1234567890-Chapter 11b_Dynamic Force Analysis.pptx

1. 1

2. 2
If a moving links in a mechanism is running with considerable amount of linear
and/or angular accelerations, inertia forces are generated and these inertia
forces also must be overcome by the driving motor as an addition to the forces
exerted by the external load or work the mechanism does.
So, and are no longer applicable. The consideration that says all
forces on components of machine systems are in balance (static and dynamic
equilibrium ) is seldom in real machines (except when the machine is stopped)
I and m are inertial (bodily) properties. At this stage we need to know the
description of the inertial properties.
0

F

0

τ

a
F


m


α
τ


I


Governing rules will be:

3. 3
Of course techniques for static force analysis are important, not only
because stationary structures must be designed to withstand their imposed
loads, but also because they introduce concepts and approaches that can be
built upon and extended to non equilibrium conditions .
Therefore the purpose of this chapter is to learn how much acceleration will
result from a system of unbalanced forces and also learn how these dynamic
forces can be assessed for systems that are not in equilibrium

4. 4
Centroid is the point where the resultant of distributed force
system is assumed to act and generate the same dynamic
results.
C
entroid
R
esultant F
orce

5. 3
2
1
3
3
2
2
1
1
m
m
m
m
x
m
x
m
x
x





5
If the distributed force is gravity force acting on each particle of mass, then
concentrated force itself is called the “weight” and the centroid is called the
“center of gravity” or “mass center”.
Mass times distance, mr, is called as the first mass moment. This concept of
first mass moment is normally used in deriving the center of mass of a system
of particles or a rigid body. In figure a series of masses are located on a line.
The center of mass or centroid is located at
x3
x1
x2
x
-
m
1 m
2 m
3
G






n
i
i
n
i
i
i
m
m
x
x
1
1

6. 6
The coordinates of the masses located on a plane can be obtained as:
m
1
m
2
m
3
x
-
y
-
G
3
2
1
3
3
2
2
1
1
1
1
m
m
m
m
x
m
x
m
x
m
m
x
x n
i
i
n
i
i
i










3
2
1
3
3
2
2
1
1
1
1
m
m
m
m
y
m
y
m
y
m
m
y
y n
i
i
n
i
i
i











7. 



 n
i
i
n
i
i
i
m
m
z
z
1
1
7
This procedure can be extended to masses concentrated in a volume by simply
writing an equation for the z axis. A more general form of mass center location
for three dimensional body can be obtained by using integration instead of
summation. The relations then become




 n
i
i
n
i
i
i
m
m
x
x
1
1




 n
i
i
n
i
i
i
m
m
y
y
1
1
m
xdm
x


m
ydm
y


m
zdm
z



8. 8
Mass moment of inertia is the name given to rotational inertia, the rotational
inertia analog of mass for linear motion. It appears in the relationships for the
dynamics of rotational motion. The Mass Moment of Inertia of a solid measures
the solid's ability to resist changes in rotational speed about a specific axis.
r
o
o
m
The moment of inertia for a point mass is just the mass times the square of
perpendicular distance to the rotation axis. The mass moment of inertia for a
single particle is given as:
m
r
I 2
00 

9. 9
When calculating the mass moment of inertia for a rigid body, one thinks of the
body as a sum of particles, each having a mass of dm. Integration is used to
sum the moment of inertia of each dm to get the mass moment of inertia of
body. The equation for the mass moment of inertia of the rigid body is

 
 dm
r
dm
r
I 2
2
00
r
o
o
dm

10. 10
The integration over mass can be replaced by integration over volume, area, or
length. For a fully three dimensional body using the density r one can relate the
element of mass to the element of volume.
x
y
dm
z
 dm
z
y
dm
r
I x
xx   

 2
2
2
 dm
z
x
dm
r
I y
yy   

 2
2
2
 dm
y
x
dm
r
I z
zz   

 2
2
2
rz
ry
rx
These three integrals are called the
principle mass moment of inertia of the
body. Another three similar integrals are



 xydm
I
I yx
xy



 yzdm
I
I zy
yz



 xzdm
I
I zx
xz
}mass products
of inertia of the
body

11. G
IC
G
, m
IC
G
, m
O
rigin
a
l B
od
y
k
M
o
de
l
11
Sometime in place of the mass moment of inertia the radius of gyration k is
provided. The mass moment of inertia can be calculated from k using the
relation
where m is the total mass of the body. One can interpret the radius of gyration as
the distance from the axis that one could put a single particle of mass m equal to
the mass of the rigid body and have this particle have the same mass moment of
inertia as the original body.
2
mk
I 
k
M
o
de
l

12. G
ICG
,m
ICG
,m
o o
d
12
The moment of inertia around any axis can be calculated from the moment of
inertia around parallel axis which passes through the center of mass. The
equation to calculate this is called the parallel axis theorem and is given as
2
md
I
I CG 


13. 13
Solution: Mass moments of inertial of a point mass about an axis passing
through itself
What are the mass moments of inertial of a point mass about an axis passing
through itself and about an axis r distance away from it ?
o
o
m
  

 0
0
2
dm
dm
r
Ioo

14. 14
Solution: Mass moments of inertial of a point mass about an axis passing
through itself
What are the mass moments of inertial of a point mass about an axis passing
through itself and about an axis r distance away from it ?
o
o
m
  

 0
0
2
dm
dm
r
Ioo
o
o
r
m
x
x
m
r
m
r
m
r
I
I xx
oo
2
2
2
0 




Mass moments of inertial of a point mass about an
axis r distance away from it. Using parallel axis
theorem

15. 15
Solution:
Find the mass moment of inertia of a
slender rod of length L (slender rod means
that it has a length, and the remaining
dimensions are negligible small) about an
axis perpendicular to the rod and passing
through its mass center.
L
L/2
o
o
L
L/2
o
o
dx
x


2
/
0
2
*
2
L
oo dm
x
I
Let density of the material is r in kg/m.
Then, infinitesimal mass dm=rdx.
Substituting this into above equation,
 

2
/
0
2
*
2
L
oo dx
x
I r
12
2
mL

2
/
3
|
3
2 L
o
x
r
3
2
3
2 






L
r

16. 16
An uniform steel bar shown in the figure is used as an oscillating cam follower.
Drive the equation of mass moment of inertia of the follower about an axis
through O. Use the density of steel r=7800 kg/m3.
k=2kN
/m
2 cm 25 cm 50 cm 25 cm
5 cm
y
x
O

17. 2
2
2
y
x
r 

17
Solution:
x
y
z
x
y
r
dx
dz
dy
t
w
l

 dm
r
Izz
2
dxdydz
dm r

 
 dxdydz
y
x
Izz r
)
( 2
2

18. x
y
z
dx
t
w
l
18
Solution:
x
y
z
dy
t
w
l

 
 dy
l
t
y
dx
w
t
x
Izz *
*
*
*
*
*
*
* 2
2
r
r

 

2
/
0
2
2
/
0
2
*
*
*
*
2
*
*
*
*
2
w
l
dy
y
l
t
dx
x
w
t r
r

19. 19
Solution:
r
*
*
* t
w
l
m 
12
12
*
12
*
2
2
2
2
w
l
m
w
m
l
m
Izz














































3
2
*
*
*
*
2
3
2
*
*
*
*
2
3
3
w
l
t
l
w
t
Izz r
r 12
*
*
*
*
12
*
*
*
*
2
2
w
w
l
t
l
l
w
t
Izz r
r 

mass moment of inertia about O can be found by parallel axis theorem
48
4
7
4
12
*
2
2
2
2
2
2
2
0
w
l
m
l
m
w
l
m
d
m
I
I zz







m
kg
I .
139
.
1
48
05
.
0
*
4
1
*
7
8
.
7
2
2
0 



20. 20

21. 21
The slender bar of mass m is released form rest in the horizontal position as
shown. At that instant, determine the force exerted on the bar by the support A.
Freebody Diagram
Equations Of Motion
G
a
m
F






I
M 


22. 22
The slender bar of mass m is
released form rest in the
horizontal position as shown. At
that instant, determine the force
exerted on the bar by the
support A.
Kinematics of the slender rod
?
? 


 y
x
G a
a
a



?



Freebody Diagram
Equations Of Motion
G
a
m
F







I
M 

x
x ma
F 

y
y ma
F 


A
A I
M 


23. 23
The slender bar of mass m is
released form rest in the
horizontal position as shown. At
that instant, determine the force
exerted on the bar by the
support A.
Kinematics of the slender rod
?
? 


 y
x
G a
a
a



?



)
x
(
x r
ax







r
ay



x


r
At the instant bar is released, its angular velocity 0



0

x
a

j
l
i
l
k
ay






2
1
2
1
x 









0

x
a

l
ay
2
1



24. 24
Freebody Diagram
Equations Of Motion

A
A I
M 

0
0 




x
x
x
x
A
m
A
ma
F















l
m
mg
A
ma
mg
A
ma
F
y
y
y
y
y
2
1
4
mg
Ay 
2
2 2
2
12 2 3
A G
ml l ml
I I md m
 
    
 
 
l
g
ml
l
mg
2
3
3
2
1 2











0

x
a

l
ay
2
1



25. 25
D’Alembert’s principle permits the reduction of a problem in dynamics to one in
statics. This is accomplished by introducing a fictitious force equal in magnitude
to the product of the mass of the body and its acceleration, and directed opposite
to the acceleration. The result is a condition of kinetic equilibrium.
fictitious force and torque
0




 a
a
a
F




m
m
m
0




 α
α
α
τ




I
I
I
The meaning of the equation; i.e. indication of a dynamic case still holds true,
but equation, having zero on right hand side becomes very easy to solve, like
that in a “static force analysis” problem.
CG
SF
m, I
a

CG
SF
m, I
a

-ma
I

26. 26
1. Do an acceleration analysis and calculate the linear acceleration of the
mass centers of each moving link. Also calculate the angular
acceleration of each moving link.
2. Masses and centroidal inertias of each moving link must be known
beforehand.
3. Add one fictitious force on each moving body equal to the mass of that
body times the acceleration of its mass center, direction opposite to its
acceleration, applied directly onto the center of gravity, apart from the
already existing real forces.
4. Add fictitious torque on each moving body equal to the centroidal inertia
of that body times its angular acceleration, direction or sense opposite to
that of acceleration apart from the already existing real torques.
5. Solve statically.

27. 4
B
2
A
3
G
3
x

27
AB=10 cm, AG3=BG3=5 cm, =60o
m2=m4=0.5 kg, m3=0.8 kg, I3=0.01 kg.m2
In the figure, a double- slider mechanism working in horizontal plane is shown.
The slider at B is moving rightward with a constant velocity of 1 m/sec. Calculate
the amount of force on this mechanism in the given kinematic state.
B
A
B
A V
V
V






?

A
V

 s
m
VB /
1
AB
to
V
B
A 
 ?
B
V
A
V
B
A
V

 s
m
VA /
5774
.
0
s
m
V
B
A /
1547
.
1


28. 4
B
2
A
3
G
3
x

28
In the figure, a double- slider mechanism working in horizontal plane is shown.
The slider at B is moving rightward with a constant velocity of 1 m/sec. Calculate
the amount of force on this mechanism in the given kinematic state.
B
A
B
A V
V
V





B
to
A
from
s
m
AB
V
a B
A
n
B
A
2
2
2
/
33
.
13
1
.
0
1547
.
1




?

A
a
B
A
B
A a
a
a





0
B
A
A a
a


 t
B
A
n
B
A a
a




AB
to
at
B
A

 ?

29. 4
B
2
A
3
G
3
x

29
B
to
A
from
s
m
AB
V
a B
A
n
B
A
2
2
2
/
33
.
13
1
.
0
1547
.
1




?

A
a
B
A
A a
a


 t
B
A
n
B
A a
a




AB
to
at
B
A

 ?
2
/
698
,
7 s
m
at
B
A

t
B
A
a
A
a
2
/
396
.
15 s
m
aA 
G3
3
G
a
n
B
A
a
2
/
698
.
7
2
/
3
s
m
a
a A
G 

3

AB
at
B
A

CCW
s
rad
AB
at
B
A
2
3 /
98
.
78
1
.
0
698
.
7





30. 4
B
2
A
3
G
3
x

30
t
B
A
a
A
a
2
/
396
.
15 s
m
aA 
G3
3
G
a
n
B
A
a
2
/
698
.
7
2
/
3
s
m
a
a A
G 

CCW
s
rad
AB
at
B
A
2
3 /
98
.
78
1
.
0
698
.
7




3
G
ma

A
ma

3

I

D’Alembert forces and moments
0
90
698
.
7
396
.
15
*
5
.
0 


 N
maA
0
90
1584
.
6
698
.
7
*
8
.
0
3



 N
maG
CW
Nm
I 7689
.
0
98
.
76
*
01
.
0
3 

 

31. 4
B
2
A
3
G
3
x

31
N
698
.
7
N
1584
.
6
Nm
7689
.
0
B
F 4
B
2
A
3
G
3
N
698
.
7
N
1584
.
6
Nm
7689
.
0
B
F
12
F
14
F

32. 32
4
B
2
A
3
G
3
N
698
.
7
N
1584
.
6
Nm
7689
.
0
B
F
12
F
14
F
 



 12
12 0
;
0 F
F
F
F
F B
B
x
N
F
F
Fy
86
.
13
0
1584
.
6
698
.
7
;
0
14
14







N
F
F
MB
11
.
15
0867
.
0
3087
.
1
0
60
sin
*
1
.
0
*
60
cos
*
05
.
0
*
1584
.
6
60
cos
*
1
.
0
*
698
.
7
7698
.
0
;
0
12
12










 N
FB 11
.
15
+
x
y

33. 4
B
2
A
3
G
3
x

33
3
G
ma

A
ma

3

I

0
90
698
.
7
396
.
15
*
5
.
0 


 N
maA
0
90
1584
.
6
698
.
7
*
8
.
0
3



 N
maG
CW
Nm
I 7689
.
0
98
.
76
*
01
.
0
3 

 
3
G
ma

3

I

3
G
ma

3
G
ma

h
h
ma
I G *
3
3 
 

3
3
G
ma
I
h


m
h 125
.
0
1589
.
6
7689
.
0


h
3
G
ma

3
G
ma


34. 4
B
2
A
3
G
3
x

34
h
N
1584
.
6
N
698
.
7
B
F 4
B
2
A
3
G
3
N
698
.
7
B
F
12
F
14
F
m
h 125
.
0
1589
.
6
7689
.
0


h
N
1584
.
6

35. 35
4
B
2
A
3
G
3
N
698
.
7
B
F
12
F
14
F
h
N
1584
.
6
 



 12
12 0
;
0 F
F
F
F
F B
B
x
N
F
F
Fy
86
.
13
0
1584
.
6
698
.
7
;
0
14
14







N
F
F
MB
11
.
15
0867
.
0
3087
.
1
0
60
sin
*
1
.
0
*
60
cos
*
1
.
0
*
698
.
7
60
cos
*
)
125
.
0
05
.
0
(
*
1584
.
6
;
0
12
12










 N
FB 11
.
15
+
x
y

36. At the crank angle shown and asuming that gravity
and friction effects are negligible, find all the
constraint forces and the driving torque required
to produce the velocity and acceleration
conditions specified.
36

37. Gaziantep University
Example 15.4:
ˆ
RAO= 60 mm, RO
2 ˆ
m4 = 5 kg, IG = 0.025 kg · m2, IG = 0.012 kg · m2, IG = 0.054 kg · m2, α 2 = 0, α 3 = -119k rad/s2, α 4 = -625k rad/s2, AG = 162∠-
2 3 4 3
73.2° m/s2, AG = 104∠233° m/s2, FC = -0.8 j kN.
4
Figure 15.9
Calculate the inertia forces and inertia torques
−m A =0
2 G2
−m A =−(1.5)(46.8
i−155j)=−70.2i+233j(N)
3 G3
−m A =−(5.0)(−62.6i −83.1j)=−313i+415j(N)
4 G4
−I α =0
G2
2
−I α =−(0.012)(
−119k) =1.43k(N ⋅m)
G3 3
−I α =−(0.054)(−625k)=33.8k(N ⋅m)
G4
4
15-12

38. Gaziantep University
Considering the free-body diagram of link 4 and 3 respectively,
M = R ×(−m A ) +(−I α ) + R ×F + R ×F = 0
∑ O4 G4O4 4 G4 G4 4 CO4 c BO4 34
M = R ×(−m A ) +(−I α ) + R ×F = 0
∑ A G3A 3 G3 G3 3 BA 43
x y
M = 25.2k +33.8k −96k + (−125F +83F )k = 0
∑ O4 34 34
x y
M =18.6k +1.43k + (70.5F − 208F )k = 0
∑ A 34 34
F =− 300i −39j
34
F = F +F +F +(−m A ) = 0
∑ i4 14 34 C 4 G
4
Summing forces on the link 2,3,4,
F =− 13i +390j(N)
14
F = F +F +(−m A ) = 0
∑ i3 23 43 3 G3
F =− 230i −238j(N)
23
F = F +F +(−m A ) = 0
∑ i2 12 32 2 G2
F = F =− F
12 23 32
M = R ×F +M +(−I α ) = 0
∑ O AO 32 12 G 2
2 2 2
M =− R ×F =18.6k(N ⋅m)
12 AO2 32
15-13

39. Gaziantep University
15.5 The Principle of Superposition
Linear System: the response or output of a system is directly proportional to the drive or input to the system.
In the absence of Coulomb or dry friction, most mechanisms are linear for force analysis purpose.
The principle of superposition: for linear systems the individual responses to several disturbances or driving
functions can be superposed on each other to obtain the total response
of the system.
Example:
nonlinear factor: static or Coulomb friction, systems with clearances or backlash
systems with springs that change stiffness as they are deflected
Complete dynamic force analysis of a planar motion mechanism:
(1) make a kinematic analysis of the mechanism. (2) make a
complete static force analysis of the mechanism.
(3) calculate the inertia forces and inertia torques for each link or element of the mechanism.
make another complete force analysis of the mechanism.
(4) add the results of steps 2 and 3 to obtain the resultant forces and torques on each link.
15-14

40. 40
Crank AB of the mechanism shown is balanced such that the mass center is at A.
Mass center of the link CD is at its mid point. At the given instant, link 4 is
translating rightward with constant velocity of 5 m/sec. Calculate the amount of
motor torque required on crank AB to keep at the given kinematics state.
1
4
2
3
A
C
B
D
=45
F
AC=10 cm, CB=BD=4 cm,
AF=2 cm CG3=4 cm,
m2=m3=m4=5 kg,
I2=I3=I4=0.05 kg-m2

41. 41
Crank AB of the mechanism shown is balanced such that the mass center is at A. Mass center of the link CD is at its mid
point. At the given instant, link 4 is translating rightward with constant velocity of 5 m/sec. Calculate the amount of motor
torque required on crank AB to keep at the given kinematics state.
1
4
2
3
A
C
B
D
=45
F
AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=4
cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2
C
B
C
B V
V
V





AB
to
VB 
 ?

 sec
/
5 m
VC
BC
to
V
C
B 
 ?
s
m
VC /
5

5 m/s
s
m
V
C
B /
05
.
5

s
m
VB /
85
.
3


42. 42
Crank AB of the mechanism shown is balanced such
that the mass center is at A. Mass center of the link
CD is at its mid point. At the given instant, link 4 is
translating rightward with constant velocity of 5 m/sec.
Calculate the amount of motor torque required on
crank AB to keep at the given kinematics state.
1
4
2
3
A
C
B
D
=45
F
AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=5
cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2
C
B
C
B a
a
a




 0

C
a
n
C
B
n
C
B
t
B
n
B a
a
a
a 




A
to
B
from
s
m
AB
V
a B
n
B
2
2
2
/
5
.
192
077
.
0
85
.
3



BC
to
at
C
B 
 ?
C
to
B
from
s
m
BC
V
a C
B
n
C
B
2
2
2
/
6
.
637
04
.
0
05
.
5



AB
to
at
B 
 ?
2
/
5
.
192 s
m
an
B 
2
/
6
.
637 s
m
an
C
B 
2
/
483 s
m
at
C
B 
2
/
776 s
m
at
B 

43. 43
Crank AB of the mechanism shown is balanced such
that the mass center is at A. Mass center of the link
CD is at its mid point. At the given instant, link 4 is
translating rightward with constant velocity of 5 m/sec.
Calculate the amount of motor torque required on
crank AB to keep at the given kinematics state.
1
4
2
3
A
C
B
D
=45
F
AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=5
cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2
?
?,
?, 3
2 

 B
a


2
/
5
.
192 s
m
an
B 
2
/
6
.
637 s
m
an
C
B 
2
/
483 s
m
at
C
B 
2
/
776 s
m
at
B 
CCW
s
rad
AB
at
B
2
2
2 /
10078
077
.
0
776
* 


 

CW
s
rad
BC
at
C
B
2
3
3 /
20000
04
.
0
800
* 


 

B
a

262
/
800 2

 s
m
aB
B
3
2

44. 44
D’Alembert forces and moments
1
4
2
3
A
C
B
D
=45
F
2
/
5
.
192 s
m
an
B 
2
/
6
.
637 s
m
an
C
B 
2
/
483 s
m
at
C
B 
2
/
776 s
m
at
B 
CCW
s
rad 2
2 /
10078


CW
s
rad 2
3 /
20000


B
a

262
/
800 2

 s
m
aB
B
I33
I22

82
4000
800
5
3 


 N
*
a
m B
CW
Nm
*
.
I 504
10078
05
0
2
2 

 
CCW
Nm
*
.
I 1000
20000
05
0
3
3 

 
m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2
-m3aB


45. 45
1
4
2
3
A
C
B
D
=45
F
I33
I22
-m3aB

4
2
3
A
C
B
D
=
45
F
C
B
A
2
2
°
504 Nm

4000 N
1000 Nm
Cy
F
Cx
F
Bx
F By
F
By
F
Bx
F
Cy
F
Cx
F
Ax
F
Ay
F
4
A
F
A4

46. 46
4
2
3
A
C
B
D
=
45
F
C
B
A
2
2
°
504 Nm

4000 N
1000 Nm
Cy
F
Cx
F
Bx
F By
F
By
F
Bx
F
Cy
F
Cx
F
Ax
F
Ay
F
4
A
F
A4
 



 Bx
Ax
Bx
Ax
x F
F
F
F
;
F 0
0
 




 By
Ay
By
Ay
y F
F
F
F
;
F 0
0
0
0714
0
02828
0
504
0
22
22
504
0











By
Bx
By
Bx
A
F
*
.
F
*
.
cos
*
AB
*
F
sin
*
AB
F
T
;
M

47. 47
4
2
3
A
C
B
D
=
45
F
C
B
A
2
2
°
504 Nm

4000 N
1000 Nm
Cy
F
Cx
F
Bx
F By
F
By
F
Bx
F
Cy
F
Cx
F
Ax
F
Ay
F
4
A
F
A4
0
69
.
556
;
0
82
cos
*
4000
;
0 






 Bx
Cx
Bx
Cx
x F
F
F
F
F
0
07
.
3961
;
0
82
sin
*
4000
;
0 






 By
Cy
By
Cy
y F
F
F
F
F
0
*
02828
.
0
*
02828
.
0
1000
;
0
45
cos
*
*
45
sin
*
1000
;
0 






 Cy
Cx
Cy
Cx
B F
F
BC
F
BC
F
M

48. 48
4
2
3
A
C
B
D
=
45
F
C
B
A
2
2
°
504 Nm

4000 N
1000 Nm
Cy
F
Cx
F
Bx
F By
F
By
F
Bx
F
Cy
F
Cx
F
Ax
F
Ay
F
4
A
F
A4
 

 0
0 Cx
x F
;
F
 


 0
0 4
Ay
Cy
y F
F
;
F
 

 0
0 4 CA
*
F
T
;
M Cy
A
A

49. 49
4
2
3
A
C
B
D
=
45
F
C
B
A
2
2
°
504
Nm

4000 N
1000
Nm
Cy
F
Cx
F
Bx
F By
F
By
F
Bx
F
Cy
F
Cx
F
Ax
F
Ay
F
4
A
F
A
4
 



 Bx
Ax
Bx
Ax
x F
F
F
F
;
F 0
0
 




 By
Ay
By
Ay
y F
F
F
F
;
F 0
0
0
*
0714
.
0
*
02828
.
0
504
0
22
cos
*
*
22
sin
*
504
;
0












By
Bx
By
Bx
A
F
F
T
AB
F
AB
F
T
M
0
69
.
556
;
0
82
cos
*
4000
;
0








Bx
Cx
Bx
Cx
x
F
F
F
F
F
0
07
.
3961
;
0
82
sin
*
4000
;
0








By
Cy
By
Cy
y
F
F
F
F
F
0
*
02828
.
0
*
02828
.
0
1000
;
0
45
cos
*
*
45
sin
*
1000
;
0








Cy
Cx
Cy
Cx
B
F
F
BC
F
BC
F
M
 

 0
;
0 Cx
x F
F
 


 0
0 4
Ay
Cy
y F
F
;
F
 

 0
0 4 CA
*
F
T
;
M Cy
A
A
N
.
FBx 69
556

 N
.
.
FCy 34
35355
02828
0
1000


N
.
.
.
FBy 41
39316
07
3961
34
35355 


CCW
Nm
.
T
.
*
.
)
.
(
*
.
93
3326
0
41
39316
0714
0
69
556
02828
0
504







50. 50

51. 51
If the boundaries (surfaces) of the volume is well defined mathemetically, we
calculate Mass Moment of Inertia by integration. if this is not possible and a
specimen exist, we move it and measure the motion. Motion is related to the
mass moment of inertia.
O
G

G

52.   


o
o I
M
52
Differential equation of motion of the compound pendulum. This is a second
order nonlinear differential equation. To ease the solution this equation must be
linearized which can be linearized by using Taylor series expansion then
O
G

mg
mgsin mgcos

sin
mg
r
M G
o 



 

o
G
o I
mg
r
M 


 sin
rG
0
sin 
 
 mg
r
I G
o



 
sin
0

 
 mg
r
I G
o



53. 53
Where A&B are arbitrary coefficient related with the initial values of the motion
t is time
n is natural circular frequency
O
G

mg
mgsin mgcos
rG
o
o
G
o
I
I
mg
r
I 0

 


0

 
 mg
r
I G
o


0

 

o
G
I
mg
r


2
n
)
sin(
)
cos( t
B
t
A n
n 

 


54. 54
When
O
G

mg
mgsin mgcos
rG
)
sin(
)
cos( t
B
t
A n
n 

 

o
G
n
I
mg
r


.
sec
.
rad
0
;
0
;
0 0








t
t
)
0
sin(
)
0
cos(
0 n
n B
A 

 

0
1
A

0

)
cos(
)
sin( t
B
t
A n
n
n
n 



 



)
0
cos(
)
0
sin(
0 n
n
n
n B
A 


 


1
0
0

B

55. 55
)
sin(
)
cos( t
B
t
A n
n 

 

0


A 0

B
)
cos(
0 t
n


 
t
n

0

0


1 cycle
Elapsed time for one cycle is called period 
n



2


2 
4

56. 56
n



2

o
G
n
I
mg
r


Procedure
Measure time; say 20 cycles by a stopwach.
Calculate one period by dividing this time into 20, which minimize the
personal mistakes.
0
2
2
I
mg
rG








 With only unknown I0

57. 57
Differential equation of motion of the Torsional pendulum.

 k




 

o
I
k 

0

 
 k
Io


o
o
o
o
I
I
k
I
I 0

 



k

0
2
2
I
k








 With only unknown I0

58. 58

sin
5
.
0
2
*
2 mg
b
MG 


mg


mg/2 mg/2

0.5mgcos
0.5mgsin
0.5mgsin
0.5mgsin
G
G


0
I
MG
 
0
sin
2

 

mbg
IG


b
l

59. 59
mg


mg/2 mg/2
G
G
0
sin
2

 

mbg
IG


b
l

l

2
b 
 l
b

2


l
b
2


 
sin
0
2

 

mbg
IG


0
4
2

 

l
g
mb
IG



60. 60
mg


mg/2 mg/2
G
G
b
l

l

2
b
0
4
2

 

l
g
mb
IG


G
G
G
G
I
lI
g
mb
I
I 0
4
2

 


0
4
2

 

G
lI
g
mb


G
lI
g
mb
4
2 2
2










61. 61
© 2007 Sadettin Kapucu

62.  Coordinate Transformation
› Reference coordinate frame
OXYZ
› Body-attached frame O’uvw
w
v
u k
j
i w
v
u
uvw p
p
p
P 



62
z
y
x k
j
i z
y
x
xyz p
p
p
P 



x
y
z
P
u
v
w
O,
Point represented in OXYZ:
z
w
y
v
x
u p
p
p
p
p
p 


T
z
y
x
xyz p
p
p
P ]
,
,
[

Point represented in O’uvw:
Two frames coincide ==>
O’

63.  Mutually perpendicular  Unit vectors
63
Properties of orthonormal coordinate frame
0
0
0






j
k
k
i
j
i






1
|
|
1
|
|
1
|
|



k
j
i



Properties: Dot Product
Let and be arbitrary vectors in and be
the angle from to , then
3
R 

cos
y
x
y
x 

x y
x y
x
y


64.  Coordinate Transformation
› Rotation only
w
v
u k
j
i w
v
u
uvw p
p
p
P 



64
x
y
z
P
z
y
x k
j
i z
y
x
xyz p
p
p
P 



uvw
xyz RP
P  u
v
w
How to relate the coordinate in these two frames?

65.  Basic Rotation
› , , and represent
the projections of onto OX,
OY, OZ axes, respectively
› Since
65
y
p
x
p
x
p
P
y
p z
p
w
v
u
x p
p
p
P
p w
x
v
x
u
x
x k
i
j
i
i
i
i 







w
v
u
y p
p
p
P
p w
y
v
y
u
y
y k
j
j
j
i
j
j 







w
v
u
z p
p
p
P
p w
z
v
z
u
z
z k
k
j
k
i
k
k 







w
v
u k
j
i w
v
u p
p
p
P 


x
y
z
P
u
v
w
z
p

66.  Basic Rotation Matrix
› Rotation about x-axis with
66








































w
v
u
z
y
x
p
p
p
p
p
p
w
z
v
z
u
z
w
y
v
y
u
y
w
x
v
x
u
x
k
k
j
k
i
k
k
j
j
j
i
j
k
i
j
i
i
i
x
z
y
v
w
P
u


















C
S
S
C
)
,
x
(
Rot
0
0
0
0
1


67.  Is it True?
› Rotation about x axis with
67








cos
p
sin
p
p
sin
p
cos
p
p
p
p
p
p
p
cos
sin
sin
cos
p
p
p
w
v
z
w
v
y
u
x
w
v
u
z
y
x





































0
0
0
0
1
x
z
y
v
w
P
u



68. › Rotation about x-axis with
› Rotation about y-axis with
› Rotation about z-axis with
68
uvw
xyz RP
P 

















C
S
S
C
)
,
x
(
Rot
0
0
0
0
1
C
S
S
C
)
,
y
(
Rot

















0
0
1
0
0









 

1
0
0
0
0
)
,
( 



 C
S
S
C
z
Rot




69.  Basic Rotation Matrix
› Obtain the coordinate of from the coordinate
of








































z
y
x
w
v
u
p
p
p
p
p
p
z
w
y
w
x
w
z
v
y
v
x
v
z
u
y
u
x
u
k
k
j
k
i
k
k
j
j
j
i
j
k
i
j
i
i
i
69
uvw
xyz RP
P 




















w
z
v
z
u
z
w
y
v
y
u
y
w
x
v
x
u
x
k
k
j
k
i
k
k
j
j
j
i
j
k
i
j
i
i
i
R
xyz
uvw QP
P 
T
R
R
Q 
 1
3
1
I
R
R
R
R
QR T


 
uvw
P xyz
P
<== 3X3 identity matrix
Dot products are commutative!

70.  A point is attached to a rotating frame,
the frame rotates 60 degree about the OZ axis of
the reference frame. Find the coordinates of the
point relative to the reference frame after the
rotation.
70
)
2
,
3
,
4
(

uvw
a






























 


2
964
.
4
598
.
0
2
3
4
1
0
0
0
5
.
0
866
.
0
0
866
.
0
5
.
0
)
60
,
( uvw
xyz a
z
Rot
a

71.  A point is the coordinate w.r.t. the
reference coordinate system, find the
corresponding point w.r.t. the rotated
OU-V-W coordinate system if it has been
rotated 60 degree about OZ axis.
)
2
,
3
,
4
(

xyz
a
uvw
a
71



































2
964
.
1
598
.
4
2
3
4
1
0
0
0
5
.
0
866
.
0
0
866
.
0
5
.
0
)
60
,
( xyz
T
uvw a
z
Rot
a

72. 72
• position vector of P
in {B} is transformed
to position vector of P
in {A}
• description of {B} as
seen from an observer
in {A}
Rotation of {B} with respect to {A}
Translation of the origin of {B} with respect to origin of {A}

73.  Two Special Cases
1. Translation only
› Axes of {B} and {A} are
parallel
2. Rotation only
› Origins of {B} and {A}
are coincident
73
1

B
A
R
'
o
A
P
B
B
A
P
A
r
r
R
r 

0
'

o
A
r

74. 74
• Coordinate transformation from {B} to {A}
• Homogeneous transformation matrix
'
o
A
P
B
B
A
P
A
r
r
R
r 




















 1
1
0
1 3
1
' P
B
o
A
B
A
P
A
r
r
R
r













 


1
0
1
0
1
3
3
3
3
1
'
P
R
r
R
T
o
A
B
A
B
A
Position
vector
Rotation
matrix
Scaling

75.  Special cases
1. Translation
2. Rotation
75









1
0
0
3
1
1
3
B
A
B
A R
T









1
0 3
1
'
3
3
o
A
B
A r
I
T

76.  Translation along Z-axis with h:
76
x
y
z
P
u
v
w
O, O’













1
0
0
0
1
0
0
0
0
1
0
0
0
0
1
)
,
(
h
h
z
Trans



















































1
1
1
0
0
0
1
0
0
0
0
1
0
0
0
0
1
1
h
p
p
p
p
p
p
h
z
y
x
w
v
u
w
v
u
x
y
z P
u
v
w
O, O’
h

77.  Rotation about the X-axis by






































1
1
0
0
0
0
0
0
0
0
0
0
1
1
w
v
u
p
p
p
C
S
S
C
z
y
x




77














1
0
0
0
0
0
0
0
0
0
0
1
)
,
(





C
S
S
C
x
Rot
x
z
y
v
w
P
u

78.  Composite Homogeneous Transformation
Matrix
 Rules:
› Transformation (rotation/translation) w.r.t
(X,Y,Z) (OLD FRAME), using pre-
multiplication
› Transformation (rotation/translation) w.r.t
(U,V,W) (NEW FRAME), using post-
multiplication
78

79.  Find the homogeneous transformation matrix
(T) for the following operation:
4
4
,
,
,
, 
 I
T
T
T
T
T x
a
x
d
z
z 

79
:
axis
OZ
about
of
Rotation
axis
OZ
along
d
of
n
Translatio
axis
OX
along
a
of
n
Translatio
axis
OX
about
Rotation
Answer


















































 

1
0
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
1
0
0
0
0
1
0
0
0
1
1
0
0
0
1
0
0
0
0
1
0
0
0
0
1
1
0
0
0
0
1
0
0
0
0
0
0








C
S
S
C
a
d
C
S
S
C

80.  A frame in space (Geometric
Interpretation)
80
x
y
z
)
,
,
( z
y
x p
p
p
P













1
0
0
0
z
z
z
z
y
y
y
y
x
x
x
x
p
a
s
n
p
a
s
n
p
a
s
n
F
n
s
a






 

1
0
1
3
3
3 P
R
F
Principal axis n w.r.t. the reference coordinate system

81.  Translation
81
y
z
n
s
a n
s
a










































1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
z
z
z
z
z
y
y
y
y
y
x
x
x
x
x
z
z
z
z
y
y
y
y
x
x
x
x
z
y
x
new
d
p
a
s
n
d
p
a
s
n
d
p
a
s
n
p
a
s
n
p
a
s
n
p
a
s
n
d
d
d
F
old
z
y
x
new F
d
d
d
Trans
F 
 )
,
,
(

82. 82

83. 83
X
Z
Y
P
x
z
y
O
O’
Inertial Frame
Body coordinate system it
rotates at the same
angular velocity as the
body
Arbitrary point in the body


Rigid body angular
velocity wrt inertial frame

84. r
R
R o





84
o
R

x
Z
Y
P
x
z
y
O
O’
r

R

Position of P


k
p
j
p
i
p
r z
y
x







The position of P wrt inertial
coordinate frame
The absolute velocity of P is
dt
r
d
dt
R
d
dt
R
d
V o








85. r
R
R o





85
o
R

X
Z
Y
P
x
z
y
O
O
’
r

R



dt
k
d
p
dt
j
d
p
dt
i
d
p
k
dt
dp
j
dt
dp
i
dt
dp
dt
r
d
z
y
x
z
y
x













The absolute velocity of P is
dt
r
d
dt
R
d
dt
R
d
V o







dt
k
d
p
k
dt
dp
dt
j
d
p
j
dt
dp
dt
i
d
p
i
dt
dp
dt
r
d
z
z
y
y
x
x













Becomes zero
because body is rigid
r
x



r
x
dt
r
d 





86. r
R
R o





86
The absolute velocity of P is
dt
r
d
dt
R
d
dt
R
d
V o







o
o
V
dt
R
d 


r
x
dt
r
d 




r
x
V
V o







o
R

X
Z
Y
P
x
z
y
O
O
’
r

R




87. r
R
R o





87
The absolute velocity of P is
r
x
V
V o







Acceleration of P wrt inertial coordinate system is
dt
r
x
d
dt
V
d
dt
V
d
a o )
(




 



dt
r
d
x
r
x
dt
d
a
a o










 
r
x
x
r
x
a
a o









 


o
R

X
Z
Y
P
x
z
y
O
O
’
r

R




88. 88

89. 89

90. 90
The 0.8 m arm OA for a remote-control mechanism is pivoted about the horizontal x-axis of
the clevis, and the entire assembly rotates about the z-axis with a constant speed
N=60rev/min. Simultaneously the arm is being raised at the constant rate . For
the position where b=60o determine (a) angular velocity of OA, (b) the angular acceleration of
OA, (c) the velocity of point A, and (d) the acceleration of point A.
s
rad /
4

b

s
rad
N
z /
283
.
6
60
/
)
60
(
2
60
/
2 

 


s
rad
k
i
z
x /
283
.
6
4








 


x
x 









 k


283
.
6


2
/
13
.
25
4
283
.
6 s
rad
j
i
x
k







z


x

 



k
j
r



4
.
0
693
.
0 

s
m
k
j
i
k
j
i
r
x
V
/
77
.
2
60
.
1
35
.
4
4
.
0
693
.
0
0
283
.
6
0
4














 
s
rad
x /
4

 b
 

91. 91
The 0.8 m arm OA for a remote-control mechanism is pivoted about the horizontal x-axis of
the clevis, and the entire assembly rotates about the z-axis with a constant speed
N=60rev/min. Simultaneously the arm is being raised at the constant rate . For
the position where b=60o determine (a) angular velocity of OA, (b) the angular acceleration of
OA, (c) the velocity of point A, and (d) the acceleration of point A.
s
rad /
4

b

s
rad
k
i
z
x /
283
.
6
4








 


2
/
13
.
25
4
283
.
6 s
rad
j
i
x
k







z


x

 



k
j
r



4
.
0
693
.
0 

V
x
r
x
r
x
x
r
x
a



















 )
(
2
/
40
.
6
44
.
38
11
.
20
77
.
2
60
.
1
35
.
4
283
.
6
0
4
4
.
0
693
.
0
0
0
13
.
25
0
s
m
k
j
i
k
j
i
k
j
i
a


















92. 92
The electric motor with an attached disk is running at a constant low speed of 120
rey/mm in the direction shown. Its housing and mounting base are initially at rest.
The entire assembly is next set in rotation about the vertical Z-axis at the constant
rate N=60 rev/min with a fixed angle g of 300. Determine (a) the angular velocity
and angular acceleration of the disk, (b) the space and body cones, and (c) the
velocity and acceleration of point A at the top of the disk for the instant shown.
s
rad /
4

b

)
sin
(cos k
j



g
g 



s
rad
o /
4
60
/
)
2
(
120 

 

s
rad /
2
60
/
)
60
(
2 
 


K
k
o
o










 


s
rad
k
j
k
j
k
j
k
j
k
o
o
o
o
/
)
0
.
5
3
(
)
30
sin
2
4
(
)
30
cos
2
(
)
sin
(
)
cos
(
)
sin
(cos




























g

g
g
g



93. 93
The electric motor with an attached disk is running at a constant low speed of 120
rey/mm in the direction shown. Its housing and mounting base are initially at rest.
The entire assembly is next set in rotation about the vertical Z-axis at the constant
rate N=60 rev/min with a fixed angle g of 300. Determine (a) the angular velocity
and angular acceleration of the disk, (b) the space and body cones, and (c) the
velocity and acceleration of point A at the top of the disk for the instant shown.
s
rad
o /
4
60
/
)
2
(
120 

 

s
rad /
2
60
/
)
60
(
2 
 


s
rad
k
j /
)
0
.
5
3
(




 


94. 94
The electric motor with an attached disk is running at a constant low speed of 120
rey/mm in the direction shown. Its housing and mounting base are initially at rest.
The entire assembly is next set in rotation about the vertical Z-axis at the constant
rate N=60 rev/min with a fixed angle g of 300. Determine (a) the angular velocity
and angular acceleration of the disk, (b) the space and body cones, and (c) the
velocity and acceleration of point A at the top of the disk for the instant shown.
s
rad
o /
4
60
/
)
2
(
120 

 

s
rad /
2
60
/
)
60
(
2 
 










x



s
rad
i
i
i
i
i
o
o
o
/
4
.
68
30
cos
)
4
)(
2
(
)
cos
(
)
cos
sin
(
)
cos
sin
cos
( 2

















g

g
g
g
g
g

 
k
j
x
k
j o





)
sin
(
)
cos
(
)
sin
(cos g

g
g
g
 







95. 95
The electric motor with an attached disk is running at a constant low speed of 120
rey/mm in the direction shown. Its housing and mounting base are initially at rest.
The entire assembly is next set in rotation about the vertical Z-axis at the constant
rate N=60 rev/min with a fixed angle g of 300. Determine (a) the angular velocity
and angular acceleration of the disk, (b) the space and body cones, and (c) the
velocity and acceleration of point A at the top of the disk for the instant shown.
s
rad
o /
4
60
/
)
2
(
120 

 

s
rad /
2
60
/
)
60
(
2 
 


s
rad
k
j /
)
0
.
5
3
(




 


96. 96
The electric motor with an attached disk is running at a constant low speed of 120
rey/mm in the direction shown. Its housing and mounting base are initially at rest.
The entire assembly is next set in rotation about the vertical Z-axis at the constant
rate N=60 rev/min with a fixed angle g of 300. Determine (a) the angular velocity
and angular acceleration of the disk, (b) the space and body cones, and (c) the
velocity and acceleration of point A at the top of the disk for the instant shown.
s
rad /
4

b

k
j
r



250
.
0
125
.
0 

s
m
i
k
j
i
r
x
V /
1920
.
0
250
.
0
125
.
0
0
5
3
0










 



V
x
r
x
r
x
x
r
x
a















 


 )
(
2
/
83
.
11
6
.
26
)
192
.
0
(
)
5
3
(
)
250
.
0
125
.
0
(
4
.
68
s
m
k
j
i
x
k
j
k
j
x
i
a
















 


97. 97
B
r

X
Z
Y
A
x
z
y
O
B
B
A
r

A
r



coordinate system rotates with
this angular velocirty


Body coordinate frame rotates
with this angular velocirty
B
F


Letting
r
x
r
V
V B
A










Denote the angular velocity of the reference wrt the body frame, the angular
velocity of the body is related to that of the coordinate system
B
F








The velocity of a point of the body may be represented by
r
x
V
V
r
x
r
x
V
V rel
B
B
F
B
A

















 

98. 98
B
r

X
Z
Y
A
x
z
y
O
B
B
A
r

A
r



coordinate system rotates with
this angular velocirty


Body coordinate frame rotates
with this angular velocirty














B
A
B
A
B
A
B
A
B
A r
x
x
r
x
r
x
r
a
a














2
Acceleration of a point of the body is obtained as:
The velocity of a point of the body may be represented by








B
A
B
F
B
F
B
A
B
F
rel r
x
x
r
x
a









 rel
B
A
B
F V
x
r
x
x













 2
2 









































B
A
B
A
B
A
F
B
B
A
F
B
F
B
B
A
F
B
B
A
B
A
B
A
rel
rel
B
A
r
x
x
r
x
r
x
x
r
x
x
r
x
a
a
r
x
x
r
x
V
x
a
a
a































 2
2
r
x
r
V
V B
A











99. 99
The motor housing and its bracket rotate about the Z axis at the constant rate
The motor shaft and disk have a constant angular velocity of spin with respect
to the motor housing in the direction shown. If g constant at 30o, determine the velocity
and acceleration of point A at the top of the disk and angular acceleration  of the disk.
s
rad /
3


K


3


J
rB


350
.
0

k
j
r
B
A



120
.
0
300
.
0 

s
m
i
I
J
x
K
r
x
V B
B
/
05
.
1
05
.
1
350
.
0
3














rel
B
A V
r
x
V
V









s
m
i
i
i
k
j
x
K
r
x
B
A /
599
.
0
)
30
sin
36
.
0
(
)
30
cos
9
.
0
(
)
120
.
0
300
.
0
(
3







 









s
m
i
k
j
x
j
r
x
p
V
B
A
rel /
960
.
0
)
120
.
0
300
.
0
(
8










s
m
i
i
i
i
VA /
689
.
0
960
.
0
599
.
0
05
.
1











s
rad
p /
8


100. 100
The motor housing and its bracket rotate about the Z axis at the constant rate
The motor shaft and disk have a constant angular velocity of spin with respect
to the motor housing in the direction shown. If g constant at 30o, determine the velocity
and acceleration of point A at the top of the disk and angular acceleration  of the disk.
s
rad /
3


2
/
899
.
0
73
.
2
)
30
sin
30
cos
(
15
.
3
15
.
3
)
350
.
0
3
(
3
)
(
s
m
k
j
k
j
J
J
x
K
x
K
r
x
x
V B
B



























 
2
/
899
.
0
557
.
1
)
599
.
0
(
3
)
120
.
0
300
.
0
(
3
3
)
(
s
m
k
j
i
x
K
k
j
x
K
x
K
r
x
x
B
A




















2
/
88
.
2
99
.
4
)
30
sin
30
cos
(
76
.
5
76
.
5
960
.
0
)
3
(
2
2 s
m
k
j
k
j
J
i
x
K
V
x rel








 








s
rad
p /
8















B
A
B
A
rel
rel
B
A r
x
x
r
x
V
x
a
a
a











2
0




2
/
68
.
7
))
120
.
0
300
.
0
(
8
(
8
)
( s
m
k
k
j
x
j
x
j
r
x
p
x
p
a
B
A
rel













2
/
086
.
8
703
.
0 s
m
k
j
aA




 2
2
2
/
12
.
8
086
.
8
703
.
0 s
m
aA 


2
/
8
.
20
)
30
cos
24
(
0 s
rad
i
i


 






)
8
3
(
3 j
K
x
K
x












 



101. 101

102. 102
x
z
y
CG


r
 
dt
L
d
F


dt
V
m
d CG )
(


dt
V
d
m CG


 
dt
H
d
M CG

  
 
dt
dm
r
x
x
r
d 






103. 103
x
z
y
CG


r
 
dt
L
d
F


dt
V
m
d CG )
(


dt
V
d
m CG


C
z
y
x
CG
V
x
k
V
j
V
i
V
dt
V
d 













z
y
x
z
y
x
V
V
V
k
j
i






 i
V
V y
z
z
y


 
   j
V
V x
z
z
x


 

 k
V
V x
y
y
x


 


104. 104
x
z
y
CG


r
 
dt
L
d
F


dt
V
m
d CG )
(


dt
V
d
m CG


C
z
y
x
CG
V
x
k
V
j
V
i
V
dt
V
d 













z
y
x
z
y
x
V
V
V
k
j
i






 i
V
V y
z
z
y


 
   j
V
V x
z
z
x


 

 k
V
V x
y
y
x


 

 
 

 y
z
z
y
x
x V
V
V
m
F 


 
 

 z
x
x
z
y
y V
V
V
m
F 


 
 

 x
y
y
x
z
z V
V
V
m
F 



105. 105
x
z
y
CG


r
 
dt
H
d
M CG

  
 
dt
dm
r
x
x
r
d 





 
 

 y
z
z
y
x
x V
V
V
m
F 


 
 

 z
x
x
z
y
y V
V
V
m
F 


 
 

 x
y
y
x
z
z V
V
V
m
F 


z
y
x
z
y
x
r
r
r
k
j
i






 i
r
r y
z
z
y


 
   j
r
r x
z
z
x


 

 k
r
r x
y
y
x


 


r
x




106. 106
x
z
y
CG


r
 
dt
H
d
M CG

  
 
dt
dm
r
x
x
r
d 





 
 

 y
z
z
y
x
x V
V
V
m
F 


 
 

 z
x
x
z
y
y V
V
V
m
F 


 
 

 x
y
y
x
z
z V
V
V
m
F 


x
y
y
x
z
x
x
z
y
z
z
y
z
y
x
r
r
r
r
r
r
r
r
r
k
j
i





 





 
r
x
x
r



 i
r
r y
z
z
y


 
   j
r
r x
z
z
x


 
  k
r
r x
y
y
x


 


r
x




107. 107
x
z
y
CG


r
 
dt
H
d
M CG

  
 
dt
dm
r
x
x
r
d 





 
 

 y
z
z
y
x
x V
V
V
m
F 


 
 

 z
x
x
z
y
y V
V
V
m
F 


 
 

 x
y
y
x
z
z V
V
V
m
F 


x
y
y
x
z
x
x
z
y
z
z
y
z
y
x
r
r
r
r
r
r
r
r
r
k
j
i





 





 
r
x
x
r




   
 i
r
r
r
r
r
r z
x
x
z
z
x
y
y
x
y




 



   
  j
r
r
r
r
r
r y
z
z
y
z
x
y
y
x
x




 



   
 k
r
r
r
r
r
r y
z
z
y
y
z
x
x
z
x




 




108. 108
x
z
y
CG


r
 
dt
H
d
M CG

  
 
dt
dm
r
x
x
r
d 





 
 

 y
z
z
y
x
x V
V
V
m
F 


 
 

 z
x
x
z
y
y V
V
V
m
F 


 
 

 x
y
y
x
z
z V
V
V
m
F 


  
 dm
r
x
x
r



    
  dm
i
r
r
r
r
r
r z
x
x
z
z
x
y
y
x
y
 








   
  dm
j
r
r
r
r
r
r y
z
z
y
z
x
y
y
x
x
 








   
  dm
k
r
r
r
r
r
r y
z
z
y
y
z
x
x
z
x
 








H


109. 109
x
z
y
CG


r
 
dt
H
d
M CG

  
 
dt
dm
r
x
x
r
d 





 
 

 y
z
z
y
x
x V
V
V
m
F 


 
 

 z
x
x
z
y
y V
V
V
m
F 


 
 

 x
y
y
x
z
z V
V
V
m
F 


   
 
   
 
   
  






































dm
r
r
r
r
r
r
dm
r
r
r
r
r
r
dm
r
r
r
r
r
r
H
H
H
H
y
z
z
y
y
z
x
x
z
x
y
z
z
y
z
x
y
y
x
x
z
x
x
z
z
x
y
y
x
y
z
y
x














110. 110
x
z
y
CG


r
 
dt
H
d
M CG

  
 
dt
dm
r
x
x
r
d 





 
 

 y
z
z
y
x
x V
V
V
m
F 


 
 

 z
x
x
z
y
y V
V
V
m
F 


 
 

 x
y
y
x
z
z V
V
V
m
F 


   
 
   
 
   
  






































dm
r
r
r
r
r
r
dm
r
r
r
r
r
r
dm
r
r
r
r
r
r
H
H
H
H
y
z
z
y
y
z
x
x
z
x
y
z
z
y
z
x
y
y
x
x
z
x
x
z
z
x
y
y
x
y
z
y
x













 
 
  






































dm
r
r
r
r
r
r
dm
r
r
r
r
r
r
dm
r
r
r
r
r
r
H
H
H
H
y
z
y
z
y
x
z
x
x
z
z
y
z
z
y
x
y
x
y
x
z
x
z
x
z
y
x
y
y
x
z
y
x
2
2
2
2
2
2














111. 111
x
z
y
CG


r
 
dt
H
d
M CG

  
 
dt
dm
r
x
x
r
d 





 
 

 y
z
z
y
x
x V
V
V
m
F 


 
 

 z
x
x
z
y
y V
V
V
m
F 


 
 

 x
y
y
x
z
z V
V
V
m
F 


 
 
  









































dm
r
r
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
H
H
H
H
y
x
z
y
z
y
x
z
x
z
y
z
z
x
y
x
y
x
z
x
z
y
x
y
z
y
x
z
y
x
2
2
2
2
2
2










 
 
  






































dm
r
r
r
r
r
r
dm
r
r
r
r
r
r
dm
r
r
r
r
r
r
H
H
H
H
y
z
y
z
y
x
z
x
x
z
z
y
z
z
y
x
y
x
y
x
z
x
z
x
z
y
x
y
y
x
z
y
x
2
2
2
2
2
2













 
 
  



















































z
y
x
y
x
y
z
x
z
z
y
z
x
x
y
z
x
y
x
z
y
z
y
x
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
H
H
H
H



2
2
2
2
2
2


112. 112
x
z
y
CG


r
 
dt
H
d
M CG

  
 
dt
dm
r
x
x
r
d 





 
 

 y
z
z
y
x
x V
V
V
m
F 


 
 

 z
x
x
z
y
y V
V
V
m
F 


 
 

 x
y
y
x
z
z V
V
V
m
F 


 
 
  









































dm
r
r
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
H
H
H
H
y
x
z
y
z
y
x
z
x
z
y
z
z
x
y
x
y
x
z
x
z
y
x
y
z
y
x
z
y
x
2
2
2
2
2
2










 
 
  



















































z
y
x
y
x
y
z
x
z
z
y
z
x
x
y
z
x
y
x
z
y
z
y
x
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
H
H
H
H



2
2
2
2
2
2


113. 113
x
z
y
CG


r
 
dt
H
d
M CG

  
 
dt
dm
r
x
x
r
d 





 
 

 y
z
z
y
x
x V
V
V
m
F 


 
 

 z
x
x
z
y
y V
V
V
m
F 


 
 

 x
y
y
x
z
z V
V
V
m
F 


 
 
  



















































z
y
x
y
x
y
z
x
z
z
y
z
x
x
y
z
x
y
x
z
y
z
y
x
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
dm
r
r
H
H
H
H



2
2
2
2
2
2







































z
y
x
zz
zy
zx
yz
yy
yx
xy
xy
xx
z
y
x
I
I
I
I
I
I
I
I
I
H
H
H
H





114. 114
x
z
y
CG


r
 
dt
H
d
M CG

  
 
dt
dm
r
x
x
r
d 





 
 

 y
z
z
y
x
x V
V
V
m
F 


 
 

 z
x
x
z
y
y V
V
V
m
F 


 
 

 x
y
y
x
z
z V
V
V
m
F 








































z
y
x
zz
zy
zx
yz
yy
yx
xy
xy
xx
z
y
x
I
I
I
I
I
I
I
I
I
H
H
H
H




H
x
k
H
j
H
i
H
dt
H
d
z
y
x















115. 115
x
z
y
CG


r
 
dt
H
d
M CG

  
 
dt
dm
r
x
x
r
d 











































z
y
x
zz
zy
zx
yz
yy
yx
xy
xy
xx
z
y
x
I
I
I
I
I
I
I
I
I
H
H
H
H




H
x
k
H
j
H
i
H
dt
H
d
z
y
x














z
y
x
z
y
x
H
H
H
k
j
i






 i
H
H y
z
z
y


 

  j
H
H x
z
z
x


 

 k
H
H x
y
y
x


 


H
x




116. 116
x
z
y
CG


r






































z
y
x
zz
zy
zx
yz
yy
yx
xy
xy
xx
z
y
x
I
I
I
I
I
I
I
I
I
H
H
H
H




H
x
k
H
j
H
i
H
dt
H
d
z
y
x














     k
H
H
j
H
H
i
H
H
H
x x
y
y
x
x
z
z
x
y
z
z
y











 





       
2
2
x
y
yz
z
y
yy
zz
y
x
z
xz
z
x
y
xy
x
xx
x I
I
I
I
I
I
M 









 








 


       
2
2
x
z
xz
z
x
zz
xx
y
x
z
yz
y
yy
z
y
x
xy
y I
I
I
I
I
I
M 









 








 


       
2
2
y
x
xy
y
x
xx
yy
z
zz
z
x
y
yz
z
y
x
xz
x I
I
I
I
I
I
M 









 








 



117. 117
x
z
y
CG


r
 
 

 y
z
z
y
x
x V
V
V
m
F 


 
 

 z
x
x
z
y
y V
V
V
m
F 


 
 

 x
y
y
x
z
z V
V
V
m
F 


       
2
2
x
y
yz
z
y
yy
zz
y
x
z
xz
z
x
y
xy
x
xx
x I
I
I
I
I
I
M 









 








 


       
2
2
x
z
xz
z
x
zz
xx
y
x
z
yz
y
yy
z
y
x
xy
y I
I
I
I
I
I
M 









 








 


       
2
2
y
x
xy
y
x
xx
yy
z
zz
z
x
y
yz
z
y
x
xz
x I
I
I
I
I
I
M 









 








 

