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Chapter 3: Dynamics of the physical point
Prof. Felice IazziDipartimento di Fisica del Polit
[email protected] : felice.iaz
ecnico di Torino
tel. 3355995361
$=1: Forces
The force is a physical vector quantity which is defined as the
result of a measurement performed through adynamometer. A
dynamometer is basically a spring whose length can be measured.
Each dynamometer has aat rest length and all elongation or
compression of the spring with respect to such at rest length
isndicated on an adjacent linear scale.
There are only 4 fundamental forces in nature (actually only 3,
because in 1984 electromagnetic and weaknteraction have been
demonstrated to be different aspects of the same interaction) and
they are:
gravitational, electromagnetic, weak and strong interactions.
They combine together to form infinite forcevectors which appear as
the usual forces measured in the most part of the visible phenomena
and in the
technology of the industrial world.As a result a small number of
different types of forces can be classified as the only present in
the phenomenaof the classical physics. They are listed below.
Gravitational Force GFr
Features: this force appears on a physical point 1 (i.e. a point
with mass m1) when another physical point 2with mass m2 is at a
distance r attracting 1 toward 2
Definition : it acts on the point 1, along the direction from 1
to 2 and is expressed by:
21
21
2
21
r
r
r
mmFG r
rr
(= 6.673 10-11 Nm2Kg-2 in MKSA) (3.1.1)
Where is the vector displacement from point 2 (the attracting
point) to point 1 (the attracted point) (being
of course
21rr
21rrr
= ).
Of course the definition is symmetrical and also point 2 is
attracted by point 1, with the same modulus butopposite
direction.
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Gravitational Force on the Earth surface gFr
Features : This force is the gravitational force evaluated on
the Earth surface, i.e. in a volume of area fewkm 2 and height less
than 1 km. Direction and intensity of the force can be calculated
from eq. 3.1.1assuming that the forces generated by all points of
the Earth which attract a point of mass m in the volume isequal to
the force generated by a single point in the centre of the Earth
containing all the Earth massM(Thisstatement will be demonstrated
in a following chapter). The gravitational force applied by the
Earth on apoint-like mass m in the volume can be written as:
21
21
2221
21
2
)1()( r
r
R
hR
Mm
r
r
hR
MmFG r
r
r
rr
+
=
+
(3.1.2)
where the vector in (3.1.2) is the vector displacement from the
Earth centre to the point m. Under the aboveassumptions the ratio h
toR is very small (negligible) with respect 1 and the direction of
the force can beapproximated by the perpendicular to the horizontal
surface (in other words in all points the direction isnearly the
same). Therefore the gravitational force on m can be written
as:
(3.1.3)jgmFgrr
where the vector in (3.1.3) is the vertical unit vector on the
surface, directed toward the sky.
Definition : it is a constant force acting on every mass m on
the Earths surface, always vertical, directedtoward the ground and
proportional throughgto the mass m.
Electrostatic (or Coulomb) Force EFr
Features: this force appears on an electrically charged physical
point 1 (i.e. a point with a charge q1) whenanother charged
physical point 2 with charge q2 is at a distance r; electric charge
is a physical quantity
measured by a class of instruments called
Definition : it acts on the point 1, along the direction from 1
to 2: 1 is attracted toward 2 if the 2 chargeshave different sign,
otherwise 1 is attracted by 2. The Coulomb force is expressed
by:
21
21
2
21
04
1
r
r
r
qqFE r
rr
(04
1
= 8.99 109 Nm2C-2 in MKSA)
Where is the vector displacement from point 2 to point 1 (being
of course21rr
21rrr
= ).
Of course the definition is symmetrical and also point 2 is
attracted by 1, with the same modulus butopposite direction.
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Elastic Force eFr
Features: this force appears at both ends of an extensible
object (e.g. a spring, an elastic cord) when it
s elongated or shortened with respect to the at rest lenght
Definition : it acts on the 2 masses connected to the ends of
the object and is expressed by:
ixkFerr
Tension Force Tr
Features: this force appears at both ends of a not extensible
object (e.g. a cord, a rigid slab or rod)
Definition : it acts on the 2 masses connected to the ends of
the cord and is:
a) equal to 0 if the distance between the ends is less than the
cord length;
b) otherwise is directed along the straight line joining the
ends, always toward the center of the cord. Theodulus of the is the
same at both endsm
Forces between two surfaces in contact
Features: when 2 plane surfaces are in contact, i.e. they are
parallel and their distance is 0, one force 12Fr
appears acting on the surface 1 due to surface 2 and one force
21Fr
appears acting on the surface 2 due to
surface 1. These forces are symmetrical. If 2 surfaces are not
planes but have (at least one of them) a largecurvature radius,
they can have a nearly planar contact in one point and the plane
tangent to both surfacesn the point is considered as contact plane
(e.g. a wheel on a planar surface).
These contact forces have 2 components: a) the normal (to the
surface) force and the parallel (to the surface)orce which is
called friction.f
Normal Force
Definition : it acts on surface 1 along the direction
perpendicular to the contact plane, from surface 2 to
surface 1 (never in opposite direction) and its value N is :
a) a suitable positive value if both surfaces are in contactb) 0
if the contact disappears (distance > 0)
Static Friction Force
Definition: it acts along the contact plane, when both surfaces
are at rest with respect to each other, in asuitable direction on
the contact plane and with a suitable absolute value ranging from 0
to a maximum
FsMAX , which is given by:
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FsMAX = s. N
The coefficient s is called static friction coefficient: it
depends on the pair of materials in contact but not on
the contact area.
Dynamical (or Kinetic) Friction Force
Definition: it acts along the contact plane, when both surfaces
are in relative motion with respect to eachother, in the direction
opposite to the relative motion, and its expression is given
by:
v
vNF dd r
rr
The coefficient d is called dynamic (or kinetic) friction
coefficient: it depends on the pair of materials in
contact but not on the contact area.
Viscous Force vFr
Features: this force appears when a body is moving inside a
fluid like water, air, oils
Definition: it acts on the body, is parallel to the velocity, in
the opposite direction: its intensity is
proportional to the velocity (as a first approximation which
holds at low velocities)
vFvrr
The coefficient is called viscosity (or viscous) coefficient and
depends on the fluid, on the shape of thebody but not on the
mass.
Lorentz Force LFr
Features: this force appears when a charged point of mass m and
charge q is moving inside a magnetic fieldwith a velocity . A
magnetic field is produced by special metals or by electric current
(as will be discussed
n Physics II) and characterized by a vector
vr
Br
which is present in every geometrical point of a region of
the
space
Definition: it acts on the charged point and is expressed
as:
BvqFLrrr
$=2: The 3 Laws (Principles) of the Dynamics
I Law
In an inertial reference frame, a point-like mass, not submitted
to any force, maintains the same velocity.
II Law
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In an inertial reference frame, the sum TFr
of all forces applied to a point-like mass m is equal to the
product
of the mass times the acceleration ar
of the point:
amFTrr
=
III Law
If a point-like mass applies a forcem1 12Fr
to a point-like mass , the mass applies to a forcem2 m2 m1
21Fr
equal and opposite:
2112 FFrr
=
$=2.1: Exercises
Exercise 2.1.1
Two boxes of masses m andMlie on a horizontal perfectly smooth
plane and are in contact to each other. A
horizontal forceFr
is applied toM, on the side opposite to that one in contact with
m, which pushesMagainst m. Find the value of the acceleration of
the boxes.
DATA: =1[Kg];m M=5[Kg]; =500 [N];F
Exercise 2.1.2
Two boxes of masses m andMlie on a horizontal rough plane and
are in contact to each other. The 2
dynamic friction coefficients are m and M respectively. A
horizontal forceFr
is applied toM, on the side
opposite to that one in contact with m, which pushesMagainst m.
Find the value of the acceleration of theboxes.
DATA: =1[Kg];m M =5[Kg]; m
=0.4; M
=0.45; =500 [N];F
y
M
mF
x
R1 R2
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Exercise 2.1.3
Three boxes of masses , and lie on a horizontal perfectly smooth
plane: is connected to by
a cord and is connected to by another cord, like 3 wagons of a
train. A horizontal force
m1 m2 m3 m1 m2
m2 m3 Fr
is applied
to , directed opposite with respect and . Find the value of the
acceleration of the masses ,
, and the tensions of the 2 cords.
m1
2 m
m2 m3 m1m 3
DATA: =1[Kg]; =2[Kg]; =3[Kg]; =200 [N];m1 m2 m3 F
Exercise 2.1.4
Three boxes of masses , and lie on a horizontal rough plane: is
connected to by a cord and
is connected to by another cord, like 3 wagons of a train. The 3
dynamic friction coefficients are
m1 m2 m3 m1 m2m2 m3 1 ,
2 and 3 respectively. A horizontal forceFr
is applied to , directed opposite with respect and .
Find the value of the acceleration of the masses , , and the
tensions of the 2 cords.
m1
3
m2 m3m1 m2 m
DATA: =1[Kg]; =2[Kg]; =3[Kg];m1 m2 m3 1=0.4; 2 =0.45; 3=0.5;
=200 [N];F
Exercise 2.1.5
On a perfectly smooth plane, inclined of with respect to the
horizontal surface, there is a mass m at aheight h. Find the time
at which m reaches the basis of the surface.
m1 m2 m3T12 T23 F
R2R1 R3
y
M
mF
x
N
R1
N
R2
m1 m2 m3T
12T
23F
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DATA: =1[Kg]; h= 4 [m];m =30
m
h
Exercise 2.1.6
On a rough plane, inclined of with respect to the horizontal
surface, there is a mass m on the basis of theplane. The static and
dynamic friction coefficients between m and the plane are s and d
respectively. The
mass is kicked up and gets an initial velocity . Find at which
time and at which height h does m stop
and determine whether it comes back to the basis or not.
v0 t0
DATA: =1[Kg];m d=0.4; s =0.5; v =10 [m/s];0 =30;
Exercise 2.1.7
On a rough plane, inclined of with respect to the horizontal
surface, there is a car of mass m travellingdown toward the bottom
of the slope. The static friction on the wheels isFopposite to the
velocity. The carhas a viscous friction coefficient and, at the
time 0, the engine is switched off, while the velocity is .
Find the velocity and the distance from the point where the
engine was switched off, at the time .
v0t0
DATA: =1000[Kg]; =100 [N]; =60 [s]; =72 [Km/h]; tm F t0 v0
an()=0.07; = 4 104
[N/s]
Exercise 2.1.8
m
h
v0
v0 m
Fd
Fs
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A cord of length lhas an end fixed in the point O on a
horizontal perfectly smooth plane. A point-like massm is connected
to the other end. The mass is kicked and gets an initial velocity
perpendicular to the cord.
Find the tension of the cord.
v0
DATA: =1[Kg]; =10 [m/s]; l=1 [m]m v0
Exercise 2.1.9
On an inclined plane of angle a there is a cube of mass m1,
initially at rest, which is connected to one endof a cord. This
cord goes upward on the inclined plane and, through a massless
pulley, falls down parallelto the vertical edge of the inclined
plane and is connected, at the other end, to another mass m2 .
There isno friction between the surfaces of the inclined plane and
the cube and the viscosity of the air is negligible.At the
beginning all masses are at rest. Find the acceleration of both
masses.
Data: a = 30 ; m1 = 1 [kg;] m2 = m1/2
Exercise 2.1.10
Repeat Exercise 2.1.9 with the hypothesis that there is friction
on the inclined plane and the static and
dynamic coefficients are s and drespectively.
Data: a = 30 ; m1 = 1 [kg;] m2 = 3m1/4; s = 0.25; d= 0.2;
m1m
s
r
y
x
T
v0
m
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Lets check whether the static friction is enough to maintain the
system at rest:
+=
==
+=
Tgmym
xm
mgNrm
TmgFsm s
22
2
1
1
0cos
sin
&&
&&
&&
&&
ys &&&& =
( to be completed)
Exercise 2.1.11
On a truck which is moving straightforward on an horizontal road
there is a big cube of concrete of mass m.The mass of the truck
isM. The truck brakes suddenly with an acceleration of modulus a.
Find: a) themaximum allowed acceleration amax in order that the
concrete doesnt slide on the truck, b) the dynamicalfriction force
acting on the truck ifa =2amax.
Data: m = 1000 [kg;] M = 10000[kg]; s = 0.6; d= 0.4;
Exercise N. 2.1.12
A cord of length l has one end fixed on an horizontal desk and
has a cube of mass m connected to the otherend. There is friction
between the desk and the cube and the static and dynamic
coefficients are ms and mdrespectively. At the initial time the
velocity of the cube is v0 . Find the elapsed time until the stop
of thecube.
Data: m = 0.5 [kg;] l = 2 [m]; s = 0.5; d= 0.45;
Exercise 2.1.13
A cord of length lhas an end fixed in the point O on a
horizontal rough plane. A point-like mass m isconnected to the
other end. The static and dynamic friction coefficients between m
and the plane are s and
d respectively. The mass is kicked and gets an initial velocity
perpendicular to the cord. Find at whichv0
m1m
r
y
sx
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time and at which distances from the starting pointdoes m stop
and find the tension of the cord when m
s in the middle point of the trajectory.
t0
DATA: =1[Kg];m d=0.4; s =0.5; v =10 [m/s]; l=1 [m];0
Exercise 2.1.14
A cord of length lhas an end fixed in the point O on a
horizontal rough plane. A mass m, having a viscousfriction
coefficient , is connected to the other end. The static and dynamic
friction coefficients between m
and the plane are s and dt0
respectively. The mass is kicked and gets an initial velocity
perpendicular to
the cord. Find at the time and at which distances from the
starting pointdoes m stop and find the tension
of the cord when m is in the middle point of the trajectory.
v0
DATA: =1[Kg];m d=0.4; s =0.5; v =10 [m/s]; l=1 [m];0
Solution
Assuming a reference system with intrinsic coordinates in the
horizontal plane and a z axis as vertical(positive toward high);
considering that the present forces are: dynamic and viscous
friction in the velocityopposite direction, tension Tperpendicular
and normalNof the plane vertical, we can write:
Ngmzm
=
=
&
&
2
Tl
sm
Nssm d
+==
&&
&&
0
, which, defining sysy &&&& == , gives: gym
y d
=&
gym
y d
=& )(
gmy
my d
+=&
ln(y dm g
)=
m t+ lnC
tmd Ce
gmys
+
==
&
d
dt(y +
dm g
)=
m (y +
dm g
)
The initial conditions are: and , which give:0)0( vs =&
s(0)= 0
tmdd e
gmv
gmts
++
=
)()( 0& , and : )1()()( 0
tmdd e
mgmvt
gmts
++
=
(2.2.10.1)
Remark:
Looking at eq.(2.2.10.1), for very high values oft,s(t) becomes
negative: of course this doesnt occurbecause at a time t0, the
velocity, which is always decreasing, becomes 0. To find t0 we
put:
0
)(0 0t
mdd egm
vgm
++
=
0
)( 0t
mdd egm
vgm
+=
)1ln( 00
gm
vmt
d
+=
.
At t the velocity is 0, therefore the viscous friction is 0 and
the dynamic friction becomes static: the II law
becomes and the value of which satisfies the law also at is
0
sFsm = && sF 0t 0=sF . At m stops.0t
Exercise 2.1.15
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A perfectly smooth plane of mass M, inclined of with respect to
the horizontal surface, can slide withoutfriction on the ground. On
the inclined plane there is a mass m at a height h. Initially
bothMand m are at rest
and a horizontal force Fr
starts to pushM. Find the accelerations ofMand m.
DATA: =1[Kg];m M =5[Kg]; =3[Kg];m3 m =0.4; M=0.45; F=500
[N];
Solution
''''
'
'
cos0
sin
cos
sin
xtgyxtgy
Xxx
yy
NMgRYM
FNXM
mgNym
Nxm
&&&&&&
&&&&&&
&&&&
&&
&&
&&
&&
==
+=
=
==+=
=
=
Exercise 2.1.16
A springAB of elastic constantK1 and rest length l1 is hung up
by one endA to the ceiling, while a point likemass m1 is hung up to
the other endB in vertical position. Another spring CD of elastic
constantK2 and restength l2 is hung up by one end Cto the mass m1 ,
while another point like mass m2 is hung up to the other
endD in vertical position. Again, another springEFof elastic
constantK3 and rest length l3 is hung up byone endEto the mass m2 ,
while another point like mass m3 is hung up to the other endFin
vertical position.
The system is in equilibrium with the gravity. Find the
elongations l1 , l2 , l3 of the springs.
DATA: m1 = 1[kg] ; K1=100 [N/m]; m2 = 2[kg] ; K2=200 [N/m]; m3 =
3[kg] ; K3=300 [N/m];
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Solution:
Exercise 2.1.17
A pulley is fixed to the ceiling and 2 masses m and M are
hanging to it by a cord. Neglecting the air find theaccelerations
of the masses and the tension of the cord.
DATA: m = 1[kg] ; M= 2[kg] ;
Solution:
Choosing a reference frame having y-axis vertical, positive
toward high, and callingy and Y theaccelerations of m and M
respectively, recalling that m andMhave equal and opposite
velocities, we canwrite:
Yy
MgTYM
mgTym
&&&&
&&
&&
=
=
=
gmM
MmT
gmM
mMy
+=
+
=&&
Exercise 2.1.18
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A pulley is fixed to the ceiling and 2 masses m andMare hanging
to it by a cord. Due to the air, both masses
have the same viscous friction coefficient. Find the
accelerations of the masses and the tension of the cordafter a time
tsince the start.
DATA: m = 1[kg] ; t= 2 [s]; M= 2[kg] ;5104 =
Solution:
Choosing a reference frame having y-axis vertical, positive
toward high, and calling y and Y theaccelerations of m and M
respectively, recalling that the masses have equal and opposite
velocities, we canwrite:
Yy
YMgTYM
ymgTym
&&&&
&&&
&&&
==
=
yMgTyM
ymgTym
&&&
&&&
+=
= [ ]gMmyMmT
ygmMyMm
)()(2
1
2)()(
++=
=+
&&
&&&
(2.1.18.1)
That can be solved changing variable as: uyuy
&&&& = ; and calling gmM
mMbg
mMa
+
+ ,
2, we get:
)(a
buabauu =+=& (2.1.18.2)
and changing again variable: awwuww
a
bu == &&&)( , we get:
)1(
)1(
atat
at
at
ea
bmbemgT
eby
ea
by
+=
=
=
&&
&
; use has been made of the I.C.: 0)0( =y&
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$= 3: Forces generated by systems of points
$= 3.1: The gravitational and electric fields
Gravitation and Coulomb forces are generated by physical points
having mass or electric charge. Theseforces act on other physical
points and their intensity and direction are depending on the
relative positions ofboth generating and forced points. If the
generating point is in a fixed position (that we can assume
asorigin O of an inertial frame) in each geometrical point of the
space a physical point of mass m (charge q)
undergoes a force ),,( zyxFr
proportional to the mass m (charge q). The ratio of the
gravitational (electric)
force to the mass m (charge q) is a vector mzyxFzyxE GG
/),,(),,(rr
( qzyxFzyxE EE /),,(),,(rr
), which
s called Gravitational (Electric) Field. The knowledge of the
field allows to find the force acting on aphysical point simply by
multiplying the field times the mass or charge of the point.
$= 3.2: Systems of points generating a field
A point-like particle, having mass or electric charge, generates
a field in every point of the space. This field
has the direction along the segment joining the point to the
particle. Several particles, lying in differentpositions, generate
in the same point a field which is the (vector) sum of the single
fields (this propertyfollows from the addition of the forces). Even
though it is in general necessary to make this sum in order
toevaluate the total field generated by a system of massive or
charged particles, in some cases the spatialdistribution of the
particles shows special symmetries, which allow an analytical
evaluation of the field andts properties. To take advantage from
the symmetries we will use the concept ofFlux of a field and
the
Gauss theorem. The Gauss theorem requires that the field has the
following properties:
1) direction along the segment joining the geometrical point to
the particle2) intensity inversely proportional to the square of
the distance
Both properties are present in the gravitational and electric
field. In the following the case of the electricfield only will be
demonstrated: the gravitational case can be easily deduced from the
electric one by simplyexchanging charges with masses and
substituting the constants.
$= 3.3: Gauss theorem
First lets define the infinitesimal flux d of a field ),,(
zyxEr
generated by a point-like charge Q on an
nfinitesimal surface dS, located around the point (x,y,z), whose
normal has unit vector (see Fig.3.2.a):nr
dSnzyxEzyxdrr
),,(),,( (3.3.1)
Before going on in demonstrating the theorem, lets describe
Fig.3.3.a:
a) Sis a closed surface and contains Qb) AB is the infinitesimal
surface dSaround point (x,y,z)c) QA, QB are distances of the
pointsA andB from the charge: the angle BQA ) is infinitesimal
because
dSis infinitesimal and therefore QBQA (unless higher order
infinitesimal difference)
d) Bis a point on QB such that: 'QB (in other words 'QA = B lies
on a sphere of centre Q and radiusQA : we shall call 'dS the
infinitesimal spherical surface containingA andB)
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e) A and B are points of a sphere of centre in Q and radius 1" :
the total surface of thissphere is
" == QBQA
4"=S and the infinitesimal spherical surface "dS containingA
andB undergoes the
same angle 'BQA)
as 'dS
Fig. (3.3.a) (symbols in bold are vectors)
Lets now recall that22
'
"
"
AQ
dS
AQ
dS= and that all infinitesimal areas can be approximated by
the
underlying planar surfaces containingAB, AB, AB: on the other
hand the angle between the field and the
normal to is equal to
",', dSdSdS
dS BAB)
' . The electric field is given by:
AQ
AQ
AQ
QE
rr
2
04
1
Since dSdS = )cos(' , from eq. (3.2.1) we can write:
"4"
"
4"
"1
4'
1
4
)cos(1
44
1),,(),,(
02
02
2
20
20
20
20
dSQ
AQ
dSQ
AQ
dSAQ
AQ
QdS
AQ
Q
dSAQ
QdSn
AQ
AQ
AQ
QdSnzyxEzyxd
==
=
===
r
rrr
(3.3.2)
If we define the total flux ),( SEr
ofEr
on a closed surface of any shape as the sum of all the
infinitesimal
fluxes
S
),,( zyxd , built on (actually this sum is an integral because
the terms are infinitesimal), we get:S
000
44
"4
),,(
QQ
dSQ
zyxd === (3.3.3)
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Remark 1
What happens if more than 1 charge is presents inside S? We can
write the infinitesimal flux on
generated by 2 charges Q and Qlocated inside Sin different
positions, which produce 2 fieldsdS ',EErr
:
),,('),,(),,('),,(
)],,('),,([),,(),,(
zyxdzyxddSnzyxEdSnzyxE
dSnzyxEzyxEdSnzyxEzyxd TOTTOT
+=+==+=
rrrr
rrrrr
(3.3.4)
The total flux TOT is the sum (integral) of all the
infinitesimal fluxes ),,( zyxd TOT on the surface Sand
therefore:
00
'),,('),,(
QQzyxdzyxdTOT +=+= (3.3.5)
Repeating the demonstration it is proved that the total flux on
a closed surface, containing a number of
charges Qi is equal to the total internal charge divided by 0
:
0
= i
i
TOT
Q
(3.3.6)
Remark 2
What happens if the charge Q is outside S? In Fig.(3.2.b) one
can see that areasAB andDCrefer to the sameareaAB on the sphere of
radius1, but the angles between fields and normal vectors have
opposite signs.Therefore in the sum (integral) of the infinitesimal
fluxes of all infinitesimal surfaces of S they cancel, as
well as all the other pairs of infinitesimal surfaces generated
by each single cone. We can conclude that thecharges outside the
surface do not give any contribution to the total flux. The Gauss
theorem must be written
n terms of internal charges only, as:)(IkQ
0
)(
= k
I
k
TOT
Q
(3.3.7)
Fig. (3.2.b) (symbols in bold are vectors)
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$= 4: Symmetries
$=5: Linear momentum and impulse
Linear momentum
The linear momentum pr
of a physical point of mass m, moving with a velocity vr
is a vector quantity
defined as:
vmprr
(5.1)
Impulse
When a force )(tFr
is acting on a physical point of mass m for an infinitesimal
time interval dt, the
nfinitesimal vector quantity )(tIdr
is called impulse of the force or also impulse applied to the
point m
by the force )(tFr
, and is defined as:
dttFtId )()(rr
(5.2)
The total impulse of the force acting for a finite time from t0
to t1 is defined as:),( 10 ttIr
1
0
)(),( 10
t
t
dttFttIrr
(5.3)
Note: It is not meaningful to define the impulse as a finite
quantity because in general a force depends on thetime and during a
finite time interval it assumes different values: during an
infinitesimal time interval itsvalue remains constant apart
infinitesimal quantities of second or higher order.
Properties of the impulse
1) The sum of the impulses of several forces ),...(),(),( 321
tFtFtF rrr is equal to the impulse of thesum of the forces
(additivity of the impulse with respect to forces)
2) The sum of the impulses of a force in 2 consecutive intervals
(t0,t1) and (t1,t2) is equal to theimpulse of the total interval
(t0,t2) (additivity of the impulse with respect to time)
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Impulse theorem
When a physical point of mass m undergoes to some forces
),...(),(),( 321 tFtFtFrrr
for a finite time interval
from t0 to t1 , the variation of its linear momentum is equal to
the total impulse of the total force
),...()()( 321 tFtFtFFTrrrr
++=
Proof:
Lets calculate the total impulses of each one of the forces and
sum them, remembering the additionproperties of the integrals and
remembering the II Newton law:
1
0
)(),( 1101
t
t
dttFttIrr
; ; . (5.4) 1
0
)(),( 2102
t
t
dttFttIrr
1
0
)(),( 3103
t
t
dttFttIrr
)()()()((
)())((())(()(),(
0101
10
1
0
1
0
1
0
1
0
1
0
1
0
tptptvtvmdtamdtam
dttFdttFdttFdttFIttI
t
t
t
t
t
t
T
t
t i
i
t
t i
i
i i
t
t
iiT
rrrrrr
rrrrrr
====
=====
(5.5)
Where )(tFTr
has been used to indicate the total force acting on m, which is
equal to the mass times
acceleration because of the II law.
$=6: Angular momentum and torque
Angular momentum
The angular momentum of a physical point of mass m and vector
displacementOLr
rr
, moving with a
velocity , with respect to an arbitrarily chosen point O (called
pole), is a vector quantity defined as:vr
vmrrL OOrrrr
)( (6.1)
where Orr
is the vector displacement of the point O .
Momentum of the force (or torque)
When a forceFr
is acting on a physical point of mass m (whose vector
displacement is rr
) the momentum Or
of the force with respect to an arbitrarily chosen point O
(called pole) of vector displacement Orr
, is defined
as:
Frr OOrrrr
)( (6.2)
Additivity of the torques
If a force is a sum of some forces: ,...321 FFFFrrrr
++= its torque is :
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......)()()(
...)()()(
321321
321
+++=+++
=+++=
rrrrrrrrrrrr
rrrrrrrrr
FrrFrrFrr
FFFrrFrr
OOO
OOO(6.3)
.e. the total torque is the sum of the torques of each
force.
Angular momentum theorem (or torque theorem)
When a physical point of mass m undergoes to some forces ,...,,
321 FFFrrr
whose sum is TFr
, the variation of
ts angular momentum with respect to a pole O is equal to the
total torque of the forces minus the velocity ofthe pole times
(vector-product) the linear momentum of the mass.
Proof:
Remembering the addition properties and the II Newton law:
T
OOTOO
OO
OO
vmrFrrvmr
vmrrvmrrdt
vmrrd
dt
Ld
rr&r
rrrr&r
&rrrr&r&rrrrr
+=+=
=+=
)(
)()(
))((
(6.4)
Remarks:
a) if the pole is at rest, eq. (6.4) reduces to: TOOdt
Ldr
r
= (6.4)
b) if the velocity of the pole is parallel to the velocity of
the point, eq. (6.4) reduces to: TOOdt
Ldr
r
=
c) Eqs. (6.4) and (6.4) are vector equations: therefore each one
splits into 3 scalar equations for thecomponents
d) If conditions a) or b) are satisfied and the total torque
0=TOr , from (6.4) we get:.0 constL
dt
LdO
O ==r
r
(6.5)
i.e. the angular momentum is a constant. It must be kept in mind
that eq. (6.4) is a vector equation: in somecases only 1 or 2
components of the torque are null and therefore only these
components are constant in (6.5)
$=7: Kinetic energy and work
Kinetic energy
The kinetic energy of a physical point of mass m, moving with a
velocitycE vr
is a scalar quantity
defined as:
22
2
1)(
2
1vmvmEc =
r(7.1)
Kinetic energy can be written using the components of the
velocity but it must be always taken into accountthat it is a
scalar quantity and has nocomponents itself:
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)(2
1 222zyxc vvvmE ++= (7.2)
Work
When a force ),,( zyxF
r
is acting on a physical point of mass m along an infinitesimal
path)() tr( ttrrd
rrr+= , the infinitesimal scalar quantity is called work of the
force along the path,
and is defined as the scalar product:
dW
rdzyxFdWrr
),,( (7.3)
In general a point is moving along a finite path which lies on a
trajectory, i.e. a curve in the space, definedby the 3 functions :
the trajectory is made by an infinite number of small
displacements)(),(),( tztytx rd
rand
the total work of the force acting from the point A to the point
B along the trajectory is defined as:B,AL
++=B
A
zyx
B
A
BA dzzyxFdyzyxFdxzyxFrdzyxFW ,,
, ),,(),,(),,(),,( rr
(7.4)
WhereFx,Fy, Fz, are the components of the force along the 3
axes.
Note: from Eq. (2.3) it is clear that the total work of a force
acting on a physical point m moving from the
geometrical point A to the geometrical point B depends not only
upon A and B but also upon the curve which joinsA andB. Therefore
it is important to discuss the geometrical aspects of Eq. (2.3) in
some detail.
The infinitesimal displacement can be written as:rdr
kdzjdyidxrd
rrrr
++= (7.5)
Recalling that along a trajectory:
,)( dttxdx &= ,)( dttydy &= (7.6),)( dttz&
f the coordinates are known as a function of the time we can
also write:
++=B
A
B
A
z
B
A
yxBA dttzzyxFdttyzyxFdttxzyxFW , ,,
, )(),,()(),,()(),,( &&& (7.7)
Where all the components of the force can be read as a function
of the time through the coordinates.
Properties of the work
1) The sum of the works of several forces ),...(),(),( 321
tFtFtF rrr from a point A to a point B along atrajectory is equal
to the work of the sum of the forces (additivity of the work with
respect to forces)
2) The sum of the works of a force in 2 consecutive paths (A,B)
and (B,C) along a trajectory is equal tothe work on the total
interval (A,C) (additivity of the work with respect to path)
Work theorem or Kinetic Energy theorem
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When a physical point of mass m moves from a point A to a point
B along a trajectory , being submitted to
some forces ),...,,(),,,(),,,( 321 zyxFzyxFzyxFrrr
),,( zyxFT
, the variation of its kinetic energy is equal to the total
work
of the sumr
of these forces.
Proof:
Lets calculate the total works of all the forces, i.e. the work
of the total force acting on a physical point,remembering the
addition properties of the integrals and that this total force is
equal to the product of masstimes acceleration:
[ ] [ ] [ ])()()(
2
1)(
2
1
222
)(
)()(
)()(
)),,(),,(),,((),,(
22222
, ,,,
,,
,,
,,
,,
AEBEAmvBmvzyx
m
zdzydyxdxmzdzmydymxdxm
dtdt
zdzmdtdt
ydymdtdt
xdxmdtdt
dz
dt
zdmdtdt
dy
dt
ydmdtdt
dx
dt
xdm
dzdt
zdmdy
dt
ydmdx
dt
xdmdzzmdyymdxxm
dzzyxFdyzyxFdxzyxFrdzyxFW
cc
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
TzTyTx
B
A
TBA
==
++=
=
+=++=
=++=++=
=++=++=
=++=
&&&
&&&&&&&&&&&&
&&
&&
&&
&&&
&&&&&&&&&
rr
(7.8)
$=8: Potential energy
$= 8.1: Definition of the potential energy
Lets come back to the expression of the work of Eq. (2.2)
dzzyxFdyzyxFdxzyxFrdzyxF zyx ++= ),,(),,(),,(),,(rr
(8.1.1)
which is an infinitesimal quantity because is a sum of 3 terms,
each one being a product of a finite quantity(Fx, Fy or Fz) times
an infinitesimal quantity (dx, dy or dz).
In some special cases, i.e. with some special kind of forces,
the following situation occurs:
x
zyxUzyxF
x
zyxUzyxF
x
zyxUzyxF
z
y
x
=
=
=
),,(),,(
),,(),,(
),,(),,(
r
r
r
(8.1.2)
In other words, there exists a scalar function U(x,y,z) of the
coordinates (and not of the time), whose partial
derivatives with respect tox,y andz are the components of the
force.
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It must be stressed that the existence of a unique function
U(x,y,z) that satisfies all the 3 equations in thesystem (4.2) is
not guaranteed for all existing forces: in fact only some forces
have been found to have sucha function.
Those forces, for which a function U(x,y,z), satisfying the
requirements (8.1.2), exists, are called
conservative forces; the other ones are called not
conservatives.The function U(x,y,z) is said potential energy of the
force.
$= 8.2: Work of conservative forces
If a force is conservative, i.e. it satisfies eq. (8.1.2), the
work of this force from A toB along a path is givenby:
)()(),,(),,(),,(
),,(),,(),,(),,(
,,
,,
,,
BUAUdUdzz
zyxUdy
y
zyxUdx
x
zyxU
dzzyxFdyzyxFdxzyxFrdzyxFW
B
A
B
A
B
A
zyx
B
A
bA
==
+
+
=
=++==
rr
(8.2.3)
In (8.2.3) use has been done of the property :
dzz
zyxUdy
y
zyxUdx
x
zyxUzyxdU
+
+
=
),,(),,(),,(),,( (8.2.4)
already used in the Error Theory: it states that the
differential variation of a function of several variables isequal
(unless higher order terms) to the sum of the partial differentials
obtained from the first order partial
derivatives.
Remarks:
a) the result (8.2.3) states that the work of a conservative
force is independent on the path as aconsequence the work along a
closed curve is null
$= 8.3: Theorem of Work in presence of conservative forces
If the total force TFr
acting on a physical point is the sum of non conservative
NCFr
and conservative CFr
forces: CT FFFNC rrr += , the work theorem is written as:
)()()()(
),,(),,(),,(
,,
,,,
,,
AEBEBUAUW
rdzyxFrdzyxFrdzyxFW
cc
NC
BA
B
A
C
B
A
NC
B
A
TBA
=+=
=+=
rrrrrr
(8.2.5)
Eq. (8.2.5) becomes of fundamental importance when all non
conservative forces are absent: in this case the
non conservative work is null and one gets:NCBAW ,,
)()()()()()()()( AUAEBUBEAEBEBUAU cccc +=+= (8.2.6)
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Equation (8.2.6) states that in absence of non conservative
forces the sum of the kinetic and potential energyof a point-like
mass is constant during the motion. The sum of kinetic and
potential energy is called total
mechanical energy: and eq. (8.2.6) becomes the famous theorem of
the conservation of the
energy.
UEE cT +
$= 8.4: Conservative and non conservative forces
(to be completed)
$= 8.5: Central forces
(to be completed)
$= 8.6: Exercises
Exercise 8.6.1
A mass is at rest at the bottom of a rough inclined plane. It is
kicked upward on the plane and receives an
nitial velocity of modulus . Using the impulse theorem, find the
time t from the start after which it stops.
m
0v
Data: ,kgm 1= 5.0=s , 4.0=d , ,smv /20= = 30 .
Solution
Choose a reference system as in figure
( ) tNgm
tgmNvm d
+=
=
cos00
sin0 0
cosgmN =
( )gv
td sincos
0
+=
st 361.=
Exercise 8.6.2
Repeat the previous exercise to find the maximum height and
whether it returns back.
( )
( )( )
==0,0,
, sincos
d
O
d
d
O
T
dO idximgmgrdFWrrrr
xy
d
m
hmax
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2
00,2
1mvEEW kkfdO ==
dh sin=
20,
21sin mvdmgW dO =
2
02
1sincos mvdmgdmg d =
( ) sincos21
2
0
+=
dg
vd
mmd 340=
Exercise 8.6.3
A car of mass is moving along a rough horizontal road with a
velocitym 0vr
. Suddenly it breaks (blocking
the wheels). Find the distance from the start of breaking and
the stop. The dynamic friction as coefficient
d .
Data: 5.0=d , .hkmv /500=
Solution
( ) mgddxmgkdzjdyidximgrdFW dd
d
d
d
d
T
dO ==++== 000
,
rrrrrr
2
0, 2
1mvEEEW
iifdO
===
g
vd
d2
2
0=
md 67.19=
Exercise 8.6.4
A rocket is going up toward the space along a radial trajectory
from the Earth surface. At an height h(outside the atmosphere) a
satellite is thrown away from the rocket at a velocity v0 inclined
of an angle
O d
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with respect to the radial trajectory. Find the angle such that
the satellite will finish into a circular orbitaround the Earth and
find the radius of the orbit.
DATA:R=6.371*10^6[m]; h=24*10^6[m]; =6.67*10^--11[m];
[N*m2/kg2]; ME=5.9824*10^24[m];v0 =
6000 [m/s];
Solution
The total force acting on the satellite after the launch is the
gravity, which is a central force. Therefore theangular momentum is
conserved since the time of the launch.
Assuming a cylindrical reference system with the origin in the
centre of the Earth positive toward up, we can
write both angular momentum )(rLr
and total mechanical energy as a function of the distance
r:)(rE
r
mMrvrvmrE
kvrmrvrvmrrL
r
r
+=
=+=
))()((21)(
))()(()(
22
rrrrr
(8.4.1.1)
CallingL0 andE0 the angular momentum and energy at the beginning
andLrandEr the same in the orbitand defining r0 = R+h, the
conservation theorems state:
r
mMrvrvm
r
mMvm
mrvvmr
r
+=
=
))()((2
1
2
1
sin
22
0
2
0
00
(8.4.1.2)
and to be circular the final orbit requires that:
2
2
0)(
r
mM
r
mv
rv Mr
=
=
r
Mv
r
Mv
r
Mv
rvsenvr
=
=
=
2
2
0
2
0
00
2
1
2
1
)(
00
2
2
0
0
2 2
vr
rvsen
v
Mr
vr
Mv
=
=
=
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Exercise 8.6.2
On a horizontal, smooth table a spring is fixed by one end to a
point O and a mass m is connected to theother end. The rest length
of the spring is negligible and the elastic constant is k.
Initially the mass is at rest
at a distance r0 from O and is kicked getting an initial
velocity v0 inclined ofwith respect to the normal to
the spring. Find the maximum distance that the spring will
reach.
Sol.
The only force is the (central) elastic force that in
cylindrical coordinates looks like: . The
velocity can be written as:
rr
= rkF
rrr
Nr vvv += , where the radial and normal components are written.
The
physical properties of the force are: 1) it is conservative,
therefore the total energy is conserved, 2) thetorque with respect
the pole O is 0 because the force is parallel to the radial vector:
therefore also the angularmomentum is conserved. We can write the
general expression for angular momentum and energy as:
222
2
1)(
2
1
)(
rkvvmE
kvrvvrL
Nr
NNro
++=
=+=rrrrr
And for initial and final times:
20
20
00
2
1
2
1
cos
rkvmE
kvrLo
+=
=rr
and22
2
1
2
1ff
ffo
rkvmE
kvrL
+=
=rr
which are equal respectively. Therefore:
2220
20
00 cos
ff
ff
rkvmrkvm
vrvr
+=+=
mk
vrm
k
rm
k
vrm
k
vrf
/2
cos4
2
0
2
0
2
0
2
0
2
0
2
02 ++
=
Maximum will be the positive square root of the solution
with+.
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$=9: System of points
$= 9.1: Definition of System of points
A set ofNphysical points of masses mi is said a System
ofNpoints.
Center of mass
Given a system ofNpoints, each one of mass mi and vector
displacement irr
, the Center of Mass of the
system (CM) is defined as the (geometrical) point whose vector
displacement CMrr
is:
N
i
i
N
i
ii
CM
m
rm
r
r
r(9.1.1)
When the points of the system are moving, their vector
displacements are function of the time: therefore alsothe CM is
moving, with velocity CMv
rand acceleration CMa
rgiven by the corresponding derivatives:
=
N
i
i
N
i
ii
N
i
i
N
i
ii
CM
m
vm
m
dt
rdm
v
rr
r (9.1.2)
=N
i
i
N
i
ii
N
i
i
N
i
ii
CM
m
am
m
dt
rdm
a
rr
r2
2
(9.1.3)
Linear momentum of a system
The linear momentum TPr
of a system is defined as the sum of the linear momenta of each
point of the
system:
=N
i
iiTCM
N
i
i
N
i
iiT amPvmvmPr&rrrr
)( (9.1.4)
From (3.6.4) and (3.6.2), defining as the total mass of the
system one gets:N
i
iT mM
CMCMTTCMCMTT PaMPPvMP&rr&rrrr
=== (9.1.5)
Kinetic energy of a system
The kinetic energyETof a system is defined as the sum of the
kinetic energies of each point of the system:
N
iiiT
vmE 2
2
1 r (9.1.6)
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Angular momentum of a system
The angular momentum of a system with respect to a pole O is
defined as the sum of the angular
momenta of each i-th point of the system with respect to the
same pole O:
T
OLr
i
OLr
N
i
iiOi
N
i
i
O
T
O vmrrLLrrrrr
)( (9.1.7)
$= 9.2: Internal and external forces in a system of points
Definition of internal and external forces
On each of theNpoints of a system some forces are applied: some
of these forces are produced by points notbelonging to the system
and we shall call them external forces; e.g. if we consider a
system of 3 particles ofmasses m1, m2, m3, on a desk, the 3 gravity
forces m1.g, m2.g, m3.g, are external forces because due to
theEarth which is outside the system. The same holds for the 3
normal forces due to the desk.
On the other hand the gravitational forces applied by each mass
to each other mass is produced by a point ofthe system on another
point of the system: we shall call this kind of forces internal
forces.
In general, on the i-th point of a system there are some
external forces whose sum we shall call ;and there
are (N-1) forces
iFr
ijFr
produced by the (N-1) other points on the i-th point; from this
definition follows that the
sum of all the internal forces acting on the i-th point is given
by:Ii
Fr
=
=N
j
ij
I
i FF1
rr(9.2.1)
Theorem of the internal forces
Its easy matter to demonstrate the following theorem: the sum
TFr
of all the forces acting on all the points of
a system is the sum EFr
of all external forces only.
Proof:
EN
i
N
j
ij
N
i
i
N
j
ij
N
i
iT FFFFFFrrrrrr
=+=+= = ==== 1 1111
)( (9.2.2)
In fact in the 3-th member of the (9.2.2) the first sum contains
only external forces, therefore is EFr
; the
second double sum contains only internal forcesand containsall
pairs ijFr
and jiFr
. Remembering the III law
f the dynamics we get:o
(9.2.3)0)()(),:,(),:,(1 1
==+= = = ijjipairs
ijij
ijjipairs
jiij
N
i
N
j
ij FFFFFrrrrr
As we wanted to proof.
Definition of internal and external torques
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If a pole O has been chosen, as for the forces, on each i-th
point of a system there are external torques (due
to external forces) whose sum we shall call Oir ;and internal
torques O
ji,r
(due to the (N-1) internal forces);
again from this definition follows that the sum iIO,
r
of all the internal torques acting on the i-th point is
given by:
(9.2.4)=
=N
j
ji
O
iI
O
1
,, rr
quesTheorem of the internal tor
te the following theorem: the sum TOr
As for the forces we can demonstra of all the torques acting on
all the
points of a system is the sum EOr
of all external torques only if the internal forces between a
point i and a
pointj act along the directio ji rrnrr
joining i withj.
Proof:
Where the 2nd term in the 4th member of Eq. (3.7.2) is equal
(9.2.6)
Its useful to stress again that in the last equation of (3.7.6)
the vector product is equal to 0 only in the
E
O
N
j
ij
N
i
Oi
N
i
iOi
N
j
ij
N
i
iOi
N
i
i
O
T
O
FrrFrr
FFrr
rrrrrrr
rrrrrr
=+=
=+==
===
==
111
11
)()(
)()(
(9.2.5)
to 0 because:
0)(
)()()(
)()()()(1 111
==
=+=
=+==
= ===
N
pairs
ijji
N
pairs
ijOjOi
N
pairs
ijOjijOi
pairs
jiOjijOi
i j
ijOi
j
ij
i
Oi
Frr
FrrrrFrrFrr
FrrFrrFrrFrr
rrr
rrrrrrrrrrr
NN NNN
rrrrrrrrrrrr
hypothesis that the internal forces between i andj act along the
direction ji rrrr
. It must be said that all
forces between two physical points studied in this Chapter 3
satisfy this requirement.
the
=10:Cardinal Equations of the Dynamics of the Systems
the systems
nt, using the definition of internal and external forces:
$
$= 10.1: The two Cardinal Equations
I cardinal equation of the dynamics of
Lets write the II law of the dynamics for each i-th poi
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(10.1.1)
............
1
2221
1
1111
=
+=
+=
=
=
N
j
j
N
j
j
FFam
FFam
rrr
rrr
If we add all left and all rigth members of the equations
(3.7.4), recalling Eqs. (3.7.2) and (3.6.4) we have:
CMT
EN
j
ij
N
i
ii
N
i
PPFFFam&r&rrrrr
===+= == 11
)( (10.1.2)
T
E PF&rr
= is said I cardinal equation of the dynamics of the systems
II cardinal equation of the dynamics of the systems
We proceed in an analogous way as done for the forces in the I
Cardinal Equation, writing:
],...2,1[, Nivmrdt
Ld iOiiO
i
O =+= rr&r
r
(10.1.3)
and summing allNequations, getting:
E
OTO
E
O
N
iiO
Ni
O
N
iiO
N i
O
T
O
Prvmr
vmrdt
Ld
dt
Ld
rr&rrr&r
rr&r
rr
+=+=
=+=
)(
)(
1
111 (10.1.4)
where use has been done of the definition (3.6.7) and of the
theorem of the internal torques (3.7.5)
E
OTO
T
O Prdt
Ldrr&r
r
+= is said II cardinal equation of the dynamics of the
systems
$= 10.2: Exercises
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Appendix 3A Linear II order Differential Equations with constant
coefficients
A linear II order differential equation with constant
coefficients is a form like:
)()()()(
2
2
tFtfcdt
tdfb
dt
tfda =++ (3A1)
where the unknown functionf(t) and its derivatives appear on the
left side as a sum with the first power andmultiplied by constant
parameters: the right side contains only a known functionF(t) .
In Physics and in Engineering equations of the type (1) are
encountered in several problems. These problemsdiffer from each
other not only for the different values of the parameters a, b, c
but also for the differentInitial Conditions of the problems.
Therefore the solution of eq. (1) should contain 2 constants
(integrationconstants) to be determined by applying the initial
conditions. In Mathematical Analysis a solution satisfyingthese
requirements (equation satisfied and initial conditions satisfied)
is called general solution and isdemonstrated to be unique.
As a further requirement of Physics and Engeneering, since in
the most cases f(t) represents physicalquantities, we shall look
for a solution in the field of the real numbers.
Lets now try to solve eq.(1).
Before solving eq.(1) lets try to solve the following equation
called associated homogeneous equation:
0)()()(
00
2
02
=++ tfcdt
tdfb
dt
tfda (3A2)
To do that we can try with a function of exponential type that
we insert in (2), getting:tetf = )(0
02 =++ cba (3A3)
Eq. (3) is a second degree algebraic equation in the unknown and
is called characteristic equation. It is
solved by the following values:
a
c
a
b
a
b= 2
2
42
(3A4)
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A linear combination of the two exponential solutions:
tteBeAtf
+ += )(0 (3A5)
s a solution of Eq. (2) which can satisfy the initial conditions
with suitable values of A and B: therefore it isthe general
solution of the homogeneous associated equation (2). It must be
stressed that it is not the generalsolution of our equation
(1).
There are 3 possible cases of the values of :
1)a
c
a
b>
2
2
4 )()(
2
2
2
2
440
ta
c
a
bt
a
c
a
bt
a
b
eBeAetf
+=
2)a
c
a
b=
2
2
4 )()(0 tBAetf
ta
b
+=
3)a
c
a
b
