JJ309 Fluid Mechanics Unit 2

7/30/2019 JJ309 Fluid Mechanics Unit 2

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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.

.


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MOMENT OF INERTIA J1

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

ca lled 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 Mk

2, then kis 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 bekg- metre

2units.


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1 1

XM


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 distancebetween the two axes.( fig. 2.1)

Let A = Area of the plane figure.Ix = moment of inertia of the area A about an axisXX 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

XG

X

Example 2.1Fig. 2.1

Find the moment of inertia of the uniform rod in the fig.2.2 about axis XY and XY.

Y Y

X

Solution 2.1

M = mass of rod

M12

IXY =3

Fig.2.2

and IXY =1

2

M 3

+1

2 =,

4M12

3


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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

IXX =

=

bd3

12

600x 2003

12

= 4 x 104

mm4.

IYY =

=

db3

12200x 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 mmThus Izz = 4 x 10

4+ 200 x 600 x 300

2

= 1.12 x 106

mm4


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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 perpendicularto 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 mr2.

z

Ox x

yr

Y P

Figure 2.4

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

r2

= x2

+y2

Mr2

= mx2

+my2

Now mx2

is the moment of inertia about OY (=B), and my2

is the moment of inertiaabout OX (=A); therefore the moment of inertia about Oz = A + B.


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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 discIxx = ! ma

2

IYY = I ( d2

+d

2

)x y

= ! m ( a2

+ a2)

= !ma2


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2.4 MOMENT OF INERTIA IN SIMPLE CASES:

Type of form Model M.I

Rectangular/square

b

d

bd3

12

Thin rod1

M = mass

13M

3

Thin ring r Mr2

Solid spherer

2Mr

2

5

Triangleh

b

b h312


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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 0 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 theproduct of one force and perpendicular distance between them, that is

Moment of couple = F pExamples 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.


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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 = F r Where F = 225N

= 225 x 0.15 r =300

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

a =d

dt

Fig.2.8 Angular Motion


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in the sense of increasing 0.Angular acceleration may be measured in rad/s

2.

If the angular acceleration is uniform, then its magnitude is

=

2 1

t

if the angular speed changes from 1

to 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

=

2

1

t

Where = 300 r/min

31.43

= 12.57

30

(300x 2 x22)= 60x 7 rad/s

= -18.30

30= 31.43 rad/s

= - 0.6287 rad/s2 o = 120 r/min

=(120x 2x22)

rad/s

60x7

= 12.57 rad/s


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Activity 2A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THENExT

INPUT!

2.1 A pulley attached to the motor shaft revolves at 1435 r/min. Determine the linearvelocity 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 in20 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 thedensity of the flywheel material is 7600 kg/m

3.

dimensions in millimeter

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

moment of inertia for the wheel and axle.


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The moment of inertia of a body, about agiven 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 mIs

2.2 O.O524 radIs2; 41.9 radians

2.3 11.8 kgm2

2.4 1.22 kgm2


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INPUT

2.6 ANGULAR MOMENTUM

Momentum = mass x velocity.

= m x v

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

When a body has motion of rotation, the momentum of the body is the product of themoment of inertia of the body and its angular velocity.

momentum of rotating body =I

and momentum of a body having a motion of translation = m v ( v = r)

= (mr2)

M = I (I = mr2)

2.7 ANGULAR IMPULSE

This is the change in momentum produced by the action of a force applied on a bodywithin an infinitely short interval of time. Donating impulse by I, we have

Impulse = Force x TimeI = F x t (2.1)Let abe the acceleration generated by the force, then by Newtons 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 inmomentum produced by it The unit of impulse is the same as that of momentum,

i.e. kg sec (kgs).


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2.8 WORK DONE BY A TORQUE

Let a force F turns a light rod OA with length r through an angle to OB as shown infig. 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. HenceS = r

Where must be measured in radians.Thus work done = Fr

Or work done = TQThe 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 isconstant 100 N. Find the torque exerted on the nut and the work done when the nut

turns through 300.


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Solution 2.6

Torque TQ = Pr

= l00 x 300 x l0-3

= 30 Nm.

Work Done = TQ

= 30 x l6 (in radians)

= l5.7 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 revlmin.

Solution 2.7

Power = 2N TQN = 2850l60 = 47.5 revls

Power = 400 x 0.8 = 320 W

Torque TQ = 320l2 x 47.5

= l.07 Nm.

2.9 ANGULAR KINETIC ENERGY

When a body has motion of rotation, it will have an energy due to this rotation. This2

kinetic energy of a body due to its motion of rotation is given by =

I2

Ior

2g

, whereI= mass moment of inertia of the rotating body about the2

axis of rotation and in the angular velocity of the body.

Power is rate doing work. Power =Work done

=Time taken

F x S

t

butS

= v Power = F x rt

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.


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2

Example 2.8

A wheel has a 5.4 m long string wrapped round its shaft. The string is pulled with aconstant force of 10 Newton, and it is observed that the wheel is rotating at 3revolutions 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 mForce P = 10 N

Speed of wheel, = 3 revisec = 2 x 3 = 6 radisecLet 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,

E =I

= I(6 )2

Nm

2g 2x 9.81= 18.1 I Nm

Now equating work done and the kinetic energy,

18.1I = 54

I =54

= 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 3minutes. Find the average torque exerted on it, if the radius of gyration of the fly

wheel is 60 cm. Takeg = 9.81 misec

2.


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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,

o = 0Final revolution = 180 rpm.

Final velocity, =2 x180

60= 6 radisec

Time taken, t= 3 min = 3 x 60 = 180 sec

Radius of gyration,

K= 60 cm = 0.6 mLet = 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,

= o + twith usual notations.

6 = 0 + x 180

=6

=180

radisec

2

30

Now using the relation,

T =I

with usual notations.g

=2,880

x

9.81 30= 30.7 kg m

Example 2.10

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

Solution 2.10

Given, Mass of bullet,

M = 25 gm = 0.025 kgVelocity of bullet, v = 400 misec

Penetration of bullet,


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s = 20 cm = 0.2 mlet, v1 = velocity of the bullet after coming out from 10 cm thick target,

E = kinetic energy of the bullet, andR = Resisting force of the target

Using the relation,2

E =my

2gwith usual notations.

2

=0.025x 400

2x 9.81= 204 kgm

A little consideration will show, that the total kinetic energy is spent in penetrating

20cm into the target.

P x 0.2 = 204

or P =204

= 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,2

E =my

2gwith usual notations

102 =

0.025x y2

1

=0.00128y

2

R =

2x 9.81

102

0.00128

1

= 282.3 misec

2.10 KINETIC ENERGY OF A TORQUE

Kinetic energy K.E = my

2

= m (r )2

(v = r )

= (m r2)

2

= I2

( I = m r2)

i.e. kinetic energy K.E = I

2

Where KE = Kinetic energy (J)


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I = moment of inertia (kg.m2)

= angular velocity ( radisec)


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Example 2.11

A flywheel whose moment of inertia is 50 kg m2

is rotating at 4 radis. Find itskinetic energy.

Solution 2.11Given,

I = 50 kg m2

and = 4 radis

Kinetic Energy = I2

= x 50 x 42

= x 800

= 400 J


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Activity 2B

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THENExT

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 radis2, what torque is acting?.

2.6. A 45 kg flywheel, revolving at 50 rimin, 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 450rimin. 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 rimin.The radius of gyration of the flywheel is 0.8 meter. If the velocity of the flywheel is

reduced to 15 rimin 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

radis to 3 radis. What is the moment of inertia of the flywheel?


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Kinetic energy ( K.E) =

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


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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 rimin. 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 rimin 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 200rimin. 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.

= 200 rimin

= 0 (stationary)

1 2


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Feedback to Self-Assessment 2

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

1. 2.456 kg m2is; 0.4912 Nm.

2. 15.7 s.

3. 121.7 rimin; 54.47 J

CONGRATULATIONS!!!!..

May success be with you

always.

JJ309 Fluid Mechanics Unit 2

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