Dynamic Analysis of Mechanical Systems Kinematics
1
3.3 3D Kinematics of Rigid Bodies
3.3.1 Fixed Axis Rotation
A shown in the figure, let’s consider a rigid body that rotates
with angular velocity about a fixed axis n-n. In the Cartesian
frame O XYZ− , the position of a point P on the body can be
located by a position vector r . As you can see, the path of
point P is an arc in the plane normal to
In a small time interval t , the body swings an small angle
t=  . This leads to the change of r
( )
magnitude
direction
sin
sin
sin
t
t t
  
 


=


= = 
r
r r
r
r
r r
r




 


According to the definition of derivative, the velocity of point P
can be expressed as
( )
0
lim
t
t
t t →
 
= = = 
 
rr
v r

 (1)
⊥ ⊥v v r，
Taking time derivative of Eq.(1) by the chain rule leads to the
acceleration of P

( )
t n
= =  +  =  +  
a a
a r r v r r     (2)
where =  ---- the angular acceleration of the body.
t = a r ----the tangential component
( )n =  a r  ----the normal component

Dynamic Analysis of Mechanical Systems Kinematics
2
3.3.2 Fixed Point Rotation
As shown in the Figure, rotation of a body about a fixed point can
be regarded as the rotation about an axis called the instantaneous
axis. Thus, α can be decomposed into the following form
 ⊥= +α α α (3)
where α is aligned with the rotation axis, representing the
change rate of  in magnitude, while ⊥α is normal to the
rotation axis, reflecting the change rate of  in direction.

3.3.3 General Motion
The general motion of a body can be visualized as a combination
of the translation along with a point (the reference point) on the
body and the rotation about an axis passing through the point.
Let A and B are two points on a body. If we take B as the
reference point, the position vector of A can be represented by
A B AB= +r r r , =AB BAr (4)

Taking time derivative of Eq.(4) gives
A B AB= +v v v
ABv --- the velocity of A relative to B.
For pure translation
A B AB=  =v v v 0
For pure rotation with angular velocity about an axis passing
through B
B A AB AB=  = = v v v r0 
For general motion
A B AB= + v v r (5)
Taking time derivative of Eq.(5) leads to the acceleration of A.
( )A B AB AB= +  +  a a r r   (6)



Dynamic Analysis of Mechanical Systems Kinematics
3
For 2D motion in the x y− plane, because kω = and kα = ,
Eqs.(5) and (6) can be simplified as
( )( )
2
A B AB
A B AB AB
B AB AB

  
 
= + 
= +  +  
= +  −
v v k r
a a k r k k r
a k r r
(7)

Eq.(7) can also be expressed by using 2D coordinate vector
notation. Let Ar and Br are the 2D coordinate vectors of points A
and B. Then， AB A B= −r r r can be expressed as
cos

sin
AB
AB AB
AB
l l


 
= = = =  
 
r
r u r u
r
， ， (8)
where = ABl r is the length of ABr , u is the unit vector of ABr ,
 is the position angle of ABr measured from the x axis. Thus,
Eq.(7) can be rewritten as
2 2
A B AB B
A B AB AB B
l
l l
 
   
= + = +
= + − = + −
v v Qr v Qu
a a Qr r a Qu u
(9)
0 1

1 0
   
− 
= = =  
 
Q， ，
whereQ is a 2 2 rotation matrix that rotates a 2D coordinate
vector an angle /2 anticlockwise about the z axis such that
⊥r Qr or T 0=r Qr .
In general, let 1r and 2r be two vectors lying in the x y− plane,
it is easy to see that the hybrid product ( )1 2k r r can be
represented by
( ) ( ) T1 2 2 1 2 1 =  k r r r k r r Qr (10)
The 2D coordinate vector notation will be used in kinematic and
dynamic modeling and analysis of planar systems

Dynamic Analysis of Mechanical Systems Kinematics
4
Sample Problem 1 (2D Problem)
For a 4R four bar linkage shown in the figure, assume link 1
rotates with a constant angular velocity ( )1 rad s . Determine:
 the angular velocity ( 2 , 3 ) and angular acceleration ( 2 , 3 )
of link 2 and link 3 in terms of 1 ;
 velocity and acceleration of point 2P on link 2 and point 3P
on link 3 in terms of 1 .

Solution:
In the Cartesian frame A xy− , the position loop closure
constraint equation can be represented by
B CB D CD+ − − =r r r r 0 (1)
We express these vectors in the form of 2D coordinate vectors
1 2
1 1 1 2 2 2
1 2
3
0 0 0 3 3 3
3
cos cos
,
sin sin
cos1
,
sin0
B CB
D CD
l l l l
l l l l
 
 


   
= =   
   
  
= = =   
   
r u = r u =
r u r u =

Taking time derivative of Eq.(1) leads to the velocity constraint
equation
B CB C+ − =v v v 0
or
1 1 1 2 2 2 3 3 3+l l l  − =Qu Qu Qu 0 (2)
It can be seen that there are two unknowns, i.e. 2 and 3 in
Eq.(2) which also contains two scalar equations.

In order to determine 2 explicitly, we take dot product on both
sides of Eq.(2) with 3u and note that 3 3⊥u Qu . Thus, we have
T T
1 1 3 1 2 2 3 1+ 0l l  =u Qu u Qu (3)
Rearranging Eq.(3) yields the explicit expression of 2 in terms
of 1
( )
( )
T
1 3 11 3 1
2 1 1T
2 3 2 2 3 2
sin
sin
ll
l l
 
  
 
−
= − = −
−
u Qu
u Qu
(4)
Similarly, we take dot product on both sides of Eq.(2) with 2u , and
note that 2 2⊥u Qu . This leads to the explicit expression of 3 in
terms of 1
( )
( )
T
1 2 11 2 1
3 1 1T
3 2 3 3 3 2
sin
sin
ll
l l
 
  
 
−
= = −
−
u Qu
u Qu
(5)


Dynamic Analysis of Mechanical Systems Kinematics
5
Taking point B as a reference point of link 2, the velocity of
point 2P can be determined by
2 2 21 1 1 2
P P B P Bl l = +v Qu Qu (6a)
( )
( )
2
2 2 2
2
2 2
2 2
cos
= , =
sin
P B
P B P B P B
P B
l
 
 
 + 
=  
+ 
r
r u
r

Substituting Eq.(4) into Eq.(6a) leads to
2 2 2
T
1 3 1
1 1 1T
2 3 2
P P B P B
l
l l
l

 
= − 
 
u Qu
v Qu Qu
u Qu
(6b)
In the same manner, taking point D as a reference point of link 3,
the velocity of point 3P can be expressed in term of 1 .
( )
( )
3 3 3 3 3
3
3 3 3
3
T
1 2 1
3 1T
3 2 3
3 3
3 3

cos
= , =
sin
P P D P D P D P D
P D
P D P D P D
P D
l
l l
l
l
 
 
 
 
= =  
 
 +
=  
+ 
u Qu
v Qu Qu
u Qu
r
r u
r
(6c)
Acceleration Analysis:
Differentiating the velocity constraint equation with respect time
gives the acceleration constraint equation
2 2 2
2 2 2 3 3 3 1 1 1 2 2 2 3 3 3l l l l l    − = + −Qu Qu u u u (7)
Taking dot product on both sides of Eq.(7) with 3u and 2u
respectively, leads to


2 T 2 T 2
1 1 1 3 2 2 2 3 3 3
2 T
2 3 2
2 T 2 2 T
1 1 1 2 2 2 3 3 2 3
3 T
3 2 3
l l l
l
l l l
l
  

  

+ −
=
+ −
= −
u u u u
u Qu
u u u u
u Qu
(8)
To this end, the acceleration of point 2P and 3P can be readily
obtained by differentiating Eq.(6) with respect time.
2 2 22 2
3 3 33 3
2 2
1 1 1 2 2
2
3 3
P B P B
P D P D
P P B P B
P P D P D
l l l
l l
  
 
= − + −
= −
a u Qu u
a Qu u
(9)
Important Remarks:
In the velocity and acceleration analysis of four bar planar
linkages, there are always two unknowns involved. It is
advantageous to solve one of the unknowns explicitly by using the
dot product technique. This technique is particularly efficient in
the case when only the output motion is of interest.


Dynamic Analysis of Mechanical Systems Kinematics
6
Sample Problem 2(3D problem)
The bar AB of a spatial mechanism as shown is connected by ball
socket joints to collars sliding on the fixed bars CD and EF. The
point B moves with a constant velocity (m/s)20iv =B .

Determine the velocity and acceleration of the collar A and the
angular velocity and angular acceleration of bar AB.

Solution:
Setup the coordinate system as shown.

Velocity Analysis
Velocity of point A can be expressed by either of two ways
DAAA v ev =
where
AD
AD
DA
rr
rr
e
−
−
= ---- the unit vector of bar CD.
A B AB= + v v r where B Bv=v i
x y z  = + +i j k ----the angular velocity of bar AB.
The velocity constraint equation can be formulated as
ABBDAA vv rie +=  (1)
Now, we have four unknowns but only three equations are
available. Note that the ball socket joint restricts the rotation about
the axis of bar AB. So, an additional constraint equation can be
generated
0AB =r (2)
In order to solve Av , we take dot product on both sides of Eq.(1)
with ABr and noting that ( ) 0AB AB =r r . This leads to
A AB DA B ABv v=r e r i
Hence, Av can be obtained by
AB
A B
AB DA
v v=
r i
r e
(3)
In order to solve ω , we rewrite Eq.(1) as
ier
BDAAAB
vv −= (4)
and we take cross product on both sides of Eq.(4) with
AB
r . This
leads to
( ) ( )ierrr
BDAAABABAB
vv −=  (5)
The use of the Lagrangian identity (see Eq.(10) in Chapter 2)
leads to
( ) ( ) ( )AB AB AB AB AB AB  = −r r r r r r   (6)
y
z
x
400 mm
300 mm
A
B
C
D
E F
( )200, 200, 400 mm−
Bv
Av
Dynamic Analysis of Mechanical Systems Kinematics
7
Substituting Eq.(2) and (6), gives
( )
2
AB AB AB  =r r r  (7)
Substituting Eq.(7) into Eq.(5), finally results in
( )ier
r
BDAAAB
AB
vv −=
2
1
 (8)
Numerical Calculation of Velocity Analysis
( )
kji
kji
rr
rr
e
3
1
3
2
3
2
1.02.02.0
)3.04.0(2.02.0
222
1.0 +−=
+−+
−+−
=
−
−
=

AD
AD
DA
Given
iv 20=B , kir 3.04.0 +−=AB

the velocity of A can be evaluated by Eq.(3)
0.4
20 48 m/s
2 1
0.4 0.3
3 3
AB
A B
AB DA
v v
−
= =  =
−  + 
r i
r e

From Eq.(8) we have
( )ier
r
BDAAAB
AB
vv −=
2
1

3
1
48
3
2
4820
3
2
48
3.004.0
3.04.0
1
22
−−
−
+
=
kji

( )rad/s2.5140.4.38
163212
3.004.04 kji
kji
++=
−
−=

Acceleration Analysis
The acceleration of point A can be expressed in two ways
A A DAa=a e
( )
ABABBA
rraa ++= 
where 0B =a because Bv is a constant vector.
x y z  = + +i j k ---- the angular acceleration of AB
Hence, the acceleration constraint equation can be formulated as
( )A DA AB ABa =  +  e r r   (9)
Since
( ) ( ) ( )
2
AB AB AB AB  = = −r r r r   −   
Then
2
A DA AB ABa =  −e r r  (10)
Taking dot product on both sides of Eq.(10) with
ABr and
noticing that ( ) 0AB AB =r r , leads to
2 2
A AB DA ABa = −r e r
Hence,
Aa can be obtained by
Dynamic Analysis of Mechanical Systems Kinematics
8
2 2
AB
A
AB DA
a = −
r
r e

(11)
In order to solve , we rewrite Eq.(10) as

2
AB A DA ABa = +r e r  (12)
We take cross product on both sides of Eq.(12) with
AB
r and note
that AB AB =r r 0 . This gives
( )AB AB A AB DAa  = r r r e
Expanding
( ) ( ) ( )AB AB AB AB AB AB  = −r r r r r r  
and noting that 0AB =r , finally results in
( )2
1
A AB DA
AB
a= r e
r
 (13)
Do the numerical calculation of acceleration by yourself.











Important Remarks:
In velocity and acceleration analysis of four bar spatial
mechanisms, there are normally four unknowns involved. One is
the output/input velocity or acceleration along/about an axis. The
others are three components of angular velocity or angular
acceleration of the connecting rod.

The procedure to solve the problem of this kind can be divided
into four steps:
(1) Formulate the loop closure constraint equation.
(2) Formulate the additional constraint equation according to the
joint type at each end of the connecting rod. This is the key
point in dealing with the problem of this kind
(3) Solve the output/input velocity or acceleration by using the
projection technique (dot product).
(4) Solve the angular velocity and acceleration of the connecting
rod by using Lagrangian identity.
Please solve the relevant the problem in the sample sheet.













Dynamic Analysis of Mechanical Systems Kinematics
9
3.4 Moving Coordinate System
According to Eq.(5) and Eq.(6) in Section 3.3, the velocity and
acceleration of point A on a body are given by
( )
A B AB
A B AB AB
= + 
= +  +  
v v r
a a r r

  

And we assume that the coordinates of these vectors are evaluated
in an inertial reference frame O XYZ− . However, it would be
more convenient if we express the coordinates of these vectors in
a moving frame as the inertial matrix of a body in the moving
frame becomes a constant matrix (see 3D Dynamics in Chapter 4).


As shown in the figure, by taking point B on the body as the
origin, let’s setup a moving frame B xyz− that rotates with
angular velocity . Note that the B xyz− maybe or maybe not
attached to the body. If the B xyz− is attached to the body,
 = , otherwise   . In the B xyz− , can be represented
by
kji zyx  ++= (1)
where i , j , k are the unit vectors of three axes of the B xyz− ,
, x y z  , are the coordinates of  expressed in the B xyz− . It
is important to note that i , j , k rotate with angular velocity ,
therefore, the time derivatives of i , j , k can be represented as
kkjjii ===   , ,
Taking time derivatives of Eq. (1) by the chain rule, leads to
angular acceleration of the body expressed in the moving frame
B xyz−
( )
d
d
x y z x y z
x y z x y z
x y z
xyzt
     
     
  
= + + + + +
= + + +  +  + 
 
= + + +  = +  
 
i j k i j k
i j k i j k
ω
i j k

  
   
(2)
Remarks:
 The first term in Eq.(2) represents the change rate of  in
amplitude, while the second term represents the change rate of
 in direction with respect to . If the moving frame is
body-fixed, then  =  0 .
 When the moving frame is not attached to the body, it is
convenient to represent by
rel += (3)
rel ---the angular velocity of the body relative to the moving
frame. Eq,(3) is referred to as Addition Theorem of angular
velocity of a rigid body.
Dynamic Analysis of Mechanical Systems Kinematics
10
According to the addition theorem, angular velocity of the end
link (link n) of an n-DOF serial kinematic chain can be expressed
as the sum of the angular velocity of link i relative to link 1i −
, 1
1
n
n i i
i
−
=
=ω (4)
where , 1 1i i i i− −= −ω ω ω .


IF the chain is composed of n revolute joints as shown in the
figure, we can also obtain the general expression of the velocity of
point 1nA + on link n by using the addition theorem (see the
complementary reading below).


1 , -1
1 1
n
n n
A i i i i i i
i i

+
= =
=  =  v q ω q s (5)
where
i ---the magnitude of , 1i i−ω
is ---the unit vector of , 1i i−ω whose direction is coincident with
thi joint axis
1i n iA A+=q --- the position vector from 1nA + to an arbitrary point
iA on the thi joint axis.


ATTENTION: The coordinates of all vectors should be expressed
in the same frame when the vector addition is made.




Dynamic Analysis of Mechanical Systems Kinematics
11
Complementary readings
Consider an DOFn − ( )6n  serial kinematic chain, which is
composed of n revolute joints and n movable links. Let i and
is be the magnitude and unit vector of the thi revolute joint
connecting link i with link 1i − , and +1i i iA A=r be the position
vector from iA to +1iA where iA is an arbitrary point on the
thi joint axis. Determine the velocity of point 1nA + on the end
link n .


Consider first the case where 2n =
2 1 11 1 1 1 1
as A A A= +  =  =v v ω r s r v 0 (1)
3 2 2
2
2 2 2
1
A A A i i
i=
 
= +  = +  
 
v v ω r v s r (2)
Substituting Eq.(1) into Eq.(2), results in
( )
( )
( )
3
2
1 2
1
2
1 1 2 1
1
1 1 1 1 2 1 2 2 2
1 2 2 2 2
2
1
A i i i i
i
i i i i
i
i i
i i
i i i
i
=
=
=
 
=  +  
 
 
=  +  + − 
 
=  +  + − + 
=  + + 
= 



v s r s r
s r s r r r
s r s r r r s r
s r r s r
q s
 
 
  
 

(3)
where
( )1 1 2 3 1A A= − + =q r r , 2 2 3 2A A= − =q r
Combining Eq.(3) with the addition theorem
2
2
1
i i
i

=
=ω s ,leads to
3
2
12
A i i
i
i i

=
   
=   
  

v q s
sω
(4)
In general
1
1 1
ˆn
n n
A i i
i i i
i iin
 +
= =
   
= = =   
  
 
v q s
ξ ξ
sω
(5)
ξ ——The twist of link n about point 1nA + .
ˆ
iξ ——The unit twist of joint i about point 1nA +
Conclusion: The twist of link n about point 1nA + can be expressed
as the linear combination of unit twists of all joints about
point 1nA + .
Think: The case where a prismatic joint is used.
Implementation of acceleration analysis.


Dynamic Analysis of Mechanical Systems Kinematics
12
Sample Problem 1
The bar OA with the length of
l rotates about the x-axis with
constant rate  . The entire
assembly rotates about the
z-axis with constant rate .
For the given angle  ,
determine the angular
velocity and acceleration of
bar OA, and the velocity and
the acceleration of point A.

Solution
Establish the moving frame
xyzO − as shown.
The angular velocity and acceleration analysis of bar OA,
The angular velocity of the moving frame
k=
The angular velocity of the bar OA relative to the moving frame
i=rel
Thus, the angular velocity of the bar can be expressed as
ki  += 
Because  and  are constants, taking time derivative of 
leads to the angular acceleration of the bar OA
( ) jkikω   =+== 
The velocity and acceleration analysis of point A
The position vector of point A can be represented by
kjr  sincos ll +=
For a fixed point rotation, the velocity of point A can be expressed
as
kji
kji
rv 

 cossincos
sincos0
0  lll
ll
+−−===
The acceleration of A can be represented by
=  + a r v 
where
( ) ikjjr  sinsincos  lll =+=
( ) kji
kji
v



sincossin
cossincos
0
222 


lll
lll
−+−=
−−
=

The acceleration of point A can finally be obtained by
( ) kjia   sincossin2 22  lll −+−=




z
O



l

x
y
Dynamic Analysis of Mechanical Systems Kinematics
13
Sample Problem 2
A shown in the figure, the
bent bar is rigidly attached to
the vertical shaft which
rotates with a constant rate
0 .The circular disk with
radius R is pinned to the bent
bar and undergoes a pure roll
on the horizontal surface.
Determine the disk’s angular
velocity disk and angular
acceleration disk

Solution
Setup a moving frame B-xyz
fixed to the bent bar where
the x axis is coincident with
horizontal part of the bar and
the y axis is coincident with
rotation axis of the vertical shaft.
Since the B xyz− is attached to the bar, then
0bar = = j 
While, the angular velocity of the disk relative to the B xyz−
can be formulated by
( )cos sinrel rel  = − +i j
rel --- the magnitude rel is the unknown. According to addition
theorem, the angular velocity of the disk can be expressed as
( )0+ cos sindisk bar rel rel rel    = = − + +i j  
Now, we need to establish a constraint equation for solving rel .
Because of the rolling without slipping, point C is the
instantaneous center of zero velocity of the disk. This means
C A disk CA= +  =v v r 0
where
( )0
0
( cos ) sin
( cos )
A bar AB h b b
h b
  
 
=  =  + −
= − +
v r j i j
k


and the position vector from A to C can be formulated by
sin cosCA R R = − −r i j
Thus
0 0( cos ) cos sin 0
sin cos 0
C A disk CA
rel relh b
R R
      
 
= +  =
− + + − + =
− −
v v r
i j k
k
 0
0

Expanding this equation and equating its k component, leads to
0 cos sinrel
h b
R R
   
 
= + − 
 

Since the coordinates of disk in the B xyz− are constants,
( )0+ cos sindisk bar rel rel rel    = = − + +i j  
the angular acceleration of the disk can finally be obtained by
( )( )0 0
0
cos sin
cos
disk disk
rel rel
rel
     
  
= 
=  − + +
=
j i j
k
  



Dynamic Analysis of Mechanical Systems Kinematics
14
Sample Problem 3
As shown in the figure,
the collar and clevis A
are given a constant
upward velocity 0.2m/s.
That causes the ball end
of the bar AB to slide in
the radial slot in a
rotating disk. Suppose
the disk rotates at
constant rate
2 rad/s = , determine
the angular acceleration
α of the bar AB when the z-coordinate of A reaches 75 mmz = .

Solution:
Establish the moving frame xyzO − attached to the disk as
shown in the figure.
The angular velocity of the xyzO − is given by
=Ω k (1)
The angular velocity of bar AB relative to the xyzO − is given by
rel rel= −ω i (2)
The angular velocity of bar AB is given by
rel rel = + = − +ω Ω ω i k (3)
In order to determine rel , the velocity constraint equation due to
the loop closure of the mechanism can be formulated by
B A BA= + v v ω r (4)
where
BA B A y z= − = −r r r j k (5)
A z=v k (6)
B y y= +v j j (7)
Substituting = j Ω j and =Ω k into Eq.(7), gives
B y y= −v j i (8)
Substituting Eqs.(5), (6) and (8) into Eq.(4) and equating the j and k
components allows rel to be expressed as
rel rel
z y
y z
 = = −， (9)
Then, substituting the first expression into Eq.(3) leads to the
expression of the angular velocity of the bar AB in terms of z , y
and
rel
z
y
= + = − +ω Ω ω i k , 2 2y l z= − (10)
Hence, taking time derivative of ω leads to the angular
acceleration of the bae AB
( )
( )
2 2
2 2
zy z zy z
y y y y
zy z zy z
y y y y
 
= − = − 
= −  = −
i i i Ω i
i k i i j

(11)
Finally, substituting ( )y z y z= − , 22 zly −= , and the
assumed date 0.25 m/sz = , 75 mmz = , 2 rad/s = ,
125 mml = into Eq.(11), results in

23 4 (rad/s )= − −i j (12)



Dynamic Analysis of Mechanical Systems Kinematics
15
Sample Problem 4
As shown in the figure,
the motor turns the disk at
the constant rate
30 rad/sp = . The motor
is also rotating about the
horizontal axis BO at a
constant rate 2 rad/s = .
Simultaneously, the entire
assembly is rotating about
the vertical axis C-C at a
constant rate 8 rad/sq = .
For the instant 30 = ,
determine the angular
velocity and angular
acceleration of the disk.

Solution
According to the addition theorem, the angular velocity of the disk
diskω can be expressed as
/ /disk shaft motor shaft disk motor= + +ω ω ω ω (1)
shaftω --- the angular velocity of the horizontal shaft BO
/motor shaftω ----the angular velocity of the motor relative to the
horizontal shaft BO
/disk motorω ------the angular velocity the disk relative to the motor

Setup the moving frame xyzO − attached to the motor as shown
such that
motor=Ω ω and /rel motor shaft=ω ω . In the xyzO −

kiω  cossin qqshaft +=
/motor shaft = −ω j , /disk motor p=ω i
Thus
sin cosmotor q q  = = − +ω Ω i j k
and
( )/ sin cosdisk disk motor p q q  = + = + − +ω Ω ω i j k (2)

It can be seen from Eq. (2) that p , q and are constants but
 varies with time. So, taking time derivative of diskω gives the
angular acceleration of the disk diskα
( ) ( )
( )
d d d
sin cos
d d d
cos sin sin cos
sin cos
cos cos sin
disk motor diskp q q
t t t
q q q q
p q q
q pq p q

 
      
  
    
    
= + − + +     
    
= − + −
+ −
= + + −
α i j k ω ω
i j k
i k
i j k
(3)

Substituting the assumed data into this equation, finally results in
( )
2
cos cos sin
(8 2 .866) 30 8 .866 2 (30 8 .5)
13.4 208 52 (rad/s )
disk q pq p q    = + + −
=   +   +  − 
= + +
α i j k
i j k
i j k
(4)




Dynamic Analysis of Mechanical Systems Kinematics
16
Sample Problem 5
The figure shows a 5-DOF serial robot. The arm 1 rotates about the
z-axis at the constant rate
1 , the arm 2 rotates about x-axis at
constant rate 2 = ，the arm 3 rotates about the axis 21 OO − at
constant rate 3 , and the arm 4 rotates about the axis which is
through 2O and momentarily parallel to the x-axis at the constant
rate  =4 . Finally, the arm 5 rotates about the axis AO −2 at the
constant rate 5 . For the given configuration as shown, evaluate
the angular velocity and angular acceleration of the arm 5 in the
moving frame 1O xyz− attached to the arm 1.

Solution:
According to the addition theorem, the angular velocity of the arm
5 can be expressed by
5
5 , 1
1
i i
i
−
=
=ω ω (1)
where , 1i i−ω denotes the angular velocity of arm i relative to arm
1i − ( 1~5i = )

At the instant shown, , 1i i−ω evaluated in the 1O xyz− can be
formulated by
1,0 1=ω k
2,1 2 = =ω i i
3 2 3 3cos sin   = +ω j k， (2)
4 3 4 = − = −ω i i，
( ) ( )5,4 5 5cos sin     = − + −ω j k
Substituting these expressions into Eq.(1) leads to
( ) ( )
( ) ( )( ) ( )( )
5 1 3 3
5 5
3 5 1 5
cos sin
cos sin
cos cos sin
      
     
          
= + + + −
+ − + −
= − + + − + + −
ω k i j k i
j k
i j k
(3)
Do the angular acceleration analysis of the robot by yourself






Dynamic Analysis of Mechanical Systems Kinematics
17
Complementary readings
Based upon the velocity and acceleration expressions of point A
on a rigid body provided that point A and point B are on the same
body,

AB
ABBA
v
rvv +=  (1)
( )
AB
A B AB AB= +  +  a a r r
a
   (2)
we now consider more general case where point A and point B are
not on the same body.


As seen from the figure, setup a moving frame B xyz− and take
point B as the reference point. The point A can be expressed as
A B AB= +r r r , AB x y z= + +r i j k (3)
where , ,x y z are the coordinates of vector
ABr expressed in the
B xyz− , , ,i j k are the unit vectors of three axes of the B xyz−
Taking time derivative of Ar , leads to
AB x y z x y z= + + + + +v i j k i j k
, , =  =  = i Ω i j Ω j k Ω k
Thus
AB AB rel=  +v Ω r v (4)
where
rel x y z= + +v i j k ——velocity of point A relative to the B xyz−
as observed in the B xyz− .

Thus, the velocity of point A can be expressed as
A B AB rel= +  +v v Ω r v (5)
Taking time derivative of
Av by the chain rule leads to the
acceleration of point A
dd d
d d d
relAB
A B AB
t t t
= +  +  +
vΩ r
a a r Ω
where
( )
d
d
AB
AB AB rel
t
 =  =   +
r
Ω Ω v Ω Ω r v
d
d
rel
rel rel
t
= + 
v
a Ω v ,
rel x y z= + +a i j k
Thus
( ) 2A B AB AB rel rel= +  +   +  +a a Ω r Ω Ω r Ω v a (6)

If =Ω ω , and rel relv a are zero vectors. Thus, Eq.(5) is identical to
Eq.(1), and Eq.(6) is identical to Eq.(2)


Dynamic Analysis of Mechanical Systems Kinematics
18
Example 6
The motor housing and its bracket rotate about the Z-axis at the
constant rate 3 rad/s= . The motor shaft and disk have a constant
rate 8 rad/sp = with respect to the motor housing in the direction
shown. Given 30= , determine the velocity and acceleration of
point A at the top of the disk at the instant shown.


Solution
Setup the moving frame B xyz− attached to the motor
cos sin = −K i j k  , cos = −K j i , sin =K k i
Note that there is no relative rotation between the motor housing
and bar OB, so Ω=Ω K

Velocity of A is given by
A B AB rel= +  +v v Ω r v
where
, ,B OB Y OB YΩ r Ωr=  = −v K J i
( )
( )
, ,
, ,cos sin
AB AB y AB z
AB y AB z
Ω r r
Ω r r 
 =  +
= − +
Ω r K j k
i

( ), , ,rel AB AB y AB z AB zp r r pr=  =  + =v p r j j k i
Thus
( )( ), , , ,cos sin
A B AB rel
AB y AB z OB Y AB zΩ r r r pr 
= +  +
= − + − +
v v Ω r v
i

Do the acceleration analysis by yourself using Eq.(6).

















O
C