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

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. 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)

4. 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

5. Acceleration of link, general case
AtPA P
APA
Length of link: p
AnPA
,  
AnPA
AtPA A AP
APA AA AA
AP=AA+APA

6. Graphical acceleration analysis
Four-bar linkage example (example 4.1)
AtBA B
AtB 3 A
AtA
Clockwise
acceleration of
crank 4 2
AnA 1

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

8. Graphical solution of equation AB=AA+ABA
AtB
AtBA 2 3 4 1
AtA
AnA B A
AtB AA
AtBA
AnBA
-AnB
Steps:
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.

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. Algebraic acceleration analysis (4.10)2 3 4 1 B A a b c R4 R2 R1 1
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

11. Loop equation
Differentiate twice:
This equation means:

12. Solution

13. 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’

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. Coriolis acceleration
APslip: acceleration of P as seen by observer moving with rod
VPslip P
APcoriolis
AP’n
AP’t
APslip O AP

16. 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

17. Application: crank-slider mechanism
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
Unknown quantities marked in blue .
B2, B3
normal to crank
Link 2, a
3, 3, 3
2, 2 2 O2
Link 3, b d

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. Position analysis
Make sure you consider the correct quadrant for 3

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

21. Velocity analysis
is the relative velocity of B3 w.r.t. B2

22. Acceleration analysis
Where:

23. Relation between accelerations of B2 (on crank) and B3 (on slider)
AB3slip
// crank
AB3
Crank .
B2, B3
AB3Coriolis
┴ crank
AB2
Rocker

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

25. 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

26. Summary of results R AC, length VC2/(R-r) C VC, length (R-r) r
AP, length VC2/(R-r)+ VC2 /r P
VP=0

27. 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

28. Inverse curvature: Summary of results
VC, length (R+r) C r
AC, length VC2/(R+r) P
VP=0 R
AP, length VC2 /r -VC2/(R+r)

29. 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