Vectors and Kinematics

Chapter – 1 Vectors and Kinematics

Text Book: AN INTRODUCTION TO MECHANICS by Kleppner and Kolenkow

Dr. Virendra Kumar VermaMadanapalle Institute of Technology and Science

(MITS)

Scalars and VectorsA scalar quantity is a quantity that has only magnitude.

A vector quantity is a quantity that has both a magnitude and a direction.

Scalar quantitiesLength, Area, Volume,

Speed, Mass, Density

Temperature, PressureEnergy, Entropy

Work, Power

Vector quantitiesDisplacement, Direction,

Velocity, Acceleration,Momentum, Force,

Electric field, Magnetic field

Volume

Velocity

Vector notationVector notation was invented by a physicist, Willard Gibbs of Yale University.

By using vector notation, physical laws can often be written in compact and simple form.

For example, Newton’s second law In old notation,

In vector notation,zz

yy

xx

maF

maF

maF

amF

Equal vectors If two vectors have the same length and the same direction they are equal.

B C

The vectors B and C are equal. B = C

The length of a vector is called its magnitude. e.g. Magnitude of vector B = |B|

Unit vectors

If the length of a vector is one unit. e.g. The vector of unit length parallel to A is Â. A = |A|Â.

A unit vector is a vector that has a magnitude of exactly 1. Ex: The unit vectors point along axes in a right-handed coordinate system.

Algebra of vectors Multiplication of a Vector by a Scalar:

If we multiply a vector a by a scalar s, we get a new vector. Its magnitude is the product of the magnitude of a and the absolute value of s.

Algebra of vectors Addition of two Vectors:

Algebra of vectors Subtraction of two Vectors:

Scalar Product (“Dot” Product)

cosabba

The scalar product of the vectors and is defined as a

b

)cos)(())(cos( bababa

)on of Projection)(( babba

)on of Projection)(( ababa

The above equation can be re written as

)cos)(( baba

OR

Scalar Product (“Dot” Product)

The commutative law applies to a scalar product, so we can write

abba

When two vectors are in unit-vector notation, we write their dot product as

zzyyxx

zyxzyx

bababa

kbjbibkajaiaba

)ˆˆˆ()ˆˆˆ(

then ,0 If ba

a = 0or b = 0or cos θ = 0 ( is perpendicular to )a

b

Note:2aaa

Example 1.1 Law of Cosines

BAC

)()( BABACC

cos2222

BABAC

cos2222 ABBAC

This result is generally expressed in terms of the angle φ

]cos)cos(cos[

Example 1.2 Work and the Dot product

The work W done by a force F on an object is the displacement d of the object times the component of F along the direction of d. If the force is applied at an angle θ to the displacement,

dFW

or

dFW

)cos(

Vector Product (“Cross” Product) The vector product of and , written , produces a third

vector whose magnitude is

A

B

BA

C

sinBABAC

ABC

)( ABBA

Vector Product (“Cross” Product) When two vectors are in unit-vector notation, we write their cross product as

zyx

zyx

xyyxxzxzzyzy

zyxzyx

bbb

aaa

kji

kabbajabbaiabba

kbjbibkajaiaba

ˆˆˆ

ˆ)(ˆ)(ˆ)(

)ˆˆˆ()ˆˆˆ(

Note: If and are parallel or antiparallel, .a

b

0ba

Example 1.4 Area as a Vector

Consider the area of a quadrilateral formed by two vectors, andThe area of the parallelogram A is given by

A = base ×height = C D sinθ

If we think of A as a vector, we have

C

D

DC

DCA

Magnitude of A is the area of the parallelogram, and the vector product defines the convention for assigning a direction to the area.

Example 1.5 Vector Algebra

Example 1.6 Construction of a perpendicular Vector

Find a unit vector in the xy plane which is perpendicular to A = (3,5,1).

LetA = (3,5,-7)B = (2,7,1)Find A + B, A – B, |A|, |B|, A · B, and the cosine of the angle between A and B.

Base Vectors

Base vectors are a set of orthogonal (perpendicular) unit vectors, one for each dimension.

Displacement and the Position Vector

kSjSiSS

kzzjyyixxS

zyxˆˆˆ

ˆ)(ˆ)(ˆ)( 121212

The values of the coordinates of the initial and final points depend on the coordinates system, S does not.

Displacement and the Position Vector

kzjyixr ˆˆˆ

Rrr

'

''

)'()'(

12

12

rr

RrRr

12 rrS

A true vector, such as a displacement S, is independent of coordinate system.

Velocity and Acceleration

Motion in One Dimension

12

12 )()(

tt

txtxv

The average velocity ‘v’ of the point between two times, t1 and t2, is defined by

)( 1tx )( 2tx

The instantaneous velocity ‘v’ is the limit of the average velocity as the time interval approaches zero.

t

txttxv

t

)()(lim

0 dt

dxv

t

tvttva

t

)()(lim

0

Velocity and Acceleration

In a similar fashion, the instantaneous acceleration is

dt

dva

Motion in Several Dimensions

),( 111 yxr

The position of the particle

At time t1, At time t2,

),( 222 yxr

The displacement of the particle between times t1 and t2 is

),( 121212 yyxxrr

Motion in Several DimensionsWe can generalize our example by considering the position at some time t, and some later time t+Δt. The displacement of the particle between these times is

)()( trttrr

This above vector equation is equivalent to the two scalar equations

)()( txttxx

)()( tyttyy


dt

rdt

rv

t

0

lim

The velocity v of the particle as it moves along the path is defined to be

dt

dy

t

yv

dt

dx

t

xv

ty

tx

0

0

lim

lim

dt

dz

t

zv

tz

0

lim

In 3D, The third component of velocity


kdt

dzj

dt

dyi

dt

dx

dt

rdv ˆˆˆ

kvjvivv zyxˆˆˆ

The magnitude of ‘v’ is

2

1222 )( zyx vvvv

Similarly, the acceleration a is defined by

dt

rd

kdt

dvj

dt

dvi

dt

dv

dt

vda zyx

2

ˆˆˆ


Let the particle undergo a displacement Δr in time Δt. In the time Δt→0, Δr becomes tangent to the trajectory.

tv

tdt

rdr

Δt→0, v is parallel to Δr

The instantaneous velocity ‘v’ of a particle is everywhere tangent to the trajectory.


dt

rd

kdt

dvj

dt

dvi

dt

dv

dt

vda zyx

2

ˆˆˆ

Similarly, the acceleration a is defined by

Example 1.7 Finding v from r

The position of a particle is given by

)ˆˆ( jeieAr tt

Where α is a constant. Find the velocity, and sketch the trajectory.

)ˆˆ( jeieA

dt

rdv

tt

tx eAv t

y eAv ==>

Solution:

The magnitude of ‘v’ is

2

122

2

122

)(

)(

tt

yx

eeA

vvv

To sketch the trajectory, apply the limiting cases.At t = 0, we get At t = ∞, we get

)ˆˆ()0(

)ˆˆ()0(

jiAv

jiAr

axis. thealong pointed vectors

ˆˆ

limit In this

0

xvandr

iAevandiAer

eande

tt

tt

Example 1.7 Finding v from r

Note:

1. Check

2. Find the acceleration.

3. Check

4. Check

5. Check the direction of , and at t = 0 and t ∞.

)0()0( vr

)0()0( va

r

v a

)()( trta

Example 1.8 Uniform Circular Motion

)ˆsinˆ(cos jtitrr

Consider a particle is moving in the xy plane according to

constant

)]sin(cos[ 2/1222

r

titrr

The trajectory is a circle.

)ˆsinˆ(cos jtitrr


dt

rdv

)ˆcosˆsin( jtitrv

0

)sincoscossin(2

ttttrrv

constant. rv is perpendicular to .v

r


dt

vda

r

jtitra

2

2 )ˆsinˆcos(

The acceleration is directed radially inward, and is known as the centripetal acceleration.

Note:

1. Check

2. Check

va

ar

Kinematical equations

ttatttatttatvtv

tvttvttvtvtv

)()2()()()(

)()2()()()(

10001

10001

Suppose - velocity at time t1 and - velocity at time t0.

Dividing the time interval (t1-t0) in n parts,

)( 1tv

)( 0tv

nttt /)( 01

For n∞ (Δt0), and the sum becomes an integral:

1

0

)()()( 01

t

t

dttatvtv


The above result is the same as the formal integration of

1

0

)()()( 01

t

t

dttatvtv

] velocityintial)([,)()(

)()()(

)()(

)()(

0001

01

1

0

1

0

1

0

1

0

vtvdttavtv

dttatvtv

dttatvd

dttatvd

t

t

t

t

t

t

t

t


If acceleration a is constant and t0 = 0, we get

tavtv

0)(

200

0

00

2

1)(

)()(

tatvrtr

or

dttavrtrt

More about the derivative of a vectorConsider some vector A(t) which is function of time. The change in A during the interval from t to t + Δt is

)()( tAttAA

We define the time derivative of A by

t

tAttA

dt

Adt

)()(lim

0

is a new vector.

Depending on the behaviour of A:

The magnitude of can be large or small.

The direction of can be point in any direction.

dt

Ad

dt

Ad

dt

Ad

More about the derivative of a vector

Case 1:

. tomagnitude thechange

but unaltereddirection

||

AA

AA

Case 2:

unalteredy practicall magnitude the

leavebut direction in change AA

Case 3: In general

direction. and magnitudeboth in change willA

Case 3


change.cannot magnitude its since rotate,must

, lar toperpendicu always is

A

AAΔ

change. magnitude its and same ofdirection

, toparallel always is

A

AAΔ


. oflocity angular ve thecalled is /

,0limit theTaking

22

getwe.2/2/sin 1,For 2

sin2

Adtd

dt

dA

dt

Ad

t

tA

t

A

and

AA

AA

AA


magnitude.in constant is if zero is / and

),0/( rotatenot does if zero is /

limit, the takingand by dividing

small,ly sufficient For

.

||

||

||

||

AdtAd

dtdAdtAd

dt

dA

dt

Ad

dt

dA

dt

Ad

t

AA

AA

AAA


rv

rtdt

dr

dt

dr

dt

rd

trA

or

)(

and Then . vector rotating thebe Let

constant. is of magnitude the

, lar toperpendicu is ifagain seecan we

2)(

Then .let relation, second In the

)(

)(

)(

2

A

A/dtAd

dt

AdAA

dt

d

BA

dt

BdAB

dt

AdBA

dt

d

dt

BdAB

dt

AdBA

dt

d

dt

AdcA

dt

dcAc

dt

d

Some formal identities

Motion in Plane Polar Coordinates

x

y

yxr

arctan

22


The lines of constant x and of constant y are straight and perpendicular to each other. The lines of constant θ and constant r are perpendicular.


.directions fixed have ˆ and ˆ whereas

position,vary with ˆ and ˆ ofdirection The

ji

r

cosˆsinˆˆsinˆcosˆˆ

jijir

Velocity in Polar Coordinates

dt

rdrrr

rrdt

d

dt

rdv

ˆˆ

)ˆ(

jyixjyixdt

d

dt

rdv ˆˆ)ˆˆ(

-scoordinatecartesian in that Recall

]ˆcosˆsinˆ[ˆ

)cosˆsinˆ(

cosˆsinˆ

)(sinˆ)(cosˆˆ

sinˆcosˆˆ

thatknow We

ji

ji

ji

dt

dj

dt

di

dt

rd

jir

ˆˆ rrrv

Velocity in Polar Coordinates

ˆˆ rrrv

direction. )ˆ (i.e., l tangentiain the is termsecond Theoutward.radially directed

velocity theofcomponent theisright on the first term The

circle. theof arc on the liesmotion the.,.l. tangentiaisvelocity

ˆ and 0 ,constant.2direction. radial fixed ain motion the.,.

radial. isvelocity ˆ and 0 constant, 1

ei

rvrrei

rrv.


]ˆsinˆcosˆ[ˆ

)sinˆcosˆ(

sinˆcosˆ

)(cosˆ)(sinˆˆ

cosˆsinˆˆ

thatknow We

rjir

ji

ji

dt

dj

dt

di

dt

d

ji

rdt

d

dt

rd

ˆˆ

ˆˆ

Acceleration in Polar Coordinates

ˆˆˆˆˆ

)ˆˆ(

dt

drrrr

dt

drrr

rrrdt

d

dt

vda

get we,/ˆ and/ˆ of value theSubstitute dtddtrd

rrrrrrra ˆˆˆˆˆ 2

ˆ)2(ˆ)( 2 rrrrra

termTangential | Radial

ˆˆ

ˆˆ

vrv

rrrv

r

termTangential | Radial

ˆˆ

ˆ)2(ˆ)( 2

ara

rrrrra

r

Velocity and Acceleration in Polar Coordinates

References

1. Fundamentals of Physics by Halliday, Resnick and Walker

2. Berkeley Physics Course Volume-1

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Vectors and Kinematics

Education

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