GENERAL DYNAMICS J3010/1/1

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.

.

GENERAL DYNAMICS J3010/1/2

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

VelocityThe 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.

AccelerationThe 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.

GENERAL DYNAMICS J3010/1/3

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 ( u + v)2

Further, since distance travel = average speed x t then s =

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

1 ( u + v)t (1.2)2

s = 1 ( u + u + at )I t2

Or s = u t + 1 at2

2

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

s = 1 ( u + v ) (v - u )I a2

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 is340 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

2

60 = 0 + ½ (9.81)2

t = 3.50s

The hammer takes 3.50 s to fall 60 m. The sound takes60

= 0.18 to travel the same340

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 will be constant and

= 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 r when is measured in radians and hence linear speed v is

V = arc QP

=t

r(1.4)

tUsing (1.3) and (1.4) leads toV = r 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 where r = 0.35 m and = 6 radls= 0.35 x 6

V = 2.1 mls

1.4 RELATIONSHIP BETWEEN LINEAR ACCELERATION AND ANGULAR

ACCELERATION

By equation ddt

and V = r

Hence

d

1 dv

dt v

r dtsince r is constant

However

r dv

is linear acceleration adt

Hence = a

r

Or a = r 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 1.3 t = 3 s

1 = 0 radls

2 = 3000 revlmin = 2πN160

= 2 X X 3000

60= 314.16 radls

= 2 1

t

= 314.16 0

3

= 104.72 radls2

a = r = 0.1 x 104.72

a = 10.47 mls2

V = r 2

= 0.1 x 314.16= 31.42 mls

Activity 1A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXTINPUT…!

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

A. Vr B. Vlr C. V2lr 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 theVelocity 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 fig.1.3. The work done by a constant force is thUs:Work done = F s cos

fig. 1.3 Notation for the work done

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

work done is Fs. Work done is zero if direction force = 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 to1

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 J Power = 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 negatiVe2

or s = v 2a

work done = force x distance

= Ps

= Ps 2

2aHoweVer P = maAnd 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. Then

Work done = aVerage load x extension= Y P x X= Y SX x X

= Y SX2

since strain energyU = Work done

ThUs U = Y P x

= Y SX2

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

2

Example 1.7

A wagon of mass 12 tone traVeling at 16 kmlh strikes a pair of parallelspring-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 =16 m3.6 s

Kinetic energy of wagon = Y Mv2

= Y x 12 x 1000 x 16

= 118,500 J

3.6

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 = 59,250Or Y x 600 x 1,000 X2 = 59,250

X = 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 body

m2 = mass of the second body

U1 = initial Velocity of the first body

U2 = initial Velocity of the second body

V1 = 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

7000V1 =

750V1 = 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 NEXTINPUT…!

1.8 A flywheel rotating at 1200 reVlmin slow down at a constant rate of 900 reVlmin in30 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 in50 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.

GENERAL DYNAMICS J3010/1/18

Power =work done

time taken= Fv

Feedback to Activity 1B

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1.8 a. 125.7 radis

b.

c.

94.2 radis

-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 of120 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?

GENERAL DYNAMICS J3010/1/20

Feedback to Self-Assessment 1

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

1. 23.6 mis; 2230 mis2

2. 2943 J

3. 844 J

4. 1.07 Nm

5. 1260 N; 102 kmih

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