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1. 1
Acceleration analysis (Chapter 4)
• Objective: Compute accelerations (linear
and angular) of all components of a
mechanism

2. 2
Outline
• Definition of acceleration (4.1, 4.2)
• Acceleration analysis using relative acceleration
equations for points on same link (4.3)
– Acceleration on points on same link
– Graphical acceleration analysis
– Algebraic acceleration analysis
• General approach for acceleration analysis (4.5)
– Coriolis acceleration
– Application
– Rolling acceleration

3. 3
• Definition of acceleration (4.1, 4.2)
– Angular = α= rate of change in angular
velocity
– Linear = A = rate of change in linear velocity
(Note: a vector will be denoted by either a bold
character or using an arrow above the
character)

4. 4
Acceleration of link in pure rotation (4.3)
A
P
At
PA
An
PA
APA
ω, α
θ
Magnitude of tangential component = pα,
magnitude of normal component = p ω2
Length of link: p

5. 5
Acceleration of link, general case
A
P
At
PA
An
PA
APA
ω, α
θ
Length of link: p
AA
AA
APA
AP
AP=AA+APA
An
PA
At
PA

6. 6
Graphical acceleration analysis
2
3
4
1
At
A
An
A
B
A
At
BA
At
B
Clockwise
acceleration of
crank
Four-bar linkage example (example 4.1)

7. 7
• Problem definition: given the positions of the links, their
angular velocities and the acceleration of the input link
(link 2), find the linear accelerations of A and B and the
angular accelerations of links 2 and 3.
Solution:
– Find velocity of A
– Solve graphically equation:
– Find the angular accelerations of links 3 and 4
n
BA
t
BAA
n
B
t
B
BAAB
AAAAA
AAA


++=+
⇔+=

8. 8
Graphical solution of equation
AB=AA+ABA
2
3
4
1
At
A
An
A
B
A
At
B
AA
An
BA-An
B
At
BA
At
BA
At
B
Steps:
•Draw AA, An
BA, -At
BA
•Draw line normal to link 3 starting from
tip of –An
B
•Draw line normal to link 4 starting from origin
of AA
•Find intersection and draw At
B and At
BA.

9. 9
• Guidelines
– Start from the link for which you have most
information
– Find the accelerations of its points
– Continue with the next link, formulate and solve
equation: acceleration of one end = acceleration of
other end + acceleration difference
– We always know the normal components of the
acceleration of a point if we know the angular velocity
of the link on which it lies
– We always know the direction of the tangential
components of the acceleration

10. 10
Algebraic acceleration analysis
(4.10)
2
3
4
1
B
A
a
b c
Given: dimensions, positions, and velocities of links and angular
acceleration of crank, find angular accelerations of coupler
and rocker and linear accelerations of nodes A and B
1
R2
R3
R4
R1

11. 11
0=−−+++ n
B
t
B
t
BA
n
BA
t
A
n
A AAAAAA
0)( 4
2
3
2
2
2 44
4
33
3
22
2
=+−−+−+− αωαωαω θθθθθθ jcecejbebejaeae jjjjjj
0=−−+ 1432 RRRR
Loop equation
Differentiate twice:
This equation means:

12. 12
Solution
4
2
4
2
2
2
4
2
43
2
32
2
222
3
4
4
2
43
2
32
2
222
3
4
4
3
44
22
sinsinsincos
cos
cos
coscoscossin
sin
sin
:where
αω
αω
θωθωθωθα
θ
θ
θωθωθωθα
θ
θ
α
α
θθ
θθ
jcece
jaeae
cbaaF
bE
cD
cbaaC
bB
cA
BDAE
BFCE
BDAE
AFCD
jj
B
jj
+−=
+−=
+−−=
=
=
−++=
=
=
−
−
=
−
−
=
A
AA

13. 13
General approach for acceleration
analysis (4.5)
• Acceleration of P = Acceleration of P’ +
Acceleration of P seen from observer moving with
rod+Coriolis acceleration of P’
P, P’ (colocated points
at some instant), P on
slider, P’ on bar

14. 14
Coriolis acceleration
Whenever a point is moving on a path and the
path is rotating, there is an extra component
of the acceleration due to coupling between
the motion of the point on the path and the
rotation of the path. This component is
called Coriolis acceleration.

15. 15
Coriolis acceleration
VPslip
P
O
AP’
t
AP
coriolis
AP’
n
AP
slip
AP
AP
slip
: acceleration of P as seen by observer moving with rod

16. 16
Coriolis acceleration
• Coriolis acceleration=2⋅ω⋅Vslip
• Coriolis acceleration is normalto the radius, OP, and it
points towards the left of an observer moving with the
slider if rotation is counterclockwise. If the rotation is
clockwise it points to the right.
• To find the acceleration of a point, P, moving on a rotating
path: Consider a point, P’, that is fixed on the path and
coincides with P at a particular instant. Find the
acceleration of P’,
and add the slip acceleration of P and
the Coriolis acceleration of P.
• AP=acceleration of P’+acceleration of P seen from observer
moving with rod+Coriolis acceleration=AP’+AP
slip
+AP
Coriolis

17. 17
Application: crank-slider mechanism
B2, B3
O2 Link 3, b
θ2
Link 2, a
ω2, α2
B2 on link 2
B3 on link 3
These points
coincide at the instant
when the mechanism
is shown.
When θ2=0, a=d-b
d
θ3, ω3, α3
Unknown quantities marked in blue
.
normal to crank

18. 18
General approach for kinematic
analysis
• Represent links with vectors. Use complex
numbers. Write loop equation.
• Solve equation for position analysis
• Differentiate loop equation once and solve
it for velocity analysis
• Differentiate loop equation again and solve
it for acceleration analysis

19. 19
Position analysis
2
222
2 sincos θθ dbda −−=
)sin(sin 2
1
3 θθ
b
a−=
Make sure you consider the correct quadrant for θ3

20. 20
Velocity analysis
B2 on crank,
B3, on slider
O2
rocker
crank
VB2 ┴
crank
VB3B2
// crank
VB3 ┴
rocker
.
VB3= VB2+ VB3B2

21. 21
Velocity analysis
)cos(
1
32
2
3
θθ
ω
ω
−
=
b
a
)cos(
)sin(
23
23
2
θθ
θθ
ω
−
−
−= aa
a is the relative velocity of B3 w.r.t. B2

22. 22
Acceleration analysis
)cos(
sincos
32
33
θθ
θθ
−
+
=
CB
a
)cos(
sincos
32
22
3
θθ
θθ
α
−−
−
=
b
BC
3
2
32
2
22222 coscossinsin2 θωθωθαθω baaaB −++= 
Where:
3
2
32
2
22222 sinsincoscos2 θωθωθαθω baaaC −+−−= 

23. 23
Relation between accelerations of B2 (on
crank) and B3 (on slider)
B2, B3
AB2
AB3
Coriolis
┴ crank
AB3
slip
// crank AB3
Slip
B
Coriolis
BBB
3323
AAAA ++=
.
Crank
Rocker

24. 24
Rolling acceleration (4.7)
R
r
ω (absolute)
P
C
O
C
O
ω (absolute)
Ω
First assume that angular acceleration, α, is zero
No slip condition: VP=0
Ω

25. 25
Find accelerations of C and P
∀ ω=-Ω⋅(R-r)/r (Negative sign means that CCW
rotation around center of big circle, O, results in
CW rotation of disk around its own center)
• VC= Ω⋅(R-r) (Normal to radius OC)
• An
C=VC
2
/(R-r) (directed toward the center O)
• An
P=VC
2
/(R-r)+ VC
2
/r (also directed toward the
center O)
• Tangential components of acceleration of C and P
are zero

26. 26
Summary of results
AC, length VC
2
/(R-r)
P
C VC, length Ω⋅(R-r)
r
VP=0
R
AP, length VC
2
/(R-r)+ VC
2
/r

27. 27
Inverse curvature
∀ ω=Ω⋅ (R+r)/r
• VC=Ω⋅(R+r) (normal to OC)
• An
C=VC
2
/(R+r) (directed toward the center O)
• An
P=VC
2
/r - VC
2
/(R+r) (directed away from the
center O)
• Tangential components of acceleration are zero

28. 28
Inverse curvature: Summary of results
AC, length VC
2
/(R+r)
P
AP, length VC
2
/r -VC
2
/(R+r)
C
VC, length Ω⋅(R+r)
r
VP=0
R

29. 29
Now consider nonzero angular
acceleration, α≠0
• The results for zero angular acceleration are
still correct, but
• AC
t
=αr (normal to OC)
• AP
t
is still zero
• These results are valid for both types of
curvature