Acceleration analysis (Chapter 4)
• Objective: Compute accelerations (linear
and angular) of all components of a
mechanism
1
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
2
• Definition of acceleration (4.1, 4.2)
– Angular = = 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)
3
Acceleration of link in pure rotation (4.3)
AtPA
P
APA
Length of link: p
AnPA

, 
A
Magnitude of tangential component = p,
magnitude of normal component = p 2
4
Acceleration of link, general case
AtPA
P
APA
Length of link: p
AnPA

, 
AP
A
AA
AnPA
AtPA
APA
AA
AP=AA+APA
5
Graphical acceleration analysis
Four-bar linkage example (example 4.1)
AtBA
3
Clockwise
acceleration of
crank
A
2
B
AtA
AtB
4
1
6
• 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:



AB  AA  ABA

t n 
t
n
AB  AB  AA  ABA  ABA
– Find the angular accelerations of links 3 and 4
7
Graphical solution of equation
AB=AA+ABA
At
BA
3 B
AtB
A
2
AtA
AtB
AA
AtBA
-An
4
B
AnBA
Steps:
1
•Draw AA, AnBA, -AtBA
•Draw line normal to link 3 starting from
tip of –AnB
•Draw line normal to link 4 starting from origin
of AA
•Find intersection and draw AtB and AtBA.
8
• 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
9
Algebraic acceleration analysis
(4.10)
R3
3
A
R4
B
b
c
2 a
1
R2
4
R1
1
Given: dimensions, positions, and velocities of links and angular
acceleration of crank, find angular accelerations of coupler
10
and rocker and linear accelerations of nodes A and B
Loop equation
R2  R3  R4  R1  0
Differentiate twice:
  2 ae j 2  ae j 2 j2   2be j3  be j3 j3  (  2 ce j 4  ce j 4 j4 )  0
2
3
4
This equation means:
t  An  A t  A t  An  0
An

A
B
B
A
A
BA
BA
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Solution
CD  AF
AE  BD
CE  BF
4 
AE  BD
where :
A  c sin  4
3 
B  b sin  3
C  a 2 sin  2  a 22 cos 2  b 32 cos 3  c 42 cos 4
D  c cos 4
E  b cos 3
F  a 2 cos 2  a 22 sin  2  b 32 sin  3  c 42 sin  4
AA   22 ae j 2  ae j 2 j 2
AB   42 ce j 4  ce j 4 j 4
12
General approach for acceleration
analysis (4.5)
P, P’ (colocated points at
some instant), P on
slider, P’ on bar
• Acceleration of P = Acceleration of P’ +
Acceleration of P seen from observer moving with
rod+Coriolis acceleration of P’
13
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.
14
Coriolis acceleration
APslip: acceleration of P as seen by observer moving with rod
VPslip
P
AP’
O
AP
APcoriolis
n
AP’t
slip
AP
15
Coriolis acceleration
• Coriolis acceleration=2Vslip
• Coriolis acceleration is normal to 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’+APslip+APCoriolis
16
Application: crank-slider mechanism
Unknown quantities marked in blue
.
B2, B3
normal to crank
Link 2, a
O2
B2 on link 2
B3 on link 3
These points
coincide at the instant
when the mechanism
is shown.
When 2=0, a=d-b
2
3, 3, 3
2, 2
Link 3, b
d
17
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
18
Position analysis
a  d cos 2  b2  d 2 sin 2  2
3  sin1( sin 2 )
b
a
Make sure you consider the correct quadrant for 3
19
Velocity analysis
VB3= VB2+ VB3B2
VB3B2
B2 on crank,
B3, on slider
VB3 ┴
// crank
rocker
.
crank
O2
rocker
VB2 ┴
crank
20
Velocity analysis
a 2
1
3 
b cos( 2  3 )
a  a 2
sin(3   2 )
cos(3   2 )
a is the relative velocity of B3 w.r.t. B2
21
Acceleration analysis
a 
B cos 3  C sin  3
cos( 2   3 )
3 
C cos 2  B sin  2
 b cos( 2  3 )
Where:
B  2a2 sin  2  a2 sin  2  a22 cos 2  32b cos3
C  2a2 cos 2  a2 cos 2  a22 sin  2  32b sin 3
22
Relation between accelerations of B2 (on
crank) and B3 (on slider)
Slip
Coriolis
AB3  AB2  AB
 AB
3
3
AB3slip
// crank
Crank
.
B2, B3
Rocker
23
Rolling acceleration (4.7)
First assume that angular acceleration, , is zero
O

O

 (absolute)
R
 (absolute)
C
r C
P
No slip condition: VP=0
24
Find accelerations of C and P
• -(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= (R-r) (Normal to radius OC)
• AnC=VC2/(R-r) (directed toward the center O)
• AnP=VC2/(R-r)+ VC2 /r (also directed toward the
center O)
• Tangential components of acceleration of C and P
are zero
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Summary of results
AC, length VC2/(R-r)
VC, length (R-r)
R
C
r
P
AP, length VC2/(R-r)+ VC2 /r
VP=0
26
Inverse curvature
•  (R+r)/r
• VC=(R+r) (normal to OC)
• AnC=VC2/(R+r) (directed toward the center O)
• AnP=VC2/r - VC2 /(R+r) (directed away from the
center O)
• Tangential components of acceleration are zero
27
Inverse curvature: Summary of results
VC, length (R+r)
C
r
VP=0
R
AC, length VC2/(R+r)
P
AP, length VC2 /r -VC2/(R+r)
28
Now consider nonzero angular
acceleration, 0
• The results for zero angular acceleration are
still correct, but
• ACt=r (normal to OC)
• APt is still zero
• These results are valid for both types of
curvature
29