Dynamics Rectilinear - Continuous and Erratic

7/28/2019 Dynamics Rectilinear - Continuous and Erratic

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Chapter 12 : Kinematics Of A Particle



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.



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



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.



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.



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.



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.



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



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




• Displacement – change in its position, vector

quantity




• 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




• 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




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




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




• 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




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




• 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 '




• 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




• 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





• 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





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.




• The positive sense for each scalar can be indicated

by an arrow shown alongside each kinematics eqn

as it is applied




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




• Remember that the three kinematics equations can

only be applied to situation where the acceleration

of the particle is constant.




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



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



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



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



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



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



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



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



When t = 4s,

s = 4.43m

mt s

2

2/1

2 60

18.0

60

1

4.0

1

EXAMPLE 12.2



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



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



)(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



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



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



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



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



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



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




vdt

ds

Slope of s-t graph = velocity




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




adt

dv

Slope of v-t graph = acceleration




• 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)






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



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



06;3010

6.06.0;100

dt

dv

av st

dt

dvat v st

The results are plotted.

EXAMPLE 12.6




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=



• 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




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

=




• 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


EXAMPLE 12 7



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



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



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



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




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



• 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





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





• 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






EXAMPLE 12 8



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



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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Dynamics Rectilinear - Continuous and Erratic

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