Chapter-2 Measurement of Physical quantities in Physics Introduction You have studied in the last chapter, the fundamentals of scientific study in relevance to the subject Physics, the science of physical phenomenon of nature. You can now proceed to frame the hypothesis/theory/law to explain any physical phenomenon of nature. But as we have already discussed that there is no meaning of words until unless our theory or law is based on certain physical parameters of study or may be called as physical quantities, those could be expressed and measured quantitatively. These quantities are the basis of experimental validation of the theory propounded. Now we shall try to understand the concept of physical quantities regarding their definition and their use in interpreting the laws of Physics. At the same time regarding measurement of these quantities, we shall glance over the different standards of measurements prevailing world wide to quantitatively measure physical quantities and how these standards make computations involving physical quantities make so simple. Unit One Now in this unit of about half an hour you will be able to learn a) Concept of fundamental and derived quantity. b) System of measurements in practice.
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Transcript
Chapter-2
Measurement of Physical quantities in Physics
Introduction
You have studied in the last chapter, the fundamentals of scientific study
in relevance to the subject Physics, the science of physical phenomenon
of nature. You can now proceed to frame the hypothesis/theory/law to
explain any physical phenomenon of nature. But as we have already
discussed that there is no meaning of words until unless our theory or
law is based on certain physical parameters of study or may be called as
physical quantities, those could be expressed and measured
quantitatively. These quantities are the basis of experimental validation
of the theory propounded. Now we shall try to understand the concept of
physical quantities regarding their definition and their use in
interpreting the laws of Physics. At the same time regarding
measurement of these quantities, we shall glance over the different
standards of measurements prevailing world wide to quantitatively
measure physical quantities and how these standards make
computations involving physical quantities make so simple.
Unit One
Now in this unit of about half an hour you will be able to learn
a) Concept of fundamental and derived quantity.
b) System of measurements in practice.
c) System of measurements in S.I. system.
d) System of units and conversion of units from one system to
another.
Concept of fundamental and derived quantity
As we have discussed that to understand the seen phenomenon of nature
theories/hypothesis/laws are framed. But there is no sense and meaning
of words until unless the laws or postulates are based on measurable
quantities. The laws of physics are expressed and experimentally
validated in terms of these physical quantities. The laws expressed in
terms of these physical quantities are invariant with space and time.
For the purpose of Physics it is very essential that quantities should be
defined clearly and precisely and should have the conceptual meaning
for the postulated law. Among these are force, time, velocity, density,
temperature, charge, magnetic susceptibility and numerous others.
Now as per the practice all the physical quantities involved in the Physics
today may be grouped in two sets. One set contains the physical
quantities as a fundamental one and other set contains quantities derived
from these fundamental one. For an example length and time may be
taken as a fundamental quantities and the quantity velocity defined as
the ratio of length upon time may be termed as a derived quantity.
However which quantity will be taken as a fundamental and which one as
derived depends on system to system. For example S.I. system of units
takes length, mass and time as a fundamental quantity and the quantity
force defined as the product of mass and acceleration (length/time2) as a
derived quantity. While the F.P.S. system of units and measurement
takes force, length and time as a fundamental quantity and mass as a
derived quantity.
Now the intention of creating physical quantities is that physical laws are
expressed in terms of these quantities and which are themselves
evaluated in terms of numerical values. These values should be such that
can be characterized as intrinsic to the experiment conducted and may
be compared to the same quantity arrived in another set of experiment.
Now the comparison of physical quantities can be done only when each
physical quantity either fundamental or derived one is based on some
standard of measurement. The standard of measurement is a set of
standard where each fundamental quantity is assigned a specific unit
value. The unit value to that quantity is such that it is invariant with
space and time and is easily accessible. For example in the S.I. system of
measurement the physical quantity mass has been given a unit value as
one kilogram and defined as mass of the international standard body
preserved at Severes, France. However it certainly involves the
procedure to account the unit value of the fundamental quantity, where
different objects having different values in the same set of conditions can
be compared. For example by the procedure to obtain mass of a standard
body, the spring balance in terms of stretch of spring may be used. The
stretch of spring is directly proportional to the mass of the body. By
comparing the different stretches of spring, the masses of different
bodies, may be compared.
So in final words the operational definition of fundamental quantity
involves two steps, first choice of a standard and second the
establishment of procedures for measuring the quantity in terms of
standard so that a number and unit are determined as a measure of
quantity. But very important aspect of choice of standard is that it should
be accessible and invariant. For example we have selected our standard
for length to measure the distance between two points as one meter,
then by a comparison of this length with a second object three times in
length as standard, we say that second bar has a length of three meters.
However most quantities cannot be measured directly in comparison to
standard and indirect approach using some involved procedure is
required and also certain assumptions are made to ascertain the matter.
For example measuring the time of sending and receiving the
electromagnetic pulse with known speed, the distance can be measured
as product of speed and one half of time interval. Here we have set half
time of sending and receiving the signals as our standard of
measurement and different distances may be compared with different
times of observations. However the speed of pulse is to be determined
through the other acclaimed procedure.
Similarly, we use an indirect method to measure very small distances
between atoms and molecules by particle scattering method.
System of measurements in practice
Now so far we have studied that quantities in physics are either
fundamental one or derived one depending upon the system of
measurement we are using and shall study the different systems of
measurement prevailing and used all over the world. The fundamental
quantities are defined in terms of a standard of measurement devised in
that particular system of measurement and procedure to measure the
quantity so that comparison of different objects may be done in terms of
that quantity. The derived quantities are derived from these fundamental
one and for the complete descriptions will have dimensions showing the
fundamental quantities involved and units of these dimensions involved
depending upon the system of measurement used.
Now we shall discuss the different system of measurements prevailing all
over the world and their merits and demerits.
Imperial System of units measurement
Before S.I. system of units adopted around the world, the British
systems of English units and later Imperial system of units were used in
Britain, the Commonwealth and the United States. The system came to
be known as U.S. Customary units in the United States and is still in use
there and in a few Caribbean countries. These various systems of
measurement have at times been called foot-pound-second systems
after the Imperial units for distance, weight and time. Many Imperial
Ex 10. The SI and C.G.S. units of energy are joule and erg respectively. How many erg are
equal to one joule.
Sol: In M.K.S. system of units Energy =Kg m2 /s 2 = 1 joule= 107 grams cm 2 /s 2 = 1 Erg in
C.G.S. system
Ex 11. Young’s modulus of steel is 19x1010 N/m2. Express it in dyne/cm2. Here dyne is the
C.G.S. unit of force.
Sol: 19 x1011 Dyne /cm2;
Ex 12. The density of a substance is 8 g/cm3. Now we have a new system where unit of
length is 5 cm and unit of mass 20 g. Find the density in the new system.
Sol: Density of substance = 8 g/cm3; Unit of length = 5 cm ; Unit of mass = 20 g.
Since dimension of density = ML-3T0
2/
1 = (M
2/M
1) (L
2/L
1)-3 = 20 (5)-3= 4/25; Density of substance in new system = 50 units
Exercise 1
Q1. If the units of force, energy and velocity in new system be 10 N, 5 J and 0.5 m s-1
respectively, find the units of mass, length and time in the new system.
Which gives M=E/v2=5J/(0.5)2 = 20 Kg; L=E/F=0.5 m; T=E/FV=1 sec
Q2. The radius of proton is about 10-9 microns and the radius of the universe is about 1028 cm.
Name a physical object whose size is approximately half way between these two on a
logarithmic scale.
Q3 Assuming that the length of the day uniformly increases by 0.001s in a century. Calculate
the cumulative effect on the measure of time over twenty centuries. Such a slowing down of
the earth‘s rotation is indicated by observations of the occurrences of solar eclipses during
this period.
Q4. The unit of length convenient on the nuclear scale is fermi ; 1 f=10-15 m. Nuclear sizes
obey roughly the following empirical relation; r=r0A10 where r is the radius of the nucleus.
A its mass number and r0 is a constant equal to 1.2 f. Find out whether mass density is nearly
constant for different nuclei.
Solutions Exercise 1
Ans1. Let M = F x E y v z and so x=0, y=1 and z=-2
Ans2: On logarithmic scale the exponent value= -15+(15+26)/2= -15 + 41/2=5.5
Hence the size of object=105.5 1x106 ( Radius of moon)
Ans3: Increase in length of day in 20 centuries = 20 x 0.001s
Average increase in length of day = 0.001x 20/2 = 0.01s
Cumulative error in time measurement = cumulative error in day length increase in 20
centuries = 20 x100 x 365 x 0.01/3600 h= 2.03 h
Ans4:Mass density of nucleus=A(mp)/(4/3 r3) 1017 Kg/m3
Introduction
So far we have defined the physical quantities in terms of their interdependence and evolved
procedures to measure the physical quantities. Now we shall try to understand the way the
physical quantities appear in the laws of physics and the rules of mathematical manipulations
followed.
Each physical quantity may be attributed a set of dimensions according to the base units
involved in the quantity and laws of physics involve mathematical manipulations of these
physical quantities. As each statement should follow certain rules involving the physical
quantities regarding their mathematical manipulations, is governed by the dimensional
analysis.
Beside that physical quantity regarding their representation numerically should follow certain
rules so that degree of accuracy of the measurement could be ascertained at a glance of the
presentation of measurement. It is governed by the rules of significant digits of measurement
of physical quantities. These two are the scope of study now.
So at the end of this unit of around one hour you will be able to learn
Dimensions of physical quantities and dimensional analysis.
Concept of significant digits for the presentation of measurement of quantities in Physics.
Dimensions of Physical Quantities
As we have stated earlier that physical quantities can be added, subtracted, multiplied or
divided in the same manner as any other algebraic quantity. The addition or subtraction of
quantities is meaningful only if the quantities have followed same standard of units and so
are dimensionally homogeneous. The dimensional consistency is must for equation to be
correct. The correctness of equation can be checked by comparing the physical dimensions of
each term in the equation under discussion. A list of various physical quantities with their
usual convention and dimensions are
:
S.No Quantity Symbol Physical formula
S.I unit Dimension formula
1 Acceleration a a=v/t m/s2 M0LT-2
2 Angular acceleration
= /t Rad/sec2 M0 L0 T -2
3 Angular Displacement
=arc/radius Radian M0 L0 T0
4 Angular Momentum
L L=m v r Kg m2/s M L2 T -1
5 Angular Velocity
=/t Rad/sec M0 L0 T -1
6 Area A l x b (Metre)2 M0L2T0
7 Capacitance C Q = CV Farad M-1L-2 T 4A2
8 Charge q q=I t Coulomb M0 L0 T A9 Current I ---- Ampereor A M0L0T0A10 Density d d =M/V Kg/m3 M L-3 T 0
11 Dielectric constant
r
r=/0 ___ M0L0T0
12 Displacement S ___ Metre or m M0LT0
13 Electric dipole
moment
P P=q 2l Coulomb metre
M0L T A
14 Energy KE/U KE=1/2mv2 Joule M L2 T -2
15 Force F F=ma Newton or N M L T -2
16 Force Constant
K K=F/x N/m M L0 T -2
17 Frequency f f=1/T Hertz M0 L0 T -1
18 Gravitational Constant
G G=F r2/m1m2 N m2/kg2 M-1L3 T -2
19 Heat Q Q=mst Joule/calorie M L2 T -2
20 Impulse - F x t N.sec M L T -1
21 Intensity of electrical field
E E = F/q N/coul M L T -3A-1
21 Intensity of electrical field
E E = F/q N/coul M L T -3A-1
22 Latent Heat L Q = mL Joule/kg M0 L2 T -2
23 Magnetic dipole
moment
M M=NIA Amp.m2 M0 L2A T 0
24 Magnetic flux E=d/dt Weber M L2 T - 2A-1
25 Magnetic intensity
H B =H A/m M L-1T 0A
26 Moment of Inertia
I I=mr2 Kg m2 ML2T0
27 Momentum P P = mv Kg m/s M L T -1
28 Mutual inductance
L E=L.di/dt Henery M L 2 T -2A-2
29 Power P P=W/t Watt M L2 T -3
30 Pressure P P =F/A Pascal M L-1 T -2
31 Resistance R V=IR Ohm M L2 T-3 A-2
32 Specific heat S Q=m s t Joule/kg.kelvin
M0 L2T -2-1
33 Strain ___ l/l; A/A; v/v
___ M0L0T0
34 Stress ___ F/A N/m2 M0 L0 T -2
35 Surface Tension
T F / l N/m M L0 T -2
36 Temperature ___ Kelvin M0 L0 T0 1
37 Torque F.d N.m. M L2 T -2
38 Universal gas constant
R PV = nRT Joule/mol.k M L 2 T -2-1
39 Velocity v v = s/t m/s M0LT -1
40 Volume V V=l x b x h(Metre)3
M0L3T0
41 Work W F.d N.m M L2 T-2
42 Young Modulus
Y Y = (F/A)/l/l
N/m2 M L-1 T -2
Application of dimensional analysis
1. To find the unit of a physical Quantity.
2. To convert units of a physical Quantity from one system to another.
3. To check the dimensional correctness of a given relation. It is to be further noted that
every dimensionally correct relation does not mean to have physically correct relationship. If
we have some idea or can make an educated guess as to how one physical quantity relates to
another we can use dimensions to derive the form of the equation.
As an example, consider the equation for the period of pendulum bob. We might suppose that
the period depends on the mass of the bob, the length of the pendulum and the acceleration
due to gravity
We can express this as T=mx l y gz. Where x, y and z are as yet undetermined indices.
To find the values of x, y and z we convert the formula into its dimensions. On the left-hand
side the dimension of the period is [T], the dimension of mass is [M], the dimension of the
length of the pendulum is [L] and the dimension of g is [LT-2].
[T]=[M]x[L]y[LT-2]z.
Equating left-hand indices with matching dimensions on the right-hand side.
[M]: x=0
[L]: y+z=0
[T]: 2z=-1
From this we can deduce that z=-1/2, while y=1/2 and x=1/2
Substituting these values into the original equation we obtain. T=
k(l/g)1/2. Where k is a constant of proportionality. Compare this with the
equation for the period of a pendulum
The form of the equation is correct, but it cannot determine the constant of proportionality.
Limitations of application of dimensional analysis
1. If the dimensions are given, physical quantity may not be unique as
many physical quantities have same dimensions.
2. Since numerical constants have no dimensions, can’t be deduced by
dimensional analysis. For example 1, etc.
3. The dimensional analysis can’t be used to derive relationship other
than the product of power functions. However the dimensional
correctness of the relation may be checked. For e.g. S=u t+1/2 a t2 , y= a sin t
4. The method of dimension can’t be applied to derive relationship when a physical quantity
depends on more than three quantities. For e.g. T=2 I/(m g l)
Ex1. If velocity, force and time are taken to be fundamental quantities find dimensional
formula for quantity mass
(A) V-1
FT-1
(B) V-1
FT (C) VF-1
T-1
(D) V-1
F-1
T
Sol: M=K (V)x (F)y (T)z ; x= -1, y =1, z =1; M = K V-1 F T
Ex2. You may not know integration. But using dimensional analysis can check or prove
results .in the integral
dx/(2ax-x2)1/2 = an sin-1 [x/a-1]
The value of n should be
(A) 1 (B) –1 (C) 0 (D) 1/2
Sol: The dimension of a is that of x and for dimensional consistency of the equation n should be equal to zero.
Ex3. In the cauchy’s formula for the refractive index n = A+B/2 the dimensions of A and B
are
(A) Both are dimensionless (B) A is dimension less , B has dimension M0 L-2 T0
(C) A is dimension less, B has dimensions M0L2T0 (D) Both A and B have dimensions
M0L2T0
Sol: Since refractive index is dimensionless hence A should be dimensionless and B should have dimensions of Length 2 that is M0L2T0
Ex4. A highly rigid cubical block A of small mass M and side L is fixed rigidly on the
another cubical block B of same dimensions and of low modulus of rigidity such that
lower face of A completely covers the upper face of B. The lower face of B is rigidly held
on a horizontal surface. A small force F is applies perpendicular to one of the side face of
A. After the force is withdrawn, block A executes small oscillation, the time period of
which is given by
(A) 2 (mL) (B) 2 (M/L) (C) 2 (ML/) (D) 2 (M/L)
Sol: Check the dimensions of the left and right side quantities.
Ex5. The frequency of oscillation of an object of mass m suspended by means of spring of
force constant k is given by f = c m x ky, where c is a dimensionless constant the value of
x and y are
(A) x = ½, y = ½ (B) x = - ½, y = ½ (C) x = ½, y = - ½ (D) x = - ½, y = - ½
Sol: Putting the dimensions on the two sides and equating the powers of the base quantities we get x= -1/2, y =1/2
Multiple Choice Type of Questions
Ex1. Suppose a quantity x can be dimensionally represented in terms of M, L and T i.e.[X] =
Ma Lb T c the quantity mass
(A) Can always be dimensionally represented in terms of L, T and x.
(B) May be represented in terms of L,T and x if a = 0
(C) May be represented in terms of L, T and x if a = 0
(D) May be represented in terms of L, T and x if a 0
Sol: Any quantity may be represented in terms of other quantities as base quantities such that the quantities are dependent upon each other with a certain relationship. Here I the case the quantity M may be represented in terms of L, T, x if a0 so that M remains related with the rest of quantities.
Ex2. Choose the correct statement (s)
(A) A dimensionally correct equation may be correct
(B) A dimensionally correct equation may be incorrect
(C) A dimensionally incorrect equation must be correct
(D) A dimensionally incorrect equation must be incorrect
Sol: A dimensionally correct equation may be correct or incorrect. But a dimensionally incorrect equation can never be correct.
Ex3. Choose the correct statement (s)
(A) All quantities may be represented dimensionally in terms of the base quantities.
(B) All base quantities cannot be represented dimensionally in terms of the rest of base
quantities.
(C) The dimension of a base quantity in other base quantity is always zero.
(D) The dimension of a derived quantity is never zero in any base quantity.
Sol: ( A ), (B),( C )
Descriptive Type of Questions:
Ex.1 (a) Can a physical quantity have no unit and dimensions? If yes give an example.
(b) Can a physical quantity have units without dimensions? If yes give an example.
Ans41: (a) Strain has neither unit nor dimensions but it is a defined physical quantity.(b) Angle measured in radians is physical quantity, which has unit of radian but no dimension. The angle measurement in a plane that is radian and angle of a solid that is steradian are supplementary units to supplement the physical quantities like angular displacement and angular velocity. No dimension has been assigned to these quantities.
Ex2. Find the dimensional formulae of e0& m0 (Where e0 is the absolute permittivity and m0
is the permeability of vacuum or free space respectively ).
Sol: The dimensions of 0= M-1 L-3 T 4A2 ; The dimensions of
0 = MLT-2A-2
Ex3. Assuming that the largest mass that can be moved by a flowing river depends on
velocity of flow density of river water and on gravity, find out how the mass varies with
velocity of flow.
Sol: M=K(v)x(d)y(g)z ; x=6, y=1, z=-3 and so M= K v6 d g –3
Ex4. The gas constant R depends upon pressure of the gas P, volume of the gas V,
temperature of the gas T and number of moles n. Derive an expression for gas constant, R
Sol: Let P= K V x n y R z T m , on solving for dimensions of base quantities M,L,T and n we
get x=-1, y=1, z=1, T=1 and so PV=n RT
Exercise 2
Matching Type of Questions:
Q1. In the following table, there are two lists A and B, but the list B is not in order of list A.
write down the list B in order of list A in each table.
(a) Moment of inertia (i) Newton /Meter 2
(b) Surface tension (ii) kg/ (metre-s)
(c) Angular acceleration (iii) kg – meter2
(d) Coefficient of viscosity (iv) Newton/meter
(e) Coefficient of elasticity (v) radian/s2
(f) Momentum (vi) MLT-1
(g) Gravitational Constant (vii) ML2T-1
(h) Plank Constant (viii) M-1 L3 T-2
Q2. Column I gives three physical quantities. Select the appropriate units for these from
choices given in column II. Some of the physical quantities may have more than one choice :
I II
Capacitance Ohm second
Inductance Coul2 joul-1
Magnetic inductance Coulomb (Volt)1
Newton(ampere-m)-1
Volt-sec (Ampere)-1
Q3. Match the physical quantities given in column I with dimensions expressed in column if
in tabular form
(a) Angular momentum (a) ML2T-2
(b) Latent heat (b) ML2Q-2
(c) Torque (c) ML2T-1
(d) Capacitance (d) ML3T-1Q-2
(e) Inductance (e) M-1L-2T2Q2
(f) Resistivity (f) L2T-2
Q4. The position of a particle at any time is given by S(t) = V0/a (1-e-at), where a>0 and V0
are constants. What are the dimensions of a and V0 ?
Q5. The equation of a wave traveling along the x axis by y = A e[x/a-t/T] 2, write the
dimensions of A , a and T.
Q6. Suppose an attractive nuclear force acts between two protons which may be written as F
= Ce-kt/r2
Write down the dimensional formula and appropriate SI units of C and K.
Q7. Find the dimensional formula of L/R, where R is the resistance and L is the coefficient of
self-inductance.
Q8. Find out the dimensions of electrical conductivity.
Q9. The equation of state of a real gas is given by [p+a/v2](v-b) = RT, where p, v and T are
pressure , volume and temperature respectively and R is the universal gas constant. Find out
the dimensions of the constant a in the above equation.
Q10. The heat produced in a wire carrying an electric current depends on the current, the
resistance and the time; assuming that the dependence is of the product of powers type, guess
an equation between these quantities using dimensional analysis. The dimensional formula
of resistance is ML2T-3 A-2 and heat is a form of energy.
Q11. The frequency of vibration of a stream depends on the length L between the nodes, the
tension F in the string and its mass per unit length M. Guess the expression for its frequency
from dimensional Analysis.
Q12. The kinetic energy E of a rotating body depends on its moment of inertia I and its
angular speed w. Assuming the relation to be E =K Ia wb where K is a dimensionless
constant, find a and b. Moment of Inertia of the sphere about its diameter is 2/5Mr2.
Q13 (a) In the formula X=3YZ2, X and Z have dimensions of capacitance and magnetic
induction respectively. What are the dimensions of Y in MKSQ system?
(b) Calculate the dimension(s) of VCR/L, where V=supply voltage, C=capacitance,
R=resistance and L=inductance.
Q14. If velocity of light c, the gravitational constant G and plank constant h be chosen as the
fundamental units, then find the dimensions of mass, length and time in the new system.
Concept of Significant digits
As we have seen every measurement pertains to the standard we are
going to use and its numerical value is read from the calibrated scale
based on that standard of measurement. The value of measurement
contains two parts
(i) One part with all digits read directly from the scale by the smallest
subdivision of the scale called as certain digits of measurement
(ii) and the second part contains doubtful digits at end corresponding to
the eye judgment within the least sub-division of the scale. For example
the length measured by the meter scale having least count of 1cm as
50.35 cm contains digits 5, 0 as certain 3 is doubtful and 5 is
insignificant. The digits 5, 0, 3 are termed as significant digits and 5 as
insignificant digit. There may be some confusion while dealing with the
measurements containing zero at their end but rule is the same.
As a general practice we report only the significant digits and magnitude
of any physical quantity is represented by proper power of 10. For
example if only 5 is significant in 500 cm then we report it as 5 x 102 cm,
with 5 as a significant digit. If 5 and first zero that is two digits are
significant then we write 5.0 x 102 cm. and if all the digits are significant
we report it as 5.00 x 10 2 cm.
If the integer part of the digit is zero then all the continuous zeros after
the decimal are treated as insignificant digits as the number 0.00003
having first digit as zero then all the continuous digits are also
insignificant. For this reason it is important to keep the trailing zeros to
indicate the actual number of significant figures.
However if first digit is nonzero as in 1.0005 then all zeros will be
counted for significant digits containing five significant digits as 1,0,0,0
and 5 respectively.
For numbers without decimal points, trailing zeros may or may not be
significant. Thus, 400 indicate only one significant figure. To indicate
that the trailing zeros are significant a decimal point must be added. For
example, 400 has three significant figures, and has one significant figure.
Exact numbers have an infinite number of significant digits. For example,
if there are two oranges on a table, then the number of oranges is
2.000... . Defined numbers are also like this. For example, the number of
centimeters per inch (2.54) has an infinite number of significant digits, as
does the speed of light (299792458 m/s).
Significant digits in arithmetical calculation
As per internationally accepted practice for finding out the significant digits in the
arithmetical calculation say division or multiplication of two physical quantities following
rule has been formulated for determination of significant:
1. In multiplication or division of two or more quantities the number of significant digits in
the answer is equal to the number of significant digits in the quantity, which has least number
of significant digits. Thus 60.0/2.0 will have only two significant digits. The insignificant
digits are dropped from the result if appear after the decimal point, and replaced to zero if
appear to the left of the decimal point. The least significant digit is rounded according to the
following rules:
(A) If the digit next to one rounded is more than 5, then the digit to be rounded is increased
by 1
(B) If next digit is less than 5 then rounding digit is left unchanged.
(C) If the digit next to be rounded is equal to 5 then rounding digit is increased by one if it is
odd otherwise left unchanged.
For example,
(2.80) (4.5039) = 12.61092
should be rounded off to 12.6 (three significant figures like 2.80).
2. For addition or subtraction of quantities all the numbers are written with the decimal point
in one line up to the number, which has the first doubtful digit counted from left. All digits
are written to that digit after rounding off at that level and addition/subtraction is performed.
For example,
89.332 + 1.1 = 90.432
should be rounded to get 90.4 (the tenths place is the last significant place in 1.1).
Ex1. Evaluate 25.2x1374/33.3 All the digits in this expression are significant.
Sol: Expression has value =1039.7838Since the expression has number with lowest number of significant digits as 3 with number
25.2 or 33.3 hence the expression will also have three significant digits and number will be
written as per rule 1.04x103
Ex2. Evaluate 24.36+0.0623+256.2
Sol: As per rule write the numbers with decimal point in a line and check where the first
doubtful digit occurs between these numbers, which in our case occurs at 256.2. Now change
the other numbers with proper rounding and than add as 24.4 + 0.1 + 256.2 = 280.7
Ex3. Given P=0.0030 m; Q=2.40 m and R=3000 m. Find out the number of significant
figures in P, Q, R respectively.
Sol: No of significant figures in P=2(3,0) since the first digit is zero so all zeros after that are insignificant;No of significant figures in Q=3(2,4,0); No of significant figures in R=4(3,0,0,0)
Ex4. The volume of one sphere is 1.76 c.c. Find out the volume of 25 such spheres
(according to idea of significant figures).
Sol: The volume of 25 spheres=25 x1.76 = 44. Since the number 1.76 has three significant
digits and so result should also be written to the three significant digits as 44.0 c.c,
Exercise 3
Q1. The length breadth and thickness of a block are measured as 12.5 cm, 5.6 cm, and 0.32
cm respectively. Which measurement is least accurate?
Q2. The radius of the circle is stated as 2.12 cm. Find out its area written as in appropriate
number of significant digits.
Q3. Round off the following numbers to three significant digits (a) 15462 (b) 14.745 (c)
14.750 (d) 14.650x1012.
Hints and Solutions Exercise 2
Ans4: The power of e should be dimensionless so the dimension of a = M0L0T-1; The
dimension of V0=M0LT-1;
Ans5: The dimension of a and T should be of such that x/a and t/T are dimensionless and
dimensions of A should be of y.
So the dimension of T= M0L0T; A= M0L T0 ; a=M0 L T 0;
Ans6: The dimension of C = M L1T -2 ; The dimension of K=M0L2T-1
Ans7: The dimension of L=V/(dI/dt)=W/I 2 =ML2T-2A-2; The dimension of R=W/I 2 t
=M L2 T-3A-2
Ans8: Electrical conductivity=1/R= M-1 L-2 T 3 A2
Ans9: a=pv2=ML5T-2;
Ans10: H= I x R y t z which on solving for dimensions of two side for base quantities M, L, T
and I we get H=I2 R t
Ans11: Let = K L x F y m z
Put the dimensions of the quantities on two sides and solve for powers x,y, and z we get = K/L (F/m)Ans12: E.= K I x y
Ans1: The breadth of the block is only up to two significant digits hence supposed to be least accurate.Ans2: Area of circle= R2= 14.112416 cm2=14.1 cm2 (three significant figures)
Ans3: (a) 1.55x 104 ; The last number has been rounded to five since digit next to 4 is six greater
than 5
(b) 1.47 x 101 The last number has been rounded to 7 only since digit next to it is less than 5
(c) 1.48 x 101 The last number has been rounded to 8 since digit next to it is 5, and itself odd number.
(d) 1.46 x 1013 The last number has been rounded to 6 only since digit next to it is equal to 5 and
itself even number.
Introduction
Now let us continue our discussion about measurement about physical quantities. We know,
that all sort of measurements are arrived at by taking measurements in some set of
experiments of the physical quantities. The authenticity of the results of experiments is
totally dependent upon the precision of the measurements taken, which itself dependent upon
the certain factors like how the instrument is calibrated with respect to reference one, least
sub division of instrument, skill of the person making measurements, secondary effect of
environment or errors in the instrument etc. Although we try to make accurate measurements
but it is also true that it is very difficult to arrive at the fictitious true value of measurement.
One measurement of the same quantity taken many times will differ from each other. So it is
a very difficult task to arrive the true value of the measurement. Now in this unit we shall try
to understand the different causes of errors in measurement and will classify the errors
according to their origination. Next we shall determine the standard procedures to arrive at
the true value and mode of representing them for all purposes.
Now in this next unit of around one hour you will be able to learn
Types of errors in measurements and level of uncertainty.
Presentation of magnitude of quantities in Physics.
Errors and Uncertainty
Errors are always the part of measurements and nothing can be done about. If a measurement
is repeated, the values obtained will differ and none of the results can be preferred over the
others. Although it is not possible to do anything about such error, it can be characterized.
For instance, the repeated measurements may cluster tightly together or they may spread
widely. This pattern can be analyzed systematically.
When we measure something the measurement is meaningless without knowing the
uncertainty in the measurement. This leads us to the idea of errors in measurement. Other
factors such as the conditions under which the measurements are taken may also affect the
uncertainty of the measurements. Thus when we report a measurement we must include the
maximum and minimum errors in the measurement.
For example, measuring the height of a person, the measure may be accurate to a scale of 1
mm. But depending on how the person being measured holds himself during the
measurement we might be accurate in measuring to the nearest cm.
Generally, errors can be divided into two broad and rough but useful classes: systematic and
random.
Systematic errors
Systematic errors are errors, which tend to shift all measurements in a systematic way so
their mean value is displaced. This may be due to such things as incorrect calibration of
equipment, consistently improper use of equipment or failure to properly account for some
effect. In a sense, a systematic error is rather like a blunder and large systematic errors can
and must be eliminated in a good experiment. But small systematic errors will always be
present. For instance, no instrument can ever be calibrated perfectly.
Other sources of systematic errors are external effects which can change the results of the
experiment, but for which the corrections are not well known. In science, the reasons why
several independent confirmations of experimental results are often required (especially
using different techniques) is because different apparatus at different places may be affected
by different systematic effects. Aside from making mistakes (such as thinking one is using
the x10 scale, and actually using the x100 scale), the reason why experiments sometimes
yield results, which may be far outside the quoted errors, is because of systematic effects,
which were not accounted for.
Random errors
Random errors are errors, which fluctuate from one measurement to the next. They yield
results distributed about some mean value. They can occur for a variety of reasons.
They may occur due to lack of sensitivity. For a sufficiently a small change an instrument
may not be able to respond to it or to indicate it or the observer may not be able to discern
it.
They may occur due to noise. There may be extraneous disturbances that cannot be taken
into account.
They may be due to imprecise definition. For example taking measurements with a
magnetic compass, the effect of improper leveling of instrument during observations.
They may also occur due to statistical processes such as the roll of dice.
Random errors displace measurements in an arbitrary direction whereas systematic errors
displace measurements in a single direction. Some systematic error can be substantially
eliminated (or properly taken into account). Random errors are unavoidable and must be
lived with.
So the systematic errors are to be removed from the measurements by rectification of the
cause or my taking measurements by several instruments otherwise the results will remain
shifted from the true value. However the random errors are very uncertain and it is very
difficult to account for them in our measurements. Random errors will always remain in our
measurements how precisely we have taken our measurements. By taking repetitive number
of measurements and taking average of large number of measurements, the average value
will be close to the true value. But still there is some uncertainty associated with the true
value.
The uncertainty associated with the mean value is determined by the standard deviation of
the measurements as detailed below:
Let x1 x2 x3 x4 x5…xN be the results of an experiment repeated N times then standard
deviation is defined as
Where is the average of all values of x. It is supposed to be the best value of x
derived from the experiments and the true value is likely to occur within a range .
However the interval of is quite often taken as the interval in which the true
value lies with 95% probability. And if the interval is chosen to be than the
probability of occurrence of true value in that interval is 99 %.
Probabilities of occurrence the true value in any range say is given by
But it is fully acceptable only if the numbers of experiments are large. In general the value of
N should be greater than 8 for a good approximation.
Fractional and percentage errors
If x is the error in the measurement in the value x then fractional and Percentage errors are
defined as :
Fractional error=x/x
Percentage error=x/x100 %
Propagation of errors (addition and subtraction)
Let error in quantity x is x and error in quantity y is y then the error in x + y or x - y
is (x+y) that is the errors add.
Prefixes and Magnitudes
To make sense of the vast range over which physical quantities are
measured, prefixes are used as a short-cut to writing the magnitude
using scientific notation. Other prefixes which are commonly used but
are not strictly part of the SI system.
Q1.When a current of 2.50.5 ampere flows through a wire, it develops a potential
difference of 201 volt. Find the resistance of the wire-
(A) 6.0 3 (B) 7.0 2(C) 8.0 2 (D) 9.0 3
Ans1: R=V/I=8Also R/R=V/V-I/IFor max R/R all terms to be positive and thereforeR/R=V/V + I/I=0.25; R=2 Ohmand so R= V/I R = 8.0 2
Q2. In an experiment the value of two resistance were measured to be as given below R1 =
5.0 0.2 ohm and R2 = 10.0 ± 0.1 ohms. Find there combined resistance in (i) series (ii)
parallel.
Ans2: When resistance are in series R=(R1+ R2)
R=R1+R2= 0.3 and R=15 0.3 Ohm
and when in parallel R=(R1R2)/(R1+ R2)
R/R=R1/R
1+R
2/R
2-R
2/(R
1+R
2)- R
2/(R
1+R
1)
For max R all terms must be positive and R/R =7%; and R=3.3 7%
Q3. In an experiment to determine acceleration due to gravity by simple pendulum, a student
commits 1% positive error in the measurement of length and 3% negative error in the
measurement of time period. Find the percentage error in the value of measurement of g.
Ans3: We have T=2 L/g; or T2= K L / g2 ln T= ln K +ln L- ln g; (2/T) dT=1/L dL-1/g dg; So actual percentage error in measurement of g value (dg/g) =7%
Q4. A naval destroyer is testing five clocks. Exactly at noon as determined by the wwv signal
on the successive days of a week the clocks read as follows
Clock Sun Mon Tue Wed Thru Fri Sat
A 12:36:40 12:36:56 12:37:12 12:37:27 12:37:44 12:37:59 12:38:14
B 11:59:59 12:00:02 11:59:57 12:00:07 12:00:02 11:59:56 12:00:03
C 15:50:45 15:51:43 15:52:41 15:53:39 15:54:37 15:55:35 15:56:33
D 12:03:59 12:02:52 12:01:45 12:00:38 11:59:31 11:58:24 11:57:17
E 12:03:59 12:02:49 12:01:54 12:01:52 12:01:32 12:01:22 12:01:12
Justify your choice.
Ans4: The standard deviations of the five clocks are increasing in order of C, D, A, B, E. Hence the clocks with minimum standard deviation is the most consistent one. So the clocks placed in the same order may be kept in terms of consistency as good timekeepers.
Q5. A wire has a mass 0.3±0.003 g, radius 0.5±0.005 mm and length
6±0.06 cm. Find out the maximum percentage error in the measurement
of its density.
Ans5: d=M/r2 L; For Max, d/d all terms should be positive so Maximum error =M/M+2/r r+L/L=0.04=4%
Vernier Calliper
The meter scale enables us to measure the length to the nearest millimeter
only. Engineers and scientists need to measure much smaller distances
accurately. For this a special type of scale called Vernier scale is used.
Vernier Calliper
The Vernier scale consists of a
main scale graduated in
centimeters and millimeters.
On the Vernier scale 0.9 cm is
divided into ten equal parts.
The least count or the smallest
reading which you can get with
the instrument can be
calculated as under:
Least count = one main scale (MS) division - one vernier scale (VS) division.
= 1 mm - 0.09 mm
= 0.1 mm
= 0.01 cm
The least count of the vernier
= 0.01 cm
The Vernier calliper consists of a main scale fitted with a jaw at one end.
Another jaw, containing the vernier scale, moves over the main scale. When the
two jaws are in contact, the zero of the main scale and the zero of the vernier
scale should coincide. If both the zeros do not coincide, there will be a positive
or negative zero error. After calculating the least count place the object
between the two jaws. Record the position of zero of the vernier scale on the
main scale (3.2 cm in figure below).
Principle of Vernier
You will notice that one of the vernier scale divisions coincides with one of the
main scale divisions. (In the illustration, 3rd division on the vernier coincides
with a MS division).
Reading of the instrument = MS div + (coinciding VS div x L.C.)
= 3.2 + (3 x 0.01)
= 3.2 + 0.03
= 3.23 cm
To measure the inner and outer diameter of a hollow cylinder or ring, inner and
outer callipers are used. Take measurements by the two methods as shown in
figure below.
Micrometer Screw-Gauge
Micrometer screw-gauge is another instrument used for measuring accurately
the diameter of a thin wire or the thickness of a sheet of metal.
It consists of a U-shaped frame fitted with a screwed spindle which is attached
to a thimble.
Screw-gauge
The screw has a known pitch such as 0.5 mm. Pitch of the screw is the distance
moved by the spindle per revolution. Hence in this case, for one revolution of
the screw the spindle moves forward or backward 0.5 mm. This movement of
the spindle is shown on an engraved linear millimeter scale on the sleeve. On
the thimble there is a circular scale which is divided into 50 or 100 equal parts.
When the anvil and spindle end are brought in contact, the edge of the circular
scale should be at the zero of the sleeve (linear scale) and the zero of the
circular scale should be opposite to the datum line of the sleeve. If the zero is
not coinciding with the datum line, there will be a positive or negative zero
error as shown in figure below.
Zero error in case of screw gauge
While taking a reading, the thimble is turned until the wire is held firmly
between the anvil and the spindle.
The least count of the micrometer screw can be calculated using the formula
given below:
Least count
= 0.01 mm
Determination of Diameter of a Wire
The wire whose thickness is to be determined is placed between the anvil and
spindle end, the thimble is rotated till the wire is firmly held between the anvil
and the spindle. The rachet is provided to avoid excessive pressure on the wire.
It prevents the spindle from further movement. The thickness of the wire could
be determined from the reading as shown in figure below.
Reading = Linear scale reading + (Coinciding circular scale x Least count)
= 2.5 mm + (46 x 0.01)
= (2.5 + 0.46) mm = 2.96 mm
Relationship in the Metric system of length
1 kilometer (km) = 103 m
1 centimeter (cm) = 10-2 m
1 millimeter (mm) = 10-3 m
Q1. The pitch of a screw gauge is 1mm and there are 100 divisions on the circular scale.
While measuring diameter of a wire the linear scale reads 1mm and 47 th division on the
circular scale coincides with the reference line. The length of wire is 5.6 cm. Find the curved
surface area (in cm2) of the wire in appropriate number of significant digits.
Q2. In a Searle’s experiment the diameter of the wire as measured by a screw gauge of least
count 0.001 cm is 0.050 cm. The length measured by a scale of least count 0.1 cm is 110.0
cm. When a weight of 50 N is suspended from the wire the extension is measured to be 0.125
cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of
Young’s modulus of the material of the wire from these data.
Solution:
Ans 1: Dia of wire=1+47/100=1.47 mm=0.147 cm; Length of wire =5.6 cm
Hence curved surface area = D L=2.6 cm2.
Ans 2: Y=F.L/A.F.L/ r 2 On differentiating two sides and dividing with Y on two sides
of the equation we get, dY/Y=dL/L-2/rdr-d/
For maximum error all terms should be positive
dY/Y=dL/L+2/rdr+d/=0.0489; dY=1.09x1010
Motion in One Dimension
Introduction
We have studied till now that the laws of physics are expressed in terms
of physical quantities. These quantities are either fundamental or derived
one from these fundamental quantities so that law of nature could be
best expressed in terms of involved quantities. These quantities are
meaning less until unless we have set certain standards for quantifying
the physical quantity. The set of standards and the units involved in
totality denotes the true value of the physical quantity and then only can
participate in mathematical manipulations of laws of physics. Beside that
these physical quantities don’t take part in mathematical operations like
ordinary numeric values but are assigned certain predefined properties
according the role to play to best describe the fundamental law of nature
and are so called as scalar, vector or tensor quantity.
Now we shall start our expedition to understand the very fundamental
aspect of physical observation of nature involving the motion of particles
or rigid bodies. These all aspects of motion are covered in the study of
classical mechanics; the oldest branch of Physics, which is further,
subdivided namely Statics and Dynamics.
(A) The statics is the branch of Physics that deals with the study of physical objects or
system of objects that are at rest.
(B) The Dynamics is the branch of Physics that deals with the study of physical objects or
system of objects that are in motion. When we are not concerned with the cause of motion
and limit our study to the involved parameters of particulate motion only then the dynamics
of particles may be termed as Kinematics.
As we have come across with two terms rest and motion and for the purpose of Physics it is
not so easy to call the term rest and motion as we usually do with. The terms rest and motion
denotes the state of motion of the object under consideration with respect to frame of
reference of observer. We have stated earlier in our previous discussions that measurement
of physical quantities depends upon the frame of reference of observer. For an example an
object at rest with respect to the observer attached to a moving train, is in motion with respect
to an observer attached to the Earth. Hence the complete description of a physical quantity
desires a proper set of standard units and complete information of frame of reference for
observance.
The mechanics of motion of objects we are dealing here is a part of Classical mechanics, also
known as Newtonian mechanics (1860), doctrine of Sir Alexander Newton. As physics is not
a static tool for explaining all the phenomenon of nature but itself a developing one to face
the challenges posed by the incident results. The Newtonian mechanics was developed for
understanding the observations regarding motions of objects in the nature and are perceived
through the naked eyes. It proves all experimental results when the speeds of objects are slow
enough in comparison to the speed of light. However it is helpless in describing the
collisions, decay and interactions of elementary particles of atom like proton, electron and
neutron moving at high speed of the order of speed of light, regarding deviation from the
results for the relative velocity of particles observed from the different frame of reference and
prediction of position and velocity at a time for the high speed moving particles which is an
essential requirement for describing any physical quantity. For explaining the above
phenomenon the new theories like Einstein theory of relativity (1905) and Quantum
mechanics (1925) been developed which satisfies all experimental results involving particles
of small mass and high speed (vc). These theories are considered as a more general theory
and Newtonian mechanics is considered as a special case of application for particles moving
with velocity very very less comparable to the light.
We now return to the classical mechanics to study the slow motions, which can be perceived
through our common sense without any intuitive effort. But before taking a leap for
understanding laws of physics involved in motion of objects, we shall introduce the physical
quantities involved in various types of motion and start our expedition with the simplest kind
of motion that is motion in a single direction also called as one-dimensional motion.
At the end of this chapter of around one hour, you will be able to learn
Concept of point object
Motion in One Dimension
Distance
Displacement
Average Speed
Average Velocity
Instantaneous Velocity
Average acceleration
Instantaneous acceleration
Motion with constant acceleration
Motion with variable acceleration
Time dependent acceleration
Position dependent acceleration
Velocity dependent acceleration
Relative motion in one dimension
Concept of point object
When we think of a motion there may be different possibilities of motion either in the choice
of path or the choice of the body itself. But to limit our discussion to the beginners we have
simplified our choices. We start our discussion with motion of objects that have assumed
physically zero dimensions, called particles or point objects. One tends to think of a particle
as a tiny object, e.g., a piece of shot, but actually no size limit is implied by the word
“particle”. If we are not interested in the rotational motion of an object, any object can be
considered as a particle. For example, sometimes we consider the motion of earth around the
sun. In this case we consider the motion of the center of the Earth in the circular path and
ignore the rotational motion of the earth on its own axis then for our treatment the Earth may
be considered as a particle. In some astronomical problems the solar system or even a whole
galaxy is treated as a particle. In other words when the size of the object is very small in
comparison to the distance it moves then the object may be treated as point object. There is a
specific nature of point object is that all the points on the object undergo same displacement,
hence the displacement of any of the points may be treated in the experiment. Hence in all
the concept of point object has its significance with reference to the type of motion of the
object is under consideration and displacement that undergo in comparison to its own
dimensions.
Motion in One Dimension
To describe the motion of the particle, we are now in a stage to develop the concepts of displacement,
velocity and acceleration. In the general motion of a particle in three dimensions, these quantities are
vectors, which have direction as well as magnitude. However at this stage we have confined our
discussion to the movement of a particle in a straight linear path, with only two possible directions,
distinguished by designating one positive and the other negative. A simple example of one-
dimensional motion is a vehicle moving along a straight, aligned road. We can choose any convenient
point on the vehicle for the location of the point mass for the discussion of motion. We shall now
define various physical quantities
associated with rectilinear motion of a
point object and then try to understand
the need of their development.
Distance
The distance is defined as the length
of actual path the particle traverses in
its motion. In our case when particle is
restricted its motion in a one
dimension then the distance is defined
as the length of actual path the particle
has traversed irrespective of its
direction of motion. For example the
vehicle traveling in one dimension in
the fig say East-West direction,
moving 6 miles East ward and 4 miles
back west ward will be defined to
have moved a distance 10 miles.
Hence it is a scalar quantity and always positive and increasing with time. Its unit are (m) in
MKS and (cm) in CGS system.
Displacement
The displacement of the particle or point object in general is defined as the actual
displacement the particle has undergone with respect to its original position in the time
interval under consideration. In other words it is the change in position vector of the object in
a scheduled time. The displacement of particle is not concerned with the path it has elapsed
and journey details. It is a unique value defined with a vector having direction as the
movement from its initial to final position and magnitude equal to the straight distance
between initial and ending positions. In our previous example for a motion in one dimension
the displacement will be 2 miles east ward since the vehicle has traversed a total distance of
10 miles but has been displaced with net amount of 2 miles from it’s initial position say at
origin. Hence it is a vector quantity and units are same as of distance.
Average Speed
The average speed of a particle is defined as the ratio of the total distance moved to the time
taken up to that instant of motion. So it is simply the time rate at which the distance moved
by the particle. So
Average speed = total distance / total time
For example, if the vehicle in the previous example moves a total distance 10 miles in what
ever direction in 0.2 h, than its average speed is
which suggest us that the driver might have moved with this speed uniformly elapsing total
distance of 10 miles in the given time interval of 0.2 hours. It is therefore a scalar quantity
having SI units of meters per second, and written as m/s.
Test your understanding
Ex1. A car covers a distance of 2 km in 2.5 minute. If it covers half of the distance with
speed 40 km/hr, the rest distance it will cover with speed-
Ex26. At time t=0, an object is released from rest at the top of a tall building. At the time t0
a second object is dropped from the same point, ignoring air resistance the time at which the
objects have a vertical separation of l is-
(A) t=l/gt0+t0/2 (B) t= l/gt0-t0/2 (C) t=l/gt0 (D) none of the above
Sol : Total displacement up to time t, S = l = ½ g t0 2 + g t0 (t- t0)
2l/gt0 = 2t-t
0;
t = l/gt0+t
0/2
Ex27. A steel ball is dropped from the roof of a building. A man standing in front of a 1 m
high window in the building notices that the ball takes 0.1s to fall from the top to bottom of
the window. The ball continues to fall and strikes the ground. On striking the ground, the ball
gets rebounded with the same speed with which it hits the ground. If the ball appears at the
bottom of the window 2 s after passing the bottom of the window on the way down, then the
height of the building is
(A) 12.40 m (B) 21.0 m (C) 24.0 m (D) 42.0 m
Sol : Let u and v be the velocities of the ball at the instant it appears at the top and bottom of window during the downward journey. So {(u+v)/2}(0.1)=1; Or u+v=20Also v=u+g(0.1)Or v=u+1; u=19/2 m/s
Height above window (h1)= u2/20=(9.50)2/20 =4.50 m;
Height below window h3 = v+1/2g= 15.5 m;
Total height of the building H = 4.50+1+15.50= 21.0 m
Ex28. The vertical height of P above the ground is twice that of Q. A particle is
projected downward with the speed of 9.8 m/s from P and simultaneously another
particle is projected upward with the same speed of 9.8 m/s from Q both particles reach
the ground simultaneously. The time taken to reach the ground is-
(A) 3 sec (B) 4 sec (C) 5 sec (D) 6 sec.
Sol : PG = 9.8t{1+0.5t}; QG = 9.8t{0.5t-1}; PG/QG = 2; t = 6 sec
Ex29. A particle is projected vertically upwards from a point x on the ground. It takes a time
t1 to reach a point A at a height h above the ground. It continues to move and takes a time t2
to reach the ground. The velocity of the particle at half the maximum height is
(A) 22/g (t1+ t
2) (B) g/22 (t
1+ t
2) (C) 22g/(t
1+ t
2) (D) (t
1+ t
2)/ 22g
Sol : Tm= (t2-t
1)/2+t
1; Tm = (t
1+t
2)/2; u = g(t
1+t
2)/2; H = u2/2g = g2(1+t
2)2/8g; V
m= g(t
1+t
2)/22
Ex30. A juggler keeps on moving four balls in the air throwing the balls after regular
intervals. When one ball leaves his hand (speed=20 m/s) the position of the other balls
hence at t = 2s the particle reverses its direction of motion.
Distance traveled in t=2s, x1= 4 ft
Distance traveled in next 2s x2 =
(2at – 3bt2) dt = - 20;
Total distance traveled= 24 ft
Ans107: Case1: When the two trains move in opposite directions
2l/u1+u
2 = 15/2;
Case 2: When two trains move in the same directions
2l/v1-v
2 = 15;
Solving for v1and v2 , we get v1 = 36 m/s; v
2 = 12 m/s
Ans108: Velocity of the man on the escalator Vm = x/90;
Velocity of the escalator Ve = x/60;
( Vm+Ve)t = x;
(x/90+x/60) t = x;
t = 36 s
Ans109: Let L be the distance between the two particles when they can meet so
L = (v1+v
2)t –1/2 (a
1+a
2)t2;
(a1+a
2)t2 – 2(v
1+v
2)t +2L = 0;
Now for t to be real b2- 4 a c 0 or L (v1+v
2)2/ 2(a
1+a
2);
Lmax
= (v1+v
2)2/2 (a
1+a
2)
Chapter 5 Motion in a Plane
Motion in Two and Three Dimensions
In this chapter, we will now extend our thought of a motion to a more general case of motion
in two and three dimensions. So far, we have learnt the definition of displacement, velocity,
acceleration and relative motion with application for a most simplified case of motion in one
dimension. Now we are in the stage to extend it to the motion in the x-y plane and later to
the motion in three dimensions.
Location of particle during motion
Whenever we are concerned with a motion of a particle it may be one, two or three
dimensional, it is very first essential to locate the position of a particle at a particular instant
of observation. The location of a particle could be only established if we choose the correct
frame of reference and which is in itself a combination of coordinate axis system. However
in this chapter we are more concerned about motion in a plane therefore only two reference
axis are required to locate its instantaneous position. To locate the instantaneous position of
a particle, we might choose either the distances from Cartesian’s coordinate system that is
two perpendicular coordinate axes x & y axis or polar coordinate system that is distance
from a origin and angle made by the x axis. In terms of vector notation the position of
particle is defined by the position vector r whose head points to the origin O and tail points
to its position. Unlike the ordinary vector it is not a position independent vector. It is related
with the position of object whose location is under consideration and origin of reference
frame and establishes the relationship between the two. The change in position vector of the
object denotes the displacement of the object from its initial position.
If the particle is not confined to a plane but moves in three dimensions three numbers are
needed to specify its location. A simple approach is to describe its location with respect to
three perpendicular coordinate axes or an alternative approach is to use spherical
coordinates r, and . Spherical coordinates are useful when there is a spherical symmetry
that is radial distance from origin is constant and only and varies. The usual relationship
between Cartesian coordinates and spherical coordinate is shown here:
Displacement of particle during motion in a plane
Figure-4.1 shows a particle moving along a curved path in this plane. At any instant of
motion, say at time t1, it has occupied a position A as shown in the figure. As discussed
earlier its instantaneous position may be defined in terms of vector algebra by the position
vector r1. Now in a time interval moving along the curved path, it occupy another position
B as defined by the position vector r2. Now the displacement of particle from one position to
another is defined by the vector , which is defined as per vector notation as r2-r1.
Hence, it clearly indicates that displacement of particle is independent of the path the
particle traverses during its motion but depends only on initial and final positions during
motion and follows vector law in terms of its position vector. If the particle moves along
straight line path than displacement of particle equals the distance moved but if another path
is chosen than displacement will be always less than the distance moved during that time
interval.
Now the one question arises in one’s mind why the concept of vector displacement has
been generated although there is no practicable relationship with the physical movement of
the particle during its motion. The matter will be clear in the next stage while defining
concept of instantaneous velocity / average velocity/ instantaneous acceleration that how
important are these tools to analyze and describe the different types of motions observed in
the nature.
Average velocity of particle during motion in a plane
Now again consider the motion of particle along curved path in a plane as shown in fig 1. On
the same lines of the definition of average velocity defined earlier during the study of motion
along a straight-line path, it is again described as the rate of change of displacement and
written as . It is a vector quantity directed along the direction of vector and
physically denotes the rate of change of straight movement of particle from its initial position
to the next position in the said time interval. In our case the particle has moved from position
A to position B, whatever may be path than its average velocity is defined by
Where is the vector displacement r2-r1 in the time interval . However, the concept of
average velocity during the physical movement of a particle in a plane in any arbitrary path
has little physical meaning when time interval of motion is considered large enough. For a
large time interval of motion, there is a distinct difference between the actual distance
traversed and the distance considered for our purpose that is displacement. Therefore, the
average velocity does not attribute to the state of motion either at the start, middle or at end.
It is only an average value for general interpretation of motion.
Instantaneous velocity during motion in a plane
Now consider the motion of particle along curved path in a plane as shown in fig 4.2. The fig
shows the instantaneous positions of particle initiating motion at time t1, at different time
intervals , , , as P1, P2, P3, P4. As the time interval of motion under
consideration is now decreased slowly than in a limiting condition the physical displacement
is just equal to the distance traversed and the direction of vector displacement is the
direction of physical motion along tangent to the instantaneous position of particle.
The instantaneous velocity of the particle is now defined as limiting rate of displacement
when . In that case, the direction of velocity vector v is tangent to the path of the
particle.
The instantaneous velocity v(t) is defined as
As per definition, the instantaneous velocity vector denotes the instantaneous rate of
displacement along any path of movement of particle and in terms of direction, it is directed
along the tangent to the path of motion that should be. Therefore it clearly depicts the
instantaneous state of motion say at time t=t1 in terms of magnitude and direction.
The acceleration vector during motion in a plane
Now we shall define a term acceleration vector that is essentially the backbone of all
fundamental laws of Newtonian mechanics. The term has the same conceptual meaning as
has been defined earlier during the discussion of motion in one dimension. The acceleration
vector is a term defining change in velocity vector with respect to time. Since velocity vector
is defined as a vector quantity, therefore change in velocity vector is also predicted to be a
vector quantity. The acceleration vector is further divided into its two forms like as velocity
vector in terms of time interval of motion under consideration during study of motion in a
plane.
Average acceleration
Now again consider the motion of a particle in a plane as shown in fig 4.3. The particle has
got as per definition instantaneous velocity vector at points P1
and P2
as v1, v2 at different
times of motion t1, t2 respectively. Now if the time interval of motion is considered large
enough than the average acceleration vector is defined as the ratio of the change in
instantaneous velocity vector Dv and time interval Dt,
Here is a vector difference of vectors v1 and v2 and therefore itself is a vector quantity..
Instantaneous acceleration
Now as shown in fig 4.3, if the time interval of motion is considered infinitesimal or tending to
zero than rate of change of velocity vector with respect to time is defined as instantaneous
acceleration. It is expressed as
Therefore, the instantaneous acceleration vector is the derivative of velocity vector with
respect to time. It is important to note here that if the particle during motion has velocity
changing in magnitude, direction or both the particle is said to be accelerating. We shall see
in the next chapter that a force is required to produce an acceleration of particle. Force is
required to be applied whether the acceleration is produced either due to change of
magnitude of velocity or change in direction of velocity vector or both. This is the reason why
the acceleration is defined in this way.
Motion with constant acceleration
Let us consider the motion of a particle in a plane with constant acceleration. In this case
when particle is moving with a constant acceleration therefore it implies that acceleration
vector is constant in magnitude and direction as well. Hence the components of vector a in
any particular reference frame will also remain constant. The situation is similar to the two
simultaneous constant acceleration motions occurring along two perpendicular directions.
Under the influence of that, the particle will traverse the curvilinear path however one of the
components may be zero.
An example of the above situation is the motion of a ball thrown into air, which follows a
curved path under the influence of gravity acceleration g acting downward. The horizontal
acceleration is zero.
Let us consider the general motion of a particle with constant acceleration:
ax= constant and ay=constant
Now under the following set of conditions, velocity vector at any instant
The above equation shows in a compact and more elaborate form that the velocity of
particle at any instant is the vector sum of initial velocity vector and at vector component
acquired in time t and also displacement vector r at any instant:
Projectile motion
An important case of curvilinear motion with constant acceleration is projectile motion. When
a body is thrown upward under the influence of gravity, then the body follows the path
known as projectile motion (fig 4.4). If the air resistance during follow up is neglected then
the body experiences the only acceleration due to gravity directed vertically downward.
As the acceleration vector is constant then instantaneous velocity vector, acceleration vector
and position vector lies in a plane. The motion is therefore a two-dimensional.
Let us choose the motion to be in the x, y plane and initial position of the particle be at origin
of the coordinate system. When the acceleration is constant then above equations can also
be applied considering motion along two axes separately. The x and y components of the
above equations are
The velocity components along each axis will be governed by the respective acceleration
component. Let us now apply these results to the motion of a projectile. The motion of
projectile is made complicated by the prevailing air resistance when body is drifted through
the air, rotation of earth on its own axis and variation in the acceleration due to gravity. For
employing lesser complexity, we neglect these complications. As said earlier the projectile
motion has a constant acceleration directed vertically downward with magnitude g=9.81
m/s2=32.2 ft/s2. If we take y-axis vertical with positive direction upward and x-axis horizontal
with positive direction in the direction of horizontal component of the projectile velocity at the
point of start, then we have
ay= -g and ax=0
Since there is no acceleration along the x-axis therefore the horizontal component, of
velocity vector remains constant and on the other hand, the motion along y-axis can be
considered to be with constant acceleration identical to that studied earlier. If the origin of
motion is considered from the origin of coordinate axis then the instantaneous velocity
components and instantaneous displacements are governed by the following equations,
where v0x and v0y are the initial velocity components or velocity components at origin of
axis ( see fig 4.4)
If the initial velocity vector makes an angle with the x-axis, the initial velocity
components will be
The general equation for the path y(x) can be obtained from equation by eliminating the
variable t between these two equations and is given by :
Which is the equation of a parabola therefore the trajectory of the particle follows a parabolic
path.
Now we shall try to find out the range (maximum horizontal distance) and time to reach the
highest point of the trajectory for the above motion:
Range of projectile
When the body reaches at the highest point P of the trajectory as shown in fig 4.4 the
vertical component of the velocity vector becomes zero therefore time of flight up to the
moment
and the range is the horizontal distance traveled in twice of this time and is given by
In terms of initial velocity and angle
Since the maximum value of is 1 when =90° or , then
maximum range is admitted when the body is projected at angle of 45° from the horizontal
and maximum range is given by vo2/g. It is to be noted also that for achieving maximum
range the two initial velocity components should be equal in magnitude. But if a body is
projected from a certain height then in this condition the body remains in air for a longer
duration therefore for maximum range the horizontal component is slightly higher than the
vertical component that is angle of projection is smaller than 45°. Studies have shown it to
be about 42°. It is also evident from the above equation that for a given range there are two
possible angles of projection q0 and 90-q0 that provide same range.
So far, in our discussion, we have not considered the effect of air resistance and the earth’s
orbital motion. As we would expect the air resistance reduces the range for a given angle of
projection and reduces slightly the optimum angle of projection. Due to earth’s orbital motion
the projectile motion doesn’t remain in the plane formed by the initial horizontal and vertical
components but gets slightly drifted to right in northern hemisphere and to left in the
southern hemisphere. It is due to the Coriolis effect, which arises due to the surface of earth,
is accelerating radially because of the earth rotation on its own axis.
Maximum height of projectile
At maximum height of projectile
So
Projectile thrown parallel to horizontal
Consider a projectile thrown from a point O at some height h from the ground with a velocity
. Now the equation of its motion in two directions may be depicted here as
This is the equation of trajectory.
Time of flight
Horizontal range=
Now if particles are thrown at an angle with horizontal then all the particles will arrive at
ground with equal velocities whatever is the time of arrival. The particle thrown in the
downward direction will take least time while thrown upward will take maximum time.
4.6.1.4 Projectile thrown from an inclined plane
Consider a particle thrown from the base of an inclined plane with a velocity v at an angle
from the horizontal. The angle of inclination of plane from horizontal is .
For the motion perpendicular to the plane
For the motion along the plane
Range is maximum when is and projectile hits the plane at right angles.
Projectile thrown with variable acceleration
Now consider the motion of a particle in the x–y plane with instantaneous velocity
where a and b are constants and particle is situated at x=0, y=0 at t=0
So
Eliminating t from these two equations
The radius of curvature of trajectory at any instant t is given by
Exercise 1
Q1. If T were the total time of flight of a current of water and H be the maximum height
attained by it from the point of projection, then H/T will be-
(u=projection velocity, =projection angle)
(A) (1/2 )u sin (B) (1/4) u sin (C) u sin (D) 2u sin
Ans1: Hmax= u2 sin2 /2g; T = 2u sin/g; H/T = u sin/4
Q2. A hunter aims his gun and fires a bullet directly at a monkey on a tree. At the instant
bullet leaves the gun, monkey drops, and the bullet.
(A) hits the monkey (B) misses to hit the monkey (C) cannot be said (D) None of these
Ans2: (A)
Q3. A projectile can have the same range r for two angles of projections. If t1 and t2 be the
times of flight in two cases, then the product of times of flight will be-
(A) t1t2 R (B) t1t2 R2 (C) t1t2 1/R (D) t1t2 1/R2
Ans3: t1t2=4 u2sincos/g2=2 u2sin2/g2 R
Q4. A body is thrown with a velocity of 9.8 m/s making an angle of 300 with the horizontal. It
will hit the ground after a time.
(A) 3 s (B) 2 s (C) 1.5 s (D) 1 s
Ans4: Time of flight = 2u sin/g = 1 s
Q5. A body A is dropped from a height h above the ground. At the same, time another body B at a distance d from the projection of A from the ground is fired at an angle a to the horizontal. If the two collide at the point of the maximum height of trajectory of B the angle of projection is given by :