Chapter II [Compatibility Mode]

7/30/2019 Chapter II [Compatibility Mode]

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Chapter TwoKinematics of particles


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Introduction

Kinematics: is the branch of dynamics which describes

the motion of bodies without reference to theforces that either causes the motion or are

generated as a result of the motion.

Kinematics is often referred to as the geometry of

motion


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Examples of kinematics problems that engage the

attention of engineers.

The design of cams, gears, linkages, and othermachine elements to control or produce certain

desired motions, and

The calculation of flight trajectory for aircraft,rockets and spacecraft.


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If the particle is confined to a specified path, as with a beadsliding along a fixed wire, its motion is said to be Constrained.

Example 1. - A small rock tied to the end of a string and whirled in a

circle undergoes constrained motion until the string

breaks.


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If there are no physical guides, the motion is

said to be unconstrained.

Example 2 - Airplane, rocket


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The position of a particle P at any time t can be

described by specifying its:

- Rectangular coordinates; X,Y,Z- Cylindrical coordinates; r,,z

- Spherical coordinates; R,,

- Also described by measurements along the tangentt and normal n to the curve(path variable).


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The motion of particles(or rigid bodies) may be

described by using coordinates measured fromfixed reference axis (absolute motion analysis)

or by using coordinates measured from moving

reference axis (relative motion analysis).


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Rectilinear motion

Is a motion in which a particle moves along a

straight line(one-dimensional motion).

Consider a particle P moving along a straight line.


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Average velocity: for the time interval t, it is defined as the ratioof the displacement s to the time interval t.

2.1

As t becomes smaller and approaches zero in the limit, theaverage velocity approaches the instantaneous velocity of the

particle.

2.2

t

s=Vav

S

dt

ds

t

s

tt 0

av

0

limVlimV


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Average acceleration

For the time interval t, it is defined as the ratio of the

change in velocity v to the time interval t.

2.3

Instantaneous acceleration

2.4(a)

2.4(b)

t

va av

v

dt

dv

t

v

t 0lima

sdt

sd

dt

ds

dt

d

dt

dv

t

v

t

2

2

0lima


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Note:-The acceleration is positive or negativedepending on whether the velocity increasing or

decreasing.

Considering equation 2.2 and 2.4(a) , we have

dsssds

s

sd

s

dsdt

adsvdv

a

dv

v

dsdt


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General representation of Relationshipamong s, v, a & t.


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1. Graph of s vs. t

By constructing tangent to the curve at any time t, we

obtain the slope, which is the velocity v = ds/dt


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2. Graph of v vs. t

The slope dv/dt of the v-t curve at any instant gives the

acceleration at that instant.

The area under the v-t curve during time dt is vdt which is

the displacement ds.


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The area under the v-t curve is the net

displacement of the particle during the

interval from t1 to t2.

(area under v-t curve )

2

1

2

1

s

s

t

t

vdtds

22 ss


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3. Graph of a vs. t

The area under the a-t curve during time dt is the net change in velocity of

the particle between t1 and t2.

v2 - v1 = (area under a-t curve)

2

1

2

1

v

v

t

t

adtdv


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4. Graph of vs. s

The net area under the curve b/n position coordinates s1

and s2 is

(areas under a-s curve)

2

1

2

1

v

v

s

s

adsvdv

)(2

1 21

2

2 vv


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5. Graph of v vs. s

dsCBvdvv

CB

ds

dv

v

CBdsdv

1

tan


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The graphical representations described are useful for:- visualizing the relationships among the several motion quantities.

approximating results by graphical integration or differentiation when a

lack of knowledge of the mathematical relationship prevents its

expression as an explicit mathematical function .

experimental data and motions that involve discontinuous relationship

b/n variables.


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Methods for determining the

velocity and displacementfunctions


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a) Constant acceleration, (a=const.)

- boundary conditions

at t=0 , s=s0 and v=v0

using integrating

dvadtdt

dva

atvv

atvvadtdv

o

t

o

v

v o

0


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These relations are necessarily restrictedto the special case where the acceleration

is constant.

The integration limits depend on the initial

and final conditions and for a given

problem may be different from those usedhere.


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b) Acceleration given as a function of time, a=f(t)

tttx

x

tttv

v

dttvxtxdttvdxdttvdxtv

dt

dx

dttfvtvdttfdvdttfdvtfadt

dv

0

0

0

0

0

0

0

0

c) Acceleration given as a function of position, a = f(x)

x

x

x

x

v

v

dxxfvvdxxfdvvdxxfdvv

xfdx

dvva

dt

dva

v

dxdt

dt

dxv

000

2

0212

21

oror


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d) Acceleration given as a function of velocity, a = f(v)

v

v

v

v

x

x

v

v

tv

v

vfdvvxx

vf

dvvdx

vf

dvvdxvfa

dx

dvv

tvf

dv

dtvf

dvdt

vf

dvvfa

dt

dv

0

00

0

0

0

0


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Example 1 Consider a particle moving in a straight line, and assuming

that its position is defined by the equation

Where, t is express in seconds and s is in meters.

Determine the velocity and acceleration of the particles at

any time t

326 tts


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Example 2 The acceleration of a particle is given by ,

where a is in meters per secondsquared and t is in seconds. Determine the

velocity and displacement as function time. The

initial displacement at t=0 is so=-5m, and the

initial velocity is vo=30m/s.

304 ta


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Example 3 The position of a particle which moves along a straight

line is defined by the relation

,where x is expressed in m and t insecond.

Determine:

a) The time at which the velocity will be zero.

b) The position and distance traveled by the particle atthat time.

c) The acceleration of the particle at that time.d) The distance traveled by the particle between 4s and

6s.

40156 23 tttx


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Example 4A particle moves in a straight line with velocity

shown in the figure. Knowing that x=-12m at

t=0


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Draw the a-tandx-tgraphs, and Determine:

a) The total distance traveled by the particle whent=12s.

b) The two values of t for which the particle passesthe origin.

c) The max. value of the position coordinate of theparticle.

d) The value of t for which the particle is at adistance of 15m from the origin.


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Example 5 the rocket car starts from rest and

subjected to a constant acceleration of

until t=15sec. The brakes are thenapplied which causes a decelerated at arate shown in the figure until the car

stops. Determine the max. speed of thecar and the time when the car stops.

26m/sa


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Example 6 A motorcycle patrolman starts from rest at A two seconds

after a car, speeding at the constant rate of 120km/h,

passes point A. if the patrolman accelerate at the rate of

6m/s2 until he reaches his maximum permissible speed of

150km/h, which he maintains, calculate the distance s

from point A to the point at which he overtakes the car.


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Example 7 The preliminary design for a rapid transient system calls

for the train velocity to vary with time as shown in the

plot as the train runs the 3.2km between stations A and B. The slopes of the cubic transition curves(which are of

form a+bt+ct2+dt3) are zero at the end points.

Determine the total run time t between the stations and

the maximum acceleration.


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Plane curvilinear motion


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Curvilinear motion of a particle

When a particle moves along a curve otherthan a straight line, we say that the particle

is in curvilinear motion.


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Plane curvilinear motion The analysis of motion of a particle along a

curved path that lies on a single plane.


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Consider the continuous motion of a particle along a

plane curve.

- At time t, the particle is at position P, which is

located by the position vector r measured fromsome convenient fixed origin O.


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- At time , the particle is at P located by theposition vector .

- The vector r j o in ing p and p rep resen ts the

change in t he pos it i on vec to r du r ing t he t im e

in te rva lt(displacement) .

tt rr


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The distance traveled by the particle as it

moves along the path from P to P is the scalar

length s measured along the path.

The displacement of the particle represents

the vector change of position and is clearly

independent of the choice of origin.


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Note: there is a clear distinction between themagnitude of the derivative and the

derivative of the magnitude.

- The magnitude of the derivative.

- The derivative of the magnitude

speedvvrdt

rd

rdt

dr

dt

rd


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Let us draw both vectors v and v from the sameorigin o. The vector v joining Q and Qrepresents the change in the velocity of theparticle during the time interval t.

v=v+v


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Note: The direction of the acceleration of aparticle in curvilinear motion is neithertangent to the path nor normal to thepath.


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Suppose we take the set of velocity vectors and traceout a continuous curve; such a curve is called ahodograph.

The acceleration vector is tangent to the hodograph,but this does not produce vectors tangent to the pathof the particle.


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Rectangular co-ordinates (x-y-z)

This is particularly useful for describing

motions where the x,y and z-components ofacceleration are independently generated.


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When the position of a particle P is defined atany instant by its rectangular coordinate x,y

and z, it is convenient to resolve the velocityv and the acceleration a of the particle intorectangular components.


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Resolving the position vector r of the particleinto rectangular components,

r=xi+yj+zk

Differentiating)( kzjyix

dt

d

dt

rdv

kzjyixv


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Similarly,

)( kvjvivdt

d

dt

vda zyx

kdt

dvj

dt

dvi

dt

dv zyx

kvjviv zyx

kzjyix

The magnitude of the acceleration vector is:222zyx aaaa


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From the above equations that the scalarcomponents of the velocity and acceleration

are

xa

xv

x

x

ya

yv

y

y

za

zv

z

z


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The use of rectangular components todescribe the position, the velocity and the

acceleration of a particle is particularlyeffective when the component ax of theacceleration depends only upon t,x and/or vx,similarly for ay and az.


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The motion of the particle in the x direction,its motion in they direction, and its motion in

the z direction can be considered separately.


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Projectile motion

An important application of two dimensional

kinematic theory is the problem of projectilemotion.


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Assumptions Neglect the aerodynamic drag, the earth

curvature and rotation, The altitude range is so small enough so

that the acceleration due to gravity can be

considered constant, therefore;


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Rectangular coordinates are useful for thetrajectory analysis.

In the case of the motion of a projectile, itcan be shown that the components of the

acceleration are

0

xa x gya y

0

za z


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Boundary conditionsat t=0 ; x=x0 ,y=y0; vx=vxo and vy=vy0

Position

Velocitytvzz

gttvyy

tvxx

ozo

y

x

2

00

00

2

1

)(222

0

0

oyoy

zoz

yy

xx

yygvv

vzv

gtvyv

vxv

In all these expressions,the subscript zerodenotes initial conditions


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But for two dimensional motion of the projectile,

2

00

00

2

1gttvyy

tvxx

y

x

)(222

0

0

oyoy

yy

xx

yygvv

gtvyv

vxv


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Example

A projectile is fired from the edge of a 150m cliff with an initial

velocity of 180m/s at angle of 300 with the horizontal. Neglect

air resistance, find

a) the horizontal distance from gun to the point where the

projectile strikes the ground.

b) the greatest elevation above the ground reached by the

projectile.


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ExampleA projectile is launched from point A with the

initial conditions shown in the figure.

Determine the slant distance s that locates thepoint B of impactand calculate the time of

flight.


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Example

The muzzle velocity of a long-range rifle at A is

u=400m/s. Determine the two angles of

elevation that will permit the projectile tohit the mountain target B.


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Curvilinear motionNormal and tangential coordinates


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Normal and tangential coordinate

When a particle moves along a curved path, itis sometimes convenient to describe its

motion using coordinates other thanCartesian.

When the path of motion is known, normal (n)and tangential (t) coordinates are often used.


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They are path variables, which aremeasurements made along the tangent t andnormal n to the path of the particle.

The coordinates are considered to move alongthe path with the particle.

In the n-t coordinate system, the origin islocated on the particle (the origin moves with

the particle).


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The t-axis is tangent to the path (curve) atthe instant considered, positive in thedirection of the particles motion.

The n-axis is perpendicular to the t-axis withthe positive direction toward the center ofcurvature of the curve.


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The coordinate n and t will now be used todescribe the velocity v and acceleration a.

Similarly to the unit vectors i and

j introduced for rectangularcoordinate system, unit vectorsfor t-n coordinate system can beused.

For this purpose we introduce

unit vectoret in the t-directionen in the n-direction.


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et - directed toward the direction ofmotion.

en-directed toward the center of curvatureof the path.


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During the differentialincrement of time dt, theparticle moves a differentialdistance ds along the curve

from A to A. With the radius of curvature

of the path at this positiondesignated by , we see that

ds =d


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velocity The magnitude of the velocity is:-

Since it is unnecessary to consider the

differential change in between A and A,

dt

d

dt

d

dt

dsvv

)1....(....................tt eevv


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Acceleration The acceleration a of the particle was defined

by:

Now differentiate the velocity by applying theordinary rule (chain rule) for the differentiation ofthe product of a scalar and a vector.

tvedt

d

dt

dva

tt

ttt

eveva

dt

deve

dt

dvve

dt

d

dt

dva


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Where the unit vector et now has a derivativebecause its direction changes.

. . . . . . . . . . . (1)dtdeve

dtdva tt


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To find the derivative of consider thefollowing figure

Using vector addition

et

= et

+ et

Since the magnitude

| et |= | et | = 1

dtde t


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The magnitude of et| et |= 2 sin /2

Dividing both sides by

As 0, is tangent to the path;i.e,

perpendicular to et .

2sin2te

te


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Taking the limit as 0

The vector obtained in the limit is a unitvector along the normal to the path of theparticle.

1

2

2sin

limlim00

te

1lim 0

d

deett


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But

Dividing both sides by dtBut d = ds/

Then

ntt

n

nt

edded

dee

edde

.

.1

nt e

dt

d

dt

de.

ntnt ev

dt

dee

dt

ds

dt

de

.1


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Equation (1) becomes

We can write

where, and

tntt e

dt

dve

vae

dt

dv

dt

deva ..

2

ttnneaeaa

22

va n

va t

22tn aaaa


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Note: an is always directed towards the center of

curvature of the path.

at is directed towards the positive t-direction of

the motion if the speed v is increasing andtowards the negative t-direction if the speed v isdecreasing.

At the inflection point in the curve, the normalacceleration, goes to zero since becomesinfinity.

2v


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Special case of motion Circular motion

but =r and

2v

a n

rv

2

ra n

nt

t

t

erera

ra

dt

drr

dt

d

dt

dva

2


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The particle moves along a path expressed asy = f(x). The radius of curvature, , at anypoint on the path can be calculated from

2

2

23

2)(1

dxyd

dx

dy

xy


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APPLICATIONSCars traveling along a

clover-leaf interchange

experience an acceleration

due to a change in speed as

well as due to a change in

direction of the velocity.


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Example 1 Starting from rest, a motorboat

travels around a circular path of

r = 50 m at a speed that increaseswith time, v = (0.2 t2) m/s.

Find the magnitudes of the boats

velocity and acceleration at the

instant t = 3 s.


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Example 2 A jet plane travels along a vertical

parabolic path defined by the equation

y = 0. 4x2. At point A, the jet has aspeed of 200 m/s, which is increasing at

the rate of 0. 8 m/s2. Find the magnitude

of the planes acceleration when it is atpoint A.


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Example 3

A race traveling at a speed of 250km/h on the

straightway applies his brakes at point A and

reduce his speed at a uniform rate to 200km/h at

C in a distance of 300m.

Calculate the magnitude of the total acceleration

of the race car an instant after it passes point B.


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Example 4 The motion of pin A in the fixed circular slot

is controlled by a guide B, which is being

elevated by its lead screw with a constantupward velocity vo=2m/s for the interval ofits motion.

Calculate both the normal and tangential

components of acceleration of pin A as itpasses the position for which .


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Curvilinear motionPolar coordinate system (r- )


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Polar coordinate(r- ) The third description for plane curvilinear

motion.

Where the particle is located by the radial

distance r from a fixed pole and by an angular

measurement to the radial line.


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Polar coordinates are particularly useful when a

motion is constrained through the control of a

radial distance and an angular position,

or when an unconstrained motion is observed by

measurements of a radial distance and an angular

position.


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An arbitrary fixed line,such as the x-axis, is

used as a reference for

the measurement

. Unit vectors er and e

are established in the

positive r and

directions, respectively.


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The position vector to the particle at A has a

magnitude equal to the radial distance r and a

direction specified by the unit vector er.

We express the location of the particle at A by

the vector

rer.r

r


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Velocity The velocity is obtained by differentiating the vector r.

Where the unit vector er now has a derivative because

its direction changes.

We obtain the derivation in exactly the same way thatwe derived for et.

rr

rr

r

ererv

dt

edre

dt

dr

dt

edr

dt

rdv

..

.


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To find the derivative of consider thefollowing figure

Using vector addition

er = er + ere

= e

+ e

Since the magnitude

|er| = |er| = |e|= |e| = 1

dtde r


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The magnitude of er and e| er|= |e| =2 sin /2

Dividing both sides by

As 0, is perpendicular to er .

2sin2ee r

re


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Note: As

0,

1. is directed towards

the positive e

direction.

2. is directed towards

the negative er direction.

Then,

re

e

1

2

2sinlimlimlim000

ee r


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Therefore;

1lim

1lim

0

0

dee

dee rr

rrr

rr

eddeeed

de

eddeeed

de

..1

..1


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Dividing both sides by dt, we have

Therefore the velocity equation becomes;

rr

rr

e

dt

ede

dt

d

dt

ed

edt

ede

dt

d

dt

ed

.

.

ererdt

edre

dt

drv r

rr

..


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Where and

rv r

.rv

r

r

v

v

vvv

1

22

tan


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The r-component of v is merely the rate at which

the vector r stretches.

The -component of v is due to the rotation of r.


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Acceleration Differentiating the expression for v to obtain

the acceleration a.

But from the previous derivation

dt

edre

dt

dre

dt

dr

dt

edre

dt

rda

ererdt

d

dt

rd

dt

vda

rr

r

2

2

rr e

dt

deande

dt

de .,.


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Substituting the above and simplifying

Where

errerra

ererererera

r

rr

22

rra

rrar

2

2

r

r

aa

aaa

1

22

tan


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Kinematics of particlesRelative motion


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Relative motion Relative motion analysis : is the motion analysis of

a particle using moving reference system

coordinate in reference to fixed reference

system.


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In this portion we will confine our

attention to:-

moving reference systems that translate but

do not rotate.

The relative motion analysis is limited to plane

motion.


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Note: in this section we need

1. Inertial(fixed) frame of reference.

2. Translating(not rotating) frame of reference.


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Consider two particles A and B that may haveseparate curvilinear motion in a given plane or in

parallel planes.

X,Y : inertial frame of reference X,y: translating coordinate system


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Using vector addition: position vector of particle B is

Where: rA, rB absolute position vectors

rB/A relative position vector of particle B (B

relative to A or B with respect to A)

ABABrrr

/


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Differentiating the above position vector once weobtain the velocities and twice to obtain

accelerations. Thus,

- Velocity - Acceleration

ABAB

ABAB

vvv

dt

rd

dt

rd

dt

rd

/

/

ABAB

ABAB

aaa

dt

vd

dt

vd

dt

vd

/

/


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Note: In relative motion analysis, we employedthe following two methods,

1. Trigonometric(vector diagram) A sketch of

the vector triangle is made to reveal the

trigonometry

2. Vector algebra using unit vector i and j,

express each of the vectors in vector form.


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Example 1 A 350m long train travelling at a constant speed of 40m/s crosses

over a road as shown below. If an automobile A is traveling at

45m/s and is 400m from the crossing at the instant the front of the

train reaches the crossing, determine

a) The relative velocityof the train with respect to the automobile,

and

b) The distance from the automobile to the end of the last car of the

train at the instant.


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Example 2 For the instant represented, car A has a speed of100km/h, which is increasing at the rate of 8km/h

each second. Simultaneously, car B also has a speed of

100km/h as it rounds the turn and is slowing down at

the rate of 8km/h each second. Determine the

acceleration that car B appear to have an observer in

car A.


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Example 3 For the instant represented, car A has an acceleration in

the direction of its motion and car B has a speed of

72km/h which is increasing. If the acceleration of B as

observed from A is zero for this instant,

Determine the acceleration of A and the rate at which the

speed of B is changing.


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Example 4 Airplane A is flying horizontally with a constant speed

of 200km/h and is towing the glider B, which is

gaining altitude. If the tow cable has a length r=60mand is increasing at the constant rate of 5 degrees

per second, determine the velocity and acceleration

of the glider for the instant when =15


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Constrained motion of connected

particles


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Constrained motion(dependent motion)

Sometimes the position of a particle will depend

upon the position of another or of several particles.

If the particles are connected together by an

inextensible ropes, the resulting motion is called

constrained motion


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Considering the figure, cable AB is subdivided into three segments:

the length in contact with thepulley, CD

the length CA

the length DB

It is assumed that, no matter how A and B move, the length incontact with the pulley is constant.

We could write:

constant ABBCDA lsls


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Differentiating with respect to time,

Differentiating the velocity equation

0vv

0

BA

dt

ds

dt

dsBA

0aaBA


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Important points in this technique:

Each datum must be defined from a fixed position.

In many problems, there may be multiple lengths like lCD that dont

change as the system moves. Instead of giving each of these

lengths a separate label, we may just incorporate them into aneffective length:


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where its understood that

l = cable length less the length in contact with the pulley = lAB lCD.

constant lss BA


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constant2 lshs BA

considering the fig, we could write:

where l is the length of the cable less

the red segments that remain

unchanged in length as A and B move.

Differentiating,

02

02

BA

BA

aa

vv


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constant)(2 lshhs BA

we could also write the length of

the cable by taking another datum:

Differentiating,

02

02

BA

BA

aa

vv


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Consider the fig.,

Since L, r2, r1 and b is

constant, the first andsecond time derivatives

are:-

bryr

xL 12 2

2

yx

yx

20

20


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Consider the

following figure

.)(.2

constyyyyLconstyyL

DCCBB

DAA


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NB. Clearly, it is impossible for the

signs of all three terms to be positivesimultaneously.


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Example 1 Cylinder B has a downward velocity of 0.6m/s and an

upward acceleration of 0.15m/s2.

Calculate the velocity and acceleration of block A.


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xvA

Example 2

Collars A and B slides along the fixed rods are

connected by a cord length L. If collar A has a

velocity to the right, express the velocity

of B in terms of x, vA, and s.

svB


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Part III

Kinetics of particles


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Kinetics of particles It is the study of the relations existing between

the forces acting on body, the mass of the body,

and the motion of the body.

It is the study of the relation between

unbalanced forces and the resulting motion.


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Newton s first law and third law are sufficient

for studying bodies at rest (statics) or bodies in

motion with no acceleration.


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When a body accelerates ( change in velocity

magnitude or direction) Newton s second law is

required to relate the motion of the body to the

forces acting on it.


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Kinetics problems Force-mass-acceleration method

Work and energy principles

Impulse and momentum method


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Force, mass and acceleration Newton s Second Law: If the resultant force

acting on a particle is not zero the particle will

have an acceleration proportional to the

magnitude of resultant and in the direction of the

resultant.


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The basic relation between force and

acceleration is found in Newton's second law of

motion and its verification is entirely

experimental.


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Consider a particle subjected to constant forces

We conclude that the constant is a measure of

some property of the particle that does not

change.

consta

F

a

F

a

F ...

2

2

1

1


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This property is the inertia of the particlewhich is its resistance to rate of change of

velocity.

The mass m is used as a quantitative measure

of inertia, and therefore the experimental

relation becomes,

F=ma


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The above relation provides a complete

formulation of Newton's second law; it expresses

not only that the magnitude F and a areproportional but also that the vector F and a have

the same direction.


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Types of dynamics problems Acceleration is known from kinematics conditions

Determine the corresponding forces

Forces acting on the particle are specified(Forces are constant or functions F( t, s, v, )

Determine the resulting motion


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Equation of motion and solution of problems

When a particle of mass m acted upon by severalforces. The Newtons second law can beexpressed by the equation

To determine the acceleration we must use theanalysis used in kinematics, i.e

Rectilinear motion Curvilinear motion

maF


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Rectilinear Motion

If we choose the x-direction, as the direction

of the rectilinear motion of a particle of mass

m, the acceleration in the y and z direction

will be zero, i.e

0

0

z

y

xx

F

F

maF


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Generally,

Where the acceleration and resultant force aregiven by

ZZ

yy

xx

maF

maF

maF

222zyx

zyx

aaaa

kajaiaa

222 )()(

zyx

zyx

FFFF

kFjFiFF


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Curvilinear motion In applying Newton's second law, we shall make

use of the three coordinate descriptions of

acceleration in curvilinear motion.


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Normal and tangential coordinate

Where

tt

nn

maF

maF

vava tn ,22


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Polar coordinates

Where and

maF

maF rr

2

rra r

rra n 2


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Examples


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Example 1 Block A has a mass of 30kg and block B has a mass of

15kg. The coefficient of friction between all plane

surfaces of contact are and .

Knowing that =300 and that the magnitude of the

force P applied to block A is 250N, determine

a) The acceleration of block A ,and

b) The tension in the cord

15.0s 10.0k


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Example 2 A small vehicle enters the top A of the circular path

with a horizontal velocity vo and gathers speed as itmoves down the path.

Determine an expression for the angle to theposition where the vehicle leaves the path andbecomes a projectile. Evaluate your expression forvo=0. Neglect friction and treat the vehicle as aparticle


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Exercise(problem 3/69) The slotted arm revolves in the horizontal plane about

the fixed vertical axis through point O. the 2kg slider Cis drawn toward O at the constant rate of 50mm/s bypulling the cord S. at the instant for which r=225mm,

the arm has a counterclocke wise angular velocityw=6rad/s and is slowing down at the rate of 2rad/s2.For this instant, determine the tension T in the cordand the magnitude N of the force exerted on the sliderby the sides of the smooth radial slot. Indicate which

side, A or B of the slot contacts the slider.


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Exercise (problem 3/43)

The sliders A and B are connected by a light rigid

bar and move with negligible friction in the slots,

both of which lie in a horizontal plane. For the

positions shown, the velocity of A is 0.4m/s to

the right. Determine the acceleration of each

slider and the force in the bar at this instant.


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Exercise (problem 3/36) Determine the accelerations of bodies A and B

and the tension in the cable due to the application

of the 250N force. Neglect all friction and the

masses of the pulleys.

Chapter II [Compatibility Mode]

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