J3010 Mechanics of Machines 1

MALAYSIANPOLYTECHNICS

MINISTRY OF EDUCATION

MODULE J3010MECHANICS OF MACHINES 1

Zainol Hashim (POLIMAS)Harisa Mohamad Saad (POLIMAS)

http://modul2poli.blogspot.com/

BIODATA OF MODULE WRITERSJ3010 Mechanics Machines

Name : Zainol Bin HashimAddress : Mechanical Engineering Department

Politeknik Sultan Abdul HalimMua’dzam Shah, Bandar Darulaman06000 JitraKedah Darulaman

Telephone No. : 04-9174701 ext.407Qualifications : B Sc Mech. Eng.(UTM)

Diploma Mech. Eng (UTM)Cert.Education

Position : Polytechnic Lecturer

Name : Harisa Bt Mohamad SaadAddress : Mechanical Engineering Department

Politeknik Sultan Abdul HalimMuadzam Shah, Bandar Darulaman06000 JitraKedah Darulaman

Telephone No. : 04-9174701 ext.822Qualifications : B Eng. Agricultural (Food and Process)

(UPM)Masters in Education (ITTHO)

Position : Polytechnic Lecturer


What Do You Think Of This Module?

Title of Module: Module Code :

Student’s Name: Registration No. :

Course:

Module Writers:

Please use the following scale for your evaluation:

4 Strongly Agree3 Agree2 Disagree1 Strongly Disagree

Instruction : Please I on the space provided.

No. How much do you agree with the following statements? SCALE

A. FORMAT 1 2 3 41 The pages are organized in an interesting manner.2 The font size makes it easy for me to read the module.

3The size and types of pictures and charts used are suitable forthe input.

4 The pictures and charts are easy to read and understand.5 The tables used are well-organised and easy to understand.6 The arrangement of the Input makes it easy for me to follow.7 All the instructions are displayed clearly.

B. CONTENTS 1 2 3 48 I understand all the objectives clearly.9 I understand the ideas conveyed.10 The ideas are presented in an interesting manner.11 All the instructions are easy to understand.12 I can carry out the instructions in this module.13 I can answer the questions in the activities easily.14 I can answer the questions in the self-assessment.15 The feedback section can help me identify my mistakes.16 The language used is easy to understand.17 The way the module is written makes it interesting to read.18 I can follow this module easily.19 Each unit helps me understand the topic better.

20I have become more interested in the subject after using thismodule.


CURRICULUM GRID

The curriculum grid of this module is based on the curriculum used by Malaysian

polytechnics.

No. TOPIC UNIT TotalHours

1 General 1 9 HoursDynamics ( 9 H)

2 Moment 2 6 HoursInertia ( 6 H)

3 Hoisting 3 6 Hourssystem ( 6 H )

4 Friction 4 6 Hours( 6 H )

5 Screw 5 6 Hours(6 H )

6 Belting 6 9 Hours( 9 H )


UNIT 1 GENERAL DYNAMICS

1.1 Velocity and Acceleration 1.2 Equation for linear, uniformly accelerated motion 1.3 Relationship between linear speed and angular speed 1.4 Relationship between linear acceleration

and angular acceleration 1.5 Work done by a constant force 1.6 Power 1.7 Energy and conservation of energy 1.8 Momentum and conservation of momentum

UNIT 2 MOMENT OF INERTIA

2.1 Moment of Inertia 2.2 Theorem of Parallel axes 2.3 Theorem of Perpendicular axes 2.4 Moment of Inertia in a few simple cases 2.5 Torque and Angular Acceleration 2.6 Angular Momentum 2.7 Angular impulse 2.8 Work done by a torque 2.9 Angular kinetic energy 2.10 Kinetic energy of a torque

UNIT 3 HOISTING SYSTEM

3.1 Inertia Couple 3.2 Accelerated Shaft 3.3 Shaft being brought to rest 3.4 Load raised and with accelerating upward 3.5 Load falling and accelerating downwards (No driving

torque acting) 3.6 Load falling and being brought to rest 3.7 Load rising: Coming to rest under friction only 3.8 Load balancing system


UNIT 4 FRICTION

4.1 Motion up the plane: pull P parallel to plane 4.2 Motion down the plane: Pull P parallel to plane 4.3 The angle of friction and total reaction 4.4 Application of angle of friction to motion on the

inclined plane

UNIT 5 SCREW

5.1 The square thread screw 5.2 The V- thread screw 5.3 Raising Load 5.4 Load being lowered 5.2 Overhauling of a screw

UNIT 6 BELTING

6.1 Length of an open belt drive 6.2 Length of cross belt drive 6.3 Power transmitted by a belt 6.4 Ratio of tensions 6.5 Centrifugal tension


MODULE GUIDELINES

To achieve maximum benefits in using this module, students must follow

the instructions carefully and complete all the activities.

1. This module is divided into 6 units. 2. Each page is numbered according to the subject code, unit and page number.

J3010 / UNIT 1 / 5

Subject Unit 1 Page Number 5

3. The general and specific objectives are given at the beginning of each unit. 4. The activities in each unit are arranged in a sequential order and the

following symbols are given:

OBJECTIVESThe general and specific objectives for each learning topic are

stated in this section.

INPUTThis section introduces the subject matter that you are going to learn.

ACTIVITIESThe activities in this section test your understanding of the

subject matter. You have to complete this section by

following the instructions carefully.


FEEDBACKAnswers to the questions in the activity section are given here

SELF-ASSESSMENTSelf-assessment evaluates your understanding of each unit.

FEEDBACK ON SELF-ASSESSMENTThis section contains answers to the activities in the self-assessment.

5. You have to follow the units in sequence. 6. You may proceed to the next unit after successfully completing the unit

and you are confident of your achievement.


GENERAL AIMS

This module is prepared for students in the third or fourth semester who are

undergoing the Certificate/Diploma programmes in Malaysian Polytechnics. It aims

to expose students to the mechanics of machines concept in each unit and to lead

them towards self-directed learning or with guidance from their lecturers.

PRE REQUISITE SKILLS AND KNOWLEDGE

The pre-requisite for this module is at least a pass in Mathematic at the SPM

level and the successful completion of the Engineering Science module in the

first semester.

GENERAL OBJECTIVES

At the end of this module, students should be able to:

1. know principles and fundamental concept of general dynamics. 2. know method and procedures to solve problem of mechanics machines. 3. understand facts and principles all of these topics. 4. solve problem in relation to hoisting and friction. 5. solve problem of raising and lowering load for square and Vee threaded screw. 6. find the values of Belting by using a calculator.

TEACHING AIDS AND RESOURCES NEEDED

1. Calculator


REFERENCES

1. John Hannah and M.J. Hillier (1971). Applied Mechanics; New

Zealand Pitman: Publishing Limited.

2. W. Madill (1984). Applied Mechanics Level 3; Longman Inc.,

New York: Longman Group Limited.

3. D Humphrey and J Topping (1971). A Shorter Intermediate

Mechanics; London: Longman Group Ltd.

4. George E. Drabble (1971). Applied Mechanics; London: W.H. Allen &

Co. Ltd.

5. R.S. Khurmi (1988). A Text Book Of Applied Mechanics;New

Delhi: S. Chand & Company (Pvt) Ltd.

6. R.K. Mullis (1983). Engineering Mechanics; Longman Cheshire Pty.

7. S.B.Mathur (1979). Applied mechanics; Delhi: Khanna Publishers.

8. L.Bostock and S.Chandler (1979). Applied Mathematics; Scotland:

Thomson Litho Ltd.


UNIT 1

GENERAL DYNAMICS

OBJECTIVES

General Objective : To understand the concept of general dynamics

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

> relate linear and angular velocity, linear and angular acceleration.

> solve problem using equation uniformly accelerated and

angular motion.

> describe tangent acceleration and centripetal acceleration, centripetal

and centrifugal force, work and power.

> explain the principle conservation of energy and momentum.

.


INPUT

1.0 INTRODUCTION.

A vector quantity requires a number and a direction to specify it completely; that is,

a vector has magnitude and direction. Examples of vectors are velocity, acceleration

and force.

Mechanics is the study of object or bodies, as we shall call them, when subjected to force.

1.1 VELOCITY AND ACCELERATION

Velocity The velocity of a body may be defined as its rate of change of displacement,

with respect to its surroundings, in a particular direction. As the velocity is

always expressed in a particular direction, it is also a vector quantity.

Acceleration The acceleration of a body may be defined as the rate of change of its velocity. It

is said to be positive, when the change in velocity of a body increases with time,

and negative when the velocity decreases with time. The negative acceleration is

also called retardation.


In general, the term acceleration is used to denote the rate at which the velocity

is changing. It may be uniform or variable.

1.2 EQUATION FOR LINEAR, UNIFORMLY ACCELERATED MOTION

Suppose a body moving in a straight line has an initial speed u and that it undergoes

uniform acceleration a for time t considering, let the final speed be v and the

distance traveled in the time t be s. The speed —time curve will be show in fig.1.1

Fig. 1.1 Uniform — accelerated linear motion.

Acceleration a is uniform, its magnitude is

a = change in speed

change in time

= v — u I tor at = v — uor v = u + at (1.1)

In this case, the average speed will be the speed at time tI2.

Hence average speed = 1

2 ( u + v)

Further, since distance travel = average speed x t then s = 1 2( u + v)t (1.2)

Substituting for v from (1.1) into (1.2) gives

s = 1

2 ( u + u + at )I t

Or s = u t + 1 at2

2


Substituting for t from (1.1) into (1.2) gives

s = 1 2( u + v ) (v - u )I a

or 2as = v2- u2

or v2 = u2 + 2as

Example 1.1

A workman drops a hammer from the top of scaffolding. If the speed of sound in air

is 340 mIs, how long does the workman have before shouting to another workman 60

m vertically below him if his warning is to arrive before the hammer. Neglect air

resistance.

Solution 1.1

For hammerInertial speed = 0 mIs.

Acceleration = 9.81 mIs2.Distance = 60 m

s = ut + 1 at2

260 = 0 + ½ (9.81)2 t = 3.50s

The hammer takes 3.50 s to fall 60 m. The sound takes 34060

= 0.18 to travel the same

distance so the workman has (3.50 — 0.18) = 3.32 s before shouting if the sound is to arrive before the hammer.


1.3 RELATIONSHIP BETWEEN LINEAR SPEED AND ANGULAR SPEED

If a point P moves round in a circle with a of radius r with constant linear speed v

then the angular speed ϖ will be constant and

ϖ = Ο

t

(1.3)where t is the time to move from Q to P along the arc QP of the curve. Fig 1.2

Fig. 1.2 Circular motion

However, arc length QP is rΟ when Ο is measured in radians and hence linear speed v is

V = arc QP = rΟ (1.4)tt

Using (1.3) and (1.4) leads to

V = r ω for circular motion. (1.5)Or linear speed = radius x angular speed.

Example 1.2

What is the peripheral speed of the tread on a tire of a motor car if the wheel spins

about the axle with an angular velocity of 6 radianl second. Diameter of tires is 0.7 m.

Solution 1.2

V = r ω where r = 0.35 m and ω = 6 radls= 0.35 x 6

V = 2.1 mls


1.4 RELATIONSHIP BETWEEN LINEAR ACCELERATION AND ANGULAR

ACCELERATION

By equation α =dχ

and V = r χdt

Hence α = d = 1 dv since r is constantr dtdtΙ v Ι

♣ •r

However dv

dt is linear acceleration a

Hence α = a

r

Or a = r α

Or linear acceleration = radius x angular acceleration

Example 1.3

A grinding wheel is accelerated uniformly from rest to 3000 revlmin in 3 seconds.

Find it angular and linear acceleration. If the wheel diameter is 200 mm, find the

final linear speed of a point on its rim.

Solution 1.3t = 3 s

χ 1 = 0 radls χ 2 = 3000 revlmin = 2πN160

= 2 X τ X 300060

= 314.16 radls

α =χ2 χ1

t

=314.16 0

3

α = 104.72 radls2


a = r α

= 0.1 x 104.72

a = 10.47 mls2

V = r χ 2

= 0.1 x 314.16 = 31.42 mls


Activity 1A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT

INPUT…!

1.1 Which of the following is a Vector quantity?.A. density B. speed C. area D. acceleration

1.2 Velocity is the rate of change with time ofA. displacement B. acceleration C. speed D. distance

1.3 When a body moVes round a circle with radius r at uniform speed V,the angular speed ω is

C. V2lrA. Vr B. Vlr D. 2πV

1.4 A 5 kg block, at rest on a smooth horizontal surface, is acted on by a

resultant force of 2.5 N parallel to the surface. The acceleration, in mls2 is

A. 0.5 B. 2 C. 12.5 D. 2000

1.5 A car traVel along a straight road at a steady speed of 13 mls, accelerates

uniformly for 15 s until it is moVing at 25 mls. Find its acceleration.

1.6 A particle moVes from rest with an acceleration of 2 mls2. Determine the Velocity and displacement of the particle after 20 seconds.

1.7 A parcel, starting at rest, is moVed by conVeyor belt with an acceleration of

1.5 mls2. What will its Velocity be after it has moVed 3 meters?


S = ½ ( u + v) tV = u + atV2 = u2 + 2asS = Ut + ½ at.

Feedback to Activity 1A

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

1.1 D. acceleration

1.2 A. displacement

1.3 B. Vlr

1.4 A. O.5

1.5 O.8 mls2

1.6 4O mls; 4OO m.

1.7 3 mls.


INPUT

1.5 WORK DONE BY A CONSTANT FORCE

Work done = Force x distance.

Unit of work is joule (J) orKilojoule (kJ)

When the point at which a force acts moVes, the force is said to haVe done work. When the force is constant, the work done is defined as work done = force x distance moVed in the direction of the force. It is a scalar qUantity.If a constant force F moVes a body from A to B then distance moVed in the

direction of F is s cos 0 fig.1.3. The work done by a constant force is thUs:

Work done = F s cos 0

fig. 1.3 Notation for the work done


If the body moVes in the same direction as the force, where by Ο = 00 and

work done is Fs. Work done is zero if direction force Ο = 900. If F is in Newton and s is in meters, the work done will be measUred in joUles (J)

Example 1.4

How mUch work is done when a force of 5 KN moVes its point of application 600

mm in the direction of the force.

SolUtion. 1.4

Work done = force x distance

= 5 X 103 X 600 X10-3 = 3000 J = 3 KJ.

1.6 POWER

Power is the rate of doing work, i.e. the work done in Unit time. The SI Unit of

power is the watt; it is 1 joUle per second and is written 1 W. The British Unit of

power Use earlier was the Horse- Power, and is eqUiValent to aboUt 746 watts . If a

force of F Newton keeps its point of application moVing in the direction of the force

with Uniform speed v meters per second, the work done per second is Fv joUles, and

is the power is Fv watts.

Example 1.5

The total mass of an engine and train is 200 Mg; what is the power of the engine if it

can jUst keep the train moVing at a Uniform speed of 100 kmlh1 on the leVel, the

resistances to friction, amoUnting to 1 of the weight of the train.200


SolUtion 1.5

Since the speed is Uniform, the pUll of the engine is eqUal to the total resistance, i.e. 1000g N.= 1000 x 9.81 = 9810 N.

The speed is 100 km1h1= 1000136 m1s

The work per second = 1000 x 9.8 x 1000136 JPower = 105 x 2.72 W

= 272 kW

1.7 ENERGY

The energy may be defined as the capacity to do work. It exists in many form e.g.

Mechanical, electrical, chemical, heat, light etc. BUt in applied Mechanics, we

shall deal in Mechanical Energy only. The Unit for energy is the same as those of

work i.e. example joUles.

1.7.1 CONSERVATION OF ENERGY

Energy cannot be created or destroyed bUt can be transformed from one

to another form of energy. For instance water stored in a dam possesses

potential energy which changes to kinetic energy as it flows downwards

throUgh a tUnnel to tUrn tUrbines, which in tUrn changes to electric

energy which can be Used to prodUce heat energy.

1.7.2 POTENTIAL ENERGY

The potential energy of a body may be defined as the amoUnt of work it can do when it moVes from its actUal position to the standard position chosen.The work done lifting a load of mass M and weight W = MgthroUgh a height h is Wh. This is known as the potential energy of the load referred to its original position and its Unit in that energy,i.e. the basic Unit is the joUle (J).Potential energy = Wh = Mgh (zero at earth’s surface)


Example 1.6

What is the potential energy of a 10 kg mass?(a) 100 m aboVe the sUrface of the earth. (b) at the bottom of a Vertical mine shaft 1000 m deep.

SolUtion 1.6

(a) Potential energy = mgh= 10 x 9.81 x 100 J = 9.81 KJ.

(b) Potential Energy = -10 x 9.81 x 1000 J

= -9.81 x 104 = -98.1 KJ.

1.7.3 KINETIC ENERGY

A body may possess energy dUe to its motion as well as dUe to its position. For

example, when a hammer is Used to driVe in a nail, work is done on the nail by

the hammer, hence it mUst haVe possessed energy. Also a rotating flywheel

possess energy dUe its motion. These are example of the form of energy

call kinetic energy.Kinetic energy may be described as energy dUe to motion. Only linear motion

will be considered. The kinetic energy of a body may be defined as the

amoUnt of work it can do before being broUght to rest.

1.7.4 FORMULA FOR KINETICS ENERGY

Let a body of mass m moVing with a speed V be broUght to rest with a

Uniform retardation by constant force P in a distance s.


v2= u2 + 2 as

0 = v2- 2 as since a is negatiVe

or s = v 2

2a

work done = force x distance= Ps

=Ps 2

2a HoweVer P = ma And Hence

Work done = mav2l2a

= Y mv2

The kinetic energy is thUs giVen

by Kinetic energy = Y mv2

1.7.5 STRAIN ENERGY

The work done in compressing or stretching a spring is stored as strain energy in the spring proVided that there is no permanent deformation (oVer stretching). The stiffness of a spring is the load per Unit extension and is approximately constant within the working range of the spring; thUs if S is the stiffness, the load P reqUired to prodUce an extension X is giVen by

P = S X

SUppose a load is gradUally applied to a spring so that it Varies from zero to maximUm ValUe P and prodUce a maximUm extension X. ThenWork done = aVerage load x extension

= Y P x X = Y SX x X

= Y SX2 since strain energy

U = Work done ThUs U = Y P x

= Y SX2

The Units of strain energy are same as those of work, i.e. joUles (J)


Example 1.7

A wagon of mass 12 tone traVeling at 16 kmlh strikes a pair of parallel

spring-loaded stops. If the stiffness of each spring is 600 KNlm, calcUlate

the maximUm compression in bringing the wagon to rest.

SolUtion 1.7

V = 16 kmlh = 3.616

m

s

Kinetic energy of wagon = Y Mv2

= Y x 12 x 1000 x Ι♣

3.616

Ι• 2

= 118,500 J

This kinetic energy may be assUmed to be absorbed eqUally by the two springs. Strain energy stored per spring is

Y x 118,500 = 59,250 J

ThUs X is the maximUm compression of the springs,

Y SX2 = 59,250

Or Y x 600 x 1,000 X2 = 59,250X = 0.446 m = 446 mm

1.8 MOMENTUM AND CONSERVATION OF MOMENTUM

1.8.1 MOMENTUM

The momentUm of a particle is the prodUct of the mass of the particle and its

Velocity. If m is the mass of the particle and v its Velocity the momentUm is m v.

The Unit of momentUm is eqUiValent, i.e. Ns = kg mls.


1.8.2 CONSERVATION OF MOMENTUM

If two bodies collide then the sUm of the momentUm before the collision is eqUal to

the sUm of the momentUm after collision measUred in the same direction.

m1U1 + m2U2 = m1V1 + m2V2

Where m1 = mass of the first bodym2 = mass of the second bodyU1 = initial Velocity of the first bodyU2 = initial Velocity of the second bodyV1 = final Velocity of the first body

V2 = final Velocity of the second body

Example 1.8

A 750 kg car collided head on with a 1 tone car. If both cars are traVel at 16 kmlh

at the time of impact and after impact the second car reboUnds at 3 kmlh, find the

Velocity of the first car after collision (assUme perfect elastic collision)

SolUtion 1.8

By the conserVation of momentUm and assUming that the first car also reboUnd.

+ m1U1 + m2U2 = m1V1 + m2V2

﴾750 x (+16)﴿ + ﴾ 100 x (-16) ) = ﴾ 750 x (-V1) ﴿ + ( 1000 + ( +3) )

12 x 103 — 16 x 103= -750 V1 + 3 x 103

(12 — 16 — 3) x 103= -750 V1

-7 x 103= -750 V1

V1 = 7000750

V1 = 9.333 kmlh

Where m1= 750 kg ; m2 = 1 tone = 1000 kg ; U1= + 16 kmlh ; U2 = -16 kmlh; V2 = + 3kmlh


Activity 1B

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT

INPUT…!

1.8 A flywheel rotating at 1200 reVlmin slow down at a constant rate of 900 reVlmin in 30 seconds. Find: a. the initial angUlar speed b. the final angUlar speed c. the angUlar acceleration

d. the initial speed of a point on the rim of the flywheel if its diameter is 1.1 m.

1.9 A constant force of 2 kN pUlls a crate along a leVel floor for a distance of 10 m

in 50 seconds. What power was Used?.

1.10 A car of mass 1000 kg traVeling at 30 mls has its speed redUced to 10 mls by constant

breaking force oVer a distance of 75 meter. Find the initial and final kinetic energy

and the breaking force.


Power = work done

= Fv time taken

Feedback to Activity 1B


1.8 a. 125.7 radis

b. 94.2 radis

c. -1.05 radis2

d. 69.1 mis

1.9 400 W

1.10 0.5 x 105 J, 5333 N


SELF-ASSESSMENT 1

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 1 giVen on the

next page. If yoU face any problems, discUss it with yoUr lectUrer. Good lUck.

1. The spin drier in a washing machine is a cylinder with a diameter of 500 mm. It spins at

900 reVimin. Find the speed and acceleration of a point on the side of the drUm.

2. Find the work done in raising 100 kg of water throUgh a Vertical distance of 3 m.

3. A cyclist, with his bicycle, has a total mass 80 kg. He reaches the top of the hill, with a

slope 1 in 2 measUred along the slope, at a speed of 2 mis. He then free-wheels to the

bottom of the hill where his speed has increased to 9 mis. How mUch energy has been

lost on the hill which is100 m long?

4. An electric motor is rated at 400 W. If its efficiency is 80%, find the maximUm

torqUe which it can exert when rUnning at 2850 reVimin.

5. The engine of a car has a power oUtpUt of 42 KW. It can achieVe a maximUm speed

of 120 kmih along the leVel. Find the resistance to motion. If the power oUtpUt and

resistance remained the same, what woUld be the maximUm speed a car coUld

achieVe Up an incline of 1 in 40 along the slope if the car mass is 900 kg?


Feedback to Self-Assessment 1

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

1. 23.6 mis; 2230 mis2

2. 2943J

3. 844 J CONGRATULATIONS!!!!…..

4. 1.07NmMay success be with you

always….

5. 1260 N; 102 kmih


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 G X

Fig. 2.1Example 2.1

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

Y’1

Y1

X XM

’ Fig.2.2

Solution 2.1

M = mass of rod ♣ 1 2 2• 4M1 2

M1 2 and IX’Y’ = Mι

+ 1λ

=IXY = Α

3, 33


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 G200 mm X

D

300 mm

Y

z z

Fig. 2.3

`

Solution 2.2

IXX = bd 3

12

=600 x 2003

12

= 4 x 104 mm4.

IYY = db 3

12

=200 x 6003

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 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

O x y r

Y P

x

Figure 2.4

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

r2 = x2 + y2

Σ Mr2 = Σmx2 + Σmy2

Now Σmx2 is the moment of inertia about OY (=B), and Σmy 2 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 = ! ma2

IYY = I ( d 2x + d y2 )

= ! m ( a2 + a2)

= !ma2


2.4 MOMENT OF INERTIA IN SIMPLE CASES:

Type of form Model M.I

b

Rectangular/squarebd 3

d 12

1

M 1

3

Thin rod3

M = mass

Thin ring r Mr2

Solid sphere r2

Mr 25

Triangle bh 3

h 12

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 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

FFig.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 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.


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 = 3002

= 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 = ddt

ω

Fig.2.8 Angular Motion


in the sense of increasing 0.

Angular acceleration may be measured in rad/s2.

If the angular acceleration is uniform, then its magnitude is

ω2ω

1

α =t

if the angular speed changes from ω1 to ω2 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

α = ω 2 Where ω = 300 r/min

ω1

t

(300 x 2 x22)= 31.43 12.57 = rad/s30 60 x 7

= - 18.30 = 31.43 rad/s30

α = - 0.6287 rad/s2 ωo = 120 r/min

= (120 x 2 x22) rad/s60 x 7

= 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 agiven axis is

I = m r2

Where I = moment of inertia (kg.m2)

Feedback to Activity 2A


2.1 7.515 mIs

2.2 O.O524 radIs2; 41.9 radians

2.3 11.8 kgm2

2.4 1.22 kgm2


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 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 = Iο

(∴ v = οr)and momentum of a body having a motion of translation = m v= (mr2) ο

(∴I = mr2)M = Iο

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 TimeI = F x t (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 Ο 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 = rΟWhere Ο must be measured in radians. Thus work done = FrΟOr work done = TQΟThe work done by constant torque T Q 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 300.


Solution 2.6

Torque TQ = Pr= l00 x 300 x l0-3 = 30 Nm.

Work Done = TQΟ= 30 x τl6 (Ο 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 = 2τN TQ

N = 2850l60 = 47.5 revlsPower = 400 x 0.8 = 320 W

Torque TQ = 320l2τ 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. This

kinetic energy of a body due to its motion of rotation is given by = Iο 2 or2 g

Iο 2 , where I = mass moment of inertia of the rotating body about the2

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

Power is rate doing work. Power = Work done

= F x S

tTime taken

but St = v ∴ Power = F x r

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 2.8Given, length of string

= 5.4 mForce P = 10 N

Speed of wheel,ο = 3 revisec = 2τ x 3 = 6τ 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 = Iο 2 = I ( 6τ )2 Nm 2 g 2 x 9.81

= 18.1 I NmNow equating work done and the kinetic energy,

18.1 I = 54

I

54 2

= = 2.98 Nm18.1

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 misec2.


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,

οo = 0Final revolution = 180 rpm.

∴ Final velocity, ο = 2τ x 180 = 6τ radisec60

Time taken, t = 3 min = 3 x 60 = 180 secRadius of gyration,

K = 60 cm = 0.6 mLet 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,

+ α t with usual notations.ο = οo

6τ = 0 + α x 180

∴ α =6τ τ 2

= radisec180 30Now using the relation,

T =I α

with usual notations.g

τ=2,880 x = 30.7 kg m309.81

Example 2.10

A machine gun bullet of mass 25 gm is fired with a velocity of 400 misec. 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 misecPenetration of bullet,


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,my 2

E = 2 g with usual notations.

=0.025 x 400 2 = 204 kgm

2 x 9.81A 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,

E =my 2

with usual notations

2 g

102 =0.025 x y 2 = 0.00128 y 1

21

2 x 9.81

∴ R =102

= 282.3 m i sec0.00128

2.10 KINETIC ENERGY OF A TORQUE

Kinetic energy K.E = ½ m

y2

= ½ m (οr )2 (∴ v = rο )= ½ (m r2)

ο2

= ½ I ο2 ( I = m r2)i.e. kinetic energy K.E = ½ I

ο2

Where KE = Kinetic energy (J)

I = moment of inertia (kg.m2)ο = angular velocity ( radisec)


Example 2.11

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

Solution 2.11Given,

I = 50 kg m2 and ο = 4 radis

Kinetic Energy = ‘ I ο2

= ‘ x 50 x 42 = ‘ 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 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?


Kinetic energy ( K.E) = ‘ mv2

Potential energy (P.E) = mgh

Feedback to Activity 2B


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 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.

ο = 200 rimin

ο = 0 (stationary)1 2


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….


UNIT 3

HOISTING

OBJECTIVES

General Objective : To understand the concept of dynamics of rotation.

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

> apply basic principle on which all these machines are based.

> recognize the effect of combining a hoist drum of moment inertia I with a hanging load of mass M and weight W = mg.

> sketch and recognize all force and torque that involve on these machines.

> use suitable concepts to solve related problem.

> calculate all these topic questions correctly.


INPUT

3.0 INTRODUCTION

We study the effect of combining a hoist drum of moment of inertia with a hanging load of mass M and Weight W = Mg.

In this topic we are concerned in dynamics of rotation.

3.1 INERTIA COUPLE

Comparing the formulae P = Ma and T = Iα , it is seen that moment of

inertia I plays the same part in a change of angular motion as mass M does in

change of linear motion. By analogy with the idea of inertia force we may

regard the torque T as being balanced by inertia couple, Iα , which sense is

opposite to that of the angular acceleration α, (Fig. 3.1). The problem is then

in effect reduced to a static one.

The reality of the effect of an inertia couple will be appreciated by anyone

who has tried to accelerate a bicycle wheel rapidly by hand. Although the

weight may carried wholly by the bearings an effort is required to set the

wheel spinning. An inertia couple is, of course, reactive.


Fig. 3.1

Example 3.1

A 30 kg flywheel, revolving at 5.24 radls has a radius of gyration of one

meter. Calculate the torque which must be applied to bring the flywheel to

rest in 10 seconds.

Solution 3.1

Moment of inertia of the flywheel, I = mk 2

= 30x12

I = 30kgm2

Deceleration, ω1 = ω0 + αt

0 = 5.24 + (−α )10

α = 0.524rad l s 2

Torque, T = Iα

= 30x0.524

T = 15.72Nm


3.2 ACCELERATED SHAFT Consider a shaft (Fig. 3.2) carrying a rotor having a moment of inertia about the shaft axis.

Fig. 3.2

If the bearing friction is equivalent to a couple Tf .Then, in order to accelerate the shaft and rotor the driving torque T must balance both the inertia couple Iα and the friction couple Tf .

Thus,

T = Iα + T f

Example 3.2

A flywheel has a moment of inertia of 10 kg.m2 . Calculate the angular acceleration of the wheel due to a torque of 8 Nm if the bearing friction is equivalent to a couple of 3 Nm.

Solution 3.2:

Given: I = 10 kgm2 T = 8 Nm Tf = Iα = 3 Nm

T = Iα + T f

Iα = T T f

Iα = 8 3 Nm

Iα = 5 Nm

α = 105

radls2


3.3 SHAFT BEING BROUGHT TO REST

If the shaft is being brought to rest by a braking torque T the friction couple Tf assist the braking action so that T and Tf together must balance the inertia couple Iα ; α is now a retardation its sense being opposite to that of the motion (Fig.3.3).

Fig. 3.3

Thus,

T + T f = Iα

If there is no braking torque, the friction couple alone brings the shaft to rest. Then,

T f = Iα

Note, in both cases, that

(a) the friction couple T f opposes the motion. (b) the inertia couple Iα opposes the change of motion.


Example 3.3

A flywheel together with its shaft has a total mass of 300 kg and its radius of gyration is 900 mm. If the effect of bearing friction is equivalent to a couple of 70 Nm, calculate the braking torque required to bring the flywheel to rest from a speed of12 revis in 8 s.

Solution 3.3

Given: N = l2 rev/s = l2 x 2 = 75.4 rad/s

Thus, Retardation, α =ωt

=75.48

= 9.42 radis2

I of flywheel and shaft = Mk2

= 300 x 0.92

= 243 kg m2

Inertia Couple = 1α

= 243 x 9.42

= 2290 Nm


Activity 3A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE

NEXT 1NPUT…!

3.1 A drum rotor has the moment of inertia 31.8 kg.m2 . Find the time taken for the rotor to reach a speed of 3600 revimin from rest if the driving force torque is 55 Nm and the friction torque is 5 Nm.

3.2 The rotating table of the vertical boring machines has a mass of 690 kg and a

radius of gyration of 700 mm. Find the torque required to accelerate the table

to 60 revimin in three complete revolutions from rest.

3.3 A light shaft carries a turbine rotor of mass 2 tonnes and a radius of gyration

of 600 mm. The rotor required a uniform torque of 1.2 kNm to accelerate it

from rest 6000 revimin in 10 min. Find:

(i) the friction couple, (ii) the time taken to come to rest when steam is shut off.

3.4 A winding drum of mass 200 tonne has a radius of gyration of 3 m. Find the constant torque required to raise the speed from 40 to 80 revimin in 60 seconds if the friction torque is 15 kNm. If the wheel is rotating freely at 80 revimin and a brake is applied bringing it to rest in 120 rev. Find the brake torque assuming uniform retardation.


Feedback To Activity 3A


3.1 240s

3.2 354 Nm

3.3 (i) 446 Nm, (ii) 16.9 min

3.4 140.5 kNm, 68.8 kNm.


INPUT

THE HOIST

Four cases will be considered: the load is rising or falling, being accelerated or brought to rest. In every case two equations can be written down:(a) the equation for the balance of couples at the hoist drum (b) the equation for the balance of forces at the load.

In every case we recall that the friction couple at the bearing or rope will oppose the rotation and the inertia couple will oppose the change of rotation.For acceleration, a is upwards, hence the inertia force is downwards. If angular acceleration, α is anticlockwise, the inertia couple is clockwise. If rotation of the drum is anticlockwise the friction couple acts clockwise.

3.4 LOAD RAISED AND WITH ACCELERATING UPWARD

For rotation of the hoist drum the driving torque T must balance the friction couple Tf, , the inertia couple Iα and the torque Pr due to the tension P in the rope at the drum.

Fig. 3.4


Thus,

Angular Motion:

T = Iα + Pr + Tf

For Linear motion of the load, the tension P in the rope at the load must balance both the dead weight and the inertia force Ma.

Thus,

Linear Motion:

P = Mg + Ma

Example 3.4

A hoist drum has a moment of inertia of 85 kgm2 and is used to raised a lift of mass 1 tonne with an upward acceleration of 1.5 mis2. The drum diameter is 1 m.Determine:

(a) the torque required at the drum (b) the power required after accelerating for 3 seconds from rest.

Solution 3.4

(a) The torque required at the hoist drum is made up of three parts.

1. torque Iα required to accelerate the drum 2. torque Wr required to hold the dead weight of the lift. 3. torque Mar required to accelerate the lift.

M = 1000kg W = Mg I = 85 kgm2

= 1000 x 9.8 = 98000 N

α = ar =

10.5

.5 = 3rad i s 2


Thus,

Total torque = Iα + Wr + Mar

= (85 x 3) + (9800 x 0.5) + (1000 x 1.5 x 0.5)

= 5905 Nm

(b) After 3 seconds, the lift speed.

v = at

= 1.5 x 3

= 4.5 mis (This is the speed of the drum circumference)

Therefore angular velocity of the drum,

ω = vr = 0.5

4.5 = 9rad i s

Power required = torque x angular velocity

= T ω

= 5905 X 9

=53.15 Kw

This the power required at the instant after 3 seconds.


3.5 LOAD FALLING AND ACCELERATING DOWNWARDS (NO DRIVING TORQUE ACTING)

The load is allowed to fall freely, resisted only by friction and inertia force and couples. The rotation of the hoist drum, the accelerating torque Pr due to rope tension must balance both the friction couple Tf and the inertia couple Iα.

Fig. 3.5

Thus,

Angular Motion:

Pr = Tf + Iα

For linear motion of the load the accelerating force due to the weight must balance the upward tension P in the rope and the inertia force Ma.

Thus,

Linear Motion:

P = Mg - Ma


Example 3.5

A hoist drum has a mass of 360 kg and a radius of gyration of 600 mm. The drum diameter is 750 mm. A mass of 1 tonne hangs from a light cable wrapped round the drum and is allowed to fall freely. If friction couple at the bearings is 2.7 k Nm. Calculate the runaway speed of the load after falling for 2 seconds from rest.

Solution 3.5

Given:

Md = 360 kg k = 600 mm = 0.6 m

Dd = 750 mm = 0.75 m Rd = 0.375 m Tf = 2700 Nm

M = 1 tonne = 1000 kg ω0 = 0 radis t = 2 s

I = Mk2 = 360(0.6)2 = 129.6 kgm2

Linear Motion: P = Mg - Ma = M(g - a) a = rα

P = 1000(9.81 - rα) = 9810 - 75α

Angular Motion:

Pr = Tf + Iα = 2700 + 129.6α

P = 2700 + 129.6α

0.375

α = 3.62 radis2

Then, ω1 = ω0 + αt = 0 + 3.62 (2)

= 7.24 radis

v = r ω1

= (0.375) 3.62 = 2.71 mis


3.6 LOAD FALLING AND BEING BROUGHT TO REST

We now consider the braking of the hoist drum as the load falls. The accelerations are therefore reversed as compared with the previous case. For rotation of the drum the braking torque T is assisted by the friction couple to balance the accelerating torque Pr due to the rope tension and the inertia couple Iα.

Fig. 3.6

Thus,

Angular Motion:

T + Tf = Pr + Iα

Linear Motion:

P = Mg + Ma

Example 3.6

The maximum allowable pull in a hoist cable is 200 kN. Calculate maximum load in tones which can be brought to rest with a retardation of 5 mis2 . The hoist drum has a moment of inertia of 840 kgm2 and a diameter of 2.4 m. What is the corresponding braking torque on the drum?


Solution 3.6

Given: Pmax = 200 x 103 Nm a = - 5 mis2

I = 840 kgm2 Dd = 2.4 m

rd = 1.2 m

a = rα

α = a

r = 15

.2 = 4.16 radis2

Linear Motion:P = Mg + Ma

200 x 103 = M (g + a)

= M (9.81 + 5)

M =200 x 10 3

14.81 kg

M = 13.5 tonne

Angular Motion:

T + Tf = Pr + Iα

T = Pr + Iα - Tf = 200 x 103 (1.2) + 840 (4.16) — Tf

There is no braking torque, Tf = Iα

T = 200 x 103 (1.2) + 3494.4 — 3494.4

T = 240 k Nm


3.7 LOAD RISING: COMING TO REST UNDER FRICTION ONLY

Since there is no braking torque applied, the drum is retarded by the torque Pr due to the rope tension and the friction couple Tf . These two couples must balance the inertia couple of the drum.

Thus,

Angular Motion:

Linear Motion:

Fig. 3.7

Iα = Tf + Pr

P + Ma = Mg


Example 3.7

In an experiment, a hoist drum has a diameter is 500 mm. It is used to raised load 50 kg and coming to rest under friction. The upward acceleration is 3.0 mis2 . The friction couple is 0.35 Nm. Find the moment of inertia of the drum.

Solution 3.7a 3

a = rα α = α = α = 12 radis2r 0.25

For the linear motion, P + Ma = Mg

P = Mg — Ma

P = M (g - a)

P = 50 (9.81 — 3)

P = 340.5 N

For the angular motion, Iα = Tf + Pr

Iα = 0.35 + 340.5 (0.25)

Iα = 0.35 + 85.12

Iα = 85.47

I = 85.47 12

I = 7.12 kgm2

NOTE:

Students are required to grasp firmly the following rules:

1. the friction couple opposes the rotation 2. the inertia couple opposes the change of rotation 3. the inertia force opposes the change of linear motion.

It may remarked also that in every case the direction of the rope tension P and the load weight W is unaltered, although their effect may be to accelerate or to retard the load.


Activity 3B

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE

NEXT INPUT…!

3.5 A load of mass 8 tonne is to be raised with a uniform acceleration of 1.1 mis2

by means of a light cable passing over a hoist drum of 2 m diameter. The drum has a mass of 1 tonne and a radius of gyration of 750 mm. Find the torque required at the drum if friction is neglected. What is the power exerted after 4 seconds from rest.

3.6 A mine cage of mass 4 tonne is to be raised with an acceleration of 1.5 mis2

using a hoist drum of 1.5 m diameter. The drum’s mass is 750 kg and its radius of gyration is 600 mm. The effect of bearing friction is equivalent to a couple of 3 kNm at the hoist drum. What is the power required when the load has reached a velocity of 6 mis? What is the power required at a uniform velocity of 6 mis?.

3.7 A hoist has a winding drum 0.9 m effective diameter and a radius of gyration of 0.35 m, the mass of the drum being 100 kg. A load of 320 kg is to be raised 36 m, the mass of the lifting rope being 1 kgim. If the

acceleration is 1.8 mis2 until a constant velocity of 6 mis is reached, find the power necessary just at the end of the acceleration.

3.8 A winding drum raises a cage of mass 500 kg through a height of 120 m. The winding drum has a mass of 250 kg and an effective radius of 0.5 m and a radius of gyration of 0.36 m. The mass of the rope is 3 kgim. The cage has at

first an acceleration of 1.5 mis2 until a velocity of 9 mis is reached after hich the velocity is constant until the cage nears the top, when the final retardation

is 6 mis2 . Find :(i) the time taken for the cage to reach the top (ii) the torque which must be applied to the drum at starting (iii) the power at the end of the acceleration period.


Feedback To Activity 3B


3.5 87.8 kN m, 386.4 kW

3.6 299.5 kW, 259 kW

3.7 25.45 kW

3.8 (i) 17.08 s (ii) 4957 Nm (iii) 82.2 kW


INPUT

3.8 LOAD BALANCING SYSTEM

We shall now consider some simple cases of the motion of two masses

connected by a light inextensible string. We note that a string connecting two

masses in motion is in a state of tension and that the string exerts forces on

the masses equal to the tensions at its ends.

If the string is light (that is, if its weight is neglected) the tension is the same

throughout its length. On the other hand, if the string is heavy the tension will

in general vary from point to point, depending upon the weight per unit

length. If the string is extensible the tension will vary with the extension.

Also, if the string passes round a pulley the tension is only the same on the

two sides if the pulley is smooth and the string is light. Otherwise the tension

in the string where it leaves the pulley depends upon the coefficient of

friction and the length of string in contact. In such an ideal case the tension

throughout the string will be constant.

Example 3.8

A load of mass 230kg is lifted by means of a rope which is wound several times round a drum and which then supports a balance mass of 140 kg. As the load rises the balance mass falls. The drum has a diameter of 1.2 m and a radius of gyration of 530 mm and its mass is 70 kg. The frictional resistance to the movement of the load is 110 N, and that to the movement of the balance mass 90 N. The frictional torque on the drum shaft is 80 Nm.

Find the torque required on the drum, and also the power required at an instant when the load has an upward velocity of 2.5 mis and an upward

acceleration of 1.2 mis2 .


Documents

J3010 Mechanics of Machines 1