COURSE TITLE MECHANICS

COURSE CODE STP 111

Moment of Inertia, I

Suppose a rigid object is rotating about a fixed axis O, and a

particle A of the object makes an angle Ɵ with a fixed line OY

dθ
in space at some instant. The angular velocity, dt or Ꞷ, of every

particle about O is the same, since we are dealing with a rigid

body, and the velocity v1 of A at this instant is given byr w , where

1

1 1
= OA. Thus the kinetic energy of A = = m1 r 1 w .
2 2 2
r1 m1 v 1
2 2

Similarly, the kinetic energy of another particle of the body =

1
m2 r 2 w .
2 2
2

, where r is its distance from O and m is its mass. In this way we

2 2

see that the kinetic energy, K.E., of the whole object is given by

1 1 1
2
K . E= m1 r 1 w
2
2
+ 2
2
m2 r 2 w
2
+= 2
2 2
m3 r 3 w + …

1
 1 2
=
2 2 2
w (m1 r 1 +m 2 r 2 +m 3 r 3+ …)
2

1 2
=
2
w (£ mr )
2

where £mr
2
represents the sum of the magnitudes of '' m r ' for all
2

the particles of the object. We shall see shortly how the quantity

£mr
2
can be calculated for a particular object. The magnitude of

£mr
2
is known as the moment of inertia of the object about the

axis concerned,

and we shall denote it by the symbol J. Thus

1 2
Kinetic energy, K.E., = 2
Iw (1)

The units of I are kg metre (kgm ). The unit of w is 'radian radian s

2 2 −1

Thus if I = 2 kgm and w = 3 rad s , then

2 −1

1 2 1
K.E. = 2
Iw = 2 x 2 x 3 joule = 9 J.
2

The kinetic energy of a particle of mass m moving with a

1 2
velocity v is 2
mv . It will thus be noted that the formula for the

kinetic energy of a rotating object is similar to that of a moving

2
particle, the mass m being replaced by the moment of inertia I

and the velocity v being replaced by the angular velocity w. As

we shall require values of I, the moment of inertia of several

objects about a particular axis will first be calculated.

Moment of Inertia of Uniform Rod

(1) About axis through middle. The moment of inertia of a small

element ∂x about an axis PQ through its centre O perpendicular

∂x
to the length = , where / is the length of the rod, M is its
2
( M) x
l

mass, and x is the distance of the small element from O,

2
Ml
moment of inertia, /, = 12

3
Thus if the mass of the rod is 60 g and its length is 20 cm, M = 6

x 10" 2

kg, / = 0-2 m, and / = 6 x 10" 2 x 0-22/12 = 2 x 10~ 4 kg m2

(2) About the axis through one end, A. In this case, measuring

distances x from A instead of O,

2
Ml
moment of inertia, /, = 3

Moment of Inertia of Ring

Every element of the ring is the same distance from the centre.

Hence the moment of inertia about an axis through the centre

perpendicular to the plane of the ring = Ma

2
, where M is the

mass of the ring and a is its radius.

4
 Moment of Inertia of Circular Disc

Consider the moment of inertia of a circular disc about an axis

through its centre perpendicular to its plane. If we take a small

ring of the disc enclosed between radii x and x +∂ x its mass =

2 πx ∂ ∂ x
2
M, where a is the radius of the disc and M is its mass.
πa

Each element of the ring is distant x from the Centre, and hence

the moment of inertia of the ring about the axis through O = (

2 πx ∂ ∂ x 2
2
M ¿×x
πa
2
Ma
.'. moment of inertia of whole disc = 2

Thus if the disc weighs 60 g and has a radius of 10 cm, M = 60 g

= 6x10 2kg, a = 01 m, so that / = 6x KT2 x0-l 2/2 = 3xl0_4

kgm2

Moment of Inertia of Cylinder

5
If a cylinder is solid, its moment of inertia about the axis of

symmetry is the sum of the moments of inertia of discs into

which we may imagine the cylinder cut. The moment of inertia

1
of each disc = where a is the radius; and hence, if M is
2❑
×mass × a
2

the mass of the cylinder

1
moment of inertia of solid cylinder =
2
Ma
2

If a cylinder is hollow, its moment of inertia about the axis of

symmetry is the sum of the moments of inertia of the curved

surface and that of the two ends, assuming the cylinder is closed

at both ends. Suppose a is the radius, h is the height of the

cylinder, and σ is the mass

per unit area of the surface. Then

mass of curved surface = 2 πa h σ

and moment of inertia about axis = mass x a = 2 πa h σ ,

2 3

6
If a hollow and a solid cylinder have the same mass M and the

same radius and height, it can be seen from (i) and (ii) that the

moment of inertia of the hollow cylinder is greater than that of

the solid cylinder about the- axis of symmetry. This is because

the mass is distributed on the average at a greater distance from

the axis in the former case.

Radius of Gyration

The moment of inertia of an object about an axis, £mr

2
, is

sometimes written as Mk
2
, where M is the mass of the object and

k is a quantity called the radius of gyration about the axis.

For example, the moment of inertia of a rod about an axis

2 2
Ml l
through one end = 3
=M ( )
√3

l
. Thus the radius of gyration, k, = √3 = 0-58/.

7
.

Relation Between Moment of Inertia About C.G. and

Parallel Axis.

Suppose / is the moment of inertia of a body about an axis CD

and IG is the moment of inertia about a parallel axis PQ through

the centre of gravity, G, distant h from the axis CD, If A is a

particle of mass m whose distance from PQ is x, its moment of

inertia about CD = m(h-x)2

;. I = £m(h-x)2 = £mh2 + £mx2 - £2mhx.

Couple on a Rigid Body

8
Consider a rigid body rotating about a fixed axis O with an

angular velocity w at some instant.

The force acting on the particle A = m1 × acceleration =

2
d dw d θ
m1 × ( r 1 w )=m1 ×r 1 =m1 r 1 2 . The moment of this force about the
dt dt dt

axis O = force x perpendicular distance from O=

2
d θ
¿ m1 r 1 2
×r 1
dt

Since the force acts perpendicularly to the line OA

2
2d θ
Moment or torque ¿ m r 1 1
dt
2

2
d θ
Total torque about O = I
dt
2

2
d θ
Couple, C = I
dt
2

Example

9
1. A heavy flywheel of mass 15 kg and radius 20 cm is mounted

on a horizontal axle of radius 1cm and negligible mass

compared with the flywheel. Neglecting friction, find

(i) the angular acceleration if a force of 4 kg is applied

tangentially to the axle,

(ii) the angular velocity of the flywheel after 10 seconds.

solution
2 2
Ma 15× 0.2
(i) Moment of inertia = 2
= 2
=0.3 kg m
2

Couple c ¿ 4 ×9.8 N ×0.01 m=0.4 Nm

0.4
Angular acceleration =
−2
=1.3 rad s
0.3

(ii) After 10 seconds, angular velocity = angular acceleration x

time.

= 1.3 x 10= 13 rad s −1

10
 Angular Momentum and Conservation

In linear or straight-line motion, an important property of a

moving object is its linear momentum. When an object spins or

rotates about an axis, its angular momentum plays an important

part in its motion.

Consider a particle A of a rigid object rotating about an axis O.

The momentum of A = mass x velocity = m x v = m v = m r 1 1 1 w

The 'angular momentum' of A about O is denned as the moment

of the momentum about O. Its magnitude is thus m1 v × p, where p

is the perpendicular distance from O to the direction of v. Thus

angular momentum of A = m 1 vp=m1 r 1 w = m1 r 1❑2 w

.". total angular momentum of whole body = Iw , where I is the

moment of inertia of the body about O.

11
Angular momentum is analogous to 'linear momentum', mv, in

the dynamics of a moving particle. In place of m we have J, the

moment of inertia; in place of v we have w, the angular velocity.

Further, the conservation of angular momentum, which

corresponds to the conservation of linear momentum, states that

the angular momentum about an axis of a given rotating body or

system of bodies is constant, if no external couple acts about that

axis. Thus when a high diver jumps from a diving board, his

moment of inertia, I, can be decreased by curling his body more,

in which case his angular velocity w is increased.

(ii). He may then be able to turn more somersaults before

striking the water. Similarly, a dancer on skates can spin faster

by folding her arms. The earth is an object which rotates about

an axis passing through its geographic north and south poles

with a period of 1 day. If it is struck by meteorites, then, since

action and reaction are equal, no external couple acts on the

12
earth and meteorites. Their total angular momentum is thus

conserved. Neglecting the angular momentum of the meteorites

about the earth's axis before collision compared with that of the

earth, then angular momentum of earth plus meteorites after

collision = angular momentum of earth before collision.

Since the effective mass of the earth has increased after collision

the moment of inertia has increased. Hence the earth will slow

up slightly. Similarly, if a mass is dropped gently on to a

turntable rotating freely at a steady speed, the conservation of

angular momentum leads to a reduction in the speed of the table.

Angular momentum, and the principle of the conservation of

angular momentum, have wide applications in physics.- They

are used in connection with enormous rotating masses such as

the earth, as well as minute spinning particles such as electrons,

neutrons and protons found inside atoms.

13
 SIMPLE HARMONIC MOTION

Simple Harmonic motion is the periodic motion of a body or

particle along a straight line such that the acceleration of the

body is directed toward a fixed point and is also proportional to

its displacement from that point.

There are many examples of such motions

i. The motion of the strings in a musical instrument e.g

Guitar

ii. The motion of the balance wheel of a watch

iii. The motion of the pendulum of a clock

iv. The motion of the pistons in a gasoline engine

v. The motion of the prongs of a sounding tuning fork

vi. The motion of a child’s swing

vii. The beating of the heart

viii. The motion of a bod suspended from a spiral spring

14
When the bob of a pendulum moves to-and-fro through a small

angle, the bob is said to be moving with simple harmonic

motion. The prongs of a sounding tuning fork, and the layers of

air near it, are moving with simple harmonic motion, and light

waves can be considered due to simple harmonic variations.

Simple harmonic motion is closely associated with circular

motion.

DEFINITION OF TERMS RELATING TO SHM

1. AMPLITUDE (A): of a SHM is the maximum

displacement of the body performing SHM from its

equilibrium or central position C.

2. PERIOD (T): is defined as the total time taken by a

vibrating body to make one complete revolution about a

reference point.
t 1
T = ∨T =
n f

15
3. FREQUENCY (f): is the number of complete revolutions

per second made by a vibrating body.

n 1
f= =
t T

SPEED AND ACCELERATION OF SHM

The rate of change of angle Ɵ with time t is known as the

angular velocity w. Angular velocity w is defined by:

angle turned t h roug h by t h e body
w=
Time taken

Ɵ
w=
t

Ɵ=wt

This is similar to the relation

Distance = uniform velocity X time

s
s=vt∨v=
t

For the motion in a straight line

The angle Ɵ is measured in radians (rad) and since 2π rad

= 360, we have that w is measured in rad per seconds(rad/s)

s
Also by definition, Ɵ in radians = r
∧h ence s=r Ɵ where

16
 r = A = radius of the circle. The angular velocity w is given

by
s
Ɵ r
w= =
t t

s 1
¿ ×
t r

s
But t
=v , t h e linear velocity of particle. h ence ,

1
w=v × ∨v=wr=wA
r

Hence the linear speed equals the product of the angular speed

and the radius or amplitude of motion.

EXAMPLE 1

A boy ties a stone to the end of a string which he then whirles

above his head round a circular path of radius 15cm. if the stone

makes 20 oscillations in 10 seconds, calculate the angular and

linear speeds of the stone.

Solution

1 complete oscillation = 360 = 2π rad

17
20 complete oscillations = 360 ×20

2 π ×20 rad
Ɵ 40 π
Angular speed,w= t = 10
rad /secs

4 π rad /secs

Linear speed, v ¿ wA=4 π × 15 cm s −1

−1
¿ 60 π cms

EXAMPLE 2

A spring makes 60 revolutions in 15secs. Find the period and the

frequency of the vibration.

Solution.
timetaken 15
T= = =0.25 s
number of vibrations 60

1 1 −1
f= = =4 s ∨4 Hz
T 0.25

The linear velocity v at any point, Q whose distance from C

the central point is given by:

v=w √ A −x
2 2

Acceleration, a¿−w 2
x

18
Hence the linear acceleration a equals the product of the square

of the angular speed and the displacement, x of the particle from

the centre of motion. The negative sign in the equation indicate

that the acceleration a is directed towards the centre of motion

while the displacement x is measured from that point outwards

Ɵ
From ¿ t , when t = T, the period of SHM or the time for one

cycle,Ɵ= 360 or 2π radians

2π
Hence, w= T

2π
T=
w

1 w
Similarly, the frequency, f is given by f= =
T 2π

w=2 π f

EXAMPLE 3

A body vibrates in SHM with a frequency of 50Hz and an

amplitude of 4cm. find

a. The period

19
b. The acceleration at the middle and at the end of the path of

oscillation

c. The velocities at the middle and at the end of path of

oscillation

d. The velocity and acceleration at a distance of 2cm from the

centre of oscillation

Solution
1 1
a. T = = =0.02 sec .
f 50

b. 2
Acceleration, a=−w x

i. At midpoint of vibration i.e, at C, x=0

2
a=−w × 0=0

ii. At the end of path, x= amplitude = 4cm

2
a=−w × 4=0

But
w=2 π f =w=2 π ×50=100 π

a=¿
2 2 −2 3 −2
¿ 4 ×10 π ms =4 × 10 ms

20
c. v=w √ A2−x 2

4
At x=0 , v=wA =100 π × 100 m

¿4 π m

at x=A , v=w √ A 2− A2= 0

d. v=w √ A −x2 2

100 π √ 4 −2 cms
2 2 −1

200 π √ 3 cms
−1

2 π √ 3 ms
−1

2 4 2 2
a=−w x=10 π ×
100
2 −2
¿ 200 π ms

21
 SURFACE TENSION

Introduction

Surface tension is the elastic tendency of a fluid surface which

makes it acquire the least surface area possible. Surface tension

is the force acting along the surface of a liquid causing the liquid

surface to behave like a stretched elastic skin. Surface tension

allows insects (e.g. water striders), usually denser than water, to

float and stride on a water surface.

At liquid–air interfaces, surface tension results from the greater

attraction of liquid molecules to each other (due to cohesion)

than to the molecules in the air (due to adhesion).

The net effect is an inward force at its surface that causes the

liquid to behave as if its surface were covered with a stretched

elastic membrane. Thus, the surface becomes under tension from

the imbalanced forces, which is probably where the term

22
"surface tension" came from .Because of the relatively high

attraction of water molecules for each other through a web of

hydrogen bonds, water has a higher surface tension (72.8 milli

newtons per meter at 20°C) compared to that of most other

liquids. Surface tension is an important factor in the

phenomenon of capillarity.

Surface tension has the dimension of force per unit length, or of

energy per unit area. The two are equivalent, but when referring

to energy per unit of area, it is common to use the term surface

energy, which is a more general term in the sense that it applies

also to solids.

In materials science, surface tension is used for either surface

stress or surface free energy.

23
Free surface of a liquid has tendency to contract in surface area

is called surface tension.

•SI unit of Surface tension: N/m. or (J/m²).

•Its Dimension is [M⁰L¹T ̄²].

•Angle of Contact: The angle measured from the side of the

liquid, between the tangent to the solid surface inside the liquid

and tangent to the free liquid surface at the point of contact

between solid and liquid surfaces.

Capillary Rise Method

•Let us consider a capillary tube of uniform bore dipped

vertically in a beaker containing water. Due to surface tension,

water rises to a height h in the capillary tube

•The surface tension T of the water acts inwards and the reaction

of the tube R outwards. R is equal to T in magnitude but

24
opposite in direction. This reaction R can be resolved into two

rectangular components.

•(i)Horizontal component R sin θ acting radically outwards

•(ii)Vertical component R cos θ acting upwards.

•The horizontal component acting all along the circumference of

the tube cancels each other whereas the vertical component

balances the weight of water column in the tube.

Total upward force = R cos θ circumference of the tube

•(i.e) F = 2πr R cos θ or F = 2πr T cos θ. This upward force is

responsible for the capillary rise. As the water column is in

equilibrium, this force acting upwards is equal to weight of the

water column acting downwards.

•F = w.

25
•T = hgrρ/2cosӨ

•h = 2TCosӨ/ρgr

Factors affecting surface tension:

•Impurities present in a liquid appreciably affect surface

tension. A highly soluble substance like salt increases the

surface tension whereas sparingly soluble substances like soap

decrease the surface tension.

•The surface tension decreases with rise in temperature. The

temperature at which the surface tension of a liquid becomes

zero is called critical temperature of the liquid.

Effect of temperature

26
•Oxygen in the atmosphere is known to decrease the surface

tension of various substances.

Temperature

•At Critical temperature Surface tension: Zero

•Critical temperature of water 3744K

•Surface tension Increase with impurity.

Application of surface tension

•Surface tension of soap solution is less, it can spread over large

areas and wash clothes more effectively, since the dirt particles

stick to the soap molecules.

•In soldering, addition of flux reduces the surface tension of

molten tin. Hence, it spreads.

27
•Antiseptics like dettol have low surface tension, so that they

spread faster.

•Surface tension prevents water from passing through the pores

of an umbrella.

•A duck is able to float on water as its feathers secrete oil that

lowers the surface tension of water.

Surface energy and work

work done in enlarging surface area = force x distance,

Thus the surface tension, y, can be defined as the work done per

unit area in increasing the surface area of a liquid under

isothermal conditions. This is also called the free surface energy.

VISCOSITY

28
Viscosity is the internal friction between layers of a liquid or gas

in motion. If we move through a pool of water we experience a

resistance to our motion. This shows that there is a frictional

force in liquids. We say this is due to the viscosity of the liquid.

If the frictional force is comparatively low, as in water, the

viscosity of the liquid is low ; if the frictional force is large, as in

glue or glycerine, the viscosity of the liquid is high. We can

compare roughly the viscosity of two liquids by filling two

measuring cylinders with each of them, and allowing identical

small steel ball-bearings to fall through each liquid. The sphere

falls more slowly through the liquid of higher viscosity.

As we shall see later, the viscosity of a lubricating oil is one of

the factors which decide whether it is suitable for use in an

engine. The

Ministry of Aircraft Production, for example, listed viscosity

values to which lubricating oils for aero-engines must conform.

29
The subject of viscosity has thus considerable practical

importance.

Coefficient of Viscosity

The basic formula for the frictional force, F, in a liquid was first

suggested by Newton. He saw that the larger the area of the

surface of liquid considered, the greater was the frictional force

F.

Definition, Units, and Dimensions of Coefficient of Viscosity

The magnitude of n is given by= F/A x velocity gradient"

The unit of F is a newton, the unit of A is m2 and the unit of

velocity

gradient is 1 m/s per m. Thus n may be defined as the frictional

force

30
per unit area of a liquid when it is in a region of unit velocity

gradient.

The 'unit velocity gradient' = 1 ms -1 change per m. Since the 'm'

cancels, the 'unit velocity gradient' = 1 per second. From n =

F/(A x velocity gradient), it follows that r may be expressed in

units of newtons m~ 2 (N s m-2

The coefficient of viscosity of water at 10°C is 1-3 x 10" 3 Nsm-

31