1. Analysis by
UNIT 5 ANALYSIS BY ANALYTICAL Analytical Methods
METHODS
Structure
5.1 Introduction
Objectives
5.2 Analytical Method for Velocity and Acceleration
5.2.1 Approximate Method for Slider Crank Mechanism
5.2.2 General Method for Velocity and Acceleration
5.3 Motion Referred to Motion Frames of Reference
5.3.1 Relative Motion of Two Points
5.3.2 Plane Motion of a Link
5.4 Method of Relative Velocity and Acceleration
5.5 Alternative Method of Determining Coriolis’ Component of Acceleration
5.6 Klein’s Construction for Determining Velocity and Acceleration of Slider
Crank Mechanism
5.7 Summary
5.8 Key Words
5.9 Answers to SAQs
5.1 INTRODUCTION
In unit 4, you have studied instantaneous centre method for determining velocity of any
point in a mechanism. This method is good for determining velocity and cannot be used
determining acceleration. Also, if lines drawn for determining instantaneous centre are
parallel, they cannot locate instantaneous centre. Even if these lines are nearly parallel,
they will meet at a large distance to locate instantaneous centre within the limited size of
the paper. In this unit, you will be explained relative velocity method and analytical
method and how they can be used to determine velocity and then acceleration of a point
in a mechanism. Acceleration is required for determining inertia forces which are
required in dynamic analysis of a mechanism.
Objectives
After going through this unit you should be able to
• analyse plane motion,
• analyse plane motion with moving frames of references,
• determine magnitude of Corioli’s acceleration and its direction,
• determine velocity and acceleration of any point in a mechanism, and
• apply Klein’s construction for determination of acceleration of slider in
slider crank mechanism.
21

2. Motion Analysis of
Planar Mechanism and 5.2 ANALYTICAL METHOD FOR VELOCITY AND
Synthesis
ACCELERATION
The analytical method can be used for determining velocity and acceleration of any point
in a mechanism and angular acceleration of any link.
5.2.1 Approximate Method for Slider Crank Mechanism
The slider crank mechanism OAB is shown in Figure 5.1. Let l and r be lengths of the
connecting rod AB and crank OA respectively. Let x be the distance of the piston pin
from its farthest position B, (for this point OA and AB are aligned).
A
l
r 2
3 B
θ C φ
O B1
4
1 x
Figure 5.1
x = (r + l ) − (OC + CB ) = (r + l ) − (r cos θ + l cos φ)
⎧ l ⎫
or x = r (1 − cos θ) + l (1 − cos φ) = r ⎨(1 − cos θ) + (1 − cos φ) ⎬
⎩ r ⎭
Also AC = r sin θ = l sin φ
r
∴ sin φ = sin θ
l
r2
Also cos 2 φ = 1 − sin 2 φ = 1 − sin 2 θ
l2
1
⎡ r2 ⎤2
∴ cos φ = ⎢1 − 2 sin 2 θ ⎥
⎢
⎣ l ⎥
⎦
By Binomial theorem
2 4 6
1 ⎛r⎞ 1 ⎛r⎞ 1 ⎛r⎞
cos φ = 1 − ⎜ ⎟ sin θ − ⎜ ⎟ sin θ −
2 4
⎜ ⎟ sin θ
6
2 ⎝l⎠ 8⎝l⎠ 16 ⎝ l⎠
2 4 6
1 ⎛r⎞ 1 ⎛r⎞ 1 ⎛r⎞
or (1 − cos φ) = ⎜ ⎟ sin 2 θ + ⎜ ⎟ sin 4 θ + ⎜ ⎟ sin θ + . . .
6
2⎝l⎠ 8⎝l⎠ 16 ⎝ l ⎠
Substituting (1 – cos φ) in expression of x you will get
⎡ 1 ⎛r⎞ 1 ⎛r⎞
3 ⎤
x = r ⎢(1 − cos θ) + ⎜ ⎟ sin 2 θ + ⎜ ⎟ sin 4 θ + . . .⎥
⎢
⎣ 2⎝l⎠ 8⎝l⎠ ⎥
⎦
r ⎛r⎞
In this mechanism, is approximately 0.25 and higher powers of ⎜ ⎟ shall have
l ⎝l⎠
negligible value.
⎡ 1 ⎛r⎞ ⎤
Therefore, x ≈ r ⎢(1 − cos θ) + ⎜ ⎟ sin 2 θ ⎥
⎣ 2⎝l⎠ ⎦
22

3. dx dx d θ Analysis by
Velocity of slider, V = = × Analytical Methods
dt d θ dt
dθ
=ω (angular speed of crank)
dt
dx
Therefore, V =ω
dθ
⎡ 1 ⎛r⎞ ⎤
or V = r ω ⎢sin θ + ⎜ ⎟ 2 sin θ cos θ ⎥
⎣ 2⎝l⎠ ⎦
⎛ 1 ⎛r⎞ ⎞
or V = r ω ⎜ sin θ + ⎜ ⎟ sin 2 θ ⎟
⎝ 2⎝l⎠ ⎠
Let crank rotate at the constant angular velocity.
dV dV d θ
Acceleration of slider, a= = ×
dt d θ dt
⎡ r ⎤
or a = r ω2 ⎢cos θ + cos 2 θ⎥
⎣ l ⎦
5.2.2 General Method for Velocity and Acceleration
We start with plane curvilinear motion using polar coordinates to denote the position
vector ‘r’ and its angular coordinate ‘θ’ measured from a fixed reference axis as shown in
Figure 5.2. Let er and eθ unit vectors in radial and transverse directions respectively.
These unit vectors have positive directions in increasing r and θ. The vectors er and eθ are
parallel to the positive senses of Vr and Vθ. Both these unit vectors will swing through the
angle in the time dt. As shown in figure, their tips move through the distances | d er | = 1
× dθ = dθ in the direction of eθ, and through d eθ = 1 × d θ = d θ in the direction opposite
to that of er. Therefore,
d er d er d θ dθ
= × = θ eθ , where →θ
dt dθ dt dt
d eθ d eθ d θ
and = × = θ ( − er ) = − θ er
dt dθ dt
V
Vθ Vr
P
deθ
der
eθ
er Y
r
dθ dθ
er
eθ
Z
θ
θ
X
Figure 5.2
The velocity of a particle P which has position vector ‘r’ is given by
dr d dr d er
V = = (r er ) = er + r
dt dt dt dt
= r er + r θ eθ = Vt er + Vθ eθ 23

4. Motion Analysis of where Vr and Vθ are the components of velocity in radial direction and transverse
Planar Mechanism and direction. The acceleration of particle P is given by
Synthesis
dV d
a= = (r er + r θ eθ )
dt dt
d d
= ( r er ) + (r θ eθ )
dt dt
dr d er dr d
= er + r + (θ eθ ) + r (θ eθ )
dt dt dt dt
⎛ de ⎞
= r er + r (θ eθ ) + r θ eθ + r ⎜ θ eθ + θ θ ⎟
⎝ dt ⎠
= r er + 2 r θ eθ + r θ eθ − r θ2 er
= ( r − r θ2 ) er + ( r θ + 2 r θ) eθ
= ar er + aθ eθ
where ar represents radial component and aθ represents transverse component of
acceleration.
ar = r − r θ2 and aα = r θ + 2 r θ
Here, r → Acceleration of slider along slotted link,
r θ2 → Centripetal acceleration,
r θ → Tangential component of acceleration, and
2r θ → Coriolis’ component of acceleration.
SAQ 1
Explain Corioli’s component of acceleration and what is its magnitude?
Example 5.1
The length of the crank of quick return crank and slotted lever mechanism is
15 cm and it rotates at 10 rod/s in counter clockwise sense. For the configuration
shown in Figure 5.3 determine angular acceleration of link BD.
D
A A
15 13
0
ω 60
B 0 θ
60 B
ω = 10 rad/sec 32.5 cm E
O
(a) (b)
0
42 D
0
48 A
eθ
er
θ′′
θ′
ω 0
B 60
18
0 C
O
(c)
Figure 5.3
24

5. Solution Analysis by
Analytical Methods
AE 13 13
In Figure 5.3(b), tan θ = = = or θ = 18o
BE 32.5 + 7.5 40
13 AB
Also, = = AB or AB = 42.06 cm
sin θ sin 90o
The motion of point A is common to both OA and BD. The velocity and
acceleration of A are given by
15
VA = OA × ω = × 10 = 1.5 m/s
100
15
a A = OA × ω2 = × 10 = 15 m/s 2
100
The components of VA along BD are perpendicular to it are given by
Vr = r = − 1.5 cos 48o = − 1.0 m/s
1.5 sin 48o
Vθ = r θ = 1.5 sin 48o ∴ θ =
r
1.5 sin 48o
or θ= = 2.65 rad/s
42.06
100
The transverse component of acceleration of A is give by
aθ = − a A sin 42o = − 15 sin 42o = − 10.037 m/s 2
Also aθ = (r θ) + 2 (r θ)
42.06
or − 10.037 = θ + 2 (− 1.0) (2.65)
100
5.30 − 10.037
or θ= = − 11.26 rad/s 2
0.4206
Therefore, angular acceleration of BD is 11.26 rad/s2 in clockwise sense.
5.3 MOTION REFERRED TO MOVING FRAMES OF
REFERENCE
The moving frame may, in general, translate and rotate as well as accelerate linearly or
angularly. The purpose of the following treatment is to arrive at a systematic procedure
for refering motion of a point with respect to a frame of reference if its motion is known
in relation to another moving with respect to the former.
5.3.1 Relative Motion of Two Points
Consier two points P1 and P2 moving with velocities V1 and V2 and accelerations a1
and a2. The velocity of P1 with respect to P2 is given and acceleration of P1 with respect
to P2 is given by
a12 = a1 − a2
V12 = V1 − V2
Z 25
a1
B B

6. Motion Analysis of
Planar Mechanism and
Synthesis
Figure 5.4
Also V21 = V2 − V2 = − V12
and a21 = a2 − a1 = − a12
5.3.2 Plane Motion of a Link
A motion is said to be a plane motion if all the points in the body stay in the same and
parallel planes. The concept of plane motion enables us to consider only one of the
parallel planes and analyse the motion of the points lying in that plane.
The parallel planes may not appear to be identical in shape and size but there is no
difficulty because any plane of the body can be hypothetically extended by a massless
extension of the rigid body for the purpose of kinematic analysis.
The mechanisms use links. Consider a link PQ which is shown displaced in Figure 5.5 to
position P′ Q′ in a general plane motion. The motion may be considered to comprise
translation of an arbitrary point on the link plus a rotation about an axis perpendicular to
the plane and passing through that point. In an example in Figure 5.5 eight combinations
are equivalent.
P P′
θ
Q′
Q Q1
Figure 5.5(a) : Translation from PQ to P′Q1 and Rotation about P′
P P1
P′
θ
Q Q′
Figure 5.5(b) : Translation from PQ to P1 Q′ and Rotation about Q′
P1
P
θ P′
26
C C′

7. Analysis by
Analytical Methods
Figure 5.5(c) : Translation from PQ to P1 Q1 and Rotation about C′
P
θ P′
O O′
θ
Q′
Q Q1
Figure 5.5(d) : Translation from PQ to P1 Q1 and Rotation about R′
P
P′
θ
Q1 Q′
Q
Figure 5.5(e) : Rotation from PQ to PQ1 and Translation to P′ Q′
P P1 P′
θ
Q′
Q
Figure 5.5(f) : Rotation from PQ to P1Q and Translation to P′ Q′
P P1 P′
θ
C C′
Q′
Q1 Q
Figure 5.5(g) : Rotation from PQ to Pi Q1 and Translation to about P′ Q′
P′
P1
P
θ
27
O′
O

8. Motion Analysis of
Planar Mechanism and
Synthesis
Figure 5.5(h) : Rotation from PQ to P1 Q1 and Translation to P′ Q′
The translation and rotation are commutative. In Figures 5.5(e) and (f) show equal
amount of translation and rotation but in Figures 5.5(a) and (b) translation is done first
and rotation later. Similarly, Figures 5.5(a) and (e), (c) and (g), (d) and (h) are
commutative.
It may also be noted that, whatever be the mode of combination, the amount of rotation is
same. The angular velocity of every point on the link is the same. It is, therefore, the
customary to use the term angular velocity of the link rather than of any particular point
on it.
The fact that a general plane motion can be thought of as a superpositon of translation
and rotation is a special case of Chasle’s theorem. The theorem, in general, states that any
general motion of a rigid body can be considered as an appropriate superposition of a
translational motion and a rotational motion.
SAQ 2
Determine velocity of top point of a rolling wheel if centre of the wheel is moving
with velocity ‘v’.
5.4 METHOD OF RELATIVE VELOCITY AND
ACCELERATION
For this purpose, a link of a general shape may be considered to start with. It is shown in
Figure 5.6. Let there by any arbitrary two points A and B moving with velocities VA
and VB. The velocity of B relative to A may be represented by VBA. Let relative velocity
B
VBA makes an angle θ with line joining A and B. VBA can be resolved into two components
VBA cos θ along AB and VBA sin θ perpendicular to AB. Since the link is rigid, distance AB
remains constant. Therefore, component of velocity of B relative to A along AB cannot
exist.
Therefore, VBA cos θ = 0 or cos θ = 0
π
or, θ=
2
This means a rigid link has rotation relative to point A, as well as translation along
velocity of A.
VBA
VB
B
aB
28
aB
B B
o′
b′
a′

9. Analysis by
Analytical Methods
Figure 5.6
Therefore, the direction of VBA is perpendicular to the line joining A and B. If VA is known
in magnitude and direction and for VB only direction is known, the magnitude can be
B
determined by drawing a polygon as stated below :
(a) First the velocity of A, i.e. VA is plotted by assuming a suitable scale from an
arbitrary selected point o which represents fixed reference. Let it be
represented by ‘oa’.
(b) Now, draw a line through o parallel to the direction of velocity of B, i.e. VB. B
(c) Next, draw a line through point ‘a’ which is parallel to the perpendicular to
the line AB which meets the direction of VB at point ‘b’.
B
In this velocity polygon oab, ‘ob’ represents velocity VB in magnitude and direction.
B
After drawing velocity polygon, we can proceed for determination of magnitude of
acceleration of point B. If direction of acceleration of B is known and acceleration of
point A is known in magnitude and direction. The vector equation can be written as
follows :
a B = a A + a BA
Since a BA = a BA + a BA ∴ a B = a A + a BA + a BA
t c c t
t c
Where aBA is acceleration of B relative to A; aBA is tangential component of aBA and aBA
is centripetal component of aBA.
2
VBA
c
aBA =
AB
t
It is directed from B to A along AB whereas aBA has direction perpendicular to AB.
Acceleration polygon can be drawn as follows :
(a) Plot acceleration of A in magnitude and direction by assuming suitable
scale. aA is represented by o′a′.
c
(b) Plot centripetal component a BA by drawing line parallel to AB and represent
c
its magnitude according to the vector equation. a BA is represented by a′a′1.
(c) ′
From a1 draw a line perpendicular to aa1 or parallel to the perpendicular to
t
AB to represent direction of aBA .
(d) Draw a line parallel to the direction of acceleration of B, i.e. aB from fixed
B
t
reference o′ to meet the line representing direction of a BA . They meet at
b′ . o′b′ represents acceleration of B in direction and magnitude.
Example 5.2
29

10. Motion Analysis of In a slider crank mechanism shown in Figure 5.7, the crank OA rotates at 600 rpm.
Planar Mechanism and Determine acceleration of slider B when crank is at 45o. The lengths of crank OA
Synthesis
and connecting rod AB are 7.5 cm and 30 cm respectively.
a
A
aB r
b′ o ω
o′ b
B
0
O 45
aA
a′1
a′
Figure 5.7
Solution
2π × 600
The angular velocity of crank OA, ω = = 6.28 rad/s . The crank OA is a
60
rotating body about fixed centre O. Therefore, the velocity of point A is given by
V A = ω × OA = 6.28 × 7.5 = 47.1 cm/s or 0.471 m/s
Select suitable position of pole o which represents fixed reference. Draw a line oa
perpendicular to OA to represented velocity of A, i.e. VA in magnitude and
direction. From point a, draw a line perpendicular to AB to represent direction of
relative velocity VBA. Now draw another line parallel to the motion of the slider B
from O to represent direction of the velocity of slider VB to meet another line
B
through a at b. Thus ob represents velocity of B in magnitude and direction.
Since crank OA rotates at uniform angular speed, therefore, acceleration of A will
be centripetal acceleration.
2
VA (0.471)2
aA = ac =
A = = 2.96 m/s 2
OA 7.5
100
It is directed from A towards O.
a B = a A + a BA = a A + a BA + a BA
c t
2
VBA
aBC =
c
along AB directed from B to A
AB
(3.425) 2
a BC =
c
= 39.17 m/s
0.3
Select a suitable position of pole O which represents fixed reference. Plot
acceleration of A, i.e. aA by drawing parallel to OA and representing its magnitude
by a suitable scale. This is represented by o′a′. Now, draw a line from ‘a′’ parallel
c
to AB for aBA and represent its magnitude. This is represented by a′a′1. Now from
t
a′1 draw a line perpendicular to a′a′1 or AB to represent direction of aBA . Draw a
line from o′ parallel to the motion of slider B to represent direction of motion of B.
This line from o′ will meet another line from a′1 at b′. The acceleration of slider B
is represented by o′b′ in magnitude and direction and it gives
aB = 215 m/s 2
Example 5.3
30

11. A four bar chain O2 AB O4 is shown in Figure 5.8. The point C is on link AB. The Analysis by
crank O2A rotates in clockwise sense with 100 rad/s and angular acceleration Analytical Methods
4400 rad/s2. The dimensions are shown in Figure 5.8(a). Determine acceleration of
point C and angular acceleration of link O4B.
O′2, O′4
A 3 B
28 mm C
80 mm 37 mm
4 i′,O′
O4
α = 4400 rad/sec
2
75 mm 2
ω = 100 rad/sec 1
0
53
O2 125 mm a′
1 b′ a′
(a) c′
b
b′
O2, O4 c
a b′
(b) (c)
Figure 5.8
Solution
Draw configuration diagram to the scale.
75
The velocity of point A, VA = O2 A × ω = × 100 = 7.5 m/s
1000
Velocity Diagram
For drawing velocity polygon the following steps may be followed :
(a) Assume suitable scale say 1 cm → 5 m/s.
(b) Plot velocity of A, VA perpendicular to O2A. It is represented by
o2a in velocity polygon Figure 5.8(b).
(c) Draw a line from point a perpendicular to AB to represent
direction of VBA.
(d) Draw a line from o2 perpendicular to O4B to represent direction
of VB to meet line representing direction of VBA at point b. o2b
B
represents VB in magnitude and direction.
B
(e) For determining velocity of C, plot point c on ab such that
AC
‘ac’ = × ab .
AB
From velocity polygon VBA = 6.5 m/s
and VB = 9 m/s
Acceleration Diagram
For drawing acceleration polygon, the following steps may be followed :
(a) Assume suitable scale depending on value of centripetal
acceleration of A which is a c = O2 A ω2 . Or
A
31

12. Motion Analysis of 75
Planar Mechanism and ac =
A × (100)2 = 750 m/s 2 . A scale 1 cm → 250 m/s2 may
Synthesis 1000
serve the purpose.
(b) Draw a line parallel to O2A and plot a c which will be
A
represented by 3 cm. It is represented by o2′ a1′ in polygon.
(c) Draw a line perpendicular to O2A from a′1 in the sense of
angular acceleration to represent tangential acceleration which
75
is atA = O2 A × α = × 4400 = 300 m/s 2 . It will be denoted
1000
by a′1 a′ in polygon. o′2 a′ can be joined to get acceleration of
A which is represented by o′2 a′ in magnitude and direction.
(d) Draw a line parallel to AB from a′ and plot centripetal
c
component of acceleration aBA which is given by
2
VBA (6.5)2
aBA =
c
= = 528.15 m/s 2 . Plot magnitude of aBA . It
c
AB 0.08
is represented by a′ b′1.
(e) Draw a line perpendicular to AB from b′1 to represent direction
t
of tangential component of acceleration aBA .
(f) Draw a line parallel to O4B from o′2 to represent centripetal
component of acceleration of B. The magnitude is given by
2
VB 92
aB =
c
= = 2189.189 m/s 2 . It will be denoted by
O4 B 0.037
8.76 cm and it is represented by o′2 b′2.
(g) Draw a line perpendicular to O4B or o′2 b′2 to represent
direction of tangential component of acceleration of B to meet
t
the line representing direction aBA at b′. Join a′b′ which
represents aBA.
Join o′2 to b′ to get acceleration of B, i.e. aB. B
(h) To determine acceleration of C, plot a point c′ on line a′b′ such
AC 28
that a′ c′ = × a ′ b′ = × 72 = 2.52 . Joint o′2 with c′ and
AB 80
o′2 c′ represent acceleration of C in magnitude and direction.
From acceleration polygon Figure 5.8(c), ac = 1400 m/s2.
(i) Angular acceleration of link O4B is given by
t
aB
αO4 B = = 3479.3 rad/s 2
O4 B
The acceleration polygon is shown in Figure 5.8(c).
Example 5.4
Determine acceleration of slider D in a combined four bar chain and slider crank
mechanism shown in Figure 5.9(a). The dimensions are shown in the figure. The
crank OA rotates at 240 rpm in counterclockwise sense.
d′1
c VD
aDB
d O,C
b′ t
aBA t
b′1 aDB
32 VDB
VB VA
aB
B B
t c
aBC aBA
a′
aA
B B
aDB
B B

13. Analysis by
Analytical Methods
(a) (b)
28 mm O
0
A 75
2
44 mm 3 65 mm
B
49 mm
C
D 11
(c)
Figure 5.9
Solution
Plot configuration diagram to the suitable scale.
Velocity Diagram
2π × 240
Velocity of A, VA = O2 A × = 0.7037 m/s
60
It is perpendicular to OA in sense of rotation.
(a) Plot velocity of A, VA by assuming a suitable scale say
1 cm → 0.2 m/s. It is perpendicular to OA and represented by
‘oa’ on velocity polygon.
(b) From a, draw a line perpendicular to AB to represent direction
of VBA.
(c) From o, draw a line perpendicular to BC to represent direction
of velocity of B. Extend it if necessary to meet the line
representing direction of relative velocity VBA. These two lines
join at b.
(d) From b, draw a line perpendicular to BD to represent direction
of VDB.
(e) From o, draw a line parallel to line of motion of slider D to
meet line representing direction of VDB at d.
(f) In velocity polygon od represents velocity of slider D
From velocity polygon VB = 0.5 m/s, VBA = 0.4 m/s,
B
VDB = 0.602 m/s
Velocity polygon is shown in Figure 5.9(b).
Acceleration Diagram
33

14. Motion Analysis of ⎛ 2π × 240 ⎞
Planar Mechanism and Acceleration of A, a A = a c = OA × ⎜
A ⎟ = 17.686 m/s Select a
2
Synthesis ⎝ 60 ⎠
suitable scale, say 1 cm → 5 m/s2.
(a) Draw a line parallel to OA to represent acceleration of A, aA. It
is represented by o′a′.
c
(b) Plot centripetal component of acceleration aBA, i.e. aBA by
drawing line parallel to AB.
2
VB
aBA =
c
= 3.636 m/s 2
AB
It is represented by a′ b′1.
(c) From b′1, draw a line perpendicular to AB to represent direction
of tangential component of acceleration aBA.
(d) From o′ draw a line o′ b′2 parallel to BC to represent centripetal
component of acceleration of B which has magnitude given by
2
VB
aB =
c
= 5.102 m/s 2
BC
(e) From b′2, draw a line perpendicular to BC or o′ b′2 to represent
t
direction of tangential component of acceleration of B, i.e. aB .
t
This line meets direction of aBA at b′.
(f) From b′, draw a line b′ d′1 parallel to BD to represent
c
centripetal component of acceleration of aDB, i.e. aDB .
2
VDB
aDB =
c
= 7.878 m/s 2
BD
(g) From d′1, draw a line perpendicular to b′ d′1 or BD to represent
direction of tangential component of acceleration of aDB, i.e.
t
aDB .
(h) From o′ draw a line parallel to the line of motion of slider D to
t
meet direction of aDB at d′.
o′ d′ represents acceleration of slider in magnitude and
direction. Magnitude of acceleration of slider
aD = 31.5 m/s 2
Example 5.5
Figure 5.10(a) shows Andreau Variable Stroke engine Mechanism in which
links 2 and 7 have pure rolling motion. The dimensions of various links are
indicated in the Figure 5.10. Determine acceleration of slider D if link 2 rotates at
1800 rpm.
Solution
Draw configuration diagram to the scale.
2π × 1800
Velocity of point A, VA = OA × = 3.581 m/s .
60
The velocity of point B will also be equal to VA.
Velocity Diagram
34

15. The vector equations are as follows : Analysis by
Analytical Methods
VC = V A + VCA
and VC = VB + VCB
VD = VC + VDC
C
0
30
76 mm 4 D 5
38 mm 0
15
62 mm
6
D
3
B
7
0
30 Q Q 2 A
38 mm dia
1 Wheel
Wheel
63 mm dia
(a)
d′1
C
c
a
b
a′
o′,q′
c′
b′
c′1
o,q
d d′
(b) (c)
Figure 5.10
Velocity polygon is shown in Figure 5.10(a).
(a) Assume suitable scale say 1 cm → 1 m/s.
(b) Assume suitable position of pole o, and plot VA and VB B
perpendicular to OA and QB respectively. They are represented
by oa and ob.
(c) Draw lines to represent directions of VCA and VCB
perpendicular to AC and BC from points a and b respectively to
meet at point c.
(d) Draw line to represent direction of VDC perpendicular to DC at
point c.
(e) Draw line parallel to motion of slider from o to represent
direction of motion of slider to meet direction of VDC at d.
VCA = 1.28 m/s
VCB = 1.3 m/s
VDC = 5.2 m/s
Acceleration Diagram
35

16. Motion Analysis of Vector equations are as follows :
Planar Mechanism and
Synthesis aC = a A + aCA + aCA
c t
Also aC = a B + aCB + aCB
c t
a D = aC + a DC + a DC
c t
Both the wheels are rotating at uniform angular speed
2
VA
∴ a A = ac =
A = 674.9 m/s 2
OA
2
VB (3.581)2
∴ aB = a B =
c
= = 407.1 m/s
OB 0.0315
2
VCA 1.282
aCA =
c
= = 26.425 m/s 2
AC 0.062
2
VCB 1.32
aCB =
c
= = 22.236 m/s 2
BC 0.076
Assuming scale for acceleration as 1 cm → 200 m/s2.
(a) Assuming suitable position for o′, plot aA and aB parallel to OA
B
and BQ respectively. They are represented in acceleration
polygon by o′ a′ and o′ b′ respectively.
c c
(b) From a′ and b′ plot aCA and aCB parallel to AC and BC. They
are represented by a c″2 and b′ c′1 respectively.
(c) Draw lines perpendicular to AC from c″2 and perpendicular to
t t
BC from c′1 to represent directions of aCA and aCB . These
lines meet at c′.
c
(d) Plot aDC by drawing line parallel to DC from c′. This is
represented by c′ d′1.
(e) Now draw a line from d′1 perpendicular to DC to represent
t
direction of aDC .
(f) From o′, draw a line parallel to motion of slider D to meet
t
direction of aDC at d′.
o′ d′ represents acceleration of slider D in magnitude and
direction.
Acceleration of slider = 1320 m/s2.
5.5 ALTERNATIVE METHOD OF DETERMINING
CORIOLIS’ COMPONENT OF ACCELERATION
In rigid body rotation, distance of a point from the axis of rotation remains fixed. If a link
rotates about a fixed centre and ,at the same time, a point moves over it along the link,
the absolute acceleration of the point is given by the vector sum of
(a) the absolute acceleration of the coincident point relative to which the point,
under consideration, is moving;
(b) acceleration of the point relative to the coincident point; and
36

17. (c) Coriolis’ component of acceleration. Analysis by
Analytical Methods
Figure 5.11 shows a link 2 rotating at constant angular speed say ω2, it moves from
position OC to OC1. During this interval of time, the slider link 3 moves outwards from
position B to B2 with constant velocity VBA (A is point on link 2 shich coincides with B
B
which is on link 3). The motion of the slider 3 from B to B2 may be considered in the
following three stages :
(a) B to A1 due to rotation of link 2,
(b) A1 to B1 due to outward velocity VBA,
B
(c) B1 to B2 due to acceleration perpendicular to the link which is the Coriolis’
B B
component of acceleration.
Arc B1 B2 = Arc DB2 − Arc DB1
= Arc DB2 − Arc AA1
Also Arc B1 B2 = A1 B1 δ θ = VBA δ t ω2 δ t
= VBA ω2 (δ t ) 2
The tangential component of the velocity perpendicular to the link is say Vt
and it is given by
Vt = r ω
In this case ω has been assumed constant and slider moves with constant velocity.
Therefore, tangential velocity of point B on the slider 3 will result in uniform increase in
tangential velocity because of uniform increase in value of r in the above equation.
C
C1
2
B1
D
2ω2,VBA
B2
VBA
B on link 3 δθ ω2
VBA ω2
3
VBA
A1 2ω2,VBA
A on link 2 (b) (c)
δθ
VBA 2ω2,VBA
VBA ω2
ω2
VBA
ω2
2ω2,VBA VBA
O
(d) (e)
1
(a)
Figure 5.11
To result in uniform increase in value of Vt, there has to exist constant acceleration
perpendicular to link 2.
1
Therefore, B1 B2 = a (δ t ) 2 (From Unit 1)
2
where a is the acceleration. 37

18. Motion Analysis of 1
Planar Mechanism and ∴ a (δ t )2 = VBA ω2 (δ t )2
Synthesis 2
a = 2 VBA ω2
cor
This is Coriolis’ component of acceleration and will be denoted by aBA .
Therefore, aBA = 2 ω2 VBA
cor
The directional relationship of VBA and 2 ω2 VBA is shown in Figures 5.11(b), (c), (d) and
(e). If slider moves towards centre of rotation o, its velocity can be transmitted to other
side so that it is directed outwards.
The direction of Coriolis’ component of acceleration is given by the direction of the
relative velocity vector for the two coincident points rotated by 90o in the direction of the
angular velocity of the rotation of the link.
If the angular velocity ω2 and the velocity VBA are varying, they will not affect the
expression of the Coriolis’ component of acceleration but their instant values will be used
in determining the magnitude and this value will be applicable at that instant.
SAQ 3
What are the necessary and sufficient conditions for the Coriolis’ component of
acceleration to exist?
Example 5.6
A cam and follower mechanism is shown in Figure 5.12(a), the dotted line shows
the path of point B (on the follower). The cam rotates at 100 rad/s. Draw the
velocity and acceleration diagram for the mechanism and determine the linear
acceleration of the follower. Minimum radius of cam = 30 mm and maximum
lift = 35 mm.
t b′
aBA
b′1
4 aB
2ω,VBA
A on cam. path
B on folower o′
c
a AO
3 VBA
45 mm
a′
O 30
o
(b)
2 VBA
2ω2 b
1
VBA VB
VA
a o
(a) (c)
Figure 5.12
Solution
38

19. Linear velocity of point A which is on the cam, VA = ω × OA. Analysis by
Analytical Methods
45
or, VA = 100 × = 4.5 m/s
1000
VB = V A + VBA
Velocity Diagram
(a) Assume suitable scale say 1 cm → 1 m/s.
(b) Plot velocity of A by drawing a line perpendicular to OA from pole o.
It is represented by oa.
(c) From o, draw a line parallel to motion of follower to represent
direction of velocity of follower.
(d) Draw a line from ‘a’, parallel to the motion of B which is parallel to
the tangent at cam profile to meet line representing direction of
velocity of follower at b. ob represents velocity of follower in
magnitude and direction
VB = 1.75 m/s and VBA = 4.85 m/s
Velocity polygon is shown in Figure 5.12(b).
Acceleration Diagram
aB = a A + aBA + a cor = a A + aBA + aBA + a cor
c t
(4.5) 2
2
VA
aA = = 45 = 450 m/s 2
OA 1000
2
VBA
aBA =
c
=0
∞
a cor = 2 VBA × ω = 2 × 4.85 × 100 = 970 m/s 2
(a) Select suitable scale say 1 cm → 200 m/s.
(b) Plot aA by drawing a line parallel to OA from o′ and length equal to
2.25 cm.
(c) Now plot acor perpendicular to profile of cam and length equal to
4.85 cm. It is represented by a′ b′1.
t
(d) Now draw a line perpendicular to a′ b′1 to represent direction of aBA .
(e) Draw another line from o′ to represent direction of motion of B to
t
meet direction of aBA at b′.
o′ b′ represents acceleration of B in magnitude and direction.
aB = 3 × 200 = 600 m/s 2
The acceleration polygon is shown in Figure 5.12(c).
Example 5.7
A quick return motion mechanism is shown in Figure 5.13(a). The crank rotates at
20 rad/s. Determine angular acceleration of the slotted link 3.
Solution
The configuration diagram is drawn to the scale according to the given dimensions. 39

20. Motion Analysis of Let point B (on the link 3) coincides point A (on the crank 2). Velocity of point A
Planar Mechanism and is perpendicular to OA and
Synthesis
15
VA = OA × ω = × 20 = 3 m/s
100
V A = VB + V AB
O
1 15 cm
2
4 b
A On Link B On Link 2
2 and 4
35 cm O,C
3
25 cm (b)
a
b′′ C
a′
VAB
cr
a AB
aBA aA (c)
b′
aB
t o′c′
aBC
c
aBC
(a)
b′1
Figure 5.13
Velocity Diagram
(a) Assume suitable scale say 1 cm → 1 m/s.
(b) Plot velocity VA by drawing line from o perpendicular to OA. It is
represented by oa.
(c) Draw a line from o perpendicular to slotted link 3 to represent
direction of velocity VB. B
(d) From a, draw a line parallel to slotted link 3 to represent VAB which
represents movement of slider. This line meets direction of VB at b. B
Here ob represents velocity of B in magnitude and direction and ba
represents velocity of slider VAB in magnitude and direction.
VB = 1.5 m/s ; V AB = 2.6 m/s
The velocity polygon is shown in Figure 5.13(b).
Acceleration Diagram
a A = a B + a AB + a cor = a B + a B + a tAB + a c + a cor
c t
AB
40

21. Analysis by
V2 32
aA = ac
A = A = = 60 m/s 2 Analytical Methods
OA 0.15
2
VB 1.52 V2
aB =
c
= = 9 m/s 2 ; a c B = AB = 0
A
BC 0.25 ∞
VB 1.5
a cor = 2 VAB × = 2 × 2.6 × = 31.2 m/s 2
BC 0.25
(a) Assume suitable scale say 1 cm → 10 m/s2.
(b) Plot aA by drawing line parallel to OA from a suitable point o′. It is
represented by o′ a′.
(c) Plot acor (Coriolis, component of acceleration) as per direction
determined in Figure 5.13(c) and according to the vector equation. It
is represented by b″ a′. It is perpendicular to link 3.
(d) From b″, draw a line perpendicular to b″ a′, i.e. parallel to link 3 to
represent direction of motion of slider ( a tAB ) .
c
(e) Starting from o′, plot aB by drawing line parallel to link 3. It is
represented by o′ b′1.
(f) From b′1, draw a line perpendicular to o′ b′1, i.e. perpendicular to
t
link 3 to represent direction of aB and meet line representing
direction of acceleration a tAB at b′.
Here o′ b′ represents acceleration of B in magnitude and direction and
b′ b″ represents acceleration of slider in magnitude and direction.
Vector represented by b′1 b′ represents tangential component of
acceleration of B
aB = 83.5 m/s 2
t
∴ BC × α BC = 83.5
83.5
or Angular acceleration of link 3 (α BC ) =
0.25
or α BC = 334 rad/s 2
Example 5.8
Figure 5.14(a) shows Whitworth Quick Return Mechanism. The dimensions are
written in the Figure 5.14. Determine velocity and acceleration of slider D when
crank rotates at 120 rpm uniformly in the sense indicated in Figure 5.14.
Solution
2π N 2 π × 120
The velocity of A, VA = OA × = 0.2 × = 2.51 m/s .
60 60
Draw configuration diagram to the scale. The point B is on the slotted link which
coincides with point A on link OA.
Velocity Diagram
V A = VB + V AB
VB
VC = × QC
QB
and VD = VC + VDC 41

22. Motion Analysis of
Planar Mechanism and c
Synthesis C
50 cm
15 cm D VO
Q o,q d
10 cm 26 cm
VB
Q
20 cm
B
A
b
VAB
VQB= ba
a
(a) (b)
b′′
b′
b′1
o′,q′ d′
a AB = 2ωaB VAB
r
ωQB
d′1 c′
(c) (d)
Figure 5.14
(a) Plot VA by selecting proper scale say 1 cm → 0.5 m/s by a line
perpendicular to OA. It is represented by oa.
(b) From o, draw a line perpendicular to OB for direction of VB. B
(c) From a, draw a line parallel to QB to represent direction of VAB, i.e.
sliding velocity of slider to meet direction of velocity of B, i.e. VB at b. B
The vector ob represents VB in magnitude and direction.
B
2.4
VB = 2.4 m/s ∴ VC = × 0.15 = 1.38 m/s
0.26
(d) Plot VC perpendicular to QC from o. It is represented by oc.
(e) Draw a line parallel to motion of slider from o.
(f) Draw a line perpendicular to CD from c to meet direction of velocity
of slider at d.
The velocity of slider in magnitude and direction is given by od.
2.512
VD = 0.75 m/s, VAB = = 31.5 m/s 2
0.20
Acceleration Diagram
2
VA 2.512
Acceleration of A, a A = = = 31.5 m/s2
OA 0.20
42

23. Analysis by
a A = a B + a AB + a cor = a B + a B + a c + a tAB + a cor
c t
AB Analytical Methods
2
VB 2.42 V
aB =
c
= = 22.15 m/s 2 ; a c B = AB = 0
A
QB 0.26 ∞
VB 2.4
a cor = 2 VAB × = 2 × 0.9 × = 16.61 m/s 2
QB 0.26
Acceleration of B can be determined as explained in Example 5.7. Assume
suitable scale say 1 cm → 5 m/s2. Extend o′ b′ to the other side of o′ (C is on
other side of B) upto c′ such that
QC
o′ c′ = × o′ b′
QB
a D = aC + a DC = aC + a DC + a DC
c t
2
VDC 0.82
aDC =
c
= = 1.28 m/s 2
CD 0.5
c
(a) Draw a line parallel to CD and plot aDC . It is represented
by c′ d′1.
(b) Draw a line perpendicular to c′ d′1, i.e. perpendicular to CD to
t
represent direction of aDC .
(c) Draw a line from o′ parallel to motion of slider D to meet
t
direction of aDC at d′.
The acceleration of slider is represented by o′ d′ in magnitude
and direction
aD = 6.5 m/s 2
Example 5.9
A swivelling point mechanism is shown in Figure 5.15. If the crank OA rotates at
200 rpm, determine the acceleration of sliding of link DE in the trunion. The
following dimensions are known AB = 18 cm, DE = 10 cm, EF = 10 cm, AD = DB,
BC = 6 cm, OC = 15 cm, OA = 2.5 cm.
Solution
The configuration diagram is plotted to the scale say 1 cm → 4 cm.
Velocity Diagram
VB = V A + VBA ; VS = VD + VSD and VF = VE + VFE
The point S is on the link DE which slides in the swivel block.
2π N 2.5 2 π × 200
VA = × OA = × = 0.52 m/s
60 100 60
Velocity of S related to D, VSD is perpendicular to SD. Velocity of S related
to fixed block is parallel to link DE. Select a suitable scale say
1 cm → 0.2 m/s.
(a) Plot VA. It is represented by oa in velocity polygon
(Figure 5.15(b)).
(b) Draw a line perpendicular to BC to represent direction of VB B
from o.
(c) Draw a line perpendicular to AB from a to represent direction
of VBA to meet direction of VB at b.
B
43

24. Motion Analysis of AD
Planar Mechanism and (d) Plot point d on ab such that ad = ab .
Synthesis AB
DE
(e) Plot de such that de = ds .
DS
(f) From e, draw ef perpendicular EF to represent direction of VFE.
(g) Draw a line parallel to motion of slider to meet direction of VFE
at f.
VSD = VS = 0.32 m/s, VDE = 0.84
E F
S
Q B
12 cm
D
7.5 cm
A
8.5 cm
0
45 O C
(a)
s′′
O′,C′
a
e
s
b′
d b′′
d′
b s′
f
O, C a′
b′
(b) (c)
S′1
Figure 5.15
Acceleration Diagram
2
VA (0.52) 2
Acceleration of A, a A = = = 10.816 m/s 2
OA 2.5
100
a B = a A + a BA + a BA
c t
Also, aB = aB + aB
c t
2
VSD (0.32)2 V2 (0.44) 2
aSD =
C
= ; aBA = BA =
C
= 1.075 m/s 2
SD 0.04 AB 18
100
(0.48) 2
a cor = 2 ωDE × VS ; aB =
C
= 3.84 m/s 2
0.1
VDE (0.84) 2
=2 × VS = 2 × × 0.32
DE 0.1
aS = aSD + a cor = aSD + a SD + a cor
C t
Select suitable scale say 1 cm → 2 m/s2.
44