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1. Physics Summary 11th Std (T.N.State Board)

1. Physics and Measurement

1. Scientific method

step-by-step approach in studying natural

phenomena and establishing laws which

govern these phenomena.

2. Features scientific method

i) Systematic observation

ii) Controlled experimentation

iii) Qualitative and quantitative reasoning

iv) Mathematical modeling

v) Prediction and verification or falsification of

theories

3. Unification and Reductionism

§ Attempting to explain diverse physical

phenomena with few concepts and laws is

unification.

§ Example: Newton’s universal law of gravitation

Explains motion of freely falling bodies

towards the Earth, motion of planets around

Sun, motion of Moon around the Earth, thus

unifying the fundamental forces of nature.

§ An attempt to explain a macroscopic system

in terms of its microscopic constituents is

reductionism.

§ Example: thermodynamics

Temperature, entropy, etc., of bulk systems

(macroscopic) in terms of molecular

constituents (microscopic) of the bulk system

by kinetic theory and statistical mechanics.

4. Branches of Physics

5. Impact of physics and technology on society

i) Discovery of wireless communication due to

Electricity and magnetism has shrunk the

world with effective communication

ii) Launching of satellite into space has

revolutionized the concept of communication.

iii) Microelectronics, lasers, computers,

superconductivity and nuclear energy have

changed living style of human beings.

6. Measurement.

Comparison of any physical quantity with its

standard unit

7. Physical quantities.

Quantities that can be measured, and in terms

of which, laws of physics

Examples: length, mass, time, force, energy

8. Types of physical quantities.

i) Fundamental or base quantities

Quantities which cannot be expressed in

terms of any other physical quantities.

Example :

length, mass, time, electric current, temperature,

luminous intensity

ii) Derived quantities.

Quantities that can be expressed in terms of

fundamental quantities

Example:

area, volume, velocity, acceleration

9. Unit of the quantity.

§ Arbitrarily chosen standard of measurement

of quantity accepted internationally

§ Units in which the fundamental quantities are

measured are fundamental or base units

§ Units of measurement of all other physical

quantities obtained by multiplication or

division of powers of fundamental units, are

called derived units.

10. Common system of units used in mechanics:

i) f.p.s. system

British Engineering system of units, which

uses foot, pound and second as basic units for

measuring length, mass and time.

ii) c.g.s system



Gaussian system, which uses centimeter,

gram and second as the three basic units for

measuring length, mass and time

iii) m.k.s system

Based on metre, kilogram and second as basic

units for measuring length, mass and time

respectively.

cgs, mks,SI are metric or decimal system of units.

fps system is not a metric system.

11. International System (Système International).

SI with a standard scheme of symbols, units

and abbreviations, were developed and

recommended by the General Conference on

Weights and Measures in 1971 for

international usage in scientific, technical,

industrial and commercial work.

12. Advantages of the SI system

i) Rational system of units

Uses of only one unit for a physical quantity,

ii) Coherent system of units.

All derived units are obtained from basic and

supplementary units

iii) Metric system

Multiples and submultiples are expressed as

powers of 10.

13. Length

metre ( m )

Length of path travelled by light in vacuum in

1/299,792,458 of a second

14. Mass

kilogram (kg)

Mass of prototype cylinder of platinum iridium

alloy at International Bureau of Weights and

Measures at Serves, near Paris, France.

15. Time

second (s)

Duration of 9,192,631,770 periods of radiation

corresponding to transition between 2 hyperfine

levels of ground state of Cesium-133 atom.

16. Electric current

§ ampere (A )

§ Constant current maintained in each of two

straight parallel conductors of infinite length

and negligible cross section, held one metre

apart in vacuum shall produce a force per unit

length of 2x10−7 N/m between them.

17. Temperature

kelvin (K )

Fraction of 1/273.16 of thermodynamic

temperature of the triple point* of the water.

18. Amount of substance

mole (mol )

Amount of substance which contains as many

elementary entities as there are atoms in 0.012 kg

of pure carbon-12. (1971)

19. Luminous intensity

candela (cd )

luminous intensity in a given direction, of a

source that emits monochromatic radiation of

frequency 5.4×1014 Hz. That has radiant intensity

of 1/683 watt/steradian in that direction.

20. Radian (rad)

Angle subtended at the centre of a circle by an arc

equal in length to the radius of the circle.

21. The Steradian (sr):

Solid angle subtended at the centre of a sphere,

by that surface of the sphere, which is equal in

area, to the square of radius of the sphere

22. Macrocosm And Microcosm

Macrocosm.

§ Large world, in which both objects and

distances are large.

§ Example

galaxy,stars, Sun, Earth, Moon and their distances

Microcosm

§ Small world in which both objects and

distances are small-sized.

§ Example :

molecules,atoms, proton, neutron, electron,

bacteria and their distances

23. Measurement of small distances:

Screw gauge



§ Instrument for measuring dimensions of

objects up to a maximum of about 50 mm.

§ Principle : Magnification of linear motion

using circular motion of a screw.

Least count : 0.01 mm

Vernier caliper

§ Versatile instrument for measuring diameter

of a hole, or a depth of a hole.

§ Least count : 0.1 mm

24. Measurement of large distances

For measuring larger distances such as the height

of a tree, distance of the Moon or a planet from

the Earth, some special methods are adopted.

i) Triangulation method

ii) Parallax method

iii) Radar method

25. Triangulation method to find height of a tree

q=

=q

=Ð

=

=

tan height

tan

ABC,triangleangledrtFrom

Catfinderrange

placing bymeasuredelevation ofangleACB

B.fromdistanceatobservatinofpointC

measured.betotreeofheight,AB

xh

x

h

x

h

26. Parallax method

§ Distance of a planet or a star from the Earth

can be measured by the parallax method.

§ Parallax

Apparent change in the position of an object

with respect to the background, when the

object is seen from two different positions.

§ Distance between the two positions is basis (b)

27. To Measure unknown distance by Parallax

§ Observer is specified by the position O.

§ Observer is holding a pen before him

§ Pen is looked at first by left eye L (closing

right eye) then by right eye R (closing left eye)

§ Shift in position of object when viewed with

two eyes, keeping an eye closed at a time is

Parallax.

§ Distance between left eye (L) and right eye (R)

is the basis.

x

x

b

b

radiusROLO

length ofarcLR

angle.cparallactiorangleparallaxLOR

=q

===

=

=Ð

28. Determination of distance of Moon from Earth

by Parallax method

telescope.using determinedstardistant

w.r.tBandAMfromMoon ofparallaxes

EarththefromMoon ofdistanceMC

Earth on placesoppositelydiametricltwoBA,

Earth.theofcentreC

=qq

=

=

=

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θ

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AB

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Earthon subtendedMoon ofparallaxTotal

2.1

=

==q

=

=+=Ð

][ MCAMQ

29. RADAR method

§ RADAR = Radio Detection and Ranging.

§ A radar can be used to measure accurately the

distance of a nearby planet such as Mars.

§ Also used to determine height, at which an

aeroplane flies from the ground.



§ Radio waves sent from transmitters, after

reflection from target are detected by receiver.

§ Time interval (t) between radio waves sent

and received are measured

d

taken timeSpeed)Distance(d

taken time

travelleddistance(v)Speed

2

tv ´=

´=

=

§ As time taken is for distance covered during

forward and backward path, it is divided by 2

30. Measurement of Mass

§ Mass is the quantity of matter contained in a body

§ The SI unit of mass is kilogram (kg).

§ Mass is a property of matter. It does not

depend on temperature, pressure and location

of the body in space.

§ Mass vary from a tiny mass of electron

(9.11x10−31kg) to huge mass universe (1055 kg)

§ Mass of an object is determined in kilograms

using a common balance

§ For measuring larger masses like planets, stars

etc., gravitational methods are used

§ To measure small masses of atomic sub

atomic particles mass spectrograph is used

§ Weighing balances commonly used

common balance, spring balance, electronic balance

31. Measurement of Time

§ A clock is used to measure the time interval.

§ An atomic standard of time, is based on

periodic vibration produced in Cesium atom.

§ Clocks developed later

electric oscillators, electronic oscillators, solar

clock, quartz crystal clock, atomic clock, decay

of elementary particles, radioactive dating etc.

1.2. Theory of Errors

1. Precision and accuracy

i. Accuracy

Refers to how far we are from the true value

ii. Precision

Refers to how well we measure.

iii. Example 1

If temperature outside is 40oC measured by a

weather thermometer and real outside

temperature is 40oC, thermometer is accurate.

If thermometer consistently registers this

exact temperature, thermometer is precise.

Example 2 Target Shooting

2. Error and its Types

Uncertainty in a measurement is called error.

Possible errors.

i) Random error

ii) systematic error

iii) gross error

3. Systematic errors

Reproducible inaccuracies consistently in same

direction

occur due to a problem that persists throughout

the experiment.

i) Instrumental errors

§ Arises when an instrument is not calibrated

properly at time of manufacture.

§ Measurement made with end worn out meter

scale will have errors.

§ Corrected by choosing instrument carefully.

ii) Imperfections in experimental technique or

procedure

§ Arise due to limitations in experimental

arrangement.

§ In experiments with calorimeter, if there is no

proper insulation, radiation losses.



§ To overcome this, correction has to be applied

iii) Personal errors

§ Individuals performing experiment, incorrect

initial setting up of the experiment

§ Carelessness of the individual making the

observation due to improper precautions.

iv) Errors due to external causes

§ Changes in temperature, humidity, or

pressure during measurements

v) Least count error

§ Least count is the smallest value that can be

measured by measuring instrument, and error

due to this measurement is least count error.

§ Instrument resolution is the cause of this error

§ Reduced by using high precision instrument

4. Random errors

§ Arise due to random and unpredictable

variations in experimental conditions like

pressure, temperature, voltage supply etc.

§ Also due to personal errors by observer who

performs the experiment.

§ Sometimes called “chance error”.

§ When different readings are obtained by a

person every time he repeats the experiment,

personal error occurs.

§ Large number of measurements are made and

arithmetic mean is taken.

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takenReadingsana3,a2,a1,

experimentin readingstrialofNo.n

5. Gross Error

§ Error caused due to carelessness of observer

§ Example

i) Reading instrument without setting it

proper

ii) Taking observations in wrong manner

without bothering about sources of errors

and precautions

iii) Recording wrong observations.

iv) Using wrong observations in calculations

§ Minimized by careful and alert observance

6. Error Analysis

i) Absolute Error

§ Magnitude of difference between true value

and the measured value of a quantity

§ If a1,a2, a3, ….an are measured values in an

experiment performed n times, A.M of these

values is called True value (am) of quantity.

§ Absolute error in measured values

|an–am| |an|

|a2–am| |a2|

|a1–am| |a1|

=D

=D

=D

ii) Mean Absolute error

Arithmetic mean of absolute errors in all

measurements

||1

1å=

=D

D++D+D=D

=D

=

n

i

im

m

an

a

na

|an|.....|a2||a1|

errorabsolutemean a

valuetruea

m

m

Magnitude of quantity lie between

am + Δam and am - Δam

iii) Relative error

§ Ratio of mean absolute error to mean value error

§ Also called as fractional error

§ Expresses how large absolute error is

compared to the total size of object measured.

mama

Mean value

errorabsoluteMean =errorRelative

D=

iv) Percentage error

§ Relative error expressed as a percentage

§ % error close to zero means one is close to

targeted value and it is good and acceptable.

§ It is necessary to understand whether error is

due to impression of equipment or a mistake

in the experimentation.

%100´D

mama

=errorPercentage



7. Eerror in the final result depends on

i) Errors in the individual measurements

ii) On the nature of mathematical operations

performed to get the final result.

8. Probagation of Errors in diferent mathematical

operations

i) Error in the sum of two quantities

Absolute errors in quantities A, B

= ΔA, ΔB

Measured value of A = A ± ΔA

Measured value of B = B ± ΔB

Sum Z = A + B

Error ΔZ in Z, Z ± ΔZ = (A ± ΔA)+(B ± ΔB)

= (A+B) ± (ΔA + ΔB)

= Z ± (ΔA + ΔB)

ΔZ = ΔA + ΔB

possible error in the sum of two quantities =

sum of absolute errors in individual quantities

ii) Error in the diference of two quantities

Absolute errors in quantities A, B

= ΔA, ΔB



Differenc Z = A − B

Error ΔZ in Z, Z ± ΔZ = (A ± ΔA) - (B± ΔB)

= (A −B)± (ΔA + ΔB)

= Z ± (ΔA + ΔB)

ΔZ = ΔA + ΔB

possible error in diference of two quantities =

sum of absolute errors in individual quantities

iii) Error in the product of two quantities



product Z = AB …(1)

Error ΔZ in Z,

Z± ΔZ = (A ± ΔA) (B ± ΔB)

= (AB)± (A ΔB) ± (B ΔA) ± (ΔA . ΔB)

Dividing by (1)

B

B

A

AZ

neglectedA

AsmallBeing

A

A

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B

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AB

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æ D+

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DD±

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D±=

D±

z

B

B

B

B

z

z

.

.1

Fractional error in product of two quantities =

sum of fractional errors in individual quantities

iv) Error in division or quotient of 2 quantities



quotient, Z = A/B …(1)

B

B

A

A

A

A

B

B

B

B

A

A

B

B

A

AZ

B

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Zbysidesboth Dividing

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ZZ

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A)(AZZ

.1

11

11

1

1

1

1<<xwhennx,+1x)+(1 n

Fractional error in quotient of two quantities =

sum of fractional errors in individual quantities

v) Error in the power of a quantity

Absolute errors in quantity A= ΔA Measured value of A = A ± ΔA

power of A, Z = An

A

An

z

Z

A

An1

z

Z1

Zbysidesboth Dividing

A

An1ZZ

1<<xwhen nx,+1x)n+(1

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n

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D=

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D±=

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æ D±=D±

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Z

An

fractional error in the nth power of a quantity = n times fractional error in that quantity



100C

ΔCr100

B

ΔBq100

A

ΔAp100

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ΔZ

Zin errorpercentage

C

ΔCr

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r

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´+´+´=´

++=

=

9. Significant Figures

§ Digits that are known reliably plus first

uncertain digit are significant digits.

§ Example:

Gravitational constant =6.67×10−11 N m2 kg−2

§ Digits 6 and 6 are reliable and certain

§ Digit 7 is uncertain.

§ Measured value has three signiicant igures

10. Rules for determining significant figures.

i) All non-zero digits are significant

1342 has four signiicant igures

ii) All zeros between two non zero digits are

significant

2008 has four signiicant igures

iii) All zeros to the right of a non-zero digit but to

the left of a decimal point are significant.

30700. has ive signiicant igures

iv) Number without a decimal point, the terminal

or trailing zero(s) are not significant

30700 has three signiicant igures

v) All zeros are significant if they come from a

measurement

30700 m has ive signiicant igures

vi) If the number is less than 1, the zero (s) on the

right of the decimal point but to let of the first

non zero digit are not signiicant.

0.00345 has three signiicant igures

vii)All zeros to the right of a decimal point and

to the right of non-zero digit are significant.

40.00 has four signiicant igures

0.030400 has ive signiicant igures

viii) Number of signiicant figures does not

depend on the system of units used

1.53 cm, 0.0153 m, 0.0000153 km, three

significant igures

11. Rounded off

Result of calculation with numbers containing

more than one uncertain digit should be

rounded off

i) If the digit to be dropped is smaller than 5,

then preceding digit should be let unchanged.

7.32 is rounded of to 7.3

ii) If the digit to be dropped is greater than 5, then

the preceding digit should be increased by 1


iii) If the digit to be dropped is 5 followed by

digits other than zero, then the preceding

digit should be raised by 1

18.159 rounded of to first decimal 18.2

iv) If the digit to be dropped is 5 or 5 followed by

zeros, preceding digit not changed if even


v) If the digit to be dropped is 5 or 5 followed by

zeros, preceding digit is raised by 1 if it is odd


Dimensions of Physical World

12. Dimensions

§ All derived physical quantities can be

expressed in terms of some combination of

the seven fundamental or base quantities.

§ These base quantities are Dimensions

denoted with square bracket [ ].

§ Example : Dimensions in mechanics

[L] for length [M] for mass [T] for time.

13. Dimensions of a physical quantity

§ Powers to which base quantities units raised

to represent derived unit of that quantity.

]LT[M

[T]

[L]=

time

ntDisplacemevelocity

2-0=

=

Dimensions of velocity

0 in mass, 1 in length, -1 in time.

14. Dimensional formula



§ Epression which shows how and which of the

fundamental units are required to represent

the unit of a physical quantity.

Example:Dimensional formula of acceleration.

[M0LT−2]

15. Dimensional equation

§ When the dimensional formula of a physical

quantity is expressed in the form of an

equation, such an equation is known as the.

§ Example: acceleration = [M0LT−2].

16. Classification On the basis of dimension

1) Dimensional variables

Physical quantities, which possess

dimensions and have variable values

Examples: length, velocity, and acceleration etc.

2) Dimensionless variables

Physical quantities which have no

dimensions, but have variable values

Examples : speciic gravity, strain, refractive index

3) Dimensional Constant

Physical quantities which possess dimensions

and have constant values

Example:Gravitational constant,Plancks constant

4) Dimensionless Constant

Quantities which have constant values and

also have no dimensions

Examples : π, e, numbers etc.

17. Principle of homogeneity of dimensions

Dimensions of all terms in a physical expression

should be the same.

Example: in v2 = u2 + 2as dimensions of v2, u2

and 2as are same equal to [L2T−2]

18. Application

i) To Convert a physical quantity from one

system of units to another.

ii) To Check the dimensional correctness of a

given physical equation.

iii) To Establish relations among various physical

quantities.

19. To convert a physical quantity from one system

of units to another

Product of numerical values (n) and its

corresponding unit (u) is a constant.

n [u] = constant

n1[u1] = n2[u2].

Consider a physical quantity whose dimension

‘a’ in mass, ‘b’ in length and ‘c’ in time.

Fundamental units in one system = M1, L1, T1

other system = M2, L2,T2

n1 [M1aL1b T1c] = n2 [M2a L2b T2c]

20. To check dimensional correctness

Eqn. of motion v = u + at

Apply dimensional formula on both sides

[LT−1] = [LT−1] + [LT−2] [T] [LT−1]

[LT−1] = [LT−1]+[LT−1]

Dimensions of both sides are same.

Hence the equation is dimensionally correct

21. To establish relation among physical quantities

If physical quantity Q depends on quantities

Q1 , Q2 and Q3

Q µ Q1, Q2 and Q3.

Q µ Q1a Q2b Q3c

Q = k Q1a Q2

b Q3c

where k = dimensionless constant.

Dimensional formula of Q, Q1, Q2 and Q3 are

substituted. Powers of M, L, T are made equal

on both sides of the equation.

From this, we get the values of a, b, c

22. Limitations of dimensional analysis

i) Gives no information about the dimensionless

constants in formula like 1, 2, ……..π,e, etc.

ii) Cannot decide whether the given quantity is a

vector or scalar.

iii) Not suitable to derive relations involving

trigonometry, exponential and logarithmic

function

iv) It cannot be applied to an equation involving

more than three physical quantities.

v) It can only check on whether a physical

relation is dimensionally correct but not the

correctness of the relation



2.1. KINEMATICS

1. Kinematics

§ Branch of mechanics which deals with motion

of objects without taking force into account.

§ Greek word “kinema” means “motion”.

2. Rest and Motion

§ A person sitting in a moving bus is at rest

with respect to a fellow passenger

§ But is in motion with respect to a person

outside the bus.

§ Concepts of rest and motion have meaning

only with respect to some reference frame.

3. Frame of Reference:

§ Imaginary coordinate system and position of

object described relative to it

§ Such coordinate system is frame of reference.

4. Cartesian coordinate system

§ At any given instant of time, the frame of

reference with respect to which the position

of the object is described in terms of position

coordinates (x, y, z)

§ That is distances of given position of an object

along x, y, and z–axes

5. Right– handed Cartesian coordinate system

§ x, y and z axes are drawn in anticlockwise

direction and such coordinate system is called

“right– handed Cartesian coordinate system”.

§ Though other coordinate systems do exist,

right–handed coordinate system is

conventionally followed

§ Place fingers in the direction of positive x-axis

§ Rotate them towards the direction of y-axis.

§ Thumb points in the direction of positive zaxis

6. Difference between left and right handed

coordinate systems.

7. Point mass

§ Mass of object is assumed to be concentrated

at a point.

§ This idealized mass is called “point mass”.

§ It has no internal structure like shape or size.

§ Mathematically a point mass has finite mass

with zero dimension.

§ In reality point mass does not exist, it

simplifies our calculations.

§ Point mass has meaning only with respect to a

reference frame and with respect to the kind

of motion

8. Examples of Point mass

§ To analyse the motion of Earth with respect

to Sun, Earth can be treated as a point mass.

Because distance between the Sun and Earth is

very large compared to the size of the Earth.

§ If we throw a small stone in the air, stone is

considered as a point mass

Because size of the stone is very much smaller

than the distance through which it travels.

9. Types of motion

a) Linear motion



Motion of Object in a straight line.

Examples

An athlete running on a straight track

Particle falling vertically downwards to the Earth.

b) Circular motion

Motion described by an object traversing a

circular path.

Examples

Whirling motion of a stone attached to a string

Motion of a satellite around the Earth

c) Rotational motion

Motion of object moving in a rotational

motion about an axis,

During rotation every point in the object

transverses a circular path about an axis

Examples

Rotation of a disc about an axis through its center

Spinning of the Earth about its own axis.

d) Vibratory motion

Motion of object or particle executing to–and–

fro motion about a fixed point also called

oscillatory motion.

Examples

Vibration of a string on a guitar

Movement of a swing

e) Other types of motion

elliptical motion and helical motion

Circular Rotational Vibratory

10. Motion in One, Two and hree Dimensions

§ Position of a particle in space is expressed in

terms of rectangular coordinates x, y and z.

§ When these coordinates change with time,

then the particle is said to be in motion.

§ It is not necessary that all the three

coordinates should together change with time.

§ If one or two coordinates changes with time,

particle is said to be in motion.

11. Motion in one dimension

§ Motion of particle along straight line.

§ Also known as rectilinear or linear motion.

§ Only one of the three rectangular coordinates

specifying the position of the object changes

with time.

§ For particle moving from position A to

position B along x–direction, variation in x–

coordinate alone is noticed

Examples

Motion of a train along a straight railway track.

Object falling freely under gravity close to Earth.

12. Motion in two dimensions

§ Motion of Particle along curved path in plane

§ two of three rectangular coordinates

specifying position of object change with time.

§ For particle moving in y – z plane, x does not

vary, but y and z vary

Examples

Motion of a coin on a carrom board.

An insect crawling over the loor of a room.

13. Motion in three dimensions

§ A particle moving in usual three dimensional

space has three dimensional motion.

§ All three coordinates specifying the position

of an object change with respect to time.

§ When a particle moves in three dimensions,

all the three coordinates x, y and z will vary.

Examples

A bird lying in the sky.

Random motion of a gas molecule.

Flying of a kite on a windy day



2.2. VECTOR ALGEBRA

1. Scalar

It is a property which can be described only by

magnitude. In physics a number of quantities can

be described by scalars.

Examples:Distance, mass, temperature, speed, energy

2. Vector

§ Quantity described by both magnitude and

direction.

§ Geometrically a vector is a directed line

segment.

§ In physics certain quantities can be described

only by vectors.

Examples

Force, velocity, displacement, position vector,

acceleration, linear momentum, angular

momentum

3. Magnitude of a Vector

§ Length of a vector also called ‘norm’ of vector

§ It is always a positive quantity.

For a vector = .®

A

§ Magnitude or norm = | .®

A |

or simply = A

4. Equal vectors:

Two vectors A and B are said to be equal when

they have equal magnitude and same direction

and represent the same physical quantity

5. Collinear vectors

Collinear vectors are those which act along the

same line. Angle between them can be 0° or 180°.

6. Parallel Vectors:

§ If two vectors A and B act in the same

direction along the same line or parallel lines

§

§ Angle between them is 00

7. Anti–parallel vectors:

§ Two vectors A and B are in opposite

directions along same line or parallel lines.

§ Angle between them is 1800

8. Unit vector:

§ A vector divided by its magnitude

§ Speciies only direction of the vector

§ Unit vector for .®

A denoted by A

||ˆ

A

AA

®

=

§ It has a magnitude equal to unity or one

9. Orthogonal unit vectors:

§ Two vectors which are perpendicular to each

other are called orthogonal vector

§ kji ˆ,ˆ,ˆ unit vectors specifying directions along

+ve x–axis, +ve y–axis+ve z–axis

§ These three unit vectors are directed

perpendicular to each other

§ Angle between any two of them is 90°.

10. Addition of Vectors

§ Since vectors have both magnitude and

direction they cannot be added by ordinary

algebra.

§ Vectors can be added geometrically or

analytically using

i) Triangular law of addition method

ii) Parallelogram law of vectors

11. Triangular Law of addition method

If two vectors are represented in magnitude direction

by two adjacent sides of a triangle taken in order, then

their resultant is the closing side of triangle taken in

reverse order.

Head of .®

A connected to tail of ®

B

Angle between ®

A and®

B . = θ

Resultant vector connecting tail of



®

A to head of ®

B = ®

R

Magnitude of resultant vector = ®

R (OB)

Thus ®®®

+= BAR .

BN is drawn ^r to extended OA, from B.

Angle between ®

A and®

B . = θ

q=Þ=q

q=Þ=q

®

®

®

®

sinsin

coscos

BBN

B

BN

BAN

B

AN

Magnitude of resultant vector

q++=

q+q+q+=

q+q+q+=

q+q+=

+=D®®®®

cos2

cos2)sin(cos

sincos2cos

)sin()cos()(

,

22

2222

222222

222

222

ABBAR

ABBA

BABBAR

BBAR

BNONOBOBNIn

Direction of resultant vectors:

÷ø

öçè

æq+

q=a

+=

=aD

a=

-

®®

cos

sintan

tan

,

1

BA

B

ANOA

BN

ON

BN

AR

OBN,In

andb/wAngle

12. Parallelogram law of vectors

If two vectors acting at a point are represented in

magnitude anddirection by the two adjacent sides of a

parallelogram, then their resultantis represented in

magnitude and direction by the diagonal passing

through common tail of the two vectors.

Vectors ®

A and®

B inclined to each other at an angle θ

§®

A and®

B are represented in magnitude and

direction by two sides OA and OB of a

parallelogram OACB.

§ Diagonal OC passing thro’ common tail O,

gives the magnitude and direction of the

resultant ®

R .

§ CN is drawn ^r to extended OA, from C.

§ Let ÐCON made by ®

R with ®

A = b

q=Þ=q

q=Þ=q

®

®

®

®

sinsin

coscos

BCN

B

CN

BAN

B

AN

Magnitude of resultant vector

q++=

q+q+q+=

q+q+q+=

q+q+=

+=D®®®®

cos2

cos2)sin(cos

sincos2cos

)sin()cos()(

,

22

2222

222222

222

222

ABBAR

ABBA

BABBAR

BBAR

CNONOCONCIn

Direction of resultant vectors:

÷ø

öçè

æq+

q=a

+=

=aD

b=

-

®®

cos

sintan

tan

,

1

BA

B

ANOA

CN

ON

CN

AR

OBN,In

andb/wAngle

Special Cases

i) When two vectors act in same direction

θ = 0o, cos 0o = 1

BAR

ABBAR

+=

++= 222

ii) When two vectors act in the opposite direction

θ = 180°, cos 180° = -1

BAR

ABBAR

-=

-+= 222

iii) When vectors are at right angles to each other

θ = 90°, cos 90o = 0

22 BAR +=



13. Components Of a Vector

In Cartesian coordinate system ®

A can be resolved

into three components along x, y and z directions

Resolution of a vector

22

,,,,

ˆˆˆ

yx

y

x

zyx

zyx

AAA

A

AA

A

AzyxAAA

kAjAiAA

+=

q=

q=

=q

=

++=

®

®

®

®

ofmagnitude

Asin

,cos

axisxwith bymadeangleLet

ofcomponent

14. Vector addition using components

Two vectors and in a Cartesian coordinate system

kBjBiBB

kAjAiAA

zyx

zyx

ˆˆˆ

ˆˆˆ

++=

++=

®

®

Addition of two vectors

Adding corresponding x, y and z components.

kBAjBAiBABA zzyyxxˆˆˆ +++++=+

®®

Subtraction of two vectors

subtracting corresponding x, y, z components

kBAjBAiBABA zzyyxxˆˆˆ -+-+-=-

®®

15. Multiplication of Vector by a scalar

Vector ®

A multiplied by scalar l = l®

A

If l is a positive number then O ®

A is also in the

direction of ®

A

If l is a negative number, l®

A is in the opposite

direction to ®

A .

16. Scalar product between two vectors.

Product of the magnitudes of both the vectors

and the cosine of the angle between them.

q=

=q

=

=

®®

®®

®®

cos·AB

thembetween angle

ofmagnitudesBA,

vectorstwo

BA

BA

BA

.

,

,

17. Properties Scalar product between two vectors.

i)®®

BA. is always a scalar.

if q is acute (< 90°) ®®

BA. is positive

if θ is obtuse (90°<θ< 180°) ®®

BA. is negative

ii) Scalar product is commutative,

®®®®

= ABBA ..

iii) Obey distributive law

®®®®®®®

+=+ CABACBA ..).(

iv) Angle between vectors

÷÷÷

ø

ö

ççç

è

æ=q

=q

®®

-

®®

AB

cos·AB

BA

BA

.cos

.

1

v) Parallel vectors: when θ = 0°, cos θ = 1

scalar product will be maximum

ABBA =÷ø

öçè

æ ®®

max

.

vi) Anti Parallel vectors: when θ = 180°, cos θ = -1

scalar product will be minimum

ABBA =÷ø

öçè

æ ®®

min

.

vii) mutually orthogonal

®®®®

®®

^=

=

=q

BAeiBA

BA

.0.

90. 0cos·AB

90When



viii)self–dot product

scalar product of a vector with itself

22

02

2

)(

0cos

.)(

AA

A

AAA

=

=

=

®

®®®

ix) In case of a unit vector

1ˆ.ˆˆ.ˆ.

1cos11ˆ.ˆ 0

===

=0´´=

kkjjii

nn

Simillarly

x) In the case of orthogonal unit vectors

090cos.1.1ˆ.ˆ.ˆ.ˆˆ.ˆ 0 ==== ikkjji

xi) In terms of components

zzyyxx

zyxzyx

BABABA

kBjBiBkAjAiABA

++=

++++=®®

ˆˆˆˆˆˆ.

xii) work done by force ®

F to move an object

through a small displacement ®

rd

W = ®

F . ®

rd

18. Write note on vector product of two vectors.

Vector product

Another vector having a magnitude equal to

the product of the magnitudes of two vectors

and the sine of the angle between them.

n

BAC

BA

BA

ˆ)(

,

,

q=

´=

=q

=

=

®®®

®®

®®

sin·AB

thembetween angle

ofmagnitudesBA,

vectorstwo

19. Direction of the product vector

Perpendicular to the plane containing the two

vectors, in accordance with the right hand

screw rule/ right hand thumb rule

20. Right Hand Thumb Rule

if curvature of fingers of right hand represents

rotation of the object, then thumb, held

perpendicular to curvature of fingers,

represents direction of the resultant ®

C

21. Properties of Vector products of two vectors

i) vector product of any two vectors is always

another vector whose direction is

perpendicular to the plane containing these

two vectors, i.e., orthogonal to both vectors

®®®

´= BAC

ii) vector product is not commutative,

q=´=´

´-=´

´¹´

®®®®

®®®®

®®®®

AB sin ofmagnitude

But.,

||

][

ABBA

ABBA

ABBA

magnitudes are equal but directions are

opposite to each other

iii) Orthogonal Vectors

maximum magnitude when θ = 90°, sin 90°= 1

when vectors orthogonal to each other.

nABBA ˆmax

=÷ø

öçè

æ´

®®

iv) Parallel or antiparallel Vectors

minimum magnitude

when θ = 00 or 1800 sin θ = 0

vector product of two non–zero vectors

vanishes, if vectors are parallel or antiparallel

®®®

=÷ø

öçè

æ´ 0

min

BA

v) self–cross product,

product of a vector with itself is null vector

0ˆ0 ==´®®

nAA 0AA sin

vi) Self–vector products of unit vectors are zero.

0ˆˆˆˆˆˆ

0sin11ˆ.ˆ 0

=´=´=´

=0´´=

kkjjii

nn

Simillarly

vii)In case of orthogonal unit vectors in

accordance with the right hand screw rule:

jikikjkji ˆˆˆ,ˆˆˆ,ˆˆˆ =´=´=´

Since cross product is not commutative,

jkiijkkij ˆˆˆ,ˆˆˆ,ˆˆˆ -=´-=´-=´



viii) In terms of components,

)(ˆ

)(ˆ)(ˆ

ˆˆˆ

xyyx

zxxzyzzy

zyx

zyx

BABAk

BABAjBABAi

BBB

AAA

kji

BA

-+

-+-=

=´®®

22. Application of Vector Product

velocityangular

VelocityLinearv)

momentumlinear

momentumAngulariv)

Force

ectorposition v

Torqueiii)

triangleaofAreaii)

sidesAdjacent

ramparallelogofAreai)

=w

w´=

=

´=

=

=

´=t

´=

=

´=

®

®®®

®

®®®

®

®

®®®

®®

®®

®®

rv

p

prL

F

r

Fr

BA

BA

BA

||2

1

,

||

23. Properties of components of vectors

If two vectors are equal, then their individual

components are also equal.

zzyyxx

zyxzyx

BABABA

kBjBiBkAjAiA

BA

===

++=++

=®®

,,

ˆˆˆˆˆˆ

24. Position Vector

It is a vector which denotes the position of a

particle at any instant of time, with respect to

some reference frame or coordinate system

position vector®

r of particle at a point P

kzjyixr ˆˆˆ ++=®

where x, y and z are components of ®

r

2. 3. MOTION ALONG ONE DIMENSION

1. Distance

§ Actual path length travelled by an object in

the given interval of time during the motion.

§ It is a positive scalar quantity.

2. Displacement

§ Diference between final and initial positions

of the object in a given interval of time.

§ Shortest distance between two positions of

the object

§ Its direction is from the initial to inal position

of the object, during the given interval of time.

§ It is a vector quantity.

3. Distance & Displacement Comparison

i) Distance travelled by an object in motion in a

given time is never negative or zero, it is

always positive.

Displacement of an object, in a given time can

be positive, zero or negative.

ii) Displacement of an object can be equal or less

than the distance travelled but never greater

than distance travelled.

Distance covered by an object between two

positions can have many values, but the

displacement between them has only one

value

4. Displacement Vector

Let a particle move from point P1 to point P2

kzzjyyixxr

rr

kzjyixr

kzjyixr

ˆ)(ˆ)(ˆ)(

ˆˆˆ,

ˆˆˆ

121212

12

2222

1111

-+-+-=D

-=

++=

++=

®

®®

®

®

vectorntdisplaceme

PofP.V

PofP.V

2

1,



5. Velocity and Speed

i) Average velocity

Ratio of displacement vector to time interval

It is a vector quantity.

t

rvav

D

D=

=

®®

intervaltime

vectorntdisplaceme velocityAverage

Direction of average velocity is in the

direction of the displacement vector

ii) Average speed

Ratio of total path length travelled by particle

in a time interval.

intervaltime

lengthpath totalspeedAverage =

iii) Instantaneous velocity or velocity

Limiting value of average velocity as Δt→0,

evaluated at time t or

Rate of change of position vector w.r.t time.

dt

rdv

t

rv

t

®®

®

®D

®

=

D

D=

0lim

iv) Component form of velocity

kvjvivv

zvdt

dz

yvdt

dy

xvdt

dx

kdt

dzj

dt

dyi

dt

dxv

kzjyixdt

dv

zyx

z

y

x

ˆˆˆ

,

,

,

ˆˆˆ

)ˆˆˆ(

++=

=

=

=

++=

++=

®

®

®

Then

velocityofcomponent

velocityofcomponent

velocityofcomponentIf

v) Speed

Magnitude of velocity v.

Speed is always a positive scalar

222zyx vvvv ++=

6. Momentum

§ Linear momentum or momentum of a particle

is product of mass with velocity.

§ Momentum is also a vector quantity

§ unit of the momentum is kg m s−1

§ Direction of momentum is in the direction of

velocity

§vmp =

=®

particle.ofxspeedmasspofMagnitude

§ Component form momentum

kmvjmvimvp

zmvp

ymvp

xmvp

kpjpipp

zyx

zz

yy

xx

zyx

ˆˆˆ

ˆˆˆ

++=

++=

®

®

momentumofcomponent,=

momentumofcomponent,=

momentumofcomponent,=If

7. Acceleration

i) Accelerated Motion

§ During non-uniform motion of object, velocity

of the object changes from instant to instant

§ Velocity of the object is no more constant but

change with time.

§ Such a motion is accelerated motion.

§ If change in velocity of an object per unit time

is same (constant) then the object is said to be

moving with uniformly accelerated motion.

§ If change in velocity per unit time is diferent

at diferent times, then the object is said to be

moving with non-uniform accelerated motion.

ii) Average Acceleration.

§ Ratio of change in velocity over a time interval

§ It is a vector quantity in the same direction

§ If an object changes its velocity from®

1v to ®

2v

in a time interval Δt= t2− t1,

t

va

tt

vva

avg

avg

D

D=

-

-=

=

®®

®®®

12

12

intervaltime

yin velocitchangeonacceleratiaverage

§ Average acceleration will give change in

velocity only over the entire time interval.



§ It will not give value of the acceleration at

any instant time t.

iii) Instantaneous acceleration

§ Ratio of change in velocity over Δt, as Δt

approaches zero Or

§ Rate of change of velocity

dt

vda

t

v

t®

®

®

=

D

D

®D=

0limonAccelerati

§ Acceleration is a vector quantity.

§ Its SI unit is ms−2

§ Dimensional formula M0L1T−2

§ Acceleration is +ve if velocity is increasing

§ Acceleration is -ve if velocity is decreasing.

§ Negative acceleration is called retardation

or deceleration.

iv) Component form of acceleration

2

2

2

2

2

2

2

2ˆˆˆ

ˆˆˆ

dt

rda

kdt

zdj

dt

ydi

dt

xda

kdt

dvj

dt

dvi

dt

dva

dt

vda

zyx

®®

®

®

®®

=

++=

++=

=

Acceleration is second derivative of position

vector with respect to time.

8. Relative Velocity

When two objects A and B are moving with

diferent velocities, then the velocity of one object

A with respect to another object B is called

relative velocity of object A with respect to B.

Case1: Two objects A and B moving with uniform

velocities VA and VB, along straight tracks

in the same direction.

ABBA

BAAB

VVV

VVV

®®®

®®®

-=

-=

A,w.r.tBof velocityRelative

B,w.r.tAof velocityRelative

if two objects are moving in the same direction

îíì

=þýü

s. velocitietwoof

diferencemagnitude

anotherw.r.tobjectoneof

velocityrelativeofmagnitude

Case2: Two objects A and B moving with uniform

velocities VA and VB, along straight tracks

but opposite in direction

)(

)(

AB

ABBA

BA

BAAB

VV

VVV

VV

VVV

®®

®®®

®®

®®®

+-=

--=

+=

--=

A,w.r.tBof velocityRelative

B,w.r.tAof velocityRelative

If two objects are moving in opposite directions

îíì

=þýü

s. velocitietwoof

magnitudeofsum

anotherw.r.tobjectoneof

velocityrelativeofmagnitude

Case 3 velocities vA and vB at an angle θ between

their directions.

q-

q=b

=b

q-+=

-=

®®

®

®®®

cos

sin

cos222

BA

B

ABB

BABAAB

BAAB

VV

V

VV

VVVVV

VVV

tanDirection

andbetweenAngleLet

ofmagnitude

B,w.r.tAofRV

Special Cases

i) When θ = 0°, the bodies move along parallel

straight lines in the same direction,

vAB= (vA − vB) in the direction of AV®

vBA = (vB + vA) in the direction of BV®

.

ii) When θ = 180°, the bodies move along parallel

straight lines in opposite directions,

vAB = (vA + vB) in the direction of AV®

.

vBA = (vB − vA) in the direction of BV®

.

iii) When θ = 90°, the two bodies are moving at

right angles to each other.

22BAAB VV +=V

9. Equations of Uniformly Accelerated Motion

.timeafteratbodyof velocity

0timeatobjectof velocity

linestraightain moving

bodyaofon acceleratiuniform

tv

tu

a

=

==

=

i) Velocity - time relation

Acceleration of body at any instant is frst

derivative of velocity w.r.t time



atuv

atuv

tav

dtdv

vu

t

adtdv

dt

dva

tvu

tv

u

+=

=-

=

=

=

=

=

=

òò0

0

:sidesboth g Integratin

tofromchanges Velocity

to0fromchangesTime

ii) Displacement – time relation

Velocity of is first derivative of displacement

w.r.t time

2

0

2

0

00

2

1

2

)(

)(

atuts

tatus

dtatuds

t

dtatuds

vdtds

dt

dsv

t

t

ts

+=

úû

ùêë

é+=

+=

=

+=

=

=

òò

timeatntdisplacemeparticles

originfromstartedparticle0,=ttimeAt

thatassuming sidesboth g Integratin

relationtime- VelocityBy

iii) Velocity – displacement relation

Acceleration is first derivative of velocity w.r.t

time

asuv

asuv

uvas

vsa

dvvdsa

vu

t

dvvdsa

vds

dva

dt

ds

ds

dv

dt

dva

v

u

s

v

u

s

2

2

2

2

.

.

.

.

22

22

22

2

0

0

+=

=-

-=

úû

ùêë

é=

=

=

=

=

=

=

=

òò

:sidesboth g Integratin

tofromchanges Velocity

to0fromchangesTime

iv) Displacement in terms of u,v,t only

2

)(

)(2

1

2

1 2

tuvs

tuv ut atuts

uvatat uv

+=

-+=+=

-=+=

sor

or

v) Kinematic equations

2

)(

2

2

1

22

2

tuvs

asuv

atuts

at uv

+=

+=

+=

+=

10. Free falling bodies

i) body falling from a height h under gravity § Near the surface of the Earth, acceleration

due to gravity ‘g’ is constant.

§ Motion of a body falling towards Earth from a small altitude, purely under force

of gravity is called free fall.

§ Object of mass m falling from height h

§ Assume there is no air resistance.

§ Downward direction is taken as +ve y-axis

gyuv

gtuty

gtu

gaga

kjgia

y

2

2

1

ˆ0ˆˆ0

22 +=

+=

+=

=þýü

==

++=®

equationKinematicFrom

ntDisplaceme

vt,timeanyatvelocity

u velocityinitialwardwith

downthrown isparticle

orcomponentscomparing

onaccelerati

2

ghV

g

hg

gT

g

hT

g

ht

gth

yh

gyv

gty

gt

2

2

2

2

2

1

2

2

1

2

2

=

´=

=þýü

=

=

=

=

=

=

=

gro

2

2

groundreachesit

when particleofspeed

groundreach totaken time

tisplacemenVertical

v

0=urest,fromstartsparticle

Body thrown vertically upwards



An object of mass m thrown vertically

upwards with an initial velocity u

Vertical direction is taken as positive y axis

Acceleration a = −g (neglect air friction)

gyuv

gtuts

gtu

2

2

1

22 -=

-=

-=

2ntDisplaceme

vt,timeanyatvelocity

MOTION IN A PLANE

1.i) Projectile Motion

§ When an object is thrown in the air with some

initial velocity, and allowed to move under

action of gravity alone, it is called a projectile.

§ Path followed by particle is its trajectory

ii) Examples of projectile

§ Object dropped from a moving train.

§ A bullet fired from a rifle.

§ A ball thrown in any direction.

§ A javelin or shot put thrown by an athlete.

§ A jet of water near bottom of a water tank.

iii) Projectile moves under efect of two velocities

§ Uniform velocity in the horizontal direction

Do not change if there is no air resistance.

§ A uniformly changing velocity

increasing or decreasing in vertical direction

iv) Types of projectile motion:

§ Horizontal projection

Projectile given an initial velocity in

horizontal direction

§ Angular projection

Projectile given an initial velocity at an angle

to the horizontal

v) Assumption for motion of a projectile

§ Air resistance is neglected.

§ Rotation and curvature of Earth is negligible

§ g is constant in magnitude and direction

2. Projectile in horizontal projection

§ A ball is thrown horizontally with initial

velocity ®

u from the top of a tower of height h

§ The ball covers a horizontal distance due to its

uniform horizontal velocity u,

§ A vertical downward distance because of

constant acceleration due to gravity g.

§ Under combined efect ball moves along OPA.

§ The motion is in a 2-dimensional plane.

i) Path of a projectile

velocityofcomponentvertical

velocityofcomponenthorizontal

(t) y ,travelleddistancevertical

(t)x ,travelleddistancehorizontal

groundreach totaken Time

y

x

u

u

y

x

t

=

=

=

=

=

Initial velocity ux remains constant

)1....(

0

2

1 2

x

x

x

u

xt

tux

a

attux

=

=

=

+=

direction,xalong

t,timeintraveleddistance

constantis

(1),using Sub

t,timeintraveleddistance

g=adirection,along y

0=componentdownward

2

2

2

2

2

2

2

2

2

2

1

2

1

2

1)0(

x

x

x

y

u

gK

Kxy

xu

gy

u

x gyt

gty

gtty

u

=

=

=

úû

ùêë

é=

=

+=

It is the equation of a parabola.

path followed by the projectile is a parabola



ii) Time of Flight (T)

§ Time taken by projectile to complete trajectory

or

§ Time taken by the projectile to hit the ground

g

hT

gTTh

sh

T

u

attus

y

y

yy

2

2

1)0(

2

1

=

+=

=

=

+=

2

2

motion, verticalofEqn

ntdisplacemeVertical

flightofTime

g=adirection,along y

0=componentdownward

distanceVertical

§ Time of flight of projectile depends on height

independent of the horizontal velocity

§ If one ball falls vertically and another ball is

projected horizontally with some velocity,

both balls will reach bottom at the same time

iii) Horizontal range

Horizontal distance between projection point

and point where projectile hits ground.

g

huR

TuTR

sR

g

hT

u

attus

x

x

xx

2

)0(2

1

2

2

1

=

+=

=

=

+=

2

2

Range,Horizontal

flightofTime

0=adirection,xalong

u=componenthorizontal

distanceHorizontal

iv) Resultant Velocity:

§ At any instant t, projectile has velocity

components along both x-axis and y-axis.

§ Resultant of these two components gives

velocity of the projectile at that instant t

gtv

gau

t a uv

uv

au u

t a uv

x

xx

yyy

x

xx

xxx

=

==

+=

=

==

+=

Thus,

axis-xalong componentvelocity

Thus,

axis-xalong componentvelocity

0,

,

0,

,

222

22,

ˆˆ

tguv

vvvt

jgtiuv

yx

+=

+=

+=®

instantanyatparticleofspeed

instantanyparticleatofvelocity

v) Speed of projectile when it hits ground:

When the projectile hits the ground ater

initially thrown horizontally

ghuv

ghv

g

hg

gTv

uv

g

hT

y

y

x

2

2

2

2

2 +=

=

=

=

=

=

groundreachesitwhen speed

componentvertical

component,horizontal

lightoftime

3. i) Oblique projectile

Projectile motion takes place when initial velocity is not horizontal, but at some angle with vertical

Examples:

Water ejected out of hose pipe held obliquely.

Cannon fired in a battle ground

ii) Projectile under an angular projection

An object thrown with initial velocity u

At an angle θ with the horizontal.

θu u

θu u

juiuu

y

x

yx

sin

cos

ˆˆ

=

=

+=®

componentVertical

component,Horizontal



Path followed by projectile

§ Acceleration due to gravity is in the direction opposite to vertical component

§ There is no acceleration along x direction throughout motion. Horizontal component remains same till object hits ground

θu

xgxy

θu

xg

θu

uxyt

gttuy

t+ a = uv

gaθuu

θu

xt

t θu x

at

θ = u u

t + a = uv

θ = u u

au

xyy

yy

x

xxx

x

xy

22

2

2

2

2

cos2

1tan

cos2

1

cos

sin,

2

1sin

,sin,

cos

cos

cos

,

cos

-q=

úû

ùêë

é-

q=

-q=

-==

=

=

+=

==

ng Substituti

velocityvertical

motion verticalFor

2

1tusdistanceHorizontal

velocityHorizontal

00,motion,horizontalFor

xx

i.e path followed by projectile is inverted parabola

Maximum height (hmax)

Maximum vertical distance travelled by projectile during its journey is called maximum height.

g

uh

ghu

sauv

= hs

v

gaθuu

yyyy

y

y

yy

2

sin

2sin0

2

,sin,

22

max

max22

22

max

q=

-q=

+=

=

-==

motionofequation From

,travelleddistancevertical

0

motion verticalFor

Time of flight (Tf)

Total time taken by projectile from the point of projection till it hits the horizontal plane is called time of flight

g

uT

gTθu

ats

y =sy

T

f

ff

y

y

f

q=

-=

+=

=

=

sin2

sin

0

2

2

2

1T0

2

1tu

,inntDisplaceme

tightfligh ofTime

y

g

uR

g

uu

Tfu= R

q=

q´q=

´q

2sin

sin2cos

cos

2

flight oftimex velocityHorizontal=Range

Circular Motion

1. Radian

§ Length of arc divided by the radius of the arc.

§ Planar angle subtended by a circular arc at the

center of a circle.

§ 1 radian

Angle subtended at the center of a circle by

an arc equal in length to radius of the circle.

p=

=p

=

1801

2 0

0

rad

radians 360

360circleabycoveredangletotal

2. Angular displacement

§ Angle described by the particle about the axis

of rotation (or center O) in a given time

§ Unit of angular displacement : radian

§ Relation between q, S and r

radius

(AB)length Sarc

nt displacemeAngular

=

=

=q

=q

r

S

r

S

3. Angular velocity

§ Rate of change of angular displacement

dt

d

t

t

t

q=w

D

qD=w

q=

®

®D

®

0lim

,

velocityangular

timein ntdisplacemeangular

§ unit of angular velocity : rad s−1



§ Direction of angular velocity is along the axis

of rotation following the right hand rule

4. Angular acceleration

§ Rate of change of angular velocity

dt

d®

® w=a

§ Angular acceleration is also a vector quantity

§ a need not be in same direction as w

5. Relation between linear and angular velocity

Consider an object moving along a circle

w=

=

w=

w=

w=

D

qD=

D

D

D

qD=

D

D

D

qD=D

D=qD

=

qD=

D=

D=

=

®®

®D®D

®D

velocityAngular

velocitylinearHere,

notationIn vector

tlimitr

t

Slimit

:sidesboth limitTake

tr

t

S

:tbysidesboth Divide

rS

radius

lengtharcntdisplacemeAngular

subtendedangle

sdistancearc

tintervalTime

rradiuscircle

00

0

v

rv

rv

rdt

ds

r

S

6. Tangential acceleration

dt

d

dt

dva

ra

dt

dr

dt

dv

rv

t

t

w=a

=

a=

w=

w=

on,acceleratiangular

on,acceleratitangentialHere,

:timew.r.tting Diferentia

:Relation velocityangular-Linear

§ Tangential acceleration at experienced by an

object is circular motion

§ at is in the direction of linear velocity

7. Uniform circular motion

§ When an object is moving on a circular path

with constant speed, it covers equal distances

on in equal time intervals

§ Object is said to be in uniform circular motion

§ In uniform circular motion, velocity is always

changing but speed remains same.

§ Magnitude of velocity vector remains constant

§ Only the direction changes continuously.

a. Non Uniform circular motion

§ Velocity changes in both speed and direction

(or) Speed is not constant in circular motion

§ Whenever the speed is not same in circular

motion, the particle will have both centripetal

and tangential acceleration

§ Example: bob attached to string moves in

vertical circle, speed of bob is not same

8. Centripetal acceleration

§ In uniform circular motion, velocity is

tangential at every point in circle

§ Acceleration is always acting towards center

of circle

§ This is called centripetal acceleration.

9. Expression for Centripetal acceleration

Directions of position and velocity vectors

shift through same angle q in a small

interval of time Dt



velocityangular

limlim

t,byDivide

(2)and(1)From

inwardradiallysign

velocityinchange

ntdisplacemeinChange

Speed

Radius

:motioncircularuniformFor

0t0t

=w

w-=

-=

´-=

D

D-=

D

D

D

D-=

D

DD

D´-=D

D-=

D

D=-

D-=q

-=D

D=q

-=D

==

==

®

®D

®

®D

®®

®®

®®

®

®®®

®

®®®

®®

®®

ra

r

va

vr

va

t

r

r

v

t

v

t

r

r

v

t

v

rr

vv

v

v

r

r

v

v

v

vvv

r

r

rrr

vvv

rrr

2

2

12

12

21

21

)2.....(

)1.....(

||||

||||

10. Resultant acceleration in non uniform circular

Motion

r

v

a

ra

r

vaa

aaa

t

r

tr

tr

2

222

tan

)

=q

q=

÷÷ø

öççè

æ+=

+=

®

®®

andbetween Angle

ofmagnitude

(on acceleratiResultant

11. Kinematic equations for angular motion

2

)(

2

2

1

22

2

tuvs

asuv

atuts

at uv

+=

+=

+=

+=

2

)(

2

2

1

0

20

2

20

0

t

tt

t

w+w=q

aq+w=w

a+w=q

a+w=w

LAWS OF MOTION

1. Early Concept of Motion

§ About 2500 years ago, Aristotle, said that

Force causes motion.

§ His statement is based on common sense.

§ Scientific answer must be endorsed with

quantitative experimental proof.

2. Galileo experiment.

§ In the 15th century, Galileo said force is not

required to maintain motion.

§ When a ball rolls from the top of an inclined

plane to its bottom, after reaching the ground

it moves some distance and continues to move

on to another inclined plane of same angle

§ By increasing the smoothness of both inclined

planes, the ball reach same height from where

it was released in the second plane.

§ Motion of the ball is then observed by varying

angle of inclination of the second plane

keeping the same smoothness.

§ If the angle of inclination is reduced, the ball

travels longer distance in the second plane to

reach the same height .

§ When the angle of inclination is made zero,

ball moves forever in the horizontal direction.

§ If Aristote’s idea were true, the ball would

not have moved in the second plane

§ Galileo proved that force is not required to

maintain motion. An object can be in motion

even without a force acting on it.

§ Aristotle coupled the motion with force

Galileo decoupled the motion and force.

3. Newton’s First Law

Every body continues in its state of rest or of

uniformmotion along a straight line unless it is

compelled by an external force tochange that state.

4. Inertia and its Types



§ Inability of objects to move on its own or

change its state of motion is called inertia.

§ Inertia means resistance to change its state.

i) Inertia of rest:

§ Inability of an object to change its state of rest

§ When a stationary bus starts to move, the

passengers experience a sudden backward

push. Due to inertia, the body (of a

passenger) will try tocontinue in the state of

rest, while the bus moves forward. h is

appears as a backward push

ii) Inertia of motion:

§ inability of an object to change its state of

uniform speed (constant speed) on its own

§ When the bus is in motion, and if the brake is

applied suddenly, passengers move forward

and hit against the front seat.

§ In this case, the bus comes to a stop, while the

§ body (of a passenger) continues to move

forward due to the property of inertia.

iii) Inertia of direction:

§ Inability of an object to change its direction of

motion on its own

§ When a stone attached to string is in whirling

motion, and if the string is cut suddenly, the

stone will not continue to move in circular

motion but moves tangential to the circle

§ Because the body cannot change its direction

of motion without any force acting on it.

5. Frame of Reference

When an object is at rest or in motion with

constant velocity, it has a meaning only if it is

specified w.r.t some reference frames.

6. Inertial Frames and non Inertial Frames

§ Newton’s first law defines an inertial frame.

§ There exists some special set of frames in which if a

body experiences no force, it moves with constant

velocity or remains at rest

§ A frame of reference in which Newton's first

law is valid is called inertial frame of reference.

§ A frame in which Newton's first law is not

valid is called a noninertial frame of reference.

7. Inertial Frame – The Earth

§ Newton’s first law deals with motion of

objects in the absence of any force and not the

motion under zero net force.

§ If an object is very far away from any other

object, then Newton’s first law will be valid.

§ Earth is treated as an inertial frame because an

object on laboratory table is at rest always.

§ This object never picks up acceleration in the

horizontal direction since no force acts on it in

the horizontal direction.

§ Gravitational force in the downward direction

and normal force in upward direction makes

the net force is zero in vertical direction.

8. Inertial Frames other than Earth

§ If an object appears to be at rest in one inertial

frame, it may appear to move with constant

velocity with respect to another inertial frame.

§ All inertial frames are moving with constant

velocity relative to each other

i) Object at rest outside a train appears to

move with constant velocity w.r.t moving

train . So train is treated as inertial frame.

ii) A car moving with constant velocity w.rt.

ground to a person at rest on the ground,

both frames w.r.t car and to ground are

inertial frames.

9. Examples for non Inertial Frames

i) If a train suddenly accelerates object at

rest on a table inside train appears to

accelerate backwards without any force

acting on it. It is violation of first law. So

train is not inertial frame

ii) A car is a non-inertial frame if it moves

with acceleration w.r.t ground

iii) A rotating frame is a non inertial since

rotation requires acceleration.

10. The Earth as inertial and non Inertial Frame

§ Earth is not really an inertial frame since it has

self-rotation and orbital motion.

§ Rotational efects is ignored for some motions

§ Earth’s self-rotation has very negligible efect

on it, when an object is thrown, to measure

time period of simple pendulum

§ So Earth can be treated as an inertial frame.



§ Earth is not inertial frame since self-rotation

has strong influence on wind patterns and

satellite motion analys

11. Newton’s Second Law

Force acting on an object is equal to the rate of

change of its momentum or

whenever the momentum of the body changes,

there must be a force acting on it.

constantremainsmass

Thus,

momentum

=

=

=

=

=

®®

®®

®®

®®

m

amF

dt

vdmF

vmp

dt

pdF

§ If there is an acceleration a on the body, there

must be a force acting on it. or

If there is a change in velocity, then there

must be a force acting on the body.

§ Force and acceleration are always in the same

direction.

§ Newton’s second law was a paradigm shfit

from Aristotle’s idea of motion.

§ According to Newton, force need not cause

motion but only a change in motion.

§ Second law is valid only in inertial frames.

§ In non-inertial frames, it requires some

modification.

12. Unit of force

newtons denoted by ‘N’.

One Newton is defined as

Force which acts on 1 kg of mass to give an

acceleration 1 m s−2 in the direction of the force.

13. Aristotle vs Newton’s on sliding object

§ Newton’s second law gives correct

explanation for inclined plane experiment

§ If friction is not negligible, object reaches the

bottom of the inclined plane, it travels some

distance and stops.

§ Because there is a frictional force acting in the

direction opposite to its velocity.

§ Frictional force reduces the velocity of the

object to zero and brings it to rest.

§ As per Aristotle, as soon as the body reaches

the bottom of the plane, it can travel only a

small distance and stops because there is no

force acting on the object.

§ Aristotle did not consider frictional force

14. Newton’s Third Law

For every action there is an equal and opposite reaction.

2112

®®

-= FF

Whenever an object 1 exerts a force 21

®

F on object 2, then object 2 must also exert equal

and opposite force 12

®

F on the object 1.

§ These forces lie along the line joining the

two objects

§ A single force cannot exist in nature.

§ Action and reaction act on two diferent

bodies.

§ Third law is valid in both frames.

§ Action-reaction forces are not cause and

effect forces because Action-reaction

happens at the same instant.

15. Discussion on Newton’s Laws

i) Newton’s laws are vector laws.

zz

yy

xx

zyxzyx

maF

maF

maF

kmajmaimakFjFiF

amF

=

=

=

++=++

=®®

:sidesboth comparing

:scoordinateCartesian

ˆˆˆˆˆˆ

§ Acceleration along x or y or z direction

depends only on component of force

acting along respective directions

§ Force acting along y direction cannot alter

the acceleration along x direction.



§ Fz cannot affect ay and ax

ii) Acceleration experienced by body at time t

)()( tamtF®®

=

§ depends on the force which acts on the

body at that instant of time.

§ does not depend on the force which acted

on the body before the time t

§ does n’t depend on past history of force

iii) Direction of force and direction of motion

1. Force and motion in the same direction

§ When an apple falls towards the Earth,

direction of motion of apple and that of force

are in the same downward direction

2: Force and motion not in the same direction

§ Moon experiences a force towards the Earth.

§ But it actually moves in elliptical orbit.

3. Force and motion in opposite direction

§ If an object is thrown vertically upward,

direction of motion is upward

§ gravitational force is downward

4. Zero net force, but there is motion

§ Raindrop detached from cloud experiences

downward gravitational force and upward

air drag force.

§ Ater a certain time, the upward air drag force

cancels the downward gravity.

§ Then raindrop moves at constant velocity and

touches Earth.

§ Raindrop comes with zero net force, but with

non-zero terminal velocity

iv) Newton’s second law is a second order

diferential equation

Acceleration is second derivative of position vector

2

2

2

2

dt

rdmF

dt

rda

®®

®®

=

=

force

onaccelerati

whenever second derivative of position vector

is not zero, a force must be acting on body

v) Second law is consistent with the first law

constant

bodytheon actsforcenoIf

=

=

®

®

v

dt

vdm 0

§ Newton’s first and 2nd laws are independent

§ They can not be derived from each other.

Second law is cause and effect relation.

§ Force is the cause; acceleration is the efect.

§ Eff ect should be written on the let

§ Cause on the right hand side of the equation.

§ Correct way of writing Newton’s second law

®®

®®

== Fdt

pdFam or

16. Free Body Diagram

Free body diagram is a simple tool to analyse the

motion of the object using Newton’s laws.

Systematic steps to develop free body diagram

i) Identify the forces acting on the object.

ii) Represent the object as a point.

iii) Draw the vectors representing the forces

acting on the object.

iv) Forces exerted by object should not be

included in the free body diagram.

17. Particle Moving in an Inclined Plane

§ An object of mass m slides on a frictionless

surface inclined at an angle q

§ Forces acting on it decides

a) acceleration of the object

b) speed of object when it reaches the bottom

§ Force acting on the object is

i) Downward gravitational force (mg)

ii) Normal force ^rto inclined surface (N)

§ To draw free body diagram, the block is

assumed to be a point mass

§ Since the motion is on the inclined surface, we

have to choose the coordinate system parallel

to the inclined surface

§ Gravitational force mg is resolved in to

i) llel component mgsinq along inclined plane

ii) ^r component mg cosq ^r toinclined surface



§ Angle made by the gravitational force with ^r

to surface is equal to angle of inclination q

§ There is no acceleration along y axis.

§ Object slides with acceleration along x axis.

q=

q+=

+=

=

q=

=q

=q

q=

=q-

=q-

sin2

sin20

2

sin

sin

ˆˆsin

cos

0cos

0ˆcosˆ

22

sgv

sgv

asuv

ga

mamg

imaimg

x

mg

mgN

jmgjN

y

bottom,reachesitwhen speed

equationkinematicFrom

0(u)speedinitial

objectsliding ofon accelerati

componentscompare

:directionin law2Apply

Nforcenormalofmagnitude

componentscompare

:directionin law2Apply

nd

nd

18. Two Bodies in Contact on a Horizontal Surface

direction.veforceofdirection

notationIn vector

forcecontactofmagnitude

ofng valueSubstituti

on byexertedforcecontact

on byexertedforcecontact

componentscomparing

directionxpositiveAlong

,lawsecondsNewton’By

masscombined

on acceleratiwith motion intoSet

FforceHorizontal

surfacessfrictionlehorizontal

smoothin blockstwoofmasses

x

mm

mFf

mm

Fmf

mm

mF

mm

FmFfa

amFf

amfF

iamifiF

fmm

fmm

mm

Fa

maF

imaiF

amF

mmm

a

, mm

-=

+-=

+=

úû

ùêë

é+

-=

+-=

-=

=-

=-

-=

=

+=

=

=

=

+=

=

=

=þýü

®

®

®

®®

21

212

21

212

21

1

21112

112

112

112

1212

2121

21

21

21

1

,

ˆˆˆ

ˆˆ

§ Magnitude of contact force depends on mass m2

which provides reaction force For mass m2 there is only one force acting on it in the x direction

directionve:forcethisofdirection

notationIn vector

forcecontacttheofmagnitude

componentsthecomparing

:massforlawsecondsNewton’

2112

21

221

21

221

221

221

2

ˆˆ

®®

®

-=

+

+=

+=

=

=

ff

x

mm

Fmf

mm

Fmf

amf

iamif

m

19. Motion of Connected Bodies

When objects are connected by strings and a

force F is applied vertically or horizontally or

along an inclined plane, it produces a tension

T in the string, which afects the acceleration.

Case 1: Vertical motion

Consider two blocks connected by a light and inextensible string that passes over a pulley.

21

21

21

212

21

2122

21

2122

21

21

1221

1221

11

11

1

22

22

2

121

2121

2

1

)()(

)2()1(

)2...(

ˆˆˆ

)1....(

ˆˆˆ

mm

mmgT

mm

mmgm

gmm

mmmgmT

gmm

mmmgmTa

gmm

mma

ammgmm

amamgmgm

amgmT

jamjgmjT

m

amgmT

jamjgmjT

m

gmmm

mmmm

+=

úû

ùêë

é+

-+=

úû

ùêë

é+

-+=

úû

ùêë

é+

-=-

úû

ùêë

é+

-=

+=-

+=--

-=-

-=-

=-

=-

=

=

=

>=

(1),in of valueSub

componentscompare

:lawforsecondsNewton’

componentscompare


lifttoon forcenalGravitatio

aon accelerati

Tstring thein Tension

)(,blocksofmasses



Case 2: Horizontal motion

§ Mass m2 is kept on a horizontal table and

mass m1 is hanging through a small pulley

§ Assume that there is no friction on surface.

§ Blocks are connected to unstretchable string

§ m1 moves with acceleration a downward

§ m2 also moves with acceleration a horizontally

gmN

jgmjN

am

mm

gmmT

gmm

ma

ammgm

amgmam

amT

iamiT

m

amgmT

jamjgmjT

m

gm

s ming on masForces act

gm

ms ing on masForces act

1

2

2

2

2

21

21

21

1

121

112

2

2

2

11

11

1

1

2

0ˆˆ

0,

)2()3(

)3...(

)(

)2()1(

)2...(

ˆˆ

)1....(

ˆˆˆ

=

=-

=

+=

úû

ùêë

é+

=

+=

-=-

=

=

-=-

-=-

=

=

=

=

=

directionalong yFor

inSub

inSub

componentscompare


componentscompare


Tupwardsacting Tension

forcenalgravitatioDownward

Tstringbytension Horizontal

NsurfacebyforcenormalUpward

forcenalgravitatioDownward

:

:

Tension in the string for horizontal motion is half of

the tension for vertical motion for same set of masses

and strings.

Application in industries.

Ropes in conveyor belts (horizontal motion) work

longer than cranes and lifts (vertical motion)

20. Concurrent Forces

§ A collection of forces is said to be concurrent,

if the lines of forces act at a common point.

§ Concurrent forces need not be in same plane.

§ If they are in the same plane, they are

concurrent as well as coplanar forces

21. Lami’s Theorem

§ If a system of three concurrent and coplanar

forces is in equilibrium, magnitude of each

force of the system is proportional to sine of

the angle between the other two forces.

§ Lami’s theorem is useful to analyse the forces

acting on objects in static equilibrium.

g=

b=

a

gµ

bµ

aµ

=

gba=

=

®®®

®

®

®

®®®

sin

||

sin

||

sin

||

sin

sin

sin

:

,,

321

3

2

1

321

FFF

F

F

F

FFF

Thus

:sameisConstantnalityPproportio

theorem,sLami’By

Opointcommon aatforcesact

,,forcesb/wAngle

forcesconcurrent&coplanarthree

22. Conservation Laws

§ Dynamics of motion of bodies can be analysed

using conservation laws.

§ Three conservation laws in mechanics.

§ Conservation of total energy

§ Conservation of total linear momentum

§ Conservation of angular momentum.



23. Conservation of Total Linear Momentum

Total linear momentum of system is conserved in time

Or

If there are no external forces acting on a system, total

linear momentum of system is always constant vector

24. Derivation of Law of conservation of total linear

momentum using Newton’s 2nd and 3rd laws

21

21

21

21

21

1221

212

121

221

212

12

21

0)(

0

,

,

®®®

®®

®®

®®

®®

®®

®®

®®

®®

®®

®

®

+=

=+

=+

=+

-=

-=

=

=

=

=

=

=

ppp

pp

ppdt

d

dt

pd

dt

pd

dt

pd

dt

pd

FF

dt

pdF

dt

pdF

pF

pF

F

F

totmomentumlineartotal

vectorconstanti.e

law,thirdsNewton’By

law,secondsNewton’By

,toduepofmomentum

,toduepofmomentum

pon pbyexertedForce

pon pbyexertedForce

2

1

12

21

25. Impulse

Very large force acts on an object for a very short

duration, then the force is called impulsive force

or impulse.

26. Effects of Impulse

i) When a cricket player catches ball, he pulls his hands

gradually in the direction of ball’s motion.

§ If he stops his hands soon after catching the ball,

the ball comes to rest very quickly.

§ Momentum of ball is brought to rest very quickly.

§ Average force acting on body will be very large.

§ Due to large average force, hands will get hurt.

ii) Cars are designed with air bags

§ When a car meets with accident, its momentum

reduces drastically in a very short time.

§ Passengers inside car will experience large force.

§ By using air bags, momentum will reduce slowly

and average force acting on them will be smaller

iii) Two wheelers are fittet with Shock absorbers

§ They play the same role as airbags in the car.

§ When there is a bump on the road, a sudden force

is transferred to the vehicle.

§ Shock absorber prolongs the period of transfer of

force on to the body of the rider.

§ Vehicles without shock absorbers will harm body

iv) Jumping on a concrete cemented floor is more

dangerous than jumping on the sand.

Sand brings body to rest slowly than the concrete

floor, so that the average force experienced by the

body will be lesser.

27. Relation of momentum, Impulse, Average Force

forceaverage

pconstantisforceIf

impulse

timeoverg Integratin

ppmomentumin change

entuminitialmom

entuminitialmom

ttimefinal

ttimeinitial

dt

dplawsecondNewtonsBy

ttimeofintervalshortvery

Fobjecton actsForce

f

f

i

t

pF

tFJP

F

tFp

tF

FdtJ

Jp

Fdtpp

Fdtdp

p

p

p

Fdtdp

F

avg

avg

avg

t

t

t

t

t

tfi

t

t

f

i

i

f

i

f

i

f

i

f

i

f

i

D

D=

D==D

=

D=D

=D

=

=D

=-

=

-=D

=

=

=

=

=

=

D=

=

ò

ò

òò

28. Graphical representation of constant and variable force impulse



FRICTION

1. Relative motion:

§ When a force parallel to surface is applied on

object, force tries to move object w.r.t surface.

§ This relative motion is opposed by surface by

exerting a frictional force on the object in a

direction opposite to applied force.

§ Frictional force always acts on the object

parallel to surface on which object is placed.

2. Frictional force

§ If a very gentle force in the horizontal

direction is given to an object at rest on the

table, it does not move.

§ Because of the opposing force exerted by the

surface on the object which resists its motion.

§ This force is called the frictional force

§ It always opposes the relative motion between

an object and the surface where it is placed.

§ If the force applied is increased, the object

moves ater a certain limit.

§ There are two kinds of friction namely

1) Static friction and 2) Kinetic friction.

3. Static Friction (fs)

§ Static frictional force opposes initiation of

motion of an object on the surface.

§ When the object is at rest on the surface, only

downward gravitational force and upward

normal force act on it

§ As the resultant of these two forces is zero

object is at rest

§ If external force Fext is applied the surface

exerts exactly an equal and opposite force to

resist its motion and tries to keep it at rest.

§ If the external force is increased, the surface

cannot provide suicient opposing frictional

force to balance external force.

§ Object starts to slide. This is the maximal

static friction that can be exerted by surface.

§ Magnitude of static frictional force

)sometimes(Onlymg

bodyon surfacebyexertedforceNormalN

contact.in surfacesofnatureon depends

friction.staticofcoeficient

=

=

=m

m££ Nf ss0

4. Value of Force of static friction

i) Object is at rest, no external force is applied

fs = 0

ii) If the object is at rest, and there is an external

force applied parallel to the surface,

Fext = fs , but < μsN.

iii) When object begins to slide,

fsmax = fs

5. Kinetic Friction

§ If the external force acting on object is greater

than fsmax, the objects begin to slide.

§ When an object slides, the surface exerts a

frictional force called kinetic friction, also

called sliding or dynamic friction.

§ To move an object at constant velocity we

must apply a force equal in magnitude and

opposite to the direction of kinetic friction

§ Starting of a motion is more difficult than

maintaining it μk < μs

§ Magnitude of kinetic friction fk = µkN

N = normal contact force

µk = coeffof kinetic friction between surfaces

6. Graphical representation of fs and fk

§ Static friction increases linearly with external

applied force till it reaches the maximum.

§ If the object begins to move kinetic friction is

slightly lesser than maximum static friction.



7. Laws Of Friction

i) If the bodies slip over each other force of

friction

fk = µkN

N = normal contact force

µk = coeffof kinetic friction between surfaces

ii) Direction of kinetic friction on a body is

opposite to velocity of this body with respect

to the body applying the force of friction.

iii) If the bodies do not slip over each other, force

of friction

bodyon surfacebyexertedforceNormalN

friction.staticofcoeficient

=

=m

m£

s

ss Nf

iv) fk or fs does not depend on the area of contact

as long as the normal force N is same.

8. Salient Features of Static and Kinetic Friction

Static friction Kinetic friction

It opposes the starting of

motion

It opposes the relative

motion of the object with

respect to the surface

Independent of surface of

contact

Independent of surface of

contact

ms depends on the nature

of materials mutual

contact

mk depends on nature of

materials and

temperature of surface

Depends on magnitude

of applied force

Independent of applied

force magnitude

It can take values from

zero to μsN

It can never be zero and

always equals to kN

whatever be the speed

fsmax> fk fk

max< fs

μs > μk μk < μs

9. To Move an Object - Push or pull

Case1: A body is pushed at an angle θ

§ Applied force F is resolved into 2 components

F sinθ parallel to the surface

F cosθ perpendicular to surface

§ Total downward force = mg + Fcosθ.

Normal force acting on the body increases.

Since no acceleration along vertical direction

§ Normal force Npush = mg +Fcos θ ..(1)

§ Maximal static friction fsmax = μs Npush

= μs (mg +Fcos θ)

Greater force is to be applied to push object into motion

Case2: A body is pulled at an angle θ

§ Applied force F is resolved into 2 components

F sinθ parallel to the surface

F cosθ perpendicular to surface

§ Total downward force = mg – Fcosθ. ..(2)

Normal force acting on the body increases.

Since no acceleration along vertical direction

§ Normal force Npull = mg – Fcosθ

§ Maximal static friction fsmax = μs Npull

= μs (mg – Fcosθ)

From (1) and (2) Npull < Npush

It is easier to pull an object than push to move it

10. Angle of Friction

Angle between normal force N and resultant force

R of normal force and maximum friction force fsmax

11. Coeff. of static friction and Angle of Friction

frictionofangleoftangentfrictionstaticofcoeff

(2),and(1)From

slidetobeginsobjectwhen

R force,Resultant

s

=

q=m

=m

m=

=q

+=

tan

)2......(

)1......(tan

)(

max

max

max

22max

N

f

Nf

N

f

Nf

ss

ss

s

s

12. Angle of Repose



§ Angle of inclined plane with the horizontal such

that an object placed on it begins to slide

§ Angle b/w inclined plane and horizontal = θ

§ For small θ , object may not slide down.

§ As θ is increased, object begins to slide down.

§ This value is called angle of repose.

13. Angle of repose is same as angle of friction.

§ Gravitational force mg is resolved into

i) parallel component mg sin θ tries to move

the object down

ii) perpendicular component mg cos θ is

balanced by the Normal force N

q=m

q=qm

q=

qm=

m=

=

q=

tan

sincos

)2(sin

)1...(cos

max

max

max

max

s

s

s

ss

ss

ss

mg mg

mgf

mgf

Nf

ff

(2)and(1)From

Also

movesobjectWhen

cosmg N

§ So angle of repose is same as angle of friction

§ But angle of repose refers to inclined surfaces

angle of friction refers to any type of surface.

14. Application of Angle of Repose

i) Antlions make sand traps

§ Angle of inclination of sand trap is made to be

equal to angle of repose

§ If an insect enters edge of the trap, it slides

towards bottom where the antilon hide itself.

ii) Playing on sliding board

§ Sliding is easier when angle of inclination of

board is greater than angle of repose.

§ If inclination angle is much larger than the

angle of repose, slider will reach the bottom at

greater speed and get hurt

15. Rolling Friction

§ In pure rolling, motion of the point of contact

with the surface should be at rest

§ Due to elastic nature of the surface, there will

be deformation on wheel or surface and

minimal friction between wheel and surface.

§ It is called ‘rolling friction’.

§ Rolling friction is smaller than kinetic friction

16. Rolling and kinetic friction

§ When object moves on surface, it slidse on it

§ Wheels move on surface thro’ rolling motion.

§ When a wheel moves on a surface, the point

of contact with surface is always at rest.

§ There is no relative motion between the wheel

and surface. Hence frictional force is very less.

§ If an object moves without a wheel, there is a

relative motion between the object and the

surface. Hece frictional force is larger.

17. Frictional force is necessary in some cases.

§ Walking is possible because of frictional force.

§ Vehicles (bicycle, car) can move because of the

frictional force between the tyre and the road.

§ In the braking system, kinetic friction plays a

major role.

18. Methods to Reduce Friction

i) Applying lubricants

§ Frictional force comes into efect whenever

there is relative motion between two surfaces.

§ In big machines used in industries, relative

motion between diferent parts produce

unwanted heat which reduces its efficiency.

§ To reduce this lubricants are used

ii) Ball bearings

§ If ball bearings are fixed between two

surfaces, during relative motion only rolling

friction comes to efect and not kinetic friction.

§ Rolling friction is smaller than kinetic friction;

§ So machines are protected from wear and tear

19. Friction At Atomic Level

§ Newton and Galileo, considered frictional force as natural forces like gravitational force.

§ Frictional force is electromagnetic force between the atoms on the two surfaces.

§ Even well polished surfaces have irregularities on at the microscopic level



DYNAMICS OF CIRCULAR MOTION

1. Linear and Circular Motion

§ A particle can be in linear motion with or

without any external force.

§ But in circular motion there must be some

force acting on the object.

§ No Newton’s 1st law in circular motion as

without force, circular motion cannot occur

2. Force can change velocity in 3 diff erent ways.

i) Linear Motion

Magnitude of velocity is changed without

changing the direction of the velocity.

Particle will move in the same direction

but with acceleration.

Examples

Particle falling down vertically,

Bike in a straight road with acceleration.

ii) uniform circular motion

Direction of motion alone can be changed

without changing the magnitude (speed).

iii) non uniform circular motion

Both direction and magnitude (speed) of

velocity can be changed.

Example

Oscillation of simple pendulum,

Elliptical motion of planets around Sun.

3. Centripetal force

§ If a particle is in uniform circular motion,

there must be centripetal acceleration r

vacp

2

=

towards the center of the circle.

§ If there is acceleration, some force acting on it

§ This force is called centripetal force.

§ Centripetal force means center seeking force.

§ According to Newton’s second law,

rrmF

rr

mvF

r

mvF

maF

cp

cp

cp

cpcp

ˆ

ˆ

2

2

2

w-=

-=

=

=

®

®

motioncircularuniformFor

notation In vector

forcelCentripeta

§ Centripetal force is not other forces like

gravitational force or spring force.

§ It can be said as ‘force towards center’.

§ Origin of centripetal force can be gravitational

force, tension in the string, frictional force,

Coulomb force etc.

4. Example for centripetal force.

i) Whirling motion of a stone tied to a string

Centripetal force is provided by the tensional

force on the string.

circular motion in amusement park

centripetal force is provided by the tension in

the iron ropes.

ii) In motion of satellites around the Earth,

Centripetal force is given by Earth’s

gravitational force on the satellites.

Newton’s second law for satellite motion

satellitetheofspeed v

satellitetheofmassm

Earthfromsatelliteofdistancer

forceonalsgravitatiEarth’

=

=

=

=

=

F

r

mvF

2

iii) When a car is moving on a circular track

centripetal force is given by the frictional force between the road and the tyres.

when the car moves on a curved track,

Car experiences the centripetal force provided by frictional force between the surface and the tyre of the car



Applying Newton’s second law

cartheofspeed v

cartheofmassm

trackofcurvatureofradiusr

forceFrictional

=

=

=

=

=

F

r

mvF

2

iv) When the planets orbit around the Sun,

they experience centripetal force towards the

center of the Sun. Gravitational force of the

Sun acts as centripetal force

planettheofspeed v

planettheofmassm

Sunfromplanetofdistancer

forceonalsgravitatiSun’

=

=

=

=

=

F

r

mvF

2

5. Vehicle on a leveled circular road

§ Consider a vehicle of mass ‘m’ moving at a

speed ‘v’ in the circular track of radius ‘r’.

§ Forces acting on the vehicle when it moves

i) Gravitational force mg acting downwards

ii) Normal force mg acting upwards

iii) Frictional force Fs acts inwards along road

§ If the road is horizontal, normal force and

gravitational force are equal and opposite.

§ Centripetal force is provided by force of static friction Fs between tyre and surface of road

r

mvFs

2

= ….(1)

§ Static friction is given by

mgF ss m£ ….(2)

i) Condition for safe turn

vg

r

vg

mgr

mv

s

s

s

³m

³m

m£

2

2

Max. velocity with which a car can go round a

level curve without skidding

roadandtyresb/wfriction oftCoefficien

curveofradiusr

s =m

=

m= rgv s

Coeficient of static friction between the tyre

and the surface of the road determines what

maximum speed car can have for safe turn.

ii) Condition for skidding

rg

v

mgr

mv

s

s

2

2

<m

m>

If the static friction is not able to provide

enough centripetal force to turn, the vehicle

will start to skid.

6. Banking of Tracks

§ In circular road, skidding depends on

coefficient of static friction µs

§ µs depends on the nature of the surface

§ To avoid skidding, outer edge of the road is

slightly raised compared to inner edge

§ This is called banking of roads or tracks.

§ This introduces an inclination, and the angle

is called banking angle.

7. Expression for Angle of Banking

§ Angle b/w Road and horizontal surface q

§ Normal force makes same angle with vertical.

§ When car takes turn, two forces acting on car:

Gravitational force mg (downwards)

Normal force N (perpendicular to surface)

§ Normal force is resolved into two components

i) N cos q

balances downward gravitational force mg

N cos q = mg ….(1)

ii) N sinq provides centripetal acceleration



r

mvN

2

sin =q ….(2)

dividing (2) by (1)

q=

=q

´=q

q

tan

tan

1

cos

sin

2

2

rgv

rg

v

mgr

mv

N

N

§ Banking angle q and radius of curvature of

track determines safe speed of car at turning.

§ If the speed exceeds safe speed, it starts to

skid outward but frictional force prevents

outward skidding.

§ If the speed of the car is little lesser than safe

speed, it starts to skid inward and frictional

force prevents inward skidding.

§ If the speed of the vehicle is sufficiently

greater than the correct speed, then frictional

force cannot stop the car from skidding.

8. Centrifugal F orce

§ Circular motion can be analysed by

i) Inertial frame

Newton’s laws are obeyed.

ii) Rotating frame of reference

non-inertial frame as it is accelerating.

§ Equal and opposite reaction to the centripetal

force is called centrifugal reaction, because it

tends to take the body away from the centre

9. Illustration of Centrifugal Force

Whirling motion of a stone tied to a string.

§ Not only the stone is acted upon by a force

(centripetalforce) along the string towards the

centre, but the stone also exerts an equal and

opposite force (centrifugal force) on the hand

§ Stone has angular velocity ω in inertial frame

§ In addition to inward centripetal force −mω2r

there must be an equal and opposite force that

acts on the stone outward with value +mω2r.

§ Total force acting on stone in a rotating frame

= −mω2r +mω2r = 0

§ Outward force +mω2r called centrifugal force

means ‘flee from center’.

10. Centrifugal F orce is a A pseudo force

§ ‘centrifugal force’ act on particle, only when

we analyse motion from a rotating frame.

§ There is only centripetal force given by

tension in the string.

§ For this reason centrifugal force is called as a

‘pseudo force’.

§ A pseudo force has no origin.

§ It arises due to the non inertial nature of the

frame considered.

11. Effects of Centrifugal Force

i) When a car takes a turn in a curved road, person

inside the car feels outward force which pushes the

person away.

§ This outward force is called centrifugal force.

§ If there is sufficient friction between the

person and the seat, it will prevent the person

from moving outwards.

ii) When a car moving in a straight line suddenly

takes a turn, the objects not fixed to the car try to

continue in linear motion due to their inertia of

direction.

§ While observing this motion from an inertial

frame, it appears as a straight line.

§ But, when it is observed from the rotating

frame it appears to move outwards.

iii) A person standing on a rotating platform feels an

outward centrifugal force and is likely to be pushed

away from the platform.



§ Frictional force between platform and person

is not sufficient to overcome outward push.

§ Outer edge of platform is inclined upwards

which exerts normal force to prevent from fall

12. Centrifugal Force due to Rotation of the Earth

§ Any object on the surface of Earth (rotational

frame)

experiences a

centrifugal force.

§ Centrifugal force

appears to act

exactly in

opposite

direction from the axis of rotation.

q=

=

w=

q=

=

w=

Rcosrdistancetriangleanglerightusing By

rotationofaxisfromman ofdistance vertical

Earth on man aon forcelCentrifuga

standing.isman wherelatitude

R EarththeofRadius

Earth of velocityangular

r

rmFc2

13. Features of Centripetal and Centrifugal Forces

Centripetal force Centrifugal force

Real force exerted on

body by external

agencies like

gravitational force, string

tension, normal force etc.

Pseudo or fictitious force

cannot arise from

gravitational force,

tensionforce, normal

force etc.

Acts in both inertial and

non-inertial frames

Acts only in rotating

frames (non-inertial frame

It acts towards axis of

rotation or center of the

circle in circular motion

Outwards from axis of

rotation oroutwards from

center of circular motion

Fcp = mω2r = mv2/r Fcf = mrω2r= mv2/r

Real force and has real

effects

Pseudo force but has real

effects

Origin of centripetal

force is interaction

between two objects.

Origin of centrifugal

force is inertia. It does

not arise from

interaction.

In inertial frames

centripetal force has to be

included when free body

diagrams are drawn.

In inertial frames there is

no centrifugal force. In

rotating frames, centri

petal and centrifugal

force have to be included

WORK, ENERGY AND POWER

1. Definitions

i) work in daily life.

§ Refers to physical as well as mental work.

§ Any activity can generallybe called as work.

ii) Work in Physics

§ A physical quantity with a precise defi nition.

§ Work is said to be done by the force when the force

applied on a body displaces it.

iii) energy

§ To do work, energy is required.

§ Energy is defi ned as the ability to do work.

§ Work and energy are equivalents

§ They have same dimension.

iv) Energy forms

§ mechanical,

§ electrical

§ thermal

§ nuclear

§ Many machines consume one form of energy

and deliver energy in a diff erent form.

v) Power

§ Th e rate of work done is called power.

§ A powerful strike in cricket refers to a hit on

the ball at a fast rate.

2. Expression for Work

q=

=

q=

=

=

®®

®

®

cos

.

FdrW

drFW

dr

F

bodyon forcebydoneWork

ntdisplacemeandforceb/wangle

ntdisplacemeabymovedbody

bodyaon acting Force

3. Units and Dimensions of Work

§ It is a scalar product (or dot product).

§ Thus,work done is a scalar quantity.

§ It has only magnitude and no direction.



§ Unit in SI system : N m (or) joule (J).

§ Dimensional formula : [ML2T-2].

4. Work done by force depends on

i) Force (F)

ii) Displacement (dr)

iii) Angle (θ) between Force and Displacement

5. Diff erent cases of zero work done

i) Force is zero (F = 0).

Body moving on horizontal smooth

frictionless surface will continue to do so as no

force is acting along the plane.

ii) Displacement is zero (dr = 0).

When force is applied on a rigid wall it does

not produce any displacement. Hence, the

work done is zero

iii) Force & displacement are ^r to each θ = 90o

§ When a body moves on a horizontal direction,

the gravitational force (mg) does no work on

since it acts at right angles to the displacement

§ In circular motion centripetal force does not

do work on the object moving on a circle as it

is always ^to the displacement

6. Negative work done by a force

§ In a football game, goalkeeper catches the ball

coming towards him by applying a force

§ Force is applied opposite to that of the motion

of ball till it comes to rest in his hands.

§ During the time of applying the force, he does

a negative work on the ball

7. Angle ( θ) a n d Nature of w ork

Angle (θ) cosθ Work

θ = 0 o 1 Positive,Maximum

0 < θ<90o

(acu te) 0 < cosθ< 1 Positive

θ = 9 0o

(rightangle) 0 Zero

90o<θ<180o -1 < cosθ< 0 Negative

θ = 1 80o -1 Negative,Maximum

8. Work done by a constant force

)(cos

cos

)cos(

if

r

r

r

rfi

rrFW

drFdWrr

ov

drFdW

dr

F

f

i

f

i

-q=

q=þýü

q=

q=

=

=

òòposition finaltoposition

initialfromemtoW.Dtotal




bodyaon acting forceConstant

9. Work done by a variable force

òò q=þýü

q=

q=

=

=

f

i

f

i

r

r

r

rfi

drFdWrr

ov

drFdW

dr

F

cos

)cos(

position finaltoposition

initialfromemtoW.Dtotal




bodyaon acting forceVariable

10. Law of conservation of energy.

§ For an isolated system, Total energy remains

the same in any process irrespective of

whatever internal changes may take place.

§ Energy disappearing in one form reappears in

another form.

11. SI unit of energy

SI unit and Dimension are same as that of work

Other units of energy and SI equivalent values



Unit Equivalent in joule

1 erg (CGS unit) 10-7 J

1 electron volt (eV) 1.6x10-19 J

1 calorie (cal) 4.186 J

1 kilowatt hour (kWh) 3.6x106 J

12. Types of Mechanical energy

i) Kinetic energy

Energy possessed by a body due to its motion

ii) Potential energy

Energy possessed by body by virtue of its position

13. Kinetic energy

§ All moving objects have kinetic energy.

§ A body in motion has the ability to do work.

§ KE is measured by amount of work the body

can perform before it comes to rest.

§ Amount of work done by a moving body

depends on mass and magnitude of velocity.

§ A body which is not in motion does not have

kinetic energy.

2

2

1mv

m

=

=

=

KEEnergy,Kinetic

vvelocity

bodyofmass

§ Example

Hammer kept at rest on a nail does not push

the nail into wood. When it strikes the nail, it

pushes the nail into the wood.

14. Work–Kinetic Energy Theorem

Statement

Work done by the force on the body changes the

kinetic energy of the body.

Proof

§ Consider a body of mass m at rest on a

frictionless horizontal surface.

KE

mumv

ss

uvmF

s

uvmFa

s

uva

asuv

amF

FsW

s

F

D=

=

-=

÷÷ø

öççè

æ -=

÷÷ø

öççè

æ -=

-=

+=

=

=

=

=

W

EnergyKineticin Change

W ,doneWork

W(1)in of valueSub

(2),in of valueSub

motion,ofeqn Kinetic

Law2sNewton'By

forcebydoneWork

ntdisplaceme

forceConstant

nd

22

22

22

22

22

2

1

2

1

2

2

2

2

)2.....(

)1....(

Work-kinetic energy Theorem implies

§ If the work done by the force on the body is

positive then its kinetic energy increases.

§ If the work done by the force on the body is

negative then its kinetic energy decreases.

§ If there is no work done by the force on body

then there is no change in its kinetic energy

body has moved at constant speed provided

its mass remains constant.

15. Relation b/w Momentum and Kinetic Energy

KEmp

m

p

m

pp

vv

mv

m

pv

vp

v

m

.2

2

2

.

.2

1

)2...(.

)1....(

2

2

=

=

=

=

=

=

=

=

=

®

®®

®®

®®

®®

®

momentumofmagnitude

KE

KE(2),and(1)Using

m

2

1KEenergy,kinetic

mmomentumlinear

citywith velomoving

objectofmass

16. Potential Energy

§ Potential energy is associated with its position

and configuration w.r.t surroundings.



§ Because the various forces acting on the body

also depends on position and configuration.

i) Potential energy of at a point P is defined as

Amount of work done by external force in moving

object at constant velocity from point O (initial

location) to point P (final location). At initial point

O potential energy can be taken as zero.

ò®®

=f

irdFU

ii) Types of potential energies.

§ Gravitational potential energy

Energy possessed due to gravitational force.

§ Elastic potential energy

Energy due to spring force and similar forces

§ Electrostatic potential energy.

§ Energy due to electrostatic force on charges

17. Potential energy near the surface of the Earth

§ Consider a body of mass m moved from

ground to height h against gravitational force

§ Gravitational force acting on the body

downwardy verticallactsforcesignve

directionin isforce vectorunit

=-

=

-=®

yj

jmgF g

ˆ

ˆ

§ An external applied force equal in magnitude

but opposite to gravitational force has to be

applied on body to move without acceleration

upwardy verticallisforceappliedsignve =+

=

+=

-=®

®®

mgF

jmgF

FF

a

a

ga

ˆ

§ When body is lifted up velocity is not changed

and kinetic energy remains constant.

§ Gravitational potential energy at height h is

amount of work required to take object from

ground to that height with constant velocity.

ò

ò

q=

=

®®

®®

h

a

f

ia

rdFU

rdFU

0cos||||

§ Displacement and applied force are in same

upward direction, angle between them = 00

mghU

rmg

drmgU

h

h

=

=

= ò0

0

§ Potential energy is stored in the object

through work done by the external force

§ If object falls from a height h, stored potential

energy is converted into kinetic energy.

18. Elastic Potential Energy

§ When a spring is elongated, it develops a

restoring force.

§ Potential energy possessed by a spring due to a

deforming force which stretches or compresses the

spring is termed as elastic potential energy.

§ The work done by the applied force against

the restoring force of the spring is stored as

the elastic potential energy in the spring.

§ Consider a spring-mass system.

§ A mass, m lying on a smooth horizontal table

§ One end of the spring is attached to a rigid

wall and the other end to the mass

§ As long as the spring remains in equilibrium

position, its potential energy is zero.

§ An external force is applied and it is stretched

by a distance in the direction of force

§ A restoring force or spring force developed in

the spring and tries to bring the mass back to

its original position.

§ Applied force and the spring force are equal

in magnitude but opposite in direction

sa FF®®

-=

§ According Hooke’s law,

acementalongdisplissign ve

forceapplied

ntdisplacemetooppositesign ve-

constantforce

forcerestoring

a

a

s

s

F

xkF

F

k

xkF

®

®®

®

®®

=+

+=

=

=

-=

Spring be stretched to a small distance dx.

Work done by force by displacement x is stored as elastic potential energy



ò

ò

ò

q=

q=

=

®®

®®

x

a

x

a

f

ia

dxFU

xdF

rdFU

0

0

cos

cos||||

Applied force and displacement are in same direction

2

0

2

0

2

1

2

kxU

xk

dxkxU

x

x

=

úû

ùêë

é=

= ò

If the initial position is not zero and mass is changed from position xi to xf

)(2

1 22if xxkU -=

§ Potential energy stored in the spring does not

depend on mass attached to the spring.

Force-displacement graph for a spring

§ Restoring spring force and displacement are related as F = – k x, and opposite in direction

§ Graph between F and x is a straight line with dwelling only in second and fourth quadrant

§ Shaded area is the work done by spring force

2

2

1

2

1

2

1

kx

kxx

Area

=

´´=

´= heightbase

Potential energy-displacement graph for spring

§ A compressed or extended spring will transfer

its stored potential energy into kinetic energy

of the mass attached to the spring

§ Energy transferred from kinetic to potential

and potential to kinetic

§ Total energy of the system remains constant.

At the mean position, ΔKE = ΔU

19. Conservative force

§ A force is said to be conservative if

Work done by or against force in moving the

body depends only on initial and final

positions of body and not on nature of path

followed between initial and final positions

§ Object at point A can be taken to another

point B at a height h by three paths

§ Work done against gravitational force is same

as long as initial and final positions are same.

dx

dUFx -=

§ Conservative force is equal to the negative

gradient of the potential energy.

§ Examples

Elastic spring force, electrostatic force,

magnetic force, gravitational force, etc.

20. Non-conservative force

§ A force is said to be non-conservative

if work done by or against the force in moving

a body depends upon the path between initial

and final positions.

§ Work done is different in different paths.

§ Frictional forces are non-conservative forces

Because work done against friction depends

on the length of the path moved by the body.

§ Air resistance, viscous force are non-conservative

Work done by or against these forces depends

upon the velocity of motion.

21. Comparison of conservative and non-

conservative forces



Conservative forces Non-conservative forces

Work done is

independent of the path

Work done depends

upon the path

Work done in a round

trip is zero

Work done in a round

trip is not zero

Total energy remains

constant

Energy is dissipated as

heat energy

Work done is completely

recoverable

Work done is not

completely recoverable.

Force is -ve gradient of

potential energy

No such relation exists.

22. Law of conservation of energy

Energy can neither be created nor destroyed. It may be

transformed from one form to another but the total

energy of an isolated system remains constant.

:kineticpurely

KE0energyTotal

0Uground,touchesObjectiii)

:hatmeasuredasSame

energykinetic

0energypotential

ydistanceatfallsii)Object

:energypotentialpurely

0mgh energyTotal

0(KE)h,atenergykinetic

h heightatstartsObjecti)

+=

=

¹

¹

=

+=

=

=

0

23. Illustration of Law of conservation of energy

§ When an object is thrown up kinetic energy

decreases and potential energy increases

§ When it reaches the highest point its energy is

completely potential.

§ When object falls back from a height kinetic

energy increases potential energy decreases

§ When it touches the ground its energy is

completely kinetic.

§ At the intermediate points the energy is both

kinetic and potential

§ When it reaches the ground kinetic energy is

dissipated into some other form of energy like

sound, heat, light, deformation of the body

§ Energy transformation takes place at every

point.

§ Total mechanical energy(kinetic energy+

potential energy) always remains constant,

i.e total energy is conserved.

24. Motion in a vertical circle

§ A body of mass m attached to one end of a

massless and inextensible string executes

circular motion in a vertical plane Other end

of the string fixed.

§ Length of string = radius r of circular path

§ r makes angle θ with instantaneous velocity in

vertically downward direction

§ Two forces acting on the mass.

Gravitational force acts downward

Tension along the string.

§ Applying Newton’s second law on the mass,

r

mvmgT

ammgT

r

va

dt

dvmmg

ammg

dt

dva

t

r

t

t

2

2

cos

cos

sin

sin

=q-

=q-

=

-=q

=q

-=

onacceleratilcentripeta

:directionradialtheIn

nretardatiotangential

:directiontangentialtheIn

§ Circle is divided into four sections A, B, C, D

§ Consider two positions, lowest point 1 and the

highest point 2



®

®

®

®

®

®

=

=

=

=

=

=

T

T

T

v

v

v

pointotheranyattension

pointhighesttheattension

pointlowesttheatstring thein Tension

pointotheranyat

2pointhighesttheat

point1lowestatbodyofvelocity

2

1

2

1

§ Velocity is tangential at all points

§ Tension at each point acts towards the center.

§ Tensions and velocities found by law of conservation of energy.

i) For the lowest point (1)

§ When body is at lowest point 1, gravitational force acts on the body (vertically downwards)

§ Tension T1 acting vertically upwards,

i.e. towards center

mgr

mvT

r

mvmgT

+=

=-

21

1

21

1

ii) For the highest point (2)

At highest point 2, gravitational force and tension T2 act downwards, i.e. towards center

mgr

mvT

r

mvmgT

-=

=+

22

2

22

2

iii) Diff erence in tension

)1....(2][2

22

121

22

21

21

mgvvr

mTT

mgr

mvmg

r

mvTT

+-=-

+-+=-

iv) Applying law of conservation of energy

mgTT

mggrr

mTT

grvv

mgrvvm

mvmgrmv

mvrgmv

6

24

)2....(4

2][2

1

2

12

2

1

2

1)2(

2

1

21

21

22

21

22

21

22

21

22

21

=-

+=-

=-

=-

+=

+=+

+=+

=

(1),in (2)Sub

m0

2atKE2atPE1atKE1atPE

2pointaatenergyTotal1pointatEnergyTotal

v) Minimum speed at the highest point (2)

§ Body must have a minimum speed at point 2

otherwise, string will slack before reaching

point 2 and body will not loop the circle.

grv

mgr

mv

mgr

mv

=

=

=-

=

2

22

22 0

0TTension i.e 2

To stay in the circular path, te body must have

a speed at point 2, grv =2

vi) Minimum speed at the lowest point 1

§ To have minimum speed grv =2 at point 2,

body must have minimum speed also at 1.

grv

grgrv

grvv

5

4

4

1

21

22

21

=

=-

=-

minimum speed at lowest point 1 should be 5

times more than minimum speed at the highest

point 2, so that body loops without leaving circle

Definitions of Power

§ Power is a measure of how fast or slow a

work is done.

i) Power is deined as

Rate of work done or energy delivered.

t

WP

taken time

donework Power

=

=

ii) Average power

Ratio of total work done to total time taken.

taken timeTotal

donework TotalPav =

iii) Instantaneous power

power delivered at an instant

dt

dWPins =

Unit of Power

§ It is a scalar quantity. Dimension [ML2T–3]

§ SI unit of power is watt (W)

i) One watt



power when 1 joule of work is done in 1 second,

(1 W = 1 J s–1).

ii) Higher units

kilowatt(kW) 1kW = 1000 W= 103 watt

megawatt(MW) 1MW = 106 watt

Gigawatt(GW) 1GW = 109 watt

iii) Horse-power (hp)

For motors, engines and automobiles this old

unit of power is still commercially in use

1 hp = 746 W

iv) Measuring electrical energy

watt second (Ws) leads to handling large

numerical values.

Hence Electrical energy is measured in kWh

J103.6kWh 1

J103.6unitelectrical1

sW103600

s3600W101

kWh 1unitelectrical1

6

6

3

3

´=

´=

´=

´´=

=

Electricity bills are generated in kWh for

electrical energy consumption.

1 unit of electrical energy =1 kWh.

25. Relation between

®®

®®

®®

®®

®®

®®

®®

=

=-

=úû

ùêë

é -

=

=

=

=

ò

òò

òò

òò

òò

vFdt

dW

vFdt

dW

dtvFdt

dW

dtvFdtdt

dW

dtdt

rdFdt

dt

dW

dt

dtrdF

dt

dtdW

dt

rdFdW

..

0..

0..

.

..

...

.

bydivideandmultiply

26. Collisions

§ Collision is phenomenon between two bodies

with or without physical contacts.

§ Linear momentum is conserved in all collision

processes.

§ When two bodies collide, mutual impulsive

forces between them during collision time (Δt)

produces change in their respective momenta.

§ First body exerts a force F12 on second body.

§ From Newton’s third law, second body exerts

a force F21 on the first body.

§ This causes change in momentum p1 and p2

of first body and second body respectively.

0)(

0lim

0

0

21

21

0

21

21

2112

2112

211221

212

121

=+

=D

÷ø

öçè

æ +D

=÷ø

öçè

æ +D

=D+D

-=

D÷ø

öçè

æ+=

D+D=D+D

D=D

D=D

®®

®®

®D

®®

®®

®®

®®

®®®®

®®

®®

dt

ppd

t

pp

pp

pp

FF

tFF

tFtFpp

tFp

tFp

t

lawthirdsNewton’By

i.e total linear momentum is a conserved quantity

Types of collisions

§ In collision total linear momentum and total

energy are always conserved but total kinetic

energy need not be conserved always.

§ Some part of the initial kinetic energy is

transformed to other forms of energy.

§ Because, impact of collisions and deformation

due to collisions produce heat, sound, light

i) Elastic collision

îíì

=þýü

collisionafter

energykineticTotal

collisionbefore

energykineticTotal

ii) Inelastic collision

Q

energyin losscollisionbefore

KETotal-

collision

afterKETotal

collision

afterKETotal

collision

beforeKETotal

D=

=îíì

þýü

îíì

¹þýü

§ Even though kinetic energy is not conserved

but the total energy is conserved.



§ Because total energy contains kinetic energy

term and also a term DQ, which includes all

the losses that take place during collision.

27. Elastic collisions in one dimension

Consider two elastic bodies

Assume that u1 is greater than u2.

A and B suffer a head on collision when they

strike and continue to move along the same straight line with velocities v1 and v2

From law of conservation of linear momentum,

...(3) v(v-uu

u v vu(2)by(1)Divide

)u-)(vu(vm v-(u v(um

)...(2)u-(vmv(um

um-vmvmum

vm2

1vm

2

1um

2

1um

2

1

collisionafterKETotal=collision beforeKETotal

:conservedalsoisbodiesofKE

)...(1)u-(vm v-(um

um-vmvm-um

vm+vm=um+um

collisionafter

momentumTotal=

collisionbefore

momentumTotal

2121

2211

2222211111

22

222

211

222

2221

211

2221

222

211

222111

22221111

22112211

)

))

)

)

21

21

21

-=-

+=+

+=+

=-

=-

+=+

=

=

îíì

þýü

For elastic head on collision, relative speed of two elastic bodies ater collision has the same magnitude as before collision but in opposite direction

21

112

21

122

21

221

21

211

22121121

2212111211

221222121111

21212111

2

1212

1221

mm

um2u

mm

m-mvSimilarly,

mm

um2u

mm

m-mv

um2u)m-m)vm(m

um2um-umvm vm

um-umumvm vm-um

)u-uu(vm v-(um

(1)in ng vSubstituti

uuvvOr

uu vv(3),From

++úû

ùêë

é+

=

++úû

ùêë

é+

=

+=+

+=+

-+=

-+=

-+=

-+=

(

)

Special cases

Case i : If masses of colliding bodies are equal

m1 = m2 v1 = u2 v2 = u1

After head on elastic collision, the velocities of colliding bodies are mutually interchanged.

Case ii : If the particle B is initially at rest,

u2 = 0 v1 = u2 v2 = 0

2nd body moves with initial velocity of 1stbody

Case iii :

1st body is much lighter than second body

0=

+

+

úúúú

û

ù

êêêê

ë

é

+

=

-=

+

+

úúúú

û

ù

êêêê

ë

é

+

=

»<<<

2

2

1

2

1

2

2

1

2

1

2

11

2

1

21

2

1

2

1

1

2

2

1

2

11

v

1m

m

m

m2

u

1m

m

m

m-1

vSimilarly,

uv

1m

m

u2u

1m

m

1-m

m

v

mbyDrandNrDividing

0m

m,1

m

mm2,m

1st body in the opposite direction with same initial velocity. 2nd body remains at rest

Case iv:

2nd body is much lighter than the 1st body

22

11

1

1

2

1

212

2uvSimilarly,

uv

mbyDrandNrDividing

0m

m,1

m

m,mm

=

=

»<<<

1st body continues to move with same initial velocity. 2nd body will move with twice initial velocity of 1st

28. Comparison of elastic and inelastic collisions



Elastic Collision Inelastic Collision

Total momentum is

conserved

Total momentum is

conserved

Total kinetic energy is

conserved

Total kinetic energy is

not conserved

Forces involved are

conservative forces

Forces involved are non-

conservative forces

Mechanical energy is not

dissipated.

Mechanical energy is

dissipated into heat,

light, sound etc.

29. Perfect inelastic collision

§ If two colliding bodies stick together after

collision are known as completely inelastic

collision or perfectly inelastic collision.

§ When Bubblegum is thrown on a moving

vehicle, it sticks to vehicle

§ They move together with same velocity.

§ Objects stick together permanently such that

they move with common velocity.

§ Let two bodies with masses m1, m2 move with

initial velocities u1, u2 before collision.

§ After perfect inelastic collision both the objects

move together with a common velocity v

§ Linear momentum is conserved in collisions

21

2211

212211

mm

umum vocitycommon vel

v)m(mumum

+

+=

+=+

30. Loss of kinetic energy collision

§ In perfectly inelastic collision, loss in kinetic

energy during collision is transformed to

another form of energy like sound, thermal,

heat, light etc.

2

1

mm

umum

2

1

2

1

KEKEQ

energykineticofloss

mm

umumv

ocity,common velwhere

K

collision.afterenergykinetictotal2

1

2

1KE

collision beforeenergykineticTotal

21

2211

if

21

2211

f

i

221

21

21

2

21222

211

221

222

211

)(

2

1

2

1

uumm

mm

mmumum

vmm

umum

-÷ø

öçè

æ+

=

úû

ùêë

é+

++-+=

-=D

+

+=

+=

+=

Coefficient Restitution (e)

§ Suppose we drop a rubber ball and a plastic

ball on the same floor.

§ Rubber ball will bounce back higher than

plastic ball.

§ Because loss of kinetic energy for an elastic

ball is much lesser than that of plastic ball.

§ Amount of kinetic energy after collision of

two bodies is measured by dimensionless

number called coefficient of restitution (COR).

§ It is defined as the ratio of velocity of separation

(relative velocity) after collision to the velocity of

approach (relative velocity) before collision

21

12

uu

vve

COR

-

-=

=collisionbeforeapproach ofvelocity

collisionafterseparationofvelocity

§ For an elastic collision

1

1221

=

-=-

=

e

vvuu

approachof velocityseparationofvelocity

§ There is no loss of KE after collision. So, body

bounces back with same kinetic energy

§ In any real collision, there will be some losses

in kinetic energy due to collision

§ Coefficient of restitution lies between 0<e <1



CENTER OF MASS1. Rigid body

§ A Rigid body Maintains its definite and fi xed

shape even when an external force acts on it.

§ Interatomic distances do not change in a rigid

body when an external force is applied.

§ In real life situation, bodies are not rigid,

because the shape and size of body change

when forces act on them

§ When a rigid body moves, all particles that

constitute body need not take the same path.

§ Depending on type of motion, different

particles of the body may take different paths.

§ When a wheel rolls on a surface, path of

center point of the wheel and the paths of

other points of the wheel are different.

2. Center of Mass (COM)

§ When a bulk object is thrown at an angle in

air only one point takes parabolic path and all

the other points take different paths.

§ The one point that takes the parabolic path is

called center of mass (COM) of the body.

§ Its motion is like the motion of a single point

§ center of mass of a body is defined as a point where

entire mass of the body appears to be concentrated.

§ This point can represent the entire body

§ For regular shape , uniform mass distribution,

COM is at geometric center of body.

i) Circle and sphere, COM is at centers

ii) Square& rectangle, point diagonals meet

iii) Cube&cuboid, point of body diagonals meet

§ For other bodies, COM determined using

some methods.

§ Center of mass could be well within the body

and in some cases outside the body as well.

3. Center of Mass for Distributed Point Masses

§ A point mass is a hypothetical point particle which

has nonzero mass and no size or shape.

§ To find COM for collection of n point masses

m1, m2, m3 . . . mn we have to first choose an

origin and an appropriate coordinate system

§ Let, x1, x2, x3 . . . xn be X-coordinates of the

positions of these point masses in the X

direction from the origin.

M

rmr

kjixr

kjir

M

zmZ

M

ymY

M

xmX

m

m

xmX

iiCM

i

CM

ii

CM

ii

CM

ii

CM

i

i

ii

CM

å

å

å

åå

åå

®

®

®

®

=

++=

++=

=

=

=

=

=

=

ˆˆˆ

ˆˆˆ

iii

CMCMCM

CMCMCM

z ymassesofP.V

z yxCOMofP.V

massespointofCOMofposition )z, y,(x

Similarly

particles,allofmassMtotal

4. Center of Mass of Two Point Masses

§ Let two point masses be m1 and m2,

§ They are at positions x1 and x2 on the X-axis.

i) Masses are on positive X-axis:

§ Origin is taken arbitrarily so that m1 and m2

are at positions x1 and x2 on positive X-axis



§ COM will also be on positive X-axis at xCM

21

2211

mm

xmxm

+

+=CMx

ii) Origin coincides with any one of masses:

Let origin coincides with mass m1,

its position x1 is zero, (i.e. x1 = 0).

21

22

21

221 )0(

mm

xm

mm

xmm

+=

+

+=

CM

CM

x

x

iii) Origin coincides with center of mass itself:

If origin of coincide with center of mass,

xCM = 0

Mass m1 is on the negative X-axis.

Its position x1 is negative, (i.e. -x1).

moments)ofprinciple

0

(

)(

)(

2211

2211

21

2211

xmxm

xmxm

mm

xmxm

=

+-=

+

+-=

5. Center of mass for uniform distribution of mass

§ If mass is uniformly distributed in a bulk

object, a small mass (Δm) of the body can be

treated as a point mass

§ Summations is done to obtain expressions for

the coordinates of center of mass.

i

iiCM

i

iiCM

i

iiCM

m

zmZ

m

ymY

m

xmX

SD

SD=

SD

SD=

SD

SD=

§ If small mass taken is infinitesimally small

(dm), summations is replaced by integrations

dm

zdmZ

dm

ydmY

dm

xdmX

CM

CM

CM

ò

ò=

ò

ò=

ò

ò=

6. Center of mass of a uniform rod

2

2

1

1

,

0

2

0

0

lx

x

l

xdxl

M

dxl

Mx

x

dxl

M

l

M

x

l

CM

l

l

l

CM

=

úû

ùêë

é=

=

÷ø

öçè

æ

=

ò

ò=

=

=l

=

=

=

=þýü

ò

ò

dm

xdmmassofcenter

(dm)elementsmallofmass

densitymasslinear

dxorigin fromdistanceatlength

dmmasssmallmallyinfinitesi

length

Moriginwith coincidesendone

roduniformofmass

7. Motion of Center of Mass

When a rigid body moves, its center of mass will

also move along with the body.

i

iiCM

ii

iCM

i

iiCM

ii

iCM

i

iiCM

m

ama

dt

vm

mv

dt

d

t

m

vmv

dt

xdm

mx

dt

d

t

m

xmx

S

S=

÷ø

öçè

æS

S=÷

ø

öçè

æ

S

S=

÷ø

öçè

æS

S=÷

ø

öçè

æ

S

S=

®®

®

®®

®

®®

r

r

1

1

w.r.t.diff.Again

velocity

w.r.t.diff.



SYSTEM OF PARTICLES AND RIGID BODIES

1. Torque

§ When a net force acts on a body, it produces

linear motion in the direction of applied force.

§ If the body is fi xed to a point or an axis, such

a force rotates the body depending on the

point of application of the force on the body.

§ Ability of force to produce rotational motion

in a body is called torque or moment of force.

§ Example

Opening and closing of a door about hinges

and turning of a nut using a wrench.

§ Extent of the rotation depends on magnitude

of the force, its direction and distance between

the fi xed point and the point of application.

§ When torque produces rotational motion in a

body, angular momentum changes w.r.t time.

2. Definition of torque

§ moment of external applied force about a point or axis of rotation.

nrF

Fr

Fr

Fr

ˆ)sin( q=t

´=t

=q

=

®

®®®

®®

®®

Torque

andbetweenangle

bodyon actforcewherepointofP.V

§ Its unit is N m.

§ Torque is called as a pseudo vector as it needs

the other two vectors ®®

Fr and for its existence.

3. Direction and Magnitude

§ Right hand rule.

If fingers of right hand are kept along p.v with

palm facing direction of force when fi ngers are

curled thumb points to direction of torque.

§ Direction of torque helps us to find type of

rotation caused by the torque.

§ If direction of torque is out of paper, rotation

produced by the torque is anticlockwise.

§ If the direction of the torque is into paper,

then the rotation is clockwise

§ Direction and magnitude found separately.

§ Magnitude t = r F sinθ

§ For direction, vector /right hand rule is used

4. Value of torque based on angle between r and F

i) When θ = 90o , sin 90o = 1, tmax = rF.

Torque is maximum when, r ^r F

ii) If θ = 0o , sin 0o = 0, t = 0

Torque is 0 when r and F are parallel

iii) θ = 180o, sin 180o = 0., t= 0.

Torque is 0 when r and F are antiparallel.

iv) If r= 0, t= 0

Torque is zero if force acts at reference point.



5. Torque about an axis.

§ Consider a rigid body rotating about axis AB.

§ Force F act at a point P on the rigid body.

§ The force F may not be on the plane ABP.

§ We can take the origin O at any random point

on the axis AB.

§ Torque about AB is the parallel component of

torque along AB

q´=t®®

cosFr

§ Torque perpendicular to axis AB

q´=t®®

sinFr

§ Torque about axis will rotate object about it

§ Torque perpendicular to the axis will turn the

axis of rotation.

§ When both exist simultaneously on a rigid

body, the body will have a precession.

§ Example : Precessional motion in a spinning

top when it is about to come to rest

§ Torque of force about an axis is independent

of the choice of the origin as long as it is

chosen on that axis itself.

6. Torque and Angular Acceleration

§ Consider a rigid body rotating about a fixed

axis.

§ A point mass m in the body will execute a

circular motion about a fixed axis.

§ Tangential force F acting on point mass

produces necessary torque for this rotation.

§ This force F is perpendicular to position

vector r of the point mass.

§ Torque produced by force on point mass m

about the axis

®®

®®

a=

=

a=

a=

a=

=

=

=

=

Iτ

mr

mrτ

mrτ

.mr

rαm r

ma .. rτ

r F

r Fτ

Thus

Iinertiaofmoment

notationvector

torque

2

2

2

2

)(

)(

90sin

§ Torque of force acting on point mass produces

angular acceleration about axis of rotation.

§ Directions of t and a are along axis of rotation

§ If direction t of is in the direction of a it

produces angular acceleration.

§ If t is opposite to a , angular deceleration or

retardation is produced on the point mass.

7. Angular Momentum

§ Angular momentum in rotational motion is

equivalent to linear momentum in

translational motion.

§ Angular momentum of a point mass is

defined as moment of its linear momentum.

§ Angular momentum L of a point mass having

a linear momentum p at a position r w.r.t a

point or axis is

p.torposition ofcomponent

rtopmomentumofcomponentwhere,

))(

))(

andbetween anglewhere,

magnitude

^^

^^

^==

^==

=q

=

´=

®®

®®®

r

p

prpθ rL

pr θp rL

pr

θ r p L

prL

(sin

(sin

sin



§ Angular momentum L = 0 , if linear

momentum p = 0

§ If particle is at origin r = 0 or if r and p are

parallel or antiparallel (θ = 00 or 1800)

8. Angular Momentum and Angular Velocity

§ Let a rigid body rotating about fixed axis

§ Point mass m in body will execute circular

motion about the fixed axis point mass m is at

a distance r from axis of rotation.

§ Its linear momentum at any instant is

tangential to the circular path.

§ Angular momentum L is perpendicular to r

and p.

§ Hence, it is directed along the axis of rotation.

§ Angle θ between r and p is 90o

®®

®®

w=

=

w=

w=

w=

=

=

IL

mr

mrL

r m

. rm rτ

v r m

vm rL

Thus

Iinertiaofmoment

notationvector

ofMagnitude

2

2

2

0

)(

)(

90sin

Directions of L, ω are along axis of rotation.

9. Torque and Angular Momentum

dt

dL

dt

Iωd

I

I

dt

dω

=t

=

a=t

w=

=a

w=

)(

Torque

Lmomentumangular

onacceleratiangular

velocityangular

§ This is the Newton’s second law in rotational

motion as it is in the form dt

dpF =

§ An external torque on a rigid body fixed to an

axis produces rate of change of angular

momentum in the body about that axis.

10. Law of Conservation of angular momentum:

In the absence of external torque, the angular

momentum of the rigid body or system of

particles is conserved.

constantIf ===t Ldt

dL,0,0

11. Translational equilibrium.

§ When a body is at rest without any motion on

a table, there is gravitational force acting on

the body downward and also the normal force

exerted by table on the body upward.

§ These two forces cancel each other and thus

there is no net force acting on the body.

§ When linear momentum remains constant,

the net force acting on the body is zero.

0=®

netF

Vector sum of diferent forces acting in

diferent directions on the body is zero.

0.....321 =++®®®®

nFFFF

12. Horizontal and vertical equilibria

§ If 321

®®®

++ FFF act in diferent directions on the

body, we can resolve them into horizontal and

vertical components and take resultant in the

respective directions.

13. Equilibrium of Rigid Bodies

§ When angular momentum remains constant,

net torque acting on the body is zero.

0=t®

net

§ Under this condition, the body is said to be in

rotational equilibrium.

§ Vector sum of diferent torques producing

diferent senses of rotation on the body is zero.

0.....321 =tt+t+t®®®®

n

§ Aigid body is in mechanical equilibrium when

net force and net torque acts on body is zero.



0=®

netF and 0=t®

net

§ A rigid body is in mechanical equilibrium when

both its linear momentum and angular momentum

remain constant.

14. Types of Equilibrium and their Conditions.

i) Translational equilibrium

§ Linear momentum is constant.

§ Net force is zero.

ii) Rotational equilibrium

§ Angular momentum is constant.

§ Net torque is zero.

iii) Static equilibrium

§ Linear momentum, angular momentum zero.

§ Net force and net torque are zero.

iv) Dynamic equilibrium

§ Linear and angular momentum are constant.

§ Net force and net torque are zero.

v) Stable equilibrium

§ Linear and angular momentum are zero.

§ Body tries to come back to equilibrium if

slightly disturbed and released.

§ Center of mass shits higher if disturbed from

equilibrium.

§ Potential energy is minimum and it increases

if disturbed.

vi) Unstable equilibrium


§ Body cannot come back to equilibrium if


§ Center of mass of the body shits slightly lower

if disturbed from equilibrium.

§ Potential energy of the body is not minimum

and it decreases if disturbed.

vii)Neutral equilibrium


§ Body remains at the same equilibrium if


§ Center of mass of the body does not shit

higher or lower if disturbed from equilibrium.

§ Potential energy remains same if disturbed.

15. Couple

§ A pair of forces equal in magnitude but opposite in

direction and separated by a perpendicular

distance so that their lines of action do not coincide

that causes a turning ef ect is called a couple

§ Consider a thin uniform rod AB.

§ Its center of mass is at its midpoint C.

§ Two forces equal in magnitude and opposite

in direction applied at ends A and B of rod

perpendicular to it.

§ Two forces are separated by a distance of 2r

§ As the two equal forces are opposite in

direction, they cancel each other and the net

force acting on the rod is zero.

§ Now the rod is in translational equilibrium.

§ But, the rod is not in rotational equilibrium.

§ Moment of force at A taken w.r.t center point

C, produces an anticlockwise rotation.

§ Moment of the force applied at the end B also

produces an anticlockwise rotation.

§ Moments of both the forcescause the same

sense of rotation in the rod.

§ Rod undergoes a rotational motion or turning

16. Turning ef ect of Couple

17. Principle of Moments

§ Light rod is pivoted at a point along its length.

§ Two parallel forces F1 and F2 act at two ends

at distances d1 and d2 from point of pivot



§ Normal reaction force N at the point of pivot

§ If the rod has to remain stationary in

horizontal position, it should be in

translational and rotational equilibrium.

§ Then, net force and net torque must be zero

fulcrumpointpivoted

leverofadvantagemechanical

effort

Fload

:momentsofprinciple

zero,betotorquenetFor

zero,betoforcenetFor

1

=

=

=

=

=

=

=-

+=

=-+-

1

2

2

1

2

2

1

2211

2211

21

21

0

0

d

d

F

d

d

F

F

FdFd

FdFd

FFN

FN F

§ when, d1< d2, F1> F2

Many simple machines work on Mechanical

Advantage principle.

§ when, d1 = d2; F1 = F2

This forms principle for beam balance used

for weighing goods with the condition

18. Center of Gravity

§ Center of gravity of a body is the point at which the

entire weight of the body acts irrespective of the

position and orientation of the body.

§ A Rigid body is made up of point masses.

§ Point masses experience gravitational force

towards the center of Earth.

§ As the size of Earth is very large compared to

rigid body, these forces appear to be acting

parallelly downwards

§ Resultant of these parallel forces always acts

through a point.

§ This point is center of gravity of body

§ Center of gravity and center of mass of a rigid

body coincide when the gravitational field is

uniform across the body

19. Determination of center of gravity

i) CG of plane lamina by pivoting

§ Uniform lamina is pivoted at various points

by trial and error.

§ Lamina remains horizontal when pivoted at

the point where net gravitational force acts,

which is the center of gravity.

§ When a body is supported at the center of

gravity, the sum of torques acting on all the

point masses of the rigid body becomes zero.

§ Moreover the weight is compensated by the

normal reaction force exerted by the pivot.

§ he body is in static equilibrium and hence it

remains horizontal

ii) CG of plane lamina by suspending

§ Lamina is suspended from diferent points P,

Q, R vertical lines PP', QQ', RR' all pass

through the center of gravity.

§ Reaction force at the point of suspension and

gravitational force acting at the center of

gravity cancel each other

§ Torques caused by them cancel each other.

20. Bending of Cyclist in Curves



§ Consider a cyclist negotiating a not banked

circular level road of radius r with a speed v.

§ Cycle and cyclist are one system with mass m.

§ Cnter gravity of the system CG goes in a

circle of radius r with center O

§ Choose OC as X-axis, vertical line through O

as Z-axis system as rotating frameabout Zaxis.

§ Forces acting on the system are,

i) gravitational force (mg),

ii) normal force (N)

iii) frictional force (f)

iv) centrifugal force r

mv2

§ As the system is in equilibrium, net external

force and net external torque must be zero.

§ Consider all torques about the point A

÷÷ø

öççè

æ=q

=q

q=q

=q+q-

q=

q=D

=+-

=t

=þýü

=þýü

-

®

rg

v

rg

v

r

mvmg

r

mvmg

BCr

mvABmg

BCr

mv

ABmg

net

21

2

2

2

2

2

tan

tan

0

0

0

cosACsinAC

cosACsinAC

cosACBC

sinACABABC,From

m,equilibriurotationalFor

turniseanticlockwcauses

forcelcentripetatodueTorque

turnclockwisecausesforce

nalgravitatiotodueTorque

§ Cyclist has to bend by angle θ from vertical to

stay in equilibrium (i.e. to avoid a fall).

21. Moment Of Inertia

§ For point mass mi at distance ri from fixed axis,

rmIobjectbulk forMI

rmImasspointforMI

2ii

2ii

å=

=

§ In translational motion, mass is a measure of

inertia; For rotationalmotion, moment of

inertia is a measure of rotational inertia.

§ Unit of moment of inertia is, kg m2

dimension is M L2

§ Moment of inertia is not invariable quantity.

§ It depends not only on mass of the body, but

also on the way the mass is distributed

around the axis of rotation.

22. MI of a uniformly distributed mass

§ Consider an ininitesimally small mass (dm) as

a point mass

§ Take its position (r) with respect to an axis.

dmr

rdm)dIobjectbulk entireofMI

rdm)dImasspointofMI

2

2

2

ò=

ò=ò

=

I

(

(

23. Moment of Inertia of a Uniform Rod

§ Consider uniform rod of mass M , length A

§ An origin is to be ixed for coordinate system

to coincides with center of mass, also

geometric center of the rod.

§ Rod is now along the x axis.

§ Take an ininitesimally small mass (dm) at a

distance (x) from the origin.



2

3

2

0

3

2/

2/

2

2

2

2

12

1

24

2

3

2

)(

)(

MlI

l

l

M

x

l

M

dxxl

MI

xdxl

M

xdm

dxl

Mdm

dxλ

l

M

xdmdI

l

l

l

=

´=

úû

ùêë

é=

=

÷ø

öçè

æ=

ò=

=

=

=l

=

ò

ò

-

:sideeitherddistributemass

IrodentireofMI

masssmallmallyinfinitesi

rodofh mass/lengt

inertiaofMoment

24. Moment of Inertia of a Uniform Ring

§ Consider a uniform ring of mass M, radius R.

§ Take infinitesimally small mass (dm) of length

(dx) of the ring.

§ (dm) is located at a distance R, from the axis

2

20

2

0

2

2

2

2

2

2

2

)(

2

2

)(

MRI

RMR

xMR

dxMR

I

RdxR

M

Rdm

dxR

Mdm

R

M

RdmdI

r

R

=

p2´p

=

p=

p=

÷ø

öçè

æp

=

ò=

p=

p=l

=

p

p

ò

ò:uniformlyddistributemass

IringentireofMI


ringofh mass/lengt

inertiaofMoment

25. Moment of Inertia of a Uniform Disc

Consider a disc of mass M and radius R.

Disc is made up of infinitesimally small rings.

Consider a ring of mass (dm), thickness (dr)

and radius (r).

2

4

2

0

4

2

0

3

2

2

2

2

2

2

2

2

1

4

2

4

2

2

2

)(

2

2

)(

MRI

R

R

M

r

R

M

drrR

MI

rdrrR

M

rdmisc

drrR

M

drrdm

R

M

rdmdI

R

R

=

´=

úû

ùêë

é=

=

÷ø

öçè

æ ´=

ò=

p´p

=

ps=

p=s

=

ò

ò:uniformlyddistributemass

IdentireofMI


mass/area

inertiaofMoment

26. General expression for moment of inertia

§ Expression for moment of inertia must take

care of not only mass, shape, size of objects,

but also its orientation to the axis of rotation.

§ It is applicable for objects of irregular shape

and non-uniform distribution of mass.

I = MK2

where, M = total mass of the object and

K = radius of gyration.

27. Radius of gyration

§ Perpendicular distance from the axis of rotation to

an equivalent point mass, which would have the

same mass as well as the same moment of inertia of

the object.

§ Its unit is m.

§ Its dimension is L.



§ A rotating rigid body with respect to any axis,

is made up of point masses m1, m2, m3, . .

.mn at perpendicular distances r1, r2, r3 . . . rn

n

rrrrK

MKI

n

rrrrmn

rrrrm

rmrmrmrmI

mmmm

I

n

n

n

n

n

ii

223

22

21

2

223

22

21

223

22

21

223

22

21

21

.....

).....

(

).....(

.....

....

++++=

=

=

=

++++=

++++=

++++=

===

+++=

S=

gyrationofradiusK

bodytheofMmasstotalnm

Take

rm......rmrm

rminertiaofMoment

2nn

222

211

2

Radius of gyration is the root mean square (rms)

distance of particles of body from theaxis of rotation

28. Expression of MI in I = MK2. form

i) Radius of gyration of uniform rod of mass M , length l perpendicular to centre of mass

lK

l

lK

lK

MlMK

MKI

MlI

(0.289)

gyrationofradiusFor

=

=

=

=

=

=

=

6

731.1

32

1

12

1

12

1

12

1

22

22

2

2

ii) Radius of gyration of a disc of mass M, radius

R rotating about axis through center of mass

perpendicular to the plane of the disc.

RRK

MRK

MRMK

(0.707)==

=

=

2

2

2

12

1

22

22

29. Parallel axis theorem:

MI of a body about any axis is equal to sum of its MI

about a parallel axis through its center of mass and the

product of the mass of the body and the square of the

perpendicular distance between the two axes.

2MdII

dI

C +=

=

=

itfromdistanceaataxisparallelaaboutMI

massofcenter

through axisan aboutMmassofbodyofMII C

2

2

2

22

22

2

2

)(

2

MdII

mM

dmII

xI

dmdx

dxm

dxm

dxm

d

C

c

C

+=

S=

S+=

=Sþýü+

S=

S+S+S=

++S=

+S=

+=

=

=

=

=þýü

^

=þýü

massEntire

0xmABaxisw.r.tvalues

ve-andivetakecan x

mmassofcenteraboutMI

xmm

2xd)(

)(IDE,aboutbodywholeofMI

.)(DEaxisaboutmasspointofMI

xCOMitsfromposition

mbodyon masspoint

I.DEaboutbodyofMI

DEABfromdistancer

aatABtoparallelAxis

IC.massofcenter

thro'ABaxisan aboutMI

30. Perpendicular axis theorem:

MI of a plane laminar body about an axis

perpendicular to its plane is equal to the sum of

moments of inertia about two perpendicular axes lying

in the plane of the body such that all the three axes are

mutually perpendicular and have a common point.

X and Y-axes lie in plane

Z-axis ^ to plane of laminar object

I of body about X, Y-axes = IX, IY

MI about Z-axis = IZ



IZ =IX +IY

§ Consider a plane laminar object of negligible

thickness on which lies the origin (O).

§ X and Y-axes lie on the plane and Z-axis is

perpendicular to it

§ lamina is considered to be made up of a large

number of particles of mass m.

§ Choose one such particle at a point P which

has coordinates (x, y) at a distance r from O.

yxz III

y

y

+=

å+å=

+å=

å=

=

2

2

2

)

mmx

m(x

mrIaxis,ZaboutlaminaentireofMI

mraxisZaboutparticleofMI

2

2

2Z

31. Effect of Torque on Rigid Bodies

A rigid body which has non zero external torque

(τ) about the axis of rotation would have an

angular acceleration (α) about that axis.

Relation between torque and angular acceleration

τ = Iα

I = moment of inertia of rigid body.

Torque in rotational motion is equivalent to force

in linear motion

32. Conservation of Angular Momentum

When no external torque acts on body, the net angular

momentum of a rotating rigid body remains constant.

momentumangularnalfimomentumangularinitial

I

:FormAnother

constantL0,τIf

dt

dLτ

i

=

w=w

==

=

iii I

If I increases w will decrease and vice-versa to

keep the angular momentum constant.

33. Situations where principle of conservation of

angular momentum is applicable.

i) ice dancer

§ Dancer spins slowly when hands are stretched

out and spins faster when hands are brought

close to the body.

§ Stretching of hands away from body increases

moment of inertia, thus angular velocity

decreases resulting in slower spin.

§ When hands are brought close to body,

moment of inertia decreases, and thus angular

velocity increases resulting in faster spin.

ii) Conservation of angular momentum of diver

A diver while in air curls body close to

decrease the moment of inertia, in turn helps

to increase number of somersaults in air.

34. Work done by Torque

Consider a rigid body rotating about a fixed axis.

A point P on body rotating about an axis

perpendicular to the plane of the page.

A tangential force F is applied on the body.

It produces a small displacement ds on the body.

Work done by force dw = Fds

Distance ds, angle of rotation dθ , radius r relation

ds = r dθ

Work done dw = F ds



dw = F r dθ

Fr = τ , torque produced on body

dw = τ dθ

Corresponding expression in translational motion

dw = Fds

35. Kinetic Energy in Rotation

§ Consider a rigid body rotating with angular

velocity ω about an axis.

§ Every particle of body will have same angular

velocity ω , different tangential velocities v

based on its positions from axis of rotation.

§ Choose a particle of mass mi situated at

distance ri from axis of rotation.

2

2

2

22

22

2

2

2

1

2

1

2

1

)(2

1

2

1

Mv

IKE

rm

rm

rmKE

rm

vm

rv

ii

ii

iii

iii

ii

ii

2

1motionnaltranslatioin KE

I.body,wholeofMI

bodywholeofenergykinetic

KEenergyKinetic

velocity,Tangential

i

=

w=

S=

wS=

w=

w=

=

w=

36. Rotational kinetic energy - Angular momentum

I

LKE

I

I

I

II

I

2

)(

2

1

2

1

2

1

2

2

2

2

=

w=

´w=

w=

w=

w=

=

KEenergykineticRotational

ILmomentumAngular

velocityangular

IbodyrigidofMI

37. Power Delivered by Torque

Power delivered is the work done per unit time.

®®

=

tw=

qt=

=

v.FPmotionnaltranslatioIn

dt

dwPpowerousInstantane

P

dt

d

38. Comparison of Translational- Rotational

Quantities

39. Rolling Motion

§ Rolling motion is common in daily life.

§ Motion of wheel is example of rolling motion.

§ Round objects like ring, disc, sphere etc. are

most suitable for rolling .

§ Consider a point P on the edge of the disc.

§ While rolling, point undergoes translational

motion along with its center of mass and

rotational motion w.r.t its center of mass.

40. Combination of Translation andRotation

§ Radius of rolling object is R, in one full

rotation, center of mass is displaced by 2pR

§ All points on disc are also displaced by same

2pR aft er one full rotation.

§ Center of mass takes a straight path; but, all

other points undergo a path of combination of

translational and rotational motion.



§ Point on edge undergoes a path of a cycloid

§ Velocity of center of mass vCM is only

translational velocity vTRANS (vCM = vTRANS).

§ All the other points have two velocities.

Translational velocity vTRANS,

Rotational velocity vROT (vROT = r w).

r = distance of the point from COM

w = angular velocity.

§ Rotational velocity vROT is perpendicular to

instantaneous p.v from the center of mass

§ The resultant of these twovelocities is v.

§ Resultant velocity v is perpendicular to p.v

from the point of contact of the rolling object

with the surface on which it is rolling as

§ In pure rolling, point of rolling object comes in

contact with surface is at momentary rest.

§ This is the case with every point that is on the

edge of the rolling object.

§ As the rolling proceeds, all the points on edge,

one by one come in contact with the surface;

remain at momentary rest at the time of

contact and then take the path of the cycloid

as already mentioned.

§ We can consider pure rolling in two ways.

i) combination of translational motion and

rotational motion about the center of mass.

ii) Momentary rotational motion about point of

contact.

§ As the point of contact is at momentary rest in

pure rolling, its resultant velocity zero (v=0).

§ At the point of contact, vTRANS is forward (to

right) and vROT is backwards (to the left).

§ vTRANS and vROT are equal in magnitude and

opposite in direction

v = vTRANS– vROT =0).

§ In pure rolling, for all points on the edge,

magnitudes of vTRANS and vROT are equal

(vTRANS=vROT).

§ As vTRANS = vCM and vROT= Rw, in pure rolling

vCM = R w

§ In rotational motion, center point will not

have any velocity as r is zero.

§ But in rolling motion, center point has a

velocity vCM

§ For topmost point, the two velocities vTRANS

and vROT are equal in magnitude and in the

same direction

§ Resultant velocity v is the sum of these two

velocities, v = vTRANS+vROT.

v = 2 vCM

41. Sliding

§ Sliding when vCM > Rw or vTRANS > vROT

§ The translation is more than the rotation.

§ This kind of motion happens when sudden

break is applied in a moving vehicles, or when

the vehicle enters into a slippery road.



§ Point of contact has more of vTRANS than vROT.

§ Resultant velocity v in the forward direction

§ Kinetic frictional force fk opposes relative

motion. It acts in opposite direction of

relative velocity.

§ Frictional force reduces translational velocity,

increases rotational velocity till they become

equal and the object sets on pure rolling.

§ Sliding is also referred as forward slipping.

42. Slipping

§ When vCM<Rω or vTRANS < vROT

Rotation is more than translation.

§ When we suddenly start the vehicl from rest

or the vehicle is stuck in mud, point of contact

has more of vROT than vTRANS.

§ Resultant velocity v in backward direction.

§ Kinetic frictional force fk opposes relative

motion. Hence it acts in the opposite direction

of the relative velocity.

§ Frictional force reduces rotational velocity and

increases the translational velocity till they

become equal and the object sets pure rolling.

§ Slipping is also referred as backward slipping.

43. Kinetic Energy in Pure Rolling

2CM

2CM

ROTTRANS

CM

CM

ωI2

1Mv

2

1

KEKE(KE)energykinetic

ω velocityangular

ICOMaboutMI

vmassofcenterofvelocity

Mobjectrolling ofmass

+=

+=

=

=

=

=

i) With center of mass as reference

÷÷ø

öççè

æ+=

+=

+=

2

22CM

2

22CM

2CM

2

2CM22

CM

R

KMv

2

1

R

KMv

2

1Mv

2

1

R

v)(MK

2

1Mv

2

1

1KE

KE

ii) With point of contact as reference:

÷÷ø

öççè

æ+=

+=

+=

+=

=

+=

+=

=

1R

KMv

2

1KE

)R

vMR

R

v(MK

2

1

R

v)MR(MK

2

1

)ωMR(MK2

1

ωI2

1KE

MRMK

MRIItheoremaxisparallelBy

IcontactofpointaboutMI

2

22CM

2

2CM2

2

2CM2

2

2CM22

222

20

22

2CM0

0

44. Rolling on Inclined Plane

§ A Round object of mass m, radius R is rolling

down an inclined plane without slipping

§ Forces acting on object along inclined plane

i) Component of gravitational force mg sin q

ii) Static frictional force (f).

§ Other component of gravitation force mg cosq

cancelled by normal force N exerted by plane

§ For translational motion, mg sinq is

supporting force and f is opposing force

mg sinθ – f = ma ….(1)

§ For rotational motion, frictional force f can set

torque of Rf.

)(1

gsinθaonaccelerati

maR

Kmasin.θmg

maR

Kma-sin.θmg

(1)in(2)Sub

.....(2)R

Kmaf

R

aMKRf

IαRf

2

2

R

K

2

2

2

2

+=

+÷÷ø

öççè

æ=

=÷÷ø

öççè

æ

÷÷ø

öççè

æ=

÷ø

öçè

æ=

=

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