DEPARTMENTAL PUBLICATION NO 37
DYNAMICS Part B
(TTB002)
by
Prof. Victor Krylov
Department of Aeronautical and Automotive Engineering, Loughborough University, Loughborough, Leicestershire LE11 3TU
September 2007 © Copyright This publication may not be reproduced in whole or in part without the written permission of the author
2
PREFACE The module Dynamics studied by all Part B students in the Department is a continuation of the Part A Module Engineering Mechanics, both Statics and Introduction to Dynamics. The main aim of the present module is to develop and deepen students’ knowledge of Dynamics, so that they would be able to analyse and understand the behaviour of complex dynamic systems. The ability to understand and predict the behaviour of such systems is extremely important for aeronautical and automotive engineers. For that reason, the module Dynamics is one of the basic components of the fundamental engineering education, and the acquired skills can be applied to studying more specific optional modules in Parts C and D. These Lecture Notes reflect the material delivered on lectures, and they are aimed to provide students with the basic information about methods and problems of Dynamics written in a concentrated form. The logical structure of these Notes is the following. Consideration starts with the dynamics of single particles, which mainly discusses the central-force motion and the associated aspects of orbital motion of planets and spacecraft. Then, the dynamics of systems of particles is considered, including the impact phenomena. The next step in the logical development of Dynamics is the dynamics of rigid bodies that are considered as systems of particles with fixed distances between each other. After that, a special consideration is given to a general and efficient approach to solving dynamic problems of complex systems – Analytical Mechanics. The main tool of Analytical Mechanics, Lagrange’s equations, are introduced in detail and are illustrated on several practical examples. The next chapters of these Lecture Notes are devoted to vibration of particles, rigid bodies, and complex systems with two or more degrees of freedom. The analytical technique based on Lagrange’s equations is applied widely to analysing many of the problems considered in these chapters. The last two chapters discuss non-linear vibrations and self-excited oscillations, including a simple theory of wing flutter. I am much grateful to Mrs. Pat Griffin for her careful word-processing of the manuscript and for her computer drawing of numerous figures.
Victor V. Krylov Loughborough, September 2007
3
CONTENTS
I Dynamics of Particles 6 1. Central -force motion 6
2. Orbital motion of planets and spacecraft 13
2.1 Worked example 17
II Dynamics of Systems of Particles 18
1. Generalised Newton’s 2nd law 18
2. Linear momentum of a system 19
3. Angular momentum of a system 20
4. Conservation of energy of a system 22
5. Impact phenomena 23
III Dynamics of Rigid Bodies (2D) 30
1. Basic definitions 30
2. Equations of motion of a rigid body 32
3. Work and energy 35
4. Impulse-momentum equations for rigid bodies 38
IV Introduction to Analytical Mechanics 41
1. Generalised coordinates 41
2. Lagrange’s equations 42
3. Some applications of Lagrange’s equations 47
3.1 Planetary motion 47
3.2 System of two inter-connected bodies 48
3.3 Double pendulum 50
4. Principle of least action (Hamilton’s principle) 52
5. Note on Hamilton’s equations 54
4
V Vibration of Particles 55
1. Free vibrations 55
2. Effect of damping 63
3. Forced vibrations 67
3.1 Force excitation 67
3.2 Base excitation 71
3.3 Vibrations due to rotating unbalance 74
VI Vibration of Rigid Bodies 76
1. Typical examples 76
1.1 Physical pendulum 76
1.2 Log suspended via two ropes 77
1.3 Log suspended via two rods 78
1.4 Bar with a torsional spring 80
2. Energy method of determining the frequency of vibration 81
2.1 Rocking motion of a semi-disk: comparative
analysis by energy method and by means of
Lagrange’s equations 82
VII Systems with Two or More Degrees of Freedom 85
1. Free vibrations 85
2. Matrix notation 88
3. Some practical examples 89
3.1 2-DOF vehicle model for bounce and pitch 89
3.2. Symmetrical 2-DOF system 92
4. Initial conditions 94
5. Forced harmonic vibrations 97
5.1 Construction of the inverse matrix 98
5.2 Representation in terms of normal mode
summation 101
5.3 Principle of tuned vibration absorber 102
5
VIII Non-linear Vibrations 104
1. Perturbation analysis of a non-linear pendulum 104
IX Self - excited Oscillations 110
1. Systems with ‘negative friction’ 110
1.1. Mass on the moving belt 110
1.2. Froude’s pendulum 113
2. Simple theory of wing flutter: 2-DOF model 114
2.1. Coupled bending and torsional vibrations 114
2.2. Effects of aerodynamic lift and moment 115
2.3. Conditions of instability 116
Recommended Literature 118
6
I. DYNAMICS OF PARTICLES
1. CENTRAL FORCE MOTION This type of motion is extremely important in nature: movement of planets,
elementary particles (electrons, protons, etc). In engineering, it defines the
movement of rockets, earth satellites and other space vehicles.
Let the central force F(r) be the attraction (pulling) force:
F(r) = - F(r) er = - F (r) rr
Here r is a radius-vector defining the position of a particle, and e r is a unit
vector in the direction of r. Then, Newton’s 2nd law for a particle with the
mass m subject to such a force can be written as:
- F(r) er = m r&& (1)
In particular, for gravity force F(r) = G 2o
rmm , where m0 is the mass of the
attracting (heavy) body, and G is the gravity constant.
M
r
X
v
F r
θ m0
dθ
O
m
7
Particle motion caused by a gravity central force
Consider particle motion caused by a gravity central force. Using polar co-
ordinates r and θ, the following expressions for the first and second time
derivatives of r can be written:
( ) ( )θ
θ
θθ
θ
erdtd
rerrr
ererr
r
r
&&&&&&
&&
22
.
1+−=
+=
(see e.g. Engineering Mechanics: Dynamics. Part A)
Then, it follows from Equation (1) that
( ) ( )
( ) ( )310:
2:
2
22
θ
θ
θ&
&&&
rdtd
rme
rrmr
mmGe o
r
=
−=−
It follows from (3) that ( ) 02 =θ&rdtd
or h const 2 ==θ&r (4)
Note that Equation ( 4) describes conservation of the angular momentum of
the particle of mass m:
θ&2mrvmxr =
8
Area swept out by the radius vector during the time dt:
( )
θθ
θ
&& 2221
21
21
rdtdr
dtdAA
rdrdA
====>
⎟⎠⎞
⎜⎝⎛=
But, according to Equation (4), const A =& .
This gives the mathematical proof of Kepler’s 2nd law (J. Kepler, 1571 -
1630), which was first established empirically. This law reads: ‘Areas swept
through in equal times are equal’.
Determination of the path of the body:
Let us determine the path of the body from the above equations by excluding
time t. Let us first introduce the useful change of variable:
uu
rThen
ur
&&2
1
1
−=
=
Using Equation (4), we can rewrite this expression as
( )θθ dduhuhr −=⎟
⎠⎞⎜
⎝⎛−= &&&
Continuing differentiation one gets
2
2θ
θ&&&
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
dudhr
Using Equation (4) again gives:
9
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
2
222
θduduhr&&
Now substitute r&& into Equation (2) The result is
422
2222 1 uh
ududuhuGmo −−=−
θ
or 22
2
h
Gmu
dud o=+
θ (5)
This is a non-homogenous linear differential equation.
Its solution is (check by substitution into Equation (5)!):
( )2cos1
h
GmC
ru o++== δθ
Here C and δ are constants of the integration. If to assume r = min at θ =0,
then δ = 0 and
2 cos 1h
GmC
ro+= θ
or eddr1 cos 11
+= θ , (6)
where we have introduced the new notations:
oGmCheCd
2 and 1 ==
10
Analysis of Equation (6)
Equation (6) describes three different types of curves (particle trajectories),
depending on the value of the parameter e (in polar coordinates) :
hyperbolae
ee
−>−=−<
1parabola1ellipse1
1. Ellipse )1( <e
Obviously,
2maxmin12
112
eed
eed
eedrra
−==
−+
+=+=
Other useful expressions for minr and maxr :
( ) ( )earear +=−= 1:1 maxmin
1e >
1e <
0e =
1e =
Perigee Apogee 2b
2a
om
θ X
1
1
11
If to express d in terms of a, then Equation (6) can be rewritten as
( )21cos11ea
er −
+=
θ (7)
The value of 2b is determined from the geometrical properties of the ellipse:
2b = 2a 21 e−
Note that Equations (6) and (7) reflect Kepler’s 1st law, which reads:
‘The planets move in elliptical orbits around the sun as a focus’.
The period of a particle motion on the elliptical orbit can be calculated as
hab
r
abAA π
θ
πτ 2
21 2
===&&
.
Using the above-mentioned geometrical expressions for a and b and
applying the well-known relationship Gm0 = gR2 for low altitude orbits, where
g is the gravity acceleration and R is the average radius of the attracting
body (e.g. planet), one can rewrite the above formula as
or gR
a 23
2πτ =
This is the mathematical expression of Kepler’s 3rd law:
‘The square of the period of motion is proportional to the cube of the
semi major axis of the orbit’.
12
2. Parabola ( )1e =
In this case, Equation (6) takes the form
( )θcos 111+=
dr
One can see that
πθ →
∞→→
when
rorr
01
This means that a particle moves away from the attracting body.
m
r
X
mo
θ
13
3. Hyperbola ( 1e > )
It follows from equation (6) in this case that
∞→→ rorr
01
when e1cos −=θ
i.e. at ⎟⎠⎞
⎜⎝⎛−±== −
e1cos 1
1θθ
2. ORBITAL MOTION OF PLANETS AND SPACECRAFT Expressions for a Particle’s Velocity
In this section we consider an important case of orbital motion of planets and
spacecraft. It is often necessary to determine the velocity of the orbiting body
in any point of the orbit. The easiest way to do so is to use energy
conservation law:
m
X
1θ
1θ− om
14
constV =+Τ=Ε .
Here E is the total energy of a particle, and T and V are its kinetic and
potential energies respectively.
Note that ( )2222
21
21 θυ && rrmm +==Τ
Considering V, we recall that near the surface of the attracting body (e.g.
near the earth surface)
( ) hmghhmgVV g Δ=−=Δ=Δ 12
As we mentioned already, near the Earth surface
2
2
gRGm
mgR
GmmF
o
og
==>
==
Then, the change in gravitational potential energy for large hΔ = r’ – r will
be:
gg
r
rg
VV
rrmgR
rdrmgRV
−′=
=⎟⎠⎞
⎜⎝⎛ −=−=Δ ∫ '
22
2 11'
2h
For large hΔ one should use a more precise expression for the gravity force:
2r
GmmF o
g = Δh
h1
15
In space problems it is customary to take Vg’ = 0 at 1r = ∞
Therefore,
Vg = -r
mgR2 (8)
Then, the energy conservation law can be written as
( ) constr
mgRrrm =−+=Ε2
22221 θ&& (9)
or
constr
mgRm =−=Ε2
221 υ
Hence
( )mr
gRr Ε+==
212 222 υυ (10)
It is convenient to express Ε via other parameters of the problem and
substitute such expressions into Equation (10) so that to obtain simple and
useful expressions for v.
In what follows we consider elliptical orbits only.
Take θ = 0 in Equation (7). Then, it follows from Equation (7) that
( ) ( )eaeae
rr −=
−
+==
11
1111
2min
Note that r& = 0 at θ = 0 and always rhrhr ==>= θθ &&2
Then it follows from Equation (9) that:
16
=−=Εr
mgRrhm
2
2
2
21
( ) ( )
( )aee
dCand
CeGm
hBut
eamgR
eahm
o2
2
2
22
2
11
1121
−===
−−
−=
Then
( ) ( )
( )( ) ( ) a
mgRea
mgRea
eagRm
eagReaGmh o
2111
21
1122
22
22
222
−=−
−−
−=Ε=>
−=−=
Substituting Ε = -a
mgR2
2 into Equation (10), we obtain the following
expression for v in the arbitrary point r of the orbit:
⎟⎠⎞
⎜⎝⎛ −=
argR
2112 22υ (11)
For some particular positions of the particle, the following useful expressions
can be easily obtained from Equation (11):
1. At r = rmin =a(1-e) (a perigee)
min
max11
rr
agR
ee
agRp =
−+
== υυ
2. At r = rmax = a(1 + e) (an apogee)
max
min11
rr
agR
ee
agRA =
+−
== υυ
17
2.1 WORKED EXAMPLE
Problem 3/275 (M & K, 5th Ed. )
1. Find υ for a circular orbit (e=0)
For e=0 => a=r=R+H
Then, using Equation (11), one obtains
HR
gRrr
gR+
=⎟⎠⎞
⎜⎝⎛ −=
12112 222υ
Hence )( HRgR +=υ
2. Escape means that a ∞→
Then, from Equation (11):
( )
( )υυυυ
υ
υ
12
2
12112 222
−=−=Δ
+=∴
+=⎟
⎠⎞
⎜⎝⎛
∞−=
esc
esc
esc
HRgR
HRgR
rgR
3. Substituting R = 6371 Km and g = 9.825 m/s2, we obtain:
skmvskmvskm esc /20.3,/92.10,/72.7 =Δ==υ
XH
R
A spacecraft is on a circular orbit around the earth at H = 320 Km. Determine the increase in velocity
υΔ necessary to escape the earth
18
II. DYNAMICS OF SYSTEMS PARTICLES
1. GENERALISED NEWTON’S SECOND LAW (aka the theorem about the motion of a centre of mass of a system;
see also Engineering Mechanics: Statics )
∑≠
+=ij
jiext
iii FFam
j
i ij
ext
ii
ii iFiFam ∑ ∑∑∑
≠+=
Let us introduce tot
ii
cm m
irmr
∑= (1)
where mtot = ∑i
im
∑
∑∑
=
=
===
i
exticmtot
cmtot
cmtoti
iii
ii
Fam
am
rmrmam &&&&
(2)
Ζ
X
y ir
then
0
19
2. LINEAR MOMENTUM OF A SYSTEM
By definition,
ii
ii
idef
iidef
i
mGG
vmG
υ∑∑ ==
= - for a single particle and for a system of particles
Using equation (1), one can easily obtain that
cmtotVmG = (3)
If 0=∑ ext
iiF , then it follows from equation (2) that
0=cmtot am
i.e. constVmG cmtot == (4)
If only a projection of ext
iiF∑ is zero, e.g. x – projection, then only
constVm cmxtot =,
Using the above definition of linear momentum of a system, it is often
convenient to use the equivalent form of the Equation (2):
ext
iiFG ∑=& (5)
20
3. ANGULAR MOMENTUM OF A SYSTEM
1) Angular momentum about a fixed point O. By definition, the angular momentum about a fixed point 0 is:
( )∑∑ ×==i
iiii
oio vmrHH
Let us consider oH&
( ) ( )∑∑ ×+×=i
iiii
iiio vmrvmrH &&&
But ( ) ( )∑∑ ×=×i
iii
iii Frvmr & ,
where ∑≠
+=ij
ijext
ii FFF
As before, only extiF give non-zero contribution.
Hence ( ) ( )∑∑ ×=×i
extii
iiii FrVmr &
Governing equation It follows from the above that:
( )∑ ∑ ×==i i
extii
extoio FrMH& (6)
2) Angular momentum about a mass centre G
)(det ∑ ×=i
iiiG rmH &ρ (7)
0
G
0
iρ
ir
Gr
iυ
0
ri i
21
But iGi rr ρ+= &&& ,
Then ( ) ( )∑ ∑ ×+×=i i
iiiGiiG mrmH ρρρ &&
Noticing that ( ) ( )iGiGii rmrm ρρ ×−=× &&
One obtains ( )43421
&&
Gforll
ii
iGi
Gii mrrm
0
ρρ ∑∑ ×−=×
Hence ( )∑ ×=i
iiiG mH ρρ & (8)
Equations (7) and (8) give two different forms of angular momentum of the
system about G – absolute angular momentum (7) and relative angular
momentum (8), and they are identical.
Let us now consider :HG&
Taking, for example, Equation (7), one can express GH& as
( ) ∑∑ ×++×= iiii
iGiiG rmrmH &&
444 3444 21
&&&& ρρρ
0
The equality of the first term to zero in the above expression can be proven
using the definition of the mass centre and some properties of cross products.
The rest can be rewritten using Newton’s 2nd Law:
( ) ( )∑∑ ×=×i
iii
iii Frm ρρ && ,
where ∑≠
+=ij
jiext
ii FFF
22
As only extiF give non-zero input, one can obtain the following governing
equation:
( )∑∑ ×==i
extii
ext
iGG FMH ρ& (9)
Note that GH in Equation (9) can be written both in ‘absolute’ and ‘relative’
forms.
Conservation of Angular Momentum of a System
If the sum of moments of external forces is zero about O or about G, then it
follows from Equations (6) or (9) that
constHorconstH Go ==
or, which is the same, 0H,0H Go =Δ=Δ
If only certain projections of the external force moments are zero, then only
the corresponding projections of H0 or HG are zero.
4. CONSERVATION OF ENERGY OF A SYSTEM
If no work is done on a conservative system then,
consti
i =Ε=Ε ∑
or, which is the same,
0=ΔΕ (10)
or
finalinitial Ε=Ε
23
More specifically, one can rewrite Equation (10) as:
0=Δ+Δ+Δ eg VVT
Here T is the kinetic energy of a system, and Vg and Ve are parts of its
potential energy associated with gravity and elasticity respectively.
5. IMPACT PHENOMENA
There are three stages in the impact phenomena:
1-D case (consider this case first for simplicity)
21 υ>υ
Before Impact
0υ
Collision and deformation
'1υ < '
2υ
After impact
m1 m2
m1 m2
m1 m2
24
Repulsive normal forces during the collision in the case of central impact
time
2-D case
Fn(t)
to
n
t
1m
1m
2m
2m
'2υ
2υ
'1υ 1υ
25
Conservation of Linear Momentum at the impact
As there are no external forces acting on the system of two colliding particles
its total linear momentum conserves in the normal direction to the contact
surface:
( ) ( ) constvmG nii
in == ∑ .
For example, for a one-dimensional impact:
constvmvmvmvmG =+=+= '22
'112211 (1)
Also, one needs to take into account the physics of the impact.
The simplest way to do so is to use the experimental law of restitution
(I. Newton)
This law states that ‘when two bodies collide their relative parting velocity in
the direction of the common normal at the point of impact is equal to –e times
their relative approach velocity in this direction’
Or
( ) ( )( ) ( )nn
nnvvvv
e12
'1
'2
−
−−= (2)
Here e≤ 1 is the restitution coefficient
Coefficient of restitution, e, varies from 0 (perfectly plastic impact) to 1
(perfectly elastic impact).
26
For One-dimensional Problems
12
'1
'2
vvvve
−−
−= (2’)
Then from Equations (1) and (2’) one can find final velocities ν1’ and ν2’ :
( ) ( )[ ]2112221
1 11' emmemmm
−+++
= ννν
(3)
( ) ( )[ ]1221121
2 11' emmemmm
−+++
= ννν
Consider an important particular case of mmm 21 == in Equations (3). Then
( ) ( )[ ]
( ) ( )[ ]ee
ee
−++=
−++=
1121'
1121'
212
121
ννν
ννν
Consider now a perfectly elastic impact (e=1). Then
1221 '' νννν == and ,
i.e. the exchange of velocities takes place.
Energy Losses During Impacts
This always takes place. Therefore, in reality e < 1.
In the idealised case of perfectly elastic impact, e =1, there is no energy
loss, i.e. the total energy conserves:
27
constm ii
i ==Τ=Ε ∑ 2)2/1( ν
In the above-considered 1-D case this takes the form:
constmmmm=+=+=Τ
2'
2'
22
222
211
222
211 νννν (4)
Assume for simplicity that mmm 21 ==
Then 22
21
22
21 '' νννν +=+ (*)
On the other hand, it follows from Equation (1) that
'' 2121 νννν +=+ (**)
Equations (*) and (**) give an alternative solution to the problem for the ideal case of perfectly elastic impact. This alternative solution does not require the knowledge of the restitution coefficient.
Let us prove that in this idealised case e = 1 indeed.
To do so let’s rewrite Equation (*) in the form
22
22
21
21 '' νννν −=− ,
which can be rewritten also as
( )( ) ( )( )22221111 '''' νννννννν +−=+− . (***)
But, it follows from Equation (**) that
2211 '' νννν −=− (****)
Therefore, using Equation (****) in (***), one obtains
2211 '' νννν +=+ (*****)
28
On the other hand, if
1''
12
12 =−−
−=ννννe ,
then 22112112 '''' νννννννν +=+−=− or .
Obviously, the latter expression coincides with Equation (*****) following from
the energy conservation law in the case of idealised perfectly elastic impact,
which proves the point!
For e < 1 energy loss takes place. In the case e = 0 corresponding to
perfectly inelastic impact or, which is the same, perfectly plastic impact the loss of energy is a maximum.
2-D case: Oblique Impact
In this case there are four unknowns:
( ) ( ) ( ) ( )tntn and '''' 22,1,1 νννν
n
t
1m 1m
2m2m
'2υ 2υ
'1υ 1υ
θ2
'1θθ1
'2θ
29
Equations to find the unknowns:
1. Linear momentum of the whole system conserves in the normal
direction: G(n) = const:
or
( ) ( ) ( ) ( )nnnn mmmm '' 22112211 νννν +=+
2. Linear momentum of each particle conserves in the tangential
direction (as there are no impact forces along t):
( ) ( ) ( ) ( )tttt mmmm ',' 22221111 νννν ==
3. Finally, the remaining fourth equation is Equation (2) for the restitution coefficient e:
( ) ( )( ) ( )nn
nnvvvv
e12
'1
'2
−
−−=
Solving the above-mentioned system of four simultaneous equations gives the
solution of an arbitrary problem of central impact at oblique incidence.
30
III. DYNAMICS OF RIGID BODIES (2D)
1. BASIC DEFINITIONS We recall that a rigid body is a system of particles with fixed distances
between all of them. Some general expressions useful to describe motion of
a rigid body are reminded below.
We recall (see Engineering Mechanics: Statics) that the arbitrary motion of a
rigid body can be described completely as the translation motion of one of its
points (say point O) and the rotation of the body around that point.
ανω rarra tn === ;22
x
y
ý
xř
R
r
O
( )rVor
V
rRr
×+=
+=
+=
ων
νν
~
~
~
o
r θ
ων
θωωα
θθω
r
dtd
dtd
=
===
==
&&&
&
31
2D – Case (Plain Motion)
A simple but important case of plain motion of a rigid body is characterised by
three coordinates (degrees of freedom): two coordinates of the mass centre
and the rotation angle θ.
We recall that the angular momentum about the centre of mass G is:
iii
iG mH ρρ &×= ∑
In 2D-case the magnitude of GH is
∑
∑
=
==
iii
iii
iG
m
mH
2
.
ρω
ωρρ
Introduce the notation
ii
i mI ∑= 2ρ - mass moment of inertia about Z-axis.
For a body made of a continuum the summation is replaced by the integration:
∫∑ == dmmI ii
i22 ρρ
Then ωIHG = (3)
y
Gρ
32
2. EQUATIONS OF MOTION OF A RIGID BODY
Let us apply the general equations for a system of particles derived earlier.
1. Generalised Newton’s second law:
∑=i
exticmtot Fam (1)
2. Momentum equation about the mass centre:
( )∑∑ ×==i
extii
i
extGiG FMH ρ& (2)
The above equations are valid in the general 3-D case.
In what follows we consider only 2- D case in detail.
The momentum equation (Equation (2)) then takes the form (see previous
section)
∑==i
extGiMII αω& (4)
This should be complemented by the generalised second Newton’s law:
(5)
where m = mtot and cmG aa =
Note that we already derived Equation (4) in ‘Eng. Mechanics: Statics’ by
another (a more direct) method.
∑=i
extiG Fam
33
Recall the example.
Rotation Around an Arbitrary Fixed Axis
But GGi
iiGiii
G mrmrmr υυυ ×=×=× ∑∑ (using definition of G)
and Giii
i Hm =×∑ υρ (in absolute form)
d
R
F
ϕ
2
2
22 ~
mRFd
FdmR
FdI
mRrmI ii
i
=
=
=
= ∑
ω
ω
ω
&
&
&
ω
G
i
ρ
O
Gr
ir~
iGi rr ρ+=~
Angular momentum about O:
( )
( )
( )iii
iiiG
iiiio
m
mr
vmrH
υρ
υ
×+
+×=
=×=
∑
∑ w
F
34
Then, GGGo mrHH υ×+= (6)
Consider magnitude of oH :
GGGo rmrHH ω.+= (6’)
If to recall that GH = Iω, where ii
i mI ∑= 2ρ , then Equation (6’) can be
rewritten as
H0 = I0ω,
where
20 GmrII += . (7)
Equation (7) describes the so-called Steiner’s theorem, which is used widely
for calculation of mass moments of inertia versus different axes.
The momentum equation in this case takes the form
ext
ioioo MII ∑== αω& (8)
where all oiM ext are now calculated about point O.
35
3. WORK AND ENERGY
Reminder
Work done by a force F
( )∫ ∫== dsFUorrdFU αcos
Work done by a couple
∫ θ= MdU
Let us now introduce the expressions for the kinetic energy of a rigid
body for different types of its motion:
1. Translation
Derivation of the corresponding expression for this case is quite obvious:
∑∑ ===Τ 22221
21
21 mvmvvm iii
i (9)
2. Fixed Axis Rotation
This case is easy as well:
36
222221
21
21 ωω o
Io
ii
iiii
Irmvm ===Τ ∑∑43421
(10)
3. General Plane Motion
In this case the derivation is a bit more complicated:
)cos2(21
21 2222 θωρωρ iGiG
iiii
ivvmvm ++==Τ ∑∑ (*)
Consider the last term in Equation (*):
∑
∑
==
=
θρω
θωρ
cos
cos221
iiG
iGii
mv
vm
0== ∑ ii
iG ymvω
Due to the definition of the centre of masses:
0==∑ Gii
i myym
Then the kinetic energy takes the form
2221
21 ωImvG +=Τ (11)
Here I is the moment of inertia of the body about its mass centre.
Thus, the contributions of translation and rotation into the kinetic energy of the
body are separate.
ω
G υΓ
υΓ
ρ/ υi
ρi
mi
yi
0
37
Instantaneous Centre of Zero Velocity
Practical Example: A Rolling Wheel.
υυυυω 22
==>== AARR
Useful expression for the kinetic energy in terms of the rotational velocity
about the instantaneous centre C of zero velocity:
221 ωcI=Τ
B
A
C
rA rB
Aυ
Bυ B
B
A
A
rrυ
=υ
=ω
A υA
R
C
υ
38
Energy Relations
Let us start from the earlier mentioned general expression valid for a system
of particles:
eg VVU Δ+Δ+ΔΤ=−21'
Here 21'
−U is the work done by external non-gravitational and non-elastic
forces on the transfer of the system from the situation 1 to the situation 2.
Specifics of Rigid Bodies
In rigid bodies the distances between all particles are constant. Then, ΔVe =
0. In the light of this,
gVU Δ+ΔΤ=−21'
If there is no work done by external forces, i.e. 021' =−U , then
0=Δ+ΔΤ gV
This is the energy conservation law for rigid bodies.
4. IMPULSE MOMENTUM EQUATIONS FOR RIGID BODIES
Consider the generalised Newton’s second law:
39
dtFVmd
FdtVd
m
Fam
extt
tiG
iv
i
exti
G
i
extiG
∫∫
∑
∑
=
=
=
2
1
2υ
or, which is the same,
{ {
1
1
2
2
2
1 GG
t
t
exti mmdtF υυ −=∫
This holds for all projections.
If dtF2
1
t
t
exti∫ = 0, then
12 GG = ,
which describes conservation of linear momentum of a rigid body.
Consider now the angular momentum equation:
∑=i
extGiG MH& (*)
or ∑=i
extGiMIω& (**)
It follows from Equation (*) that
dtMHHt
t i
extGiGG ∫∑=−
2
112
40
Similarly, it follows from (**) that
( ) dtMIt
t i
extGi∫∑=−
2
112 ωω
if 0dtM2
1
t
t i
extGi =∫∑ , then
1212 ωω IIorHH GG ==
which describes conservation of the angular momentum.
Worked Example. (Problem 6/114, M & K 5 Ed.)
Use energy conservation law:
( )( )0
21
0
22 −=ΔΤ
=Δ + ΔΤ
G
g
vm
V
(there is only translation movement and no rotation)
( )
( )o
g
Hence
g
mgV
2.33
0cos15421
cos55
2
=
=−−∴
−−=Δ
θ
θ
θ
5m 5m θ θ
υ = 4 m/s
The suspended log is used as a battering ram. Find θ providing the final velocity υ = 4 m/s
41
IV. INTRODUCTION TO ANALYTICAL MECHANICS
This is a generalised theory of mechanics (dynamics) established largely by
works of J.-L. Lagrange (1736 – 1813) and some other scientists. It greatly
facilitates the analysis of complex dynamic systems.
1. Generalised Coordinates
Consider a system of N particles. To describe the positions of all
particles at any instant t one should know 3N coordinates.
The number S of independent co-ordinates (or the number of
degrees of freedom) can be smaller than 3N, i.e.
S ≤ 3N
Thus, a system can be described by any S independent variables qi ,
that are called generalised coordinates
Siqi ..........2,1; =
42
Example: Spherical Pendulum
Thus, the system has only two degrees of freedom.
On the other hand, the same system is described completely by two
spherical coordinates ( angles ϕ and θ ). The third coordinate , r, is not
needed as always r = l.
In this example, the coordinates ϕ and θ can be considered as generalised
coordinates.
2. Lagrange’s Equations
Lagrange’s equations are differential equations of motion expressed in terms
of generalised coordinates.
Consider first a conservative system, i.e. a system in which the sum of kinetic
and potential energy conserves. Thus,
( ) 0=+Τ Vd (1)
X
Z
y
ϕ l θ
This system can be described traditionally by three Cartesian coordinates: x, y, z. Only two of them are independent due to the obvious constraint 2222 lzyx =++
43
Let us represent Τ and V as functions of the generalised coordinates qi
and their time derivatives iq& (that are called generalised velocities):
( )( )s
ssqqqVV
qqqqqq...,
...,,...,
21
2121=
Τ=Τ &&& (2)
Consider the differential of Τ:
Differential of Τ
is
i ii
s
i iqd
qdq
qd &
&∑∑== ∂
Τ∂+
∂Τ∂
=Τ11
(3)
Now consider the general expression for the kinetic energy Τ in generalised
coordinates (the so-called quadratic form):
jis
i
s
jij qqm &&∑ ∑
= ==Τ
1 121
(4)
Here ji
m are coefficients. It can be proven (Euler’s theorem) that the following
relationship holds for Τ defined by equation (4):
Τ=∂
Τ∂∑=
21
is
i iq
q&
& (5)
Let us use Equation (5) to exclude iqd & in equation (3)
First consider differential of Equation (5):
is
i ii
i
s
iqd
qdd &
&&
& ∑∑== ∂
Τ∂+⎟⎟
⎠
⎞⎜⎜⎝
⎛∂
Τ∂=Τ
112 (6)
44
Now subtract Equation (3) from Equation (6):
is
i ii
i
s
idq
qdd ∑∑
== ∂Τ∂
−⎟⎟⎠
⎞⎜⎜⎝
⎛∂
Τ∂=Τ
11&
& (7)
But
ii
i
ii
i
dqqdt
d
dtdq
qdq
qd
⎟⎟⎠
⎞⎜⎜⎝
⎛∂
Τ∂=
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂
Τ∂=⎟⎟
⎠
⎞⎜⎜⎝
⎛∂
Τ∂
&
&&
&
Then, it follows from Equation (7) that
∑= ⎥
⎥⎦
⎤
⎢⎢⎣
⎡
∂Τ∂
−⎟⎟⎠
⎞⎜⎜⎝
⎛∂
Τ∂=Τ
s
ii
iidq
qqdtdd
1 & (8)
Using the obvious expression for dV:
is
i idq
qVdV ∑
= ∂∂
=1
, (9)
the expression for energy conservation, d(Τ + V) = 0, can be rewritten as
01
=⎥⎥⎦
⎤
⎢⎢⎣
⎡
∂∂
+∂
Τ∂−⎟⎟
⎠
⎞⎜⎜⎝
⎛∂
Τ∂∑=
is
i iiidq
qV
qqdtd
& (10)
Because the s generalised coordinates iq are independent and, therefore
the idq can take arbitrary values, Equation (10) is satisfied only if
siqV
qqdtd
iii......2,1
0
=
=∂∂
+∂
Τ∂−⎟⎟
⎠
⎞⎜⎜⎝
⎛∂
Τ∂& (11)
Equation (11) is Lagrange’s equation (or Lagrange’s equations) for a
conservative system (all forces are potential ones).
45
It is useful to introduce Lagrange’s function L (or the Lagrangian):
VL −Τ=
Then, because 0/ =∂∂ iqV & , Equation (11) can be rewritten as
si
qL
qL
dtd
ii......2,1
0
=
=∂∂
−⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂& (12)
For Non - conservative Systems
( ) npUVd δ=+Τ (13)
where npUδ is the work done by non-potential forces.
Let us express npUδ in terms of generalised coordinates:
is
iinp qQU δδ ∑
==
1
where iQ are known as generalised forces associated with the
corresponding generalised coordinates iq
Then, with the account of Equation (13), the Lagrange’s equation takes the
form
Ni
QqV
qqdtd
iiii
.......2,1=
=∂∂
+∂
Τ∂−⎟⎟
⎠
⎞⎜⎜⎝
⎛∂
Τ∂& (14)
46
Or, if to use Lagrange’s function,
Ni
QqL
qL
dtd
iii
.......2,1=
=∂∂
−⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂& (15)
Example
Let us illustrate the application of Lagrange’s equation to a one-dimensional
movement of a particle of mass m subjected to the force F.
Then,
( )
0
0
221
21
,
1
1
1
1
2
11
=∂∂
=∂∂
=∂Τ∂
=∂
Τ∂
==⎟⎟⎠
⎞⎜⎜⎝
⎛∂
Τ∂
==∂Τ∂
=∂
Τ∂
=Τ
==
xV
qV
xq
xmxmdtd
qdtd
xmxmxq
xm
FQxq
&&&&
&&&&
&
Substituting the above expressions into Equations (14) or (15), we obtain:
Fxm =&& ,
which is the familiar Newton’s second law for a particle, as expected.
The advantage of using Lagrange’s equations shows in the case of complex
dynamic systems.
47
3. Some Examples of Application of Lagrange’s Equations
3.1 Planetary Motion
L.E. : 2,1
0
=
=∂∂
+∂
Τ∂−⎟⎟
⎠
⎞⎜⎜⎝
⎛∂
Τ∂
iqV
qqdtd
iii&
( )
rmm
GV
rrmm
qrq
o−=
+==Τ
==
2222
2
1
21
21 θυ
θ
&&
( )
21
2
1
1
1
rmmG
rV
qV
mrrq
rmrmdtd
qdtd
rmrq
o=∂∂
=∂∂
=∂Τ∂
=∂
Τ∂
==⎟⎟⎠
⎞⎜⎜⎝
⎛∂
Τ∂
=∂Τ∂
=∂
Τ∂
θ&
&&&&
&&&
mo
m
x
r θ
r&
θ&rυ
2o gRGm =
48
( )
0
0
2
2
2
2
2
2
=∂∂
=∂∂
=∂
Τ∂=
∂Τ∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂
Τ∂
=∂
Τ∂=
∂Τ∂
θ
θ
θ
θθ
VqVq
mrdtd
qdtd
mrq
&&
&&&
Then, it follows from LE that
( ) 0
0
2
22
=
=+−
θ
θ
&
&&&
rdtdm
r
mmGmrrm o
As expected, these equations are equivalent to the ones derived in Chapter 1.
3.2 System of Two Interconnected Bodies
(Problem 7.3-2 from the book of Thomson)
( )22
22
21
21
21
21
xrkkxV
Ixm
−+=
+=Τ
θ
θ&&
M(t)
k k
m R
r
I
xq1 = 2q=θ
for small x
49
To define the generalised non-potential force we recall that in this case
δθδδ MqMUnp == 2
Hence, the generalised force is M
Then,
xmqdt
d
xmxq
&&&
&&&
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂
Τ∂
=∂Τ∂
=∂
Τ∂
1
1
( )( )1
0
1
1
−−+=∂∂
=∂∂
=∂Τ∂
=∂
Τ∂
xrkkxxV
qV
xq
θ
( )rxrkVqVq
Iqdt
d
Iq
−=∂∂
=∂∂
=∂
Τ∂=
∂Τ∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂
Τ∂
=∂
Τ∂=
∂Τ∂
θθ
θ
θ
θθ
2
2
2
2
0
&&&
&&&
Then, it follows from LE that
( )tMkrkrxI
krkxxm
=+−
=−+
θθ
θ2
02&&
&&
50
N.B. In engineering practice, the values of mass moment of inertia, I, for
different bodies are often expressed in the form 2gmrI = , where rg is
the so-called radius of gyration. For example, for a thin ring of radius
R obviously rg = R. Generally, the radius of gyration is smaller than
the geometrical dimensions of the body.
3.3 Double Pendulum
(Example from the book of Landau & Lifshits)
Find Τ, V and Lagrange’s function L for the system shown in the
Figure.
11112
12
111 cos21 ϕϕ glmVlm −==Τ &
To find Τ2 and V2 , we express x2 and y2 in terms of ϕ1 and ϕ2 :
22112
22112coscossinsin
ϕϕϕϕ
llyllx
+=+=
x
y 2ϕ
1ϕ 1m
2m
2l
1l
22
11
ϕ=ϕ=
51
Then,
( )22
22
22 2
yxm
&& +=Τ
( ) ( )( ) ( ) 2221112
2221112sinsin
coscosϕϕϕϕ
ϕϕϕϕ&&&
&&&
llyllx
−−=+=
Hence
( ) ([ ) ( )( ) ) ( ) ]
[ ( ) ]212121212
22
22
12
12
222
222212121
211
221
222
222212121
211
221
22
sinsincoscos22
sinsinsin2sin
coscoscos2cos2
ϕϕϕϕϕϕϕϕ
ϕϕϕϕϕϕϕϕ
ϕϕϕϕϕϕϕϕ
+⋅++=
=+++
++⋅+=Τ
&&&&
&&&&
&&&&
llllmllll
llllm
But
( )212121 cossinsincoscos ϕϕϕϕϕϕ −=+
Then
( )[ ]2121212
22
22
12
12
2 cos22
ϕϕϕϕϕϕ −++=Τ &&&& llllm
( )221122 coscos ϕϕ llgmV +−=
( )
( )
( )
VL
glmglmmVVV
llm
lmlmm
−Τ=
−−+−=+=
−+
+++
=Τ+Τ=Τ
222
112121
2121212
22
22
2211
2121
coscos
cos22
ϕϕ
ϕϕϕϕ
ϕϕ
&&
&&
Substitution of the above expressions for Τ and V (or for L) into Lagrange’s
equations gives the two equations versus ϕ1 and ϕ2 that describe
completely the dynamics of the system under consideration.
52
4. Principle of Least Action (or Hamilton’s Principle)
Consider two moments of time, t1 and t2, in which the dynamic system is
characterised by the coordinates ( ) ( )2i
1i qandq , where i=1, 2……..s.
Now consider the integral
( ) ,,,2
1
dttqqLS ii
t
t&∫= (1)
where L = Τ - V is the Lagrangian of the system. The integral S is called
action. Its dimension is Energy x Time.
The principle of least action (P. L. Action) states that the dynamic system
moves in such a way that the integral (1) takes the minimum value.
Let us derive the differential equations that solve the problem of finding a
minimum of the integral (1). This is one of the general problems of the
Calculus of Variations.
Consider first a system with one degree of freedom, i.e. with one q(t). Let q(t)
is the function for which the integral in Equation (1) has a minimum. If to
replace q(t) by q(t) +δq(t), where δq(t) is a small function in the interval (t1,
t2) (it is called variation of the function q(t)), then the integral S will increase.
All functions q(t) + )t(qδ to be compared have to take equal values at t = t1
and t = t2 , i.e. it must be that ( ) 0)( 21 == tqtq δδ . Let us express ΔS from (1):
( )
( )dttqqL
dttqqqqLS
tt
tt
,,
,,
21
21
&
&&
∫
∫−
−++=Δ δδ (2)
53
Let us now expand ΔS in the series of powers of δq and qS& .
The necessary condition of S having a minimum is the equality to zero of the
linear terms of ΔS expansion (these terms are called the first variation or
simply variation δS of the integral (1)).
Thus, the principle of least action can be written in the form:
( ) 0dtt,q,qLS2
1
t
t=δ=δ ∫ & (3)
or
∫ =⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
= 21
0tt
dtqqLq
qLS &
&δδδ (4)
Note that qdtdq δ=δ& and integrate the second term in Equation (4) by parts:
=∂∂
=∂∂
∫∫ qdtdtd
qLdtq
qL t
ttt
δδ 21
21 &
&&
(5)
qdtqL
dtdq
qL t
t
t
tδδ&& ∂
∂−
∂∂
= ∫ 21
2
1
Now substitute Equation (5) into Equation (4):
∫ =⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−∂∂
+∂∂
= 21
2
1
0tt
t
tqdt
qL
dtd
qLq
qLS δδδ
&& (6)
The first term in (6) is zero because ( ) ( ) 021 == tqtq δδ . The remaining integral
in equation (6) can be equal to zero only if
0=∂∂
−∂∂
qL
dtd
qL
& (7)
because δq are arbitrary between t1 and t2.
u dυ
u υ υ
du
0
54
For a system with s degrees of freedom s different functions qi, where
i=1,2…s, should be considered independently. This gives us s already
familiar Lagrange’s equations:
0=∂∂
−∂∂
ii qL
dtd
qL
&, (8)
which proves the Principle of Least Action.
5. Note on Hamilton’s Equations
Let us introduce a generalised linear momentum pi associated with qi ,
i
i qLp&∂
∂= , (9)
and define Hamilton’s function (or the Hamiltonian):
LqpH iS
ii −= ∑
=&
1
For conservative systems
Ε=+Τ= VH (10)
Hamilton’s equations are:
siqHp
pHq
ii
ii
...2,1
,
=∂∂
−=∂∂
= && (11)
Equations (11) are differential equations of the first order. There are 2s
Hamilton’s equations; they are equivalent to s Lagrange’s equations.
55
V. VIBRATION OF PARTICLES 1. FREE VIBRATIONS
Consider motion of the following simple system.
Apply Newton’s second law: ∑ = xmFx &&
As Fx = - kx, we obtain
xmkx &&=−
or
0=+ kxxm && (1)
or
02 =+ xx nω&& , (2)
where mkn /=ω is called the natural circular frequency.
kx
k
x
m
Equilibrium position
Simplest 1- DOF System
56
Solution of Equation (2)
tinBtAx nn ωω s+= cos (3)
or
( )ψω += tCx usin (4)
Constants A, B or C, ψ are determined from the initial conditions, i.e. the
initial displacement xo and initial velocity ox& .
For example, for the solution in the form of Equation (3), it follows from
Equation (3) that at t=0:
noo BxandAx ω== &
Then Equation (3) can be rewritten in terms of oo xandx & as
tx
txx nn
ono ω
ωω sincos
&+= (5)
For the solution in the form of Equation (4) at t=0:
ψωψ cossin noo CxandCx == &
To find out C and ψ let us do the following:
a). 2222 ..sin no Cx ωψ=
ψω 2222 cosno Cx =&
( )22
22222
/ noo
nono
xxCHence
Cxx
ω
ωω
&
&
+=
=+
+
57
b). ψsinCxo =
ψω cosno Cx =&
( )ono xxHence &/tan 1 ωψ −=
Then, Equation (4) can be rewritten in terms of oo xandx & as
( )[ ]ononnoo xxtxxx && /tansin/ 1222 ωωω −++= (6)
Very often, instead of Equation (4) (with ‘sin’), the ‘cos’ form of equation is
used:
( )ϕω += tCx ncos (7)
In this case ϕ = −tan-1⎟⎟⎠
⎞⎜⎜⎝
⎛
no
ox
xω&
and
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−+= −
no
onnoo x
xtxxx
ωωω
&& 1222 tancos/ (8)
One period of vibration τ is obviously defined from the condition.
nn
n
fei 12..
,2
==
=
ωπτ
πτω,
where fn = ωn/2π is the natural frequency [Hz].
In Equations (4) and (7) or (6) and (8), the parameters C or
222 / noo xx ω&+ are called the amplitudes of vibration. The arguments of
‘sin’ or ‘cos’ functions in the same equations are called the phases of
vibration, ψ or ϕ being the initial phases.
:
58
For illustration:
)cos( ϕω += tCx n
nnf ωπτ 21
==
2π is the period in this plot
22π
=π
ωnt [rad]
X
C
ϕ
(ϕ<0)
π 2
π 3π 2
2π
τ is the period in this plot X
C
t [s.]
2τ
nωϕ
59
A more convenient way of description of vibration processes using complex numbers
Consider Equation (7):
( )ϕω += tx ucos
Using the well-known relationship
( ),'sincos formulasEulerzize iz ±=±
one can rewrite Equation (7) as
( )[ ] [ ]tnitni eCCex ωϕω ~ReRe == +
Here C~ = ϕiCe is the so-called complex amplitude. Operation “Re” can be
made at the very end. Before that, the expression ( )ϕω += tniCex is often
being used, assuming that x is a complex number.
Vibration of a vertically suspended particle.
m
k
x Equilibrium position
60
In this case
( )
xmxmkxkxmg
xxkmgFxxx
xmF
stst
stx
sttot
totx
&&&&43421
&&
+=−−∴
+−=Σ+=
=Σ
0
m
kxxor
kxxmHence
nn ==+
=+
ωω ,0
02&&
&&
Thus, the final differential equation is the same as for horizontally vibrating
particle.
Another example: a ‘mathematical’ pendulum.
Pt = mg sinα
For small α
Pt = mgα
For small α the displacement is
lxlx =∴= αα
Then, applying Newton’s 2nd law, one can get
0=+
−=
xlgxor
lxmgxm
&&
&&
(9)
or 02 =+ xx nω&& (10)
α
l
x m
mg
Pn Pt
61
gl
flgf
lgHere
n
nn
n
πππ
ω
ω
21,21
2==Τ==
=
General case of not small α
For a general case of not small α one should use polar coordinates : r,
α. Note that there is no movement along r. Therefore, r = l.
We recall that αααα &&&&&& lrra =+= 2
Then, Newton’s second law gives:
αα sinmgml −=&&
or 0sin =+ ααlg
&& (11)
Equation (11) is a non-linear equation, which requires a special approach
that will be considered in one of the following chapters. For small angles α:
sin α ~α, and Equation (11) can be ‘linearised’:
0
0
2 =+
=+
αωα
αα
norlg
&&
&&, (12)
with the same lg
n =ω as in the start of this example (see Equations (9)
and (10)). Indeed, multiplying Equation (12) by l and using lα =x, we can
reduce Equation (12) to the initially derived Equation (10).
0
62
Alternative derivation using Lagrange’s equation.
Let us now use Lagrange’s equation to describe large vibrations of a
mathematical pendulum. In the case under consideration the kinetic and
potential energies are:
0.
cos,21 2.2
=∂∂
+∂
Τ∂−⎟⎟
⎠
⎞⎜⎜⎝
⎛∂
Τ∂
−==Τ
iii qV
qqdtdisELThen
mglVml
&
& αα
In our case i = 1 and q1 = α. Continuing in a familiar way, we obtain:
αα
α
αα
αα
sin
0
2
2
mglV
mldtd
ml
=∂∂
=∂
Τ∂
=⎟⎠⎞
⎜⎝⎛
∂Τ∂
=∂
Τ∂
&&&
&&
It follows from L.E. that
0sin2 =+ αα mglml &&
or
0sin =+ ααlg
&& ,
i.e. , we have derived the same Equation (11), as expected.
63
3. EFFECT OF DAMPING
xmxckx &&& =−−
02
02 =++
=++
xxxor
kxxcxmor
nn ωζω &&&
&&&
Here we have introduced the new parameter (‘zeta’):
kmc
mc
n 22==
ωζ
We recall that
mk
n =ω
Let us use the general rule of solving differential equations with constant
coefficients and seek the solution in the form
tAex λ= (2)
Now substitute Equation (2) into Equation (1)
xc&
k
m
x kx
Let us add a linear viscous dashpot describing damping
xmFx &&=Σ
(1)
c
64
As a result, we obtain the characteristic equation:
02 22 =++ nn ωλζωλ (3)
The roots of the characteristic equation (3) are:
⎟⎠⎞
⎜⎝⎛ −±−= 12
2,1 ζζωλ n
Then, the general solution of Equation (1) is
tntn
tt
eAeA
eAeAx
ωζζωζζ
λλ
⎟⎠⎞
⎜⎝⎛ −−−⎟
⎠⎞
⎜⎝⎛ −+−
+=
=+=
12
212
1
2211 (4)
Analysis of the Equation (4).
Note that generally∞≤≤ ζ0 . Let us consider the following cases:
1. ζ > 1 - overdamping
λ1 and λ2 are real and negative ( no oscillations)
overdamping (ζ>1)
critical damping (ζ=1)
t,s 0
x xo
( )0x0 =&
65
2. ζ = 1 - critical damping
nωλλ −== 21 are equal and negative
For this special case
( ) tnetAAx ω−+= 21 (no oscillations)
3. ζ < 1 - underdamping
In this case ζ2 – 1 < 0
22 11 ζζ −=− iand
[ ] tntditdi eeAeAxthen ζωωω −−+= 21 , (5)
where we have introduced 21 ζωω −= nd .
The parameter isnd21 ζωω −= often called the damped natural
frequency.
If one is interested in real values of x, or in physical solution, then, taking a
real part of Equation (5) (i.e. by applying operation “Re”) and keeping in mind
that A1 and A2 are in general complex numbers, one can obtain the following
expression for x, which is also used widely:
( )ϕωζω += − tCex dtn cos , (6)
Here C is the vibration amplitude, and ϕ is the initial phase.
66
For illustration:
21
221
ζω
πω
πτ−
===ndd
d f
Useful definition – the logarithmic decrement δ
( ) ===∴ dndne τζωδ τζωln
πζζ
πζ
ζω
πζω 21
2
1
222
≈−
=−
=n
n
0
τd
t , s
tnCe ξω−
tnCe ξω−−
envelope
x2
t , s
x1
t1 t2
x
τd
⎟⎟⎠
⎞⎜⎜⎝
⎛=
2
1lnxx
δ
( )
dn
dtn
tn
e
CeCe
xx
τζω
τζω
ζω
=
==+−
−
1
1
2
1
67
3. FORCED VIBRATIONS
3.1. Force excitation
xmFx &&=Σ , then
( ) xmxckxtF &&& =−−
or
( )tFxxx nn~2 2 =++ ωζω &&& , (1)
( ) ( ) mtFtFandkmc
mcm
kwhere
n
n
/~22
=
==
=
ωζ
ω
Consider a harmonic force:
( ) [ ] tFeFtF oti
o ωω cos~~Re~ ==
Then, the general solution of Equation (1) is:
+
k
xc&
m
xkx F(t)
c
F(t)
x(t) = General solution of Equation (1) with ( ) 0tF~ =
Particular solution of Equation (1) with
( ) 0tF~ ≠
68
1. The first term in the right-hand side of the above expression has been
studied by us in the previous section. It decays with time t and is
important only for relatively small t (the so-called transient processes).
2. Consider the second term and choose the following form for it: tioeFH ω~ .
Ignoring transients, i.e. for large t, we can assume that
tioeFHtx ω~)( = (2)
Let us substitute Equation (2) into (1).
Then
( ) ( ) tio
tino
tion
tio eFeFHeHFieFH ωωωω ωωζωω /=/+//+/−/
~~~2~ 22
Hence
( )ωζωωω
ωnn i
H2
122 +−
= (3)
Function H(ω) is called the frequency response function.
Thus, the solution for large t (the steady –state solution) is
( )ωζωωω
ω
nn
ti
i
eFtx
2
~
220
+−= (4)
Strictly speaking, we have to take the real part of the right-hand side of
Equation (4) to obtain the final physical solution.
Let us analyse the function H(ω):
69
Let us introduce the non-dimensional frequency ratio nω
ωγ = .
Then
( )ζγγω iH
n 21122 +−
= (5)
or
( )( ) =−−+−
−−=
ζγγζγγω
ζγγ
ii
iHn 2121
)21(1222
2
( ) ( )
( ) ( )
( ) ( ) ⎥⎦⎤
⎢⎣⎡ +−
⎥⎦⎤
⎢⎣⎡ +−
=
⎥⎦⎤
⎢⎣⎡ +−
−−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−−
2222
21
21tan21
222
2222
2
21
21
21
21
ζγγω
ζγγ
ζγγω
ζγγγ
ζγ
n
i
n
ei
Hence
( ) ( )
( )
44 344 21
44444 344444 21
θ
γ
ζγ
γ
ζγγω
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−−
⎥⎦⎤
⎢⎣⎡ +−
=
21
21tan
2/12222 21
1i
H
n
eH (6)
or
( ) θγ ieHH =
70
It is convenient to introduce the parameter |)0(||)(|
HHG γ
=
Then it follows from Equation (6) that ( ) ∞→==>→ 10 γζ Gfor
Some other cases are illustrated in figures below.
1 2 3 4
0.5
0.25
0.15
ξ = 1
uωωγ =
-18
-90
0
θ
0
1 2
ζ=
ωωγ =
2
1
3
71
3.2. Base excitation.
In addition to a directly applied force considered above, the movement of the
base (to which the mass is connected by springs) is also an important
mechanism of forced vibrations.
Examples: structures shaken by earthquakes:
vehicle vibrations due to road roughness.
xmFx &&=Σ
( ) xmxcxxk B &&& =−−−
or
m
tkbxxx nnωωζω cos2 2 =++ &&& (7)
Equation (7) is the same as Equation (1), but 0F~ is replaced by mkb .
m
Bk
xc &
x
XB = bcosωt
k(x -xB)
B k
x
72
Another example of base excitation: a vehicle suspension.
Let us discard static forces due to static equilibrium. Then
( ) ( ) xmxxcxxk BB &&&& =−−−− (8)
Let us now make the substitution: z = x – xB. Then
tbmxmkzzczm B ωω cos2=−=++ &&&&& (9)
The solution of Equation (9) is (see Equations (4) – (6))
tieHbtz ωωω )()( 2= (10)
( ) ( ) ( ) ( )ζγγωωωω θ
iHoreHH
n
i
21122 +−
== ,
where γ = nωω
Now let us find x = z + xB .
m m X
XB =bcosωt ( )Bxxc && − ( )Bxxk −
73
( )
( )
( )( )
ti
ti
titi
ei
ib
eHb
beeHbx
ω
ω
ωω
ζγγ
ζγω
ωω
ωω
2121
1
2
22
2
+−
+=
=⎟⎟⎠
⎞⎜⎜⎝
⎛+=
=+=
(11)
( ) ( )θ
ζγγ
γζ
ζγγ
ζγ
i
B
B
exxor
ii
xx
222
22
2
21
41
2121
+−
+=
+−
+=
, (12)
⎟⎟⎠
⎞⎜⎜⎝
⎛
+−
−= −
222
31
412tan
γζγ
ζγθwhere (13)
One can see that
21 >=<nB
ifxx
ωωγ
1
2
2 431 2
0.25
1=ζ
Bxx
nωωγ =
74
Consider the simplest model of a vehicle on a rough road.
[ ]Hz
Lvfor
Lvwhere
tbL
vtbxB
==
=⎟⎠⎞
⎜⎝⎛=
πω
ωπ
2
cos2cos
For a smooth ride it should be:
mk
Lvor
n222 >> π
ωω
3.3. Vibrations due to rotating imbalance.
The vertical displacement of m is x + e sinωt
m υ
b
k c
y
υt L
m ω e
2k
x
M
C 2k
An eccentric mass m is present, with the eccentricity e M is the whole mass, including m.
75
Then, applying Newton’s second law, one can get:
( ) ( ) xckxtexdtdmxmM &&& −−=++− ωsin2
2,
or tmekxxcxM ωω sin2=++ &&& , (14)
or tMmexxx nn ωωωζω sin2 22 =++ &&& (15)
Mkwhere n =ω
The solution of Equation (15):
( ) ( )
( )θωζγγ
γ+
+−= tM
me
x sin21 222
2
(16)
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−= −2
1
12tan
γζγθwhere
Note that vibrations are low at low angular velocities ω of rotation. They
achieve maximum value at ω slightly higher than ωn.
1
1 2 3 4
2
0.05
0.25
0.5
1=ζ
nωωγ =
⎟⎠⎞
⎜⎝⎛
Mmex
76
VI. VIBRATION OF RIGID BODIES.
This is a general situation when consideration of a body as a particle is no
longer valid. This means that both translation and rotation of a body should be
taken into account.
1. TYPICAL EXAMPLES
1.1. Physical pendulum.
Then 0sin2
=+ θθog
cm
r
gr&& . 22
og
cmn
rgr
=∴ ω
If to assume that the whole mass m is concentrated in G, then cmog rr =
and we arrive to the already familiar equation of mathematical pendulum:
ωn2 = g/rcm .
G
0
θ
mg
rcm
0sin
sin
=+
=−∴
=Σ
θθ
θθ
θ
o
cm
ocm
oo
Imgr
or
Imgr
IM
&&
&&
&&
As was mentioned earlier, it is convenient to represent 2
ogo mrI = , where ogr is the radius of gyration of a body about 0
77
1.2. Log suspended via two ropes
Let us use Lagrange’s equation:
( )2221
21 ϕ&lmmv ==Τ ϕ=1q
ϕϕϕ
ϕϕ
ϕϕ
ϕ
sin;0
;
cos
22
mglV
mldtdml
mglV
=∂∂
=∂
Τ∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂
Τ∂=
∂Τ∂
−=
&&&
&&
0sin
0sin:.. 2
=+∴
=/+/
ϕϕ
ϕϕ
lg
glmlmEL
&&
&&
For small ϕ : sinϕ ~ ϕ
lg
nn ==+∴ ωϕωϕ ,02&&
mg
m
ϕ
Τ Τ
l l
This is the same equation as the equation of Mathematical Pendulum, because there is no rotation in this example.
78
1.3. Log suspended via two rods (Problem 8/124, M&K, 5ed)
Use Lagrange’s equation again:
( )
ϕϕϕϕ
ϕϕϕ
ϕϕ
ϕϕ
&&&&&&&
&&&
&&
222
22
222
~32~
32
2~31
cos2
~2cos
~31
212
21
lmmlmmldtd
lmml
lgmmglV
lmlm
⎟⎠⎞
⎜⎝⎛ +==+=⎟⎟
⎠
⎞⎜⎜⎝
⎛∂
Τ∂
+=∂
Τ∂
−−=
⎟⎠⎞
⎜⎝⎛+=Τ
( )
( ) 0sin~~32
:..
sin~sin~sin
0
2 =++⎟⎠⎞
⎜⎝⎛ +
∴
+=+=∂∂
=∂
Τ∂
ϕϕ
ϕϕϕϕ
ϕ
glmmlmm
EL
glmmglmmglV
&&
or
mg
m
ϕ
m~ l l m~ 2~
31 lmI R =
79
0sin~
32
~=
+
++ ϕϕ
lg
mm
mm&&
For small ϕ : sinϕ ~ ϕ,
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
+
+=
=+∴
mm
mmlgwhere n
n
~32
~
,02
ω
ϕωϕ&&
If m~ = 0, then ,lg
n =ω as in the previous example.
If m = 0, then lg
n 23
=ω
Compare the latter expression with ωn for a single rod.
Indeed, let us recall the Example 1.1: (Physical pendulum).
But for a rod, rcm = 2l and 2~31 lmIr =
Hence lg
lm
mlgn 2
3~
31
~2
.
2=
/
//
=/
ω ,
i.e. the values of ωn are the same, as expected.
m~
ϕ
l cmr
If we assume that I0 = Ir, then it follows from Equation (1) that for small ϕ
r
cmn I
mgr ~=ω G
80
1.4. Bar with a torsional spring.
( )( ) 2
2
2
312
0
312
~sin:31sin
2
mlmglk
ml
kmglsmallFor
mlmglk
IM
n
o
−=∴
=+−
+∴
=+−
=Σ
ω
ϕϕ
ϕϕϕ
ϕϕϕ
ϕ
&&
&&
&&
Thus, the minimum k is ( 2l )mg. Let us look at what it means.
Consider the potential energy of the system under consideration:
ϕ+ϕ= cosmg2lk
21V 2
Let us find a minimum of V, which, as we know, corresponds to the condition
of stability of the system:
0,0sin2
=∴=−= ϕϕϕϕ
mglkddV
For 0=ϕϕddV to have a minimum (and not a maximum!) it should be:
k
m ϕ
l
Equilibrium position at ϕ=0. Find natural frequency of small vibrations. Show that there is a minimum k below which the natural frequency is not a real number.
81
or 02
cos2
Statics): Mech. Eng. (see00
002
2
2
2
>−=−=
>=
==
mglkmglkd
Vd
dVd
ϕϕ
ϕϕ
ϕϕ
Thus, k > mg2l is the condition of stability. As we have seen above, this is
reflected also in the requirement for ωn to be real.
2. ENERGY METHOD OF DETERMINING THE FRQUENCY OF VIBRATION
For conservative systems:
constV =+Τ=Ε
Since Τ changes from 0 to Τmax = Ε, whereas V changes from Vmax = Ε
to 0, it is clear that Ε==Τ maxmax V . The equality Tmax = Vmax can be
used for determining the natural frequency of the system.
Example of a simple harmonic oscillator
( )
( )2maxmax
2maxmax
2121
xkV
xm
=
=Τ &
But for harmonic oscillations
( )ψω += txx ncos.max it follows that maxmax xx nω=& .
Then, using the equality Tmax = Vmax, we obtain:
( ) ( )
mk
xkxm
n
n
=∴
=
ω
ω 2max
2max 2
121
as expected.
82
2.1. Rocking motion of a semi-disk: comparative analysis by energy method and by means of Lagrange’s equations
Using energy method ( ) ( )[ ]
( )θπ
θ
cos134
cosmax
−⎟⎠⎞
⎜⎝⎛=
=−−−=
rmg
rrrrmgV
( ) ( )
( )3D Table see21
21
2
222
02
max
mrI
rrmrmIrrmII
I
o
Iorc
c
r
=
−+−=−+=
=Τ=
43421
&θ
θ
( ) 22maxmax
222
38
23
21cos1
34:
38
232
21
θπ
θπ
π
&rmrgmV
mrrmrmrmrIc
/⎟⎠⎞
⎜⎝⎛ −=−⎟
⎠⎞
⎜⎝⎛
/=Τ
⎟⎠⎞
⎜⎝⎛ −=−+=∴
gr
rgHence
rg
nn
n
n
78.72;807.0
34
43
211
34
: nsoscillatio harmonicFor 21~cos : smallFor
222
max
2
===
⎟⎠⎞
⎜⎝⎛ −=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−−∴
=
−
ωπτω
ωθπ
θπ
θωθ
θθθ&
c mg
r
G r
0
θ− cosrr
Find the period τ of small oscillations of the semi-disk shown in the Figure ( )3,&
34 DableKMrr Τ=π
θ
83
A more general approach to the solution of this problem using Lagrange's equation.
( )θcosrrmgV −= q1 = θ
( ) 2.21 θθ &cI=Τ
Use Taylor’s series expansion for Ic (θ):
1 ( ) ( ) ( ) ...00 ' ++= θθ ccc III
( )
( )[ ] ( ) ( )θθθθθθ
θθθ
θθθ
θθθ
&&&&&&
&&&&&
&&
0 smallfor ~..00)0(~
~.)
)()
''cccc
cc
c
IIII
II
dtdb
Ia
++
∂∂
+=⎟⎠⎞
⎜⎝⎛
∂Τ∂
=∂
Τ∂
( )
θθθ
θθθ
θθ
.~sin)
smallfor 0~.0
21~.
)0(21) 2'2
rmgrmgVd
II
c cc
=∂∂
∂∂
=∂
Τ∂ &&
L. E. :
( )
( )0
0
0.02
cn
n
c
Irmgwhere
or
rmgI
=
=+
=+
ω
θωθ
θθ&&
&&
Now use the expression for r and Ic(0) from the first part of this example:
π3
4rr = and ( ) 238
230 mrII cc ⎟
⎠⎞
⎜⎝⎛ −==
π
0 0
for small θ
84
Then rg0.807
38
23
34
38
23
34
2=
⎟⎠⎞
⎜⎝⎛ −
=/⎟
⎠⎞
⎜⎝⎛ −
⎟⎠⎞
⎜⎝⎛
/=
π
π
π
πωrg
rm
rgmn
Hence
gr
n78.72
==ω
πτ
As expected, the results for ωn and τ are the same. However, the L. E.
approach provides a more comprehensive insight into the problem.
85
VII SYSTEMS WITH TWO OR MORE DEGREES OF FREEDOM
1. FREE VIBRATIONS
Consider the following 2 –DOF System:
Use second Newton’s Law for each mass:
( )1221111 xxkxkxm −+−=&& (1)
( ) 2312222 xkxxkxm −−−=&& (2)
Let us seek the solution in the form:
titi eAxeAx ωω2211 ; == (3)
Substitute equations (3) into (1) and (2):
( )[ ] 022112
21 =−−+ AkAmkk ω (4)
( )[ ] 0222
3212 =−++− AmkkAk ω (5)
m2 m1
k2 k1 k3
x1 x2
86
For the system of homogeneous equations (4), (5) to have a non-trivial
solution ( 0A,0A 21 ≠≠ ), the determinant of the system must be zero:
( )[ ]
( )[ ]22
322
212
21
mkkk
kmkk
ω
ω
−+−
−−+ = 0 (6)
or
( )( ) ( ) ( ) 02221
42
221321
23221 =−++−+−++ kmmmkkkkmkkkk ωωω
(7)
or
021
3231212
2
32
1
214 =++
+
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡+
++
−444 3444 21
444 8444 76
c
b
mmkkkkkk
mkk
mkk
ωω (8)
Introducing the notation λω =2 , we can rewrite Equation (8) as:
cbb
cb
−±=∴
=+−
42
02
2,1
2
λ
λλ
For illustration purposes, let us assume that
m2m,mmandkkkk 21321 ===== (see book of Thomson, Section 5.1)
Then 2
23,3 ⎟
⎠⎞
⎜⎝⎛==
mkc
mkb , and
mk
mk
mk
mk
366.223
23
634.023
23
2
1
=⎟⎟⎠
⎞⎜⎜⎝
⎛+=
=⎟⎟⎠
⎞⎜⎜⎝
⎛−=
λ
λ
87
Hence
( )mk634.011 == λω (9)
( )mk366.222 == λω (10)
These are the natural frequencies of the system. Now return back to
Equations (4) and (5) and substitute ω1 and then ω2 into any of them,
e.g. into Equation (4):
( ) mk
kmkk
kAA
21
221
2
2
1
2 ωω −=
−+=
Then, for ( )
731.0634.01
2
121 =⎟⎟
⎠
⎞⎜⎜⎝
⎛=>=
AA
mkω
for 73.2366.2)2(
2
122 −=⎟⎟
⎠
⎞⎜⎜⎝
⎛=>=
AA
mkω
Thus, each of the two natural frequencies of the system, ω1 and ω2,
corresponds to the specific ratio of the amplitudes, ( )( ) ( )( )221
121 A/AandA/A
respectively. Such specific distributions of amplitudes associated with the
corresponding natural frequencies are called the normal modes of the
system. It is convenient to choose 1A2 = . Then, one can show the modes as
follows:
A1 A2
1 0.731
mk634.0w2
1 = -2.73
A2
A1
1
mk366.2w2
2 =
88
Note that in mathematical terminology, instead of natural frequencies and
normal modes, the more general definitions are used – eigenvalues and
eigenvectors respectively.
2. MATRIX NOTATION
It is convenient to rewrite the initial equations, Equations (1) and (2), in the
form:
( )
( ) ⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡+−
−++
⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡00
00
2
1
322
221
2
1
2
1xx
kkkkkk
xx
mm
&&
&& (11)
or
}0{}]{[}]{[ =+ XKXM && (12)
[M] is called the mass matrix, and [K] is called the stiffness matrix. The matrix notation, Equation (12), is especially useful for systems with many
degrees of freedom (more than two)
Using the amplitude vector {A} = ⎭⎬⎫
⎩⎨⎧
2
1AA
, one can rewrite Equation (12):
( ) }0{}{][][2 =+− AKMω (13)
Then the equation for natural frequencies is
0][][2 =+− KMω (14)
89
3. SOME PRACTICAL EXAMPLES 3.1. 2-DOF vehicle model for bounce and pitch.
Derive Lagrange’s equations:
( ) ( )
0,,
21
21
21
21
222
211
22
=∂Τ∂
=⎟⎠⎞
⎜⎝⎛
∂Τ∂
=∂Τ∂
++−=
+=Τ
zzm
zdtdzm
z
lzklzkV
Izm
&&&
&&
&&
θθ
θ
( ) ( )
0,,
2211
=∂
Τ∂=⎟
⎠⎞
⎜⎝⎛
∂Τ∂
=∂
Τ∂
++−=∂∂
θθ
θθ
θ
θθ
&&&
&&
IdtdI
lzklzkzV
( )( ) ( ) 222111 llzkllzkV θθθ
++−−=∂∂
bounce
pitch G
k1 k2 1l 2l
G
Z
θ k1 k2
θ− 1lz θ+ 2lz equilibrium
mNkmNk
mlml
mr
mrI
kgm
g
g
4000035000
8.1;5.1
2.1
1500
2
1
21
2
==
==
=
=
=
θ==
2
1q
zq
90
L. E. :
( ) ( )( ) ( ) 0
0
222111
2211
=++−−
=++−+
llzkllzkI
lzklzkzm
θθθ
θθ&&
&&
Let us seek the solution in the form:
tio
tio eeZz ωω θθ == , . Then
( ) ( )( ) ( ) ⎭
⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−+−−−−−+
00
2222
2112211
22112
21
o
ozIlklklklk
lklkmkkθω
ω
Equalising the determinant of the matrix to zero, we obtain:
( ) 022211
2222
211
221
2
=−−⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛−+
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛−+
44342143421321cba
lklkIlklkmkk ωω
Using the notations a, b and c2, we rewrite this equation in the form
( )( )
0
0
2422
222
=−+−−
=−−−
cmIIambab
orcIbma
ωωω
ωω
Let us solve this equation versus ω.
Simplify it first:
0
~
22
~
4 =−
++
−4342143421
cb
mIcab
mIaImb ωω
Introduce Then.2 λω =
91
cbb
cbb
cbb
cb
~4
~
2
~
~4
~
2
~
~4
~
2
~0~~
22
21
22,1
2
−+=
−−=
−±=∴
=+−
λ
λ
λ
λλ
and 22
11
λω
λω
=
=
Substitute the numerical values of the parameters. Then the results for the
natural frequencies are:
[ ]
[ ]Hzf
Hzf
582.12
098.12
22
11
==
==
πωπ
ω
The amplitude ratios corresponding to the frequencies 21 and ωω
:are)fandfor( 21
⎥⎦⎤
⎢⎣⎡=
−+
−=⎟⎟
⎠
⎞⎜⎜⎝
⎛
⎥⎦⎤
⎢⎣⎡−=
−+
−=⎟⎟
⎠
⎞⎜⎜⎝
⎛
radm
mkk
lklkZ
radm
mkk
lklkZ
o
o
o
o
266.0
412.5
2221
2211
2121
2211
2
1
ωθ
ωθ
ω
ω
92
Mode shapes:
Mostly translation
Mostly rotation
3.2. Symmetrical 2-DOF system. It is instructive to consider a symmetrical 2-DOF system. For example, let us
consider the system shown in the Figure.
The conditions of symmetry in this case are:
mmmkkk ==== 2131 ,
equilibrium
ω1:
equilibriumω2
m2 m1
k2 k1 k3
x1 x2
93
The equations of this system for the general case of arbitrary m1, m2 and k1,
k2 , k3 have been derived in the previous section. In the symmetric case
under consideration we have for the coefficients:
2
222 2,2m
kkkcm
kkb +=
+=
mk
mkk
m
kkkkkkkm
kk
cbb
22
222
222
22
22
2,12,1
22
42
±+
=
=/−//−+//+/
±+
=
=−±== ωλ
Thus,
mkk
mk
mk
mk
mk
22
2222
1211
2;2
;
+=+==
===
ωωλ
ωωλ
Now determine :212
1 ωω andforAA
( )( )
( )1
2:2
1:
22
22
2
122
2
21
2
11
22
2
2
1
−=−−+
=⎟⎟⎠
⎞⎜⎜⎝
⎛+=
=/
/−+
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
=>−+
=
mmkm
mkkk
kAA
mkk
mmkkk
kAA
mk
mkk
kAA
ω
ω
ω
Obviously, the first mode (frequency ω1) is anti-symmetric, and the second
mode (frequency ω2) is symmetric. Generally, in all symmetric systems
there are symmetric and anti-symmetric modes.
94
4. INITIAL CONDITIONS
Let us return to the example considered in Section 1 for the particular case
m2 = 2m1 = 2m, k1 = k2 = k3 = k.
⎩⎨⎧−
=Φ=
⎩⎨⎧
=Φ=
000.1732.2
,366.2
000.1731.0
,634.0
22
11
mk
mk
ω
ω
For general initial conditions a free vibration of the system contains both
modes simultaneously:
( ) ( )2221112
1 cos000.1732.2
cos000.1731.0
ϕωϕω +⎭⎬⎫
⎩⎨⎧−
++⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
tctcxx
(1)
Equations (1) are the two equations giving the solutions for x1 and x2.
Parameters 2121 and,c,c ϕϕ are the four necessary constants required
for two differential equations of the second order.
To find all four constants we need another two equations for 21 xandx && ,
which are obtained by differentiating Equations (1).
m2 m1
k2 k1 k3
x1 x2
95
( ) ( )222211112
1 sin000.1732.2
sin000.1731.0
ϕωωϕωω +⎭⎬⎫
⎩⎨⎧−
−+⎭⎬⎫
⎩⎨⎧
−=⎭⎬⎫
⎩⎨⎧
tctcxx&
& (2)
Specifying the initial conditions at t=0, one can find from Equations (1) and (2)
the values of 2121 and,c,c ϕϕ .
For example, let the initial conditions are
( )( )( ) ⎭
⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
00
0x0x
and00.1
c0x)0(x
2
11
2
1
&
&
Then, it follows from Equations (1), (2) that
( )( ) 2211 cos
000.1732.2
cos000.1731.0
00.1
III
ϕϕ⎭⎬⎫
⎩⎨⎧−
+⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
cc
( )( ) 222211 sin
000.1732.2
sin000.1731.0
00
IVIII
ϕωϕω⎭⎬⎫
⎩⎨⎧−
−⎭⎬⎫
⎩⎨⎧
−=⎭⎬⎫
⎩⎨⎧
cc
These are the four equations to determine 2121 ,,c,c ϕϕ .
( ) ( )( )
( ) ( )( )( )
( )( ) ( )
( ) 222
111
22
11
sin731.0732.20:III731.0(IV)
sin732.2731.00:III732.2(IV)
cos731.0732.20.1:I)(731.0II
cos732.2731.00.1:I732.2II
ϕω
ϕω
ϕ
ϕ
−−−=+−
+−=+
−−=+−
+=+
cx
cx
cx
cx
It follows from the last two resulting equations that
96
1coscos0
0sinsin
2121
21
====
==
ϕϕϕϕ
ϕϕ
andor
Then from the first two equations:
289.0
463.31
732.2731.00.1
289.0463.31
732.2731.00.1
2
1
−=−=+
−=
==+
=
c
c
Thus
0,0
289.0,289.0
21
21
==−==
ϕϕcc
N.B. Even if only one of the masses (on the left) has been displaced initially,
the resulting free vibration involves movements of both masses.
Typical behaviour of a 2-DOF system
x1,x2
0
t
x1
x2
97
5. FORCED HARMONIC VIBRATIONS
Consider a 2-DOF system described by the matrix equation.
tF
xx
kkkk
xx
mm
ωcos00
0 1
2
1
2221
1211
2
1
2
1
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡+
⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡&&
&& (3)
As before, for simplicity, we consider a system without damping.
Let us look for a steady-state solution of Equations (3) in the form.
tXX
xx
ωcos2
1
2
1
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧ (4)
Substitute Equation (4) into (3):
( )( ) ⎭
⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−−
01
2
12
22221
122
111 FXX
mkkkmk
ωω (5)
If to introduce the notation
( )[ ] ( )( )⎥⎥⎦
⎤
⎢⎢⎣
⎡
−−= 2
22221
122
111ω
ωωmkk
kmkz ,
then we can rewrite Equation (5) as:
( )[ ]⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
01
2
1 FXX
z ω (6)
98
5.1. Construction of the inverse matrix.
Equation (6) can be solved versus X1, X2 using the following formal
operation: multiplying both sides of Equation (6) by the inverse matrix
( )[ ] 1−ωz , so that
( )[ ]⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧ −
011
2
1 Fz
XX
ω (7)
In deriving Equation (7) we used the property
( )[ ] ( )[ ] [ ] [ ] ⎥⎦
⎤⎢⎣
⎡==−
1001
I,I1 wherezz ωω
How to find ( )[ ] 1wz − ? According to matrix algebra,
( )[ ] ( )[ ][ ])(det
1ωωω
zzadjz =− ,
where ( )[ ]ωzadj is the adjoint matrix of the square matrix ( )[ ]ωz
The adjoint matrix ( )[ ]ωzadj is defined as a transpose of the matrix of
cofactors of ( )[ ]ωz . If
( )[ ] ⎥⎦
⎤⎢⎣
⎡=
2221
1211zzzz
z ω , (8)
then the cofactor ji
c is defined as
( )jj i
jii Mc +−= 1 ,
where ji
M is a minor of the element ji
z .
99
For the matrix ( )[ ]ωz defined by Equation (8):
( )( )( )( ) 1111
2222
121212
21
212121
12
222211
11
1
1
1
1
zzC
zzC
zzC
zzC
=−=
−=−=
−=−=
=−=
+
+
+
+
Thus [ ]cofactorsof
matrixthezzzz
C ⎥⎦
⎤⎢⎣
⎡−
−=
1112
2122
and [ ]cofactorsofmatrix
transposedthezzzz
C T⎥⎦
⎤⎢⎣
⎡−
−=
1121
1222
Then
( )[ ] [ ] ( )( )⎥⎥⎦
⎤
⎢⎢⎣
⎡
−−−−=⎥
⎦
⎤⎢⎣
⎡−
−== Τ
211121
122
2221121
1222
ωωω
mkkkmk
zzzz
Czadj
It is convenient also to transform det ( )[ ]ωz as a product
( )[ ] ( )( )222
22121det ωωωωω −−= mmz ,
where ω12 and ω2
2 are the solutions of the algebraic equation ( )[ ] 0det =ωz ,
or
( )[ ] ( )( )
0
det
21124
212
211222
12211
21122
2222
111
=−+−−=
=−−−=
kkmmmkkmkk
kkmkmkz
ωωω
ωωω
Solving the above equation, we can find 21 ωω and , i.e. the normal mode
frequencies:
100
21
221122122
1
11
2
22
1
11
2
2222,1 4
121
mmkkkk
mk
mk
mk
mk −
+⎟⎟⎠
⎞⎜⎜⎝
⎛+±⎟⎟
⎠
⎞⎜⎜⎝
⎛+=ω
Finally, using Equation (7), we obtain:
( )[ ]( )
( ) ⎭⎬⎫
⎩⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−−−−=
⎭⎬⎫
⎩⎨⎧
0det1 1
211121
122
2222
1 F
mkkkmk
zXX
ωω
ω
or
( )
( )222
22121
12
2221
)( ωωωω
ω
−−
−=
mm
FmkX
(9)
( )( )222
22121
1122
ωωωω −−
−=
mm
FkX
1
121
Fkx
1
122
Fkx
1ωω
3 2 1
1
121
Fkx
1
122
Fkx
101
5.2 Representation in terms of normal mode summation. Let us rewrite Equations (9) for amplitudes of forced vibrations in the form:
( )( )( ) 22
2
222
1
122
222
121
12
2221
ωωωωωωωω
ω
−+
−=
−−
−=
cc
mm
FmkX (10)
( ) 222
422
1
322
222
121
1122
)( ωωωωωωωω −+
−=
−−
−=
cc
mm
FkX (11)
Find 4321 c,c,c,c
(a) Multiply Equation (10) by 221 ωω − and then let 1ωω = :
( )
( ) 121
2221
121222 c
mm
Fmk=
−
−
ωω
ω
(b) Multiply Equation (10) by 222 ωω − and then let 2ωω = :
( )
( ) 222
2121
122222 c
mm
Fmk=
−
−
ωω
ω
(c) Multiply Equation (11) by 221 ωω − and then let 1ωω = :
( ) 321
2221
112 cmm
Fk=
−
−
ωω
(d) Multiply Equation (11) by 222 ωω − and then let 2ωω = :
422
2121
112
)(c
mm
Fk=
−
−
ωω
Thus
( )( )
( )( )
( ) ( ) 222
22
2121
11222
121
2221
1122
222
22
2121
122222
221
21
2221
121222
1
1.1.
1.1.
ωωωωωωωω
ωωωω
ω
ωωωω
ω
−−
−+
−−
−=
−−
−+
−−
−=
mm
Fk
mm
FkX
mm
Fmk
mm
FmkX
102
5.3. Principle of tuned vibration absorber
To understand the principle of tuned vibration absorber we can use the
general matrix equation (Equation (3)) in which we have to specify
22211211 k,k,k,k . To do so, let us derive the equations for the above
system.
( )
( )( ) :
cos~:
222
11212111
111
xmF
xmxxktFxk
xmF
x
x
&&
&&
&&
=Σ
=−−+−
=Σ
ω (12)
( ) 22122 xmxxk &&=−−
or
( )0
cos~
221222
12212111=+−
−=−++xkxkxm
tFxkxkkxm&&
&& ω (13)
Compare Equations (13) with Equation (3). Then
11222221
2122111 F~F:kk,kkkk,kkk
−==−=
−=+=
m2 m1
k2 k1
x1 x2 tF ωcos~1
x
103
Now we can use the general solution (Equation (9) ) for the amplitudes:
( )( )( )
( )( )( )( )22
222
121
12
2222
222
121
12
2221
~
ωωωω
ω
ωωωω
ω
−−
−−=
−−
−=
mm
Fmk
mm
FmkX (14)
( )( )( )
( )( )222
22121
1222
222
121
1122
~
ωωωωωωωω −−
−=
−−
−=
mm
Fk
mm
FkX (15)
It follows from Equation (14) that X1 = 0 if k2 – m1ω2 = 0, i.e. this happens
at frequency 2222 ωω == mk . Note that the frequency 2222 mk=ω is
the partial resonant frequency of the second mass, m2. Of course, 02 ≠X
(see Equation (15)). Thus, by changing k2 and m2, one can change the
resonant frequency of maximum absorption.
104
VIII NON-LINEAR VIBRATIONS
So far we have considered linear vibrations associated with small amplitudes,
for which the dynamic equations can be linearised.
Here we discuss a more general situation of non-linear dynamic equations.
1. PERTURBATION ANAYLSIS OF NONLINEAR PENDULUM Example: Mathematical pendulum.
The equation of motion of mathematical pendulum has
been derived in one of the previous sections:
0sin20 =+ θωθ&& (1)
where gl /20 =ω
For simplicity, let us introduce the notation x = θ and rewrite Equation (1) as
0sin20 =+ xx ω&& (2)
Now let us expand sin x in Taylor’s series:
...61sin 3 +−= xxx
θ
l
m
105
Assuming that x<<1, we neglect higher order terms in this expansion (higher
than x3) and rewrite Equation (2) as
06
3202
0 =−+ xxxω
ω&& (3)
In further consideration, it is convenient to deal with a more general equation.
0320
20 =++ xxx αωω&& (4)
where α is a constant. Equation (3) thus corresponds to the particular case
of α = - 61 .
Let us seek the solution of Equation (4) by perturbation method:
...22
10 +++= xxxx αα (5)
and
...122
02 +++= gg ααωω (6)
Here x0 and ω0 correspond to the ‘zero’ approximation
00200 =+ xx ω&& ,
and yet unknown functions ....g,g,...x,x 121 describe higher order
corrections.
Keeping only the first order correction in Equation (6) and substituting it into
Equation (4), we can rewrite the latter in the form:
106
0322 =+−+ xgxxx αωαω&& (7)
The zero approximation for Equation (7) is
002
0 =+ xx ω&& (8)
For the initial conditions at t=0: 0, == xax & , the solution of Equation (8) is
tax ωcos~0 = (8’)
Let us substitute Equation (5) into Equation (7):
0...30
21
201
20
210 =++−−+++ xgxgxxxxx αωαααωωα &&&&
Obviously, the equation of the first approximation is
30
201
21 xgxxx ωω −=+&& (9)
Substituting Equation (81) for x0 into Equation (9) and using the well known
trigonometrical formula
ttt ωωω cos433cos
41cos3 +=
we can rewrite Equation (9) as
tatatgaxx ωωωωωω cos~433cos~
41cos~ 3232
12
1 −−=+&& (10)
107
In Equation (10), terms with cosωt would cause a resonant growth of x1
proportional to t, which cannot be. Therefore, the terms with cosωt have to
be eliminated.
We can eliminate the terms with cosωt if we choose g in such a way that
0~43~ 32 =− aga ω (11)
Then Equation (10) becomes
taxx ωωω 3cos~41 32
12
1 −=+&& (12)
It follows from Equation (11) that
22 ~43 ag ω= (13)
Substituting Equation (13) into Equation (6), we obtain
...~43 222
02 aαωωω += ,
from which it follows that
2
202
~431 aα
ωω
−= (14)
The particular solution of Equation (12) is
tax ω3cos32
~3
1 =
108
Then the general solution of Equation (12), i.e. the first approximation solution
is
tatCtCx ωωω 3cos32
~sincos
3
211 ++=
And the complete solution, x = x0 + x1, is
tatCtCtax ωαωαωαω 3cos32
sincoscos3
21 +++= (15)
Constants C1 and C2 can be found from the same initial conditions at t = 0:
x = a, x& =0
Then
tataax ωαωα 3cos32
~cos
32
~1
32
+⎟⎟⎠
⎞⎜⎜⎝
⎛−= , (16)
where (see Equation (14))
2
202
~431 aα
ωω
−=
Now recall that for our particular case of a pendulum 61
−=α . Then, it
follows from Equations (16) and (14) that
tataax ωω 3cos192
~cos
192
~1
32
−⎟⎟⎠
⎞⎜⎜⎝
⎛+= (17)
and
2
202
~811 a+
=ω
ω (18)
109
One can see that:
1. ω decreases for a larger ã
2. Free vibrations are not harmonic (presence of term with 3ω)
Note that the above solution is valid for x<1 (rad) .
ω0
ω
ã
1 2
110
IX SELF EXCITED OSCILLATIONS
1. SYSTEMS WITH ‘NEGATIVE FRICTION’ 1.1. A mass on the moving belt
Consider a simple example of a mass m placed on the moving belt. Let us
assume that the kinetic friction force F is a function of the relative velocity
( )xvFFeivxv && −==− 00 .., . Apply Newton’s second law:
( )xvFkxxcxm &&&& −=++ 0 (1)
Here 2121 , ccckkk +=+= (for clarity dashpots are not shown in the
Figure).
m
x k1 k2
F v0
belt
( )xvF 0 &−
0 xv0 &−
Exp. data
111
Let us assume that 0vx <<& . Then
( ) ( ) ( ) ...' 000 +−=− vFxvFxvF &&
Rewrite Equation (1):
( )[ ] ( )00' vFkxxvFcxm =+++ &&& .
The term ( )0vF = const, and it is only causing a static displacement
( ) kvFxst /0= . If v0 is not too large, then ( ) 0' 0 <vF and it is possible that
( ) 0' 0 <+ vFc as well, which means that the system has a ‘negative friction’.
Then for a dynamic displacement x we have
0=++ kxxxm &&& α , (2)
where
( )[ ] 0' 0 <+= vFcα
Equation (2) with α<0 describes the increase of the amplitude of vibration
with time t. Let us introduce
mkwhere
m nn
== ωω
αξ ,2
~
Then we can rewrite Equation (2) as
0~2 2 =++ xxx nn ωωξ&&
( )ϕω
ξξ
ωξ+=
<<
tCex
thenandIf
ntn cos
,1~0~
~
(3)
112
In real situations the amplitude becomes stable for larger t due to
nonlinearity (we did not consider that range of t)
The mechanism of ‘negative friction’ is responsible, in particular, for
generating string vibrations by bow movement and for the associated
structure-borne sound of violins, cellos, etc. It also causes brake squeal and
many other types of noise and vibration.
Instability
t
x
t
x
113
1.2. Froude’s pendulum
This is a similar example for rotational movement:.
The value of 0ϕ for equilibrium position ( 0=ϕ=ϕ &&& ) can be obtained from
the equation:
( )Ω=ϕ Fsinmgl 0 .
Let us assume that
( ) ,.cossin~sin;
000
0ψϕϕψϕ
ψϕϕ+−+
+= then
and the resulting equation takes the form
( )[ ] 0.cos' 0 =+Ω++ ψϕψψβ
mglFbI &43421
&& (4)
or
020 =++ ψψγψ wI &&& , (5)
where I
mglI
020
cos,
ϕωβγ ==
If ( ) 0' <ΩF it may be that 0<γ as well. Then we have the same time
behaviour as in the previous example.
Ω
ϕ
G
l Let the moment of the friction force
( ) )( ϕϕ && −Ω=−Ω FM F Then ( )ϕϕϕϕ &&&& −Ω=++ FmglbI sin , where b is the damping coefficient
114
2. SIMPLE THEORY OF WING FLUTTER: 2-DOF MODEL.
Flutter is a type of dynamic instability of coupled bending and torsional
vibrations of a wing that can develop at high speeds of air flows.
Let us consider a simple 2-DOF model of wing flutter (following R. Granger
1991)
2.1. Coupled bending and torsional vibrations. Let us first consider free vibrations (no air flow) of this 2-DOF system.
Lagrange’s equations for free vibrations are:
0
0
=++
=++
αα
α
αkIhS
hkShM h
&&&&
&&&&. (1)
Let us introduce the partial frequencies:
IkandMkhh // ααωω == ,
kh
G h
e.a.
v
α dMrI
rdMSdMMhere
khkV
hSIhMT
S
SS
h
∫
∫∫
=
==
+=
++=
2
22
2.2.
,21
21
21
21
α
αα
α
&&&&
kα
115
and let us seek the solution in the form:
titi AeHeh ωω α == , .
Substituting the above expressions into Equations (1) and equalising the
resulting determinant to zero,
( ) ( ) 0242 =++−− αα ωω kkIkMkSMI hh ,
one can obtain the two natural frequencies of the system: ω1 and ω2 , and
the corresponding mode shapes: ( )( )1/ AH and ( )( )2/ AH .
2.2. Effects of aerodynamic lift and moment
Now let us take into account the aerodynamics. This can be done by adding
the appropriate generalised forces to the right-hand side of Equations (1)
MkIhS
LhkShM h
=++
=++
αα
α
α&&&&
&&&& (2)
where L and M are aerodynamic lift and moment respectively.
The simplest expressions for L and M are:
...
...
++++=
++++=
αα
αα
αα
αα
&&
&&
&&
&&
MMhMhMM
LLhLhLL
hh
hh (3)
Here αααα &&&& MMMMLLLL hhhh ,,,,,,, are the coefficients that
depend on aircraft speed v and wing parameters.
116
Now Equation (2) can be rewritten in the form
( )( ) 0
0
=−−+−−+
=−−+−−+
hMMkMhMIhS
LhLkLhLShM
hh
hhh
ααα
ααα
ααα
αα
&&&&&&
&&&&&&
&&
&& (4)
Equations (4) describe the equivalent ‘free’ 2-DOF system in which the
inertial, stiffness and damping coefficients are modified by aerodynamic
influence.
2.3. Conditions of instability.
Now we are in a position to investigate the conditions of instability of the
solutions of Equations (4) which correspond to the development of flutter.
Note that always M > 0 , I > 0, whereas S may be positive, negative or
zero (for a mass balanced wing).
Other coefficients: ( ) 0,0,0,0,0 =<>−≤≤ hhhh MandLLkML αα&& .
The stability of the system described by Equations (4) (a system of two
simultaneous equations) can be analysed using the general approach (the
Routh method). This approach is convenient because it does not require
direct solution of a system of simultaneous equations.
According to the Routh method, the following expression is to be analysed:
214
230321 AAAAAAAR −−= , (5)
where
117
( )
( )αα
ααα
ααα
α
MkkA
MkMkLA
SLIkMLMkA
ILMMASMIA
h
hh
hh
h
−=
−−−=
+++=
−−=−=
4
3
2
1
20
&&
&&
&&
(6)
If R > 0, the system is stable.
If R < 0 the system is unstable, i.e. vibration amplitudes grow exponentially
in time.
Thus the condition R = 0 is the borderline case. The corresponding speed
of the air flow (or aircraft) is called the critical flutter speed, cfv .
The problem for an aircraft designer is to design the wings in such a way that
R > 0 in the range of operating speeds v, so that v < cfv . This can be
achieved, in particular, by making the frequency of wing torsional vibrations
much larger than the frequency of bending vibrations.
118
RECOMMENDED LITERATURE
1. Meriam, J. L. and Kraige, L. G. , Engineering Mechanics, Volume 2
Dynamics (5th Edition), Wiley, 2003.
2. Thomson, W. T., Theory of Vibration with Applications,
Chapman and Hall, 1993.
3. Landau, L. D. and Lifshitz, E. M., Mechanics,
Butterworth- Heinemann, 1995.
4. Beer, F.P., Johnston, E.R., Eisenberg, E.R. and Clausen, W.E.,
Vector Mechanics for Engineers: Statics and Dynamics (8th Edition),
McGraw Hill, 2006.
5. Curtis, H.D., Orbital Mechanics for Engineering Students, Elsevier,
2005.
