Measurements and Dimensional Analysis.pdf

PHY112
GEOMETRICAL OPTICS AND
MECHANICS

MECHANICS (MC)
Measurements
• Used to describe natural phenomena
• Need defined standards
• Characteristics of standards for measurements
➢Readily accessible
➢Possess some property that can be measured
reliably
➢Must yield the same results when used by anyone
anywhere
➢Cannot change with time

Fundamental quantities in MC
• Three basic (fundamental) quantities in mechanics
➢Length (L)
➢Mass (M)
➢Time (T)
➢Other fundamental quantities (not used in mechanics)
Temperature, Electric current, Luminous intensity,
amount of substance.
• All other quantities (in mechanics) can be
expressed in terms of the three fundamental
quantities.
➢Examples : Acceleration, Energy, Force

SYSTEM OF UNITS
➢Standard units for fundamental quantities
dictates system of units.
➢Units should be clearly and accurately defined
➢One system of units used universally in the
scientific community is the SI system of units
➢SI system – Standard fundamental units are
metre (m) {for length}, kilogram (kg) {for
mass}, and seconds (s) {for time}.
➢For definitions and history – see textbook(s)

Prefixes
• Prefixes correspond to powers of 10
• Each prefix has a specific name
• Each prefix has a specific
abbreviation

MOST COMMON PREFIXES
Prefix Abbreviation Power
Tera T 1012
Giga G 109
Mega M 106
kilo k 103
centi c 10-2
milli m 10-3
micro m 10-6
nano n 10-9
pico p 10-12
femto f 10-15

Derived and Pseudo Units
• Sometimes use derived (quantities) units.
➢Derived quantities can be expressed in terms of
fundamental quantities.
➢Example: Volume is one of the derived quantities.
Can be expressed in terms of Length.
• Pseudo Units also used in place of SI units
➢Example: Force, SI units kg m s-2, Pseudo units N

Conversion of units
• Sometimes need to convert units.
• Could be:
– From one prefix to another within same system
– From one system to another
Example 1
Convert 75 mi h-1 to (a) ft s-1 (b) m s-1
1. What is the relation between what you convert
to and what you convert from?
2. What can you multiply (or divide) by without
changing the quantity?

ft
mi 5280
1 =
mi
ft
1
5280
1=

ft
mi
5280
1
1
& =
min
60
1 =
h
h
1
min
60
1=

min
60
1
1
&
h
=
)
(a
s
60
min
1 =
min
1
60
1
s
=

s
60
min
1
1
& =
75 𝑚𝑖 ℎ−1
=
75 𝑚𝑖
1 ℎ
=
75 𝑚𝑖
1ℎ
×
5280 𝑓𝑡
1 𝑚𝑖

min
60
1
1
5280
75 h
h
ft


=
s
ft
60
min
1
min
60
5280
75


= 1
110 −
= s
ft
)
(b
in
ft 12
1 =
ft
in
1
12
1=

in
ft
12
1
1
& =
cm
in 54
.
2
1 =
in
cm
1
54
.
2
1=

cm
in
54
.
2
1
1
& =

ft
in
s
ft
h
mi
1
12
1
110
75 1

=
−
cm
m 100
1 =
m
cm
1
100
1=

cm
m
100
1
1
& =
in
cm
s
in
1
54
.
2
1
12
110


=
cm
m
s
cm
100
1
1
54
.
2
12
110



= 1
5
.
33 −
= s
m

Example 2
Convert 9 kg m-3 to mg cm-3.
g
kg 1000
1 =
kg
g
1
1000
1=

g
kg
1000
1
1
& =
g
g m
6
10
1 =
g
g
m
6
10
1
1=

g
g
1
10
1
&
6
m
=
m
cm
1
10
1
2
=

cm
m
2
10
1
1
& =
cm
m 100
1 =

3
3
1
9
9
m
kg
m
kg =
−
kg
g
m
kg
1
1000
1
9
3

=
g
g
m
g
1
10
1
9000 6
3
m

=
3
2
3
9
10
1
1
10
9










=
cm
m
m
g
m
3
000
9 −
= cm
g
m

Dimensions and Dimensional Analysis
Cuboid: What are the dimensions?
Length, Height, Width
Note: Whether Length, height or width, all
are distances.
Fundamental quantity – Length

Dimensions in Physics
• Denotes the physical nature of a quantity.
➢Remember fundamental quantities
• Dimensions denoted by square brackets [ ]
➢Dimension of Length [Length]: L
➢Dimension of Mass [Mass]: M
➢Dimension of Time [Time]: T
• What would be [Volume]? L3
• What are the [Speed]? LT-1
• What are the [Force]? MLT-2

DIMENSIONAL ANALYSIS
• Technique used to check the correctness
of an equation (if dimensions of all
parameters are known), Derive
dimensions of a parameter (if it is the
only parameter with unknown
dimensions) or to derive an equation
(where parameters with known
dimensions are related).

DIMENSIONAL ANALYSIS CONT.
Suppose have two cubes that have the same
volume. How will their dimensions compare?
The Dimensions have to be the same.
• Dimensions (length, mass, time,
combinations) can be treated as
algebraic quantities [add*, subtract*,
multiply, divide and raised to some
power]

DIMENSIONAL ANALYSIS CONT.
•Whenever have a relation, the two sides of
the equation must have the same dimensions
•Any relationship can be correct only if the
dimensions of the two sides of the equation
are the same
•Dimensions of any quantity in Mechanics
can be expressed as
  c
b
a
T
L
M
Q =
E.g.,   2
2 −
= T
ML
E   2
−
= MLT
F
  3
−
= ML
   3
L
V =

Example 1
where v is the final velocity, u is the initial
velocity, a is the acceleration, t is the time
and k1 and k2 are dimensionless constants is
dimensionally correct.
Show that the expression at
k
u
k
v 2
1 +
=
Solution
  1
−
= LT
v
      1
1
1
1 1 −
−
=

=

= LT
LT
u
k
u
k

          T
LT
t
a
k
t
a
k 

=


= −2
2
2 1
Since dimensions of the LHS are the same as
the dimensions of the two parts of the RHS,
the expression is dimensionally correct.
  1
2
−
= LT
at
k
       
at
k
v
u
k
v 2
1 &
, =
=


Example 2
The equation for the change of position of a
train starting at x = 0 m is given by
3
2
2
1
Bt
At
x +
=
where x is distance, t is time and A and B are
constants.
Determine the dimensions of A and B.
  2
]
][
[ t
A
x =
 
 2
]
[
t
x
A =
  2
2
−
=
= LT
T
L
A
AND
   
 
3
3
3
−
=
=
= LT
T
L
t
x
B
    3
t
B
x =

Another method
Make use of   c
b
a
T
L
M
Q =
Let
  2
]
][
[ t
A
x =
2
*T
T
L
M
L c
b
a
=
2
0
1
0 +
= c
b
a
T
L
M
T
L
M
M: a = 0
L: b = 1
T: c + 2 = 0, c = -2
  3
]
][
[ t
B
x =
[B] = LT-3
  c
b
a
T
L
M
A =
  2
,
T
L
A =

3
*T
T
L
M
L c
b
a
=
M: a = 0
L: b = 1
T: c + 3 = 0, c = -3
and determine a, b and c.

EXAMPLE 3
)
,
,
( m

 L
f
=
z
y
x
L
k m

 =
The frequency is given in functional form as
where L is length and the dimensions of ,  and m are
T-1, M2L2T-2 and ML-1, respectively. Use dimensional
analysis to deduce the formula for frequency.
)
,
,
( m

 L
f
=
Solution
1
:
]
[ −
T
 2
2
2
:
]
[ −
T
L
M

1
:
]
[ −
ML
m
z
y
x
L
k ]
[
]
[
]
[
]
[
]
[ m

 =

z
y
x
ML
T
L
M
L
T )
(
)
(
*
1 1
2
2
2
1 −
−
−
=
z
z
y
y
y
x
L
M
T
L
M
L
T −
−
−
= 2
2
2
1
y
z
y
z
y
x
T
M
L
T 2
2
2
1 −
+
−
+
−
=
y
z
y
z
y
x
T
M
L
T
M
L 2
2
2
1
0
0 −
+
−
+
−
=
T: -2y = -1
y = 0.5
M: 2y + z = 0 L: x + 2y – z = 0
1+ z = 0
z = -1
x +1 – (-1) = 0
x + 2 = 0
x = -2
1
5
.
0
2 −
−
= m

 kL 2
L
k
m

 =

Some Limitations of Dimensional Analysis
• Cannot use dimensional analysis alone to
determine the value of a dimensionless
constant.
• Dimensional analysis can only show that
the relation is wrong or there is a possibility
that the relation is correct.
2
2
1
at
ut
s +
=
2
4
1
at
ut
s +
= }Both dimensionally
correct
Cannot use dimensional analysis when
dealing with trigonometric functions.

Additional problems on Dimensional Analysis
x
m
k
a


−
=
x

1. Acceleration for simple harmonic motion is
, where m is mass and
is displacement. Determine [k].
Given by
2. The lift off speed, v, of a boat is a function
of the mass of the boat, m, the acceleration due
to gravity, g, the surface area of the boat, A,
and the density of water, . Use dimensional
analysis to determine the formula for lift off
speed of the boat. [One of the Limitations]

d
g
A
h
A
t
2
'
=
 '
A
3. Check the dimensional correctness of
, where A and
is height, d is distance, and g is acceleration.
are areas, h
4. Check the dimensional correctness of
h
g
m
P
P
v o
2
)
(
2
+
−
=

pressures, v is velocity,  is density, m is mass,
, where P and Po are
g is acceleration due to gravity and h is height.

5. The angular momentum, L, is a function of
the mass, m, radius, r, and angular velocity,
w. The dimensions of L and w are ML2T-1
and T-1, respectively. Use dimensional
analysis to determine the formula for angular
momentum.
6. Determine the dimensions of constants A
and B given that h
g
B
t
m
A
F −
=
is force, m is mass, g is acceleration, and h is
, where F
height.

7. The time that an object moves at constant
acceleration in one direction is a function of
the acceleration, a, and the displacement, s.
Use dimensional analysis to determine the
formula for the time that the object moves at
constant acceleration.
8. Determine the dimensions of the constant
3 2
t
a
D
s =
displacement, a is acceleration, and t is time.
D, given that , where s is

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