Unit 1
Measurement, Vectors and Motions
1.1
Dimensional analysis
1.2
Order of magnitudes of some quantities
1.3
Vectors
1.4
Reference frames of one and two dimensions
1.5
Relative motions
1.1
Dimensional analysis
The fundamental quantities used in physical descriptions are called dimensions. Length, mass,
and time are examples of dimensions. You could measure the distance between two points
and express it in units of meters, centimeters, or feet. In any case, the quantity would have the
dimension of length.
It is common to express dimensional quantities by bracketed symbols, such as [L], [M], and
[T] for length, mass, and time, respectively. Dimensional analysis is a procedure by which the
dimensional consistency of any equation may be checked. If I say, x = at, where x, a, and t
represent the displacement, acceleration and time respectively. Is it correct?
•
[x] = [L]
•
[at] = ( [L][T] −2 ) ( [T] ) = [L][T] −1
So the dimension for the sides are not equal, i.e. x = at is invalid.
Now look at the equation x = at2
•
The L.H.S. is x, i.e. [L]
•
The R.H.S. is at2, i.e. ( [L][T] −2 ) ( [T]2 ) = [L].
Hence the equation is dimensionally correct, but you should know that it is physically
incorrect. The correct equation is, as you should remember, x =
1 2
at , where x is the
2
displacement of an object moving under an acceleration a from rest. The fraction
1
is a
2
constant and has no dimension, just like π.
1
Example
The pressure P in a fluid in motion depends on the density ρ and its speed v. Find a simple
expression that relates pressure, density and speed.
Answer
We assume that a power law is valid to relate the quantities, e.g. P = k ρ x v y, where k is a
constant and x and y are to be determined. We see that pressure is the ratio of an applied force
to the exerted area, and thus its dimension is
[M][L][T]−2
= [M][L]−1 [T]−2 .
2
[L]
The dimension of density ρ is [M][L]−3 , whereas the dimension of v is [L][T]−1 . Hence, we
−1
have [ P] [M][L]
=
=
[T]−2 ([M]x [L]−3 x ) ([L] y [T]− y ) . Simplifying it, we obtain
[M][L]−1 [T]−2 = [M]x [L]−3 x + y [T]− y .
Equate the powers of each dimension on either side of the equation which gives
M:
x=1
L:
−3x+ y = −1
T:
y=2
Obviously, we have x = 1 and y = 2 and thus P ∝ ρ v2.
Example
The period τ of a simple pendulum is the time for one complete swing. How does τ depend
on the mass m of the bob, the length l of the string, and the acceleration due to gravity g?
Answer
We begin by expressing the period τ in terms of the other quantities as follows:
τ = k mx l y g z,
where k is a constant and x, y, and z are to be determined. The dimension of τ is [T]. Next,
we consider the dimensions of each quantity in the R.H.S. of the above equation. Obviously,
it is [M]x [L]y ( [LT−2] )z, and thus [τ] = [T] = [M]x [L]
y+ z
[T]−2z . Equate the powers of each
dimension, we have
T:
−2 z = 1
M:
x=0
2
L:
y + z =0
1
1
These equations are easily solved and yield x = 0, z = − , and y = . Thus,
2
2
τ =k
1.2
l
.
g
Order of magnitudes of some quantities
The below tables are references to the order of magnitudes of some quantities. The last table
shows the prefixes for powers of ten.
Approximate Values of Some Measured Lengths
One light-year
9.46×1015 m
Mean radius of the Earth
6.37×104 m
Length of a football field
9.1×101 m
Size of smallest dust particles
~10−4 m
Size of cells of most living organisms
~10−5 m
Diameter of a hydrogen atom
~10−10 m
Diameter of an atomic nucleus
~10−14 m
Approximate Masses of Various Objects
Sun
1.99×1030 kg
Earth
5.98×1024 kg
Moon
7.36×1022 kg
Human
~102 kg
Mosquito
~10−5 kg
Hydrogen atom
1.67×10−27 kg
Electron
9.11×10−31 kg
Approximate Values of Some Time Intervals
One year
3.2×107 s
One day
8.6×104 s
Time interval between normal heartbeats
8×10−1 s
Period of audile sound waves
~10−3 s
Period of vibration of an atom in a solid
~10−13 s
Period of audile visible light waves
~10−15 s
Duration of a nuclear collision
~10−22 s
3
Prefixes for Powers of Ten
Power
10−15
10−12
10−9
10−6
10−3
10−1
Prefix
femto
pico
nano
micro
milli
deci
Abbreviation
f
p
n
µ
m
d
Power
103
106
109
1012
1015
Prefix
kilo
mega
giga
tera
peta
1.3
Vectors
(a)
Definition of a vector

A = A = Aaˆ , where A is the magnitude of A and a is
an unit vector with direction pointed. Vector A is

sometimes written as A . The negative of A points in
Abbreviation
k
M
G
T
P

A
â

−A
opposite direction (i.e. −A).
(b)
Properties of vectors on addition
Given that A and B are two vectors, if A = Ax iˆ + Ay ˆj and =
B Bx iˆ + By ˆj we have
A + B = ( Ax + Bx )iˆ + ( Ay + By ) ˆj .
A + B= B + A (Commutative law)
A + ( B + C=
) ( A + B ) + C (Associative law)
(c)
Cartesian coordinate system
•
2-D (i.e. a plane)
ĵ
=
A Ax iˆ + Ay ˆj
Unit vectors
iˆ
Ax: Component of A in x-direction
Ay: Component of A in y-direction
y
The magnitude of A is given by
=
A
A
=
Ax2 + Ay2 .
( Ax , Ay )
Ay

A
Note that we have the components of A
Ax = A cosθ and Ay = A sin θ .
â
O
origin
θ
x
Ax
4
•
3-D
A = Ax i + Ay j + Az k
k̂
z
A = A = Ax2 + Ay2 + Az2
ĵ
Az
iˆ
If there is a vector B = B x i + B y j + Bz k , we have

A
A + B = ( Ax + Bx )iˆ + ( Ay + By ) ˆj + ( Az + Bz )kˆ ,
A − B = ( Ax − Bx )iˆ + ( Ay − By ) ˆj + ( Az − Bz )kˆ .
(d)
Ay
y
Ax
x
Mathematics of vectors
y
y

C

B
O

A

A

B
O
x

−B
  
C= A + B
x

D

−B
  
D= A − B
Remark:
(1) Sum of vectors is often used to represent the resultant of vectors, e.g. sum of forces on an
object, F = F1+F2 and resultant velocity of an object, v = vx + vy.
(2) For the difference of two vectors, it represents either the relative position or the relative
motion of two objects, i.e. the relative position vector and relative velocity, etc. The
position vector of object A relative to object B is rAB = rA− rB, where rA and rB are the
position vectors of A and B respectively. The velocity vector of B relative to A is vBA =
vB− vA, where vA and vB are velocity vectors of A and B respectively.
(e)
(1)
z
Examples of vectors
t = t2
Position vector

r2
r = xi + yj + zk ,
r =| r |= x 2 + y 2 + z 2 (the magnitude of r from O).
O

r1

∆r
Trajectory of
a particle
t = t1
y
x
5
(2)
Displacement (change in position)
∆r = r2 − r1 = r (t 2 ) − r (t1 )
(3)
Average velocity during time ∆t, where ∆t = t 2 − t1 .
vavg
=
(4)
∆r r (t1 + ∆t ) − r (t1 )
=
∆t
∆t

v
Instantaneous velocity
∆r
= vx iˆ + v y ˆj + vz kˆ
∆t →0 ∆t
v ≡ lim
v = lim
∆t →0
∆r dr
(first order derivative)
=
∆t dt
tangent to curve

r
O
v = v =speed.
Note the difference of velocity and speed: Velocity is a vector but speed is a scalar!
(5)
Acceleration
aavg
=
∆v v (t1 + ∆t ) − v (t1 )
=
∆t
∆t
∆v dv d 2 r
(second order derivative)
a = lim
=
=
∆t → 0 ∆t
dt dt 2
Remark:
Dot product of two vectors is defined as follows.
A ⋅ B ≡ A B cos θ= Ax Bx + Ay By + Az Bz ,
where A and B are vectors. When A ⋅ B = 0 , it means that A and B
are normal to each other. Dot product is commonly used to define
d d
physical quantities, e.g. work done, W= F ⋅ d .

B
θ

A
Cross product of two vectors is defined as follows
iˆ
A × B ≡ A B sin θ=
nˆ Ax
Bx
ˆj
Ay
By
kˆ
A=
( Ay Bz − Az By )iˆ − ( Ax Bz − Az Bx ) ˆj + ( Ax By − Ay Bx )kˆ ,
z
Bz
where n̂ is a unit vector that is perpendicular to the plane formed by vectors A and B using
the right hand rule. Notice that A × B = − B × A . When A × B = 0 , it means that A and B are
parallel or anti-parallel to each other. Cross product is commonly used to define physical
  
quantities, e.g. torque, τ = r × F .
6
Example (Challenging)
A particle is moving along a curved path and it is located by a position vector r. The velocity
vector and acceleration vector is v and a respectively. Given that the distance traveled is s and
at is the magnitude of acceleration of particle along the path. Comment on the validity of the
following relations.
(A)
dr
dv
ds
dv
= v , (B)
= a , (C)
=v.
= at and (D)
dt
dt
dt
dt
Answer
(A) is incorrect. Because =
v v=
(B) is incorrect. Because =
a a=
dr d r
dr d r
dr
and
, but
.
≠
=
dt
dt
dt
dt
dt
dv d v
dv d v
dv
dv
and
, but
. In fact,
=
≠
= at .
dt
dt
dt
dt
dt
dt
dv
= a= a , but at is a magnitude of the tangential component of a.
dt
(C) is incorrect. Because
(D) is correct. Because
=
v
ds dr
dr ds
. Note that=
.
v
≠
=
dt dt
dt
dt
1.4
Reference frames of one and two dimensions
(a)
One dimension (1D)
Each observer – such as you standing on the ground - defines a reference frame. A reference
frame requires a coordinate system and a set of clocks, which enable an observer to measure
positions, velocities, and accelerations in his or her particular frame.
x BA
x PB
x PA = x PB + x BA
Frame A
Frame B
x
P
x
O
O
As shown in the figure, we have two different frames to watch an object P.
Obviously, x PA = x PB + x BA .
In words: “ The position of P as measured by frame A is equal to the position of P as
measured by frame B plus the position B as measured by A.”
7
dx
, we take the time derivative xPA and obtain the relation of velocities.
dt
As v =
v PA = v PB + v BA .
If two frames move at constant speed with respect to each other, i.e. v BA = const . , take
both sides again, we have a PA = a PB (
(b)
dv BA
= 0 ). Generally, a PA = a PB + a BA .
dt
y
Two dimensions (2D)
y
We have
P
rPA = rPB + rBA ,
and
v PA = v PB + v BA ,
rPA
a=
a PB + a BA .
PA
rBA
rPB
Frame B
Frame A
1.5
d
on
dt
x
x
Relative motions
Two objects A and B are located by position vectors rA and rB respectively. The position
vector of A relative to B is rAB = rA− rB. Differentiating both sides, we have vAB = vA− vB,
where vAB is the velocity of A relative to B (relative velocity of A to B). Note that the relative
motion is represented by the difference of two vectors. Let’s take a look on an example. The
position vectors of objects A and B are rA and rB, respectively, then the position vector of A
relative to B is rAB = rA− rB, as shown in the figure. The position vector of B relative to A is
rBA = rB− rA. When the velocity vectors of objects A and B are vA and vB, respectively, then the
velocity vector of A relative to B is vAB = vA− vB. The velocity vector of B relative to A is vBA
= vB− vA.
B
y
rB
O
rA
B
y
rB
rBA= rB − rA
A
x
O
rA
rAB= rA − rB
A
x
8
Example
A ship A is steaming due north at 16 km/hr and a ship B is steaming due west at 12 km/hr.
Find the relative velocity of A with respect to B.
Answer
N
16 km/hr

v AB
A
12

− vB
B
W

vA
16
12 km/hr

 
The relative velocity of A with respect to B is v AB which equals to v A − vB . Referring to the

vector diagram, the relative velocity has magnitude v AB =
122 + 162 = 20 . The direction of
12

v AB is N tan −1 ( ) E, i.e. N 36o52’ E.
16
Example
A man traveling East at 8 km h−1 finds that the wind seems to blow directly from the North.
On doubling his speed he finds that it appears to come NE. Find the velocity of the wind.
Answer
− 8 iˆ
Let the velocity of the wind be

=
w x iˆ + y ˆj .
Then the velocity of the wind relative to the man is

w − 8 iˆ = ( x − 8) iˆ + y ˆj

w
Parallel to
− ĵ
Notice that the man is the observer and he feels the wind. All about such feeling (i.e. blowing
directly from the North) is relative to him. The vector subtraction represents this relative

velocity. Therefore w − 8 iˆ = ( x − 8) iˆ + y ˆj is parallel to − ĵ . Hence, we obtain x − 8 = 0.
9
− 16 iˆ
When the man doubles his speed, the velocity of the
wind relative to him is

w − 16 iˆ =( x − 16) iˆ + y ˆj .

w
Parallel to
− ( iˆ + ˆj )
But this is from the NE and is therefore parallel
to −( iˆ + ˆj ) . Hence y = x − 16 = −8.
The velocity of the wind is 8 iˆ − 8 ˆj , which is equivalent to 8 2 km h−1 from NW.
Example
A swimmer wishes to across a swift, straight river of width d. The speed of the swimmer in
still water is u and that of the water is v, where v > u. If the swimmer wants to cross the swift
with a minimum time, find the minimum time and the position that he reaches in the opposite
bank.
Answer
A
s
B
θ
ĵ
d
C

u

u
O
θ

v
Q
iˆ

The minimum time of travel can be obtained if the swimmer directs in the direction of j . The

magnitude of u is contributed completely in the direction of crossing the swift. It is the
fastest way. Hence the minimum time t = d/u. The distance downstream s = vt = (vd)/u.
10
Example (Challenging)
As an extension of the last example, keeping all information above, what is the direction
along which the swimmer should proceed such that the downstream distance he has traveled
when he reaches the opposite bank is the smallest possible? What is the corresponding time
required?
Answer
B
s
A
θ
ĵ
d
C

u

u
θ
O
θ

v
iˆ
Q
The downstream distance is smallest when OB is tangent to the circle, centered Q, radius u.
u
d 1 − ( )2
d cosθ
d v2 − u2
u
v
.
=
=
sin θ = , and AB = d cot θ =
u
sin θ
u
v
v
The time taken is t =
d
=
u cosθ
d
u
u 1 − ( )2
v
=
dv
u v2 − u2
.
11