This chapter discusses acceleration analysis of linkages using vector mathematics, equations of relative motion, and complex numbers. It presents methods to calculate the acceleration of points on moving links and objects sliding on rotating links. Key equations are derived and terms like Coriolis acceleration are defined. Examples apply the analysis to specific linkages and sliding blocks to solve for acceleration values.

1. Mechanisms of Machinery
(MEng 3301)
Chapter 4
Acceleration Analysis of Linkages
1

2. 2
4. Acceleration Analysis of
Linkages
4.1. Acceleration Analysis by Vector
Mathematics
• The acceleration of a point P moving in the x-y-z
system relative to the X-Y-Z system, is obtained by
differentiating the velocity equation
• Differentiating equation (1) yields
the acceleration Equation is
)2(RRVVV op


×+×++= ωω
)1(RVVV op

×++= ω

3. 3
• Each term on the right hand side of the acceleration
equation is evaluated as follows:
• Substituting for
• Note that
( )
( ) )4(
)3(




 +++++=
++=
=
kzjyixkzjyix
kzjyix
dt
d
V
aV oo

















kji

,,
( )
( ) )6(
)5()()()(
akzjyix
let
kzjyixkzjyixV








=++
×+×+×+++= ωωω
( )
)7(
)()()(
V
kzjyixkzjyix













×=
++×=×+×+×
ω
ωωωω

4. 4
∴ The acceleration component can be written as:
• The last term on the right hand side of the acceleration
equation is
• Substituting the corresponding values in the acceleration
equation
• The different acceleration components are:
is Coriolis’ component of acceleration, sense normal to ;
is acceleration of the origin of x-y-z system relative to X-Y-Z
system;
is acceleration of P relative to x-y-z system;
V

)8(VaV

×+= ω
)9()(
)(
RV
RVR


××+×=
×+×=×
ωωω
ωωω
)10()(2 RVRaaa op


××+×+×++= ωωωω
a
ao


V

V

×ω2

5. 5
: is the tangential acceleration of a point fixed
on the x-y-z system coincident with P as the
system rotates about O;
: is the normal component of acceleration
of the point coincident with P
Where:
: is angular velocity of x-y-z system related to
X-Y-Z system
: is velocity of P relative to x-y-z system; and
: is position vector of P.
R
V



ω
R


×ω
R

××ωω

6. 6
4.2. Acceleration Analysis Using Equations of Relative
Motion
4.2.1 Acceleration of points on a common link
• Consider a link AB rotating with an angular velocity ω and
angular acceleration α as shown.
• The relative acceleration equation is:
Where acceleration of point A
acceleration of point B
Where the acceleration term
has two components:
along the link from A to B
in the direction perpendicular to AB
)11(/ BABA aaa

+=
=
=
B
A
a
a


BAa /

( )
( ) ABtBA
ABnBA
Ra
Ra


×=
××=
α
ωω
/
/

7. 7
4.2.2 Acceleration analysis of a block sliding on a
rotating link
• Block A slides on the rotating link O2B as shown.
• At time t angular position of link 2 is θ and at t+dt, θ+dθ.
• The acceleration of the block is found by considering the
radial and tangential components of the change in velocity
of the block.
• The rotating coordinate system r - θ is attached to link O2B.
• The radial component of the velocity is V at time t, and

8. 8
• The components of dV in the radial and tangential
directions are:
(dV)r in the radial direction,
Vd θ in the tangential (transverse) direction.
• Similarly the change in the transverse component of the
velocity in time dt is dVθ as shown.
• The components of dVθ are:
-ωrdθ in the radial direction, and
ωdr + rdω in the transverse direction, neglecting
higher order differentials
• Thus the total change of velocity is
dV - ωrdθ in the radial direction, and
Vdθ + ωdr + rdω in the tangential direction.
∴ the radial component of acceleration of block A is
( )
)12(
1
2
ω
θω
ra
rddV
dt
ar
−=
−=

9. 9
or, in vector notation,
The tangential component of the acceleration is
Vectorially, the tangential acceleration is written as
∴ the acceleration of the block sliding on the rotating link is
given by:
is the sliding velocity of the block along link O2B,
and
is the sliding acceleration of the block along the
link O B.
( )13raar

××+= ωω
( )
)14(2
1
αω
ωωθθ
rV
rddrVd
dt
a
+=
++=
( )152 rVa

×+×= αωθ
( ) ( )162 aVrraA

+×+×+××= ωαωω
a
Vwhere



10. 10
• If a point A2 fixed on link O2B is coincident with A for the
position shown, the acceleration equation can be written as:
• The term is the relative acceleration b/n two
moving points,
• If the link were a curved link:
• In general the relative velocity equation is
)17(2/2 AAAA aaa

+=
2/ AAa

)18(22/ aVa AA

+×= ω
)19(22/ tnAA aaVa

++×= ω
( ) ( )
( )
onacceleratiofcomponentcoriolis’theis2,
2
2
Vand
ra
rawhere
tA
nA



×
×=
××=
ω
α
ωω
( ) ( ) ( ) ( ) )20(222 tntAnAA aaVaaa

++×++= ω

11. 11
4.3. Acceleration Analysis by Complex
Numbers
• Consider the mechanism shown below.
• Replace the elements of the mechanism by position vector
such that their sum is zero yields the acceleration equation
on two successive differentiations with respect to time.
• Differentiating the vector sum with respect to time, and
introducing the complex notation yield the equation
)25(0RRR 421 =−+
)26(0442
44422 =−− θθθ
ωω iii
ereireir 

12. 12
• Again differentiating equation (26) yield the acceleration
equation
• Separating the real and imaginary parts of the equation (27)
yield
• Solving equations (28) simultaneously, the required
accelerations are obtained to be:
)27(02 44442 2
4444444
2
22 =−+++ θθθθθ
ωωωω iiiii
ereirerierer 
)28(0sincoscos2sinsin
0cossinsin2coscos
4
2
44444444442
2
22
4
2
44444444442
2
22
=−−−+
=−−−+
θωθωθωθθω
θωθωθωθθω
rrrrr
rrrrr


[ ] )30(
sin2
cos)cos(
)29()cos(
44
224
2
22
4
24
2
22
2
444
θ
θθθω
α
θθωω
r
r
rrr
−−
=
−−=

13. Example 1
13

14. 14

15. 15

16. 16

17. 17

18. 18

19. Example 2
19

20. 20

21. Solution
21

22. 22

23. 23

24. 24

25. 25

26. 26

27. 27

28. 28

29. 29

30. 30

31. 31

32. Example 3
32

33. 33

34. 34

35. 35

36. 36

37. 37