MOMENT OF INERTIA

MOMENT OF INERTIA J3010/2/1

UNIT 2

MOMENT OF INERTIA

OBJECTIVES

General Objective : To understand the concept of moment inertia

Specific Objectives : At the end of this unit you should be able to :

 define moment inertia of mass

 describe definition torque and angular acceleration

 explain moment inertia for thin ring and rectangular.

 explain the moment of couple and kinetic energy.

.


INPUT

2.0 INTRODUCTION.

The moment of inertia of a body, about a given axis, is a measure of its resistance to
Angular. An acceleration and is given by the product of its mass times radius squared.

The second moment of area or
second moment of mass is also
called moment of inertia

2.1 MOMENT OF INERTIA:

Moment of inertia is the product of mass and the square of a distance. The unit which
it is measured is one kilogram meter squared (kgm2). It should also be noted that
∑ m r2 is a scalar quantity.
The moment of inertia is also called the second moment of area of the body.
If the moment of inertia be equal to Mk2, then k is called the radius of gyration of the
body about the axis.

2.1.1 UNIT OF MOMENT INERTIA (M.I).

The moment of inertia of an area is measured in metre4 or ft4. If the body is
measured in kilograms and distances in meter, the M.I of mass will be
kg- metre2 units.


2.2 THEOREM OF PARALLEL AXES

The moment of inertia of a lamina about any axis in the plane of the lamina equals the
sum of the moments of inertia about a parallel centrically axis in the plane of lamina
together with the product of the area of the lamina and the square of the distance
between the two axes.( fig. 2.1)
Let A = Area of the plane figure.
Ix = moment of inertia of the area A about an axis XX in the plane of the area
passing through G, the C.G ( Centre of Gravity) of the area.
Iy = moment of inertia of the area A about an axis YY in the plane of the area
parallel to XX.
r = distance between XX and YY.

Then Iy = Ix + Ar2

.P

Y Y

x r

X X
G

Fig. 2.1
Example 2.1

Find the moment of inertia of the uniform rod in the fig.2.2 about axis XY and X’Y’.
Y’ Y
l l

M
X’ X
Fig.2.2
Solution 2.1
M = mass of rod
Ml 2 l2  4 Ml 2
IXY = and IX’Y’ = M  + l2  =
3 
3   3


Example 2.2

Find the moment of inertia for the rectangular section shown in fig.2.3 about (i) the
axis XX, (ii) axis YY, (iii) the value of Izz
`

Y

B

600 mm

X G 200 mm X

D

300 mm
Y

Z Z

Fig. 2.3

Solution 2.2

bd 3
IXX =
12
600 x 200 3
=
12
= 4 x 104 mm4.

db 3
IYY =
12
200 x 600 3
=
12
= 3.6 x 109 mm4

IZZ = I CG + Ac2
In this case I CG = IXX = 4 x 104 mm4 and c = 300 mm
Thus IZZ = 4 x 104 + 200 x 600 x 3002


= 1.12 x 106 mm4
2.3 THEOREM OF PERPENDICULAR AXES

If the moments of inertia of lamina about two perpendicular axes in its plane which
meet at O are A and B the moment of inertia about an axis through O perpendicular
to the plane of the lamina is A + B.
Let OX, OY (figure 2.4) be the two perpendicular axes n the plane of the lamina,
and OZ an axis perpendicular to the lamina.
If m is the mass of a particle of the lamina at P, where as OP = r, the moment of
inertia of the lamina about OZ is Σ mr2.

Z

x x
O
y r
Y P

Figure 2.4

But if (x, y) are the coordinates of P referred to OX, OY as axes,

r2 = x 2 + y 2
Σ Mr2 = Σmx2 + Σmy2

Now Σmx2 is the moment of inertia about OY (=B), and Σmy2 is the moment of inertia
about OX (=A); therefore the moment of inertia about OZ = A + B.


Example 2.3

Find the moment of inertia of a uniform disc of radius a about an axis perpendicular to
its plane passing through a point on its circumference fig.2.5.

Y

a
X X

Y
Fig.2.5

Solution 2.3

m = mass of uniform disc
IXX = ⅓ ma2
IYY = I ( d x + d y )
2 2

= ⅓ m ( a2 + a2)
= ⅔ma2


2.4 MOMENT OF INERTIA IN SIMPLE CASES:

Type of form
Model M.I

b
bd 3
Rectangular/square
d 12

l l3
Thin rod M
3
M = mass

Thin ring r Mr2

r 2
Solid sphere Mr 2
5

Triangle b 3
h
h 12

b
b


2.5 TORQUE AND ANGULAR ACCELERATION

2.5.1 TORQUE

Torque is the turning moment of tangential applied force (F) acting at distance (r)
from the axis rotation. The unit of torque is the Newton meter (Nm)

F

d

O

Fig.2.6 Moment of a force.

In the fig.2.6 the moment of F about the point O is Moment of a force = F d
A couple is a pair of equal and parallel but unlike forces as shown in fig 2.7.

F

P

F
Fig.2.7 Moment of a couple

It can easily be proved that the moment of a couple about any point in its plane is the
product of one force and perpendicular distance between them, that is

Moment of couple = F p
Examples of a couple include turning off a tap with finger and thumb and winding up a
clock with a key. The moment of a force or couple may be measured in Newton meter


(Nm). In engineering, the moment of a force or couple is called a torque.

Example 2.4

Determine the torque created by the 225 N force acting on the gear teeth as shown.
Pitch Circle Diameter (P.C.D) 300 mm.

Solution 2.4

T=Fr Where F = 225N
300
= 225 x 0.15 r =
2
= 33.75 Nm. = 150 mm
= 0.15 m

2.5.2 ANGULAR ACCELERATION

If the angular velocity of the point P in fig. 2.8 is changing with time,
then the angular acceleration a of P is the rate of change of its angular
velocity, that is

dω
a =
dt


Fig.2.8 Angular Motion

in the sense of increasing θ.
Angular acceleration may be measured in rad/s2.

If the angular acceleration is uniform, then its magnitude is

ω 2 − ω1
α =
t
if the angular speed changes from ω1 to ω 2 in time t.

Example 2.5

The speed of flywheel is increased from 120 r/min to 300r/min in 30 seconds.
Calculate the angular acceleration of the flywheel before coming to rest.

Solution 2.5

ω 2 − ω1
α = Where ω = 300 r/min
t
31.43 − 12.57 ( 300 x 2 x 22)
= = rad/s
30 60 x 7
18.30
= - = 31.43 rad/s
30

α = - 0.6287 rad/s2 ωo = 120 r/min
(120 x 2 x 22)
= 60 x 7
rad/s
= 12.57 rad/s


Activity 2A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INPUT…!

2.1 A pulley attached to the motor shaft revolves at 1435 r/min. Determine the linear
velocity of pulley belt given the effective diameter of the pulley is 100 mm.

2.2 The angular velocity of a gear wheel uniformly increase from 15 r/min to 15 r/min in
20 seconds. Determine the angular acceleration and angular displacement of the
gear teeth.

2.2 Calculate the moment of inertia, about the axis of rotation of the flywheel shown if the
density of the flywheel material is 7600 kg/m3.

dimensions in millimeter


2.4 A wheel and axle has the 8 kg mass attached to the axle by a light cord as show. The
mass is allowed to fall freely a vertical distance of 2 meters in 10 seconds. Calculate the
moment of inertia for the wheel and axle.

The moment of inertia of a body, about a
given axis is
I = m r2
Where I = moment of inertia (kg.m2)

Feedback to Activity 2A

Have you tried the questions????? If “YES”, check your answers now

2.1 7.515 m/s

2.2 0.0524 rad/s2; 41.9 radians

2.3 11.8 kgm2

2.4 1.22 kgm2


INPUT

2.6 ANGULAR MOMENTUM

Momentum = mass x velocity.
= mxv

Angular momentum of a solid is given as the product of the moment of inertia of the
solid about axis of rotation and angular velocity.

When a body has motion of rotation, the momentum of the body is the product of the
moment of inertia of the body and its angular velocity.
∴ momentum of rotating body = Iω
and momentum of a body having a motion of translation = m v (∴ v = ωr)
= (mr ) ω
2

M = Iω (∴I = mr2)

2.7 ANGULAR IMPULSE


This is the change in momentum produced by the action of a force applied on a body
within an infinitely short interval of time. Donating impulse by I, we have
Impulse = Force x Time
I =Fxt (2.1)
Let a be the acceleration generated by the force, then by Newton’s second law, we
have F = ma
∴ Equation (2.1) becomes I = mat = m(v – u) or Ft = m(v – u) ( v = u + at )
Hence, when a force is constant, its impulse can be measured by the change in
momentum produced by it The unit of impulse is the same as that of momentum,
i.e. kg sec (kgs).

2.8 WORK DONE BY A TORQUE

Let a force F turns a light rod OA with length r through an angle θ to OB as shown in
fig. 2.9.

Fig.2.9 Work done by a torque

The torque TQ exerted about O is force times perpendicular distance from O or TQ = Fr

Now work done by F is F times distance moved. Hence Work Done = Fs
But s is the arc of a circle radius r. Hence
S = rθ
Where θ must be measured in radians.
Thus work done = Frθ
Or work done = TQθ
The work done by constant torque TQ is thus the product of the torque and the angle
turned through in radians. The work done will be in joules if TQ is in Nm.


Example 2.6

The force exerted on the end of a spanner 300 mm long used to tighten a nut is
constant 100 N. Find the torque exerted on the nut and the work done when the nut
turns through 30°.

Solution 2.6

Torque TQ = Pr
= 100 x 300 x 10-3
= 30 Nm.

Work Done = TQθ
= 30 x π/6 (θ in radians)
= 15.7 J.

Example 2.7

An electric motor is rated 400 W. If its efficiency is 80 %, find the maximum torque
which it can exert when running at 2850 rev/min.

Solution 2.7

Power = 2πN TQ
N = 2850/60 = 47.5 rev/s
Power = 400 x 0.8 = 320 W

Torque TQ = 320/2π x 47.5
= 1.07 Nm.

2.9 ANGULAR KINETIC ENERGY

When a body has motion of rotation, it will have an energy due to this rotation. This


Iω 2
kinetic energy of a body due to its motion of rotation is given by = or
2g
Iω 2
, where I = mass moment of inertia of the rotating body about the
2
axis of rotation and ω in the angular velocity of the body.
Work done F xS
Power is rate doing work. Power = =
Time taken t
S
but = v ∴ Power = F x r
t
Power of any times is equal to the product of the force and the velocity of the point of
application is the direction of force.

Example 2.8

A wheel has a 5.4 m long string wrapped round its shaft. The string is pulled with a
constant force of 10 Newton, and it is observed that the wheel is rotating at 3
revolutions per second when the string leaves the axle. Find the moment of inertia of
the wheel about its axis.

Solution 2.8
Given, length of string
= 5.4 m
Force P = 10 N
Speed of wheel,ω = 3 rev/sec = 2π x 3 = 6π rad/sec
Let I = moment of inertia of the wheel about it axis.
We know that work done in pulling the string
= Force x Distance
= 10 x 5.4 = 54 Nm

and kinetic energy of the wheel,
I ( 6π )
2
Iω 2
E = = Nm
2g 2 x 9.81
= 18.1 I Nm
Now equating work done and the kinetic energy,
18.1 I = 54
54
I = = 2.98 Nm2
18.1


Example 2.9

A fly wheel weighing 8 tones starts from rest and gets up a speed of 180 rpm in 3
minutes. Find the average torque exerted on it, if the radius of gyration of the fly
wheel is 60 cm. Take
g = 9.81 m/sec2.

Solution 2.9

Given, weight of the fly wheel
= 8 t = 8,000 kg
∴ mass of the fly wheel, m
= 8,000 kg
Initial revolution, No = 0
∴ Initial velocity,
ωo = 0
Final revolution = 180 rpm.
2π x 180
∴ Final velocity, ω = = 6π rad/sec
60
Time taken, t = 3 min = 3 x 60 = 180 sec
Radius of gyration,
K = 60 cm = 0.6 m
Let α = Constant angular acceleration, and
T = Average torque exerted on the fly wheel.
We know that the mass moment of inertia of the fly wheel,
I = mK2 = 8,000 x 0.62 = 2,880 kgm2.
Using the relation,
ω = ωo + α t with usual notations.
6π = 0 + α x 180
6π π
∴ α= = rad/sec2
180 30
Now using the relation,
Iα
T = with usual notations.
g
2,880 π
= x = 30.7 kg m
9.81 30


Example 2.10

A machine gun bullet of mass 25 gm is fired with a velocity of 400 m/sec. The bullet
can penetrate 20 cm in a given target. If the same target is 10 cm thick, what will be
the velocity of the bullet, when it comes out of the target?

Solution 2.10

Given, Mass of bullet,
M = 25 gm = 0.025 kg
Velocity of bullet, v = 400 m/sec
Penetration of bullet,
s = 20 cm = 0.2 m
let, v1 = velocity of the bullet after coming out from 10 cm thick target,
E = kinetic energy of the bullet, and
R = Resisting force of the target
Using the relation,
mv 2
E= with usual notations.
2g
0.025 x 400 2
= = 204 kgm
2 x 9.81
A little consideration will show, that the total kinetic energy is spent in penetrating
20cm into the target.
∴ P x 0.2 = 204
204
or P = = 1020 kg.
0.2
The energy spent in penetrating 10 cm (i.e. 0.1 m) thick target
= P x s = 1020 x 0.1 = 102 kg m
∴ Balance kinetic energy in the bullet after coming out from 10 cm thick target
= 204 – 102 = 102 kg m
Again using the relation,
mv 2
E = with usual notations
2g
0.025 x v12
102 = = 0.00128 v12
2 x 9.81
102
∴ R = = 282.3 m / sec
0.00128


2.10 KINETIC ENERGY OF A TORQUE

Kinetic energy K.E = ½ m v2
= ½ m (ωr )2 (∴ v = rω )
= ½ (m r2) ω2
= ½ I ω2 ( I = m r2)
i.e. kinetic energy K.E = ½ I ω2
Where KE = Kinetic energy (J)
I = moment of inertia (kg.m2)
ω = angular velocity ( rad/sec)

Example 2.11

A flywheel whose moment of inertia is 50 kg m2 is rotating at 4 rad/s. Find its kinetic
energy.

Solution 2.11
Given,
I = 50 kg m2 and ω = 4 rad/s
Kinetic Energy = ½ I ω2
= ½ x 50 x 42
= ½ x 800
= 400 J


Activity 2B

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INPUT…!

2.5. Calculate the moment of inertia, taken around the axis of rotation of the flat metal disc.
If the 11 kg disc revolves around its axis of rotation with an angular acceleration of
10 rad/s2, what torque is acting?.

2.6. A 45 kg flywheel, revolving at 50 r/min, has a radius of gyration of one meter. Calculate
the moment of inertia and torque which must be applied to bring the flywheel to rest in
10 seconds.


2.7 A 20 kg flywheel is revolving at 450r/min. If the radius of gyration is 0.65 meter,
calculate the torque which must be applied to the flywheel to bring it to rest in 20
seconds.

2.8. Calculate the kinetic energy stored in a 2.5 tones flywheel which is rotating at180 r/min.
The radius of gyration of the flywheel is 0.8 meter. If the velocity of the flywheel is
reduced to 15 r/min in one minute find the rate at which the flywheel gives out energy
(i.e. the power output).

2.9 A flywheel loses kinetic energy amounting to 640 J when its angular speed falls from 7
rad/s to 3 rad/s. What is the moment of inertia of the flywheel?

Kinetic energy ( K.E) = ½ mv2

Potential energy (P.E) = mgh

Feedback to Activity 2B

Have you tried the questions????? If “YES”, check your answers now

2.5 0.3438 kg m2; 2.438 Nm

2.6 45 kg m2; 23.57 Nm

2.7 19.9 Nm.

2.8 284.5 k J; 4.708 k W.


2.9 32 kg m2

SELF-ASSESSMENT 2

You are approaching success. Try all the questions in this self-assessment section and
check your answers with those given in the Feedback on Self-Assessment 2 given on the
next page. If you face any problems, discuss it with your lecturer. Good luck.

1. A 75 kg flat disc, with a diameter of 0.5 meter revolves about an axis perpendicular to its
circular surface at10 r/min. What is the angular momentum of the disc and the retarding
torque needed to bring the disc to rest in 5 seconds?

2. Calculate the time taken to bring a flywheel from rest to velocity of 450 r/min given the
moment of inertia is 8 kg.m2 and the applied torque is 24 N m.

3. A 7 kg gear wheel with radius of gyration of 0.3 meter is rotating at 200r/min. This gear
wheel meshes with a stationary 4.5 kg gear wheel. If the radius of gyration of the second
gear wheel also 0.3 meter, calculate the common speed of rotation after connection and
loss in kinetic energy of the system.


ω = 200 r/min

ω = 0 (stationary)
1 2

Feedback to Self-Assessment 2

Have you tried the questions????? If “YES”, check your answers now.

1. 2.456 kg m2/s; 0.4912 Nm.

2. 15.7 s.

3. 121.7 r/min; 54.47 J

CONGRATULATIONS!!!!
…..May success be with
you always….

MOMENT OF INERTIA

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