This document provides an introduction to dynamics and mechanics. It discusses two models used in dynamics - particles and rigid bodies. It describes the motion of each, including their degrees of freedom and parameters. The document then covers equations of motion using vector mechanics and analytical mechanics approaches. It introduces concepts like the center of mass and linear momentum. Key theorems discussed include the linear momentum theorem, conservation of linear momentum, and the center-of-mass motion theorem. Examples are provided to demonstrate applying these theorems to problems involving particles and rigid bodies.

1. Engineering
Mechanics:
Dynamics
• ME2211E
• Prof.Dr. Dinh Van Phong
• Dept.of Applied Mechanics, School of
Mechanical Engineering
• C3/307-308 (C1/224)
• Hanoi University of Science and
Technology
• Email: phong.dinhvan@hust.edu.vn
• 0903200960/ (024) 36230949

2. Chapter 1. Introduction to Dynamics -2-
Department of Applied Mechanics-SME
Two models using in Dynamics: Particle and Rigid body
P
rP
x
z
y
O
A
rA
x
z
y
O
B
Motion of a particle:
Motion of a rigid body:
• Motion in plane, motion in space
• Rectilinear & Curvilinear motion
• Velocity and Acceleration of a
particle
• Maximum 3 position parameters (3
Degree of Freedom)
• Translation
• Rotation about a fixed axis / fixed point
• General spatial motion
• General planar motion
• Helical motion (screw motion)
• Velocity and Acceleration of a point and angular
velocity vector, angular acceleration vector
• Maximum 6 position parameters (6 Degree of
Freedom)
Particle model Rigid body model

3. Equations of
Motion
Model of particles or
rigid bodies
• Force and acceleration
• Work and energy
• Impulse and
momentum
3 methods:
2 Approaches:
-Vector Mechanics
- Analytical Mechanics

4. Applied Mechanics - Department of Mechatronics - SME
Chapter 3.
A Method of Vector Mechanics:
Momentum Method

5. Applied Mechanics - Department of Mechatronics - SME
-5-
Content
I. SOME CONCEPTS
1. Mechanical system, Internal Force, External Force
2. The center of mass of mechanical system
• The center of mass of system of particle
• The center of mass of rigid body, system of rigid bodies
II. LINEAR MOMENTUM THEOREM
1. Linear momentum of mechanical system
2. Linear momentum theorem
3. Conservation Linear momentum
III. THE CENTER-OF-MASS MOTION THEOREM
1. The center-of-mass motion theorem
2. Conservation of the center-of-mass motion

6. Applied Mechanics - Department of Mechatronics - SME
-6-
Content
I. SOME CONCEPTS
1. Mechanical system, Internal Force,
External Force
2. The center of mass of mechanical system
• The center of mass of system of particles
• The center of mass of rigid body, system of
rigid bodies

7. Applied Mechanics - Department of Mechatronics - SME
-7-
1. Mechanical system – Internal, External force
Mechanical system includes particles and rigid bodies that interact to
each other (forces and constraints of motion)
Boundary, Inside and outside of the system.
mk
m1
e
k
F
i
k
F
i
j
F
e
l
F
Boundary
outside
internal
x
y
z
O
k
m
g
j
m
e
j
F
e
l
F
e
k
F
boundary

8. Applied Mechanics - Department of Mechatronics - SME
-8-
1. Mechanical system – Internal, External force
Internal force: forces interact between
particles / rigid bodies in a system.
External force: forces from outside the
system act on the system, such as gravity,
force associated with the environment, ..
mk
m1
e
k
F
i
k
F
i
j
F
e
l
F
Boundary
outside
internal
i i
k O k
k k
F m F O
0, ( ) 0,
= = 
 
e
k
F k
, 1,2,...
i
k
F k
, 1,2,...
Properties of the internal force system

9. Applied Mechanics - Department of Mechatronics - SME
-9-
2. Center of mass of a system of particles
1
1
1
1
,
1
,
1
n
C k k
k
n
C k k
k
n
C k k
k
x m x
m
y m y
m
z m z
m
=
=
=
=
=
=



For a system of n particles with masses mk (k=1,…,n), the
Center of mass is a geometric point C that satisfies:
1
0
n
k k
k
m u
=
=

x
y
z
mk
O
m1
C
k
x
k
r k
y
k
z
C
r
k
u
k
u - vector from C to point Mk
1 1
( ) 0
n n
k k C k k k k C
k k
u r r m u m r r
= =
= −  = − =
 
In Oxyz:
- Position vectors of particles
- Position vector of center of mass C
1 2
, ,..., n
r r r
n n
C k k k
k k
r m r m m
m 1 1
1
,
= =
 = = 
 
In case, g = const, center of mass C  center of gravity G.
C
r

10. Applied Mechanics - Department of Mechatronics - SME
-10-
2. Center of mass of a rigid body
C C C
x m x dm y m ydm z m z dm
1 1 1
, ,
− − −
= = =
  
The Center of mass of a rigid body is point C:
x
y
z
O
C
M
x
y
z
dm
r
C
r
u

11. Applied Mechanics - Department of Mechatronics - SME
-11-
2. Center of mass: mechanical system
x
y
z
O
C
Ci
i
C
r
k
r
k
u
Ci
u
C
r
k
m
i
m Cj
1 1 1 1
1 1
1 1
, ,
1
,
p p
n n
C k k i Ci C k k i Ci
k i k i
p
n
C k k i Ci
k i
x m x m x y m y m y
m m
z m z m z
m
= = = =
= =
   
= + = +
   
   
 
= +
 
 
   
 
1 1
1 1
1
( ),
p
n
C k k i Ci
k i
p
n
k i
k i
r m r m r
m
m m m
= =
= =
= +
= +
 
 
The Center of mass of system including n particles and p rigid bodies

12. Applied Mechanics - Department of Mechatronics - SME
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2. Center of mass: Rigid body and rigid body system
If the system is on or near the earth’s surface, the gravitational acceleration is
constant, the center of mass coincides with the center of gravity. Thus, the
methods used in calculation of the center of gravity can be applied to the
calculation of the center of mass.
The center of mass of the body always exists but the center of gravity exists
only when the system is in the gravity field (of the Earth).
Some remarks:
• If the homogeneous rigid body has a symmetric center (axis, plane), its
center of mass lies on that symmetric center (axis, plane).
• If the rigid body consists of several parts that have center of mass lie on an axis
(plane), then its center of mass lies on that axis (plane)

13. Applied Mechanics - Department of Mechatronics - SME
-13-
Content
II. LINEAR MOMENTUM THEOREM
1. Linear momentum of mechanical system
2. Impulse of a force
3. Linear momentum theorem
4. Conservation Linear momentum

14. Applied Mechanics - Department of Mechatronics - SME
-14-
1. Linear momentum of a mechanical system
• Linear momentum of a particle
• Linear momentum of a system of particles
p mv
= [kg.m/s]
1 1
( )
n n
k k k C
k k
p m v m v
= =
= =
 
• Linear momentum of a rigid body
• Linear momentum of a system
C C
p vdm rdm mr mv
= = = =
 
,
k C k
p p Mv M m
=  = = 
The linear momentum of a system is equal to the
product of the total mass M of the system and the
velocity of the center of mass. x
y
z
O
C
vdm
C
r
u
C
p mv
=
mk
m1
i i
m v
k k
m v
C
p mv
=

15. Applied Mechanics - Department of Mechatronics - SME
-15-
2. Impulse
• Impulse of a force during infinitesimal time dt is defined by:
(elementary impulse)
( )
dS F t dt
=
• Impulse of force F(t) during the length of time t1 to t2
2
1
( )
t
t
S F t dt
= 
t
t1
t2
F(t)
S
• The unit of impulse is Newton.second [N.s] ,
(equivalent unit: the kilogram meter per second
[kg⋅m/s]).

16. Applied Mechanics - Department of Mechatronics - SME
-16-
3. Linear momentum theorem
• In derivative form: The time rate of change of the momentum of a
particle is equal to the net force acting on the particle and it is in the
direction of that force.
For a particle
p mv ma F
= = = Newton’s second law. (mass is constant)
• In integral form: The change in the linear momentum of a particle is
equal to the impulse that acts on that particle.
dp
F dp Fdt
dt
=  =
1
( )
mv t
2
1
t
t
Fdt
 2
( )
mv t
=
+
2
1
2 1
( ) ( ) ( )
t
t
p t p t S F t dt
− = = 

17. Applied Mechanics - Department of Mechatronics - SME
-17-
3. Linear momentum theorem
• In derivative form: The time rate of change of the momentum of a system
is equal to the net of all external forces acting on the system:
For a system
e
k k k k k
p m v F
=  = 
• In integral form: The change in the linear momentum of a system is equal
to the impulse of all external forces acting on the system:
e e
k k k k
dp
F dp F dt
dt
=   =  
2
1
2 1
( ) ( )
t
e
k k k
t
p t p t Se F dt
− =  =  
Proof.
, 0
e i i
k k k k k k k
e i
k k k k k k k
p m v F F F
p p m v F F
= = +  =
  = =  =  + 

18. Applied Mechanics - Department of Mechatronics - SME
-18-
3. Conservation of Linear momentum
0 ,
e
k C
k
If F p const mv const
=  = =

• If sum of all external forces act on a system is zero, then the total
momentum of the system is a constant vector.
0 ,
e
kx x C
k
If F p const mx const
=  = =

• If the component of the net external force on a closed system is zero
along an axis, then component of the linear momentum along that
axis cannot change:
The linear momentum theorem is valid not only for particle and system of particles,
but also valid for a rigid body and system of rigid bodies. This is one of general
theorems in dynamics of the system in which the 3. Newton law is valid.

19. Applied Mechanics - Department of Mechatronics - SME
-19-
Examples
Force depending on time P(t)
Example 1. A car of mass m starts
from the rest on a straight road by
repulsion force P = P0(1-exp(-a.t)),
in which P0 and a are const. Find
the velocity expression as function
of time.
m
P
0
( ) (0) ( ) , (0) 0
t
mv t mv F t dt v
− = =

Solution.
Force depends on time, apply the
momentum theorem, on axis x.
t t
at at
at
P e dt P t e a
P t e a P a
0 0 0
0
0 0
(1 ) ( / )
( / ) /
− −
−
− = +
= + −

0
0
0
( ) (0)
(0) (1 )
t
x
t
at
mv t mv F dt
mv P e dt
−
= +
= + −


( )
0 0
1
( ) (0) ( / ) /
at
v t v P t e a P a
m
−
= + + −
x
y

20. Applied Mechanics - Department of Mechatronics - SME
-20-
Examples
Example 2. Conservation of momentum. A system
includes body A (mass m1) which is put on an inclined
plane of a prism (mass m2). The angle between the
inclined plane of the prism and the horizontal plane is
. The prism is put on a smooth horizontal plane (as
shown in Figure). At first, the body stays still on the
prism plane relatively, the prism slides to the right with
velocity v0. After that, body A slide on the inclined plane
with relative velocity u=at. Find the velocity of the
prism.
v0
u
A

B
m1
m2
x
y

21. Applied Mechanics - Department of Mechatronics - SME
-21-
Example
v0
u
A

B
m1
m2
m
v v u
m m
1
0
1 2
cos
Solution
External forces of the system: P1, P2, N1, N2
e
k x x x
F p const p
,
0 (0)
x x x
p m v m v
m m v
1 1 2 2
1 2 0
(0) (0) (0)
( )
= +
= +
(1)
(2)
x x x
p t m v t m v t
m v t u m v t
1 1 2 2
1 2
( ) ( ) ( )
( ( ) cos ) ( ) (3)
x
y
P2
P1
N1
N2
( ) (0)
x x
p t p

22. Applied Mechanics - Department of Mechatronics - SME
-22-
III. THE CENTER-OF-MASS MOTION THEOREM
1. The center-of-mass motion theorem
2. Conservation of the center-of-mass motion

23. Applied Mechanics - Department of Mechatronics - SME
-23-
1. The center of mass motion theorem
The center of mass motion theorem
Apply Newton's second law to the particle mk that belongs to system:
e i
k k k k
m a F F k n
, 1,2,...,
Take the sum of the two sides with all particles of the system,
paying attention to the properties of the internal force system
n n n n
e i e
k k k k k
k k k k
m a F F F
1 1 1 1
n n n
k k C k k C k k C
k k k
m r m r m r m r m a m a
1 1 1
n n
e
C k k
k k
m a F m m
1 1
,
Note: The center of mass motion
theorem is valid for both body
and body system.
The sum of all external forces acting on the system is
equal to the total mass of the system multiplied by the
acceleration of the center of mass of the system.

24. Applied Mechanics - Department of Mechatronics - SME
-24-
2. The center of mass motion theorem / conservation case
The center of mass motion theorem
n
e
C k
k
m a F
1
e
C kx
e
C ky
e
C kz
m x F
m y F
m z F
,
,
Some cases of conservation
1
If 0 (0)
(0) 0 (0)
n
e
k C C
k
C C C
F m v const m v
If m v m r const m r
1
If 0 (0)
(0) 0 (0)
n
e
kx C C
k
C C C
F m x const m x
If m x m x const m x

25. Applied Mechanics - Department of Mechatronics - SME
-25-
Examples
Example 1. The electric motor is fixed on the
horizontal ground as shown in the figure. The
fixed part of the motor (stator) has mass m1,
the rotating part (rotor) has mass m0 with the
center of mass at point A and OA = e.
(a)Find the maximum shear force acting on
the bolts if the motor rotates with angular
velocity .
(b)Assume that the motor is freely placed on
a smooth ground (no bolts, no friction), find
the allowable rotation speed so that the
motor is not bounced off the ground.
Solution:
a) We examine the whole motor. External forces act on the system as figure.
With center of mass C, apply the center of mass motion theorem
t
O
A
x
y
G
P1
P0
R
N
n
e
C k C
k
m a F m m a P P N R
0 1 0 1
1
( )

26. Applied Mechanics - Department of Mechatronics - SME
-26-
Examples
2
1 0 0
( ) cos
C
R m m x R m e t
 
= +  = −
A G A
C C A
m x m x m x m
x x x
m m m m m m
0 1 0 0
0 1 0 1 0 1
+
= =  =
+ + +
A A
x e t x e t
2
cos cos
  
=  = −
C
m e t
x
m m
2
0
1 0
cos
 
= −
+
R m e 2
max 0

=
Note: If the rotor rotates at high speed (), the force Rmax will be large even though e is small.
In engineering, it is common to find ways to reduce the eccentricity as small as possible,
ideally, e = 0.
The maximum value of the shear force on bolts
Determine the center of mass acceleration in the x axis
t
O
A
x
y
G
P1
P0
R
N
0 1
,
C C
m x R m y N P P

27. Applied Mechanics - Department of Mechatronics - SME
-27-
Examples
0 1 0
0 1 0 1
A G A
C C
m y m y m y
y y
m m m m
+
=  =
+ +
A A
y e t y e t
2
sin sin
  
=  = −
C
y m m m e t
1 2
1 0 0
( ) sin
 
−
= − +
C
N P P m m y m m g m e t
2
0 1 1 0 1 0 0
( ) ( ) sin
 
= + + + = + −
m m
N m m g m e g
m e
2 0 1
min 0 1 0 max
0
( ) 0
  
+
= + −    =
b) Where the motor is freely placed on a non-friction ground without shear forces
from bolts.
C C
m y N P P N m y P P
0 1 0 1
The motor does not bounce off the ground, if the maximum allowable angular velocity needs to satisfy:
Calculating yC t
O
A
x
y
G
P1
P0
R
N

28. Applied Mechanics - Department of Mechatronics - SME
-28-
Examples / conservation case
A

B
m
M
x
y
Example 2.
Determine the horizontal displacement
of the ship carrying the crane, when
crane AB is needed to be lifted vertically
from the original angle = 30 as shown
in figure. Givens : M = 20 tons, m = 2
tons, and AB = L = 8 m. Neglect the
water resistance and the mass of AB
crane.

29. Applied Mechanics - Department of Mechatronics - SME
-29-
Examples / conservation case
1 1 2
( ) ( ) ( sin )
C
Mx t M x m x d L 
=  + + +  + −
1
sin
(0) ( ) 0,36 m
C C
mL
Mx Mx t
M m

=   = =
+
A

B
m
M
x
y
Solution
By neglecting horizontal resistance, all
external forces are vertical.
Case (= 0): Suppose the ship moves a distance of  horizontally to the
right:
1 2
(0) ,
C
Mx Mx mx
= + 2 1
sin
x x d L 
= + +
= 30
Determine the center of mass of system
at two cases : = 30 and = 0
( ) 0
=const (0) 0
=const
C
C C
C
M m x
x x
x
+ = 
= =


30. Applied Mechanics - Department of Mechatronics - SME
-30-
IV. MOMENT OF INERTIA OF A RIGID BODY
ABOUT AN AXIS
1. Mass moment of inertia
2. Radius of gyration
3. Parallel-axis theorem
4. Moment of Inertia of some simple rigid body
5. Method of composite bodies

31. Applied Mechanics - Department of Mechatronics - SME
-31-
Mass moment of inertia
Definition: Moment of Inertia of rigid body about
z-axis, Iz, is scalar quantity calculated by
x
y
z
O
x
y
z dm
h
u
2 2 2
( )
z
I h dm x y dm
= = +
 
z
z
M
z
I
Moment of Inertia is a measure of the resistance of a body in rotation
z z z
I I M
 
= =
m
ma F
=
F
Moment of Inertia of rigid body about an axis

32. Applied Mechanics - Department of Mechatronics - SME
-32-
Mass moment of inertia
Moment of Inertia of rigid body about a point
x
y
z
O
x
y
z dm
u
Definition: Moment of Inertia of rigid body about a point O, IO
O
I u dm x y z dm
2 2 2 2
( )
= = + +
 
O
I y z dm x z dm x y dm
2 2 2 2 2 2
1
( ) ( ) ( )
2
 
= + + + + +
 
  
O x y z
I I I I
1
( )
2
= + +
2 2
2 2
( ) ,
( )
x
y
I y z dm
I x z dm
= +
= +


Radius of gyration of a body about
an axis of rotation z is defined
2
/
z z z z
I m I m
 
=  =
Similarly, Ix and Iy z
m
z


33. Applied Mechanics - Department of Mechatronics - SME
-33-
Parallel-Axis Theorem
Huygens-Steiner Theorem
z Cz
I I md2

= +
z Cz
I r dm I r dm
2 2
, 

= =
 
2 2 2
2 2 2
2 2 2 2
2 2
, ,
2( )
2( )
C C
C C C C
C C
x x x y y y
r x y
r x y
x y x y x x y y
r d x x y y
 
= + = +
  
= +
= +
   
= + + + + +
  
= + + +
x y
z
O
C
d
r’
r
C
x
C
y
y
z
x
dm
Proof
C-center of mass
x dm y dm
0, 0
 
 =  =
z C C z
I r dm r d x x y y dm I md
2 2 2 2
'
[ 2( )] .
  
=  =  + + + = +
C-center of mass
As shown in figure, we have

34. Applied Mechanics - Department of Mechatronics - SME
-34-
Moment of Inertia of homogeneous rigid body
a) Slender rod AB with mass m and length L.
z
I x dm
2
= 
m
dm dx
l
=
z
dm
C x
x dx
A
B
z’
With constant cross-section:
l
l
z l l
m
I x m l dx x ml
l
/2
/2
2 3 2
/2 /2
1
( / )
3 12
− −
= = =

l
l
Az
m
I x m l dx x ml
l
2 3 2
' 0 0
1
( / )
3 3
= = =

Cz Az
I ml I ml
2 2
'
1 1
&
12 3
= =
Axis Cz ⊥ AB, CA=CB
Axis Az’ ⊥ AB

35. Applied Mechanics - Department of Mechatronics - SME
-35-
Moment of Inertia of homogeneous rigid body
b) Thin ring (m, r)
2 2
2 2
Cz
O
I dm r dm
r dm mr I

= =
= = =
 

x y
I I
=
2 2
1
,
2
x y O Cz
I I mr I I mr
= = = =
dm
O=C x
y
m
r
1
( )
2
O x y z z Cz
I I I I I I
= + + = =
Note: IO and Ix = Iy
Similarly, for a thin hollow round
cylinder (m, r)
Cz
I mr2
=
z

36. Applied Mechanics - Department of Mechatronics - SME
-36-
Moment of Inertia of homogeneous rigid body
c) Thin Disk (m, r)
Cz O
dI dm dI
2

= =
2
( / )2
dm m r d
  
= : 0 r
 →
r
r
Cz O
m m
I d mr I
r r
2 4 2
2 2
0 0
2 1
2
2
4
   

= = = =

dm
O=C x
y

d
Elementary ring (dm, , d). Moment of Inertia
of this element about Cz
x y Cz
I I mr I mr
2 2
1 1
,
4 2
= = =
O x y z z
I I I I I
1
( )
2
= + + =
Note: IO and Ix = Iy
Similarly, for a cylinder (m, r)
Cz
I mr2
1
2
=
z

37. Applied Mechanics - Department of Mechatronics - SME
-37-
Moment of Inertia of homogeneous rigid body
Disk with a hole (m, r, R)
Cz O
dI dm dI
2

= =
2 2
,
( )
mdA
dm dA d d
R r
  

= =
−
r R
: , : 0 2
  
→ →
R
R
Cz r
r
m m
I d d
R r R r
m R r m R r
R r
4
2
2
2 2 2 2
0
4 4 2 2
2 2
2
4
( ) ( )
( ) ( )
2
2( )
  
   
 
= =
− −
− +
= =
−
 
Element of mass (dm, , d, , d). Moment of
Inertia of this element about Cz-axis :
Similarly, for a thick hollow round
cylinder (m, r, R)
Cz
m R r
I
2 2
( )
2
+
=
dm
O=C x
y

d
R
r
d

z

38. Applied Mechanics - Department of Mechatronics - SME
-38-
Moment of Inertia of homogeneous rigid body
Rectangle plate (m, a, b)
To determine Moment of Inertia of plate about Cy-axis:
2
2
/2
2 2
/2
1
( ) ,
12
1
12
1 1
12 12
Cy
Cy
b
Cy
b
m
dI dm a dm dy
b
m
dI a dy
b
m
I a dy ma
b
−
= =
 = 
= =

Similarly, Moment of Inertia of plate about Cx-axis : 2
1
12
Cx
I mb
=
For thin plate:
Cz C
I I
=
( )
1
2
C Cx Cy Cz
I I I I
= + +
2
1
12
Cy
I ma
=
m
C x
y
a
b
dm dy
y
2 2
1
( )
12
Cz Cx Cy
I I I m a b
= + = +

39. Applied Mechanics - Department of Mechatronics - SME
-39-
Summary
z
C
A
z
L/2 L/2
2
2
1
12
1
3
Cz
Az
I ml
I ml
=
=
m
O=C x
y
r
Huygens-Steiner theorem 2
z Cz
I I md
= + C-center of mass
m
O=C x
y
r
Cz
I mr2
1
2
=
Slender Rod Thin ring Disc
2
Cz
I mr
=

40. Chapter 5 Planar Kinetics of a Rigid Body
-40-
Composite Bodies
If a body is constructed of a number of simple shapes, such as
disks, spheres, or rods, the mass moment of inertia of the body
about any axis can be determined by algebraically adding
together all the mass moments of inertia, found about the same
axis, of the different shapes.
COMPOSITE BODIES
Algorithm:
1. Calculation of the mass moment of inertia for
each shape (about arbitrary axis)
2. Converting all moments of inertia of shapes to
common axis (using Parallel Axis Theorem)
3. Algebraic adding

41. Chapter 5 Planar Kinetics of a Rigid Body
-41-
The pendulum can be divided into a
slender rod (r) and a circular plate (p).
Calculate the moment of inertia of the
composite body. Then, determine the
radius of gyration.
Given: The pendulum consists of a slender
rod with a mass 2 kg and a circular
plate with a mass of 4 kg.
Find: The pendulum’s radius of gyration
about an axis perpendicular to the
screen and passing through point O.
Plan:
EXAMPLE

42. Chapter 5 Planar Kinetics of a Rigid Body
-42-
The center of mass for rod is at point Gr, 1 m from
Point O. The center of mass for circular plate is
at Gp, 2.5 m from point O.
IO = IG + (m) (d) 2
For the rod: IOr = (1/12) (2) (2)2 + 2 (1)2 = 2.667 kg·m2
For the plate: IOp = (1/2) (4) (0.5)2 + 4 (2.5)2 = 25.5 kg·m2
3. Now add the two MMIs about point O.
IO = IOr + IOp = 28.17 kg·m2
1. Calculate MMI for a slender rod and a circular
plate (or using handbook).
2. Using those data and the parallel-axis theorem,
calculate the following:
SOLUTION
R
P

43. Chapter 5 Planar Kinetics of a Rigid Body
-43-
SOLUTION (continued)
R
P
4. Total mass (m) equals 6 kg
Thus the radius of gyration about O:
ρO = 𝐼O/𝑚 = 28.17/6 = 2.17 m

44. Applied Mechanics - Department of Mechatronics - SME
-44-
Moment of inertia of composite bodies
Determine the moment of inertia about an
axis perpendicular to the page and passing
through the pin at O. The thin plate has a
hole in its center. Its thickness is h = 50 mm,
and the material has a density  = 50 kg/m3.
O
r
a
a
a
a
C

= + =
(p) 2 2 2
1
( ),
12
C p p
I m a a m a h
For a square plate
For the hole

= =
(h) 2 2
1
,
2
C h h
I m r m r h
= − =
= + − =
(p) (h)
2
...
( ) , 2 / 2
C C C
O C p h
I I I
I I m m d d a

45. Homework
• Read Textbooks and slides
• Do exercises : TEAMS: Force-Acceleration: slides 6-14
• Send your homework to my email
( only in 1 file and in Subject of Email specify:
your name, class, Force_Acceleration_slidenumber_.....)
• See you soon