vectors and kinematics
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Text Book: An Introduction to Mechanics by Kleppner and Kolenkow Chapter 1: Vectors and Kinematics -Explain the concept of vectors. -Explain the concepts of position, velocity and acceleration for different kinds of motion. References: Halliday, Resnick and Walker Berkley Physics Volume-1TRANSCRIPT
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
Motion in Several Dimensions
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
Motion in Several Dimensions
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
Motion in Several Dimensions
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.
Motion in Several Dimensions
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
Example 1.8 Uniform Circular Motion
dt
rdv
)ˆcosˆsin( jtitrv
0
)sincoscossin(2
ttttrrv
constant. rv is perpendicular to .v
r
Example 1.8 Uniform Circular Motion
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
Kinematical equations
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
Kinematical equations
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
More about the derivative of a vector
change.cannot magnitude its since rotate,must
, lar toperpendicu always is
A
AAΔ
change. magnitude its and same ofdirection
, toparallel always is
A
AAΔ
More about the derivative of a vector
. 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
More about the derivative of a vector
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
More about the derivative of a vector
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
Motion in Plane Polar Coordinates
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.
Motion in Plane Polar Coordinates
.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.
Motion in Plane Polar Coordinates
]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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