Rigid body kinematics

1. Planar Kinematics of Rigid Bodies and
Mechanisms
Velocity and Acceleration Analysis
1

2. Unit Vectors (i j k)
10/16/202
2

3. 3

4. Rotating Unit Vector
e
e 
 

3/6/2024
4

5. 5
Introduction
• Kinematics of rigid bodies: relations
between time and the positions, velocities,
and accelerations of the particles forming
a rigid body.
• Classification of rigid body motions:
- general motion
- motion about a fixed point
- general plane motion
- rotation about a fixed axis
• curvilinear translation
• rectilinear translation
- translation:

6. 6
Motion of the plate: is it translation or rotation?
Curvilinear translation Rotation

7. 7
Translation
• Consider rigid body in translation:
- direction of any straight line inside the
body is constant,
- all particles forming the body move in
parallel lines.
• For any two particles in the body,
A
B
A
B r
r
r





• Differentiating with respect to time,
A
B
A
A
B
A
B
v
v
r
r
r
r














All particles have the same velocity.
A
B
A
A
B
A
B
a
a
r
r
r
r


















• Differentiating with respect to time again,
All particles have the same acceleration.

8. 8
Rotation About a Fixed Axis: Representative Slab
• Consider the motion of a representative slab in
a plane perpendicular to the axis of rotation.
• Velocity of any point P of the slab,



r
v
r
k
r
v










• Acceleration of any point P of the slab,
r
r
k
r
r
a









2













• Resolving the acceleration into tangential and
normal components,
2
2




r
a
r
a
r
a
r
k
a
n
n
t
t












9. 9
General Plane Motion
A combination of translation & rotation

10. 10
General Plane Motion
Pure translation, followed by rotation about A2 (to move B'1 to B' 2)
Motion of B w.r.t. A is pure rotation, i.e. B draws a circle centered at A
Any plane motion can be represented as a translation of an
arbitrary reference point A and a rotation about A.

11. 11
Absolute and Relative Velocity
A
B
A
B r
k
v
v






 
B A B A
v v v
 
For any two points lying on the same rigid body:
Note: B A
v r

 r = distance from A to B
/
B A B A
v k r

 
Equation can be represented graphically by a velocity diagram

12. 12
Absolute and Relative Velocity
Assuming that the velocity vA of end A is known, determine
the velocity vB of end B and the angular velocity .
The direction of vB and vB/A are known. Complete the velocity diagram.
tan tan
B
B A
A
v
v v
v
 
  
B A B A
v v v
 
Locus for vB
vA
vB/A

vB
Locus
for vB/A
cos
cos cos
B A B A A
B A
v v v
v l
l l l

 
 
 
    
 
 

13. 13
Absolute and Relative Velocity in Plane Motion
• Selecting point B as the reference point and solving for the velocity vA of end A
and the angular velocity  leads to an equivalent velocity triangle.
• vA/B has the same magnitude but opposite sense of vB/A. The sense of the
relative velocity is dependent on the choice of reference point.
• Angular velocity  of the rod in its rotation about B is the same as its rotation
about A. Angular velocity is not dependent on the choice of reference point.

14. 14
Rolling Motion
Consider a circular disc that rolls without slipping on a flat surface

r
s
A1
A2
O1 O2
s r

From geometry:
s = displacement of center
v r r
 
 

15. 15
Sample Problem #1
The crank AB has a constant clockwise angular velocity of
2000 rpm. For the crank position indicated, determine (a) the
angular velocity of the connecting rod BD, and (b) the velocity
of the piston P.

16. 16
B
D
B
D v
v
v





• The velocity is obtained from the crank rotation data.
B
v

    
s
rad
4
.
209
in.
3
s
rad
4
.
209
rev
rad
2
s
60
min
min
rev
2000
























AB
B
AB
AB
v 


• The direction of the absolute velocity is horizontal.
The direction of the relative velocity is
perpendicular to BD.
D
v

B
D
v

Locus for vD
Locus for vD/B

17. 17
• Determine the velocity magnitudes
from the vector triangle drawn to scale.
B
D
D v
v and
B
D
B
D v
v
v





s
in.
9
.
495
s
ft
6
.
43
s
in.
4
.
523



B
D
D
v
v
s
rad
0
.
62
in.
8
s
in.
9
.
495




l
v
l
v
B
D
BD
BD
B
D



18. 18
Instantaneous Center of Rotation
For any body undergoing planar motion, there always exists a
point in the plane of motion at which the velocity is
instantaneously zero (if it were rigidly connected to the body).
This point is called the instantaneous center of
rotation, or C.
It may or may not lie on the body!
If the location of this point can be determined, the velocity
analysis can be simplified because the body appears to rotate
about this point at that instant.

19. 19
Instantaneous Center of Rotation
To locate the C, we use the fact that the velocity of a point on a
body is always perpendicular to the position vector from C to that
point.
If the velocity at two points A and B are
known, C lies at the intersection of the
perpendiculars to the velocity vectors
through A and B .
If the velocity vectors at A and B
are perpendicular to the line AB,
C lies at the intersection of the
line AB with the line joining the
extremities of the velocity
vectors at A and B.
If the velocity vectors are equal & parallel, C is at infinity and the angular
velocity is zero (pure translation)

20. 20
If the velocity vA of a point A on the body and
the angular velocity  of the body are
known, C is located along the line drawn
perpendicular to vA at A, at a distance
r = vA/ from A. Note that the C lies up and
to the right of A since vA must cause a
clockwise angular velocity  about C.
Instantaneous Center of Rotation

21. 21
The velocity of any point on a body undergoing general plane
motion can be determined easily if the instantaneous center
is located.
Since the body seems to rotate about
the IC at any instant, the magnitude of
velocity of any arbitrary point is v =  r,
where r is the radial distance from the IC
to that point. The velocity’s line of action
is perpendicular to its associated radial
line. Note the velocity has a direction
which tends to move the point in a
manner consistent with the angular
rotation direction.
Velocity Analysis using Instantaneous Center

22. 22
Instantaneous Center of Rotation
C lies at the intersection of the
perpendiculars to the velocity
vectors through A and B .


cos
l
v
AC
v A
A


   




tan
cos
sin
A
A
B
v
l
v
l
BC
v



The velocity of any point on the rod can be obtained.
Accelerations cannot be determined using C.

23. 23
Sample Problem #1 using instantaneous center
The crank AB has a constant clockwise angular velocity of
2000 rpm. For the crank position indicated, determine (a) the
angular velocity of the connecting rod BD, and (b) the velocity
of the piston P.
Crank-slider mechanism

24. 24
C is at the intersection of the perpendiculars to the velocities through B and D.
   
B AB BD
B
BD
v AB BC
v
BC
 

 
  
D BD
v CD



25. 25
Absolute and Relative Acceleration
Absolute acceleration of point B: A
B
A
B a
a
a





Relative acceleration includes tangential and normal components:
A
B
a

    2
B A B A
t n
a r a r
 
 

26. 26
Absolute and Relative Acceleration
• Given determine
,
and A
A v
a


.
and


B
a
   t
A
B
n
A
B
A
A
B
A
B
a
a
a
a
a
a











• Vector result depends on sense of and the
relative magnitudes of  n
A
B
A a
a and
A
a

• Must also know angular velocity .

27. 27
Absolute and Relative Acceleration
Draw acceleration diagram to scale:

28. 28
Sample Problem #1 Acceleration Analysis
Crank AG of the engine system has a constant clockwise
angular velocity of 2000 rpm.
For the crank position shown, determine the angular
acceleration of the connecting rod BD and the acceleration
of point D.

29. 29
   n
B
D
t
B
D
B
B
D
B
D a
a
a
a
a
a











  
AB
2
2 2
3
12
2000 rpm 209.4rad s constant
0
ft 209.4rad s 10,962ft s
AB
B AB
a r



  

  
From Sample Problem #1, BD = 62.0 rad/s, b = 13.95o.

30. 30
   n
B
D
t
B
D
B
B
D
B
D a
a
a
a
a
a











Draw acceleration diagram:
       2
2
12
8
2
s
ft
2563
s
rad
0
.
62
ft 

 BD
n
B
D BD
a 
      BD
BD
BD
t
B
D BD
a 

 667
.
0
ft
12
8 


Drawn to scale

31. 31
Sample Problem #2
In the position shown, crank AB has a constant angular velocity
1 = 20 rad/s counterclockwise. Determine the angular
velocities and angular accelerations of the connecting rod BD
and crank DE.
Four-bar mechanism

32. 32
B
D
B
D v
v
v





   k
k DE
BD




s
rad
29
.
11
s
rad
33
.
29 

 

Velocities
1( )
B
v AB


vD
vB
vD/B vB
vD
vD/B
( )
D B BD
v BD


( )
D DE
v DE


Velocity diagram
Shown here not to scale

33. 33
B
D
B
D a
a
a





   k
k DE
BD



 2
2
s
rad
809
s
rad
645 

 

Accelerations
aB
aD/B n
aD/B t aD/E t
aD/E n
Acceleration diagram
aB
  2
( )
D B BD
n
a BD


aD/B n
2
( )
B AB
a AB


aD/E n
aD/E t
  2
( )
D E DE
n
a DE


  ( )
D E DE
t
a DE


Shown here not to scale

34. Velocity Image
It may also be noted that
the scale factor between the
link BCD and its velocity
image is the rotational
velocity of the link i.e.
3




BD
bd
CD
cd
BC
bc
bcd is similar to BCD
bcd is obtained by
rotating BCD by 90o in the
sense of 3 and scaled by
3
bcd is the velocity
image of link-3 BCD

35. Motion of a particle inside a moving rigid body-
Coriolis Acceleration
BP
A
B
BP
A
B
V
r
ω
r
ω
V
a
V
r
ω
V
V



 









e1
e2
y
x 2
1 e
e
r 

y
x 
 2
1
BP e
e
V 

3/6/2024
35

36. Motion of a particle inside a moving rigid body-
Coriolis Acceleration
BP
A
B
BP
A
B
V
r
ω
r
ω
V
a
V
r
ω
V
V



 









BP
BP
2
1
2
1
2
1
2
1
V
ω
r)
(ω
ω
V
ω
y)
e
ω
x
e
(ω
ω
)
y
e
x
(e
ω
y)
e
x
e
(
ω
y)
e
x
(e
dt
d
ω
r
ω



























e1
e2
y
x 2
1 e
e
r 

3/6/2024
36

37. Motion of a particle inside a moving rigid body-
Coriolis Acceleration
BP
A
B
BP
A
B
V
r
ω
r
ω
V
a
V
r
ω
V
V



 









BP
BP
2
1
2
1
2
1
2
1
V
ω
r)
(ω
ω
V
ω
y)
e
ω
x
e
(ω
ω
)
y
e
x
(e
ω
y)
e
x
e
(
ω
y)
e
x
(e
dt
d
ω
r
ω



























 
BP
BP
BP
2
1
2
1
2
1
2
1
BP
a
V
ω
a
)
y
e
x
(e
ω
)
y
e
x
(e
)
y
e
x
e
(
y
e
x
e
dt
d
V


























BP
BP
A
B a
V
2ω
r)
(ω
ω
r
ω
a
a 







 
e1
e2
3/6/2024
37

38. Acceleration Vectors
BP
BP
A
B a
V
2ω
r)
(ω
ω
r
ω
a
a 







 
Coriolis
acceleration
VBP
3/6/2024
38

39. An Example of Coriolis acceleration
3/6/2024
39
Determine the acceleration of the slider.
n
O
A
t
O
A
O
A a
a
a
a 2
2
2
2
2
2



0 0
2
2
2
2
2
2 /
1250
5
.
0
50
. s
m
A
O 



n
A
t
A
c
A
A a
a
a
a
a 2
/
2
/ 3
3
2
3




0
0
2
2
A
A
c 3000m/s
50
30
2
ω
2V 2
3





a
2
2
2
/
3250
3
3
s
m
a
a
a c
A
A 



40. Inverted QR Mechanism: follow the steps
Oa
an
A
A
O
an
A
2
2
2


(//)
/)
(
(//)
)
(
3
3
3
3
n
A
B
z
t
A
B
n
A
t
A
A
B
A
xy
B a
a
a
a
a
a
a 





0
(//)
/)
(
)
(
4
3
4
3
3
c
B
B
w
t
B
B
xy
B a
a
a 

4
3
4
3 3
2 B
B
c
B
B V
a 


41. (//)
/)
(
(//)
)
(
3
3
3
3
n
A
B
z
t
A
B
n
A
t
A
A
B
A
xy
B a
a
a
a
a
a
a 





(//)
/)
(
4
/
)
(
4
3
3
3
c
B
B
w
t
B
xy
B a
a
a 

0

42. Sample Problem #3
B A
C
P

43. VP2=4m/s
VP4=3.179m/s
4=3.179/0.13937=22.8 rad/s (CW)
V3/4=VP2/P4=2.427m/s (away from B)
an
P3=42/.08=200m/s2
ac
P3/P4=2x2.427x22.8=110.67m/s2=-ac
P4/P3
an
P4=3.1792/0.13937=72.5m/s2

44. 3 4 3 4
3 3 4 4 3 4
(//) (//) ( ,/) (//) (//) ( /)
P P c P P
t n t n c
P P P P P P
a a a a
a a a x a a a y
  
    
at
P4=10.76m/s2
4=10.76/0.13937=77.2 rad/s2
0
c
a
3
n
P
a
Oa
4
n
P
a
4
t
P
a p4 3 4
P P
a

45. Animation of WhitWorth QR mechanism

46. Sample Problem #4
Quick Return Mechanism
VA3
V
A
3
/
4
VA4
0v
VB
VC
VCB
A2, A3, A4 are coincident
points on 2, 3 and 4,
respectively.
VA3=*O2A2
4=VA4/O4A4,
VB=4*O4B,
/)
,
(
3
/
(//)
/)
,
(
4
3
4
y
A
A
x
A V
V
V 

VC
(z,/) = VB
(/,/)+VCB
(w,/)

47. 2=100 rad/s, link-2=13 cm., link-4 =35 cm., O2O4=20 cm., link-5=20 cm.
Determine the velocity of slider 6.
s
m
V
ccw
s
rad
s
m
V
B
A
/
4765
.
11
79
.
32
35
.
0
)
(
/
79
.
32
2592
.
0
5
.
8
/
5
.
8
4
4







s
m
V
V
V
V
A
y
A
A
x
A
/
13
3
4
3
4
/)
,
(
3
/
(//)
/)
,
(




48. /)
,
(
(//)
/)
,
( u
CB
B
z
C V
V
V 
 Vc=9.95m/s