dynamics rectilinear - continuous and erratic
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Chapter 12 : Kinematics Of A Particle
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Chapter Objectives
• To introduce the concepts of position, displacement, velocity, and acceleration.
• To study particle motion along a straight line and
represent this motion graphically.
• To investigate particle motion along a curved path
using different coordinate systems.
• To present an analysis of dependent motion of two
particles.
• To examine the principles of relative motion of two
particles using translating axes.
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Chapter Outline
• Introduction
• Rectilinear Kinematics: Continuous Motion
• Rectilinear Kinematics: Erratic Motion
• Curvilinear Motion: Rectangular Components
• Motion of a Projectile
• Curvilinear Motion: Normal and Tangential Components
• Curvilinear Motion: Cylindrical Components
• Absolute Dependent Motion Analysis of Two Particles
• Relative Motion Analysis of Two Particles Using Translating Axes
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Introduction
• Mechanics – a branch of science which studies onhow bodies react to forces acting on them
– Statics – a branch of mechanics which deals withanalysis of loads on a particle or body inequilibrium.
– Dynamics – a branch of mechanics which dealswith particle in motion.
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Introduction
• Difference of Kinematics and Kinetics.
1) Kinematics – deals with geometry in motion
without consideration of forces involved.
2) Kinetics – relates the force acting on a particle to
its mass and acceleration.
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Introduction
• Difference of Particle to Body
– Particle – refers to the item that is being analyze
to be point in size.
– Body – it is a system of particle with appreciable
size.
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Motion of Particle
• Translation – motion of a rigid body in a
straight or curvilinear path where particles
stay in parallel and adjacent in referral to their
initial position.
– Rectilinear – motion lies in a straight line.
– Curvilinear – motions path is in a curve.
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Rectilinear Kinematics: Continuous Motion
• Rectilinear Kinematics – specifying at any instant, the
particle’s position, velocity, and acceleration
• Position
1) Single coordinate axis, s
2) Origin, O
3) Position vector r – specific location of particle P atany instant
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4) Algebraic Scalar s in metres
Note : - Magnitude of s = Dist from O to P
- The sense (arrowhead dir of r) is defined byalgebraic sign on s
=> + = right of origin, - = left of origin
Rectilinear Kinematics: Continuous Motion
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• Displacement – change in its position, vector
quantity
Rectilinear Kinematics: Continuous Motion
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• If particle moves from P to P’
=>
is +ve if particle’s position is right of its initial
position
is -ve if particle’s position is left of its initial position
s s s r r r
s
s
Rectilinear Kinematics: Continuous Motion
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• Velocity Average velocity ,
(total change in position divided by the total elapsed time)
Instantaneous velocity is defined as
(differential/infinitesimal change in position per differential change in time)
t
r vavg
t r v
t ins
/lim0
dt
dr vins
Rectilinear Kinematics: Continuous Motion
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Representing as an algebraic scalar,
Velocity is +ve = particle moving to the right
Velocity is –ve = Particle moving to the left
Magnitude of velocity is the speed (m/s)
insv
dt
dsv
Rectilinear Kinematics: Continuous Motion
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Average speed is defined as total distancetraveled by a particle, sT , divided by theelapsed time .
The particle travels along
the path of length sT in time
=> (average speed; magnitude only)
(average velocity; includes direction (sign)
t
t
sv T
avg sp
t
t
sv
t sv
avg
T avg sp
Rectilinear Kinematics: Continuous Motion
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• Acceleration – velocity of particle is known atpoints P and P’ during time interval Δt ,
average acceleration is
• Δv represents difference in the velocityduring the time interval Δt , ie
t
vaavg
vvv '
t
vaavg
Rectilinear Kinematics: Continuous Motion
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Instantaneous acceleration at time t is found bytaking smaller and smaller values of Δt and
corresponding smaller and smaller values of Δv ,
t vat
/lim0
2
2
dt
sd a
dt
dva
Rectilinear Kinematics: Continuous Motion
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• Particle is slowing down, its speed isdecreasing => decelerating =>
will be negative.
• Consequently, a will also be negative,
therefore it will act to the left , in the oppositesense to v
• If velocity is constant,
acceleration is zero
vvv '
Rectilinear Kinematics: Continuous Motion
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• Velocity as a Function of Time
Integrate ac = dv/dt, assuming that initially v
= v 0 when t = 0.
t
c
v
vdt adv
00
t avv c0
Constant Acceleration
Rectilinear Kinematics: Continuous Motion
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• Position as a Function of Time
Integrate v = ds/dt = v 0 + ac t , assuming that
initially s = s0 when t = 0
2
00
00
2
1
0
t at v s s
dt t avds
c
t c
s
s
Constant Acceleration
Rectilinear Kinematics: Continuous Motion
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• Velocity as a Function of PositionIntegrate v dv = ac ds, assuming that initially v =
v 0 at s = s0
0
2
0
22
00
s savv
dsavdv
c
s
s c
v
v
Constant Acceleration
Rectilinear Kinematics: Continuous Motion
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PROCEDURE FOR ANALYSIS
1) Coordinate System
• Establish a position coordinate s along the path
and specify its fixed origin and positive direction.• The particle’s position, velocity, and acceleration,
can be represented as s, v and a respectively and
their sense is then determined from their algebraic
signs.
Rectilinear Kinematics: Continuous Motion
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• The positive sense for each scalar can be indicated
by an arrow shown alongside each kinematics eqn
as it is applied
Rectilinear Kinematics: Continuous Motion
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2) Kinematic Equation• If a relationship is known between any two of the
four variables a, v, s and t, then a third variable
can be obtained by using one of the three the
kinematic equations• When integration is performed, it is important that
position and velocity be known at a given instant
in order to evaluate either the constant of
integration if an indefinite integral is used, or thelimits of integration if a definite integral is used
Rectilinear Kinematics: Continuous Motion
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• Remember that the three kinematics equations can
only be applied to situation where the acceleration
of the particle is constant.
Rectilinear Kinematics: Continuous Motion
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The car moves in a straight line such that for a short
time its velocity is defined by v = (0.9t 2 + 0.6t ) m/s
where t is in sec. Determine its position and
acceleration when t = 3s. When t = 0, s = 0.
EXAMPLE 12.1
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Solution:
Coordinate System. The position coordinate
extends from the fixed origin O to the car, positive tothe right.
Position. Since v = f(t), the car’s position can be
determined from v = ds/dt , since this equation relates
v , s and t . Noting that s = 0 when t = 0, we have
t t dt
dsv 6.09.0 2
EXAMPLE 12.1
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23
0
23
0
0
2
0
3.03.0
3.03.0
6.09.0
t t s
t t s
dt t t ds
t s
t s
When t = 3s,
s = 10.8m
EXAMPLE 12.1
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Acceleration. Knowing v = f(t), the acceleration is determinedfrom a = dv/dt , since this equation relates a, v and t.
6.08.1
6.09.0 2
t
t t dt d
dt dva
When t = 3s,
a = 6m/s2
EXAMPLE 12.1
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A small projectile is fired downward into a
fluid medium with an initial velocity of 60m/s.
Due to the resistance of the fluid the
projectile experiences a deceleration equal to a =(-0.4v 3)m/s2, where v is in m/s.
Determine the projectile’s
velocity and position 4s
after it is fired.
EXAMPLE 12.2
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Solution:Coordinate System. Since the motion is
downward, the position coordinate is downwards
positive, with the origin located at O.
Velocity. Here a = f(v), velocity is a function of
time using a = dv/dt , since this equation relates v ,
a and t .
34.0 vdt dva
EXAMPLE 12.2
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smt v
t v
dt v
dt v
dv
t v
t v
sm
/8.0
60
1
60
11
8.0
1
1
2
1
4.0
1
4.0
2/1
2
22
0602
0/603
When t = 4s,
v = 0.559 m/s
EXAMPLE 12.2
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Position. Since v = f(t), the projectile’s position can bedetermined from v = ds/dt , since this equation relates v , s
and t . Noting that s = 0 when t = 0, we have
2/1
2 8.060
1
t
dt
dsv
t
t s
t s
dt t ds
0
2/1
2
0
2/1
20
8.060
1
8.0
2
8.060
1
EXAMPLE 12.2
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When t = 4s,
s = 4.43m
mt s
2
2/1
2 60
18.0
60
1
4.0
1
EXAMPLE 12.2
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A rocket travels upward at75m/s. When it is 40m from
the ground, the engine fails.
Determine max height sB
reached by the rocket and itsspeed just before it hits the
ground.
EXAMPLE 12.3
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Solution:Coordinate System. Origin O for the position coordinate at
ground level with positive upward .
Maximum Height. Rocket traveling upward , v A = +75m/s
when t = 0. s = sB when v B = 0 at max ht. For entire motion,acceleration aC = -9.81m/s2 (negative since it act opposite
sense to positive velocity or positive displacement)
EXAMPLE 12.3
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)(222 A BC A B s savv
sB = 327 m
Velocity.
sm smv
s savv
C
BC C BC
/1.80/1.80
)(2
2
22
The negative root was chosen since the rocket is moving
downward
EXAMPLE 12.3
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A particle moves along a horizontal path with a velocity of v =(3t 2 – 6t ) m/s. If it is initially located at the origin O,
determine the distance traveled in 3.5s and the particle’s
average velocity and speed during the time interval.
EXAMPLE 12.4
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Solution:
Coordinate System. Assuming positive motion to the right,measured from the origin, O
Distance traveled. Since v = f(t), the position as a function
of time may be found integrating v = ds/dt with t = 0, s = 0.
EXAMPLE 12.4
mt t s
tdt dt t ds
dt t t
vdt ds
s t t
23
0 00
2
2
3
63
63
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mt t s
tdt dt t ds
dt t t vdt ds
s t t
23
0 00
2
2
3
63
63
0 ≤ t < 2 s -> -ve velocity -> the particle is moving to the left, t > 2a -> +ve velocity ->
the particle is moving to the right
m s st
125.65.3
m s st
0.42
00
t
s
EXAMPLE 12.4
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The distance traveled in 3.5s is
sT = 4.0 + 4.0 + 6.125 = 14.125m
Velocity. The displacement from t = 0 to t = 3.5s is Δs =
6.125 – 0 = 6.125m
And so the average velocity is
smt
s
vavg /75.105.3
125.6
Average speed, smt
sv T
avg sp /04.405.3
125.14
EXAMPLE 12.4
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Rectilinear Kinematics: Erratic Motion
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Rectilinear Kinematics: Erratic Motion
• When particle’s motion is erratic, it is best described
graphically using a series of curves that can be
generated experimentally from computer output.
• a graph can be established describing the relationship
with any two of the variables, a, v, s, t
• using the kinematics equations a = dv/dt, v = ds/dt,
a ds = v dv
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Given the s-t Graph, construct the v-t Graph
•The s-t graph can be plotted if the position of the
particle can be determined experimentally during aperiod of time t.
•To determine the particle’s velocity as a function of
time, the v-t Graph, use v = ds/dt
•Velocity as any instant is determined by measuring the
slope of the s-t graph
Rectilinear Kinematics: Erratic Motion
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vdt
ds
Slope of s-t graph = velocity
Rectilinear Kinematics: Erratic Motion
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Given the v-t Graph, construct the a-t Graph
•When the particle’s v-t graph is known, the
acceleration as a function of time, the a-t graph can bedetermined using a = dv/dt
•Acceleration as any instant is determined by
measuring the slope of the v-t graph
Rectilinear Kinematics: Erratic Motion
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adt
dv
Slope of v-t graph = acceleration
Rectilinear Kinematics: Erratic Motion
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• Since differentiation reduces a polynomial of
degree n to that of degree n-1, then if the s-t
graph is parabolic (2nd degree curve), the v-t
graph will be sloping line (1st degree curve), and
the a-t graph will be a constant or horizontal line
(zero degree curve)
Rectilinear Kinematics: Erratic Motion
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Solution:
v-t Graph. The v-t graph can be determined by differentiating the eqns
defining the s-t graph
6306;3010
6.03.0;100 2
dt
dsvt s st s
t dt
dsvt s st
The results are plotted.
EXAMPLE 12.6
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We obtain specific values of v by measuring the slope of the s-t
graph at a given time instant.
smt
sv /6103030150
a-t Graph. The a-t graph can be determined by
differentiating the eqns defining the lines of the v-t graph.
EXAMPLE 12.6
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06;3010
6.06.0;100
dt
dv
av st
dt
dvat v st
The results are plotted.
EXAMPLE 12.6
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Rectilinear Kinematics: Erratic Motion
Given the a-t Graph, construct the v-t Graph
• When the a-t graph is known, the v-t graph may be constructed using a = dv/dt
dt av
Change in
velocity
Area under a-t graph=
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• Knowing particle’s initial velocity v 0, and add to this small
increments of area ( Δv )
• Successive points v 1 = v 0 + Δv , for the v-t graph
• Each eqn for each segment of the a-t graph may be
integrated to yield eqns for corresponding segments of the v-
t graph
Rectilinear Kinematics: Erratic Motion
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Given the v-t Graph, construct the s-t Graph
• When the v-t graph is known, the s-t graph may be constructed using v = ds/dt
dt v s
Displacement Area under v-t
graph
=
Rectilinear Kinematics: Erratic Motion
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• knowing the initial position s0, and add to this area
increments Δs determined from v-t graph.
• describe each of there segments of the v-t graph by a
series of eqns, each of these eqns may be integrated to
yield eqns that describe corresponding segments of
the s-t graph
Rectilinear Kinematics: Erratic Motion
EXAMPLE 12 7
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EXAMPLE 12.7
A test car starts from rest and travels along a
straight track such that it accelerates at a
constant rate for 10 s and then decelerates
at a constant rate. Draw the v-t and s-t
graphs and determine the time t’ needed to
stop the car. How far has the car traveled?
EXAMPLE 12 7
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Solution:
v-t Graph. The v-t graph can be determined by integrating the straight-line
segments of the a-t graph. Using initial condition v = 0 when t = 0,
t vdt dva st t v
10,10;1010000
EXAMPLE 12.7
EXAMPLE 12 7
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When t = t’ we require v = 0. This yield t’ = 60 s
s-t Graph. Integrating the eqns of the v-t graph yields the corresponding eqns
of the s-t graph. Using the initial conditions s = 0 when t = 0,
2
005,10;10;100 t sdt t dst v st
t s
When t = 10s, v = 100m/s, using this as the initial condition for the next time
period, we have
1202,2;2;10 10100
t vdt dvat t s
t v
EXAMPLE 12.7
EXAMPLE 12 7
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When t’ = 60s, the position is s = 3000m
When t = 10s, s = 500m. Using this initial condition,
600120
1202;1202;6010
2
10500
t t s
dt t dst v st st s
EXAMPLE 12.7
R tili Ki ti E ti M ti
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Rectilinear Kinematics: Erratic Motion
Given the a-s Graph, construct the v-s Graph
• v-s graph can be determined by using v dv = a ds, integrating
this eqn between the limit v = v 0 at s = s0 and v = v 1 at s = s1
1
0
2
0
2
12
1 s
sdsavv
Area under
a-s graph
R tili Ki ti E ti M ti
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• determine the eqns which define the segments of the a-s graph
• corresponding eqns defining the segments of the v-s graph can
be obtained from integration, using vdv = a ds
Rectilinear Kinematics: Erratic Motion
Rectilinear Kinematics: Erratic Motion
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Given the v-s Graph, construct the a-s Graph
• v-s graph is known, the acceleration a at any position s can be
determined using a ds = v dv
ds
dvva
Acceleration = velocity times slope of v-s graph
Rectilinear Kinematics: Erratic Motion
Rectilinear Kinematics: Erratic Motion
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• At any point (s,v ), the slope dv/ds of the v-s graph is measured
• Since v and dv/ds are known, the value of a can be calculated
Rectilinear Kinematics: Erratic Motion
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EXAMPLE 12 8
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Time. The time can be obtained using v-s graph and v = ds/dt . For the first
segment of motion, s = 0 at t = 0,
3ln5)32.0ln(5
32.0
32.0;32.0;600
0
st
s
dsdt
dsv
dsdt svm s
st
o
At s = 60 m, t = 8.05 s
EXAMPLE 12.8
EXAMPLE 12 8
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For second segment of motion,
05.415
15
15;15;12060
6005.8
s
t
dsdt
ds
v
dsdt vm s
st
At s = 120 m, t = 12.0 s
EXAMPLE 12.8
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