218 39 Solutions Instructor Manual Chapter 3 Velocity

19

Chapter 3 Velocity

3.1 The position vector of a point is given by the equation 100 j te π=R , where R is in inches. Find the velocity of the point at 0.40 s.t =

( ) 2.5 j tt e π=R

( ) 2.5 j tt j e ππ=R

( )( )

0.400.40 2.5

2.5 cos 0.40 sin 0.402.5 sin 72 2.5 cos 72

jj e

j jj

ππ

π π ππ π

=

= +

= − °+ °

R

( )0.40 7.46 2.427 7.85 162 m/sj= − + = ∠ °R Ans.

3.2 The equation ( )2 104 -j t/R t e π= + defines the path of a particle. If R is in meters, find the

velocity of the particle at 20 s.t =

( ) ( )2 /104 j tt t e π−= +R

( ) ( )( )/10 2 /102 /10 4j t j tt te j t eπ ππ− −= − +R

( ) ( )( )20/10 2 20/10

2 2

20 40 /10 20 4

40 40.4

j j

j j

e j e

e j e

π π

π π

π

π

− −

− −

= − +

= −

R

( )20 40.000 126.920 133.074 72.5 m/sj= − = ∠− °R Ans.

3.3 If automobile A is traveling south at 88 km/h and automobile B north 60° east at 64 km/h, what is the velocity difference between B and A? What is the apparent velocity of B to the driver of A?

ˆ88 90 88 km/hA = ∠− ° = −V j ˆ ˆ64 30 55.42 32 km/hB = ∠ ° = +V i j ˆ ˆ55.42 120 km/hBA B A= − = +V V V i j

132.16 65.2 =132.16 N 24.8 E km/hBA = ∠ ° °V Ans Naming B as car 3 and A as car 2, we have

2B A=V V since 2 is translating. Then

3 3 2/ 2B B B BA= − =V V V V

3 / 2 132.16 65.2 =132.179 N 24.8 E km /hB = ∠ ° °V Ans.

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3.4 In the figure, wheel 2 rotates at 600 rev/min and drives wheel 3 without slipping. Find the velocity difference between points B and A.

( )( )2

600 rev/min 2 rad/rev60 s/min

20 rad/s cw

πω

π

=

=

2 22 2000 6275mm /sAO AOV Rω π= = =

B A BA= +V V V 5575 90 mm/sB = ∠ °V Ans.

3.5 Two points A and B, located along the radius of a wheel (see figure), have speeds of 80 and 140 m/s, respectively. The distance between the points is 300 mmBAR = . (a) What is the diameter of the wheel? (b) Find ,,AB BAV V and the angular velocity of the wheel.

( ) ( )ˆ ˆ ˆ140 80 60 m/sBA B A= − = − − − = −V V V j j j Ans.

( ) ( )ˆ ˆ ˆ80 140 60 m/sAB A B= − = − − − =V V V j j j Ans.

260 m/s 200 rad/s cw

0.300 mBA

BA

VR

ω = = = Ans.

2

22

140 m/s 0.700 m200 rad/s

BOBO

VR

ω= = =

( )2

Dia 2 2 0.7 m 1.4 m 1 400 mmBOR= = = = Ans.

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3.6 A plane leaves point B and flies east at 560 km/h. Simultaneously, at point A, 320 km southeast (see figure), a plane leaves and flies northeast at 625 km/h. (a) How close will the planes come to each other if they fly at the same altitude? (b) If they both leave at 6:00 p.m., at what time will this occur?

ˆ ˆ625 45 km/h 442 442A = ∠ ° = +V i j ; ˆ560B =V i ˆ ˆ118 442BA B A= − = −V V V i j

At initial time ( ) ˆ ˆ0 320 120 km 160 277BA = ∠ ° = − +R i j At later time

( ) ( ) ( )ˆ ˆ( ) 0 160 118 277 442BA BA BAt t t t= + = − + + −R R V i j To find the minimum of this:

( ) ( )2 22 160 118 277 442BAR t t= − + + −

( )( ) ( )( )2 2 160 118 118 2 277 442 442 0BAdR dt t t= − + + − − = 418 576 282 628 0t − = 0.677 h 41 mint = = or 6:41p.m. Ans.

( ) ˆ ˆ0.677 80.17 22.23 83 165 kmBA = − − = ∠− °R i j Ans.

3.7 To the data of Problem 3.6, add a wind of 48 km/h from the west. (a) If A flies the same heading, what is its new path? (b) What change does the wind make in the results of Problem 3.6?

With the added wind ˆ ˆ490 442 659.89 42 km/hA = + = ∠ °V i j Since the velocity is constant, the new path is a straight line at N 48º E. Ans. Since the velocities of both planes change by the same amount, the velocity difference

BAV does not change. Therefore the results of Problem 3.6 do not change. Ans.

3.8 The velocity of point B on the linkage shown in the figure is 40 m/s. Find the velocity of point A and the angular velocity of link 3.

A B AB= +V V V 48.99 165 m/sA = ∠− °V Ans. 14.64 120 m/sAB = ∠− °V

314.64 36.60 rad/s ccw0.400

AB

AB

VR

ω = = = Ans.

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3.9 The mechanism shown in the figure is driven by link 2 at 2 45 rad/s ccw.ω = Find the angular velocities of links 3 and 4.

( )( )2 22 45 rad/s 100 mm 4500 mmAO AOV Rω= = =

4 4B A BA O BO= + = +V V V V V 358.5 mm /sBAV = ;

44619 mm/sBOV = .

3358.5 1.43 rad/s ccw250

BA

BA

VR

ω = = = Ans.

4

4

44619 15.40 rad/s ccw300

BO

BO

VR

ω = = = Ans.

3.10 Crank 2 of the push-link mechanism shown in the figure is driven at 2 60 rad / s cw.ω = Find the velocities of points B and C and the angular velocities of links 3 and 4.

( )( )2 22 60 rad/s 0.150 m 9.0 m/sAO AOV Rω= = =

4 4B A BA O BO= + = +V V V V V 13.020 m/sBAV = ; 11.360 41 m/sB = ∠ °V Ans.

23

C A CA B CB= + = +V V V V V 3.830 60 m/sC = ∠ °V Ans.

313.020 43.40 rad/s cw0.300

BA

BA

VR

ω = = = ; 4

4

411.360 37.87 rad/s cw0.300

BO

BO

VR

ω = = = Ans.

3.11 Find the velocity of point C on link 4 of the mechanism shown in the figure if crank 2 is driven at 2 48 rad / s ccw.ω = What is the angular velocity of link 3?

( )( )2 22 48 rad/s 200 mm 9600 mm/sAO AOV Rω= = =

4 4B A BA O BO= + = +V V V V V

267.5 mm/sBAV = ; 3267.5 0.335 rad/s ccw800

BA

BA

VR

ω = = = Ans.

4 4C O CO B CB= + = +V V V V V ; 7117.5 75.8 mm/s 7.11 < 284 m/sC = ∠− ° = °V Ans.

3.12 The figure illustrates a parallel-bar linkage, in which opposite links have equal lengths. For this linkage, show that 3ω is always zero and that 4 2.ω ω= How would you describe the motion of link 4 with respect to link 2?

Refering to Fig. 2.19 and using 1 3r r= and 2 4r r= , we compare Eqs. (2.26) and (2.27) to see that ψ β= . Then Eq. (2.29) gives 3 0θ = and its derivative is 3 0ω = . Ans. Next we substitute Eq. (2.25) into Eq. (2.33) to see that 2γ θ= . Then Fig. 2.19 shows that, since link 3 is parallel to link 1 ( )3 0θ = , then 4 2θ γ θ= = . Finally, the derivative of this gives 4 2ω ω= . Ans. Since 4/ 2 4 2 0ω ω ω= − = , link 4 is in curvilinear translation with respect to link 2. Ans.

24

3.13 The figure illustrates the antiparallel or crossed-bar linkage. If link 2 is driven at 2 1ω = rad/s ccw, find the velocities of points C and D.

( )( )2 22 1 rad/s 0.300 m 0.300 m/sAO AOV Rω= = =

4 4B A BA O BO= + = +V V V V V Construct the velocity image of link 3.

0.402 151 m/sC = ∠ °V Ans. 0.290 249 m/sD = ∠ °V Ans.

3.14 Find the velocity of point C of the linkage shown in the figure assuming that link 2 has an angular velocity of 60 rad/s ccw. Also find the angular velocities of links 3 and 4.

( )( )2 22 60 rad/s 150 mm 9000 mm/s 9 m/sAO AOV Rω= = = =

4 4B A BA O BO= + = +V V V V V 4235 mm/sBAV = ;

47940 mm/sBOV =

34235 28.24 rad/s cw150

BA

BA

VR

ω = = = Ans.

4

4

47940 31.76 rad/s cw250

BO

BO

VR

ω = = = Ans.

Construct the velocity image of link 3; 12677.5 156.9 mm/sC = ∠ °V 1.267 156.9 m/s= ∠ ° Ans.

25

3.15 The inversion of the slider-crank mechanism shown in the figure is driven by link 2 at

2 60 rad/s ccw.ω = Find the velocity of point B and the angular velocities of links 3 and 4.

( )( )

2 22 60 rad/s 0.075 m 4.500 m/sAO AOV Rω= = =

3 3 4 3 / 4P A P A P P= + = +V V V V V

3

3

34.259 22.0 rad/s ccw0.194

P A

P A

VR

ω = = = Ans.

Construct the velocity image of link 3; 4.789 96.5 m/sB = ∠ °V Ans.

3.16 Find the velocity of the coupler point C and the angular velocities of links 3 and 4 of the mechanism shown if crank 2 has an angular velocity of 30 rad/s cw.


4 4B A BA O BO= + = +V V V V V Construct the velocity image of link 3;

2250 126.9 mm/sC = ∠ °V Ans.

32250 18.00 rad/s ccw125

BA

BA

VR

ω = = = ; 4

4

40.00 0.00150

BO

BO

VR

ω = = = Ans.

26

3.17 Link 2 of the linkage shown in the figure has an angular velocity of 10 rad/s ccw. Find the angular velocity of link 6 and the velocities of points B, C, and D.


B A BA= +V V V 289.25 180 mm/sB = ∠ °V Ans.

Construct velocity image of link 3. C A CA B CB= + = +V V V V V 605.5 207.6 mm/sC = ∠ °V Ans.

6 6D C DC O DO= + = +/V V V V V 604.5 206.2 mm/sD = ∠ °V Ans.

6

6

6604.5 4.03 rad/s ccw150

DO

DO

VR

ω = = = Ans.

3.18 The angular velocity of link 2 of the drag-link mechanism shown in the figure is 16 rad/s cw. Plot a polar velocity diagram for the velocity of point B for all crank positions. Check the positions of maximum and minimum velocities by using Freudenstein’s theorem.

The graphical construction is shown in the position where 2 135θ = ° where the result is

5.76 7.2B = ∠− °V m/s. It is repeated at increments of 2 15θΔ = ° . The maximum and minimum velocities are 9.13 146.6B,max = ∠− °V m/s at 2 15θ = ° and 4.59 63.7B,min = ∠ °V m/s at

2 225θ = ° , respectively. Within graphical accuracy these two positions approximatly verify Freudenstein’s theorem.

A numeric solution for the same problem can be found from Eq. (3.22) using Eqs. (2.25) through

(2.33) for position values. The accuracy of the values reported above have been verified this way.

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3.19 Link 2 of the mechanism shown in the figure is driven at 2 36 rad/s cw.ω = Find the angular velocity of link 3 and the velocity of point B.


4 4B A BA O BO= + = +V V V V V 5070 56.3 mm/sB = ∠− °V Ans.

3647.5 3.23 rad/s ccw200

BA

BA

VR

ω = = = Ans.

3.20 Find the velocity of point C and the angular velocity of link 3 of the push-link mechanism shown in the figure. Link 2 is the driver and rotates at 8 rad/s ccw.

( )( )2 22 8 rad/s 0.150 m 1.200 m/sAO AOV Rω= = =

4 4B A BA O BO= + = +/V V V V V Construct velocity image of link 3.

C A CA B CB= + = +V V V V V 3.848 223.2 m/sC = ∠ °V Ans.

33.784 15.14 rad/s ccw0.250

BA

BA

VR

ω = = = Ans.

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3.21 Link 2 of the mechanism shown in the figure has an angular velocity of 56 rad/s ccw. Find the velocity of point C.

( )( )2 22 56 rad/s 0.150 m 8.400 m/sAO AOV Rω= = =

4 4B A BA O BO= + = +V V V V V Construct velocity image of link3.

C A CA B CB= + = +V V V V V 9.028 137.8 m/sC = ∠ °V Ans.

3.22 Find the velocities of points B, C, and D of the double-slider mechanism shown in the figure if crank 2 rotates at 42 rad/s cw.

( )( )2 22

42 rad/s 50 mm 2100 mm/sAO AOV Rω=

= =

B A BA= +V V V 1634 180 mm/sB = ∠ °V Ans. Construct velocity image of link 3.

C A CA B CB= + = +V V V V V 1696.5 154.2 mm/sC = ∠ °V Ans. 530.75 90 mm/sD C DC= + = ∠ °V V V Ans.

29

3.23 The figure shows the mechanism used in a two-cylinder 60° V engine consisting, in part, of an articulated connecting rod. Crank 2 rotates at 2000 rev/min cw. Find the velocities of points B, C, and D.

( )2

2000 rev/min 2 rad/rev209.4 rad/s

60 min/sπ

ω = =

( )( )2 22

209.4 rad/s 50 mm 10470 mm/sAO AOV Rω=

= =

10647.5 120 mm/sB A BA= + = ∠− °V V V Ans. Construct velocity image of link 3.

12172.5 93.3 mm/sC A CA B CB= + = + = ∠− °V V V V V Ans. 9467.5 60 mm/sD C DC= + = ∠− °V V V Ans.

3.24 Make a complete velocity analysis of the linkage shown in the figure given that

2 24ω = rad/s cw. What is the absolute velocity of point B? What is its apparent velocity to an observer moving with link 4?

30


Using the path of P3 on link 4, we write 3 3 4 3 / 4P A P A P P= + = +V V V V V

3

3

34045 6.16 rad/s cw

656.75P A

P A

VR

ω = = =

From this, or graphically, we complete the velocity image of link 3, from which 3

3945 39.7 mm/sB = ∠− °V Ans. Then, since link 4 remains perpendicular to link 3, we have 4 3ω ω= and we find the velocity image of link 4.

3 / 4 2582.5 12.4 mm/sB = ∠− °V Ans.

3.25 Find BV for the linkage shown in the figure if 300 mm/s.AV =

300 mm/sAV = Using the path of P3 on link 4, we write

3 3 4 3 / 4P A P A P P= + = +/V V V V V Next construct the velocity image of link 3 or

3 3B P BP A BA= + = +V V V V V 312.5 23.0 mm/sB = ∠− °V Ans.

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3.26 The figure shows a variation of the Scotch-yoke mechanism. It is driven by crank 2 at

2 36ω = rad/s ccw. Find the velocity of the crosshead, link 4.

( )( )2 22 36 rad/s 0.250 m 9.0 m/sAO AOV Rω= = =

Using the path of A2 on link 4, we write 2 4 2 / 4A A A= +V V V

(Note that the path is unknown for4 / 2AV !)

44.658 180 m/sA = ∠ °V Ans.

All other points of link 4 have this same velocity; it is in translation.

3.27 Make a complete velocity analysis of the linkage shown in the figure for 2 72ω = rad/s ccw.


32

4 4B A BA O BO= + = +V V V V V Construct velocity image of link3.

3C A CA B CB= + = +V V V V V

31908 203.2 mm/sC = ∠ °V Ans.

Using the path of C3 on link 6, we next write 3 6 3 6 6 6/ 6 and C C C C C O= + =V V V V V , from which

61076.75 241.4 mm/sC = ∠ °V Ans.

From this, graphically, we can complete the velocity image of link 6, from which 1935 98.9 mm/sE = ∠− °V Ans.

Since link 5 remains perpendicular to link 6, 6 6

6 6

5 6 9.67 rad/s cwC O

C O

VR

ω ω= = = Ans.

From these we can get 5 5 5 5

1613 210.1D C D C= + = ∠ °V V V Ans.

3.28 Slotted links 2 and 3 are driven independently at 2 30ω = rad/s cw and 3 20ω = rad/s cw,

respectively. Find the absolute velocity of the center of the pin 4P carried in the two slots.

( )( )2 22 30 rad/s 0.054 9 m 1.647 m/sP P AV Rω= = =

( )( )3 33 20 rad/s 0.102 6 m 2.052 m/sP P BV Rω= = =

Identifying the pin as separate body 4 and, noticing the two paths it travels on bodies 2 and 3, we write

4 2 4 3 4/ 2 /3 2.355 15.6 m/sP P P P P= + = + = ∠ °V V V V V Ans.

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3.29 The mechanism shown is driven such that VC = 250 mm/s to the right. Rolling contact is assumed between links 1 and 2, but slip is possible between links 2 and 3. Determine the angular velocity of link 3. Using the path of C2 on link 3, we write

2 3 2 /3C C C= +V V V and 3 3 3 3C D C D= +V V V

3 3

3 3

3105.65 1.569 rad/s cw67.25

C D

C D

VR

ω = = = Ans.

3.30 The circular cam shown is driven at an angular velocity of 2 15ω = rad/s ccw. There is rolling contact between the cam and the roller, link 3. Find the angular velocity of the oscillating follower, link 4.

( )( )2 22 15 rad/s 31.25 mm 468.75 mm/sAO AOV Rω= = =

4 2 4 / 2D D D= +V V V and 4 4D E D E= +V V V

4

4

4381 4.355 rad/s ccw87.5

D E

D E

VR

ω = = = Ans.

34

3.31 The mechanism shown is driven by link 2 at 10 rad/s ccw. There is rolling contact at point F. Determine the velocity of points E and G and the angular velocities of links 3, 4, 5, and 6.

( )( )2 10 rad/s 25 mm 250 mm/sBA BAV Rω= = =

C B CB D CD= + = +V V V V V

3333.25 3.333 rad/s ccw

100CB

CB

VR

ω = = = Ans.

4166.675 3.333 rad/s ccw

50CD

CD

VR

ω = = = Ans.

Construct velocity image of link 3. 3 3 3

251.5 220.9 mm/sE B E B C E C= + = + = ∠ °V V V V V Ans.

Using the path of E3 on link 6, we write 3 6 3 / 6E E E= +V V V and

6 6E H E H= +V V V

6

6

6121.45 3.774 rad/s cw

32.2E H

E H

VR

ω = = = Ans.

Construct velocity image of link 6. 6 6

298.25 57.1 mm/sG E GE H GH= + = + = ∠− °V V V V V Ans.

5 6F F=V V ; 5 3E E=V V ; 5

5

5319.5 25.56 rad/s cw12.5

F E

F E

VR

ω = = = Ans.

35

3.32 The figure shows the schematic diagram for a two-piston pump. It is driven by a circular eccentric, link 2, at 2 25ω = rad/s ccw. Find the velocities of the two pistons, links 6 and 7.

( )( )2 2 2 25 rad/s 25 mm 625 mm/sF F E FEV V Rω= = = =

Using the path of F2 on link 3, we write 2 3 2 /3F F F= +V V V and

3 3F G F G= +V V V Construct velocity image of link 3.

3 3C F CF G CG= + = +V V V V V and

3 3D F DF G DG= + = +V V V V V . Then 32.55 180 mm/sA C AC= + = ∠ °V V V Ans. 144.5 180 mm/sB D BD= + = ∠ °V V V Ans.

36

3.33 The epicyclic gear train shown is driven by the arm, link 2, at 2 10ω = rad/s cw. Determine the angular velocity of the output shaft, attached to gear 3.

( )( )2 10 rad/s 75 mm 750 mm/sB BA BAV V Rω= = = = Using 0D =V construct the velocity image of link 4 from which 1500 0 mm/sC = ∠ °V .

31500 30.00 rad/s cw

50CA

CA

VR

ω = = = Ans.

3.34 The diagram shows a planar schematic approximation of an automotive front suspension. The roll center is the term used by the industry to describe the point about which the auto body seems to rotate with respect to the ground. The assumption is made that there is pivoting but no slip between the tires and the road. After making a sketch, use the concepts of instant centers to find a technique to locate the roll center.

By definition, the “roll center” (of the vehicle body ,link 2, with respect to the road, link 1,) is the instant center P12. It can be found by the repeated application of Kennedy’s theorem as shown.

37

In the automotive industry it has become common practice to use only half of this construction, assuming by symmetry that P12 must lie on the centerline of the vehicle. Notice that this is true only when the right and left suspension arms are symmetrically positioned. It is not true once the vehicle begins to roll as in a turn.

Having lost sight of the relationship to instant centers and Kennedy’s theorem, and remembering only the shortened graphical construction on one side of the vehicle, many in the industry are now confused about “the movement of the roll center along the centerline of the vehicle”. They should be thinking about the fixed and moving centrodes (Section 3.21) which are more horizontal than vertical!

3.35 Locate all instant centers for the linkage of Problem 3.22.

Instant centers 12P , 23P ,

34P , 14P (at infinity), 35P ,

56P , and 16P (at infinity) are found by inspection. All others are found by repeated applications of Kennedy’s theorem until

46P .

One line can be found for 46P ; however no second

line can be found since no line can be drawn between

14P and 16P (in finite space). Now it must be seen that 46P must be infinitely remote because the relative motion between links 4 and 6 is translation; the angle between links 4 and 6 remains constant.

38

3.36 Locate all instant centers for the mechanism of Problem 3.25.

Instant centers 12P (at infinity), 23P ,

34P (at infinity), and 14P are found by inspection. All others are found by repeated applications of Kennedy’s theorem.


Instant centers 12P , 23P , 34P (at infinity), and 14P (at infinity) are found by inspection. All others are found by repeated applications of Kennedy’s theorem except 13P .

One line ( 12 23 P P ) can be found for 13P ; however no second line can be found since no line can be drawn (in finite space) between

14P and 34P . Now it must be seen that 13P must be infinitely remote because the relative motion between links 1 and 3 is translation; the angle between links 1 and 3 remains constant.

39


Instant centers 12P , 23P , 34P , 14P , 35P , 56P (at infinity), and 16P are found by inspection. All others are found by repeated applications of Kennedy’s theorem.


Instant centers 12P and 13P are found by inspection. One line for 23P is found by Kennedy’s theorem. The other is found by drawing perpendicular to the relative velocity of slipping at the point of contact between links 2 and 3.


Instant centers 12P , 23P , 34P and 14P are found by inspection. The other two are found by Kennedy’s theorem

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218 39 Solutions Instructor Manual Chapter 3 Velocity