Measurements and Dimensional Analysis.pdf

Physics dimensional analysis

1. PHY112
GEOMETRICAL OPTICS AND
MECHANICS

2. 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

3. 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

4. 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)

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

6. MOST COMMON PREFIXES
PrefixAbbreviationPower
Tera T 10
12
Giga G 10
9
Mega M 10
6
kilo k 10
3
centi c 10
-2
milli m 10
-3
micro m 10
-6
nano n 10
-9
pico p 10
-12
femto f 10
-15

7. 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

8. 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?

9. ftmi52801= mi
ft
1
5280
1= ft
mi
5280
1
1&= min601=h h1
min60
1= min60
1
1&
h
= )(a s60min1= min1
60
1
s
= s60
min1
1&= 75????????????ℎ
−1
=
75????????????
1ℎ
=
75????????????
1ℎ
×
5280????????????
1????????????

10. min60
1
1
528075 h
h
ft


= s
ft
60
min1
min60
528075


= 1
110
−
= sft )(b inft121= ft
in
1
12
1= in
ft
12
1
1&= cmin54.21= in
cm
1
54.2
1= cm
in
54.2
1
1&=

11. ft
in
s
ft
hmi
1
12
1
110
75
1
=
− cmm1001= m
cm
1
100
1= cm
m
100
1
1&= in
cm
s
in
1
54.2
1
12110


= cm
m
s
cm
100
1
1
54.212110


= 1
5.33
−
= sm

12. Example 2
Convert 9 kg m
-3
to mg cm
-3
.gkg10001= kg
g
1
1000
1= g
kg
1000
1
1&= gg m
6
101= 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&= cmm1001=

13. 3
3
1
9
9
m
kg
mkg=
− kg
g
m
kg
1
1000
1
9
3
= g
g
m
g
1
10
1
9000
6
3
m
= 3
23
9
10
1
1
109










=
cm
m
m
gm 3
0009
−
= cmgm

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

15. 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]? L
3
•What are the [Speed]? LT
-1
•What are the [Force]? MLT
-2

16. 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).

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

18. 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 
cba
TLMQ=
E.g.,
22−
=TMLE 
2−
=MLTF 
3−
=ML 
3
LV=

19. Example 1
where v is the final velocity, uis the initial
velocity, ais the acceleration, tis the time
and k
1and k
2are dimensionless constants is
dimensionally correct.
Show that the expression atkukv
21+=
Solution
1−
=LTv 
11
11
1
−−
=== LTLTukuk

20.  TLTtaktak ==
−2
22
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
−
=LTatk atkvukv
21
&, ==

21. Example 2
The equation for the change of position of a
train starting at x = 0 m is given by32
2
1
BtAtx +=
where x is distance, t is time and Aand B are
constants.
Determine the dimensions of A and B.
2
]][[tAx= 

2
][
t
x
A= 
2
2
−
== LT
T
L
A
AND


3
33
−
=== LT
T
L
t
x
B 
3
tBx=

22. Another method
Make use of 
cba
TLMQ=
Let
2
]][[tAx= 2
*TTLML
cba
= 2010 +
=
cba
TLMTLM
M: a = 0
L: b = 1
T: c + 2 = 0, c = -2
3
]][[tBx=
[B] = LT
-3
cba
TLMA= 
2
,
T
L
A= 3
*TTLML
cba
=
M: a = 0
L: b = 1
T: c + 3 = 0, c = -3
and determine a, b and c.

23. EXAMPLE 3),,(m Lf= zyx
Lkm=
The frequency is given in functional form as
where L is length and the dimensions of , and mare
T
-1
, M
2
L
2
T
-2
and ML
-1
, respectively. Use dimensional
analysis to deduce the formula for frequency.),,(m Lf=
Solution1
:][
−
T 222
:][
−
TLM 1
:][
−
MLm zyx
Lk ][][][][][ m=

24. zyx
MLTLMLT )()(*1
12221 −−−
= zzyyyx
LMTLMLT
−−−
=
2221 yzyzyx
TMLT
2221 −+−+−
= yzyzyx
TMLTML
222100 −+−+−
= T: -2y = -1
y = 0.5
M: 2y + z = 0L: x + 2y –z = 0
1+ z = 0
z = -1
x +1 –(-1) = 0
x + 2 = 0
x = -215.02 −−
= mkL 2
L
k
m

=

25. 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
atuts+= 2
4
1
atuts +=
}
Both dimensionally
correct
Cannot use dimensional analysis when
dealing with trigonometric functions.

26. Additional problems on Dimensional Analysisx
m
k
a

−= x

1.Acceleration for simple harmonic motion is
, where mis 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]

27. dgA
hA
t
2'
= 'A 3. Check the dimensional correctness of
, where Aand
is height, d is distance, and g is acceleration.
are areas, h
4. Check the dimensional correctness ofhgm
PP
v
o
2
)(2
+
−
=

pressures, v is velocity, is density, mis mass,
, where Pand P
oare
gis acceleration due to gravity and his height.

28. 5. The angular momentum, L, is a function of
the mass, m, radius, r, and angular velocity,
w. The dimensions of Land ware ML
2
T
-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 thathgBtmAF −=
is force, mis mass, gis acceleration, and his
, where F
height.

29. 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
taDs=
displacement, ais acceleration, and tis time.
D, given that , where sis