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Chapter 11 KINEMATICS OF PARTICLES
x
PO
x
The motion of a particle along astraight line is termed rectilinearmotion. To define the positionPof the particle on that line, we
choose a fixed origin O and apositive direction. The distancex from O toP, with the
appropriate sign, completely defines the position of the particle
on the line and is called the position coordinateof theparticle.
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x
PO
x
The velocity v of the particle is equal to the time derivative ofthe position coordinatex,
v =dx
dtand the accelerationa is obtained by differentiating v withrespect to t,
a =dv
dt
or a =d2x
dt2
we can also express a as
a = vdv
dx
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x
PO
x
v =dx
dta =
dv
dt
or a = d2
xdt2 a = v dvdxor
The velocity v and acceleration a are represented by algebraic
numbers which can be positive or negative. A positive value for
v indicates that the particle moves in the positive direction, anda negative value that it moves in the negative direction. A
positive value fora, however, may mean that the particle is truly
accelerated (i.e., moves faster) in the positive direction, or that
it is decelerated (i.e., moves more slowly) in the negativedirection. A negative value fora is subject to a similar
interpretation.
+-
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Two types of motion are frequently encountered: uniform
rectilinear motion, in which the velocity v of the particle is
constant and
x =xo + vt
and uniformly accelerated rectilinear motion, in which the
acceleration a of the particle is constant and
v = vo + at
x =xo
+ vot + at2
1
2
v2 = vo + 2a(x -xo )2
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x
O
xA
xB
xB/A
A B
When particlesA andB move along the same straight line, the
relative motion ofB with respect toA can be considered.
Denoting byxB/Athe relative position coordinate ofB with respect
toA , we have
xB =xA +xB/A
Differentiating twice with respect to t, we obtain
vB = vA + vB/A aB = aA + aB/A
where vB/A and aB/A represent, respectively, the relative velocityand the relative acceleration ofB with respect toA.
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A
B
C
xA
xB
xC
When several blocks are are connected by inextensible cords,
it is possible to write a linear relation between their position
coordinates. Similar relations can then be written between
their velocities and their accelerations and can be used toanalyze their motion.
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Sometimes it is convenient to use a graphical solution for
problems involving rectilinear motion of a particle. The graphical
solution most commonly involvesx - t, v - t, and a - tcurves.
a
tv
tx
t
t1 t2
v1
v2
t1 t2
v2 - v1 = a dtt1
t2
x1
x2
t1 t2
x2 -x1 = v dtt1
t2
At any given time t,
v = slope ofx - tcurve
a = slope ofv - tcurve
while over any given time interval
t1 to t2,
v2 - v1 = area undera - tcurve
x2 -x1 = area underv - tcurve
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x
y
r
P
Po
O
v
s
The curvilinear motion of a particle
involves particle motion along a
curved path. The positionPof the
particle at a given time is definedby theposition vectorr joining the
origin O of the coordinate system
with the pointP.
The velocity v of the particle is defined by the relation
v =dr
dtThe velocity vector is tangent to the path of the particle, and
has a magnitude v equal to the time derivative of the lengths ofthe arc described by the particle:
v =ds
dt
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x
y
r
P
Po
O
v
s
v =dr
dt
In general, the acceleration aof the particle is not tangent
to the path of the particle. It
is defined by the relation
v =ds
dt
a =dv
dtx
y
r P
Po
O
a
s
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x
y
zi
j
k
vx
vy
vz
xiyj
zk
P
x
y
z
i
j
k
r
ax
ay
az
P
Denoting byx,y, andzthe rectangular
coordinates of a particleP, the
rectangular components of velocity and
acceleration ofPare equal, respectively,to the first and second derivatives with
respect to tof the corresponding
coordinates:
vx =x vy =y vz=z. . .
ax =x ay =y az=z.. .. ..
r
The use of rectangular components is
particularly effective in the study of the
motion of projectiles.
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x
y
z
x
y
z
A
B
rA
rB rB/A
For two particles A andB moving
in space, we consider the
relative motion ofB with respect
toA , or more precisely, withrespect to a moving frame
attached toA and in translation
withA. Denoting by rB/Athe
relative position vector ofB with
respect toA , we have
rB = rA + rB/A
Denoting by vB/Aand aB/A, respectively, the relative velocityand
the relative acceleration ofB with respect toA, we also have
vB = vA + vB/A
aB = aA + aB/Aand
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x
y
C
P
an = en
O
v 2
r
at = etdv
dt
It is sometimes convenient to
resolve the velocity and acceleration
of a particlePinto components other
than the rectangularx,y, andz
components. For a particlePmoving
along a path confined to a plane, we
attach toP the unit vectors et
tangent to the path and en normal to
the path and directed toward thecenter of curvature of the path.
The velocity and acceleration are expressed in terms of tangential
and normal components. The velocity of the particle is
v = vet
The acceleration is
a = et + env2
r
dv
dt
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v = vet
In these equations, v is the speed of the particle and r is theradius of curvature of its path. The velocity vectorv is directed
along the tangent to the path. The acceleration vectora
consists of a component at directed along the tangent to thepath and a component an directed toward the center of
curvature of the path,
a = et + env2r
dvdt
x
y
C
P
an = en
O
v 2
r
at = etdv
dt
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x
P
O
eq
q
r = rer
erWhen the position of a particle moving
in a plane is defined by its polar
coordinates r and q, it is convenient to
use radial and transverse componentsdirected, respectively, along the
position vectorr of the particle and in
the direction obtained by rotating r
through 90
o
counterclockwise. Unitvectors er and eq are attached toPand are directed in the radial
and transverse directions. The velocity and acceleration of the
particle in terms of radial and transverse components is
v = rer+ rqeq. .
a = (r- rq2)er + (rq + 2rq)eq... .. . .
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x
P
O
eq
q
r = rer
erv = rer+ rqeq
. .
a = (r- rq2)er + (rq + 2rq)eq... .. . .
In these equations the dots represent differentiation withrespect to time. The scalar components of of the velocity
and acceleration in the radial and transverse directions are
therefore
vr= r vq= rq. .
ar= r- rq2 aq = rq + 2rq... .. . .
It is important to note that ar is not equal to the time derivative
ofvr, and that aq is not equal to the time derivative ofvq.
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Chapter 12 KINETICS OF PARTICLES:
NEWTONS SECOND LAW
Denoting by m the mass of a particle, by S F the sum, orresultant, of the forces acting on the particle, and by a the
acceleration of the particle relative to a newtonian frame of
reference, we write
S F = maIntroducing the linear momentum of a particle,L = mv,
Newtons second law can also be written as
S F = L.
which expresses that the resultant of the forces acting on a
particle is equal to the rate of change of the linear momentum
of the particle.
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To solve a problem involving the motion of a
particle, S F = ma should be replaced byequations containing scalar quantities. Using
rectangular components ofF and a, we have
SFx = max SFy= may SFz= maz
x
y
P
an
O
at
x
y
z
ax
ay
az
P
x
P
aq
O
ar
Using tangential and normal components,
SFt = mat= m dvdtv2
rUsing radial and transverse components,
...
.. . .SFr = mar= m(r- rq2)
qr
SFn = man= m
SFq = maq= m(rq + 2rq)
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x
y
z
Pr
f
HO
O
The angular momentum HO of a
particle about point O is defined as
the moment about O of the linear
momentum mv of that particle.
HO = rxmv
We note that HO is a vector
perpendicular to the plane containing r and mv and of magnitude
HO = rmv sin f
Resolving the vectors r and mv into rectangular components,
we express the angular momentum HO in determinant form as
HO =i j k
x y z
mvx mvy mvz
mv
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x
y
z
Pr
fHO
O
HO =i j k
x y z
mvx mvy mvzIn the case of a particle moving
in thexy plane, we havez= vz= 0.
The angular momentum is perpendicular to thexy plane and is
completely defined by its magnitudeHO =Hz= m(xvy -yvx)
Computing the rate of change HO of the angular momentum HO ,
and applying Newtons second law, we writeS MO = HO
.
.
which states that the sum of the moments aboutO of the
forces acting on a particle is equal to the rate of change of the
angular momentum of the particle aboutO.
mv
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f0
mv0
r0
O r
P0
Pf
mv When the only force acting on a
particlePis a force F directed
toward or away from a fixed
point O, the particle is said to bemoving under a central force.
Since S MO = 0 at any giveninstant, it follows that HO = 0 for
all values of t, and
HO = constant
We conclude that the angular momentum of a particle moving
under a central force is constant, both in magnitude and
direction, and that the particle moves in a plane perpendicularto HO.
.
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f0
mv0
r0
O r
P0
Pf
mv
rmv sin f = romvo sin fo
Using polar coordinates and recalling that vq = rqandHO = mr2q,we have
r2q = h.
where h is a constant representing the angular momentum per
unit massHo/m, of the particle.
Recalling thatHO = rmv sin f, wehave, for pointsPO and P
..
for the motion of any particle under
a central force.
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A i t t li ti f th ti
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M
m
F
-F
r
An important application of the motion
under a central force is provided by the
orbital motion of bodies under gravitational
attraction. According to Newtons law of
universal gravitation, two particles at adistance rfrom each other and of masses
Mand m, respectively, attract each other
with equal and opposite forces F and -F directed along the line
joining the particles. The magnitudeF of the two forces is
F= GMm
r2
where G is the constant of gravitation. In the case of a body of
mass m subjected to the gravitational attraction of the earth, theproduct GM, whereMis the mass of the earth, is expressed as
GM=gR2
whereg= 9.81 m/s2 = 32.2 ft/s2 andR is the radius of the earth.
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A particle moving under a central
force describes a trajectory defined
by the differential equation
d2u
dq 2+ u =
F
mh 2u 2
whereF> 0 corresponds to an attractive force and u = 1/r. In the
case of a particle moving under a force of gravitational attraction,
we substituteF= GMm/r2into this equation. Measuring q from theaxis OA joining the focus O to the pointA of the trajectory closest
to O, we find
1
r= u = + C cos qGM
h2
Aq
O
r
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1
r= u = + C cos qGM
h2
This is the equation of a conic ofeccentricity e= Ch2/GM. Theconic is an ellipse ife < 1, a
The values of the initial velocity corresponding, respectively, toa parabolic and circular trajectory are
vesc =2GM
r0vcirc =
GM
r0
parabola ife =1, and a hyperbola
ife > 1. The constants Cand h canbe determined from the initial conditions; if the particle isprojected from pointA with an initial velocity v0 perpendicular to
OA, we have h = r0v0.
Aq
O
r
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vesc=2GM
r0vcirc=
GM
r0
vesc is the escape velocity, which isthe smallest value ofv0 for which
the particle will not return to its
starting point.
Theperiodic time t of a planet or satellite is defined as the timerequired by that body to describe its orbit,
2pab
h
t =
whereh = r0v0and where a and b represent the semimajor and
semiminor axes of the orbit. These semiaxes are respectively
equal to the arithmetic and geometric means of the maximum
and minimum values of the radius vectorr.
Aq
O
r
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s2
A1
A2
A
s1s
dr
F
ads
dU= F dr =Fds cosa
The work ofF during a finite
displacement fromA1 toA2 ,denoted by U1 2 , is obtained
U1 2 = F drA1
A2
by integrating along the path described by
the particle.
For a force defined by its rectangular components, we write
U1 2 = (Fxdx +Fydy +Fzdz)A
1
A2
A
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U1 2 = - Wdy = Wy1 - Wy2
y1
y2
y
dy
W
A1
A2
The work of the weight
W of a body as its
center of gravity movesfrom an elevationy1to
y2 is obtained by setting
Fx =Fz= 0 and
Fy = - W.
y1
y2
The work is negative when the elevation increases, andpositive when the elevation decreases.
A
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r1
r2
M
m
F
A
-F
O
A
A1
A2
dr
r
q
dq
The work of the gravitational force
F exerted by a particle of massM
located at O on a particle of mass
m as the latter moves fromA1 toA2 is obtained from
r2
r1U1 2 =
GMm
r2 dr
GMm
r2
GMm
r1
= -
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The kinetic energy of a particle of mass m moving with a
velocity v is defined as the scalar quantity
T= mv212
From Newtons second law theprinciple of work and energy
is derived. This principle states that the kinetic energy of a
particle atA2can be obtained by adding to its kinetic energy
atA1 the work done during the displacement fromA1 toA2 by
the force F exerted on the particle:
T1 + U1 2 = T2
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The power developed by a machine is defined as the time rate
at which work is done:
Power = = F vdUdt
where F is the force exerted on the particle and v is the velocity
of the particle. The mechanical efficiency, denoted by h, isexpressed as
h =power output
power input
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When the work of a force F is independent of the path followed,
the force F is said to be a conservative force, and its work is
equal to minus the change in the potential energy Vassociated
with F :
U1 2 = V1 - V2
The potential energy associated with each force considered
earlier is
Force of gravity (weight):
Gravitational force:
Elastic force exerted by a spring:
Vg = Wy
Vg = - GMmr
Ve = kx2
1
2
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U1 2 = V1 - V2
T1 + V1 = T2 + V1
This relationship between work and potential energy, when
combined with the relationship between work and kineticenergy (T1 + U1 2 = T2) results in
This is theprinciple of conservation of energy, which states that
when a particle moves under the action of conservative forces,
the sum of its kinetic and potential energies remains constant.
The application of this principle facilitates the solution ofproblems involving only conservative forces.
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O
r
v
r0
P0
v0
P
f0
f When a particle moves under acentral force F, its angular momentum
about the center of force O remains
constant. If the central force F is alsoconservative, the principles of
conservation of angular momentum
and conservation of energy can be
used jointly to analyze the motion of
the particle. For the case ofoblique
launching, we have
(HO)0 =HO : r0mv0 sin f0 = rmv sin f
T0 + V0 = T+ V: mv2 - = mv2 -0GMm
r0
GMm
r
1
2
1
2
where m is the mass of the vehicle andMis the mass of the earth.
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The linear momentum of a particle is defined as the product mv
of the mass m of the particle and its velocity v. From Newtons
second law, F = ma, we derive the relation
mv1 + Fdt= mv2t1
t2
where mv1 and mv2 represent the momentum of the particle at atime t1 and a time t2 , respectively, and where the integral defines
the linear impulse of the force F during the corresponding time
interval. Therefore,
mv1 + Imp1 2 = mv2
which expresses the principle of impulse and momentum for a
particle.
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When the particle considered is subjected to several forces, the
sum of the impulses of these forces should be used;
Since vector quantities are involved, it is necessary to consider
theirx andy components separately.
mv1
+ SImp1 2
= mv2
The method of impulse and momentum is effective in the studyof impulsive motion of a particle, when very large forces, called
impulsive forces, are applied for a very short interval of timeDt,since this method involves impulses FDtof the forces, ratherthan the forces themselves. Neglecting the impulse of any
nonimpulsive force, we write
mv1 + SFDt= mv2
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Smv1
+ SFDt= Smv2
In the case of the impulsive motion of several particles, we write
where the second term involves only impulsive, external forces.
In the particular case when the sum of the impulses of the
external forces is zero, the equation above reduces to
Smv1 = Smv2
that is, the total momentum of the particles is conserved.
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Line of
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mAvA + mBvB = mAvA + mBvB
The second equation relates the
relative velocities of the two bodiesbefore and after impact,
vB - vA = e (vA - vB )
The constant e is known as thecoefficient of restitution; its value lies
between 0 and 1 and depends on the
material involved. When e = 0, the
impact is termedperfectly plastic; whene = 1 , the impact is termedperfectly
elastic.
A
B
vA
vB
Line of
Impact
A
B
vA
vB
Before Impact
After Impact
In the case of oblique central impactLine of
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A
B
vA
vB
In the case ofoblique central impact,
the velocities of the two colliding
bodies before and after impact are
resolved into n components along
the line of impact and tcomponents
along the common tangent to the
surfaces in contact. In the tdirection,
mA (vA)n + mB (vB)n =
mA (vA)n + mB (vB)n
Line of
Impact
A
BvA
vB
Before Impact
After Impact
n
t
vA
vB
(vA)t= (vA)t (vB)t= (vB)t
(vB)n - (vA)n = e [(vA)n - (vB)n]
while in the n directionn
t
Line of
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A
B
vA
vB mA (vA)n + mB (vB)n =mA (vA)n + mB (vB)n
Line of
Impact
A
BvA
vB
Before Impact
After Impact
n
t
vA
vB
(vA)t= (vA)t (vB)t= (vB)t
(vB
)n
- (vA
)n
= e [(vA
)n
- (vB
)n]
n
tAlthough this method was
developed for bodies moving freely
before and after impact, it could beextended to the case when one or
both of the colliding bodies is
constrained in its motion.
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m v Consider the motion of the particles
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x
y
z
Ox
y
z
G
ri Pi
miv iConsider the motion of the particles
of a system with respect to a
centroidal frame Gxyzattached to
the mass centerG of the system and
in translation with respect to thenewtonian frame Oxyz. The angular
momentum of the system about its
mass centerG is defined as the
sum of the moments about G of the momenta mivi of theparticles in their motion relative to the frame Gxyz. The same
result is obtained by considering the moments about G of the
momenta mivi of the particles in their absolute motion. Therefore
HG = S (rixmivi) = S (rixmivi)i =1n
i =1
n
m vyn
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miv i
x
y
z
Ox
y
z
G
ri Pi
HG = S (rixmivi)i =1
i =1
n
= S (rixmivi)We can derive the relation
S MG = HG.
which expresses that the moment resultant aboutG of the
external forces is equal to the rate of change of the angular
momentum aboutG of the system of particles.
When no external force acts on a system of particles, thelinear momentum L and the angular momentum Ho of the
system are conserved. In problems involving central forces,
the angular momentum of the system about the center of force
O will also be conserved.
m v The kinetic energy T of a system ofy
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miv i The kinetic energy Tof a system of
particles is defined as the sum of
the kinetic energies of the particles.
T= S mivi2
i = 1
n12
x
y
z
Ox
y
z
G
ri Pi
Using the centroidal reference
frame Gxyzwe note that the
kinetic energy of the system can also be obtained by adding the
kinetic energy mv2 associated with the motion of the mass
centerG and the kinetic energy of the system in its motion
relative to the frame Gxyz:
1
2
T= mv 2 + S mivii = 1
n1
2
21
2
m vy n1
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miv i
x
y
z
Ox
y
z
G
ri Pi
T= mv 2 + S mivii = 1
n1
2
21
2
Theprinciple of work and energy
can be applied to a system of
particles as well as to individual
particles
T1 + U1 2 = T2where U1 2 represents the work of all the forces acting on the
particles of the system, internal and external.
If all the forces acting on the particles of the system areconservative, theprinciple of conservation of energycan be
applied to the system of particles
T1 + V1 = T2 + V2
y (mAvA)1 y (m v )y t
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x
y
O
(mAvA)1
(mBvB)1
(mCvC)1
x
y
O
(mAvA)2
(mBvB)2
(mCvC)2
x
y
t1
t2S Fdt
t1
t2S MOdt
O
Theprinciple of impulse and momentum for a system of particles
can be expressed graphically as shown above. The momenta of
the particles at timet1and the impulses of the external forces
from t1 to t2 form a system of vectors equipollent to the system of
the momenta of the particles at time t2 .
y (mAvA)1 y (m v )
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x
y
O
(mAvA)1
(mBvB)1
(mCvC)1
x
y
O
(mAvA)2
(mBvB)2
(mCvC)2
If no external forces act on the system of particles, the systemsof momenta shown above are equipollent and we have
L1 = L2 (HO)1 = (HO)2
Many problems involving the motion of systems of particles canbe solved by applying simultaneously the principle of impulse
and momentum and the principle of conservation of energy or
by expressing that the linear momentum, angular momentum,
and energy of the system are conserved.
(D m)vB
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(D m)vA
Smivi
AS
B
( ) B
Smivi
AS
B
Forvariable systems of particles, first consider a steady stream
of particles, such as a stream of water diverted by a fixed vane orthe flow of air through a jet engine. The principle of impulse and
momentum is applied to a system Sof particles during a time
interval Dt, including particles which enter the system atA duringthat time interval and those (of the same mass Dm) which leavethe system atB. The system formed by the momentum (Dm)vAofthe particles enteringSin the time Dtand the impulses of theforces exerted on Sduring that time is equipollent to the
momentum (Dm)vBof the particles leavingSin the same time Dt.
SSF Dt
SM Dt
(D m)vB
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(D m)vA
Smivi
AS
B
( ) B
Smivi
AS
B
Equating thex components,y components, and moments about
a fixed point of the vectors involved, we could obtain as manyas three equations, which could be solved for the desired
unknowns. From this result, we can derive the expression
SF = (vB - vA)dm
dt
where vB - vArepresents the difference between the vectors vB
and vAand where dm/dtis the mass rate of flow of the stream.
SSF Dt
SM Dt
v va
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Consider a system of particles gaining
mass by continually absorbing particles
or losing mass by continually expelling
particles (as in the case of a rocket).
Applying the principle of impulse and
momentum to the system during a time interval Dt, we take careto include particles gained or lost during the time interval. The
action on a system Sof the particles being absorbed by S isequivalent to a thrust
mv
Dm
(Dm) va
u = va - v
m
SS
S F Dt
S
(m + Dm)
(m + Dm)(v + Dv)
P = udm
dt
v va
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P = udm
dt
where dm/dtis the rate at which mass is being absorbed, and u
is the velocity of the particles relative to S. In the case of
particles being expelled by S, the rate dm/dtis negative and P isin a direction opposite to that in which particles are being
expelled.
mv
Dm
(Dm) va
u = va - v
m
SS
S F Dt
S
(m + Dm)
(m + Dm)(v + Dv)
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Chapter 15 KINEMATICS OF RIGID BODIES
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Chapter 15 KINEMATICS OF RIGID BODIES
In rigid body translation, all points of
the body have the same velocity and
the same acceleration at any given
instant.
Considering the rotation of a rigid
body about a fixed axis, the positionof the body is defined by the angle qthat the line BP, drawn from the axis
of rotation to a pointPof the body,
x
yA
r
P
O
A
qf
B
z
v = = rq sin f
forms with a fixed plane. The magnitude of the velocity ofP is
ds
dt
.
where q is the time derivative ofq..
ds .
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v = = rq sin fds
dt
.
The velocity ofP is expressed as
v = = wxrdrdt
where the vector
w = wk= qk.
is directed along the fixed axis of rotation and represents theangular velocityof the body.
x
yA
r
P
O
A
qf
B
z
dr .z
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v = = wxrdrdt
The vectora represents the angular acceleration of the bodyand is directed along the fixed axis of rotation.
Denoting by a the derivative dw/dtofthe angular velocity, we express the
acceleration ofP as
a =a
x r +w
x (w
xr)
w = wk= qk.
differentiating w and recalling that kis constant in magnitudeand direction, we find that
a = ak= wk= qk. ..
x
yA
r
P
O
A
qf
B
y v = wk x r Consider the motion of a representative
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x
y
O
w = wk
v = wkxr
r
Consider the motion of a representative
slab located in a plane perpendicualr to
the axis of rotation of the body. The
angular velocity is perpendicular to the
slab, so the velocity of pointPof the
slab is
x
y
w = wk
v = wkxr
where v is contained in the plane ofthe slab. The acceleration of pointP
can be resolved into tangential and
normal components respectively equal
to
at= akxr at= ra
an= -w2r an = rw2a = ak
at= akxr
O an= -w2r
P
P
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vA vA yk
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A
B
vB
Plane motion = Translation withA + Rotation aboutA
A
B
vA
The most general plane motion of a rigid slab can beconsidered as the sum of a translation and a rotation. The slab
shown can be assumed to translate with pointA, while
simultaneously rotating aboutA. It follows that the velocity of
any pointB of the slab can be expressed asvB = vA + vB/A
where vA is the velocity ofA andvB/Ais the relative velocity ofB
with respect toA.
B
x
vB/A
rB/AA
(fixed)wk
yk
vA vA vB/A
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B
x
vB/A
rB/AA
(fixed)wk
A
B
vB
Plane motion = Translation withA + Rotation aboutA
A
B
vA vA vB
vB = vA + vB/A
Denoting by rB/Athe position ofB relative toA, we note that
vB/A = wkx rB/A vB/A = (rB/A )w = rw
The fundamental equation relating the absolute velocities of
pointsA andB and the relative velocity ofB with respect toA
can be expressed in the form of a vector diagram and used to
solve problems involving the motion of various types of
mechanisms.
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C
A
B
vA
vB
vA
vB
CAnother approach to the solution of
problems involving the velocities of
the points of a rigid slab in plane
motion is based on determination of
the instantaneous center of rotation
Cof the slab.
aAy
wk
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Plane motion = Translation withA + Rotation aboutA
aB = aA + aB/A
The fact that any plane motion of a rigid slab can be consideredthe sum of a translation of the slab with reference to pointA and
a rotation aboutA is used to relate the absolute accelerations of
any two pointsA andB of the slab and the relative acceleration
ofB with respect toA.
A
B
A
aBA
B
aA
aA
A
B
x
(aB/A)n
akwk
(aB/A)t
aB/A
where aB/Aconsists of a normal component(aB/A)n of magnitude
rw2 directed towardA, and a tangential component (aB/A)tof
magnitude ra perpendicular to the line AB.
aAy
wk
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Plane motion = Translation withA + Rotation aboutA
aB = aA + aB/A
The fundamental equation relating the
absolute accelerations of pointsA andB
and the relative acceleration ofB with
respect toA can be expressed in the form of
a vector diagram and used to determine the
accelerations of given points of various
mechanisms.
A
B
A
aBA
B
aA
aA
A
B
x
(aB/A)n
akwk
(aB/A)t
aB/A
(aB/A)n
(aB/A)t
aA
aB aB/A
aAy
wk
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The instantaneous center of rotation C
cannot be used for the determination of
accelerations, since point C, in general,
does not have zero acceleration.
Plane motion = Translation withA + Rotation aboutA
aB = aA + aB/A
A
B
A
aBA
B
aA
aA
A
B
x
(aB/A)n
akwk
(aB/A)t
aB/A
(aB/A)n
(aB/A)t
aA
aB aB/A
Y The rate of change of a vector is
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X
Z
x
y
z
O
ij
k
Q
WA
The rate of change of a vector is
the same with respect to a fixed
frame of reference and with
respect to a frame in translation.The rate of change of a vector
with respect to a rotating frame
of reference is different. The
rate of change of a general
vectorQ with respect a fixed frame OXYZandwith respect to a frame Oxyzrotating with an
angular velocity W is
(Q)OXYZ= (Q)Oxyz+ WxQ. .
The first part represents the rate of change ofQ with respect to
the rotating frame Oxyzand the second part, W x Q, is inducedby the rotation of the frame Oxyz.
Y ( ).
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X
Y
x
y
O
r
W
vP= WxrP P
vP/F= (r)Oxy.
Consider the two-dimensional
analysis of a particleP movingwith respect to a frame F
rotating with an angular
velocity W about a fixed axis.The absolute velocity ofPcan
be expressed as
vP= vP+ vP/F
where vP = absolute velocity of particleP
vP
= velocity of pointPof moving frame Fcoinciding
withP
vP/F = velocity ofPrelative to moving frame F
The same expression forvP is obtained if the frame is in
translation rather than rotation.
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aP = aP+ aP/F + acY ( ).
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P P P/F c
aP = absolute acceleration of
particleP
aP= acceleration of pointPofmoving frame Fcoinciding
withP
aP/F = acceleration ofPrelative to
moving frame F
ac = 2Wx (r)Oxy= 2WxvP/F.
Since W andvP/Fare perpendicular to each other in the case ofplane motion, the Coriolis acceleration has a magnitudeac = 2WvP/F. Its direction is obtained by rotating the vectorvP/Fthrough 90o in the sense of rotation of the moving frame. The
Coriolis acceleration can be used to analyze the motion of
mechanisms which contain parts sliding on each other.
X
Y
x
y
O
r
W
vP= WxrP P
vP/F= (r)Oxy.
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P
r
a w
O
In three dimensions, the most general
displacement of a rigid body with a fixed
point O is equivalent to a rotation of the
body about an axis through O. The
angular velocity w and the instantaneousaxis of rotation of the body at a given
v = = wxrdrdt
Differentiating this expression, the acceleration is
instant can be defined. The velocity of a pointPof the body can
be expressed as
a = axr + wx (wxr)Since the direction ofw changes from instant to instant, theangular acceleration a is, in general, not directed along theinstanteneous axis of rotation.
Y The most general motion of a rigid
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B
rB/A
a w
O
A
X
Y
Z
XZ
rA
body in space is equivalent, at any
given instant, to the sum of a
translation and a rotation. Considering
two particlesA andB of the body
vB = vA + vB/AwherevB/Ais the velocity ofB relative
to a frameAXYZattached toA andof fixed orientation. Denoting by rB/A
vB = vA + wxrB/Awhere w is the angular velocity of the body at the instantconsidered. The acceleration ofB is, by similar reasoning
the position vector ofB relative toA, we write
aB = aA + aB/A aB = aA + axrB/A + wx (wxrB/A)or
C id th th di i lY
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Consider the three-dimensional
motion of a particleP relative to a
frame Oxyzrotating with an angular
velocity W with respect to fixedframe OXYZ. The absolute velocity
vPofPcan be expressed as
vP= vP+ vP/F
where vP = absolute velocity of particleP
vP
= velocity of pointPof moving frame Fcoinciding
withP
vP/F = velocity ofPrelative to moving frame F
X
Z
x
y
z
O
i
j
k
P
WA r
Y
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aP= aP+ aP/F+ ac
where aP = absolute acceleration of particleP
aP= acceleration of pointPof movingframe Fcoinciding withP
aP/F = acceleration ofPrelative to movingframe F
ac
= 2Wx (r)Oxy
= 2WxvP/F
.
The absolute acceleration aPofP
is expressed as
= complementary (Coriolis) acceleration
The magnitude acof the Coriolis acceleration is not equal to
2WvP/F except in the special case when Wand vP/Fareperpendicular to each other.
X
Z
x
y
z
O
i
j
k
P
WA r
Y The equations
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X
Y
Z
x
y
z
O
P
AX
Z
rA
rP
rP/A
aP= aP+ aP/F+ ac
vP= vP+ vP/F
q
and
remain valid when the frame
Axyzmoves in a known, butarbitrary, fashion with
respect to the fixed frame
OXYZ, provided that themotion ofA is included in the terms vPand aPrepresenting the
absolute velocity and acceleration of the coinciding pointP.
Rotating frames of reference are particularly useful in the study
of the three-dimensional motion of rigid bodies.
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Chapter 16 PLANE MOTION OF RIGID BODIES:
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FORCES AND ACCELERATIONS
The relations existingbetween the forces acting
on a rigid body, the shape
and mass of the body,
and the motion producedare studied as the kinetics
of rigid bodies. In general,
our analysis is restricted
to the plane motion of
G
F1
F2
F3
F4HG
ma
G
.
rigid slabs and rigid bodies symmetrical with respect to thereference plane.
The two equations for theH.
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q
motion of a system of
particles apply to the most
general case of the motionof a rigid body. The first
equation defines the
motion of the mass center
G of the body.
G
F1
F2
F3
F4
HG
ma
G
SF = ma
where m is the mass of the body, and a the acceleration ofG.The second is related to the motion of the body relative to a
centroidal frame of reference.
SMG = HG.
HG
.SF = ma
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G
F1
F2
F3
F4
HG
ma
G
SF = ma
SMG = HG.
where HG is the rate of
change of the angular
momentum HG of the
body about its masscenterG.
These equations express that the system of the external forces
is equipollent to the system consisting of the vectormaattached
atG and the couple of momentHG.
.
.
.
FHG.
For the plane motion of
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G
F1
F2
F3
F4G
ma
G
HG= Iw
rigid slabs and rigid
bodies symmetrical with
respect to the referenceplane, the angular
momentum of the body is
expressed as
whereIis the moment of inertia of the body about a centroidal
axis perpendicular to the reference plane and w is the angularvelocity of the body. Differentiating both members of this
equation
HG= Iw = Ia. .
F1F4
For the restricted case
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G
F2
F3
For the restricted case
considered here, the rate
of change of the angular
momentum of the rigidbody can be represented
by a vector of the same
direction as a (i.e.
The plane motion of a rigid body symmetrical with respect to
the reference plane is defined by the three scalar equations
ma
G
Ia
perpendicular to the plane of reference) and of magnitudeIa
.
SFx = max SFy = may SMG =IaThe external forces acting on a rigid body are actuallyequivalentto the effective forces of the various particles forming the body.
This statement is known as dAlemberts principle.
F1F4 dAlemberts principle can
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G
F2
F3
be expressed in the form
of a vector diagram, where
the effective forces arerepresented by a vector
maattached at G and a
coupleIa. In the case of aslab in translation, the
effective forces (partb ofthe figure) reduce to a
(a) (b)
single vectorma ; while in the particular case of a slab in
centroidal rotation, they reduce to the single coupleIa ; in anyother case of plane motion, both the vectorma andIa shouldbe included.
ma
G
Ia
F1F4 Any problem involving the
plane motion of a rigid slab
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G
F2
F3
plane motion of a rigid slab
may be solved by drawing
a free-body-diagram
equation similar to that
shown. Three equations of
of motion can then be
obtained by equating the x
components,y components, and moments about an arbitrary pointA, of theforces and vectors involved.
This method can be used to solve problems involving the
plane motion of several connected rigid bodies.
Some problems, such as noncentroidal rotation of rods and
plates, the rolling motion of spheres and wheels, and the plane
motion ofvarious types of linkages, which move under
constraints, must be supplemented by kinematic analysis.
ma
G
Ia
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Chapter 17 PLANE MOTION OF RIGID BODIES:
ENERGY AND MOMENTUM METHODS
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ENERGY AND MOMENTUM METHODS
The principle of work and energy for a rigid body is expressed in
the form
T1 + U1 2 = T2
where T1 and T2 represent the initial and final values of the
kinetic energy of the rigid body and U1 2 the work of theexternal forces acting on the rigid body.
The work of a force F applied at a pointA is
U1 2 = (Fcos a) dss1s2
whereFis the magnitude of the force, a the angle it forms withthe direction of motion ofA, ands the variable of integration
measuring the distance traveled byA along its path.
The work of a couple of momentM applied to a rigid body during
f
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U1 2 = M dsq1q2
a rotation in q of the rigid body is
The kinetic energy of a rigid body in plane motion is
T= mv 2 + Iw212
12
where v is the velocity of the mass centerG of
the body, w the angular velocity of the body,andI its moment of inertia about an axis
through G perpendicular to the plane of
reference.
G
w v
T = mv 2 + Iw21 1
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T= mv 2 + Iw22 2
The kinetic energy of a rigid body in plane
motion may be separated into two parts:(1) the kinetic energy mv 2associated
with the motion of the mass centerG of the
1
2
1
2body, and (2) the kinetic energy Iw2 associated with the rotation
of the body about G.For a rigid body rotating about a fixed axis through O with an
angular velocity w,
T= IOw21
2
O
w
where IOis the moment of inertia of the body
about the fixed axis.
G
w v
When a rigid body, or a system of rigid bodies, moves under
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the action of conservative forces, the principle of work and
energy may be expressed in the form
which is referred to as theprinciple of conservation of energy.
This principle may be used to solve problems involving
conservative forces such as the force of gravity or the force
exerted by a spring.
T1 + V1 = T2 + V2
The concept of power is extended to a rotating body subjected
to a couple
Power = = =MwdUdt
M dqdt
whereMis the magnitude of the couple and w is the angular
velocity of the body.
Theprinciple of impulse and momentum derived for a system of
particles can be applied to the motion of a rigid body
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(Dm)v
particles can be applied to the motion of a rigid body.
Syst Momenta1+ Syst Ext Imp1 2= Syst Momenta2
P
mv
Iw
For a rigid slab or a rigid body symmetrical with respect to the
reference plane, the system of the momenta of the particles
forming the body is equivalent to a vectormv attached to the
mass centerG of the body and a coupleIw. The vectormv is
associated with translation of the body with G and represents thelinear momentum of the body, while the coupleIw corresponds
to the rotation of
the body about G
and represents theangular momentum
of the body about
an axis through G.
The principle of impulse and momentum can be expressed
hi ll b d i th di ti
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x
y
O
Iw1
mv1
graphically by drawing three diagrams representing
respectively the system of initial momenta of the body, the
impulses of the external forces acting on it, and the system of
the final momenta of the body. Summing and equating
respectively thex components, they components, and the
moments about any given point of the vectors shown in the
figure, we obtain three equations of motion which may be
solved for the desired unknowns.
x
y
O x
y
O
Iw2
mv2Fdt
GG
y y ymv2Fdt
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x
y
O
Iw1
mv1
x
y
O x
y
O
Iw2
In problems dealing with several connected rigid bodies each
body may be considered separately or, if no more than threeunknowns are involved, the principles of impulse and
momentum may be applied to the entire system, considering
the impulses of the external forces only.
When the lines of action of all the external forces acting on a
system of rigid bodies pass through a given point O, the angular
momentum of the system about O is conserved.
G G
The eccentric impactof two rigid
b di i d fi d i t i
n
B
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bodies is defined as an impact in
which the mass centers of the
colliding bodies are not locatedon the line of impact. In such a
situation a relation for the impact
involving the coefficient of
restitution e holds, and the
velocities of points A and Bwhere contact occurs during the
impact should be used.
n
A
vA
vB
B
(a) Before impact
(vB)n - (vA)n = e[(vA)n - (vB)n] n
A
vA
vB
n
(b) After impact
B
n
B
n
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(a) Before impact (b) After impact
(vB)n - (vA)n = e[(vA)n - (vB)n]
where (vA)nand (vB)nare the components along the line of impact
of the velocities ofA andB before impact, and (vA)nand (vB)n
their components after impact. This equation is applicable notonly when the colliding bodies move freely after impact but also
when the bodies are partially constrained in their motion.
n
A
vA
vB
B
n
A
vA
vB
B
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Chapter 18 KINETICS OF RIGID BODIES IN
THREE DIMENSIONS
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THREE DIMENSIONS
SF = ma SMG = HG.
The two fundamental equations for the motion of a system ofparticles
X
Y
ZO
x
y
z G
wHG
provide the foundation for three
dimensional analysis, just as they
do in the case of plane motion of
rigid bodies. The computation of
the angular momentum HG and itsderivative HG , however, are now
considerably more involved.
.
The rectangular components of the angular momentum HG of a
rigid body may be expressed in terms of the components of its
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rigid body may be expressed in terms of the components of its
angular velocity w and of its centroidal moments and productsof inertia:
Hx = +Ixwx -Ixywy -IxzwzHy = -Iyxwx +Iywy -Iyzwz
Hz= -Izxwx -Izywy +IzwzIf principal axes of inertia Gxyz
are used, these relations reduce to
Hx=IxwxHy=IywyHz=IzwzX
Y
ZO
x
y
zG
wHG
In general, the angular momentum
H and the angular velocity do not
Y y w
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HGand the angular velocitywdo nothave the same direction. They will,
however, have the same direction if
w is directed along one of theprincipal axes of inertia of the body.
X
Y
Z O
G
HG
The system of the momenta of theparticles forming a rigid body may be
reduced to the vectormv attached at G
and the couple HG. Once these are
determined, the angular momentum
HO of the body about any given point Omay be obtained by writing
mv
r
HO = rxmv + HG
XZ O
x
zG
HG
In the particular case of a rigid body constrained to rotate
about a fixed point O the components of the angular
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x
y
zO
wHO
about a fixed point O, the components of the angular
momentum HO of the body about O may be obtained
directly from the components of its angular velocity and
from its moments and products of inertia with respect to
axes through O.
Hx = +Ixwx -Ixywy -IxzwzHy = -Iyxwx +Iywy -IyzwzHz= -Izxwx -Izywy +Izwz
Theprinciple of impulse and momentum for a rigid body in three-
dimensional motion is expressed by the same fundamental
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p y
formula used for a rigid body in plane motion.
Syst Momenta1+ Syst Ext Imp1 2= Syst Momenta2
X
Y
Z O
G
HG
mv
r
Hx = +Ixwx -Ixywy -IxzwzHy = -Iyxwx +Iywy -IyzwzHz= -Izxwx -Izywy +Izwz
Hx=Ixwx Hy=Iywy
Hz= Izwz
The initial and final system momenta should be represented
as shown in the figure and computed from
or
The kinetic energy of a rigid body in three-dimensional motion
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gy g y
may be divided into two parts, one associated with the motion
of its mass centerG, and the other with its motion about G.
Using principal axesx, y, z, we write
T= mv 2 + (Ixwx +Iywy +Izwz )2 2 21
2
1
2
where v= velocity of the mass center
w = angular velocitym = mass of rigid body
Ix,Iy,Iz= principal centroidal moments of inertia.
In the case of a rigid body constrained to rotate about a fixed
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In the case of a rigid body constrained to rotate about a fixed
pointO, the kinetic energy may be expressed as
T= (Ixwx +Iywy +Izwz )2 2 21
2
wherex,y, andzaxes are the principal axes of inertia
of the body at O. The equations for kinetic energy makeit possible to extend to the three-dimensional motion of a
rigid body the application of theprinciple of work and
energyand of theprinciple of conservation of energy.
YY
yThe fundamental equations
SF = ma SM H.
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X
ZO
X
Z
G
wHG x
z
SF = ma SMG = HGcan be applied to the motion of a
rigid body in three dimensions. We
first recall that HG represents the
angular momentum of the body
relative to a centroidal frame
GXYZof fixed orientation andthat HG represents the rate of
change ofHG with respect to that frame. As the body rotates, its
moments and products of inertia with respect to GXYZ change
continually. It is therefore more convenient to use a frame Gxyzrotating with the body to resolve w into components and tocompute the moments and products of inertia which are used to
determine HG.
.
SF = ma SMG = HG.
YY
y
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HG represents the rate of change
ofHG with respect to the frameGXYZof fixed orientation,
therefore
.
HG = (HG )Gxyz+W xHG
. .
where HG = angular momentum of the body with respect to the
frame GXYZof fixed orientation
(HG)Gxyz= rate of change ofHG with respect to the rotatingframe Gxyz
W= angular velocity of the rotating frame Gxyz
.
X
ZO
X
Z
G
wHG x
z
SF = ma SMG = HG.
YY
y
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HG = (HG )Gxyz+ W xHG. .
SubstitutingHGabove into S MG,
SMG = (HG )Gxyz+ W xHG.
If the rotating frame is attached to the body, its angular velocity
W is identical to the angular velocity w of the body.Setting W = w , using principal axes, and writing this equationin scalar form, we obtain Eulers equations of moton.
.
X
ZO
X
Z
G
wHG x
z
Y
y
In the case of a rigid body
constrained to rotate about a fixed
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SMO = (HO )Oxyz+W xHO
.
X
Z
wHO xy
z
O
constrained to rotate about a fixed
pointO, an alternative method of
solution may be used, involving
moments of the forces and the rateof change of the angular momentum
about point O.
where SMO = sum of the moments about O of the forces appliedto the rigid body
HO = angular momentum of the body with respect to the
frame OXYZ(HO)Oxyz= rate of change ofHO with respect to the rotating
frame Oxyz
W= angular velocity of the rotating frame Oxyz
.
When the motion ofgyroscopes and
h i i l b di
Z q
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otheraxisymmetrical bodies are
considered, the Eulerian angles f, q,
and yare introduced to define theposition of a gyroscope. The time
derivatives of these angles
represent, respectively, the rates of
precession, nutation, and spin of the
gyroscope. The angular velocity wis expressed in terms of these
derivatives as
w = -fsin qi + qj + (y+ fcos q)k. . . .
DB
BD
A
A
C
C
YO
fy
Z q
w = -fsin qi + qj + (y+ fcos q)k. . . .
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DB
BD
A
A
C
C
YO
fy
B
B
A
A
C
C
Z
O
q
x
y
z
fK.
yk.
qj.
The unit vectors are
associated with theframe Oxyzattached
to the inner gimbal of
the gyroscope (figure
to the right)androtate, therfore, with
the angular velocity
W = -fsin qi + qj + fcos qk. . .
Z q zfK.Denoting byI the
moment of inertia of
A
Z q
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B
B
A
A
C
C
O
x
y
yk.
qj.
moment of inertia of
the gyroscope with
respect to its spin
axiszand byIits
moment of inertia
with respect to a
transverse axis
through O, we write
HO = -Ifsin qi +Iqj +I(y+ fcos q)k. . . .
W = -fsin qi + qj + fcosqk. . .
SMO = (HO )Oxyz+ W xHO.Substituting forH
O and into
leads to the differential equations defining the motion of the
gyroscope.
DB
B
D
A
A
C
C
YO
fy
Z q z In the particular case of the steady
precession of a gyroscope, the angle
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B
BO
x
y
fK. yk
.
SMO
p gy p , g
q, the rate of precession f, and the
rate of spin y remain constant. Such
motion is possible only if the moments
of the external forces about O satisfy
the relation
.
.
SMO = (Iwz-Ifcos q)fsin qj. .
i.e., if the external forces reduce to a couple of moment equal to
the right-hand member of the equation above and applied about
an axis perpendicular to the precession axis and to the spinaxis.
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Chapter 19 MECHANICAL VIBRATIONS
C f f
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Consider the free vibration of a
particle, i.e., the motion of a particleP
subjected to a restoring forceproportional to the displacement of the
particle - such as the force exerted by
a spring. If the displacementx of the
particlePis measured from itsequilibrium position O, the resultant F
of the forces acting onP(including its
weight) has a magnitude kx and is
directed toward O. Applying Newtons
second law (F= ma) with a =x, thedifferential equation of motion is
O
+xm
-xm
PEquilibrium
+ mx + kx = 0..
..
x
-x
mx + kx = 02
..
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O
+xm
-xm
PEquilibrium
+
setting wn2 = k/m
x + wn2x = 0..
The motion defined by this expression
is called simple harmonic motion.
x =xm sin (wnt+ f)
The solution of this equation, whichrepresents the displacement of the
particle P is expressed as
wherexm= amplitude of the vibration
wn = k/m = naturalcircularfrequency
f = phase angle
x
-x
x + wn2x = 0..
i ( t + f)
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O
+xm
-xm
PEquilibrium
+
x =xm sin (wnt+ f)
Theperiod of the vibration (i.e., thetime required for a full cycle) and its
frequency(i.e., the number of cycles
per second) are expressed as
Period = tn = 2pwn
Frequency =fn = =wn2p
1
tnThe velocity and acceleration of the particle are obtained bydifferentiating x, and their maximum values are
vm =xmwn am =xmwn2
x
The oscillatory motion of the
particleP may be represented
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O
P
p y p
by the projection on thex axis of
the motion of a point Q
describing an auxiliary circle of
radiusxm with the constant
angular velocity wn. Theinstantaneous values of the
velocity and acceleration ofPmay then be obtained by
projecting on thex axis the
vectors vm and am representing,
respectively, the velocity andacceleration ofQ.
xm
am=xmwn2
f
wnt
vm=xmwnwnt+ f
x QO
Q
v
a
x
While the motion of a simple pendulum is not truely a simple
harmonic motion the form las gi en abo e ma be sed ith
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harmonic motion, the formulas given above may be used with
wn2 =g/lto calculate the period and frequency of the small
oscillations of a simple pendulum.
The free vibrations of a rigid bodymay be analyzed by
choosing an appropriate variable, such as a distancex or an
angle q, to define the position of the body, drawing a diagramexpressing the equivalence of the external and effective forces,
and writing an equation relating the selected variable and its
second derivative. If the equation obtained is of the form
x + wn2x = 0 or q+ wn2q= 0....
the vibration considered is a simple harmonic motion and its
period and frequency may be obtained.
Theprinciple of conservation of energymay be used as an
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alternative method for the determination of the period and
frequency of the simple harmonic motion of a particle or rigid
body. Choosing an appropriate variable, such as q, to definethe position of the system, we express that the total energy
of the system is conserved, T1 + V1 = T2 + V2, between the
position of maximum displacement (q1 = qm) and the position
of maximum velocity (q2 = qm). If the motion considered issimple harmonic, the two members of the equation obtained
consist of homogeneous quadratic expressions in qm and qm ,
respectively. Substituting qm = qmwn in this equation, we mayfactor out qm and solve for the circular frequencywn.
2
..
. .
The forced vibration of a mechanical system
occurs when the system is subjected to a
i di f h it i l ti ll
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x
Equilibrium
P =Pm sin wft
wft= 0
dmdm sin wft
wft
xEquilibrium
periodic force or when it is elastically
connected to a support which has an
alternating motion. The differentialequation describing
each system is
presented
below.
mx + kx =Pm sin wf t
mx + kx = kdm sin wf t
..
..
dmdm sin wft
mx + kx =Pm sin wft..
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x
Equilibrium
P =Pm sin wft
wft= 0wft
xEquilibrium
mx + kx = kdm sin wft..
The general solution
of these equations is
obtained by adding
a particular solution
of the form
xpart =xm sin wft
to the general solution of the correspondinghomogeneous equation. The particular solution represents the
steady-state vibration of the system, while the solution of the
homogeneous equation represents a transient free vibration
which may generally be neglected.
mx + kx =Pm sin wft..
mx + kx = kdm sin wft..
x = x sin w t
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xEquilibrium
P =Pm sin wft
wft= 0
dmdm sin wft
wft
xEquilibrium
xpart =xm sin wft
Dividing the amplitudexm of the steady-
state vibration byPm/kin the case of a
periodic force, or bydmin the case of anoscillating support, the magnification factor
of the vibration is defined by
Magnification factor = =xmPm/k
xmdm
1
1 - (wf /wn )2=
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The equation of motion describing the damped free vibrations
of a system with viscous dampingis
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y p g
mx + cx + kx = 0.. .
where c is a constant called the coeficient of viscous damping.
Defining the critical damping coefficientcc as
cc = 2m = 2mwnk
m
wherewnis the natural frequency of the system in the absenceof damping, we distinguish three different cases of damping,
namely, (1) heavy damping, when c > cc, (2) criticaldamping,when c = cc, (3) light damping, when c < cc. In the first two cases,
the system when disturbed tends to regain its equilibrium
position without oscillation. In the third case, the motion is
vibratory with diminishing amplitude.
The damped forced vibrations of a mechanical system occurs
when a system with viscous damping is subjected to a periodic
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force P of magnitudeP=Pm sin wftor when it is elasticallyconnected to a support with an alternating motion d= d
m
sin wf
t.
In the first case the motion is defined by the differential equation
mx + cx + kx =Pm sin wft.. .
mx + cx + kx = kdm sin wft
.. .
In the second case the motion is defined by the differential
equation
The steady-state vibration of the system is represented by a
particular solution of mx + cx + kx =Pm sin wftof the form.. .
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f
xpart =xm sin (wft - f)
Dividing the amplitudexm of the steady-state vibration byPm/kin
the case of a periodic force, or by dm in the case of an oscillatingsupport, the expression for the magnification factor is
xm
Pm/k= =
xm
dm
1
[1 - (wf/ wn)2]2 + [2(c/cc)(wf/ wn)]2
where wn = k/m = natural circular frequency of undampedsystem
cc = 2m wn = critical damping coefficientc/cc = damping factor
x x 1
xpart =xm sin (wft - f)
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In addition, the phase differencej between the impressed force
or support movement and the resulting steady-state vibration of
the damped system is defined by the relationship
tan j=2(c/cc) (wf/ wn)
1 - (wf/ wn)2
The vibrations of mechanical systems and the oscillations of
electrical circuits are defined by the same differential equations.
xm
Pm/k= =
xm
dm
1
[1 - (wf
/ wn
)2]2 + [2(c/cc
)(wf
/ wn
)]2
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