10 cm 4 D Position nalysis Draw the linkage to scale. Start by locating the pivots and. Then locate point Velocity nalysis onsider the points at location

32 1v = 1 v = 1 v / 1v = 1 v v / 4 (1) Where 1v / =1ω r/ ( to r/ ) 1v 4 = 1v 4 / 4 = 1ω 4 r / ( to r / ) 1v / 4 = 5cm / s along r/ Solve Eq. (1) using a velocity polygon, and determine the velocity of D by image. 1v D = 5.79 cm / s and 1ω = 1 v / = 5.81 = 0.4 rad / s W r / 17 1ω 4 = 1 v 4 / 4 =.018 = rad / s W r / 5 cceleration nalysis gain, consider the points at location. 1α = 1 α = 1 α / = 1 r α / + 1 t α / 1α = 1 α 4 / α / 4 = 1 r α 4 / t α 4 / t α / n α / c α / 4 ombining the equations, Where r 1α / + 1 t α / = 1 r α 4 / t α 4 / t α / n α / c α / 4 () 1αr / = 1ω r / = 0.4(17) =1.99 cm / s (opposite to r / ) t 1α / = 1 α r/ ( to r/)

33 D b 4 4 t a /4 c a /4 r 1a 4 / 4 a o' b b a c 4 o t 1a 4 / 4 t 1a / r 1a / d b Velocity Scale cm/s cceleration Scale 1 cm/s d 1α r 4 / 4 = 1ω4 r / = 0.086(5) = 0.60 cm / s (opposite to r/) t 1α 4 / 4 4αt / 4 = 1 α4 r/ ( to r/) =10 cm / s (along r/) 4αn / 4 = 1 v / 4 = 0 c 1α / 4 = 1 ω4 4 v / 4 1 αc / 4 = 1 ω4 4 v / 4 = (0.086)(5) =

34 The direction for 1 c α / 4 is perpendicular to and in the direction defined by rotating 4 v / 4 90 in the direction of 1ω4. This direction is generally down and to the left. Solve Eq. () using an acceleration polygon, and determine the acceleration of D by image. 1a D = 4.9 cm / s 1α = 1 t α / = r. / 17 = rad / s W Problem.1 In the mechanism shown, link 4 moves to the left with a velocity of 8 in/s and the acceleration is 80 in/s to the left. Draw the velocity and acceleration polygons and solve for the angular velocity and acceleration of link. 4.0" Position nalysis Draw link at 45 to the horizontal lineand 4" long. onstruct link at an angle of 10 to the horizontal and through point. Velocity nalysis: 1 v = 1 v v / 4 = 1 v v / 4 (1) Now, 1v = 1 ω r/ 1 v = 1 ω r/ 1 v 4 = 8 in / sec in the horizontal direction to the left

35 4 v / 4 = 1 v / 4 along the link 4 1v = 1 ω r/ ( to r/ ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, 1v / = 4 v/ =15. in / sec b b Velocity Polygon 8 in/sec = 4" 4 b 4 b4 O a Ov a cceleration Polygon 50 in/sec 1at b lso, 1ω = 1 v / = r 0.6 = 5.157rad / sec / 4 From the directions given in the position and velocity polygons 1 ω = rad / sec W cceleration nalysis:

36 1 a = 1 a4 + 1 a /4 = 1 a4 + 4 a /4 + 1 ω4 4 v /4 () ut, 1 ω 4 = 0. Therefore, 1a = 1 a a /4 lso, Therefore, Now, 1a = 1 a / = 1 t a / + 1 r a / t 1a / + 1 r a / = 1 a a /4 1 a4 = 80 in / sec in the horizontal direction to the left 4a /4 is along the link 4 1 ar / = 1ω r / (opposite to r / ) t 1a / = 1 α r/ and is to r/ From the acceleration polygon, Therefore, t 1a / = 89 in / sec 1α = 1 at / r/ = 89 4 =.5 rad / sec From the directions given on the acceleration and position polygons, 1α =.5 rad / sec W

37 Problem.1 In the mechanism below, the angular velocity of Link is rad/s W and the angular acceleration is 5 rad/s W. Determine the following: v4, vd4, ω4, a4, ad4, α 4,. D = 5. in = 6.0 in D = 10.0 in α 4 ω 75 Position nalysis Draw the mechanism to scale. Locate the pivots and. Draw link and locate point. Then draw line D. Velocity nalysis: 1v = 1 v = 1 v / = 1 v4 + 1 v/4 (1) 1 v 4 = 1 v 4 / 4 Now, 1v / = 1 ω r/ 1 v / = 1 ω r/ = 6 =1 in / sec ( to r/ ) 1v4/4 = 1 ω4 r/ 1 v4 /4 = 1 ω4 r/ ( to r/) 1v /4 in the direction of r/ Solve Eq. (1) graphically with a velocity polygon. From the polygon,

38 b 4 d 4 O a 1 r a 4 / 4 t 1a 4 / 4 D 4a /4 1 r a / cceleration Scale 10 in/sec 5.44 in E t 1a / 1 c a /4 4 b Velocity Scale 10 in/sec b o d 4 b 4 1v4 = in / sec or 1ω4 = 1 v4/4 = = 1.191rad / sec W r / lso, 1v /4 =8.80 in / sec lso,

39 1vD4 =11.91 in / sec cceleration nalysis: 1 a = 1 a = 1 a / = 1 a a / 4 1a4 = 1 a4 /4 Now, a r / + 1 at / = 1 a r 4 / a t 4 / a / ω4 4 v 4 / 4 () 1 ar / = 1 ω ( 1 ω r/ ) 1 a r / = 1 ω r/ = 6 = 4 in / sec in the direction opposite to r / 1at / = 1 α r/ 1 at / = 1 α r/ = 5 6 = 0 in / sec ( to r/) 1 ar 4 / 4 = 1 ω4 ( 1 ω4 r/) 1 a r 4 / 4 = 1 ω4 r/ = = 9.71 in / sec in the direction opposite to r / 1at 4 / 4 = 1 α4 r/ 1 at 4 / 4 = 1 α4 r/ ( to r/ ) 4 a / 4 in the direction of r 4 / 4 1 ω4 4 v / 4 = = in / sec Solve Eq. () graphically with an acceleration polygon. From the polygon, or lso, lso, 1 at =.75 in / sec 4/4 1 α4 = 1 at 4/4 =.75 r 4 / =.469 rad / sec W 1 a 4 = 5.66 in / sec 1aD 4 = 7.47 in / sec

40 To find the center of the curvature of the path that 4 traces on link, we must find an expression which involves that the radius of curvature of the path. This term is a n 4 / and it can be evaluated from the following: therefore, and lso, 1a / 4 = 1 a 4 / 1 at / 4 = 1 a t 4 / 1 an / 4 = 1 a n 4 / an / ω4 1 v /4 = an 4 / 1 ω 1 v4 / () For our purpose, we should arrange Eq. () as Now, a 4 / ( ) n = an / ω 1 v4/ + 1 ω4 1 v /4 a / 4 n n = v /4 = 0 1 ω4 1 v / 4 = = 0.97 in / sec ( to D) down and to the left. 1 ω 1 v 4 / = 8.80 = 5.0 in / sec ( to D) up and to the right Let E be the location of the center of curvature of 4 on link. If we choose up and to the right as the positive direction, a n 4/ = 1 v 4 / r / E = ( ) = 14.4 in / sec ecause a n 4 / is negative, a n 4 / points down and to the left which is the direction of E. The magnitude of the distance is given by r/ E = 1 v4 / 14.4 = = 5.44 in The direction of E is shown on the drawing

41 Problem.14 Resolve Problem.1 if ω = rad/sec (constant) Position nalysis Draw the mechanism to scale. Locate the pivots and. Draw link and locate point. Then draw line D. Velocity nalysis: 1v = 1 v = 1 v / = 1 v v / 4 (1) 1 v 4 = 1 v 4 / 4 Now, 1v / = 1 ω r/ 1 v / = 1 ω r/ = 6 =1 in / sec ( to r/ ) 1 v 4 / 4 = 1 ω4 r/ 1 v 4 / 4 = 1 ω4 r/ ( to r/) 1v / 4 in the direction of r/ Solve Eq. (1) graphically with a velocity polygon. From the polygon, or lso, lso, 1 v 4 = in / sec 1ω4 = 1 v 4 / 4 = r = 1.191rad / sec W / v / 4 =8.80 in / sec 1vD 4 =11.91 in / sec cceleration nalysis: 1a = 1 a = 1 a / = 1 a a / 4 1 a 4 = 1 a 4 / 4 a r / + 1 at / = 1 a r 4 / a t 4 / a /4 + 1 ω4 4 v4/4 () - 7 -

42 D O a r 1a 4 / 4 cceleration Scale 10 in/sec 1 r a / b 4 1 t a 4 / 4 4ad 4 / in E 4 b 1 c a /4 Velocity Scale 10 in/sec b o d 4 b 4 Now, 1a r / = 1 ω ( 1 ω r/ ) 1 a r / = 1 ω r/ = 6 = 4 in / sec in the direction opposite to r / 1at / = 1 α r/ 1 at / = 1 α r/ = 0 6 = 0 1a r 4 / 4 = 1 ω4 ( 1 ω4 r/) 1 a r 4 / 4 = 1 ω4 r/ = = 9.71 in / sec in the direction opposite to r / 1at 4 / 4 = 1 α4 r/ 1 at 4 / 4 = 1 α4 r/ ( to r/ ) 4a / 4 in the direction of r 4 / 4 1 ω4 4 v / 4 = = in / sec

43 Solve Eq. () graphically with an acceleration polygon. From the polygon, or lso, lso, 1 at 4 / 4 =.515 in / sec 1 α4 = 1 at 4/4 =.515 r 4 / = 0.51 rad / sec W 1a 4 =10.9 in / sec 1 ad 4 = in / sec To find the center of the curvature of the path that 4 traces on link, we must find an expression which involves that the radius of curvature of the path. This term is a n 4 / and it can be evaluated from the following: therefore, and lso, 1 a / 4 = 1 a 4 / 1 at / 4 = 1 a t 4 / 1 an / 4 = 1 a n 4 / an / ω4 1 v /4 = an 4 / 1 ω 1 v4 / () For our purpose, we should arrange Eq. () as Now, a 4 / ( ) n = an / ω 1 v4/ + 1 ω4 1 v /4 a / 4 n n = v /4 = 0 1 ω4 1 v / 4 = = 0.97 in / sec ( to D) down and to the left. 1 ω 1 v 4 / = 8.80 = 5.0 in / sec ( to D) up and to the right

44 Let E be the location of the center of curvature of 4 on link. If we choose up and to the right as the positive direction, a n 4/ = 1 v 4 / r / E = ( ) = 14.4 in / sec ecause a n 4 / is negative, a n 4 / points down and to the left which is the direction of E. The magnitude of the distance is given by r/ E = 1 v 4 / 14.4 = = 5.44 in The direction of E is shown on the drawing. Note that the location of E does not depend on the acceleration of link. Problem.15 In the mechanism below, the velocity and acceleration of Point are given. Determine the angular velocity and acceleration of Links and 4. On the velocity and acceleration diagrams, locate the velocity and acceleration of Point E on Link. Y D 5 X 4 =.8" = 5." D = 1.0" DE =.4" v = -55 in/s a = in/s E Position nalysis Draw the linkage to scale. Locate both pivots and start with link. Locate point and draw line. Then locate point D. onstruct point E on a line perpendicular to the line D. Velocity nalysis: 1v = 1 v Now, 1v = 1 v v / 4 1v = 1 v + 1 v / (1)

45 1v = 1 v / 4 in the direction of r / 1v / = 1 ω r / 1 v / = 1 ω r / ( to r / ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, or 1v / = 1.6 in / s 1ω = 1 v / = r 1.6 =.51rad / sw / 8.59 lso, D 4 e d d' c b e ' 1 t a / E o c 4a /4 1ar / 1a c 1a /4 Velocity Scale 10 in/s o' cceleration Scale 50 in/s 1ω =1ω 4 b The velocity of E can be found by image. The magnitude is 1vE =18.6 in / s

46 cceleration nalysis: 1a = 1 a 1a = 1 a a / 4 1a = 4 a / ω4 4 v / 4 Now, 1a = 1 a + 1 a / 4a / ω4 4 v / 4 = 1 a + 1 r a / + 1 t a / () 4a / 4 in the direction of r / 1 ω4 1 v / 4 = =108 in / s r r 1a / =1ω ( 1ω r / ) 1a / t t 1a / = 1α r / 1a / = 1α r / ( to r / ) = 1ω r / = = 54.1 in / s Solve Eq. () graphically with an acceleration polygon. From the polygon, t 1a / = 4. in / s or 1α = 1 t a / = r 4. / 8.59 = 5.0 rad / s lso, 1α = 1α 4 The acceleration of point E is given by image. The magnitude is 1aE = 149 in / s

47 Problem.16 In the figure below, ω = 500 rad/s W (constant). Find ω 4, ω 4, ω, 6 ω 5, ω 5, v D, α 4 α 4, α, 6α 5, and a D. D E = 6 cm = cm E = 8 cm D =. cm 4 ω E Position nalysis Locate the relative position of points and E and the line of motion of point D. Next locate point. Draw the line E and locate point. Then locate point D by drawing an arc centered at and with a radius of.. Velocity nalysis: 1v = 1 v = 1 v / = 1 v v / 4 (1) 1v 4 =1v 4 /E 4 1vD 5 = 1 vd 6 = 1 v vd 5 / 5 () 1v 5 = 1 v 4 = 1 v 4 / E 4 Now, 1v / = 1 ω r / 1 v / = 1 ω r / = =15 m / s ( to r / ) 1v 4 /E 4 =1ω 4 r 4 /E 4 1v 4 /E 4 = 1ω 4 r 4 /E 4 ( to r 4 /E 4 ) 1v / 4 in the direction of r 4 /E

48 D 5 b 4 c 5 o d 5 Velocity Polygon.5 m/s 4 E b r 1a D 5/5 cceleration Polygon 000 m/s c 4 4a /4 b r 1a / c 1a /4 b' 4 t 1a D 5/5 r 1a 4 /E 4 o' t 1a 4 /E4 1a D5 d 5 ' Solve Eq. (1) graphically with a velocity polygon. From the polygon, lso, 1v / 4 =14.4 m /s or 1v 4 /E 4 = 4. m / sec 1ω 4 = 1 v 4/E4 = 4. = 95.9 rad / s W r 4 /E lso, 1ω =1ω4 Now, 1v 4 /E 4 =1ω 4 r 4 /E 4 1v 4 /E 4 = 1ω 4 r 4 /E 4 = = 7.66 m / s ( to r 4 /E 4 )

49 1vD 5 in horizontal direction 1vD 5 / 5 = 1 ω5 rd 5 / 5 1 vd 5 / 5 = 1 ω5 rd 5 / 5 ( to rd 5 / 5 ) Solve Eq. () graphically with a velocity polygon. From the polygon, or nd, 1vD 5 / 5 =.88 m / s 1ω 5 = 1 v D5 / 5 = r.88 = 11rad / s W D5 / vD 5 = 1 vd 6 =.18 m / s For the relative angular velocities, ω 4 =1ω 4 1ω = 95.8W 500 W = W 6ω 5 = 1ω 5 1ω 6 = 11W 0 =11W ω5 =1ω5 1ω =11W 95.8 W = 16.8 W cceleration nalysis: 1a = 1 a = 1 a / = 1 a a / 4 1a 4 =1a 4 /E 4 r 1a / + 1 t a / = 1 r a 4 /E t a 4 / E a / ω4 4 v / 4 () 1aD 5 = 1 ad 6 = 1 a ad 5 / 5 Now, 1a 5 = 1 a 4 = 1 a 4 /E 4 1aD 5 = 1 r a 4 /E t a 4 /E r a D5 / t a D5 / 5 (4) r r 1a / = 1ω ( 1ω r / ) 1a / = 1ω r / = = 7500 m / s in the direction opposite to r / t t 1a / =1α r / 1a / = 1α r / = 0 1a r 4 /E 4 =1ω 4 ( 1ω 4 r 4/E4 ) 1a r 4 /E 4 = 1ω 4 r 4 /E4 = = 404 m / s in the direction opposite to r 4 /E 4 1at 4 /E 4 =1 α 4 r 4/E4 1at 4 /E 4 = 1α 4 r 4 /E4 ( to r 4/E4 )

50 4a / 4 in the direction opposite to r 4 /E 4 c 1a / 4 = 1 ω4 4 v / 4 = = 760 m / s ( to r 4 /E 4 ) Solve Eq. () graphically with an acceleration polygon. From the polygon, t 1a 4 /E 4 = 9960 m / s or 1α4 = 1 at 4/E4 = 9960 r 4 /E = 6,000 rad / s W lso, α 4 =1α 4 1α = 6,000 0 = 6,000 rad / s W nd, 1α4 = 1α lso, using acceleration image 1a 4 =18,100 m /s Now, 1aD 5 in horizontal direction r r 1a D5 / 5 =1ω 5 ( 1ω 5 r D5 / 5 ) 1a D5 / 5 = 1ω 5 r D5 / 5 = = 470 m / s in the direction of -rd 5 / 5 t t 1a D5 / 5 = 1α 5 r D5 / 5 1a D5 / 5 = 1α 5 r D5 / 5 ( to r D5 / 5 ) Solve Eq. (4) graphically with an acceleration polygon. From the polygon, t 1a D5 / 5 = 8640 m / s or 1α 5 = 1 t a D5 / 5 = r 8640 D5 / = 70,000 rad / s W lso, 6α 5 = 1α 5 1α 6 = 70,000 W 0 = 70,000 rad / s W nd, α5 =1α5 1α = 70,000 W 6,000W = 496,000 rad /s W lso, 1aD 5 =19,400 m / s Problem.17 In the mechanism below, the angular velocity of Link is 60 rpm W (constant). Determine the acceleration of Point 6 and the angular velocity of Link

51 ω =.6 ft D = 9. ft = 6 ft D 4 Position nalysis First locate pivot and the line of action (through ) of the slider. Next draw link and locate point. Then locate point D and finally locate point. Velocity nalysis: 1v = 1 v / = 1 v = 1 ω r/ 1vD = 1 vd 4 = 1 v + 1 vd / (1) ompute 1 v by inage. Now, 1v 6 = 1 v 5 = 1 v 6 / 6 = 1 v + 1 v 6 / () 1ω = 60rpm = π rad / s = 6.8 rad / s 1v = 1ω r / = =.6 ft / s( to r / ) 1vD in horizontal direction 1vD / = 1 ω rd/ 1 vd / = 1 ω rd/ ( to rd/) 1v 6 / 6 = 1ω 6 r / 1v 6 / 6 = 1ω 6 r / ( to r / ) 1v 6 / along D Solve Eqs. (1) and () graphically with a velocity polygon. From the polygon, 1v 6 = 5.85 ft / s 1vD / =16.5 ft / s 1v 6 / =16.7 ft / s - 8 -

52 and 1ω = 1 v D / = 16.5 = 1.77 rad / s W r D/ 9. cceleration nalysis: 1a = 1 a / = 1 a = 1 r a / + 1 t a / 1aD = 1 ad 4 = 1 a + 1 ad / = 1 a + 1 r a D / + 1 t a D / () ompute 1 a by image. or 1a 6 = 1 a 5 = 1 a 6 / 6 = 1 a + 1 a 5 / r 1a 6 / t a 6 / 6 = 1 a + 1 c a 5 / + t a 5 / + n a 5 / (4) Now, 1aD in horizontal direction

53 6 d ' 1ar 6/ 6 o' D c b c 6 c 1a 1a 6 d Velocity Scale 10 ft/s o c 1a 6/ b t 1a 6/6 a 6 / cceleration Scale 50 ft/s c 6 r r 1a / = 1ω ( 1ω r/) 1a / in the direction opposite to r / = 1ω r/ = =14 ft / s t t 1a / = 1α r/ 1a / = 1α r/ = 0.6 = 0 ft/s t t 1a D / =1α r D/ 1a D / = 1α r /D ( to r / D ) r r 1a D / =1ω (1ω r D/ ) 1a D / = 1v D / = r D/ 9. = 9.1 ft / s in the direction opposite to rd/ t t 1a 6 / 6 =1α 6 r / 1a 6 / 6 = 1α 6 r / ( to r / )

54 r r 1a 6 / 6 =1ω 6 (1ω 6 r / ) 1a 6 / 6 = 1v 6 / 6 = r 5.85 / 6 = 5.70 ft / s in the direction opposite to r / c c 1a 5 / = 1ω v 5 / 1a 5 / = 1ω v 5 / = (1.77)(16.66) = ft / s in the direction perpendicular to D and in the direction obtained by rotating v 5 / 90 in the direction of 1ω. The direction is shown on the acceleration polygon. t a 5 / is along the slide(line D) an 5 / = 1 v 5 / = 0 Solve Eqs. () and (4) graphically with an acceleration polygon. From the polygon, 1a 6 =17 ft / s lso, 1α 6 = 1 t a 6 / 6 = 171 r / 6 = 8.5 rad / s W Problem.18 In the position shown is horizontal. Draw the velocity diagram to determine the sliding velocity of link 6. Determine a new position for point (between and D) so that the velocity of link 6 would be equal and opposite to the one calculated for the original position of point..85" ω = 5 rad/s 1.1"." =.9" D =.1" D Position nalysis: Draw the linkage to scale

55 in b, b Velocity Polygon 5 in/s 6 c D d 6 d, d 5 o Velocity nalysis: The equations needed for the analysis are: 1v = 1 v / = 1 ω r / 1 v = 1 ω r / = 5(.9) =14.5 in / s 1v = 1 v 1v = 1 v + 1 v / (1) 1v = 1 v v / 4 1v 4 = 0 1v = 1 v / 4 = 4 v / 4 = 4 v 1vD = 1 vd 5 = 1 v + 1 vd / 1vD 5 = 1 vd vd 5 /D 6 () Now, 1v =14.5 in / s ( to r/) 1v / = 1 ω r/ ( to r/) 1v is on line of r / Solve Eq. (1) graphically with a velocity polygon. From the polygon,

56 1v / = 6.90 in / s or 1ω = 1 v / = r 6.90 = 5.04 rad / s / 1.7 From the directions given in the position and velocity polygons lso, 1ω = 5.04rad / sw 1v = 1 v / 4 =1.6 in / s Using velocity image theorem, Now, 1vD = 1 vd 5 =15.5 in / s 1vD 6 is on the vertical axis, 1vD 5 /D 6 is on the horizantal axis. Solve Eq. () graphically with a velocity polygon. From the polygon, also, 1vD 6 = 6.85 in / s 1vD 5 /D 6 = 1.6 in / s To find the new location of point which will make the velocity of link 6 change signs, plot a new velocity polygon with the velocity of d 6 in the direction indicated. b, b c New Velocity Polygon 10 in/s O v d 6 d, d

57 using velocity image theorem, r / = r D/ 1 v / = (.1) 1v 6.90 = 0.48 in D / 4.7 The location of is shown on the following figure. ' D Problem.19 The scotch-yoke mechanism is driven by crank at ω = 6 rad/s (W). Link 4 slides horizontally. Find the velocity of point on Link 4. ω =.0" = 1.7" Position nalysis: Draw the linkage to scale

58 1 in =.0" = 1.7" a, a Velocity Polygon 50 in/s a 4 Ov Velocity nalysis: 1v = 1 v / = 1 ω r / 1 v = 1 ω r / = 6 () = 7 in / s 1v = 1 v v / 4 (1) 1v = 1 v Now, 1v = 7 in / s ( to r/) 1v / 4 is along the slider 1v 4 moves on a horizontal axis, and because 4 is a rigid body 1v 4 = 1v 4 Solve Eq. (1) graphically with a velocity polygon. From the polygon, 1v 4 = 1v 4 = 7.5 in / s

59 Problem.0 The circular cam shown is driven at an angular velocity ω = 15 rad/s (W) and α = 100 rad/s (W). There is rolling contact between the cam and the roller, link. Find the angular velocity and angular acceleration of the oscillating follower, link 4. 4 D E (.0",.0") Y 15 X = 1." DE =.50" =.00" D = 0.50" Position nalysis: Draw the linkage to scale. Note that because of rolling contact and because we are to find the velocity and acceleration of link 4 only, we can model the system as a four-bar linkage. If we were asked for the kinematic properties of link, we would have to model the system using rolling contact directly. Velocity nalysis: 1v = 1 v / = 1 ω r / 1 v = 1 ω r / =15 (1.) =18. in / s 1 v 5 = 1 v 1 v D5 = 1 v v D5 / 5 (1) Now, 1v D5 = 1 v D4 = 1 ω 4 r D/E 1 v D4 = 1 ω 4 r D/E 1v D5 / 5 = 1 ω 5 r D/ ( to r D/ ) 1v D5 = 1 ω 4 r D4/E4 ( to r D/E ) Solve Eq. (1) graphically with a velocity polygon. The velocity directions can be gotten directly from the polygon. The magnitudes are given by:

60 D 4 5 E d 5, d 4 d Velocity Polygon, 8 in/s b 15 o' o cceleration Polygon 100 in/s d' 4 b' 1v D5 / 5 =11.4 in / s 1 ω 5 = 1 v D5 / 5 = r 11.4 =4.56 rad / s D/.5 From the directions given in the position and velocity polygons lso, 1ω 5 = 4.56 rad /sw 1v D4/E4 =14.6 in / s 1ω 4 = 1 v D4 /E 4 = r 14.6 = 4.17 rad / s D/E.5 From the directions given in the position and velocity polygons 1ω 4 = 4.17 rad / s W cceleration nalysis: For the acceleration analysis, use the same points in the same order as was done in the velocity analysis

61 1a = 1 a / = 1 r a / + 1 t a / 1 a 5 = 1 a Therefore, Now, 1a D5 = 1 a a D5 / 5 = 1 r a / + 1 t a / + 1 r a D5 / t a D5 / 5 1 a D5 = 1 a D4 = 1 a D4 /E 4 = 1 r a D4 /E t a D4 /E 4 1 ar D4 /E at D4 /E 4 = 1 ar / + 1 at / + 1 ar D5 / at D5 / 5 () r 1a / = 1 ω r / =15 1. = 74.5 in / s in a direction opposite to r/. 1ar D4 /E 4 r 1a D5 / 5 = 1ω 4 r D/E = = 60.8 in / s in a direction opposite to rd/e. = 1 ω 5 r D/ = = 5.0 in / s in a direction opposite to rd/. 1 at / = 1 α r / 1 t a / = 1 α r / = =1 in / s ( to r / ) 1 t a D4 /E 4 = 1 α 4 r D/E 1 at D4 /E 4 = 1 α 4 r D/E ( to r D/E ) 1 at D5 / 5 = 1 α 5 r D/ 1 t a D5 / 5 = 1 α 5 r D/ ( to r D/ ) Solve Eq. () graphically with an acceleration polygon. The acceleration directions can be gotten directly from the polygon. The magnitudes are given by: 1a t D4/ E4 =10.6 in / s 1α4 = 1 at D4/ E4 r = 10.6 D/E.5 =4.5 rad / s From the directions given in the position and acceleration polygons t 1a D4 /E 4 =4.5 rad / s W - 9 -

62 Problem.1 For the mechanism shown, assume that link rolls on the frame (link 1) and link 4 rolls on Link. ssume that link is rotating W with a constant angular velocity of 100 rad/s. Determine the angular acceleration of link and link " Y 0." = 1.0" 1 X 1.0" 0.5" 4 Position nalysis Draw the linkage to scale. Start with link and locate point. Locate point and draw link 4. Then draw a line corresponding to the path of point. Then locate point and draw a circle 1." in radius. Draw a line from point tangent to the circle centered at. Then locate point on the radial line from the tangent point to. Velocity nalysis: Find angular velocity of link, Now, 1v = 1 v /D = 1 v = 1 ω r/d 1v = 1 v + 1 v / = 1 v 4 = 1 v 4 / 4 (1) 1v = 1ω r /D = = 50 in / s ( tor /D ) 1v / = 1 ω r/ 1 v / = 1 ω r/ ( to r/) 1v 4 / 4 =1ω 4 r / 1v 4 / 4 = 1ω 4 r / ( to r / )

63 D Velocity Scale 0 in/sec b b 4 cceleration Scale 100 in/sec 1a r / o' 1a t 4 / 4 o c 4 a 1at / Solve Eq. (1) graphically with a velocity polygon. From the polygon, Then 1v / = 7.5 in / s 1v 4 / 4 = 44.6 in / s 1ω = 1 v / = r 7.5 = 7.0 rad / s /.91 1ω 4 = 1 v 4 / 4 = r 44.6 = 44.6 rad / s / 1 To determine the direction of 1ω, determine the direction that r/ must be rotated to be parallel to 1 v /. This direction is clearly counterclockwise

64 To determine the direction of 1ω 4, determine the direction that r/ must be rotated to be parallel to 1v 4 / 4. This direction is clearly clockwise. cceleration nalysis: 1a = 1 a = 1 ad /D a /D ecause of rolling contact on a flat surface, lso, 1aD /D 1 = 1 n a D /D 1 = 1 r a D / 1a /D = 1 r a /D + 1 t a /D ombining the equations, 1a = 1 r a D / + 1 r a / D + 1 t a /D = 1 t a /D = 1 α r/d = 0 Going to point, 1a = 1 a + 1 a / = 1 a + 1 r a / + 1 t a / also, 1a = 1 a a / 4 = 1 a 4 / a / 4 Then, 1a 4 / a / 4 = 1 a + 1 a / Expanding the terms, r 1a 4 / t a 4 / n a / 4 = 1 a + 1 r a / + 1 t a / Expanding 1 n a / 4 recognizing that there is rolling contact between a circle and flat surface, and that 1 a = 0, 1 a r 4 / a t 4 / an 4 / 4 = 1 a r / + 1 at / which simplifies to Now, 1 at 4 / 4 = 1 a r / + 1 at / () r r 1a / = 1ω ( 1ω r / ) 1a / = 1ω r / = =194 in / s in the direction opposite to r/ t 1a / = 1 α r/ ( to r /) t 1a 4 / 4 = 1 α4 r/ ( to r/)

65 Solve Eq. () graphically with an acceleration polygon. From the polygon, 1at / =17. in / s Then and 1a t 4 / 4 =194 in / s 1α = 1 t a / = 17. = 4.41 rad / s r /.91 1α4 = 1 t a 4 / 4 = r 194 = 194 rad / s / 1 To determine the direction of 1 α, determine the direction that r/ must be rotated to be parallel to 1 t a /. This direction is clearly counterclockwise. To determine the direction of 1 α4, determine the direction that r/ must be rotated to be parallel to 1 t a 4 / 4. This direction is clearly counterclockwise. Problem. For the mechanism shown, assume that link 4 rolls on the frame (link 1). If link is rotating W with a constant angular velocity of 10 rad/s, determine the angular accelerations of links and 4 and the acceleration of point E on link. 00 Y X = 0.95" =.5" D = 1." E = 1." E =.75".45" 4 E D Position nalysis Draw the linkage to scale. Start with link and locate point. Then draw a line corresponding to the path of point. Then locate point and draw link 4. Then locate point E. Velocity nalysis: Find angular velocity of link, 1v = 1 v / = 1 v = 1 ω r/

66 1v = 1 v + 1 v / (1) 1v = 1 v 4 E D c o e' 1a r / b' cceleration Scale 50 in/s b Velocity Scale 5 in/s 1a t / 1a c' o' c' 4 d' 1 Now, 1v = 1 ω r/ = = 9.5 in / s( to r/) 1v along the line of motion of. 1v / = 1 ω r/ 1 v / = 1 ω r/ ( to r/) Solve Eq. (1) graphically with a velocity polygon. From the polygon, 1v = 1 v 4 = 7.5 in / s lso, 1v / = 4.81in / s

67 or 1ω = 1 v / = r 4.81 =1.48 rad / s /.5 To determine the direction of 1ω, determine the direction that r/ must be rotated to be parallel to 1 v /. This direction is clearly counterclockwise. For link 4, or 1v 4 =1v D4 + 1v 4 /D 4 = 1v 4 /D 4 = 1ω 4 r / D 1ω 4 = 1 v 4 /D 4 = r 7.5 = 6.8 rad / s /D 1. To determine the direction of 1ω4, determine the direction that r/d must be rotated to be parallel to 1v 4 /D 4. This direction is clearly counterclockwise. cceleration nalysis: 1a = 1 a = 1 r a / + 1 t a /E 1a = 1 a 4 = 1 a + 1 a / 1a = 1 r a / + 1 t a /E + 1 r a / + 1 t a / () 1aD 4 = 1 a ad 4 / 4 = 1 ad 4 / D 1 = 1 n a D4 /D 1 n 1a D4 /D 1 = 1 a r a D4 / t a D4 / 4 = 1 n a D4 / n a 4 /F n a F1 /D 1 Where F is the center of curvature of the line. onsequently, F is at infinity and both 1 n a 4 /F 1 and n 1a F1 /D 1 are zero. lso, 1 n a D4 / 4 and 1 r a D4 / 4 cancel. Therefore, the acceleration equation reduces to or 1a t a D4 / 4 =0 t 1a D4 / 4 = 1 a 4 Now, 1a along the line of motion of point. 1a r / = 1ω ( 1ω r / ) 1a r / = 1ω r / = = 95 in / s in the direction opposite to r /

68 t t 1a / =1α r/ 1a / = 1α r/ = 0 1= 0in/s r r 1a / =1ω ( 1ω r / ) 1a / = 1ω r / = = 7.1 in / s in the direction opposite to r / t t 1a / = 1α r / 1a / = 1α r / ( to r / ) Solve Eq. () graphically with an acceleration polygon. From the polygon, 1a = 65.8 in / s and by image, 1aE = 1 in / s in the directions shown. lso, or t 1a / = 84. in / s 1α = 1 t a / = 84. r /.5 = 5.9 rad / s To determine the direction of 1α, determine the direction that r / must be rotated to be parallel to 1 t a /. This direction is clearly clockwise. lso, or 1a t D4 / 4 =1α 4 r D/ = 1a 4 1a t D4 / 4 = 1α 4 r D/ 1α4 = 1 t a D4 / 4 = r 65.8 = 54.8 rad / s D/ 1. To determine the direction of 1α4, determine the direction that r D/ must be rotated to be parallel to 1 t a D4 / 4. This direction is clearly counterclockwise

69 Problem. If v = 10 in/s (constant) downward, find ω, α, v, and a. am ontact 1" E Rolling ontact F = 1.0" FE =." FD = 0.75" DE =.4" D F " Velocity nalysis 1v = 1 v + 1 v / = 1 v + 1 ω r / 1 v = 1 v + 1 v / = 1 v /F = 1 ω r /F Therefore, 1 v /F = 1 v + 1 v / + 1 v / (1) ecause of rolling contact, 1 v / = 0 lso, 1v /F = 1 ω r /F ( to r /F ) 1v / = 1 ω r / ( to r / ) Therefore, we can solve Eq. (1) using the velocity polygon. Using the velocity polygon,

70 D Velocity Scale 10 in/s o f b, b F cceleration Scale 100 in/s c' o' a, f a c t 1a / F b' n 1a / F 1ω = 1 v / = r =14.14 rad / s / 1.0 To determine the direction of 1 ω, determine the direction that we must rotate r / 90 to get the direction of 1 v /. This is W. lso, 1ω = 1 v /F = r =10.14 rad / s /F 1.0 To determine the direction of 1 ω, determine the direction that we must rotate r/ F 90 to get the direction of 1 v /F. This is W. The velocity of point is found by image. The magnitude is 1v =10 in / s and the direction is given by the velocity polygon

71 cceleration nalysis and 1 a = 1 a + 1 a / = 1 a + 1 at / + 1 ar / = 1 a + 1 α r / + 1 ω ( 1 ω r / ) 1a = 1 a + 1 n a / 1 a = 1 a /F = 1 at /F + 1 ar /F = 1 α r /F + 1 ω ( 1 ω r /F ) ompute the normal component of acceleration at the rolling contact point. n 1a / = 1 r a /O + 1 r a O / + 1 r a / where O is at infinity in the direction of. Then, 1 at /F + 1 ar /F = 1 a + 1 at / + 1 ar / + 1 ar /O + 1 a r O / + 1 ar / = 1 a + 1 at / + 1 ar /O + 1 ar O / The known information can be summarized as follows: 1a = 0 1 at / = 1 α r / ( to r / ) 1a /O r = 1 v /O r /O = 1 v /O = 0 r = 1 v O/ = 1 v O/ = 0 1a O / r O / t 1a /F = 1 α r /F ( to r /F ) 1 r a /F = 1 v /F = =10 in / s r /F 1 opposite to r /F Note that 1 ar / and 1 ar / are in opposite directions and cancel each other. Therefore, the acceleration equations can be combined into the following simple equation 1 t a /F + 1 r a /F = 1 t a / () and solved for the unknown magnitudes of 1 t a /F and 1 t a /. Using values from the acceleration polygon, 1α = 1 at /F = r 104 =104 rad / s /F

72 To determine the direction of 1 α, determine the direction that we must rotate r /F 90 to get the direction of 1 t a /F. This is W. The acceleration of point is found by image. The magnitude is 1a =148 in / s and the direction is given by the polygon. Problem.4 In the figure shown below, points,, and are collinear. If v downward, find v, and a. = 10 in/s (constant) am ontact 1" 45 Rolling ontact 0.5" 0.45"

73 am ontact Rolling ontact o b, c a Velocity Scale 10 in/sec Velocity nalysis 1v = 1 v = 1 v = 1 v + 1 v / (1) This can be solved using velocity polygons as shown since we know the directions of 1 v and 1v. Therefore, 1 v = 10 in/sec. cceleration nalysis Differentiate Eq. (1) to solve for 1 a. 1a = 1 a and 1 a = 1 a + 1 ar / + 1 at / + 1 ar /O + 1 ar O / + 1 ar / () where O is at infinity in the direction of. 1a = 0 1ar / = 1 v / r / opposite to r/

74 1 at / = 1 α r/ ( to r/ ) 1a r /O = 1 v /O = 1 v /O r /O = 0 1a r O / = 1 vo / = 1 vo / r O / 1 ar / = 1 v / r / = 0 Note that 1 ar / and 1 ar / are in opposite directions and cancel each other. Therefore, all of the terms on the right hand side of Eq. () are either zero or cancel each other except for t. Therefore, 1 a / 1 a = 1 at / However, 1 a must be horizontal and 1 at / is perpendicular to r/ which is not horizontal. ecause the directions are different, the magnitudes must be zero. Therefore, 1 a must be zero. Problem.5 Part of an eight-link mechanism is shown in the figure. There is rolling contact at location and the velocity and acceleration of points 6 and 5 are as shown. Find ω 8 and α 7 for the position given. lso find the velocity of E 7 by image. E D =.5" D = 1.0" =.85" 7 D 7 8 v 5 = 10 0 in/s v 6 = in/s a 5 = 0 70 in/s a 6 = in/s

75 Position nalysis Draw the mechanism to scale. Vectors are: r/ = in r/ = in rd/ =1in Velocity nalysis E 7 D 8 a a 7 d 7 a 6 O a b 7, b 8 1 r a 7/ 7 O v c 5 1 r a 8 / 8 c 5 1 V E 7 Velocity Scale 5 in/sec e 7 1 r a D7 /7 1 t a 8 / 8 1 t a 7 / 7 cceleration Scale 0 in/sec

76 Velocity analysis: The basic equations are: 1v 7 =1v 7 / 7 +1v 7 1v8= 1 v8/8+ 1 v8 1v 8 = 1 v v 8 / 7 ombining the equations, where 1v 8 / v 8 = 1 v 7 / v v 8 / 7 1 v 8 / 8 = 1 ω8 r/ ( to ) 1v 7 / 7 = 1 ω7 r/ ( to ) 1 v 8 / 7 = 0 (rolling contact) 1v 8 = 1 v 5 =10 0Þ 1v 7 = 1 v 6 =10 60Þ The solution is given on the polygon. The velocity of E 7 is found by image. Then, Therefore, 1vE 7 =18.89 in / sec 1v 8 / 8 =.687 in / sec 1 v 7 / 7 = 9.48 in / sec 1ω8 = 1 v 8/8 = r.687 =1.4 rad / sec / To determine the direction of 1 ω8, determine the direction that we must rotate r/ 90 to get the direction of 1 v 8 / 8. This is W. cceleration nalysis 1a 7 = 1a 7 / 7 + 1a 7 1a8 = 1 a8/8 + 1 a8 1a 8 = 1 a a 8 / 7 ombining the equations, 1a 8 / a 8 = 1 a 7 / a a 8 /

77 In component form: 1 ar 8 / at 8 / a8 = 1 ar 7 / at 7 / a7+ 1 ar 8 / ar 8 /D ar D7 / 7 omputing the individual terms, 1 a 8 = 1 a 5 = 0 70Þ 1a 7 = 1 a 6 = 0 180Þ 1ar 8 / 8 = 1 v 8 / 8 = (.687) =.610 r / in sec to 1ar 7 / 7 = 1 v 7/7 = (9.48) 1 r 7 / 7 1ar 8 / 8 = 1 v 8 / 8 = 0 1 ar 8 /D7 = 1 v 8 /D 7 = 0 = in sec to 1v 1 ar D7/7 D7/7 = = (4.544) =0.65 r D/ 1 in sec Dto 1 at 8 / 8 1 at 7 / 7 = 1 α8 r/ ( to ) = 1 α7 r/ ( to ) From the acceleration polygon, Then, 1 at 7 / 7 = 1 α7 r/ = 7.81 in / sec 1α7 = 1 at 7/7 = r 7.81 =.91 rad / sec / To determine the direction of 1 α7, determine the direction that we must rotate r/ 90 to get the direction of 1 at 7 / 7. This is W

78 Problem.6 In the mechanism shown below, link is turning W at the rate of 0 rad/s, and link rolls on link. Draw the velocity and acceleration polygons for the mechanism, and determine a and α. 4 D = 4.0" E =.8" E = 4.0" D = 10" E 6.4" 10.0" 0 Position nalysis Draw the linkage by scale. Locate the relative positions of and D. Locate point and then draw the circle arc centered at. Locate point by finding the intersection of a circle arc of 10 inches centered at D and a second circle of 6.8 inches centered at. Finally draw the circle centered at and of radius 4.0 inches. Velocity nalysis: The equations required for the velocity analysis are: 1v E = 1v E / 1vE = 1 ve + 1 ve /E 1vE = 1 v + 1 ve / (1) 1v = 1 v 4 = 1 v 4 /D 4 ecause 1 ve /E = 0, 1 ve = 1 ve and 1vE / = 1 ω re / 1 ve / = 1 ω re/ = =118.8 in / s( to re/ )

79 1vE / = 1 ω re / 1 ve / = 1 ω re / ( to re / ) 1 v 4 /D 4 = 1 ω4 r 4 /D 4 1 v 4 /D 4 = 1 ω4 r 4 /D 4 ( to r 4 /D 4 ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, 4 in 4 D E cceleration Polygon 000 in/s r 1a E / t 1a 4/ D4 t 1a E / c Velocity Polygon 100 in/s b e r 1a 4 /D 4 o' c n 1a E /E r 1a E / e ' o 1vE / =104 in / s or 1ω = 1 v E / = r 104 = 6.0 rad / s W E / 4 lso, 1v 4 /D 4 = 69. in / s or 1ω 4 = 1 v 4 /D 4 = r 69. = 6.9 rad / s W 4 /D

80 cceleration nalysis: The equations required for the acceleration analysis are: 1a E =1a E / 1aE = 1 ae + 1 ae /E 1aE = 1 a + 1 ae / 1a = 1 a 4 = 1 a 4 /D 4 r 1a E / + 1 n a E /E = 1 r a 4 /D t a 4 /D r a E / + 1 t a E / () Now, r r 1a E / =1ω ( 1ω re / ) 1a E / = 1ω re / = = 60 in / s in the direction opposite to re / n 1a E /E = 1 n a E / + 1 n a / + 1 n a /E 1an E/E = 1 v E / 1 v / + 1 v /E = r E / r / r 104 /E = 500 in / s in the direction opposite to re / 1a r 4 /D 4 = 1ω 4 ( 1ω 4 r 4/D4 ) 1a r 4 /D 4 = 1ω 4 r 4 /D4 = = 480 in / s in the direction opposite to r 4 /D 4 1at 4 /D 4 = 1 α4 r4/d4 1 at 4 /D 4 = 1 α4 r4/d4 ( to r4 /D4) r r 1a E / =1ω ( 1ω r E/ ) 1a E / = 1ω r E/ = = 700 in / s in the direction opposite to re / t t 1a E / = 1α r E/ 1a E / = 1α r E/ ( to r E/ ) Solve Eq. () graphically with an acceleration polygon. From the polygon, or t 1a E / = 770 in / s 1α = 1 t a E / = r 770 E / 4 =19 rad / s To find the direction, determine the direction that re / must be rotated 90 to get the direction of 1 t a E /. The direction is clearly W

81 lso from the polygon, 1a = 190 in / s Problem.7 In the mechanism shown below, Link is turning W at the rate of 00 rpm. Draw the velocity polygon for the mechanism, and determine v and ω. 4 D = 1.0" E = 0.7" E = 1.0" D =.5" E 0 1.6".5" Position nalysis: Draw the linkage to scale. 4 =1.0" E=0.7" E=1.0" D=.5" D c 10 in/sec e e E 0.0 Velocity Polygon O v Velocity nalysis: 1ω = π = 0.94 rad / s 1v = 1 v / = 1 ω r / 1 v = 1 ω r / = 0.94 (1) = 0.94 in / s

82 1vE = 1 ve / = 1 ω re / 1 ve = 1 ω re / = 0.94 (1.46) = 0.57 in / s 1vE = 1 ve 1v = 1 ve + 1 v /E (1) Now, 1vE = 0.57 in / s ( to re/) 1v = 1 v 4 = 1 ω4 r/d ( to r/d) 1vE / = 1 ω re/ ( to re /) tangent to two circles. Solve Eq. (1) graphically with a velocity polygon. The velocity directions can be gotten directly from the polygon. The magnitudes are given by: 1v E/ = 7 in / s 1ω = 1 v E / = r 7 = 7 rad / s E / 1 From the directions given in the position and velocity polygons 1ω = 7rad / s W lso, from the velocity polygon, 1v =18.9 in / s

83 Problem.8 ssume that link 7 rolls on link without slipping and find ω 7. = 1.8" G = 0.85" GF = 1.7" D =.9" DE =.5" E =.0" G Y 1.0" " 5 D 4 6 F (-.0", 0.95") 90 E X ω = rad/s Problem.9 In the two degree-of-freedom mechanism shown, ω is given as 10 rad/s W. What should the linear velocity of link 6 be so that ω 4 = 5 rad/s W? 1.75" = 0.5" = 1.0" D = 0.8" DE = 1.65" 6 E 4 D 10 Y Rolling Only X ω =10 rad/s 5 " 1.5"

84 Position nalysis: Draw the linkage to scale. Locate and first. Then draw the two circles and locate point D. Draw the horizontal line on which E is located and locate E 1.65" from D. Velocity nalysis: ompute the velocity of points and D. 1v = 1 ω r / 1 v = 1 ω r / = = 5in/s 1v = 1 v = 1 ω r / 1 ω = 1 v r / = 5 1 = 5 rad / s 1vD = 1 vd 4 = 1 ω rd / 1 vd 4 = 1 ω rd / = = 4in/s Next consider the coincident points at E. 1vE 4 = 1 ve 5 = 1 vd ve 4 /D 4 1vE 4 = 1 ve ve 4 /E 6 1vE ve 4 /E 6 = 1 vd ve 4 /D 4 (1) d, d 4 Velocity Polygon in/s o' e 6 e 4 4 D 6 E Position Polygon 1 in

85 Now, 1vD 4 = 4in/ s( to rd / ) 1vE 4 / D 4 = 1 ω4 re 4 /D 4 = = 8.5 in / s ( to re 4 /D 4 ) 1vE 6 is in the vertical direction, 1 ve 4 /E 6 is in the horizontal direction. Solve Eq. (1) graphically with a velocity polygon. From the polygon 1vE 6 =1.8 in / s in the direction shown on the polygon. Problem.0 In the mechanism shown, ω = 10 rad/s. Determine v / and v Equivalent linkages and ) oincident points at using two approaches: 1) ω 5 = 0.5" = 1.0" Solution (Equivalent Linkage) The equivalent linkage is shown below. For the equivalent linkage, we need only find the velocity and acceleration of point. The velocity equations which must be solved are: and 1v = 1v / = 1ω r / 1v = 1 v + 1 v / (1)

86 Here we have written the velocity equation in terms of the velocity of relative to rather than vice versa because we can easily identify the direction of the velocity of relative to. We also know the direction for the velocity (and acceleration) of. 1v = 10(0.5) = 5in/ s( to ) 1v is along the slide direction between link and the frame 1v / is along the slide direction between link 4 and link From the velocity polygon, 1v =.61 in/s in the direction shown. 1 ω 4 a, a 4 a, c o Velocity Scale 5 in/s Solution (Direct pproach) c To analyze the problem, we can determine the velocity and acceleration of any point on link because all points on link have the same velocity and the same acceleration. The point to choose is the contact point. To solve for the velocity and acceleration of, first find the velocity of point. Then write the relative velocity expression between points and and solve for the velocity of. Velocity nalysis The relevant equations are: and 1v = 1v / = 1ω r /

87 1v = 1 v + 1 v / 1v / = 10(0.77) = 7.7 in / s( to ) 1v is along the slide direction between link and the frame 1v / is along the face of link 1 ω c o Velocity Scale 5 in/s c From the velocity polygon, v = 5 in/s in the direction shown

88 Problem.1 In the mechanism shown, ω = 0 rad/s W. t the instant shown, point D, the center of curvature of link, lies directly above point E, and point lies directly above point. Determine v / and ω using: 1) equivalent linkages and ) oincident points at. D = 0.75" = 1.5".5" Y E ω X 1.".0" Position nalysis Draw the linkage to scale. First locate pivots and E. Then locate point and draw link. Next locate point D and draw link. Solution (Equivalent Linkage) The equivalent linkage is a fourbar linkage involving points,, D, and E. For the equivalent linkage, we need only find the velocity of point. The velocity equations which must be solved are: 1v = 1v / = 1ω r / 1vD x = 1 v x + 1 vd x / x = 1 vd = 1 ve + 1 vd /E Where x is the imaginary link between points and D. Simplifying and combining terms 1vD /E = 1 v + 1 vd x / x (1)

89 D E c Velocity Scale 0 in/s b o d e c Now 1v = 1ω r / = 0(0.75) =15 in / s( to ) 1vD /E = 1 ω rd/e is perpendicular to DE 1vD x / x is perpendicular to D From the velocity polygon, Therefore, 1vD /E = 1 v = 15 in / s( to ) 1ω = 1 v D/E = 15 = 4.8 rad / s r D/E.5 To determine the direction of 1ω, determine the direction that we must rotate r D/E 90 to get the direction of 1 vd /E. This is counterclockwise To find 1 v /, compute 1v and 1 v and use

90 1v / = 1 v 1 v oth 1v and 1 v may be determined by velocity image (or computed directly). From the polygon, 1v / = 44.8 in / s in the direction indicated by the polygon. Solution (Direct pproach) First find the velocity of point. Then write the relative velocity expression between points and and solve for the velocity of. The relevant equations are: and 1v = 1v / = 1ω r / 1v = 1 v + 1 v / = 1 v /E Where 1v / = 0(.158) = 4.16 in / s ( to ) 1v /E is perpendicular to E 1v / is along the tangent to the contact point (perpendicular to D) From the velocity polygon, and 1v / = 44.8 in / s 1v /E = 9.84 in / s Therefore, 1ω = 1 v /E = 9.84 = 4.8 rad / s r /E.0 To determine the direction of 1ω, determine the direction that we must rotate r /E 90 to get the direction of 1 v /E. This is counterclockwise - 1 -

91 Problem. In the position shown, find the velocity and acceleration of link using: 1) equivalent linkages and ) oincident points at = 0.5" = 1.1" ω 45 ω = 100 rad/ s W α = 0,000 rad /s W Solution (Equivalent Linkage) This problem is similar to Example.11 except for a nonzero value for the acceleration. The equivalent linkage is shown below. For the equivalent linkage, we need only find the velocity and acceleration of point. Velocity nalysis The velocity equations which must be solved are: and 1v = 1 v / = 1 ω r/ 1 v = 1 v + 1 v / (1) Here we have written the velocity equation in terms of the velocity of relative to rather than vice versa because we can easily identify the direction of the velocity of relative to. We also know the direction for the velocity (and acceleration) of. 1 v = 100(0.5) = 50 in / s ( to ) 1v is along the slide direction between link and the frame 1v / is along the slide direction between link 4 and link - 1 -