Mechanism Synthesis & Analysis unit 2…..

A presentation on how to create kinematic motions in mechanical systems

1. ME 262 –Unit 2
Mechanism Synthesis and Analysis
Dr. F. W. ADAM
MECHANICAL ENGINEERING DEPARTMENT
KNUST

2. Mechanism Synthesis and Analysis
Mechanism: is a mechanical device that
has the purpose of transferring motion
from a source to an output.
•Analysis –determination of position,
velocity, acceleration, etc. for a given
mechanism
•Synthesis –design of mechanism to do a
specific task.

3. Logical synthesis of mechanisms
•Type, number and dimensional synthesis
Type synthesis: Kind of mechanism
selected
Number synthesis: Number of joints or
links
Dimensional synthesis: The determination
of the proportions (lengths) of the links
necessary to accomplish the desired
motions

4. Terminology and Definitions
•Kinematics: the study of the motion of bodies without reference to
mass or force
•Links: one of the rigid bodies or members joined together to form a
kinematic chain considered as rigid bodies
•Kinematic pair/ mechanical joint : a connection between two bodies
that imposes constraints on their relative movement.
•Ground/frame: a fixed or stationary link in a mechanism.
•Lower pair: the two lower pair have surface contact with one on
another.
•Higher pair: the two lower pair have point or line contact with one
another.
•Degree of freedom (DOF)/Mobility: of a mechanical system is the
number of independent parameters that define its configuration.
•Frame: the fixed link in a mechanism.

5. Terminology and Definitions
•Closed-Loop Kinematic Chains: a kinematic
chain is an assembly of links and joints. Each
link in a closed-loop kinematic chain is
connected to two or more other links.
•Open-Loop Kinematic Chains: at least one link is
connected to only on other link.

6. Terminology and Definitions
•Manipulators: manipulators designed to
simulate human arm and hand motion are an
example of open-loop kinematic chains.
•Robots: programmable manipulators

7. Terminology and Definitions
•Linkage/mechanism/machine/engine: an
assemblage of rigid bodies connected by
kinematic pairs.
•Planar motion: motion of all linkages is 2D
(projected onto a common plane).
•Spacialmotion: at least the motion of one link
is 3D.

8. Terminology and Definitions
Inversion:iftwo
otherwiseidentical
linkageshavedifferent
fixedlinks,theneach
isaninversionofthe
other.

9. Link Types
A link is a rigid connection between two or more elements
of different kinematic pairs.

10. Mechanisms for specific applications
•Drafting Instrument
•Pantograph Linkages
•Slider-Crank
Mechanism
•Rotating
Combustion(Wankel)
Engine

11. Mechanisms for specific applications
Fluid Links
Gear trains
Lamination-Type Impulse Drive
Swash Plate

12. Mechanisms for specific applications
Power Screws
Differential Screws
Ball Screws

13. Mechanisms for specific applications
Universal Joints Special-Use Clutches
Sprag-Type Reverse-Locking Clutches
One-Way Clutches
Universal Joints
Automotive Steering Linkage
Computer Controlled Industrial Robots

14. Scissor mechanisms

15. Classification of mechanisms
NoDescription
1Screw
2Roller
3Wheel
4Cam
5Linkage
6Ratchet

16. Kinematic Pair(Joints)
•Lower pair(six types)
•Higher pair(cam and its follower have line or
point contact between the elements)
•Wrapping(chains and belts)

17. Degree of freedom
ForexamplethebeadshowninAhasonly
onecoordinateneededtospecifyits
location.
ThebarshowninBneedstwo
coordinates(complete,independentand
withadmissiblevariation)tolocatethe
endpoint.(x,y)or(r,θ)
A
B
C
ThebarshowninDneedsthree
coordinates(complete,independentand
withadmissiblevariation)tolocatethe
endpoint.(x,y,θ)

18. Lower Pairs(surface contacts)

19. Higher Order Joints

20. Degree of freedom(dof)/Mobility
•Any unconstrained rigid body has six dof
•A body moving in a plane has three dof
•A cluster of n bodies not connected have dof=3n
•A fixed link(frame) has no degree of freedom
•Each lower pair reduces total dofby 2
•Each higher pair reduces total dofby 1
•Grubbler’sCriterion: the total dofof a planar
mechanism is
dof=3(n-1)-2j-h
Where n=number of links
j=number of lower pairs
h=number of higher pairs

21. Examples

22. Classification of Closed Planar Four-Bar
Linkages-The GrashofCriterion
•Closed planar linkages of four pin-connected
rigid links are usually identified as four-bar
linkages.
•If one of the links can perform a full rotation
relative to the other three links, the linkage is
calledGrashofmechanism

23. Four-Bar Linkages
Grashofcondition
Asimplerelationshipwhich
predictsthebehaviorofa
four-linkmechanisminversion
basedonlyonlinklengths.
S=lengthofshortestlink
L=lengthoflongestlink
P=lengthofonelink
Q=lengthofotherlink
IfS+L≤P+Qthenatleastone
linkwillcrank,andthe
mechanismiscalledaGrashof
mechanism.

24. Criteria of motion for each class of
Four-Bar Linkages
Type of Mechanism Shortest link Relationship between link
lengths
Grashof Any S+L≤P+Q
crank rocker Driver crank S+L<P+Q
drag link Fixed link S+L<P+Q
doublerocker coupler S+L<P+Q
crossover
position(change point)
S+L=P+Q
Non-Grashof S+L<P+Q
double rockerof the
second kind(triple rocker)
S+L>P+Q

25. Straight line mechanisms

26. Straight line mechanisms
watt’s linkage as used in his steam engine

27. Slider-crank mechanism2
2
22
sin1cos
sin1cos1sincos
coscos
sinsin




















l
r
lrx
l
r
lrx
lrh

28. Geneva Wheel

29. Scotch yoke

30. EXAMPLE OF MECHANISM
Ken Youssefi MECHANICAL ENGINEERING DEPT. 30
Can crusher
Simple press
Rear-window wiper

31. EXAMPLE OF MECHANISMS
31
Moves packages from an assembly
bench to a conveyor
Lift platform
Microwave carrier to assist
people on wheelchair

32. EXAMPLE OF MECHANISMS
32
Lift platform
Front loader
Device to close the
top flap of boxes

33. EXAMPLE OF MECHANISMS
33
Stair climbing mechanism
A box that
turns itself off
Airplane landing
gear mechanism

34. EXAMPLE OF MECHANISMS
34
Conceptual design for an
exercise machine
Rowing type exercise machine

35. EXAMPLE OF MECHANISMS
35

36. EXAMPLE OF MECHANISMS
36
Six-bar linkage prosthetic
knee mechanism
Extension position
Flexed position

37. Dimensional Synthesis
•The determination of the proportions (length)
of the links necessary to accomplish the
desired motions.
•Function generation, path generation and
Motion Generation (body guidance).

38. Function generation, path generation
and Motion Generation(body
guidance).
•FunctionGeneration: the correlation of an input
motion with an output motion in a mechanism
•Path Generation: Control of a point in the plane
such that it follows some prescribed path
•Motion Generation: Control of a line in the plane
such that is follows some prescribed set of
sequential positions

39. 39
Mechanism Categories
Function Generation Mechanisms
A function generator is a linkage in which the relative
motion between links connected to the ground is of interest.
A four-bar hand actuated wheelchair
brake mechanism

40. 40
Mechanism Categories
Function Generation Mechanisms
A four-bar drive linkage for a lawn sprinkler

41. 41
Mechanism Categories
Function Generation Mechanisms
A four-bar function generation mechanism to
operate an artificial hand used for gripping.
A four-bar function generation
mechanism to lower an attic stairway.

42. 42
Mechanism Categories
Motion Generation Mechanisms
In motion generation, the entire motion of the coupler
link is of interest (rigid body guidance).
New Rollerblade brake system

43. 43
Mechanism Categories
Motion Generation Mechanisms
Four-bar automobile hood linkage design

44. 44
Motion Generation Mechanisms
Rotating a monitor into a storage position
Moving a storage bin from an
accessible position to a stored position

45. 45
Motion Generation Mechanisms
Lifting a boat out of water
Moving a trash pan from the floor up over
a trash bin and into a dump position

46. 46
Mechanism Categories
Path Generation Mechanisms
In path generation, we are concerned only with the path of a tracer
point and not with the motion (rotation) of the coupler link.
Crane –straight line motion

47. 47
Mechanism Categories
Path Generation Mechanisms
A four-bar path generation mechanism as
part of an arm-actuated propulsion system
for a wheelchair
A four-bar path generation mechanism to guide a
thread in an automatic sewing machine

48. Slider-Crank Mechanism
•The slider-crank mechanism below has stroke
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2
•The extreme positions &#3627408437;
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by constructing circle arcs through ??????
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length &#3627408479;
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3+&#3627408479;
2respectively.
Centered slider-crank mechanism General/ offset slider-crank mechanism

49. Rocker Output-Two Position with
Angular Displacement
Example1
DesignafourbarGrashofcrank-rocker
mechanismtogive45°ofrockermotionwith
equaltimeforwardandback,fromaconstant
speedmotorinput.

50. Example 1
•DrawtheoutputlinkO
4Binbothextreme
positions,B
1andB
2inanyconvenientlocation,
suchthatthedesiredangleofmotionθ
4is
subtended.
•DrawthechordB
1B
2andextenditineither
direction.
•SelectaconvenientpointO
2onthelineB
1B
2
extended.
•BisectlinesegmentB
1B
2,drawacircleofthat
radiusaboutO
2.
•LabelthetwointersectionofthecircleandB
1B
2
extended,A
1andA
2.
•MeasurethelengthofthecouplerasA
1toB
1or
A
2toB
2.
•Measuregroundlength1,cranklength2,and
rockerlength4.
•FindtheGrashofcondition.Ifnon-Grashof,redo
steps3to8withO
2furtherfromO
4.
•Make model of the linkage and check its
function and transmission angles.

51. Example 2
•Rocker Output-Two Position with Complex
Displacement (Motion)
–Design a fourbarlinkage to move link CD from
C
1D
1to C
2D
2.

52. Example 2
•DrawthelinkCDinitstwodesired
positions,C
1D
1andC
2D
2inplaneas
shown.
•Draw construction line from point C
1to
C
2and from point D
1to D
2.
•Bisect line C
1C
2and line D
1D
2and
extend their perpendicular bisectors to
intersect at O
4.Their intersection is the
rotopole.
•Select a convenient radius and draw an
arc about the rotopoleto intersect both
lines O
4C
1and O
4C
2.Label the
intersection B
1and B
2.
•Do steps 2 to 8 of example 1 to
complete the linkage.
•Make a model of the linkage and
articulate it to check its function and its
transmission angles.

53. Coupler Output-Two Position with
Complex Displacement (Motion)
Example 3: Design a fourbarlinkage
to move link CD from C
1D
1to C
2D
2
(with moving pivots at C and D).

54. Example 3
•Draw the link CD in its two desired positions, C
1D
1
and C
2D
2in plane as shown.
•Draw construction line from point C
1to C
2and from
point D
1to
D
2.
•Bisect line C
1C
2and line D
1D
2and extend their
perpendicular bisectors in convenient directions. The
rotopolewill not be used in this solution.
•Select any convenient point on each bisector as the
fixed pivots O
2and O
4, respectively.
•Connect O
2with C
1and call it link 2. Connect O
4with
D
1and call it link 4.
•Line C
1D
1is link 3. Line O
2O
4is link 1.
•Check the Grashofcondition, and repeat steps 4 to 7
if unsatisfied. Note that any Grashofcondition is
potentially acceptable in this case.
•Construct a model and check its function to be sure
it can get from the initial to final position without
encountering ant limit (toggle) positions.
•Check the transmission angles.

55. Review Example 3
Design a dyad to control and limits the
extremes of motion of the linkages in the
previous example to its two design positions

56. Review Example 3
•Selectaconvenientpointonlink2ofthe
linkagedesignedinExample3.Notethat
itneednotbeonthelineO
2C
1.Label
thispointB
1.
•DrawanarcaboutcenterO
2throughB
1
tointersectthecorrespondinglineO
2B
2
inthesecondpositionoflink2.Label
thispointB
2.ThechordB
1B
2provides
uswiththesameprobleminExample1.
•Dosteps2-9ofExample1tocomplete
thelinkage,exceptaddlinks5and6and
centerO
6ratherthanlinks2and3and
centerO
2.Link6willbethedriver
crank.Thefourbarsubchainofthelinks
O
6,A
1,B
1,O
2mustbeaGrashofcrank-
rocker.

57. Three-Position Synthesis
•Example 4 –Coupler Output –3 position with
Complex Displacement
•Design a fourbarlinkage to move the link CD
shown from position C
1D
1to C
2D
2and then to
position C
3D
3. Moving pivots are C and D. Find
the fixed pivot locations.

58. Example 4
•Draw link CD in its three position
•C
1D
1,C
2D
2, C
3D
3in the plane as shown.
•Draw construction lines from point C
1to C
2and
from C
2to C
3.
•BisectlineC
1C
2andlineC
2C
3andextendtheir
perpendicularbisectoruntiltheyintersect.Label
theirintersectionO
2.
•Repeat steps 2 and 3 for lines D
1D
2and D
2D
3. Label
the intersection O
4.
•Connect O
2with C
1and call link 2. Connect O
4with
D
1and call link 4.
•Line C
1D
1is link 3. Line O
2O
4is link 1.
•Check the Grashofcondition. Note that any
Grashofcondition is potentially acceptable in this
case.
•8.Construct a model and check its function to be
sure it can get from initial to final position without
encountering any limits positions.

59. Three-Position Synthesis –Example 5
•Coupler Output –3 position
with Complex Displacement
(Alternate Attachment
Points for Moving Pivots)
•Design a fourbarlinkage to
move the link CD shown
from position C
1D
1to C
2D
2
and then to position C
3D
3.
Use different moving pivot
than CD. Find the fixed
pivot locations.

60. Example 5
•Draw link CD in its three position C
1D
1,C
2D
2, C
3D
3in the
plane as shown.
•Define new attachment points E
1and F
1that have a fixed
relationship between C
1D
1and E
1F
1within the link. Now
use E
1F
1to define the three position of the link.
•Draw construction lines from point E
1to E
2and from E
2to
E
3.
•Bisect line E
1E
2and line E
2E
3and extend their
perpendicularbisector until they intersect. Label their
intersection O
2.
•Repeat steps 2 and 3 for lines F
1F
2and F
2F
3. Label the
intersection O
4.
•Connect O
2with E
1and call link 2. Connect O
4with F
1and
call link 4.
•Line E
1F
1is link 3. Line O
2O
4is link 1
•Check the Grashofcondition. Note that any Grashof
condition is potentially acceptable in this case.
•Construct a model and check its function to be sure it can
get from initial to final position without encountering any
limits positions. If not, change locations of point E and F
and repeat steps 3 to 9.

61. Example 6
•Example 6 –Three –Position Synthesis with
Specified Fixed Pivots -Inverting the 3-position
problem
•Invert a fourbarlinkage which move the link CD
shown from position C
1D
1to C
2D
2and then to
position C
3D
3. Use specified fixed pivots O
2and O
4.

62. Example 6
•Draw link CD in its three position C
1D
1,C
2D
2, C
3D
3in the plane as
shown.
•Draw the ground link O
2O
4in its desired position in the plane with
respect to the first coupler position C
1D
1
•Draw construction arc from point C
2to O
2and from D
2to O
2
whose radii define the side of triangle C
2O
2D
2This define the
relationship of the fixed pivot O
2to the coupler line CD in the
second coupler position.
•Draw construction arc from point C
2to O
4and from D
2to O
4
whose radii define the side of triangle C
2O
4D
2This define the
relationship of the fixed pivot O
4to the coupler line CD in the
second coupler position.
•Now transfer this relationship back to the first coupler position
C
1D
1so that the ground plane position O
2’O
4’ bears the same
relationship to C
1D
1as O
2O
4bore to the second coupler position
C
2D
2. In effect, you are sliding C
2along the dotted line C
2C
1and D
2
along the dotted D
2D
1. By doing this, we have pretended that
ground plane moved from O
2O
4to O
2’O
4’ instead of the coupler
moving from C
1D
1to C
2D
2. We have inverted the problem.
•Repeat the process for the third coupler position as shown in the
figure and transfer the third relative ground link position to the
first, or reference, position.
•The three inverted position of the ground plane that correspond
to the three desired coupler positions are labeledO
2O
4,O
2’O
4’ ,
and O
2’’O
4’’ and have also been renamed
•E
1F
1, E
2F
2and E
3F
3as shown in the figure.

63. Example 6

64. Example 7
•Example 7 –Finding the
Moving Pivots for Three
Positions and Specified
Fixed Pivots
•–Design a fourbarlinkage
to move the link CD shown
from position C
1D
1to C
2D
2
and then to position C
3D
3.
Use specified fixed pivots
O
2and O
4. Find the
required moving pivot
location on the coupler by
inversion.

65. Example 7
•Start with inverted three positions plane as shown in the figures. Lines E
1F
1, E
2F
2and E
3F
3define
the three positions of the inverted link to be moved.
•Draw construction lines from point E
1to E
2and from point E
2to E
3.
•Bisect line E
1E
2and line E
2E
3extend the perpendicular bisector until they intersect. Label the
intersection G.
•Repeat steps 2 and 3 for lines F
1F
2and line F
2F
3. Label the intersection H.
•Connect G with E
1and call it link 2. Connect H with F
1and call it link 4.
•In this inverted linkage, line E
1F
1is the coupler, link 3. Line GH is the “ground” link1.
•We must now reinvert the linkage to return to the original arrangement. Line E
1F
1is really the
ground O
2O
4and GH is really the coupler. The figure shows the reinversion of the linkage inwhich
points G and H are now the moving pivots on the coupler and E
1F
1has resumed its real identity as
ground link O
2O
4.
•The figure reintroduces the original line C
1D
1in its correct relationship to line O
2O
4at the initial
position as shown in the original example 3. This form the required coupler plane and defines a
minimal shape of link 3.
•

66. Example 7
•The angular motions required to reach the second and third position of
line CD shown in the figure are the same as those defined in figure b for
the linkage inversion. The angle F
1HF
2in the figure b is the same as the
angle H
1O
4H
2in the figure and F
2HF
3is the same as angle H
2O
4H
3. The
angular excursions of link 2 retain the same between figure b and e as
well. The angular motions of links 2 and 4 are the same for both inversion
as the link excursions are relative to one another.
•Check the Grashofcondition. Note that any Grashofcondition is
potentially acceptable in this case provided that the linkage has mobility
among all three position. This solution is a non-Grashoflinkage.
•11. Construct a model and check its function to be sure it can get from
initial to final position without encountering any limit (toggle) positions.
•In this case link 3 and 4 reach a toggle position between points H
1and H
2.
This means that this linkage cannot be driven from link 2 as it will hang up
at that toggle position. It must driven from link 4.

67. Quick –Return Mechanism

68. Quick –Return Mechanism
•FourbarQuick-Return
•Time ratio (T
R
) defines the degree of quick-return of the linkage.
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•Works well for time ratios down to about 1.5

69. Synthesis of Crank-and-Rocker
Mechanism with unit time ratio180
minmax


70. Example 1
•FourbarCrank-Rocker Quick-
Return Linkage for Specified
Time Ratio
•Redesign Example 1 (two
position) to provide a time
1:1.25 with 45°output rocker
motion.

71. Example 1
•Draw the output link O
4B in both
extreme position, in any convenient
location, such that the desired angle
of motion, θ
4, is subtended.
•Calculate α, β, δ using the equations.
In this example α=160°, β=200°,
δ=20°.
•Draw a construction line through
point B
1at any convenient angle.
•Draw a construction line through
point B
2at an angle δ from the first
line.
•Label the intersection of the two
construction line O
2.
•The line O
2O
4define ground.

72. Example 1
•Calculate the length of crank and coupler by
measuring O
2B
1and O
2B
2and solve
simultaneously;
•coupler + crank=O
2B
1; coupler -crank=O
2B
2
or you can construct the cranklength by
swinging an arc centeredat O
2from B
1to cut
line O
2B
2extended.
•Label that intersection B
1’. The line B
2B
1’ is
twice the crank length. Bisect this line
segment to measure crank length O
2A
1.
•Calculate the Grashofcondition.
•If non-Grashof, repeat steps 3 to 8 with O
2
further O
4.
•Make a model of the linkage and articulate it
to check its function.
•Check the transmission angle.

73. Example 2
•SixbarDrag Link Quick-Return Linkage for
Specified Time Ratio.
•Provide a time ratio of 1:1.4 with 90 degree
rocker motion.

74. Example 2
•Calculate α, β, δ. In this example α=150°, β=210°.
•Draw a line of centersXX at any convenient location.
•Choose a crank pivot location O
2on line XX and draw
an axis YY perpendicular to XX.
•Draw a line of convenient radius O
2A about centerO
2
•Lay out angle α with vertex at O
2,symmetrical about
quadrant one
•Label points A
1and A
2at the intersection of the lines
subtending angle α and the circle of radius O
2A.

75. Example 2
•Set the compass to a convenient radius AC long enough to cut XX in two places
on either side of O
2when swung from A
1and A
2. Label the intersection C
1and
C
2.
•The line O
2A
1is the driver crank, link 2, and line A
1C
1is the coupler, link 3.
•The distance C
1C
2is twice the driven (dragged) crank length. Bisect it to locate
the fixed pivot O
4.
•The line O2O4 now defines the ground link. Link O4C1 is the driven crank, link 4.
•Calculate the Grashofcondition. If Non-Grashof, repeat steps 7-11 with a shorter
radius in step 7.
•invert the method of example 1 (two positions) to create the output dyad using
XX as the chord and O
4C
1as the driving crank. The point B
1and B
2will lie on line
XX and be spaced apart a distance 2O
4C
1. The pivot O
6will lie on the
perpendicular bisector of B
1B
2, at a distance from line XX which subtends the
specified output rocker angle.
•Check the transmission angle.

76. Example 2

77. Synthesis of Rocker-and-Rocker
Mechanism
Three-Position Synthesis
•Given ψ
12,ψ
23,ψ
13,∅
12,∅
23,∅
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2??????&#3627408475;&#3627408465;(??????
4,&#3627408479;
4,&#3627408479;
3?)
•Draw ??????
2&#3627408436;in its three specified positions and locate ??????
4
•Draw a ray from ??????
4&#3627408481;&#3627408476;&#3627408436;
2and rotate it through ∅
12to locate &#3627408436;
2
′
•Draw a ray from ??????
4&#3627408481;&#3627408476;&#3627408436;
3and rotate it through ∅
13to locate &#3627408436;
3
′
•Draw mid-normalsto the lines &#3627408436;
1
′
&#3627408436;
2
′
and &#3627408436;
2
′
&#3627408436;
3
′
these intersect at &#3627408437;
1

78. Coupler Curves
For the mechanisms considered, the
displacement of the links joined with the fixed
link was the input or output of the simple
mechanisms. In great number of applications
the output from a simple mechanism is the path
traced by one of the points on the coupler link.
These paths are generally called “coupler point
curves” or “coupler paths”.

79. Coupler Curves

80. Coupler Curves

81. MECHANISM ANALYSIS

82. Review of Vector Analysis
i.Cartesian/rectangular coordinate system
i
kj

83. Example
The motion of a particle is described in
Cartesian coordinates as
&#3627408485;&#3627408481;=2&#3627408481;
2
+4&#3627408481;
&#3627408486;&#3627408481;=0.1&#3627408481;
3
+cos&#3627408481;&#3627408487;&#3627408481;=3&#3627408481;
Find the velocity and acceleration of the
particle.

84. ii. Cylindrical coordinate system
&#3627408531;&#3627408481;=????????????
&#3627408531;+&#3627408487;&#3627408524;=??????&#3627408464;&#3627408476;&#3627408480;??????&#3627408522;+??????&#3627408480;??????&#3627408475;??????&#3627408523;+&#3627408487;&#3627408524;
??????&#3627408481;=ሶ&#3627408531;&#3627408481;=ሶ????????????
&#3627408531;+??????ሶ????????????
??????+ሶ&#3627408487;&#3627408524;??????&#3627408481;=ሷ??????−??????ሶ??????
2
??????
&#3627408531;+??????ሷ??????+2ሶ??????ሶ????????????
??????+ሷ&#3627408487;&#3627408524;
??????
&#3627408531;=&#3627408464;&#3627408476;&#3627408480;??????&#3627408522;+&#3627408480;??????&#3627408475;??????&#3627408523;??????&#3627408475;&#3627408465;??????
??????=−&#3627408480;??????&#3627408475;??????&#3627408522;+&#3627408464;&#3627408476;&#3627408480;??????&#3627408523;

85. Example
Attheinstantshownthelengthof
theboomABisbeingdecreasedat
theconstantrateof0.2m/sandthe
boomisbeingloweredatthe
constantrateof0.08rad/s.Using
cylindricalcoordinatesystem,
Determine(a)thevelocityofpointB,
(b)theaccelerationofpointB.

86. Solution
??????=6&#3627408474;;ሶ??????=−0.08&#3627408479;??????&#3627408465;/&#3627408480;;
ሶ??????=−0.2&#3627408474;/&#3627408480;;
ሶ&#3627408487;=ሷ&#3627408487;=0;ሷ??????=0;ሷ??????=0
∴??????&#3627408481;=(−0.2??????
&#3627408531;−0.48??????
??????)m/s
??????&#3627408475;&#3627408465;??????&#3627408481;=−??????ሶ??????
2
??????
&#3627408531;+2ሶ??????ሶ????????????
??????
??????&#3627408481;=−38.4??????
&#3627408531;+32??????
??????&#3627408474;&#3627408474;/&#3627408480;
2

87. iii. Spherical coordinate system
??????
??????×??????
??????=??????
??????
&#3627408531;&#3627408481;=????????????
&#3627408531;
&#3627408485;=??????&#3627408480;??????&#3627408475;??????&#3627408464;&#3627408476;&#3627408480;??????&#3627408486;=??????&#3627408480;??????&#3627408475;??????&#3627408480;??????&#3627408475;??????&#3627408487;=??????&#3627408464;&#3627408476;&#3627408480;??????
??????
??????=&#3627408480;??????&#3627408475;??????&#3627408464;&#3627408476;&#3627408480;??????&#3627408522;+&#3627408480;??????&#3627408475;??????&#3627408480;??????&#3627408475;??????&#3627408523;+&#3627408464;&#3627408476;&#3627408480;??????&#3627408524;
??????
??????=−&#3627408480;??????&#3627408475;??????&#3627408522;+&#3627408464;&#3627408476;&#3627408480;??????&#3627408523;
??????
??????=&#3627408464;&#3627408476;&#3627408480;??????&#3627408464;&#3627408476;&#3627408480;??????&#3627408522;+&#3627408480;??????&#3627408475;??????&#3627408523;−&#3627408480;??????&#3627408475;??????&#3627408524;
&#3627408465;??????
??????
&#3627408465;&#3627408481;
=ሶ??????&#3627408480;??????&#3627408475;????????????
??????+ሶ????????????
??????
&#3627408465;??????
??????
&#3627408465;&#3627408481;
=−ሶ??????&#3627408480;??????&#3627408475;????????????
??????−ሶ??????????????????&#3627408532;????????????
??????
&#3627408465;??????
??????
&#3627408465;&#3627408481;
=ሶ??????&#3627408464;&#3627408476;&#3627408480;????????????
??????−ሶ????????????
??????
??????&#3627408481;=ሶ????????????
??????+??????ሶ??????&#3627408480;??????&#3627408475;????????????
??????+??????ሶ????????????
??????
??????&#3627408481;=(ሷ??????−??????ሶ??????
2
&#3627408480;??????&#3627408475;
2
??????−??????ሶ??????
2
)??????
??????+??????ሷ??????&#3627408480;??????&#3627408475;??????+2ሶ??????ሶ??????&#3627408480;??????&#3627408475;??????+2??????ሶ??????ሶ??????&#3627408464;&#3627408476;&#3627408480;????????????
??????+
(??????ሷ??????+2ሶ??????ሶ??????−??????ሶ??????
2
&#3627408480;??????&#3627408475;??????&#3627408464;&#3627408476;&#3627408480;??????)??????
??????

88. Example
Atagiveninstant,thesatellite
dishhasanangularmotion??????
1
=6rad/sandሶ??????
1=3&#3627408479;??????&#3627408465;/&#3627408480;
2
aboutthezaxis.Atthissame
instant??????=25
??????
,theangular
motionaboutthexaxisis??????
2=
2rad/s,andሶ??????
2=1.5&#3627408479;??????&#3627408465;/&#3627408480;
2
.
Usingsphericalcoordinate
system,determinethevelocity
andaccelerationofthesignal
hornAatthisinstant.

89. Solution
??????=1.4&#3627408474;;??????=90
??????
;??????=65
??????
ሶ??????=ሷ??????=0;
ሶ??????=??????
2=2&#3627408479;??????&#3627408465;/&#3627408480;;ሷ??????=ሶ??????
2=1.5&#3627408479;??????&#3627408465;/&#3627408480;
2
;
ሶ??????=??????
1=6&#3627408479;??????&#3627408465;/&#3627408480;;ሷ??????=ሶ??????
1=3&#3627408479;??????&#3627408465;/&#3627408480;
2
∴??????&#3627408481;=(2.8??????
??????+8.16??????
??????)m/s
??????&#3627408475;&#3627408465;??????&#3627408481;=−??????ሶ??????
2
??????
&#3627408531;+2ሶ??????ሶ????????????
??????
??????&#3627408481;=−47??????
&#3627408531;+34.25??????
??????+21.4??????
??????&#3627408474;/&#3627408480;
2

90. Transport Theorem
•ConsideravectorAobserved
fromamovingcoordinate
systemxyz,whichisrotating
withangularvelocityΩ
90
Figure a
Figure b

91. General Motion
91ABAB
rrr
/

PositionVector
Velocity
ABAB
r
dt
d
r
dt
d
r
dt
d
/
 
B/A/
r x 
xyzABAB
vvv
Acceleration  
B/A/
r x 
dt
d
v
dt
d
v
dt
d
v
dt
d
xyzABAB    
B/AB/AB/AB/A r x x v x 2r x a 
xyzxyzABaa
  
B/A
2
B/AB/AB/A r v x 2r x a 
xyzxyzABaa

For2Dmotions,theaboveequationreducesto

92. Example 1
Crank AB of the engine system shown has a constant
clockwise angular velocity of 2000 rpm. For the crank
position shown, determine the angular velocity as well
as the angular acceleration of the connecting rod BD
and the linear velocity and acceleration of point D.

93. solution
Since the crank rotates about A with constant ??????
&#3627408436;&#3627408437;=2000&#3627408479;&#3627408477;&#3627408474;=
209.4&#3627408479;??????&#3627408465;/&#3627408480;, The velocity of B can be calculated from
&#3627408483;
&#3627408437;=&#3627408483;
&#3627408436;+&#3627408483;
&#3627408437;/&#3627408436;+Ω
&#3627408436;&#3627408437;x&#3627408479;
&#3627408437;/&#3627408436;
&#3627408483;
&#3627408437;/&#3627408436;=&#3627408483;
&#3627408436;=0
&#3627408483;
&#3627408437;=Ω
&#3627408436;&#3627408437;x&#3627408479;
&#3627408437;/&#3627408436;=209.4&#3627408524;x0.03cos40&#3627408522;+0.03sin40&#3627408523;
&#3627408483;
&#3627408437;=−4.04&#3627408522;−4.81&#3627408523;&#3627408474;&#3627408480;
−1
&#3627408483;
&#3627408439;=&#3627408483;
&#3627408437;+&#3627408483;
&#3627408439;/&#3627408437;+Ω
&#3627408439;&#3627408437;x&#3627408479;
&#3627408439;/&#3627408437;
&#3627408483;
&#3627408439;/&#3627408437;=0
&#3627408483;
&#3627408439;&#3627408522;=−4.04&#3627408522;+4.81&#3627408523;+Ω
&#3627408439;&#3627408437;&#3627408524;x0.03cos40&#3627408522;+0.03sin40&#3627408523;
Expanding and comparing coefficients
&#3627408483;
&#3627408439;=−4.04−0.019Ω
&#3627408439;&#3627408437;
0=4.81+0.023Ω
&#3627408439;&#3627408437;
Ω
&#3627408439;&#3627408437;=−6.28&#3627408479;??????&#3627408465;/&#3627408480;; &#3627408483;
&#3627408439;=−209.13&#3627408474;&#3627408480;
−1

94. Solution
Since &#3627409148;
&#3627408436;&#3627408437;=0. The acceleration of B can be calculated from
??????
&#3627408437;=??????
&#3627408436;+??????
&#3627408437;/&#3627408436;+2Ω
&#3627408436;&#3627408437;x&#3627408483;
&#3627408437;/&#3627408436;+&#3627409148;
&#3627408436;&#3627408437;x&#3627408479;
&#3627408436;&#3627408437;−Ω
&#3627408436;&#3627408437;
2
&#3627408479;
&#3627408437;/&#3627408436;
&#3627408483;
&#3627408437;/&#3627408436;=??????
&#3627408436;=??????
&#3627408437;/&#3627408436;=&#3627409148;
&#3627408436;&#3627408437;=0
??????
&#3627408437;=−Ω
&#3627408436;&#3627408437;
2
&#3627408479;
&#3627408437;/&#3627408436;=−209.4
2
0.03cos40&#3627408522;+0.03sin40&#3627408523;
??????
&#3627408437;=−1007.63&#3627408522;+845.56&#3627408523;&#3627408474;&#3627408480;
−2
??????
&#3627408439;=??????
&#3627408437;+??????
&#3627408439;/&#3627408437;+2Ω
&#3627408439;&#3627408437;x&#3627408483;
&#3627408439;/&#3627408437;+&#3627409148;
&#3627408439;&#3627408437;x&#3627408479;
&#3627408439;/&#3627408437;−Ω
&#3627408439;&#3627408437;
2
&#3627408479;
&#3627408439;/&#3627408437;
&#3627408483;
&#3627408439;/&#3627408437;=??????
&#3627408439;/&#3627408437;=0
??????
&#3627408439;
=−1007.63&#3627408522;+845.56&#3627408523;+&#3627409148;
&#3627408439;&#3627408437;&#3627408524;x0.03cos40&#3627408522;+0.03sin40&#3627408523;
−6.28
2
0.03cos40&#3627408522;+0.03sin40&#3627408523;
Expanding and comparing coefficients
??????
&#3627408439;&#3627408522;
=−1007.63−0.019&#3627409148;
&#3627408439;&#3627408437;−0.91&#3627408522;
+−845.56+0.0236&#3627409148;
&#3627408439;&#3627408437;−0.75&#3627408523;
0=−846.33+0.023&#3627409148;
&#3627408439;&#3627408437;
??????
&#3627408439;=−1037.84−0.019&#3627409148;
&#3627408439;&#3627408437;
;

95. Example 2
Determinethevelocityandacceleration
ofthecollarCatthisinstant
Solution
&#3627408483;
&#3627408439;=&#3627408483;
&#3627408437;+&#3627408483;
&#3627408439;/&#3627408437;+Ω
&#3627408439;&#3627408437;x&#3627408479;
&#3627408439;/&#3627408437;
&#3627408483;
&#3627408439;=3&#3627408523;+5&#3627408524;x0.2&#3627408523;
=−&#3627408522;+3&#3627408523;&#3627408474;/&#3627408480;
??????
&#3627408439;
=??????
&#3627408437;+??????
&#3627408439;/&#3627408437;+2Ω
&#3627408439;&#3627408437;x&#3627408483;
&#3627408439;/&#3627408437;
+&#3627409148;
&#3627408439;&#3627408437;x&#3627408479;
&#3627408439;/&#3627408437;−Ω
&#3627408439;&#3627408437;
2
&#3627408479;
&#3627408439;/&#3627408437;
??????
&#3627408439;
=2&#3627408523;+10&#3627408524;x3&#3627408523;−6&#3627408524;x0.2&#3627408523;
−25(0.2&#3627408523;)
??????
&#3627408439;=−188&#3627408522;−3&#3627408523;&#3627408474;&#3627408480;
−2

96. Relative Velocity Method
Thecrankofaslidercrankmechanismrotatesclockwiseata
constantspeedof300rpm.Thecrankis150mmandthe
connectingrodis600mmlong.Determine:
1.Linearvelocityandaccelerationofthemidpointoftheconnecting
rod,and
2.Angularvelocityandangularaccelerationoftheconnectingrod,ata
crankangleof45°frominnerdeadcentreposition.

97. solution
Velocity analysis
&#3627408483;
&#3627408437;=??????&#3627408479;
&#3627408437;=0.15×10??????=4.713&#3627408474;/&#3627408480;
From the velocity diagram
&#3627408483;
&#3627408436;&#3627408437;=3.4&#3627408474;/&#3627408480;
&#3627408483;
&#3627408436;=4&#3627408474;/&#3627408480;
&#3627408483;
&#3627408439;=4.1&#3627408474;/&#3627408480;
Acceleration analysis
??????
&#3627408437;
??????
=
&#3627408483;
&#3627408437;
2
??????&#3627408437;
=
4.713
2
0.15
=148.1&#3627408474;/&#3627408480;
2
??????
&#3627408436;&#3627408437;
??????
=
&#3627408483;
&#3627408436;&#3627408437;
2
&#3627408436;&#3627408437;
=
3.4
2
0.6
=19.3&#3627408474;/&#3627408480;
2
From the acceleration diagram
&#3627409148;
&#3627408436;&#3627408437;=
??????
&#3627408436;&#3627408437;
??????
&#3627408436;&#3627408437;
=
103
2
0.6
=171.67&#3627408479;??????&#3627408465;/&#3627408480;
2

98. Example 2
Inthemechanismshown,theslideQDisdrivenwithauniform
angularvelocityof10rad/sandtheblockChasamassof1.5kg.
Drawthevelocityandaccelerationdiagramsforthemechanismin
thepositionshown.OQ=1m;OC=2m.

99. solution
Velocity analysis
&#3627408483;
??????=28.8&#3627408474;/&#3627408480;
From the velocity diagram
&#3627408483;
????????????′=7.5&#3627408474;/&#3627408480;
Acceleration analysis
??????
??????
??????
=
&#3627408483;
??????
2
??????&#3627408438;
=414&#3627408474;/&#3627408480;
2
??????
&#3627408438;
′
&#3627408452;
??????
=288&#3627408474;/&#3627408480;
2
??????
&#3627408438;
′
&#3627408438;
&#3627408438;
=2&#3627408483;
????????????′??????
&#3627408438;
′
&#3627408452;
??????
&#3627408438;
′
&#3627408438;
&#3627408438;
=150m/&#3627408480;
2
From the acceleration diagram
??????
&#3627408438;=417&#3627408474;/&#3627408480;
2

100. Assignment 1
Determine the angular velocity and
angular acceleration of the rod DE
at the instant shown. The collar at
C is pin connected to AB and slides
over rod DE.

101. Assignment 2
The rocker of a crank-rocker linkage is to have a length
of 500 mm and swing through a total angle of 45°with
a time ratio of 1.25. Determine a suitable set of
dimensions for &#3627408479;
1,&#3627408479;
2??????&#3627408475;&#3627408465;&#3627408479;
3.

102. Assignment 3
Thefigureshowstwopositionsofafoldingseatusedinthe
aislesofbusestoaccommodateextraseatedpassengers.Design
afour-barlinkagetosupporttheseatsothatitwilllockinthe
openpositionandfoldtoastableclosingpositionalongtheside
oftheaisle.
3. Determine the link lengths of a slider-crank linkage to have
a stroke of 600 mm and a time ratio of 1.2.

103. Assignment 4
Designafourbarlinkage
tomovetheobjectbelow
fromposition1to2using
pointsAandBfor
attachment.Addadriver
dyadtolimititsmotionto
therangeofpositions
shownmakingitasixbar.
Allfixedpivotsshouldbe
onthebase.

104. Assignment 5
OwusuHouse of Flapjacks wants to automate
their flapjack production. They need a
mechanism which will automatically flip the
flapjacks “on the fly” as they travel through the
griddle on a continuously moving conveyor. This
mechanism must track the constant velocity of
the conveyor, pick up a pancake, flip it over and
place it back onto the conveyor.

105. Assignment 6
??????
2&#3627408436;=8&#3627408464;&#3627408474;
&#3627408436;&#3627408437;=32&#3627408464;&#3627408474;
??????
4&#3627408437;=16&#3627408464;&#3627408474;
??????
2??????
4=16&#3627408464;&#3627408474;
??????
4&#3627408438;=12&#3627408464;&#3627408474;
&#3627408436;&#3627408439;=16&#3627408464;&#3627408474;
Draw the coupler curve of point D at the middle of the
coupler AB for a complete revolution of link 2.

106. Assignment 7
IflinkCDhasanangularvelocityof=6rad/s,determine
thevelocityandaccelerationofpointEonlinkBCandthe
angularvelocityandangularaccelerationoflinkABatthe
instantshown.

107. Assignment 8
AttheinstantshownrodABhasanangularvelocity??????
&#3627408436;&#3627408437;=4
rad/sandanangularacceleration??????
&#3627408438;&#3627408439;=2&#3627408479;??????&#3627408465;/&#3627408480;
2
.Determine
theangularvelocityandangularaccelerationofrodCDatthis
instant.ThecollaratCispinconnectedtoCDandslidesfreely
alongAB.

108. Assignment 9
The"quick-return"mechanism
consistsofacrankAB,slider
blockB,andslottedlinkCD.If
thecrankhastheangular
motionshown,determinethe
angularmotionoftheslotted
linkatthisinstant.

109. Assignment 10
BallCmoveswithaspeedof3m/s,
whichisincreasingataconstant
rateof1.5m/&#3627408480;
2
,bothmeasured
relativetothecircularplateand
directedasshown.Atthesame
instanttheplaterotateswiththe
angularvelocityandangular
accelerationshown.Determinethe
velocityandaccelerationofthe
ballatthisinstant.