### Kinematics of Rigid Bodies

```Plane Kinematics of Rigid Bodies

Kinematics
study of body motion without reference to force.

Rigid Body
 It has dimensions. (particle doesn’t have it).
 distance between 2 points in the body remains unchanged.
 assumption validity? i.e. there is no real rigid body.

Plane Motion

Definition: All parts of the body move in parallel planes.
1
Plane Motion
Definition:
All parts (points) of the body move in parallel planes.
Motion Plane
Movement of one cutting face (corresponding
to its motion plane) describes movement of
the whole body.
Treat the body as
thin slab object.
Any slab object (cutting face)
is okay,
But we usually use the plane
where the object’s C.G.is in.
C.G.
All corresponding points in other motion plane
have the same velocity and acceleration
2
3
Rigid-Body Plane Motion
I (Pure) Translation
II (Pure) Rotation
III General Motion
4
Type of Plane motion
C.G.

(Pure) Translation
Definition: A line between two points in the body remains
parallel through out the motion.
Motion of one
point can be used
to describe
motion of the
whole body.
Rectilinear
Translation
Curvilinear
Translation
C.G.
Treat it as a particle
5
Types of Plane Motions (cont.)

Fixed-axis (Pure) Rotation:
All , move along
circular path with
center at
Not having the same velocity
and acceleration (depends on
Use techniques for particle (circular) motion (n-t, r-)
Important Property of Fixed-Axis Rotation
Rotation axis
motion plane
Rotation Axis
Motion Plane
Types of Plane Motions (cont.)

General Plane motion:
 need new techniques in this chapter
7
5/2 Rigid Body’s Rotation
“Rotating” concept is basically
a concept on rotation of “line”.
a concept on (whole) rigid body.
same
,?


Angle between any two lines on “rigid” object
does not change during the time dtd   0
Define
+
any Reference
    
  1  
0
for rigid body

angular position (+/-)
( ) 
angular velocity
( ) 
angular acceleration
Any lines on a rigid body in its plane of
motion have the same angular
displacement, velocity and acceleration
   ,    ,
8
(a) Angular Motion Relations
Define

( ) 
angular position (+/-)
angular velocity
( ) 
angular acceleration
+
any Reference
Similar to rectilinear motion
s 
v 
a 

d

dt

 
d    d 
 , ,  relation
(or  d   d )
Sign convention of all variables must be consistent!
9
Angular Motion Relations

Observed the similarity with the linear motion
d


dt
s 
v 
a 

d
dt
2) Graphical meanings
d    d 
1) Integrals Calculation
   (, , t )  (t ),  (t ) ?
 : const
Area :
1 2
(20  02 )
2
  o   t
1
   0  0t   t 2 2   2  2 (  )
2
o
o
10

d
dt

d
dt
CCW:+
2
60
=4t
o  1800
 = - 60
=4t
=0
=4t
= + 203.50
d    d
CCW
A flywheel rotating freely
at 1800 rev/min clockwise is subjected to a variable
counterclockwise torque which is first applied at time t = 0. The torque
produces
C.Wa counterclockwise angular acceleration  = 4t rad/ss, Determine the
total number of revolutions, clockwise plus counterclockwise, turned by the
flywheel during the first 14 seconds of torque application.

+

60
t
d    dt
  60  2t 2

t
0
0
2
d


(

60


2
t
)dt


0
2
 |t 14  60 (14)  143  809.6 rad
3
0  60  2t22 t2  9.71 s
 |t 14
809.6
N
 128.85 rev  60  2(14)2
2
2
C .W  60 t2  t2 3  1216.96 rad
3
CCW   |t 4  |t 4.91  {60 (14)  2 (14)3}  (1219.96)
3
2
3
  60 t  t 3
 203.50
| 1216.96 |  | 410.36 |
1627.32
= 258.996 rev
11
2
Fixed-Axis (Pure) Rotation
(scalar notation)
Can we find?
r
Rigid
body
rOP
 vP aP
,
P
points
whole body (rigid body)
r
r
rOP
O


Point P
n-t coord:
Fixed-Axis
(Pure) Rotation Only !
 r
 
v  
 r
at  v    r  r
an 
v2

 r 2  v
13
 , ,

r, vt , at , an
The pinion A of the hoist motor drives gear B, which is attached to the
the hoisting drum. The load L is lifted from its rest position and
acquires an upward velocity of 2 m/s in a vertical rise of 0.8 m with
constant acceleration. As the load passes this position, compute
(a) the acceleration of point C on the cable in contact with the drum
(b) the angular velocity and angular acceleration of the pinion A.
v
C  C  5
rc
(aC )t  aL
,
(a )
C  C t  6.25
rc
rBB 
rB B 
vA 
vB  rBB
(a A )t  (aB )t  rBB
rBB
A 
 15
rA
CCW
rB B
A 
 18.75
rA
CCW
C and L has
(vC )t  vL =2 the same v a ?
vC 2
22
(aC )n =
=
 10
rc
0.4
aC  102  2.52  10.31
vL =2
vL 2
22
aL 

2 s 2(0.8)
 2.5 m/s2
14
n
n
t
t
floor- fixed
Non-slipping
share
the
same
Non-slipping
vt (= v)
Why?
at
an
share
the
same
v t (=v) = 0
a t = 0 an
Without gear teeth, the ball is not guaranteed to roll without slipping.
Two possible motion:
- roll without slipping
- roll with slipping.
If it do really roll without slipping,
It should have some “motion constraint”.
[ fixed relation between  and v ]
(Next session: we will find out this)
In case of No gear-teeth
15
17
Fixed-Axis (Pure) Rotation (vector notation)

|  | 


The body rotating about the O axis
Represent “angular velocity”  as
Vector


Direction: “Right-hand rule”
From the last section, the magnitude of
the velocity is v  r
Think in 3D Vector
We consider only Plane
motion of rigid body
  r
 // 

The cross product can be used to
establish the direction:
v r
( r )
18
Fixed Axis Rotation (vector notation)

Differentiating the velocity with respect to time:
a  v  r r

d
(  r )
dt
  r
We consider
only “Plane
motion” of
rigid body
r 2
 // 
r
   (  r )    r

r
Direction
depends
on  , r
r
Direction: r
  r // (  r ) // v
an    (  r )
at    r
v r
  (  r ) : - r direction
kˆ  ( kˆ  ( iˆ  ˆj )) 
(when r   )

2
( iˆ   ˆj)
19
Fixed-Axis (Pure) Rotation (vector notation)
v r
a      r     r
20
r  0.3iˆ  0.28 ˆj
The rectangular plate rotates clockwise.
If edge BC has a constant angular
velocity of 6 rad/s. Determine the vector
expression of the velocity and
acceleration of point A, using coordinate
as given.
Rigid body
v r
  BC
 = BC
v  6kˆ  (0.3iˆ  0.28 ˆj)  1.8 ˆj  1.68iˆ
a    (  r )    r
a  6kˆ  (1.8 ˆj  1.68iˆ))  0  (0.3iˆ  0.28 ˆj)
  (  r ) : - r direction (when r   )
kˆ  ( kˆ  ( iˆ  ˆj )) 

2
( iˆ   ˆj)
 10.8iˆ  10.08 ˆj
21
Equations Review:
(Pure) Translation
Movement of one point describes movement of the whole body.
dv
ds
v
 v
v   s a 
dt
dt


a
 dv 
 dr 
a
v
v
r
dt
dt
(Pure) Rotation
rotation of one line describes rotation of the whole body.
Whole body shares the same “angular” quantities.

+


d
 
dt

v r
d
 
dt
d  d
a    (  r )    r
Reference
General Plane Motion
Absolute motion
Relative motion
23
5/3 Absolute Motion
(of General Plane Motion)

Use geometric constraint (which define the configuration of the body
to obtain the velocity and acceleration (in general motion)

Idea: write a (position) constraint equation which always applies
regardless of the system’s configuration, then differentiate the
equation to get velocity and acceleration.
?
y  x?
x 2  y 2  b2
2 xx  2 yy  0
y
x y
x
y   (?  )?
y  b sin 
y   b cos
y   b cos   2b sin 
for complex constraint method of relative motion may be easier.
24
?
y  x?
x 2  y 2  b2
2 xx  2 yy  0
y   (?  )?
y  b sin 
y   b cos
y
x y
x
Define the displacement and its positive direction.
The variable must be measured from the fixed
reference point or line.
Find the equation of constraint motion.
Equation must be true all during the motion.
Differentiate it to find (angular) velocity and acceleration.
25
5/56. Express the angular velocity AB and angular acceleration  AB
of the connecting rod AB in terms of the crank angle 0 for a
given constant  .
l
r

sin  sin 
AB ,  AB

l sin   r sin 
l  cos   r cos
0
l  cos   l  sin   r cos   r 2 sin 
2
 AB
l  2 sin   r 2 sin 
 
l cos 

r
sin 
l
2
0
r2
1
2
l
r 2 2 3/ 2
(1  2 sin  )
l
r cos 
l cos 
AB    
r
 0
l
cos 
r
1  ( sin  ) 2
l
26
motion constraint
SP5/4 A wheel of radius r rolls on a flat surface without slipping.
Determine angular motion of the wheel in terms of the linear
motion of its center O. Also determine the acceleration of a point
on the rim of the wheel as the points comes into contact with the
surface on which the wheel rolls 
vO aO   
s  r
vO

s  r
Point O:
rectilinear motion
vO  r
aO ( v  r)  r
Point C’s trajectory
x  r  r sin 
y  r  r cos 27
SP5/4 A wheel of radius r rolls on a flat surface without slipping.
Determine angular motion of the wheel in terms of the linear
motion of its center O. Also determine the acceleration of a point
on the rim of the wheel as the points comes into contact with the
surface on which the wheel rolls s  r v  r
a  r
O
x  r  r sin 
y  r  r cos 
x  r  r cos  vO (1  cos )
  2 :
x0
y0
velocity
=0
x0
y  r
2
Acceleration in the
direction of axis x = 0
O
D
C’’
x  aO (1  cos )  r2 sin
y  aO sin   r2 cos
y  r sin  vO sin
Not depend on t, t are!
O
When Point D comes to contact
the surface,
It also has a velocity (=0), and 28
acc. as above.
x  r  r sin 
y  r  r cos 
y  aO sin   r2 cos
y  r sin  vO sin
x0
  2 :
x  aO (1  cos )  r2 sin
x  r  r cos  vO (1  cos )
y0
velocity
=0
x0
y  r
Acceleration in the
direction of axis x = 0
2
O
O
v B  v A ( aB ) t  a A
C’’
vA
Analogy
aA
No slipping
vB  v A
aB  aA
vA
Rel. vel. =0
Rel acc. = 0
aA
vB  0
aB  0
vA  0 a A  0
29
Non-slipping Condition
vO  r
aO  r
O
t
floor- fixed
C’’
“no slipping” implies:
1) Contact point { C , C’ } on two body has no
relative velocity.
2) Contact point { C, C’ } on two body has same
tangential component of acceleration
(aC )t  aL
(vC )t  vL
31
Each cables do not slip. Load-supporting pulleys are rigid body.
r1, r2 , r3 , r4 1, 1 2 , 2: known
Find: o ,o , v L , aL
 r4
r3 
( AB )(  )
 ( v B t  v A t )

ds A
( AB)()  (vB  vA )
r1, 1, 1 => vA ,(a A )t
vA  r11
(aA )t  r11
r2 , 2 ,2 => vB ,(aB )t
vB  r22 (aB )t  r22
d v B  v A
o 

dt
r3  r4
r3
r4
O
r22  r11
A

dsB
B
r3  r4
( AB)( )  (aB )t  (aA )t
d  (aB )t  (a A )t
o 

dt
r3  r4

r2 2  r11
r3  rr
O and L has same vertical velocity & acceleartion
vL  v A  r3
r22  r11
r3  rr
r22  r11
aL  a A  r3
r3  rr 32
x  L cos 
x  ( L sin  )
x  L{(cos ) 2  (sin ) }

v0
L2  x 2
   cot    2
x  v0
x0
v0 2

 2
2
2
2
L

x
L x
x
33
General Plane Motion of Rigid Body
-Introduction
นิยามการเคลื่อนที่ของวัตถุเกร็ ง
-Rotation
วิธีอธิบายการเคลื่อนที่แบบหมุน
- Calculation methods
usually for any t
-Absolute motion
ใช้คานวณ
V
-Relative velocity
ใช้คานวณ
V
A
using constraint equation
Observer is
at the point
of rigid body
where its
velocity = 0
Instantaneous Center of
Zero Velocity (ICZV)
ใช้คานวณ
V
-Relative acceleration
ใช้คานวณ
V
A
Motion relative to
rotating axes
ใช้คานวณ
V
A
Translati
ng-only
observer
using its
geometric
shape at
that
instant
usually for
Translating some t
and rotating (instant)
Observer
35
5/4
Relative
Velocity

relative VA  VB  VA B
velocity
concept
General Plane Motion = Translation + Rotation
Motion of point (observer) B, detected by O
= Motion of plate moving “translationally”
Different
viewpoint
“simultaneous”
Since the distant between the two
points on a rigid body is constant,
an observer at one point will see
the other point move in a circular
motion around it!
Wait!
B really sees A moving circularly?
O
B sees A has no movement !?!?!?
36
Applying the
relative concept


VA
Observer B is on the plate
Rotating observer (attached to B)
Rotating frame


 VB  VA B
Which one?
Only this case

Observer B is sitting on the
magic carpet.
non-rotating observer (attached to B)
non-rotating frame
A
A
B
A
B
B see A no moving at all
B see A
having a velocity
perpendicular with its distance. 37
Relative Velocity

(non-rotating observer)
We use: non-rotating observer (frame) attached to B
Absolute world:
Relative world:

v A  vB

Line
AB
vA  vB  vA / B
Observer O detects:
same?
Only when
non-rotating observer
(see next page)
vB
Line / B

v A  vB
r
v A / B    rA / B
Observer at B see A moving
in a circle around it
Observer B detects:
 rA 38/ B
Translating observer see same , as absolute Observer
The rotating of rigid body
= The rotating of line compared with
“fixed” reference axis
B
1
  2  1
   2  1
O
2
same 
B’s reference
line
B
1
same 
2
O’s reference line
39
Translating-Rotating observer sees different ,
with absolute Observer
The rotating of rigid body
= The rotating of line compared with
“fixed” reference axis
B
1
1
different

   2  1
different 
B’s reference
line
B
O
  2  1
2
O’s reference line
2
The rotating B’s “reference line”,
observed by absolute observer.
40
Understanding the equation
absolute
absolute
vA 
vB
absolute

Non-rotating,
Moving with B
Non-rotating,
Moving with B
 vA/ BrA/ B
Hint on solving problems

Any 2 points
the and unknown
Identify
the on
known
same rigid body


A
Above equation (2D: “3D-fake”) can be
solved when there are at most 2 unknown
scalar quantities
B
Above equations usually contains 5 scalar
quantities (not including position vector r)
Important key:
vA / B
vA / B always
perpendicular to line
AB. Its direction can
be deduced from 
Also works with A as the observer
(vB / A  v A / B41)
Fixed-Axis (Pure)
Rotation
Relative Vector Analysis
on General Plane Motion

vA  vB  vA / B
vG  rOP  vP
vP    r
vP / G    r
P
vP

P
G
vG vP rOP  
 of observer at G
=  (of rigid body)
of observer at O
| vP  vG |

| rOP |
vG
vG
vP / G    r
vG
vP

G
P
vG
42
Velocity at A
is key point to find 
AB
OA
 ?
A, B on the same rigid body (bar AB)
vA
=
OA  rA
OA kˆ
Solved by
Vector Analysis
 ?
vB
+ vA / B
CB  rB
AB  rA B
2 kˆ
AB kˆ
rA  0.1 ˆj
rB  0.75 iˆ
rA B  0.175 iˆ  0.05 ˆj
44


OA  rA
From




 CB  rB  AB  rA B
OAkˆ  0.1 ˆj  2 kˆ  (0.075iˆ )  ABkˆ  (0.175iˆ  0.050 ˆj )
i
k
+
j
0.1OA iˆ   0.15 ˆj  0.175AB ˆj  0.050AB iˆ
kˆ
OA
A

y
O
x
3
 
7
AB
6
 
7

C
B
+
3D vector calculation (i,j,k):
(right hand rule)
45
AB  ?
OA  ?
  tan 1
100  50
 15.95o
250  75
A, B on the same rigid body (bar AB)
VA


VB
 VA B

D
M
?
rBCB
?
Solved by Graphical Method
vA
vB
vA B

 0.15
v A  vB tan 
 0.0429
v A / B  vB / cos 
 0.156
vA
A   0.429
rA
CW
vA/ B
 0.857
rBC
CW
B 
need to find the angular direction from the figure
46
47
48
Note: relative velocity technique
Direction is simply found:

vA B
vA
vB
simplest, be careful about sign / 1
 graphical solution:
 3D (i, j, k) vector:
 : same in absolute and relative world
vA  vB    rA / B

To know

vB
B
A
vB
1
2
vA B
Non-rotating
 of the rigid body
vA
r
vA
(only) direction
of vB is enough
vA
B
r
A
1
direction of v A / B
49
50
 VB  VA B
VA

0.250
  cos
 60o
0.500
1
  0.8

20o
D
?
M
ABC

rOB
?
vB
vA 
sin 60  0.226
sin50
Graphical solution
vA
20
70o
50o
vA / B
 rABABC
  60o
o
vB  rOB
 0.2
vB
vA/ B 
sin 70  0.245
sin50
ABC 
vA/ B
 0.49 CCW
rAB
vC  ABC  rBC
  
vC  vB  vC / B  vB  ABC  rC / B
  37.316o
vc  0.175
1
vC / B  v A / B  0.1225
2

vB  0.2
60o
51
vB


VA
VB 
VA B

  cos 1
0.250
 60o
0.500
20
D
M

o
rOB
?
?
vA

vA  vA cos 20o iˆ  sin 20o ˆj

vB  0.8kˆ  0.25iˆ  0.2 ˆj
Vector solution

  
vC  vB  vC / B  vB  CB  rC / B

vC  0.1746 cos127.5o iˆ  sin127.5o ˆj



vA/ B  ABC kˆ  0.5  cos iˆ  sin  ˆj


vA  0.226m / s
ABC  0.491
Direction?
CW
52
53
Relative Velocity (Part 2)
Another usage of the
relative velocity equation

B
vb  OD
D
vA  vB  vA / B
A on OD
B on
screw
A
2 points need not be in the same rigid body
parerell

For constrained sliding contact
between two links in a mechanism.
 Pick points A and B as coincident points,
vA
 OD c.c.w.
vA B
(the points may be imaginary).

vB
some reason later!
The observer on B no longer see A moving
around it in a circle.
54
55
vP / Q
vP
40o 
Q on OA
B
P
0.175
PO 


 0.4  0.175cos   0.263
cos 
O
0.175sin 
0.4  0.175cos 
 19.402o
  tan 1
0.175cos 0.4-0.175cos
 30o
vQ
vP  vQ  vP / Q
M
D



3(0.263)
?

 
vP  vP  sin  iˆ  cos  ˆj  vP 0.5iˆ  0.866 ˆj

vQ  3(0.263)  sin iˆ  cos  ˆj

vP / Q  vP / Q  cos  iˆ  sin  ˆj
 

vP / Q
vP
 0.262iˆ  0.745 ˆj


 vP / Q 0.766iˆ  0.6428 ˆj
?
vP  vQ  vP / Q
vQ


vP  2.161 m/s
BC 

 
vC  vB  vC / B
 0  BC  rC / B


vP / Q  1.06 m/s
vP
rBP

 12.348kˆ  0.35 cos  iˆ  sin  ˆj


 1.21 cos300o iˆ  sin 300o ˆj


Ans
56
57
=?
  tan 1
Q on slot C
V=1.5
=30
vo  OP
vO  vD    r
y

D

vO O
  
vP  vO  vP / O
vO
 15 CW
OP


 1.5iˆ  OP  rP / O

x
vQ
vQ/P
Note : vP not // OP
( vP / O // OP )
vQ  vP  vQ / P

?
2.8iˆ



 2.8iˆ  0.75 ˆj


  2.8iˆ  0.75 ˆj 
 v  sin  iˆ  cos  ˆj 
vQ cos  iˆ  sin  ˆj
Q/P
0.75 ˆj
C 
Non-slipping
condition
OP 
 1.5iˆ  15kˆ  0.1 sin  iˆ  cos  ˆj
vP
200
vO  OP | OP |
vo , OP  vP
0
C
0.1sin 
 23.79o
0.2  0.1cos 
?
vQ
rQC

vQ  2.26m / s
2.26
0.150sin 
sin 
Plus
CCW
58
5/6
a A  aB  a A B
relative
Relative
acceleration
Acceleration concept

General Plane Motion = Translation + Rotation
Motion of point (observer) B, detected by O
= Motion of plate moving “translationally”
Different
viewpoint
“simultaneous”
Since the distant between the two
points on a rigid body is constant,
an observer at one point will see
the other point move in a circular
motion around it!
Wait!
B really sees A moving circularly?
O
B sees A has no movement !?!?!?
61
Relative Acceleration

(non-rotating observer)
We use: non-rotating observer (frame) attached to B
Absolute world:
Relative world:
same?
Line
Line
Only when
non-rotating observer
(see the proof at relative
velocity part)
a A  aB  a A / B
Observer O detects:
Line / B
Line / B
a A / B    (  rA / B )
   rA / B
Observer at B see A moving in a circle
around it
vB
Observer B detects:
  rA / B
62
Understanding Equations
Non-rotating
aA  aB
   (  aA / BrA / B )    rA / B
, : the same both in absolute and relative (translation-only) world
Hint on solving problems

Identify the known and unknown

Above equation (2D: “3D-fake”) can be solved
when there are at most 2 unknown scalar quantities

Above equations usually contains 6 scalar quantities
(not including position vector r)
63
OA  rA
vA  vB  vA / B = CB  rB + AB  rA B
OA kˆ
2 kˆ
AB kˆ
6
7
OA  
3
7
Find : OA  ?  AB
Given : CB  2 rad s (constant)
aA
 AB  

aB
OA  ( OA  rA ) CB  ( CB  rCB )
 OA  rA
 CB  rB

?
aA / B
AB  ( AB  rA / B )
  AB  rA / B
OA  OAkˆ
 AB   AB kˆ
 2 kˆ  (2 kˆ  (75iˆ ))  (2)2 (75iˆ)  300 iˆ mm sec2
 AB  (AB  rA B )   AB  rA B   6 kˆ  (  6 kˆ  ( 175 iˆ  50 ˆj ))   AB kˆ  ( 175 iˆ  50 ˆj )
7
7
aB  CB  (CB  rB )  CB  rB
aA B

aA  OA  ( OA  rA )  OA  rA
900 ˆ 1800 ˆ
i
j   AB (50iˆ  175 ˆj ) mm sec 2
7
49
900 ˆ
  100 OA iˆ 
j mm sec 2
49
64
Given : CB  2 rad s (constant)
Find : OA  ?  AB  ? (at this instant)

aA
aB
OA  ( OA  rA ) CB  ( CB  rCB )
 OA  rA
 CB  rB

aA / B
AB  ( AB  rA / B )
  AB  rA / B
OA  OAkˆ
100 OA iˆ 
100 OA  300 

900 ˆ
1800 ˆ
900 ˆ
i 
j
j  300ˆi 
7
49
49
900
 50  AB
7
900
1800

 175 AB
49
49
 AB   AB kˆ
 AB (50iˆ  175 ˆj)
 AB  0.105 rad sec2
ANS
ANS
direction :  kˆ
65
Find : OA  ?  AB

aA
 ? (at this instant)

aB
OA  ( OA  rA ) CB  ( CB  rCB )
 OA  rA
AB  ( AB  rA / B )
  AB  rA / B
 CB  rB
OA  OAkˆ
vA
=
OA  rA
OA kˆ
vB
CB  rB
2 kˆ
+ vA / B
aA / B
 AB   AB kˆ
Find velocity first,
Before acceleration
AB  rA B
AB kˆ
  (  r )
kˆ  ( kˆ  ( iˆ  ˆj ))
 
2
( iˆ   ˆj)
66
5/124 The center O of the disk has the velocity and acceleration shown in
the figure. If the disk rolls without slipping on the horizontal surface,
determine the velocity of A and the acceleration of B for the instant
represented.
Non-slipping condition
a

r

v

r

s  r
o
o


  

 
vA  vO  vA / O  vO    rA / O

 
You can
calculate using
point O and D.
 3iˆ   7.5kˆ  0.4  cos45o iˆ  sin 45o ˆj



 9.85 cos32.6o iˆ  sin 32.6o ˆj m / s
D
I.C.Z.V
vO
r
a
  O  12.5 rad/s2
r





aB  aO  aB / O n  aB / O t
You can calculate
using point O and
and D.
 aO    (  rB / O )     rB / O 




 5iˆ   7.5kˆ   7.5kˆ  0.2iˆ  12.5kˆ  0.2iˆ
 5iˆ 11.25iˆ  2.5 ˆj



 16.44 cos171.3o iˆ  sin 171.3o ˆj m / s 2
69
70
vA vE
OA
 4 m/s (constant)
OA 
O
vE
4

 20 CCW
rOE 0.2
vA  OA  rOA  2.5
vA
vA  2.5 ˆj
vD  vA  vD/ A
B
BD
vD



vD  BD  rD / B

 BDkˆ   0.25 ˆj

vD/ A    r


= AD kˆ  0.2iˆ  0.15 ˆj


BD  7.5kˆ
71
OA
(constant)
OA=20
( aE )t  0
OE 
 aE t
rOE

0
0
0.2
O
aA  OA  (OA  rA / O )  OA  rA / O
 20kˆ  (20kˆ  0.125iˆ)  0  0.125iˆ  50iˆ
BD  (BD  rD/ B )   BD  rD/ B

 7.5kˆ  (7.5kˆ  0.25 ˆj )   BD kˆ  .250 ˆj
 0.25BDiˆ  14.06 ˆj

 12.5kˆ  (12.5kˆ  (0.2iˆ  0.15 ˆj))   ADkˆ  rD/ A
14.06  23.44  0.200 AD
0.25BD  0.15 AD  18.75
( CW )
72
73
General Plane Motion
(non-rotating observer)
any point: A, B (on same rigid body moving in GPM)
vA  vB  v
rA/ B
A/ B
Graphical Approach
Cross-Vector Approach
New technique
I.C.Z.V
B is special point : I.C.Z.V
aA  aB  aA/ B rA/ B 
   rA/ B
75
Note: relative velocity technique
Direction is simply found:

vA B
vA
vB
simplest, be careful about sign / 1
 graphical solution:
 3D (i, j, k) vector:
 : same in absolute and relative world
vA  vB    rA / B

To know

vB
B
A
vB
1
2
vA B
Non-rotating
 of the rigid body
vA
r
vA
(only) direction
of vB is enough
vA
B
r
A
1
direction of v A / B
76
Checkpoint: circular motion
Don’t know it is
fixed-point rotation
or not
(General Motion)
Fixed-point rotation
(Rotation)

?

v  r
from rotation point

*
v
r
valid method?
I.C.Z.V concept
78
General
Plane
motion
5/5 Instantaneous Center of Zero Velocity
vB  (rB / Z )
vA

B
A
direction
of vB
Extension theory using relative velocity.
vA  vZ  vA / Z
vA 
vB  vZ  vB / Z



vA
rA / Z
vB 
(at this instant)
vA/ C    rA/ Z
P
Z
If vz  0
vB / C
 of observer at C
=  of rigid body
(in Absolute Observer’s
perception)
   rB / C
Z is called I.C.Z.V (the point where its velocity at that instant is zero)
- can find  easily by geometry
- can find velocity and its direction
of any points easily by geometry
each point on the body can
be though of as rotating
around point Z.
For calculating v and  only79
Finding
an I.C.Z.V.
not a
rigid body
vA
vB
z
B
General Plane motion
“instantaneous”
Translational motion
 ?

A
v v
 0
r 
I.C.Z.V
at Inf.
A
vA
vA
vA
vB
A
vB
B
vB
B
z
vA

rA / C
A
vB
D
B

z
z
A
vA
A
B
z
vA    rA/ Z (where Z = I.C.Z.V)
vA
vB
B
 of observer at C
=  of rigid body
vD  vD/ Z    rD/ Z
vD2
rD / Z
(aD )t   rD/ Z 
I.C.Z.V for calculating
instantaneous velocity only
az usually  0
(Even vz = 0)
vD2 /C
vD2

rD /C rD /C
Arm OB of the linkage has a clockwise angular velocity of 10 rad/s in the
position shown where  = 45°. Determine the velocity of A, the velocity of
D, and the angular velocity of link AB for the instant shown.
 of what? OA, AB, BO
I.C.Z.V


VA
vB
30
=
7
350 2


 VB  VA B
solved by
relative
velocity

D
350 2
350
OB rOB
381
45o
 1500 2
mm / s

You have to find
Direction yourself
VA  CB (CA) 
30
(350 ) mm/s
7
?
M
rBCB
?
Direction of (absolute) velocity
of two point in the same rigid body
Thus, we can locate the
instantaneous center of velocity,
which is point C
vD  CB (CD ) 
30
(381) mm/s
7
82
83
0.075cos 
  tan
 16.1o
2rB
1
vC
y

x
vB

vA
rB
  60
o
vB
0.9

rB 0.065
ICZV

Vector Diagram
ABC 

rC
 0.9 m/s
 0.075 sin 
 0.065
CCW
2rB
rC 
 0.135
cos 
vC  ABC rC  13.86  0.135  1.873 m/s
84
R  0 (fixed)
s  P A ?
Non-slipping motion
ICZV of P1
P 1 =
Rs
Rs
2r
CW
Rs
2
ICZV of sun
ICZV of A
Rs
P 2 =
ICZV of P2
Sun Gear: Fixed-Axis (Pure) rotation
Planet Gear: general Plane motion
Rs
A = 2
( R  r)
CCW
Rs
2r
CW
85
vA
vA  ? to make vF  2
  60
( vD ) y
4
vD 

m/s
o
cos 30
3

30
ICZV
100 mm
vB
ICZV
vD
vF  2 m/s
vD  2
vD
40

o
2(0.1) cos30
3
8
v A  CD  0.2  m/s
3
(vD ) y  2
86
I.C.Z.V
Non-slipping condition
Absolute
motion
s  r
vo  r
vO  r
ao  r
aO 
dvO
dt
rectilinear
 r
aO 
dvO
dt
I.C.Z.V
an 
r
v

Rr
r*
2
2
2
n
r*  R  r
circle
vO  r
at  vt  r
5/140
I.C.Z.V
t
87
88
VD1
vC1
I
vC1
=8r
VD0
vC1  r1  r (8)  8r
vH3
vH 2  2r OA  8r
vH 3  4r OA  16r
II
ICZV
VD2
vH2
2 
vC 2
vH2
vC1
8r

 16
r/2 r/2
CW
III
r

   r 16  24r
2

= 8r
vC2
vC2
d
ICZV
vH3
VD3
vC 2 r  d

vH 3
d
 3 

24r r  d

16r
d
vH 3 16r

d
2r
d  2r
CCW
Practice Before Rotating Observer
91
5/134 The sliding collar moves up and down the shaft, causing an
oscillation of crank OB. If the velocity of A is not changing as it passes the
null position where AB is horizontal and OB is vertical, determine the
angular acceleration of OB in that position.
aA
OB  0

0
 AB 
aA  0
vA
CCW
L

aB
OB  ( OB  rOB ) AB  ( AB  rA / B )
 OB  rB
ICZV
of AB
aA / B
  AB  rA / B
vB  0
vA ˆ  vA ˆ

ˆ
ˆ
0  OB k  ( rj )  k   k  liˆ    ABkˆ   Liˆ
L
L

2
v 
iˆ :  rOB   A  L  0
L
ˆj :  L AB  0
 AB  0
OB
Lv A2

r
CW
92
5/134 The sliding collar moves up and down the shaft, causing an
oscillation of crank OB. If the velocity of A is not changing as it passes the
null position where AB is horizontal and OB is vertical, determine the
angular acceleration of OB in that position.
aA
OB  0
0
 AB 
vA
CCW
L

aB
OB 
( rA / B )
rOB
(rA / B ) AB
(rA/ B )
 AB  0
  AB  rA / B
(rOB )OB
(rOB )OB
aA  0
aA / B
OB  ( OB  rOB ) AB  ( AB  rA / B )
 OB  rB
ICZV
of AB

2
AB
=0
2
(rA/ B )AB
2
AB

L(
vA 2
)
2
Lv
A
L 
r
r
CW
93
94
vA  vB  DAB  rA / B
Find a D
vA  vB  0.2
vA ˆj  0.2 ˆj  DABkˆ  (0.2sin  iˆ  0.2cos  ˆj)
DAB  0 vA  vB  0.2
ICZV of DAB
is at infinite
a DAB DAB
 AC 
3D vector solution
vA
DAB  0
0.2
 1.6
0.125
CW
0
aA  aB  DAB  (DAB  rA / B )  DAB  rA / B
aA  AC  (AC  rA / C )   AC  rA / C
  14.48o
 DAB
OB  120 rev/min
(constant)
Graphical solution
(0.8  0.32) 2

 2.479 2
0
0.2cos14.48
(a A )t  rA AC
CW
(aA / B )t  rA / BDAB
(0.2 ) 2
aB 
 0.8 2
0.05
2 (120)
vB 
0.05  0.2
60

(aA )n  vAAC  0.32 2
(aB )  0.8 2
aD  aB  DAB  rD / B
 0.8 2iˆ  2.479 2kˆ  0.3( sin  iˆ  cos  j )
95
 0.081 2iˆ  0.186 ˆj
96
97
98
99
5/134 The sliding collar moves up and down the shaft, causing an
oscillation of crank OB. If the velocity of A is not changing as it passes the
null position where AB is horizontal and OB is vertical, determine the
angular acceleration of OB in that position.
aA
OB  0
0
 AB 
vA
CCW
L

aB
OB 
( rA / B )
rOB
(rA / B ) AB
(rA/ B )
 AB  0
  AB  rA / B
(rOB )OB
(rOB )OB
aA  0
aA / B
OB  ( OB  rOB ) AB  ( AB  rA / B )
 OB  rB
ICZV
of AB

2
AB
=0
2
(rA/ B )AB
2
AB

L(
vA 2
)
2
Lv
A
L 
r
r
CW
100
101
 AB  ? aA  ?
vA  vB  vA/ B
AB  0
ICZV at inf.


AC AB
CA 
vA  vB  6 m/s
AB translational.
vA 6
CA 3
aA  aB  aA/ B  aB  AB  AB  rBA    AB  rBA
AC  AC  rCA    AC  rCA  aB  AB  AB  rBA    AB  rBA
 AC (3)

 AB (5)
 AC CD
 (3)(22 )  tan    2
rCA
2
CA
(3)(22 )
 AB
3(22 )(tan  )  2 7


3
3
CW
 AB
1  (3)(22 ) 
 
3
5  sin  
CW
 (3)(2 2 )
(3)(22 )

sin 
a A   AC   AC  rCA    AC  rCA

aB  2
  
 7 
=2kˆ  2kˆ  3 ˆj    kˆ   3 ˆj
 3 
a A  -7iˆ 12 ˆj
 
102
AC  ?  AC  ?
vC  vA  AB  rAC  vrel:C / A
 AB  0

CD vrel:C / A
vrel:C / A  (3)(0.75) tan   3.897
30.75

CD 
vC
9
0.5
CCW
vC  (3)(0.75)  4.5
cos 
CD  CD  rDC   CD  rDC
aC  aA  AB  AB  rAC    AB  rAC  2 AB  vrel:C / A   arel:C / A
2(3)(3.897)

2(3)(3.897)
sin 
 0.75 32
0.5 9
2



2(3)(3.897) 
  40.5  sin   2(3)(3.897)


  0.75 32   
cos  
cos 
sin 
rel:C / A




a
CD 
 40.5
CDrDC  CD (0.5)
2(3)(3.897) 

  40.5 
 tan 
sin  

 128.24881
CD arel:C / A
1 
2(3)(3.897) 
 40.5 
 tan   233.824
0.5 
sin  
CCW
103
104
```