Dynamics Lecture1

7/29/2019 Dynamics Lecture1

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Introduction to Dynamics


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1. Introduction to Dynamics

Time

Mass

Force

Particle

Rigid Body

Newtons Law

1st Law: Stay at rest or move with uniform velocity

2nd Law: F =ma

3rd Law: Each action has an equal magnitude but oppositedirection reaction


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Units and Dimensions

Units: SI Dimension:

Mass (kg) (M)

Length (m) (L)Time (sec) (T)

Force (N) (F)

F = ML/T2

Fx = mv2

[MLT-2] [L] = [M][LT-1]2


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Gravitation

Mutual attraction between bodies:2

21

r

mmGF

Accuracy, Limits andApproximations


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Introduction to Dynamics

Mechanics the state of rest of motion ofbodies subjected to the action of forces

Static equilibrium of a body that is either at

rest or moves with constant velocity Dynamics deals with accelerated motion of

a body

1) Kinematics treats with geometric aspects

of the motion

2) Kinetics analysis of the forces causing

the motion


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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


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Curvilinear Motion: Cylindrical Components

Absolute Dependent Motion Analysis of Two

Particles

Relative Motion Analysis of Two Particles

Using Translating Axes

Chapter Outline


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Rectilinear Kinematics:

Continuous Motion

Rectilinear Kinematics specifying at any

instant, the particles position, velocity, and

acceleration

Position

1) Single coordinate axis, s

2) Origin, O

3) Position vectorr specific location of

particle P at any instant


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4) Algebraic Scalars in metres

Note : - Magnitude ofs = Dist from O to P- The sense (arrowhead dir ofr) is defined

by algebraic sign on s

=> + = right of origin, - = left of origin


Continuous Motion


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Displacement change in its position, vector

quantity


Continuous Motion


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If particle moves from Pto P

=>

is +ve if particles position is right of its

initial position

is -ve if particles position is left of its initial

position

sss

rrr

s

s


Continuous Motion


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Velocity

Average velocity,

Instantaneous velocityis defined as

t

rvavg

trvt

ins

/lim0

dt

drvins


Continuous Motion


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Representing as an algebraic scalar,

Velocity is +ve = particle moving to the right

Velocity isve = Particle moving to the left

Magnitude of velocity is the speed(m/s)

insv

dt

dsv


Continuous Motion


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Average speedis defined as totaldistance traveled by a particle, sT, dividedby the elapsed time .

The particle travels along

the path of length sTin time

=>

t

ts

v Tavgsp

t

t

sv

tsv

avg

T

avgsp


Continuous Motion


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Acceleration velocity of particle is known at

points P and P during time interval t,

average acceleration is

vrepresents difference in the velocity

during the time intervalt, ievvv '

t

vaavg


Continuous Motion


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Instantaneous acceleration at time t is found

by taking smaller and smaller values oftand

corresponding smaller and smaller values of

v, tvat

/lim0

2

2

dt

sda

dt

dva


Continuous Motion


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Particle is slowing down, its speed is

decreasing=> decelerating=>

will be negative.

Consequently, a will also be negative,therefore it will act to the left, in the opposite

sense to v

Ifvelocity is constant,acceleration is zero

vvv '


Continuous Motion


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Velocity as a Function of Time

Integrate ac= dv/dt, assuming that initially v

= v0when t = 0.

t

c

v

vdtadv

00

tavv c 0

Constant Acceleration


Continuous Motion


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Position as a Function of Time

Integrate v = ds/dt= v0+ act, assuming that

initially s = s0when t = 0

200

00

2

1

0

tatvss

dttavds

c

t

c

s

s



Continuous Motion


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Velocity as a Function of Position

Integrate v dv = acds, assuming that initially v =

v0at s = s0

02

0

2 2

00

ssavv

dsavdv

c

s

sc

v

v



Continuous Motion


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PROCEDURE FOR ANALYSIS

1) Coordinate System

Establish a position coordinate s along thepath and specify its fixed origin and positive

direction.

The particles position, velocity, and

acceleration, can be represented as s, v and arespectively and their sense is then

determined from their algebraic signs.


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


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 inorder to evaluate either the constant of integration

if an indefinite integral is used, or the limits of

integration if a definite integral is used


Continuous Motion


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Remember that the three kinematics equations

can only be applied to situation where the

acceleration of the particle isconstant.


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 = (0.9t2+0.6t) m/s where t is in sec. Determine it position

and acceleration when t= 3s. When t= 0, s = 0.

EXAMPLE


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Solution:

Coordinate System. The position coordinate

extends from the fixed origin O to the car,

positive to the right.

Position. Since v = f(t),the cars position can

be determined from v = ds/dt, since this equationrelates v, s and t. Noting that s = 0 when t= 0, we

have

ttdtds

v 6.09.0

2


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23

0

23

0

0

2

0

3.03.0

3.03.0

6.09.0

tts

tts

dtttds

ts

ts

When t= 3s,

s = 10.8m


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Acceleration. Knowing v = f(t), the acceleration

is determined from a = dv/dt, since this equationrelates a, vand t.

6.08.1

6.09.02

t

tt

dt

d

dt

dva

When t= 3s,

a = 6m/s2


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A small projectile is forced 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 =(-0.4v3)m/s2, where vis in m/s2.

Determine the projectiles

velocity and position 4safter it is fired.

EXAMPLE


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Solution:

Coordinate System. Since the motion isdownward, the position coordinate is downwards

positive, with the origin located at O.

Velocity. Here a = f(v), velocity is a function oftime using a = dv/dt, since this equation relates v,

a and t.3

4.0 vdt

dv

a


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smtv

tv

dtv

dtv

dv

tv

tv

sm

/8.060

1

60

11

8.0

1

1

2

1

4.0

1

4.0

2/1

2

22

0602

0/60 3

When t= 4s,

v= 0.559 m/s


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Position. Since v = f(t),the projectiles position

can be determined from v = ds/dt, since this equation

relates v, s and t. Noting that s = 0 when t= 0, wehave

2/1

28.0

60

1

t

dt

dsv

t

ts

ts

dttds

0

2/1

2

0

2/1

20

8.0601

8.02

8.060

1


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When t= 4s,

s = 4.43m

mts

60

18.0

60

1

4.0

12/1

2


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A rocket travel upward at75m/s. When it is 40m from

the ground, the engine fails.

Determine max height sB

reached by the rocket andits speed just before it hits

the ground.

EXAMPLE


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Solution:Coordinate System. Origin O for the position

coordinate at ground level withpositive upward.

Maximum Height. Rocket traveling upward, vA =

+75m/s when t= 0. s = sB when vB = 0 at max ht. Forentire motion, acceleration aC= -9.81m/s

2 (negative

since it act opposite sense to positive velocity or

positive displacement)


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)(222 ABCAB ssavv

sB = 327 m

Velocity.

smsmv

ssavv

C

BCCBC

/1/80/1.80

)(2

2

22

The negative root was chosen since the rocket is

moving downward


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A metallic particle travels downward

through a fluid that extends from

plate A and plate B under the

influence of magnetic field. If

particle is released from rest at

midpoint C, s = 100 mm, and

acceleration, a = (4s) m/s2, where s

in meters, determine velocity when

it reaches plate B and time need to

travel from C to B

EXAMPLE


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Solution:Coordinate System. It is shown that s is taken

positive downward, measured from plate A

Velocity. Since a = f(s), velocity as a function of

position can be obtained by using v dv = a ds.Realising v = 0at s = 100mm = 0.1m


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21

2

1.0

2

0

2

1.00

01.02

2

4

2

1

4

sv

sv

dssdvv

dsadvv

Sv

sv

At s = 200mm = 0.2m,

smmsmvB /346/346.0


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tss

tss

dts

ds

dts

dtvds

ts

ts

233.201.0ln

201.0ln

201.0

01.02

2

01.0

2

01.0 5.02

5.0

2

At s = 200mm = 0.2m, t= 0.658s


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A particle moves along a horizontal path with a velocityofv= (3t2 6t) m/s. if it is initially located at the origin O,

determine the distance traveled in 3.5s and the particles

average velocity and speed during the time interval.

EXAMPLE


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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 afunction of time may be found integrating v = ds/dtwith

t= 0, s = 0.


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mtts

tdtdttds

dttt

vdtds

s tt

23

0 00

2

2

3

63

63


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mtts

tdtdttds

dttt

vdtds

s tt

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

msst

125.65.3

msst

0.42

00

t

s


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The distance traveled in 3.5s is

sT= 4.0 + 4.0 + 6.125 = 14.125mVelocity. The displacement from t = 0 to t = 3.5s is

s = 6.125 0 = 6.125m

And so the average velocity is

sm

t

saavg /75.1

05.3

125.6

Average speed, smt

sv T

avgsp/04.4

05.3

125.14


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Rectilinear Kinematics

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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Graphical Solution of Rectilinear-Motion

Problems

Given thex-tcurve, the v-tcurve is equal to thex-tcurve slope.

Given the v-tcurve, the a-tcurve is equal to the v-tcurve slope.


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Graphical Solution of Rectilinear-Motion

Problems

Given the a-tcurve, the change in velocity between t1 and t2 isequal to the area under the a-tcurve between t1 and t2.

Given the v-tcurve, the change in position between t1 and t2 is

equal to the area under the v-tcurve between t1 and t2.

Oth G hi l M th d


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Other Graphical Methods

Moment-area methodto determine particle

position at time tdirectly from the a-tcurve:

1

0

110

01 curveunderarea

v

v

dvtttv

tvxx

using dv = a dt ,

1

0

11001

v

v

dtatttvxx

1

01

v

v

dtatt first moment of area undera-tcurvewith respect to t= t1 line.

Ct

tta-ttvxx

centroidofabscissa

curveunderarea 11001


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Other Graphical Methods

Method to determine particle acceleration

from v-xcurve:

BC

AB

dx

dvva

tan

subnormalto v-x curve

Curvilinear Motion: Position, Velocity &


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Acceleration Particle moving along a curve other than a straight line

is in curvilinear motion.

Position vectorof a particle at time tis defined by a

vector between origin O of a fixed reference frame and

the position occupied by particle.

Consider particle which occupies position Pdefinedby at time tand Pdefined by at t + t,r

r

dt

ds

t

sv

dt

rd

t

rv

t

t

0

0

lim

lim

instantaneous velocity (vector)

instantaneous speed (scalar)



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Acceleration

dt

vd

t

va

t

0lim

instantaneous acceleration (vector)

Consider velocity of particle at time tand velocity

at t + t,

v

v

In general, acceleration vector is not tangent to

particle path and velocity vector.

Derivatives of Vector Functions


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Derivatives of Vector Functions

uP

Let be a vector function of scalar variable u,

u

uPuuP

u

P

du

Pd

uu

00limlim

Derivative of vector sum,

du

Qd

du

Pd

du

QPd

du

PdfP

du

df

du

Pfd

Derivative of product of scalar and vector functions,

Derivative ofscalarproductand vectorproduct,

du

QdPQ

du

Pd

du

QPd

du

QdPQ

du

Pd

du

QPd

Rectangular Components of Velocity &


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Rectangular Components of Velocity &

Acceleration

When position vector of particle Pis given byits rectangular components,

kzjyixr

Velocity vector,

kvjviv

kzjyixkdt

dzjdt

dyidt

dxv

zyx

Acceleration vector,

kajaia

kzjyixkdt

zdj

dt

ydi

dt

xda

zyx

2

2

2

2

2

2


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Given the s-tGraph, construct the v-tGraph

The s-tgraph can be plotted if the position of the

particle can be determined experimentallyduring aperiod of time t.

To determine the particles velocity as a function of

time, the v-tGraph, use v = ds/dt

Velocity as any instant is determined by

measuring the slope of the s-tgraph



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vdt

ds

Slope ofs-tgraph =

velocity



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Given the v-tGraph, construct the a-tGraph

When the particles v-tgraph is known, the

acceleration as a function of time, the a-tgraph canbe determined using a = dv/dt

Acceleration as any instant is determined by

measuring the slope of the v-t graph



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adt

dv

Slope ofv-tgraph =

acceleration



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Since differentiation reduces a polynomial of

degree n to that of degree n-1, then if the s-tgraph

is parabolic (2nd degree curve), the v-tgraph will be

sloping line (1st degree curve), and the a-tgraphwill be a constant or horizontal line (zero degree

curve)


EXAMPLE


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EXAMPLE

A bicycle moves along a straight road such that it

position is described by the graph as shown.Construct the v-tand a-tgraphs for 0 t 30s.


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Solution:

v-tGraph. The v-tgraph can be determined by

differentiating the eqns defining the s-tgraph

6306;3010

6.03.0;100 2

dt

dsvtssts

tdt

dsvtsst

The results are plotted.


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We obtain specify values ofvby measuring the

slope of the s-tgraph at a given time instant.

smt

sv /6

1030

30150

a-tGraph. The a-tgraph can be determined by

differentiating the eqns defining the lines of thev-tgraph.


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06;3010

6.06.0;100

dt

dv

avst

dt

dvatvst

The results are plotted.


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Given the a-tGraph, construct the v-tGraph

When the a-tgraph is known, the v-tgraph may

be constructed using a = dv/dt

dtavChange in

velocity

Area under

a-tgraph=


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Knowing particles initial velocity

v0, and add to this small increments

of area (v)

Successive points v1 = v0+v, for

the v-tgraph

Each eqn for each segment of the

a-tgraph may be integrated to yield

eqns for corresponding segments

of the v-tgraph



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Given the v-tGraph, construct the s-tGraph

When the v-tgraph is known, the s-tgraph may

be constructed using v = ds/dt

dtvs

DisplacementArea under

v-tgraph=




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knowing the initial position s0,

and add to this area increments

s determined from v-tgraph.

describe each of there

segments of the v-tgraph by a

series of eqns, each of these

eqns may be integratedto yieldeqns that describe

corresponding segments of the

s-tgraph


EXAMPLE


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EXAMPLE

A test car starts from rest

and travels along a

straight track such that it

accelerates at a constant

rate for 10 s and thendecelerates at a constant

rate. Draw the v-t and s-t

graphs and determine the

time t needed to stop thecar. How far has the car

traveled?


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Solution:

v-tGraph. The v-tgraph can be determined by

integrating the straight-line segments of the a-tgraph. Using initial conditionv= 0 when t= 0,

tvdtdvast

tv

10,10;10100 00

Wh t 10 100 / i hi i i i l


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When t = twe require v= 0. This yield t= 60 s

s-tGraph. Integrating the eqns of the v-tgraph

yields the corresponding eqns of the s-tgraph.

Using the initial conditionss = 0 when t= 0,

2

005,10;10;100 tsdttdstvst

ts

When t= 10s, v= 100m/s, using this as initial

condition for the next time period, we have

1202,2;2;1010100

tvdtdvattstv


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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

tts

dttdstvststs


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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 = v0at s =s0and v = v1at s = s1

1

0

2

0

2

12

1 s

sdsavv

Area undera-s graph

R tili Ki 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



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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 ofv-s graph




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At any point (s,v), the slope dv/ds of the v-s graphis measured

Since vand dv/ds are known, the value ofa can

be calculated


EXAMPLE


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EXAMPLE

The v-s graph describing the motion of a motorcycle

is shown in Fig 12-15a. Construct the a-s graph ofthe motion and determine the time needed for the

motorcycle to reach the position s = 120 m.

Solution:


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Solution:

a-s Graph. Since the eqns for the segments of the

v-s graph are given, a-s graph can be determined

using a ds = v dv.

0

;15;12060

6.004.0

32.0;600

ds

dvva

vmsm

sds

dvva

svms


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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

ds

v

dsdtsvms

st

o

At s = 60 m, t= 8.05 s


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For second segment of motion,

05.415

15

15;15;12060

6005.8

s

t

dsdt

ds

v

dsdtvms

st

At s = 120 m, t= 12.0 s

G l C ili M ti


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General Curvilinear Motion

Curvilinear motion occurs when the particle movesalong a curved path

Position. The position of the particle, measured

from a fixed point O, is designated by thepositionvectorr= r(t).


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Displacement. Suppose during a small time

intervaltthe particle moves a distances along

the curve to a new position P`, defined by r` = r+

r. The displacementrrepresents the change inthe particles position.


G l C ili M ti


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Velocity. During the timet, the average velocity

of the particle is defined as

t

rvavg

The instantaneous velocity is determined from

this equation by lettingt 0, and consequently

the direction ofrapproaches the tangentto the

curve at point P. Hence,

dt

drvins




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Direction ofvins is tangent to the curve

Magnitude ofvins is the speed, which may be

obtained by noting the magnitude of the

displacementr is the length of the straight line

segment from P to P`.

dt

dsv



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Acceleration. If the particle has a velocity vat

time t and a velocity v` = v +vat time t` = t +t.

The average acceleration during the time interval

t is

t

vaavg

2

2

dt

rd

dt

dv

a


Curvilinear Motion: Rectangular


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Components

Position. Position vector is defined by

r=xi + yj + zk

The magnitude ofris always positive and definedas

222 zyxr

The direction of r is specified

by the components of the

unit vectorur= r/r



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Velocity.

zvyvxv

kvjvivdt

drv

zyx

zyx

whereThe velocity has a magnitude

defined as the positive value of

222

zyx vvvv and a direction that is specified by the components

of the unit vectoruv=v/vand is always tangent to

the path.

g

Components



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Acceleration.

zva

yva

xva

kajaiadt

dva

zz

yy

xx

zyx

The acceleration has a magnitude defined as thepositive value of

222

zyx aaaa

where


Components



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The acceleration has a direction specified by the

components of the unit vectorua = a/a.

Since a represents the time rate ofchange in

velocity, a will notbe tangent to the path.


Components



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Coordinate System

A rectangular coordinate system can be used tosolve problems for which the motion can

conveniently be expressed in terms of itsx, yand z

components.


Components



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Kinematic Quantities

Since the rectilinear motion occurs along each

coordinate axis, the motion of each component is

found using v = ds/dtand a = dv/dt, ora ds = v ds

Once thex, y, zcomponents ofvand a have

been determined. The magnitudes of these vectors

are found from the Pythagorean theorem and theirdirections from the components of their unit

vectors.


Components

EXAMPLE


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EXAMPLE

At any instant the horizontal

position of the weather balloonis defined byx = (9t) m, where

tis in second. If the equation

of the path is y = x2/30,

determine the distance of the

balloon from the station at A,

the magnitude and direction of

the both the velocity andacceleration when t= 2 s.


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Solution:

Position. When t= 2 s,x= 9(2) m = 18 m and

y= (18)2/30 = 10.8 mThe straight-line distance fromA to B is

218.101822

r m

Velocity.

smxdt

dyv

smtdt

d

xv

y

x

/8.1030/

/99

2


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When t = 2 s, the magnitude of velocity is

smv /1.148.109 22

The direction is tangent to the path, where

2.50tan 1

x

y

vv

v

Acceleration.

2/4.5

0

smva

va

yy

xx


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222 /4.54.50 sma

The direction ofa is

900

4.5tan 1a

EXAMPLE


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The motion of box B is defined

by the position vector r ={0.5sin(2t)i + 0.5cos(2t)j

0.2tk} m, where t is in seconds

and the arguments for sine and

cosine are in radians (rad =

180). Determine the location

of box when t = 0.75 s and the

magnitude of its velocity andacceleration at his instant.


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Solution:

Position. Evaluating rwhen t= 0.75 s yields

mkjradiradr st })75.0(2.0)5.1cos(5.0)5.1sin(5.0{75.0

mkji }150.00354.0499.0{

The distance of the box from the origin is

mr 522.0)150.0()0354.0()499.0( 222


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The direction ofris obtained from the components

of the unit vector,

107

1.86

2.17)955.0(cos

287.00678.0955.0

522.0

150.0

522.0

0352.0

522.0

499.0

1

kji

kjir

rur

V l it


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Velocity.

smkjtitdt

rdv /}2.0)2sin(1)2cos(1{

Hence at t= 0.75 s, the magnitude of velocity, is

smvvvv zyx /02.1222

Acceleration. The acceleration is not tangent

to the path.2/})2cos(2)2sin(2{ smjtit

dtvda

At t = 0.75 s, a = 2 m/s2

Motion of a Projectile


101/234

Free-flight motion studied in terms of rectangular

components since projectiles acceleration always

act vertically

Consider projectile launched at (x0, y0) Path defined in thex-yplane

Air resistance neglected

Only force acting on the projectile is its weight,

resulting in constant downwards accelerationac= g= 9.81 m/s

2




102/234




103/234

Horizontal Motion Since ax= 0,

);(2

;21

;

0

2

0

2

200

0

ssavv

tatvxx

tavv

c

c

c

xx

x

xx

vv

tvxx

vv

)(

)(

)(

0

00

0

Horizontal component of velocity remain constantduring the motion




104/234

Vertical. Positive yaxis is directed upward, then

ay= - g

);(2

;2

1

;

0

2

0

2

2

00

0

yyavv

tatvyy

tavv

c

c

c

)(2)(

2

1)(

)(

0

2

0

2

00

0

yygvv

gttvyy

gtvv

yx

y

yy




105/234

Problems involving the motion of a projectile have

at most three unknowns since only three

independent equations can be written:

- one in the horizontal direction- two in the vertical direction

Velocity in the horizontal and vertical direction are

used to obtain the resultant velocity

Resultant velocity is always tangent to the path




106/234

PROCEDURE FOR ANALYSISCoordinate System

Establish the fixedx, y, zaxes and sketch the

trajectory of the particle

Specify the three unknowns and data between

any two points on the path

Acceleration of gravity always acts downwards

Express the particle initial and final velocities inthex, ycomponents




107/234

Positive and negative position, velocity and

acceleration components always act in accordance

with their associated coordinate directions

Kinematics Equations Decide on the equations to be applied between

the two points on the path for the most direct

solution



108/234

Horizontal Motion

Velocity in the horizontal orxdirections is

constant (vx) = (vo)x

x = xo + (vo)x t

Vertical Motion

Only two of the following three equations should

be used



109/234

)(2)(2

1)(

)(

0

2

0

2

00

0

yygvv

gttvyy

gtvv

yx

y

yy


EXAMPLE


110/234

A sack slides off theramp with a horizontal

velocity of 12 m/s. If the

height of the ramp is 6 m

from the floor, determinethe time needed for the

sack to strike the floor

and the range Rwhere

the sacks begin to pileup.


111/234

Coordinate System. Origin of the coordinates is

established at the beginning of the path, pointA.Initial velocity of a sack has components (vA)x= 12

m/s and (vA)y= 0

Acceleration between point A and B ay= -9.81 m/s2

Since (vB)x= (vA)x= 12 m/s, the three unknown are(vB)y, Rand the time of flight tAB


112/234

Vertical Motion. Vertical distance fromA to B is

known

st

tatvyy

AB

ABcABy

11.1

;2

1)( 200

The above calculations also indicate that if a sack

is released from restat A, it would take the same

amount of time to strike the floor at C


113/234

Horizontal Motion.

mR

tvxx ABx

3.13

)( 00

EXAMPLE


114/234

The chipping machine is designed to eject wood at

chips vO = 7.5 m/s. If the tube is oriented at 30from the horizontal, determine how high, h, the

chips strike the pile if they land on the pile 6 m

from the tube.


115/234

Coordinate System. Three unknown h, time of

flight, tOA and the vertical component of velocity

(vB)y. Taking origin at O, for initial velocity of a chip,

(vA)x= (vO)x= 6.5 m/s and ay= -9.81 m/s2

smv

smv

yO

xO

/75.3)30sin5.7()(

/5.6)30cos5.7()(

Horizontal Motion.


116/234

sttvxx

OA

OAxA

9231.0)( 00

Vertical Motion.

Relating tOA to initial and final elevation of the chips,

mh

tatvyhy OAcOAyOA

38.1

2

1)(1.2 20

EXAMPLE


117/234

The track for this racing event

was designed so that theriders jump off the slope at

30, from a height of 1m.

During the race, it was

observed that the riderremained in mid air for 1.5 s.

Determine the speed at which he was traveling offthe slope, the horizontal distance he travels before

striking the ground, and the maximum height he

attains. Neglect the size of the bike and rider.

Coordinate System Origin is established at point


118/234

Coordinate System. Origin is established at point

A. Three unknown are initial speed vA, range Rand

the vertical component of velocity vB.

Vertical Motion. Since time of flight and the vertical

distance between the ends of the paths are known,

smv

tatvss

A

ABCAByAyAyB

/4.13

2

1)()()( 2


119/234

Horizontal Motion

For maximum height h, we consider path AC

Three unknown are time of flight, tAC, horizontal

distance from A to C and the height h

At maximum height (vC)y = 0

m

R

tvss ABAxAxB

4.17

)5.1(30cos38.130

)()()(


120/234

Curvilinear Motion: Normal and


121/234

Tangential Components

When the path of motion of a particle is known,

describe the path using n and t coordinates which

act normal and tangent to the path

Consider origin located at the particle

Planar Motion

Consider particle P which is moving in a plane

along a fixed curve, such that at a given instant itis at position s, measured from point O



122/234

Consider a coordinate

system that has origin at a

fixed pointon the curve on

the curve, and at the instant,considered this origin

happen to coincide with the

location of the particle

taxis is tangentto the curve at Pand is positive in

the direction ofincreasing s




123/234

Designate this positive position direction with unit

vectorut For normal axis, note that geometrically, the curve

is constructed from series differential arc segments Each segment ds is

formed from the arc of an

associated circle having

a radius of curvature(rho) and center of

curvature O




124/234

Normal axis n is perpendicular to the taxis and is

directed from Ptowards the center of curvature O

Positive direction is always on the concave side of

the curve, designed by un Plane containing both the n and taxes is known

as the oscillating plane and is fixed on the plane of

motion



T i l C


125/234

Velocity.

Since the particle is moving, s is a function of time

Particles velocity v has direction that is always

tangent to the path and a magnitude that isdetermined by taking the time derivative of the path

function s = s(t)

sv

uvv t

where




126/234

Acceleration

Acceleration of the particle is the time rate of

change of velocity

tt uvuvva





127/234

As the particle moves along the arc ds in time dt,

utpreserves its magnitude of unity

When particle changes direction, it becomes ut

ut = ut + dut dut stretches between the arrowhead ofut and

ut, which lie on an infinitesimal arc of radius ut = 1

nnnt uvusuu





128/234

2

va

vdvdsava

uauaa

n

tt

nntt

Magnitude of acceleration is the positive value

of

22nt aaa

where

and





129/234

Consider two special cases of motion

If the particle moves along a straight line, then

and an = 0. Thus , we can

conclude that the tangential component ofacceleration represents the time rate of change in

the magnitude of velocity.

If the particle moves along the curve with aconstant speed, then and

vaa t

0 vat /2vaa n



T ti l C t


130/234

Normal component of acceleration represents the

time rate of change in the direction of the velocity.

Since an always acts towards the center of

curvature, this component is sometimes referred toas the centripetal acceleration

As a result, a particle moving along the curved

path will have accelerations directed as shown





131/234

Three Dimensional Motion

If the particle is moving along a space curve, at a

given instant, taxis is completely unique

An infinite number of straight lines can beconstructed normal to tangent axis at P




132/234

For planar motion,

- choose positive n axis directed from Ptowards

paths center of curvature O

- The above axis also referred as principle normalto curve at P

-utand un are always perpendicular to one another

and lies in the osculating plane





133/234

For spatial motion,

a third unit vectorub, defines a binormalaxis b

which is perpendicular to utand un

Three unit vectors are related by vector crossproduct

ub = utX un un is always on the concave side



134/234




135/234

Velocity

Particles velocityis always tangent to the path

Magnitude of the velocity is found from the

derivative of the path function

Tangential Acceleration

Tangential component of acceleration is the resultof the time rate of change in the magnitude of

velocity

sv





136/234

Tangential component acts in the positive s

direction if the particles speed is increasing and in

the opposite direction if the seed is decreasing

For rectilinear motion,vdvdsava tt

If at is constant,

)()(2

)(

)(2

1

020

2

0

2

00

ssavv

tavv

tatvss

cc

cc

cc





137/234

Normal Acceleration

Normal component of acceleration is the result of

the time rate of change in the direction of the

particles velocity Normal component is always directed towards the

center of curvature of the path along the positive n

axis

For magnitude of the normal component,

2van





138/234

If the path is expressed as y = f(x), the radius of

the curvature at any point on the path is

determined from

22

2/32

/

])/(1[

dxyd

dxdy


EXAMPLE


139/234

When the skier reaches the

point A along the parabolicpath, he has a speed of 6m/s

which is increasing at 2m/s2.

Determine the direction of his

velocity and the direction and

magnitude of this

acceleration at this instant.

Neglect the size of the skierin the calculation.

Coordinate System. Establish the origin of the n,

t t th fi d i t A th th d


140/234

taxes at the fixed pointA on the path and

determine the components ofvand a along theseaxes.

Velocity. The velocity is directed tangent to the

path.1,

201

10

2 xdx

dyxy

v make an angle of = tan-1 = 45 with thexaxis

smvA /6

Acceleration. Determined from t uvuva

)/( 2


141/234

Acceleration. Determined from nt uvuva )/(

mdxyd

dxdy28.28

/

])/(1[22

2/32

The acceleration becomes

2

2

/}273.12{ smuu

uv

uva

nt

ntA


142/234

5.57

327.1

2tan

/37.2237.12

1

222

sma

Thus, 57.5 45 = 12.5

a = 2.37 m/s2

EXAMPLE


143/234

Race car C travels round the horizontal circular

track that has a radius of 90 m. If the car increasesits speed at a constant rate of 2.1 m/s2, starting

from rest, determine the time needed for it to reach

an acceleration of 2.4 m/s2. What is its speed at

this instant?

Coordinate System. The origin of the n and t


144/234

Coordinate System. The origin of the n and t

axes is coincident with the car at the instant. The t

axis is in the direction of the motion, and thepositive n axis is directed toward the center of the

circle.

Acceleration. The magnitude of accelerationcan be related to its components using 22

nt aaa

t

tavv ct

1.2

)(0

222

/049.0 smtv

an Thus,


145/234

n

us,

The time needed for the acceleration to reach

2.4m/s2 is22nt aaa

Solving fort= 4.87 s

Velocity. The speed at time t= 4.87 s is

smtv /2.101.2

EXAMPLE


146/234

The boxes travels alone

the industrial conveyor. If

a box starts from rest at A

and increases its speed

such that at= (0.2t) m/s2

,determine the magnitude

of its acceleration when it

arrives at point B.

Coordinate System. The position of the box at

any instant is defined by s from the fixed point A


147/234

any instant is defined by s, from the fixed pointA.

The acceleration is to be determined at B, so theorigin of the n, taxes is at this point.

Acceleration. Since vA when t = 0

2

00

1.0

2.0

2.0

tv

dttdv

tva

tv

t

(1)

(2)


148/234

The time needed for the box to reach point B can

be determined by realizing that the position ofB issB= 3 + 2(2)/4 = 6.142 m, since sA = 0 when t= 0

st

dttds

tdt

ds

v

B

tB

690.5

1.0

1.0

0

2142.6

0

2

Substituting into eqn (1) and (2),


149/234

smvsmva

B

BtB

/238.3)69.5(1.0/138.1)690.5(2.0)(

2

2

At B, B = 2 m,

22

/242.5)( smv

aB

BnB

2

22

/36.5

)242.5()138.1(

sm

aB

Curvilinear Motion: Cylindrical

Components


150/234

Components

If motion is restricted to the plane, polar

coordinates rand are used

Polar Coordinates Specify the location ofPusing both the radial

coordinater, which extends outward from the fixed

origin O to the particle and a traverse coordinate,

which is the counterclockwise angle between afixed reference line and the raxis


Components


151/234

Angle usually measured in

degrees or radians, where 1 rad

= 180

Positive directions of the randcoordinates are defined by

the unit vectors urand u urextends from Palong

increasing r, when is heldfixed

Components


Components


152/234

uextends from Pin the direction that occurs

when ris held fixed and is increased

Note these directions are perpendicular to each

otherPosition

At any instant, position of the particle defined by

the position vector

rurr

Components


Components


153/234

Velocity

Instantaneous velocity v is obtained by the time

derivative ofr

rr ururrv

To evaluate , note that urchanges only its

direction with respect to time since magnitude of

this vector = 1During time t, a change rwill not cause a

change in the direction ofu

ru

Components


Components


154/234

However, a change will cause ur to become ur

where

ur = ur+ ur

Time change is urFor small angles , vector has a magnitude of 1

and acts in the udirection

uu

utt

u

u

r

t

r

tr

00 limlim

Components


Components


155/234

rv

rv

uvuvv

r

rr

where

Radical componentvris a measure of the rate of

increase or decrease in the length of the radial

coordinateTransverse componentv is the rate of motion

along the circumference of a circle having a radius

r

Components


Components


156/234

Angular velocity indicates the rate of

change of the angle

Sincevrandvare mutually perpendicular, the

magnitude of the velocity or speed is simply thepositive value of

Direction ofv is tangentto the path at P

dtd /

22 rrv

Components


Components


157/234

Acceleration Taking the time derivatives, for the instant

acceleration,

urururururvarr

During the time t, a change r will not change

the direction ualthough a change in will cause

u to become u For small angles, this vector has a magnitude =

1 and acts in theurdirection

u= - ur

p


Components


158/234

rra

rra

uauaa

r

rr

2

2

where

The term is called the angular

acceleration since it measures the change made

in the angular velocity during an instant of time Use unit rad/s2

22 / dtd

Components


Components


159/234

Since arand aare always perpendicular, themagnitude of the acceleration is simply the positive

value of

222

2

rrrra Direction is determined from the vector addition

of its components

Acceleration is not tangent to the path

p


Components


160/234

Cylindrical Coordinates

If the particle Pmoves along

a space, then its location may

be specified by the threecylindrical coordinates r, , z

zcoordinate is similar to that

used for rectangular

coordinates

Components


Components


161/234

Since the unit vector defining its direction, uz, is

constant, the time derivatives of this vector are

zero

Position, velocity, acceleration of the particle canbe written in cylindrical coordinates as shown

zr

zr

zrp

uzurrurra

uzururv

uzurr

)2()( 2

Components


Components


162/234

Time Derivatives

Two types of problems usually occur

1) If the coordinates are specified as time

parametric equations, r = r(t) and = (t), thenthe time derivative can be formed directly.

2) If the time parametric equations are not given, it

is necessary to specify the path r = f() and findthe relationship between the time derivatives

using the chain rule of calculus.

Components


Components


163/234


Coordinate System

Polar coordinate are used to solve problem

involving angular motion of the radial coordinate r,used to describe the particles motion

To use polar coordinates, the origin is established

at a fixed point and the radial line r is directed to

the particle The transverse coordinate is measured from a

fixed reference line to radial line

p


Components


164/234

Velocity and Acceleration

Once r and the four time derivatives have

been evaluated at the instant considered, their

values can be used to obtain the radial andtransverse components ofv and a

Use chain rule of calculus to find the time

derivatives ofr = f()

Motion in 3D requires a simple extension of theabove procedure to find

,,,rr

p

Th t k i t f h i th t i

EXAMPLE


165/234

The amusement park consists of a chair that is

rotating in a horizontal circular path of radius rsuchthat the arm OB has an angular velocity and

angular acceleration. Determine the radial and

transverse components of velocity and acceleration

of the passenger.

Coordinate System Since the angular motion of


166/234

Coordinate System. Since the angular motion of

the arm is reported, polar coordinates are chosenfor the solution. is not related to r, since radius is

constant for all .

Velocity and Acceleration. Since ris constant,

00 rrrr

rvr 0


167/234

rrra

rrrarv

r

r

2

22

This special case of circular motion happen to be

collinear with rand axes

Th d OA i i i h h i l l h

EXAMPLE


168/234

The rod OA is rotating in the horizontal plane such

that = (t3) rad. At the same time, the collarB issliding outwards along OA so that r= (100t2)mm. If

in both cases, tis in seconds, determine the

velocity and acceleration of the collar when t= 1s.

Coordinate System. Since time-parametric

equations of the particle is given it is not


169/234

equations of the particle is given, it is not

necessary to relate rto .

Velocity and Acceleration.

22

2

31

2

/66/200200

/23/200200

73.51100100

11

11

1

sradtsmmr

sradtsmmtr

radtmmtr

stst

stst

stst

ururv r


170/234

smmuur /}300200{

The magnitude ofv is

1143.57

3.56200

300tan

/361300200

1

22

smmv

2

2

/}1800700{

)2()(

smmuu

urrurra r


171/234

/}1800700{ smmuur

The magnitude ofa is

1693.57)180(

7.68700

1800tan

/19301800700

1

222

smma

The searchlight casts a spot of light along the face

EXAMPLE


172/234

of a wall that is located 100m from the searchlight.

Determine the magnitudes of the velocity and

acceleration at which the spot appears to travel

across the wall at the instant = 45. The

searchlight is rotating at a constant rate of 4 rad/s

Coordinate System. Polar coordinates will be

used since the angular rate of the searchlight is


173/234

g g

given. To find the time derivatives, it is necessary to

relate rto .

r= 100/cos = 100sec


)tan(sec100sec100tansec100

)tan(sec100

2322

r

r

Since = 4 rad/s = constant, = 0, when = 45,


174/234

2.6788

7.5654.141

r

rr

smv

smuu

ururv

r

r

/800

/}7.5657.565{

2

2 )2()( urrurra r


175/234

2

2

/6400

/}5.45255.4525{

smma

smmuur

Due to the rotation of the

EXAMPLE


176/234

Due to the rotation of the

forked rod, ballA travelsacross the slotted path, a

portion of which is in the shape

of a cardioids, r= 0.15(1 cos

)m where is in radians. Ifthe balls velocity is v = 1.2m/s

and its acceleration is 9m/s2 at

instant = 180, determine the

angular velocity and angularacceleration of the fork.

Coordinate System. For this unusual path, use


177/234

Coordinate System. For this unusual path, use

polar coordinates.


)(sin15.0)()(cos15.0

)(sin15.0

)cos1(15.0

r

r

r

Evaluating these results at = 180

215.003.0 rrmr

Since v = 1.2 m/s


178/234

srad

rrv/4

22

2

222

/18

)2()(

srad

rrrra

Absolute Dependent MotionAnalysis of Two Particles


179/234

Motion of one particle will dependon the

corresponding motion of another particle

Dependency occur when particles are

interconnected by the inextensible cords which arewrapped around pulleys

For example, the movement of blockA downward

along the inclined plane will cause a corresponding

movement of block B up the other incline Specify the locations of the blocks usingposition

coordinate sAand sB



180/234

Note each of the coordinate axes is (1)

referenced from a fixedpoint (O) or fixed datum

line, (2) measured along each inclined plane in the

direction of motion of block A and block B and (3)has a positive sense from C to A and D to B

If total cord length is lT, the position coordinate are

elated by the equation

TBCDA lsls

Absolute Dependent Motion

Analysis of Two Particles


181/234

Here lCD is the length passing over arc CD

Taking time derivative of this expression, realizing

that lCD and lTremain constant, while sA and sB

measure the lengths of the changing segments ofthe cord

ABBA vv

dt

ds

dt

ds 0 or

The negative sign indicates that when blockA hasa velocity downward in the direction of position sA, it

causes a corresponding upward velocity of block B;

B moving in the negative sB direction

y



182/234

Time differentiation of the velocities yields the

relation between accelerations

aB = - aA

For example involvingdependent motion of two blocks

Position of blockA is specified

by sA, and the position of the end

of the cord which block B issuspended is defined by sB




183/234

Chose coordinate axes which are (1) referencedfrom fixed points and datums, (2) measured in the

direction of motion of each block, (3) positive to the

right (sA

) and positive downward (sB)

During the motion, the red colored segments of

the cord remain constant

Iflrepresents the total length of the cord minus

these segments, then the position coordinates canbe related by

lshs AB 22

y



184/234

Since land h are constant during the motion, thetwo time derivatives yields

ABAB aavv 22

When B moves downward(+sB),A moves to left (-sA) with

two times the motion

This example can also be

worked by defining the position

of block B from the center of the

bottom pulley ( a fixed point)




185/234

ABAB

AB

aavv

lshsh

22

)(2

Time differentiation yields

y



186/234

PROCEDURE FOR ANALYSISPosition-Coordinate Equation

Establish position coordinates which have their

origin located at a fixedpoint or datum

The coordinates are directed along the path ofmotion and extend to a point having the same

motion as each of the particles

It is not necessarythat the origin be the same for

each of the coordinates; however, it is important

that each coordinate axis selected be directed

along thepath of motion of the particle


187/234




188/234

Time Derivatives

Two successive time derivatives of the position-

coordinates equations yield the required velocity

and acceleration equations which relate motions of

the particles

The signs of the terms in these equations will be

consistent with those that specify the positive andnegative sense of the position coordinates

y

EXAMPLE

Determine the speed of blockA if block B has an


189/234

upward speed of 2 m/s.

Position Coordinate System. There is one cordin

this system having segments which are changing


190/234

this system having segments which are changing

length. Position coordinates sAand sB will be usedsince each is measured from a fixed point (CorD)

and extends along each blockspath of motion. In

particular, sBis directed to point Esince motion of

B and Eis the same. The red colored segments of

the cords remain at a constant length and do not

have to be considered as the block move.

The remaining length of the cord, l, is also

considered and is related to the changing position

di t d b th ti


191/234

coordinates sA

and sB

by the equation

lss BA 3

Time Derivative. Taking the time derivativeyields 03 BA vv

so that when vB = -2m/s (upward)vA= 6m/s

Determine the speed of blockA if block B has an

EXAMPLE


192/234

p

upward speed of 2m/s.

Position Coordinate Equation. Positions ofA


193/234

and B are defined using coordinates sA and sB.Since the system has two cords which change

length, it is necessary to use a third coordinate sC

in order to relate sAto sB. Length of the cords can

be expressed in terms ofsA and sC, and the lengthof the other cord can be expressed in terms ofsB

and sC. The red colored segments are not

considered in this analysis.

For the remaining cord length,

21 )(2 lssslss CBBCA


194/234

21 )(2 lssslss CBBCA

Eliminating sC yields,

1224 llss BA

Time Derivative. The time derivative gives

04 BA vv

so that vB= -2m/s (upward)

smsmvB /8/8

Determine the speed with which block B rises if the

d f th d t A i ll d d ith d f

EXAMPLE


195/234

end of the cord atA is pulled down with a speed of

2m/s.


196/234

Excluding the red colored segments of the cords,

the remaining constant cord lengths l1 and l2(along


197/234

the hook and link dimensions) can be expressed as

12 24 llss BC

2

1

lsssss

lss

BCBCA

BC

Eliminating sCyields

Time Derivative. The time derivative give


198/234

smsmv

vv

B

BA

/5.0/5.0

04

when vA = 2m/s (downward)

A man atA s hoisting a safe

EXAMPLE


199/234

g

S by walking to the rightwith a constant velocity vA =

0.5m/s. Determine the

velocity and acceleration of

the safe when it reaches theelevation at E. The rope is

30m long and passes over

a small pulley at D.


200/234

Using Pythagorean Theorem,


201/234

yIxI CDDA 151522

15225

151530

2

22

xy

yx

lll CDDA

(1)

Time Derivative. Taking time derivative,

using the chain rule where, vS = dy/dtand vA =

dx/dt


202/234

dx/dt

A

S

vx

x

dt

dx

x

x

dt

dyv

2

2

225

225

2

2

1

At y= 10 m,x= 20 m, vA

= 0.5 m/s, vS=

400mm/s

(2)

The acceleration is determined by taking the time

derivative of eqn (2),


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2/32

22

22/322

2

225

225

225

1

225

1

)225(

)/(

x

vdt

dvxx

vdt

dx

xxv

x

dtdxx

dt

yda

AA

AAS

Atx= 20 m, with vA = 0.5 m/s,

2/6.3 smmaS

Relative Motion Analysis of TwoParticles Using Translating Axes


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There are many cases where the path of themotion for a particle is complicated, so that it may

be feasible to analyze the motions in parts by using

two or more frames of reference

For example, motion of a particle located at the tipof an airplane propeller while the plane is in flight,

is more easily described if one observes first the

motion of the airplane from a fixed reference and

then superimposes (vectorially) the circular motionof the particle measured from a reference attached

to the airplane



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Position. Consider particleA and B,

which moves along the

arbitrary paths aa and bb,

respectively

The absolute position of

each particle rA and rB, is

measured from the commonorigin O of the fixedx, y, z

reference frame

Relative Motion Analysis of Two

Particles Using Translating Axes


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Origin of the second frame of referencex, yandz is attached to and moves with particleA

Axes of this frame onlypermitted to translate

relative to fixed frame

Relative position of B with respect toA is

designated by a relative-position vectorrB/A Using vector addition

ABAB rrr /


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Since thex, yand zaxes translate,the components ofrB/A will notchange

direction and therefore time derivative

o this vector components will only

have to account for the change in the

vector magnitude

Velocity ofB is equal to the velocity

ofA plus (vectorially) the relativevelocity of B relative toA as

measured by the translating observer

fixed in thex, yand zreference



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Acceleration. The time derivative yields a similar relationship

between the absolute and relative accelerations of

the particles A and B

Here aB/A is the acceleration ofB as seen by the

observer located atA and translating with thex, y

and zreference frame

ABAB aaa /



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PROCEDURE FOR ANALYSIS When applying the relative position equations, rB

= rA + rB/A, it is necessary to specify the location of

the fixedx, y , zand translatingx, yand z

Usually, the originA of the translating axes is

located at a point having a known positionrA A graphical representation of the vector addition

can be shown, and both the known and unknownquantities labeled on this sketch




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Since vector addition forms a triangle, there can

be at most two unknowns, represented by the

magnitudes and/or directions of the vector

quantities These unknown can be solved for either

graphically, using trigonometry, or resolving each of

the three vectors rA, rB and rB/A into rectangular or

Cartesian components, thereby generating a set ofscalar equations




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The relative motion equations vB = vA + vB/A andaB = aA + aB/A are applied in the same manner as

explained above, except in this case, origin O of

the fixed axesx, y, zaxes does not have to be

specified

A train, traveling at a constant speed of 90km/h,

EXAMPLE


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crosses over a road. If automobile A is traveling t67.5km/h along the road, determine the magnitude

and direction of relative velocity of the train with

respect to the automobile

Vector Analysis. The relative velocity is

measured from the translatingx, yaxes attached

to the automobile Since v and v are known in


214/234

to the automobile. Since vT

and vA

are known in

both magnitude and direction, the unknowns

become thexand ycomponents ofvT/A. Using the

x, yaxes and a Cartesian vector analysis,

hkmjiv

vjii

vvv

AT

AT

ATAT

/)~

7.47~

3.42{

)~

45sin5.67~

45cos5.67(~

90

/

/

/

The magnitude ofvT/A is

222


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hkmv AT /8.63)7.473.42(

222

/

The direction ofvT/Adefined from the xaxis is

40.48

3.42

7.47tan/

/

xAT

yAT

v

v

PlaneA is flying along a straight-line path, while

plane B is flying along a circular path having a

EXAMPLE


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plane B is flying along a circular path having a

radius of curvature ofB = 400 km. Determine the

velocity and acceleration ofB as measured by the

pilot ofA.

Velocity. Thex, yaxes are located at an

arbitrary fixed point. Since the motion relative to

plane A is to be determined, the translating frame


217/234

of reference x. yis attached to it. Applying therelative-velocity equation in scalar form since the

velocity vectors of both plane are parallel at the

instant shown,

hkmhkmv

v

vvv

AB

AB

ABAB

/100/100

700600

/

/

/)(

Acceleration. Plane B has both tangential

and normal components of acceleration, since it is

flying along a curved path. Magnitude of normal


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acceleration,

22

/900 hkmv

a BnB

Applying the relative-acceleration equation,

2/

/

/

/~

150~

900

~50

~100

~900

hkmjia

ajji

aaa

AB

AB

ABAB


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From the figure shown, the magnitude and directionof ABa /

46.9

900

150tan/912 12/

hkma AB

At the instant, carA and B

EXAMPLE


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are traveling with the speedof 18 m/s and 12 m/s

respectively. Also at this

instant, A has a decrease in

speed of 2 m/s2, and B hasan increase in speed of 3

m/s2. Determine the velocity

and acceleration ofB with

respect to A.

Velocity. The fixedx, yaxes are established at a

point on the ground and the translatingx, yaxes


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are attached to carA. Using Cartesian vectoranalysis,

smv

smjivvjij

vvv

AB

AB

AB

ABAB

/69.9588.39

/~

588.3~

9

~

60sin18

~

60cos18

~

12

22/

/

/

/

Thus,

Its direction is

588.3/ yABv


222/234

7.21

9tan /

xAB

y

v

Acceleration. The magnitude of the normal

component is

22

/440.1 smv

a B

nB

Applying the equation for relative acceleration

yields


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2

/

/

/

/

~

732.4

~

440.2

~60sin2

~60cos2

~3

~440.1

smjia

ajiji

aaa

AB

AB

ABAB

Magnitude and direction is

7.62

/32.5 2/

sma AB

Chapter Review

Rectilinear Kinematics. Rectilinear


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kinematics refers to motion along a straight line. Aposition coordinate s specifies the location of the

particle on the line and the displacement s is the

change in this position.

The average velocity is a vector quantity, defined

as the displacement divided by the time interval.

trvavg

This is the different than the average speed, which

Chapter Review


225/234

is a scalar and is the total distance traveled dividedby the time of travel.

t

sv T

avgsp

The time, position, instantaneous velocity and

instantaneous acceleration are related by the

differential equations

dvvdsadtdvadtdsv //

Chapter Review


226/234

If the acceleration is known to be constant, then theintegration of these equations yields

tavv c 02

0021 tatvss c

020

2 2 ssavv c

Graphical Solutions. If the motion is erratic, then

Chapter Review


227/234

it can be described by a graph. If one of thesegraphs is given, then the others can be established

using the differential relations, v = ds/dt, a = dv/dt,

or a ds = v dv.

Curvilinear Motion, x, y, z. For this case, motion

along the path is resolved into rectilinear motion

along the x, y, z axes. The equation of the path is

used to relate the motion along each axis.

Projectile Motion. Free flight motion of a

Chapter Review


228/234

projectile follows a parabolic path. It has a constantvelocity in the horizontal direction and constant

acceleration of g = 9.81 m/s2 in the vertical

direction. Any two of the three equations for

constant acceleration apply in the vertical direction,and in the horizontal direction only

tvxx x)( 00

Curvilinear Motion n, t. If normal and tangential

axes are used for the analysis, then v is always in

Chapter Review


229/234

a es a e used o e a a ys s, e s a ays

the positive tdirection. The acceleration has two

components. The tangential components, at,

accounts for the change in the magnitude of the

velocity; a slowing down is in the negative tdirection,and a speeding up is in the positive tdirection. The

normal component, an accounts for the change in the

direction of velocity. The component is always in the

positive n direction.

Curvilinear Motion r, , z. If the path of motion is

expressed in polar coordinates, then the velocity

Chapter Review


230/234

p p , y

and acceleration components can be written as

rrarv

rrarv rr

2

2

To apply these equations, it is necessary to determine

at the instant considered. If the path r

= f() is given, then the chain rule of calculus must be

used to obtain the time derivatives.

,,,, rrr

Once the data is substituted into the equations,

then the algebraic sign of the results will indicate

Chapter Review


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g g

the direction of components ofvora along each

axis.

Absolute Dependent Motion of Two Particles.

The dependent motion of blocks that aresuspended from pulleys and cables can be related

by the geometry of the system.

This is done by first establishing position

Chapter Review


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y g

coordinates, measured from a fixed origin to each

block so that they are directed along the line of

motion of the blocks. Using geometry and/or

trigonometry, the coordinates are then related tothe cable length in order to formulate a position

coordinate equation. The first time derivative of this

equation gives a relationship between the velocities

of the blocks, and a second time derivative givesthe relationship between their accelerations.

Relative Motion Analysis Using Translating Axes.

Chapter Review


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If two particlesA and B undergo independentmotions, then these motions can be related to their

relative motion. Using a translating set axes

attached to one of the particles (A), the velocity and

acceleration equations become

ABAB

ABAB

aaa

vvv

/

/

Chapter Review


234/234

For planar motion, each of these equationsproduces two scalar equations, one in thex, and

the other in the ydirection. For solution, the vectors

can be expressed in Cartesian form or thexand y

scalar components can be written directly.

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