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7/29/2019 Eng. Dynamics as Easy as 123
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ENT142ENGINEERING DYNAMICS
ZOL BAHRI RAZALI
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HOW to understand Engineering DYNAMICS
Friendly notes to understand as simple as 123
Zol Bahri Razali
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ABSTRACT
Is an Engineering Mechanics : DYNAMICS is a tough subject?
Obviously most of engineering students at college and university
expect that the subject is tough and very difficult to understand. Why?
Maybe they have experienced with a pre-requisite subject : STATICS.
The DYNAMICS is a practical subject and an expansion of theoretical
knowledge of Statics. If really students understand and can imagine
the Free Body Diagram (FBD) in Statics, they are easy to imagine the
movement of the particle in FBD in DYNAMICS. Therefore they will
understand the concept of movement in this subject, and understand
what is DYNAMICS.
This book will guide students to understand the concept of
DYNAMICS through a method of 123, as simple as to understand
123. For engineering students at college and university, this simple
book is expected to give them to understand basic theoretical
knowledge on DYNAMICS especially in Free Body Diagram (FBD).
For individual who work at higher level of engineering such as
Engineers, Designers or Technicians, this book might be not suitable
for references because the contents is not in depth and breadth, i.etoo simple and very basic theoretical knowledge.
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ACKNOWLEDGEMENT
Main references of this note are from the references:
1. Hibbeler, R.C. 2002. Engineering Mechanics : Dynamics. 2ed.
Prentice Halls, Pearson Education Asia Pte. Ltd., Singapore.
2. Hibbeler, R.C. 2001. Engineering Mechanics : Statics. 2ed.
Prentice Halls, Pearson Education Asia Pte. Ltd., Singapore.
3. Hibbeler, R.C. 2010. Engineering Mechanics : Dynamics. Twelfth
Edition in SI Unit. Prentice Halls, Pearson Education South Asia
Pte. Ltd., Singapore..
4. Meriam, J.L. and Kraige, L.G. 2001. Engineering Mechanics:
Dynamics, John Willey & Sons, Inc.
5. Beer, F.P., Johnston, E.R. and Clausen, W.E. 2004. Vector
Mechanics for Engineers: Dynamics, Mc Graw Hill.
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CHAPTER 12
INTRODUCTION TO DYNAMICS
Chapter ObjectivesTo introduce the concepts of position, displacement, velocity and accelerationTo study particle motion along a straight line and represent this motion graphicallyTo investigate particle motion along a curved path using different coordinate
systemsTo present an analysis of dependent motion of two particlesTo examine the principles of relative motion of two particles using translating axes
12.1 Introductions
Mechanics branch of the physical science that is concerned with the state of
rest or motion of bodies subjected to the action of the forces
Mechanics of rigid body - divided into statics and dynamics
Statics - concerned with the equilibrium of the body that is either at the rest or
moves with constant velocity
Dynamics - concerned with the accelerated motion of a body. Presented in 2 parts:
a) Kinematicsgeometric aspect of motion
b) Kineticsanalysis of the force causing the motion
diff diff
BASICALLY: s v a intg intg
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12.2 Rectilinear Kinematics: Continuous Motion
Rectilinear Kinematicsat any given instant, the particles position, velocity and
acceleration.
Positionthe straight line path of a particle. From the origin (o), position vector r specifythe location of the particle (p).
Convenient (r) represent by (s)
Displacementthe change in its position
Eg : If the particle moves from P to P, the displacement is r = r- r
s = s s
s is positive particles final position is to the right of its initial position, ie :s>s.
Displacement of a particlevector quantity Distance traveled is a positive vector.
r
Os
r
r r
s s
s
Displacement
P P
Os
s
Position
P
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{Velocity}
If the particle moves through a displacement r, from P-P during the time
interval t, the average velocity
Vavg = r , V = dr : instantaneous velocityt dt
V as an algebraic scalar, V = ds
dt
t or dt always positive:1. particle moving the right, velocity is positive2. particle moving to the left velocity is negative.
The magnitude of the velocity is known as the speed.( units : m/s )
vavg = stt
v
P P
O s
s
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{ Acceleration }
Provided the velocity of the particle is known at two point P P, the average
acceleration
aavg =tv
v difference in the velocity during the time interval v = v - v
Acceleration: a =dt
dv
a
acceleration
a =2
2
dt
sd
adeceleration
decelerationwhen the particle is slowing down- speed decreasing
- vvv 1 is negative
acceleration is zerowhen velocity is constant.- ovvv
( unit = m/s 2 )
a ds = v dv a = dtdv
& v = dt
ds
a
P P
O s
v v
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constant accelerationeach of three kinematics equations a c =dt
dv
a = ac
v =
dt
ds, a
cds = vdv
maybe integrated to obtained formula
that related a,v,s,t.
The three formulas of constant acceleration :
1) Velocity as a function of time v
vo
dv t
o
ca dt
v = vo
+ a c t
2) Position as a function of time
s
o
t
ocvds ( + a c t) dt tavdtdsv co
s = s o + v o t + 21 a c t
2
3) Velocity as a function of position
v.dv = a c .ds o
v
s
sc
o o
dsavdv
=> v 2 = v 20
+ 2a c ( s-s o )
This formula only useful when the acceleration is constant and when t = o ,
s = s o , v = v o
e.g.a body fall freely toward the earth.
+
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See Example:
- 12.1
- 12.2
- 12.3
- 12.4
- 12.5
Exercise : 12.1
- 12.2
- 12.3
- 12.4
- 12.5
0194579207
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12.3 Rectilinear Kinematics: Erratic Motion
When particles motion during a time is erratic, may best be describes graphically using aseries of curves.
Using the kinematics equations:
dt
dva v
dt
ds dvvdsa ..
a) Given s-t graph, construct the v-t equations
s
t
b) tv graph ta graph
v = ,dtds
By experimentally, if the position can
be determined during the time of
period, graph s-t can be plotted.
By v = ,dtds the graph v-t can be
plotted.
( slope of s-t graph = velocity ).
dtdva
dtdva
(slope of tv graph = acceleration)
See example 12.6, page 19
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c) ta graph , tv graph
using dtdsa ,
adtv
( change in velocity = area
under ta )
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d) tv graph ts graph
dtdsv vdts
(displacement = area under tv graph)
See example 12.7, page 21
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e) sa graph sv graph.
dvvdsa ..
between the limits ovv at oss
2
1 ( )20
2
1vv =
1
.s
so
dsa
1vv at
1ss = area under sa graph
f) sv graph sa graph
dvvdsa ..
)(ds
dvva
acceleration = velocity x slope of sv graph.
See Example 12.8.
Exercise:
- 12.42- 12.43- 12.44- 12.45- 12.46
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12.4 General Curvilinear Motion
- curvilinear motion occurs when the particle moves along a curved path.
- position- considered a particle located
at point p on a space curve defined bythe path function s.
position vector r = r ( t ) magnitude and direction change as the
particle moves along the curve.
- displacement- during small line t, theparticle moves a distance s along thecurves.
r = r + r
the displacement r represent thechange in the particles position.
r = r - r
- velocityduring the time t , the averagevelocity.
V avg =t
r
, V =
dtdr
dr will be tangent to the curve at p, thedirection of V is also tangent to the
curve.
Path
P
r s
O
s
Position
P
s
r Pr
r
O
s
Displacement
v
P
r
O
s
Velocity
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The magnitude ofv, called speed. V =dt
ds.
V
t
vavg
, where v = v1 - v
Instant a new acceleration, t 0
dt
dva
2
2
dt
rda
Velocity vector is always directedtangent to the path.
a tangent to the hodograph, not tangentto the path of motion.
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12.5 Curvilinear Motion: Rectangular Components
Displacement
r = x i + y j + zk
magnitude of r always positive
r = (x 2 + y 2 + z 2 )
unit vector ur = (1/r)r
Velocity
v = vx
i v y j vZ k
v =dt
dr=
dt
d( x i ) +
dtd ( y j ) +
dtd ( z k)
v =dt
dr= v
xi + v
yj + v
zk,
where : v x =.
x
v y =.
y
v z =.
z
The velocity has a magnitude defined as the positive value of
v = (v x2 + v y
2 + v z2)
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Acceleration :
a =
dt
dv= a x i + a y j + a z k
where : a x =.
xv =..
x
a y =.
yv =..
y
a z =.
zv =..
z
The acceleration has a magnitude defined by the positive value of
a = (a x2 + a y
2 + a z2 )
See Example 12.9 and 12.10.
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12.6 Motion of a Projectile
The free-flight motion of a projectilestudied in terms of its rectangular
components. The projectiles acceleration always acts in the vertical direction.
Projectile launched at point ( ox , oy ) , initial velocity is V o , having two components
( Vo
)x and ( V o )y . The projectile has a constant downward acceleration,
a c = g = 9.812sm .
Horizontal motion : -
Since a x = 0 ; v = vo + ac t ; vx = ( vo ) x
x = xo+ vo t + 21
at2
; x = xo + ( vo ) x t
v 2 = vo 2 + 2ac ( s-so ) ; vx = ( vo ) x
First and last equation indicated that the horizontal component of velocity always remains
constant during the motion.
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Vertical motion : -
Since ay = -g. + v = vo + ac t ; vy = ( vo )ygt.
y = yo + vo t + 21 ac t
2 ; y = yo + (vo) yt - 21 g
v 2 = vo + 2 ac ( y- y2 ) ; vy
2 = (vo )2 -2g (y)
Only two of the above three equations are independent of one another.
Problems involving the motion of projectile can have at most three unknowns
since only three independent equations can be written.
- one equations in the horizontal direction.- two equations in the vertical direction.
Once vx and vy are obtained, the resultant velocity v which is always tangent to thepath.
See Example:
- 12.11
- 12.12
- 12.13
Exercise:
- 12.71
- 12.72
- 12.73
- 12.74
- 12.75
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12.7 Curvilinear motion: Normal and tangential components
When the path along which a particle is moving is known, it is convenient todescribe the motion using n and t components (normal and tangent) to the path, and at the
instant considered here their origin located at the particle.
Planer motion :-
( at instant considered )o - center of curvature.
s - radius of curvature.
taxis - tangent to the curve at P.
n-axis - perpendicular to the t- axis,directed from P towards the center of
curvature.
Positive direction , will be designatedby the unit vector, u n ( normal ) and u t
( tangent ).
Velocity :-
Since the particle moving , s is a function of time. The particle velocity v has a
direction that is always tangent to the path, and the magnitude that is determined bytaking the time derivative of the path function s = s(t) .
v =dt
ds
v = vu t
where v =.
s
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Acceleration :-
The acceleration of the particle is the time rate of the change of the velocity.
a =.
v =.
v u t + v.
u t
by formulation ,
.
u t =.
o u n =s
s.
u n =s
vu n
substitute to the above equation
a = a t u t + a n u n
where a t =.
v
or a t ds = v.dv
and a n =s
v2
magnitude of acceleration is the
positive value of a = 22 nt aa
Two special cases of motion :
1) The particle moves along a straight line , s .
oan , taa =.
v
The tangential components of acceleration represents the time rate of change in the
magnitude of the velocity.
2) The particle moves along a curve with a constant speed then
.,2
.
svaaova nt
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12.8 Relativemotion analysis of two particles using Translating Axes
Position :
The axes of this frame are only persuitted to
translate relative to the fixed frame . The
relative position of Bwith respect to A
is designate by a relative position vector
r AB .
rB
= rA
+ rAB
Velocity : An equation that related the velocities of the particle can be determined by
taking the time derivative.
vB
= vA
+ v AB
where v B =dt
drB
v A =dt
drA
v AB =dt
dr AB
v B and v A - refer to absolute velocities
-observed from the fixed frame.
v AB - relative velocity
-observed from the translating frame.
vB/AvB
vA
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Acceleration :
ABAB aaa
ABa is the acceleration ofB as seen by the observer located at A and translating with
the x,y,z reference frame.
aB/A
aA aB
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CHAPTER 13
KINEMATICS OF A PARTICLE
(FORCE & ACCELERATION)
Chapter ObjectivesTostate Newtons Law of Motion and Gravitational Attraction and to define mass
and weightTo analyze the accelerate motion of a particle using the equation of motion with
different coordinate systems.To investigate central force motion and apply it to problems in space mechanics
13.1 Newtons Law of Motion
Galileo (1590) - experiment to study the motion of pendulum and falling bodies.
- the effects of forces acting on bodies in motion.
Isaac Newton (1687)
o 1st
Law - A particle originally at rest, or moving in the straight linewith a constant velocity, will remain in this state provided
the particle is not subjected to an unbalanced force.
o 2nd LawA particle acted upon by an unbalanced force F,experiences an acceleration a that has the same direction as the
force and a magnitude that is directly proportional to the force.
o 3rd LawThe mutual forces of action and reaction between twoparticles are equal, opposite and collinear.
Note : - The unbalanced force acting on the particle is proportional to the time rate of
the change of the particles linear momentum.
Newtons 2nd law of motion relates the accelerated motion of a particle to the force
that act on it.
F = ma where m : mass of the particle.
( equation of motion )
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Since m is constant , we can also write F =dt
mvd )(, where mv is the particles
linear momentum.
Albert Einstein ( 1905 )developed the theory of relativity and place limitation on
on the use of Newtons second law.
~ Newtons Law of Gravitational Attraction ~
F = Gr
mm 21 where :
F force of attraction between two particles
G - universal constant of gravitation
G = 66.73 + 10
12
2
3
kgsm
m 1 ,m 2 - mass of each of the two particles
r - distance between centre.
~ Mass and weight ~
Property of matter by which we can compare the response of one body with that of
another.
w = m.g , wweight
by comparison F = ma , we term g the acceleration due to gravity.
m (kg)
a = g (m/s2)
W= m.g (N)
SI system
[W= m.g (N)]
[g = 9.81 2s
m ]
Weight w = kg 2s
m
= N
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13.2 The equation of motion
The resultant force FR = FF = am.
The resultant of these forces produce the vector am. , its magnitude and direction can be
represented on the kinetic diagram.
F2 F2
FR = FP ma
F1 PF1
a
F.B.D. Kinetic diagram
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13.3 Equation of motion for a system of particles
F i + F i = ii am ( F = ma )
where F i - resultant ext. force.
fi - resultant int. force.
ama = ii am F = ... gam
If r a is position vector which locates at the center of mass G.
SpringIf the particle is connected to an elastic spring.
F s = ks
s = oll
where : l deformed length
ol = unreformed length.
KinematicsIf the velocity are position of the particle is to be found, apply the
kinematics equation.
acceleration is a function of time.
dtdva ,
dtdsv
acceleration is a function of displacement : integrate ... dvvdsa
acceleration is constant , to determined the velocity or position of the particle.
- v = vo + act
- s = so + vot + 21 a ct
2
- v 2 = vo
2 + 2 ca ( s-s o )
In all cases, make sure the positive inertial coordinates directions.
miai
Kinetic diagram
Fi
fi
F.B.D.
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13.4 Equation of motion : rectangular coordinates
When the particle moving relative to a inertial x, y, z frame, the force actingmaybe expressed in term of their i, j, k. components.
F = am. F ix + F yj + F z k= )( kajaiam zyx
F x = xam. F
y= yam ..
F z = zam.
Procedure for analysis.
1) FBD :- Draw the FBD, it provides a graphical representation that accounts for all
the forces (F) which act on the particle.
2) Equation of motion :- use scalar or Cartesian vector analysis (3-D)for the solution.
a) Friction :
If the particle contacts the rough surface, it maybe necessary to use the frictional
Equation.
F f = .Nk where :
[ Ff
always act on the FBD such k
- coefficient of kinetic friction.
that it opposes the motion of the F f - magnitude of the friction
particle relative to the surface it N - Normal forces.
contacts ]
See Example:
- 13.1
- 13.2
- 13.3
- 13.4
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Exercise : 13.1
- 13.2
- 13.3
- 13.4
- 13.5
13.5 Equation of motion : Normal and tangential coordinates
When the particle moves over a curved path, the equation of motion for the
particle maybe written in the tangential, normal and binormal directions.
F = ma
F tu t + F n u n + F b u b = nt amam ..
where Ft,F
n, F
b- force components acting on the particle in the tangential, normal and
binormal.
There is no motion of the particle in the binormal direction, since the particle is
constrained to move along the path.
F tt am. dt
dvat ( time rate of change in the
magnitude of velocity )
F nn am.
F .ob
F t acts in the direction of motion, the particles speed will be increase. Opposite direction, the speed will slow down.
g
van
2
( time rate of change in the velocity direction )
vector always acts in the positiven direction ( toward the paths centre of
curvature), F n which cause na , also acts in this direction.
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Procedure for Analysis
The motion of a particle along a known curved path, normal and tangential coordinates
should be considered.
FBDDraw the FBD for the particle
Equation of motionas mention
Kinematics g
vands
dvvat
dtdvat
2
,.,
If the path is defined as y = f (x), the radius of curvature at thepoint where the particle located can be obtained:
= [1 + (dx
dy ) 2 ] 3/2 /d2y / dx2
See Example:
- 13.6
- 13.7
- 13.8
- 13.9
Exercise : 13.48
- 13.49
- 13.50
- 13.51
- 13.52
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CHAPTER 14
KINEMATICS OF A PARTICLE
(WORK AND ENERGY)
Chapter ObjectivesTo develop the principle of work and energy and apply it solve problems that
involve force, velocity and displacementTo study problems that involves power and efficiencyTo introduce the concept of a conservative force and apply the theorem of
conservation of energy to solve kinetics problems
14.1 The work of a force
A force F does work when the particle undergoes a displacement in the directionof the force.
Eg : particles moves along the paths, from
position r to r, dr = r r, where dr = ds. If
the angle , the work dU is done by F is ascalar quantity
dU = F.ds cos
Definition of the dot product dU = F.dr
dU = F.ds cos = F.dr
dU = 0 when F perpendiculars.
dr = ds cos
Work of a variable force.
If the particle undergoes finite displacement
along its path, s1
to s2
, work is determined
by integration.
u21 =
2
1
.s
rdrF
2
1
coss
sF .
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If ( F cos vs s ) is plotted, workingcomponents as the area under the curve,
from position s 1 to s 2 .
Work of constant force moving along a straight line
If the Fc
has a constant magnitude and acts at a constant angle from its straight line
path, so work done F c is
u21 = F c cos
2
1
s
sd s
= F c cos ( s 2 - s 1 ).
Work F c represents the area of the rectangle.
Work of weight
The particle moves along the path s, from s1
to s2
.At the immediate point, the
displacement dr = dx i + dy j + dz k.
Since w = -w j
u21 = drf.
= 2
1
)).((r
rdzkdyjdxiwj
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= 2
1
11 12
)( yywwdy
u21 = -w (y 2 - y1 )
This, work done is equal to the magnitude of
the particles weight times its vertical displacement.
Work of spring
The magnitude of force developed in a linear elastic spring when the spring is displaced a
distance s from its unstreched position is Fs = ks , where k is the spring stiffness. If the
spring is elongated or compressed from a position s1 to a further position s2 , the work
done on the spring by Fs is positive, since in each case, the force and displacement are inthe same direction.
U1-2 = 2
1
s
sFsds =
2
1
s
sksds
= .2
12
1 21
2
2ksks
When the particle is attach to a spring, then the force Fs exerted on the particle isopposite to that exerted on the spring.
The force will do negative work on the particle when the particle is moving so as afurther elongate ( or compress ) the spring.
U1-2 = - ( ).21
21 2
1
2
2ksks
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If both same directionpositive work .opposite directionnegative work.
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14.2 Principle work and energy
If the particle has a mass, m and is subjected to a system of external forces, represented
by the resultant FR = Ft = ma t.
Applying the kinematics equation:
dsdvvat .
2
1
s
stdsF =
2
1
.v
vmvdv
=2
1
2
2 21
21 mvmv
U21 = the sum of the work done by all the forces acting as the particle
moves from point 1 -2.
T = 22
1 mv T = particle final kinetic energy
mv2
1 2 = particle initial kinetic energy
T1 + U1-2 = T2
Note : i ) Ft = tam. , to obtain ta , integrate ta = .. dsdvv
ii ) F n = nam. cannot be used, since these force do no work on the particle onthe forces directed normal to the path.
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Procedure for analysis
The principle of work and energy is used to solve the kinetic problems that involve
velocity , force and displacement.
FBD : Draw FBD in order to account for all the forces.
Principle of w E :
T1 + U1-2 = T2
Kinetic energy at the initial / final point always positive T = 22
1 mv
A forces does work when it moves through its displacement in the direction of the
force.( +ve same direction )
Force that are functions of displacement must be integrate to obtain the work.
The work of a weightweight magnitude and the vertical displacement.
Uw = wy ( + ).
The work of spring , s = 21 ks
2.
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14.3 Principle of Work and Energy for a system of particles
thi particles
m i massF i resultant external forcefi resultant internal force
Which applies in the tangential direction, the principle of work and energy:
2
1
2
1
2
2
2
21)()(
21 i
i
i
i
s
s
s
siittii vmdsfidsFivm
For all of the thi particles.
2
1
2
1
2
2
2
1 21)()(
21 i
i
i
i
s
s
i
siittii vmdsfidsFivm
T1+ U 21 = T 2 systems finalkinetic energy
work done by all the
external or internal forces.
systems initial kinetic energy.
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Work of Friction Caused by Sliding.
The cases where a body is sliding over thesurface on another body in the presence of
friction.
Applied force P just balance the resultant
friction force ..Nk
22
21
21 mvNPsmv sk
is satisfied when P = k N
The sliding motion will generate heat, a form of energy which seems not to be accounted
for in the W&E equation.
See Example:
- 14.2
- 14.3
- 14.4
- 14.5
- 14.6
Quiz:- 14.114.41 (each student has to answer one question)
v v
P P
s
W
P
F = kN
N
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14.4 Power and Efficiency
Defined as the amount of work performance per unit of time.
P =dtdu where du = F.dr
=dt
drF.
= F.dt
dror P = F.V.
Power is a scalar, where in the formulation; V represents the velocity of the point which
is acted upon by the force, F.
I W a H = I sJ = I N. sm = I w
Note:
The term power provides the useful basis for determining the type of motor ormachine which is required to do a certain amount of work in a given time. For example,
two pumps may each be able to empty a reservoir if given enough time, however thepump having the larger power will be complete the job sooner.
The mechanical efficiency of a machine is defined as the ratioof the output of useful power produce by the machine to the
input of power supplied to the machine.
inputpower
outputpowerE
If the energy applied,inputenergy
outputenergyE
Since machines consist of a series of moving parts, frictional forces will always bedeveloped within the machine, and as result, extra energy or power is needed to
overcome these forces.
The efficiency of a machine is always less than 1
Power
Efficiency
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Procedure for Analysis
o Determine the external force F
o If accelerating , amF . o One F and V have been found, P = F.V = Fr cos
o In some problem F per unit time, P = dtdu
See Example:
- 14.7
- 14.8
Exercise
- 14.42
- 14.43
- 14.44
- 14.45
- 14.46
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14.5 Conservative forces and potential energy
Conservative force
When the work done by a force in moving a particle from one point to another isindependent of the path followed by the particle, the force is called a conservativeforce.
e.g. : - weight of the particle
: - the force of an elastic spring
weightdepends on the particles vertical displacement
spring depends only on the extension/compression.
in contrastforce of friction exerted on a moving objectsdepends on the
path/neoconservative.
Potential Energy
Energy - capacity for doing works.- from the motion of particlekinetic energy.
- from the position of particle ( fixed datum / reference ) - potential
energy
(potential energy due to gravity ( weight ) and elastic spring is important.)
Gravitational potential energy: Vg = Wy
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Elastic potential energy: V e =2
21
sk
[ V e always + ve, the spring has the
capacity for always doing positivework ] when the spring back tooutstretch position.
Potential Function
If a particle is subjected to both gravitational and elastic forces, - potential function.
V = V g + Ve
U21 = V 1 - V 2 work done by a conservative force.
E.g. : potential function for a particle ofweight W, suspended from a spring can be
expressed in term of its position, s ,
measured from a datum.
V = Vg + Ve
= - W(s) + 22
1 ks
If the particle moves from s1 to lower s2,
U )2
1()2
1(2
22
2
112121ksWsksWsVV
= )2
12
1()(2
1
2
212ksksssW
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14.6 Conservation of Energy
When the particle acted upon by a system of both conservative and neoconservative,
work done by conservative position.
222111 )( VTnonconsUVT
If noncons is zero =>2211
VTVT
Note : conservative force (not follow the path )weight / springneoconservative force (follow exactly the path )friction.
2211VTVT conservation of mechanical energy, conservation energy.
Note : during the motion, sum of potential and kinetic energies remains constant.
For this occur, kinetic energy must be transformed into potential energy andvice versa.
E.g. : The ball of weight w is dropped from a height, h above the ground (datum).
at initial position, mechanical energy
E = T1
+ V1
= )(
2
1 21 ge vvmv
= whmv 212
1
= 0 + wh
= .wh
When the ball has fallen a distance ,2
h its speed can be determined by using:
)(22
0
2
oc yyavv Therefore:
=)(2
oc yya
= )
2(2 hg 22 TVE
v = )2
(2 hg = 2)(2
1)2
( ghmhw
= gh2 = 2)(2
1)2
( ghg
whw
= .wh
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When the ball strikes the ground, P.E. (0) and
)(222
oco yyavv Total energy:
= ).(.2 hg
= gh2 33 TVE
= 2)2(2
10 gh
g
w
= .wh
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CHAPTER 15
KINETICS OF A PARTICLE:
IMPULSE & MOMENTUM
Chapter ObjectivesTo develop the principle of linear impulse and momentum for a particleTo study the conservation of linear momentum for particlesTo analyze the mechanics of impactTo introduce the concept of angular impulse and momentumTo solved problems involving steady fluid streams and propulsion with variable
mass
15.1 Principle of Linear Impulse & Momentum
Equation of motion for a particle of mass m:
dt
dvmmaF
Rearranging:
2
1
1
2
t
t
v
v dvmdtF
or
2
1
12
t
t
mvmvdtF Eq. 1
Linear Momentum
principle of linear impulse &
momentum(time integration of the
equation of motion)
linear particlesimpulse linear
momentum
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Vector of L = mv in Eq. 1 is referred to as the particles linear momentum Magnitude of mv units of mass-velocity
Integral I = F dt in Eq. 1 is referred to as the linear impulse Measures the effect of a force during the time the force acts. Magnitude: force-time If forceexpressed as a function of time, impulse = direct evaluation of the integral If forceconstant direction during the time period t1 to t2, impulse = area under the
curve of force vs. time:
2
1
t
t
dtFI
If forceconstant in magnitude & direction:
)( 122
1
ttFdtFI c
t
t
c shaded rectangular area
21
2
1
mvdtFmv
t
t
Initial momentum of the particle at t1 plus the vector sumof all the impulses applied to the particle during the time
interval t1 to t2 is equivalent to the final momentum of the
particle at t2.
Linear Impulse
Principle of Linear Impulse & Momentum
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The scalar x, y, z components of the previous equation are:
21
21
21
)()(
)()(
)()(
2
1
2
1
2
1
z
t
t
zz
y
t
t
yy
x
t
t
xx
vmdtFvm
vmdtFvm
vmdtFvm
To solve a linear impulse and momentum:
a) FBD - establish the x, y, z inertial frame of reference
- draw the particles FBD account for all the forces that produce impulses on
the particle
- establishdirection & sense of the particles initial & final velocities
b) Principle of Impulse & Momentum
- apply the principle:21
2
1
mvdtFmv
t
t
- if motion occurs on x-y planethe 2 scalar component equation can be formulated
by:
1. resolving the vector components of F from FBD
2. using the data on the impulse and momentum diagrams
Scalar Equation
Procedure for Analysis
See Example:
15.1 15.2 15.3
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15.2 Principle of Linear Impulse & Momentum for a System of Particles
dt
dvmF iii
Multiplying both sides by dt and integrating between the limits t = t 1, vi = (vi)1 &t = t2, vi = (vi)2,
21 )()(2
ii
t
t
iii vmdtFvm
The initial linear momenta of the system
added vectorially to the impulses of allthe external forces acting on the system
during the time period t1 to t2 are equalto the systems final linear momenta.
Location of mass centre G of the system:
iiiG mmrmrm ,
Taking the time derivatives:
iiG vmvm
Total linear momentum of the system of particles plus (vectorially)
the external impulses acting on the system of particles during the
time interval t1 to t2 equal to the aggregate particles final linear
momentum.
Sum of
external
forces
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15.3 Conservation of Linear Momentum for a System of Particles
When the sum of the external impulses acting on a system of particles is zero:
21 )()( iiii vmvm
In other form:
21)()( GG vv
Applied when particles collide or interact.
Nonimpulsive forces: - causing negligible impulses- including any force that is very small compared to otherlarger (impulsive) forces
Impulsive forces: - forces which are very large & act for a very short periodof time; produce a significant change in momentum
- normally occur due to an explosion or the striking of one
body against another
conservation of linear momentum
See Example:
15.4 15.5 15.6 15.7 15.8
Exercise:
15.32 15.33 15.34 15.35 15.36
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15.4 Impact
Impact occurs when two bodies collide with each other during a very short periodof time causing relatively large (impulsive) forces to be exerted between twobodies.
E.g.: striking of a hammer on a rail golf club on a ball
Central impactthe direction of motion ofthe mass centers of the two collidingparticles is along a line passing through the
mass centers of the particles
Oblique impactwhen the motion of one orboth of the particles is at angle with the lineof impact
Particle have the initial momenta, (vA)1 > (vB)1 During the collisionthe particles will undergo a period of deformationequal but
opposite deformation impulse P dt At maximum deformationboth particles move with common velocity v Then a period of restitution occurs (particles will either return to original shape or
remain deform)restitution impulse R dt pushes particles apart from another,where P dt > R dt
Just after separation, particles final momenta = (vB)2 > (vA)2
Analysis of Central Impact
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To determine the velocities, apply conservation of momentum for the system of
particles by referring to the figure above:
+ mA (vA)1 + mB (vB)1 = mA (vA)2 + mB (vB)2
To solve for final velocities (initial values of particles will be known in mostcases), consider deformation phase:
+ mA (vA)1 - P dt = mA v
For restitution phase:
+ mA (vA)1 - R dt = mA ( vA)2
Ratio of the restitution impulse to the deformation impulse = coefficient ofrestitution, e:
vv
vv
dtP
dtRe
A
A
1
2
)(
)(
For particle B:
1
2
)(
)(
B
B
vvvv
dtP
dtR
e
If remove unknown v:
11
22
)()(
)()(
BA
AB
vv
vve
+
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In general, value of e = between zero and one:
1. Elastic Impact (e = 1): If the collision between the two particles is perfectlyelastic, deformation impulse P dt = equal and opposite to restitution impulseR dt
2. Plastic Impact (e = 0): No restitution impulse given to the particles (R dt = 0);after collision, both particles couple or stick together and move with common
velocity.
Use the following two equations:1. The conservation of momentum applies to the system of particles,
mv1 = mv22. The coefficient of restitution, e = [(vB)2(vA)2] / [(vA)1(vB)1], relates the
relative velocities of the particles along the line of impact, just before andjust after collision.
In determining (vAx)2, (vAy)2, (vBx)2 and (vBy)2,consider these four equations:
1. Momentum of the system is conserved along
the line of impact, x axis, so that m(vx)1 =m(vx)2
2. The coefficient of restitution, e = [(vBx)2(vAx)2] / [(vAx)1(vBx)1], relates the relative
velocity components of the particles along theline of impact (x axis)
3. Momentum of particle A is conserved alongthe y axis, perpendicular to the line of impact,
since no impulse acts on particle A in thisdirection.
4. Momentum of particle B is conserved alongthe y axis, perpendicular to the line of impact,
since no impulse acts on particle B in thisdirection.
Coefficient of Restitution
Procedure for Analysis
(Central Impact)
Procedure for Analysis
(Oblique Impact)
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See Example:
15.9 15.10 15.11
15.5 Angular Momentum
Angular momentum of a particle about point O = moment of the particleslinear momentum about O.
Also being referred to as the moment of momentum.
If a particle is moving along a curve lying in thex-y plane, magnitude ofHo:
(Ho)z = (d)(mv)
Common unit = kg m2/ s Directionright-hand rule
If the particle is moving along a space curve, angular momentum Ho:
Ho = rx mv
In Cartesian components:
zyx
zyx
mvmvmv
rrr
kji
Ho
Scalar Formulation
Vector Formulation
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CHAPTER 16
PLANAR KINEMATICS OF
A RIGID BODY
Chapter ObjectivesTo classify the various types of rigid-body planar motionTo investigate rigid-body translation and show how to analyze motion about a
fixed axisTo study planar motion using an absolute motion analysisTo provide relative motion analysis of velocity and acceleration using a translating
frame of referenceTo show how to find the instantaneous center of zero velocity and determine the
velocity of a point on a body using this methodTo provide a relative motion analysis of velocity and acceleration using a rotating
frame of reference
16.1 Rigid-Body Motion
Particles of a rigid body move along paths equidistant from a fixed plane Has 3 types:
1. Translation- every line segment on the body remains parallel to its original direction during the
motion- rectilinear translation: path of motionalong equidistant straight lines- curvilinear translation: path of motionalong curved lines which are equidistant
Planar motion
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2. Rotation about a fixed axis- all particles of the body (except those lie on the
axis of rotation) move along circular paths
3. General plane motion- undergoes a combination of translation
and rotation
16.2 Translation
Position:- location of pointsA andBdefined from fixedx, y reference frameusing
position vectors rA and rB
- x,ycoordinate systemfixed in the body where origin =A (base point)- position ofB with respect toA = relative position vector rB/A (r ofB with respect to
A)
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- vector addition: rB = rA + rB/A
Velocity:- relationship between instantaneous velocities of A and Bobtained by taking the
time derivative of the position equation:
vB = vA + drB/A/dt
Since drB/A/dt= 0 due to the magnitude ofrB/A = constant, and vB = vA = absolute
velocities,
vB = vA
Acceleration:- time derivative of velocity equation:
aB = aA
- velocity and acceleration equation indicates that all points in a rigid bodysubjected to either rectilinear or curvilinear translation move with the same velocityand acceleration.
16.3 Rotation About a Fixed Axis
Angular motion
- only lines or bodies undergo angular motion- angular motion of a radial line r located within the shaded plane and directed from
point O on the axis of rotation to point P
1. Angular position
- angular position or r = defined by angle - measured between a fixed reference line and r
2. Angular displacement
- defined by the change in the angular position,
measured as a differential d- has a magnitude of d, measured in degrees, radians
or revolutions, where 1 rev = 2 rad.
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- Since motion is about a fixed axis, direction of dwhich always along the axis
- Directiondetermine by the right hand rule
3. Angular velocity
- defined as the time rate of change in angular position,
, where = d / dt +
- has a magnitude measured in rad/s- directionalways along the axis of rotation where
the sense of rotation being referred as clockwise or
counterclockwise- arbitrarily chosen counterclockwise as positive
4. Angular acceleration
- measures the time rate of change of the angular velocity
- magnitude: = d / dt or = d2 / dt2 +
- directiondepends on whether is increasing or decreasing- e.g.: if is decreasing, = angular deceleration, directionopposite to
- by eliminating dt from the above equation,
d = d +
5. Constant angular acceleration
- when angular acceleration of the body is constant,
= c
+ = o + c t
+ = o + o t + c t2
+ 2 = o2 + 2c ( - o)
where o = initial angular positiono = initial angular velocity
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Motion of point P- as rigid body rotates, point P travels along a circular
path of radius r and center at point O.
1. Position- defined by the position vector r, which extends from
O to P
2. Velocity
- has a magnitude of rvrvr ,
- since r= constant, rvvrvr ,0
- since rv ,
- direction ofv = tangent to the circular path- magnitude and direction of vaccounted from:
prv
where rp: directed from any point on the axis of rotation
to point P
- to establish the direction of vright hand rule- by referring to the figure,
since
rv
rvrr
rv
p
p
,sin
sin
3. Acceleration
- can be expressed in terms of its normal and tangential components:
where
rara
dtdrvr
vadtdva
nt
nt
2
2
,
,/,,
/,/
- tangential componentsrepresents the time rate of change in the velocitysmagnitude
- normal componenttime rate of change in the velocitys direction
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- acceleration in terms of vector cross product:
since
)(
&
::
pp
p
p
p
p
rra
rvdt
dr
dt
d
dt
drr
dt
d
dt
dva
- by referring to the next figure, rra pt sin
- applying right hand rule yields pr in the direction
ofat
- hence obtainrr
aaa nt2
- magnitude:22
tn aaa
Procedure for Analysis:
To determine velocity and acceleration of a point located on a rigid body that is rotating
about a fixed axis:
a) Angular Motion1. Establish positive sense of direction along the axis of rotation and show it
alongside each kinematics equation as it is applied.2. If a relationship is known between any two of the 4 variables , , and t, then a
third variable can be obtained by using one of the following kinematics equation
which relates all 3 variables:
dddt
d
dt
d ,,
3. For constant angular acceleration, use:
)(20
2
0
2
2
2
100
0
c
c
c
tt
t
4. ,, - determine from algebraic signs of numerical quantities.
b) Motion of P1. Velocity of P and components of acceleration can be determine from:
ra
ra
rv
n
t
2
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2. If geometry of problem is different to visualize, use:
rra
rra
rrv
pn
pt
p
2)(
Note:- rpdirected from any point on the axis of rotation to point P- rlies in the plane of motion P- vectorsexpressed in terms of its i, j, k components.
See Example 16.1 and 16.2.
16.5 Relative-Motion Analysis: Velocity
General motion: combination of translation androtation
To view motions separatelyuse relative-motion analysis, involving 2 sets of coordinateaxes
Fixed referencemeasures the absolute positionof 2 points A & B on the body
Translating referencedo not rotate with thebody; only allowed to translate with respect to
the fixed frame; originattached to the selected
base point A
Position vector rAspecifies the location of base point A Relative position rB/Alocates point B with respect to point A by vector addition, position of B: rB = rA + rB/A
Points A & Bundergo displacements drA & drB during an instant of time dt Consider general plane motion by its component parts:
- entire bodytranslates by drAA moves to its final position and B to B
Position
Displacement
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- rotated about A by d - B moves to its final position (relative displacementdrB/A)
- displacement of B:
ABAB drdrdr /
to determine the relationship between the velocities of points A and B take thetime derivative (divide displacement equation by dt):
dt
dr
dt
dr
dt
dr ABAB /
ABAB vvv /
absolute
velocities of
points A & B
Velocity
due to rotation about A
due to translation of A
due to translation & rotation
relative
velocity
vB/A
relative velocity of B with
respect to A
velocity of base point A
velocity of point B
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since vB/A also representing the effect of circular motion about A:
ABAB
ABAB
rvv
rv
/
//
A) Vector Analysis
1. Kinematics Diagram
Establish the directions of the fixed x,y coordinates and draw a kinematicsdiagram of the body
Indicate vA, vB, , rB/A If magnitudes of vA, vB or are unknown, the sense of direction can be
assumed
2. Velocity Equation
To apply ABAB rvv / , express the vectors in Cartesian vector form and
substitute them into the equation.
Evaluate the cross product and then equate the i andj components to obtaintwo scalar equations.
If negative answer obtained for an unknown magnitude, direction of vectoropposite to that shown on the kinematics diagram.
B) Scalar Analysis
1. Kinematics Diagram
Draw a kinematics diagram to show the relative motion Consider body to be pinned momentarily at base point A, magnitude: vB/A =
rB/A Direction of vB/Aestablished from the diagram
relative-position vector drawn
from A to B
angular velocity of the body
velocity of base point A
velocity of point B
Procedure for Analysis
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2. Velocity Equation
From equation vB = vA + vB/A, represent each vectors graphically by showingmagnitudes and directions.
Scalar equationdetermine from x & y components of these vectors.
See Example:
Exercise
16.4 16.37 16.58 16.63
16.6 16.7 16.8 16.9
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CHAPTER 17
KINETICS OF A RIGID BODY
(FORCE AND ACCELERATION)
Chapter ObjectivesTo introduce the methods used to determine the mass moment of inertia of a bodyTo develop the planar kinetic equation of motion for a symmetric rigid bodyTo discuss applications of these equations to bodies undergoing translation,
rotation about a fixed axis and general plane motion
17.1 Moment of Inertia
A body has a definite size and shape.F = m.a (mass is a measure of the body resistance to acceleration)
Rotational aspect, caused by moment, M
M =I where Imoment inertia- moment inertia is a measure of the resistance of a body to
angular acceleration (M =I)
Moment inertiaas the integral of the second moment about an axis of all the element ofmass, dm which compose the body.
dmrIm
2
moment arm, r perpendicular distance from z axis Value of I, different for each axis If material having variable density, , which dm = dV,
dVrIV
2
When being a constant,
dVrIV
2
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Procedure for analysis.
For integration, consider only symmetric bodies having surface which are generated by
revolving a curve about an axis.
Shell element Dish element
- height, z, radius, r = y radius, y
- thickness, dy - thickness, dz- dV = (2y) (z) dy - dv = (y2)dz
See Example:
- 17.1- 17.2
ParallelAxis Theorem
If the moment of inertia of the body about an axis passing through the bodys mess centre
is known, then moment of inertia about any other parallel axis may be determined byparallel axis theorem.
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Using Phythagorean theorem,
r2 = (d+x)
2 +y2
Hence, moment of inertia,
m m m
mm
dmddmxddmyxI
dmxddmrI
222
22
'2)''(
])'[(
WhereIGmoment inertia about z axis passing through the mass center, G
mmass of the body
dperpendicular distance.
Radius of Gyration
Moment of inertia of a body about a specified axis, using the radius of gyration, k.
m
IkmkI ,2
Similarity betweenk
&r, from
dI=
r2
dm, moment of inertia of an elemental mass, dm ofthe body about an axis.
Composite bodies
The body of constructed of a number of simple shape such as disk, spheres and rods, the
moment of inertia of the body about any axis, z can be determined by adding
algebraically the moment of inertia of all the composite shape.
)( 2mdII G
See Example:
- 17.3- 17.4
IG zero through total mass, m
(since r2
= x2+y
2) mass
center
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17.2 Planar Kinetic Equations of Motion
Limit of studyplanar kinetic to rigid bodies, which along with their loadings, are
considered to be symmetrical.
The inertial frame of reference x, y, z has its
origin coincident with the arbitrary point P inthe body. By definition, these axes do not rotate
and are either fixed or translate with constant
velocity.
Equation of translation motion.
The external forces represent the effect of gravitation electrical, magnetic or contact forcebetween adjacent bodies.
The analysis of a system of particles:
GmaF
(the translation equation of motion far the mass centre of a rigid body)
The sum of all the external forces acting on the body is equal to the bodys mass timesthe acceleration of its mass centre, G
For x-y plane,
yGy
xGx
amF
amF
)(
)(
Equation of rotational motion
Particle FBD Particle kinetic diagram
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where : Firesultant external force
firesultant internal force far I particle cause by interactions with adjacentparticle
mimass of particle
aiinstant acceleration
If moments of the force acting on the particle are summed about point P:
iiii amrfrFr
or iiiP amrM )(
The moment about P can be expressed in term of acceleration of point P.
If the body has an angular acceleration and angular velocity w,
)()(2rrarmM piiP
)()([
2
rrrrarm pi
Cross product operation with Cartesian component,
)]}([)(])()[(){()( yjxikyjxijaiayjximkM yPxPiiP
kyxaxaym yPxPi ])()([22
])()([)(2raxaymM yPxPiip
Letting mi dm,
))()())((2
dmraxdmaydmMm
yP
m
xP
m
p
MPrepresent only the moment of the external forces acting on the body about P.
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Note: the resultant of moment of the internal force is zero, since for the entire body, these
forces occur in equal and opposite collinear, thus the moment of each pairs of forcesabout P cancels.
The integral of first and second term are wed to locate the bodys centre mass G
Since y dm = m and x dm =x m, and if point P coincides with the mass centre a forthe body,x == 0,
Gp IM
This rotational equation of motion state that the sum of the moment of all the external
forces computed about the bodys mass center G, is equal to the product of moment of
inertia of the body about an axis passing through G and the bodys angular acceleration.
PkpM )(
where kkinetics momentWhen moments of the external force shown, on the free body diagram are summed about
point P, they are equivalent to the sum of the `kinetic-moments of the component ofmaG
about P plus the kinetic moment ofIG.
Equation of motion:
GG
yGy
xGx
IM
amF
amF
)(
)(
or PkpM )(
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17.3 Equation of Motion :Translation
When the rigid body undergoes a translation, all the particles of the body have the sameacceleration,
aG = a
= 0
Rotational equation of motion applied at point G,
MG = 0
Rectilinear translation:All the particles of the body travel along parallel straight line path.
SinceIG = 0, only maG on the kinetic diagram:
0
)(
)(
G
yGy
xGx
M
amF
amF
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CHAPTER 18
PLANAR KINEMATICS OF
A RIGID BODY:
WORK AND ENERGY
Chapter ObjectivesTo develop formulation for the kinetic energy of a body and define the various
ways a force and couple do workTo apply the principle of work and energy to solve rigid-body planar kinetics
problems that involve force, velocity and displacementTo show how the conservation of energy can be used to solve rigid-body planar
kinetic problems
18.1 Kinetic Energy
Consider the rigid body shownwith an arbitraryith particle of the body, having a mass dm, is
located at r from the arbitrary point P.
If at the instant shown the particle has a velocity vi,then the particles kinetic energy, Ti = dm vi
2
Kinetic energy of the entire body:
m
ivdmT2
2
1
In terms of velocity of point P,
jxvijv
yjxikjviv
vvv
ypxp
ypxp
pipi
])[(])[(
)()()(
/
Square of magnitude of vi:
222
222222
222
)(2)(2
)(2)()(2)(
])[(])[(
rxvyvv
xxvvyyvv
xvyvvvv
ypxpp
ypypxpxp
ypxpiii
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Substitute the equation of K.E.:
mm m
ypxp
m
p dmrdmxvdmyvvdmT )()()()()()(22
212
21
2
212
21 )()( Pypxpp ImxvmyvmvT
For ,0 yx 2
2
12
2
1
GG ImvT
For a body which having either rectilinear orcurvilinear translation, = 0,
2
21
GmvT
Body has both translation & rotational kinetic energy:
2
2
12
2
1
GG ImvT
Note that vG = rG,22
21 )( GG ImrT
entire
mass m
of the
body
bodys center of mass G with
respect to P
bodys
moment of
inertia IP
Translation
Rotation about a
fixed axis
moment of
inertia, Io
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2
021 IT
Kinetic Energy: 2212
21 GG ImvT
18.2 The Work of a Force
a) Work of a Variable Force
s
F dsFU cos
where:
UFwork done by force
Fexternal force
spath
- angle between the tails ofthe forcevector & the differential
displacement
b) Work of a Constant Force
sFU CFC )cos(
where:
UFCwork done by forceFCexternal force
Stranslation
- angle of direction
General Plane
Motion
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c) Work of a Weight
- undergoes a vertical displacement y
- upward directionnegative work(weight and displacementoppositedirections)
yWUW
d) Work of a Spring Force
)(2
1212
221 skskUS
for12
ss
where:ks = Fs = spring force
s1 = initial compression position
s2 = further position
e) Forces That Do No Work
- act at a fixed points on the body orhaving direction perpendicular to the
displacement- refer to figurerolling resistance force
Frdoes no work since acting on a
round body as it rolls without slipping
over a rough surface
- due to Fracts at a point which has zerovelocity (instantaneous center, IC)during any instant of time dt.
- work of Fr = 0
See Example 18.1.
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18.3 The Work of a Couple
when body translatespositive work of one force
cancels the negative work of the other body undergoes different rotation d about an axis,
each force undergoes displacement ds = (r/2) d total work done:
dM
dFr
dFdFdU rrM
)(
)()(22
when body rotates through finite angle (rad), work
of a couple:
2
1
dMUM
for constant magnitude,
UM= M (2 - 1)
work is positive for M and (2 - 1) are having thesame direction.
See Example 18.2.
18.4 Principle of Work and Energy
2211 TUT
bodys initial translational & work done by all the external bodys final
rotational kinetic energy forces & couple moments translational &
rotational kinetic energy
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Work of a weight Magnitude of work Vertical displacement
stretch / compression of
spring
spring stiffness
Procedure for Analysis
Kinetic Energy (Kinematic Diagrams)
- translation -2
21 GmvT - rotation - 2
21 GIT
- special case (rotation about a fixed axis) - 202
1 IT
- use kinematic diagramdetermine vG, and relationship between them
Work (Free-Body Diagram - FBD)
- draw FBDcount for all forces and couple moments- integrate forcesobtain work
- graphicallywork = area under force-displacement curve- work of a weight:
UW = W y
- work of spring:
US = k s2
- work of coupleproduct of couple moment and angle (rad) through whichit rotates
- workpositive when force (couple moment)same direction asdisplacement (rotation)
Principle of Work & Energy
2211 TUT
bodys initial translational & work done by all the external bodys final
rotational kinetic energy forces & couple moments translational &
rotational kinetic energy
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See Example 18.2.
See Example:
- 18.3
- 18.4
- 18.5
- 18.6
18.5 Conservation of Energy
Determine by knowing the height ofthe bodys center of gravity
Vg = W yG P.E.positive when yG = positive
Ve = + k s2
In deformed position, the spring forceacting on the body always has thecapacity for doing positive work when
spring is returned back to its original
undeformed position
GravitationalPotential Energy
Elastic
Potential Energy
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For body subjected to both gravitational and elastic forces,
Total P.E., V = Vg +Ve
or
Principle of work energy:
222111 )( VTUVT noncons
For ,0)( 21 nonconsU
2211VTVT conservation of mechanical energy
Procedure for Analysis
Potential Energy
- draw initial and final position diagram
- if center of gravity, G, performing a vertical displacement, establish a
horizontal datum to measure Vg.
- use V = Vg +Ve, where Vg = W yG (+/-),
Ve = + k s2
Kinetic Energy
- translation -2
21
GmvT
- rotation - 221 GIT
Conservation of
Energy