Eng. Dynamics as Easy as 123

7/29/2019 Eng. Dynamics as Easy as 123

1/78

ENT142ENGINEERING DYNAMICS

ZOL BAHRI RAZALI

1 |

HOW to understand Engineering DYNAMICS

Friendly notes to understand as simple as 123

Zol Bahri Razali


2/78


ZOL BAHRI RAZALI

2 |

ABSTRACT

Is an Engineering Mechanics : DYNAMICS is a tough subject?

Obviously most of engineering students at college and university

expect that the subject is tough and very difficult to understand. Why?

Maybe they have experienced with a pre-requisite subject : STATICS.

The DYNAMICS is a practical subject and an expansion of theoretical

knowledge of Statics. If really students understand and can imagine

the Free Body Diagram (FBD) in Statics, they are easy to imagine the

movement of the particle in FBD in DYNAMICS. Therefore they will

understand the concept of movement in this subject, and understand

what is DYNAMICS.

This book will guide students to understand the concept of

DYNAMICS through a method of 123, as simple as to understand

123. For engineering students at college and university, this simple

book is expected to give them to understand basic theoretical

knowledge on DYNAMICS especially in Free Body Diagram (FBD).

For individual who work at higher level of engineering such as

Engineers, Designers or Technicians, this book might be not suitable

for references because the contents is not in depth and breadth, i.etoo simple and very basic theoretical knowledge.


3/78


ZOL BAHRI RAZALI

3 |

ACKNOWLEDGEMENT

Main references of this note are from the references:

1. Hibbeler, R.C. 2002. Engineering Mechanics : Dynamics. 2ed.

Prentice Halls, Pearson Education Asia Pte. Ltd., Singapore.

2. Hibbeler, R.C. 2001. Engineering Mechanics : Statics. 2ed.

Prentice Halls, Pearson Education Asia Pte. Ltd., Singapore.

3. Hibbeler, R.C. 2010. Engineering Mechanics : Dynamics. Twelfth

Edition in SI Unit. Prentice Halls, Pearson Education South Asia

Pte. Ltd., Singapore..

4. Meriam, J.L. and Kraige, L.G. 2001. Engineering Mechanics:

Dynamics, John Willey & Sons, Inc.

5. Beer, F.P., Johnston, E.R. and Clausen, W.E. 2004. Vector

Mechanics for Engineers: Dynamics, Mc Graw Hill.


4/78


ZOL BAHRI RAZALI

4 |

CHAPTER 12

INTRODUCTION TO DYNAMICS

Chapter ObjectivesTo introduce the concepts of position, displacement, velocity and accelerationTo study particle motion along a straight line and represent this motion graphicallyTo investigate particle motion along a curved path using different coordinate

systemsTo present an analysis of dependent motion of two particlesTo examine the principles of relative motion of two particles using translating axes

12.1 Introductions

Mechanics branch of the physical science that is concerned with the state of

rest or motion of bodies subjected to the action of the forces

Mechanics of rigid body - divided into statics and dynamics

Statics - concerned with the equilibrium of the body that is either at the rest or

moves with constant velocity

Dynamics - concerned with the accelerated motion of a body. Presented in 2 parts:

a) Kinematicsgeometric aspect of motion

b) Kineticsanalysis of the force causing the motion

diff diff

BASICALLY: s v a intg intg


5/78


ZOL BAHRI RAZALI

5 |

12.2 Rectilinear Kinematics: Continuous Motion

Rectilinear Kinematicsat any given instant, the particles position, velocity and

acceleration.

Positionthe straight line path of a particle. From the origin (o), position vector r specifythe location of the particle (p).

Convenient (r) represent by (s)

Displacementthe change in its position

Eg : If the particle moves from P to P, the displacement is r = r- r

s = s s

s is positive particles final position is to the right of its initial position, ie :s>s.

Displacement of a particlevector quantity Distance traveled is a positive vector.

r

Os

r

r r

s s

s

Displacement

P P

Os

s

Position

P


6/78


ZOL BAHRI RAZALI

6 |

{Velocity}

If the particle moves through a displacement r, from P-P during the time

interval t, the average velocity

Vavg = r , V = dr : instantaneous velocityt dt

V as an algebraic scalar, V = ds

dt

t or dt always positive:1. particle moving the right, velocity is positive2. particle moving to the left velocity is negative.

The magnitude of the velocity is known as the speed.( units : m/s )

vavg = stt

v

P P

O s

s


7/78


ZOL BAHRI RAZALI

7 |

{ Acceleration }

Provided the velocity of the particle is known at two point P P, the average

acceleration

aavg =tv

v difference in the velocity during the time interval v = v - v

Acceleration: a =dt

dv

a

acceleration

a =2

2

dt

sd

adeceleration

decelerationwhen the particle is slowing down- speed decreasing

- vvv 1 is negative

acceleration is zerowhen velocity is constant.- ovvv

( unit = m/s 2 )

a ds = v dv a = dtdv

& v = dt

ds

a

P P

O s

v v


8/78


ZOL BAHRI RAZALI

8 |

constant accelerationeach of three kinematics equations a c =dt

dv

a = ac

v =

dt

ds, a

cds = vdv

maybe integrated to obtained formula

that related a,v,s,t.

The three formulas of constant acceleration :

1) Velocity as a function of time v

vo

dv t

o

ca dt

v = vo

+ a c t

2) Position as a function of time

s

o

t

ocvds ( + a c t) dt tavdtdsv co

s = s o + v o t + 21 a c t

2

3) Velocity as a function of position

v.dv = a c .ds o

v

s

sc

o o

dsavdv

=> v 2 = v 20

+ 2a c ( s-s o )

This formula only useful when the acceleration is constant and when t = o ,

s = s o , v = v o

e.g.a body fall freely toward the earth.

+


9/78


ZOL BAHRI RAZALI

9 |

See Example:

- 12.1

- 12.2

- 12.3

- 12.4

- 12.5

Exercise : 12.1

- 12.2

- 12.3

- 12.4

- 12.5

0194579207


10/78


ZOL BAHRI RAZALI

10 |

12.3 Rectilinear Kinematics: Erratic Motion

When particles motion during a time is erratic, may best be describes graphically using aseries of curves.

Using the kinematics equations:

dt

dva v

dt

ds dvvdsa ..

a) Given s-t graph, construct the v-t equations

s

t

b) tv graph ta graph

v = ,dtds

By experimentally, if the position can

be determined during the time of

period, graph s-t can be plotted.

By v = ,dtds the graph v-t can be

plotted.

( slope of s-t graph = velocity ).

dtdva

dtdva

(slope of tv graph = acceleration)

See example 12.6, page 19


11/78


ZOL BAHRI RAZALI

11 |

c) ta graph , tv graph

using dtdsa ,

adtv

( change in velocity = area

under ta )


12/78


ZOL BAHRI RAZALI

12 |

d) tv graph ts graph

dtdsv vdts

(displacement = area under tv graph)

See example 12.7, page 21


13/78


ZOL BAHRI RAZALI

13 |

e) sa graph sv graph.

dvvdsa ..

between the limits ovv at oss

2

1 ( )20

2

1vv =

1

.s

so

dsa

1vv at

1ss = area under sa graph

f) sv graph sa graph

dvvdsa ..

)(ds

dvva

acceleration = velocity x slope of sv graph.

See Example 12.8.

Exercise:

- 12.42- 12.43- 12.44- 12.45- 12.46


14/78


ZOL BAHRI RAZALI

14 |

12.4 General Curvilinear Motion

- curvilinear motion occurs when the particle moves along a curved path.

- position- considered a particle located

at point p on a space curve defined bythe path function s.

position vector r = r ( t ) magnitude and direction change as the

particle moves along the curve.

- displacement- during small line t, theparticle moves a distance s along thecurves.

r = r + r

the displacement r represent thechange in the particles position.

r = r - r

- velocityduring the time t , the averagevelocity.

V avg =t

r

, V =

dtdr

dr will be tangent to the curve at p, thedirection of V is also tangent to the

curve.

Path

P

r s

O

s

Position

P

s

r Pr

r

O

s

Displacement

v

P

r

O

s

Velocity


15/78


ZOL BAHRI RAZALI

15 |

The magnitude ofv, called speed. V =dt

ds.

V

t

vavg

, where v = v1 - v

Instant a new acceleration, t 0

dt

dva

2

2

dt

rda

Velocity vector is always directedtangent to the path.

a tangent to the hodograph, not tangentto the path of motion.


16/78


ZOL BAHRI RAZALI

16 |

12.5 Curvilinear Motion: Rectangular Components

Displacement

r = x i + y j + zk

magnitude of r always positive

r = (x 2 + y 2 + z 2 )

unit vector ur = (1/r)r

Velocity

v = vx

i v y j vZ k

v =dt

dr=

dt

d( x i ) +

dtd ( y j ) +

dtd ( z k)

v =dt

dr= v

xi + v

yj + v

zk,

where : v x =.

x

v y =.

y

v z =.

z

The velocity has a magnitude defined as the positive value of

v = (v x2 + v y

2 + v z2)


17/78


ZOL BAHRI RAZALI

17 |

Acceleration :

a =

dt

dv= a x i + a y j + a z k

where : a x =.

xv =..

x

a y =.

yv =..

y

a z =.

zv =..

z

The acceleration has a magnitude defined by the positive value of

a = (a x2 + a y

2 + a z2 )

See Example 12.9 and 12.10.


18/78


ZOL BAHRI RAZALI

18 |

12.6 Motion of a Projectile

The free-flight motion of a projectilestudied in terms of its rectangular

components. The projectiles acceleration always acts in the vertical direction.

Projectile launched at point ( ox , oy ) , initial velocity is V o , having two components

( Vo

)x and ( V o )y . The projectile has a constant downward acceleration,

a c = g = 9.812sm .

Horizontal motion : -

Since a x = 0 ; v = vo + ac t ; vx = ( vo ) x

x = xo+ vo t + 21

at2

; x = xo + ( vo ) x t

v 2 = vo 2 + 2ac ( s-so ) ; vx = ( vo ) x

First and last equation indicated that the horizontal component of velocity always remains

constant during the motion.


19/78


ZOL BAHRI RAZALI

19 |

Vertical motion : -

Since ay = -g. + v = vo + ac t ; vy = ( vo )ygt.

y = yo + vo t + 21 ac t

2 ; y = yo + (vo) yt - 21 g

v 2 = vo + 2 ac ( y- y2 ) ; vy

2 = (vo )2 -2g (y)

Only two of the above three equations are independent of one another.

Problems involving the motion of projectile can have at most three unknowns

since only three independent equations can be written.

- one equations in the horizontal direction.- two equations in the vertical direction.

Once vx and vy are obtained, the resultant velocity v which is always tangent to thepath.

See Example:

- 12.11

- 12.12

- 12.13

Exercise:

- 12.71

- 12.72

- 12.73

- 12.74

- 12.75


20/78


ZOL BAHRI RAZALI

20 |

12.7 Curvilinear motion: Normal and tangential components

When the path along which a particle is moving is known, it is convenient todescribe the motion using n and t components (normal and tangent) to the path, and at the

instant considered here their origin located at the particle.

Planer motion :-

( at instant considered )o - center of curvature.

s - radius of curvature.

taxis - tangent to the curve at P.

n-axis - perpendicular to the t- axis,directed from P towards the center of

curvature.

Positive direction , will be designatedby the unit vector, u n ( normal ) and u t

( tangent ).

Velocity :-

Since the particle moving , s is a function of time. The particle velocity v has a

direction that is always tangent to the path, and the magnitude that is determined bytaking the time derivative of the path function s = s(t) .

v =dt

ds

v = vu t

where v =.

s


21/78


ZOL BAHRI RAZALI

21 |

Acceleration :-

The acceleration of the particle is the time rate of the change of the velocity.

a =.

v =.

v u t + v.

u t

by formulation ,

.

u t =.

o u n =s

s.

u n =s

vu n

substitute to the above equation

a = a t u t + a n u n

where a t =.

v

or a t ds = v.dv

and a n =s

v2

magnitude of acceleration is the

positive value of a = 22 nt aa

Two special cases of motion :

1) The particle moves along a straight line , s .

oan , taa =.

v

The tangential components of acceleration represents the time rate of change in the

magnitude of the velocity.

2) The particle moves along a curve with a constant speed then

.,2

.

svaaova nt


22/78


ZOL BAHRI RAZALI

22 |

12.8 Relativemotion analysis of two particles using Translating Axes

Position :

The axes of this frame are only persuitted to

translate relative to the fixed frame . The

relative position of Bwith respect to A

is designate by a relative position vector

r AB .

rB

= rA

+ rAB

Velocity : An equation that related the velocities of the particle can be determined by

taking the time derivative.

vB

= vA

+ v AB

where v B =dt

drB

v A =dt

drA

v AB =dt

dr AB

v B and v A - refer to absolute velocities

-observed from the fixed frame.

v AB - relative velocity

-observed from the translating frame.

vB/AvB

vA


23/78


ZOL BAHRI RAZALI

23 |

Acceleration :

ABAB aaa

ABa is the acceleration ofB as seen by the observer located at A and translating with

the x,y,z reference frame.

aB/A

aA aB


24/78


ZOL BAHRI RAZALI

24 |

CHAPTER 13

KINEMATICS OF A PARTICLE

(FORCE & ACCELERATION)

Chapter ObjectivesTostate Newtons Law of Motion and Gravitational Attraction and to define mass

and weightTo analyze the accelerate motion of a particle using the equation of motion with

different coordinate systems.To investigate central force motion and apply it to problems in space mechanics

13.1 Newtons Law of Motion

Galileo (1590) - experiment to study the motion of pendulum and falling bodies.

- the effects of forces acting on bodies in motion.

Isaac Newton (1687)

o 1st

Law - A particle originally at rest, or moving in the straight linewith a constant velocity, will remain in this state provided

the particle is not subjected to an unbalanced force.

o 2nd LawA particle acted upon by an unbalanced force F,experiences an acceleration a that has the same direction as the

force and a magnitude that is directly proportional to the force.

o 3rd LawThe mutual forces of action and reaction between twoparticles are equal, opposite and collinear.

Note : - The unbalanced force acting on the particle is proportional to the time rate of

the change of the particles linear momentum.

Newtons 2nd law of motion relates the accelerated motion of a particle to the force

that act on it.

F = ma where m : mass of the particle.

( equation of motion )


25/78


ZOL BAHRI RAZALI

25 |

Since m is constant , we can also write F =dt

mvd )(, where mv is the particles

linear momentum.

Albert Einstein ( 1905 )developed the theory of relativity and place limitation on

on the use of Newtons second law.

~ Newtons Law of Gravitational Attraction ~

F = Gr

mm 21 where :

F force of attraction between two particles

G - universal constant of gravitation

G = 66.73 + 10

12

2

3

kgsm

m 1 ,m 2 - mass of each of the two particles

r - distance between centre.

~ Mass and weight ~

Property of matter by which we can compare the response of one body with that of

another.

w = m.g , wweight

by comparison F = ma , we term g the acceleration due to gravity.

m (kg)

a = g (m/s2)

W= m.g (N)

SI system

[W= m.g (N)]

[g = 9.81 2s

m ]

Weight w = kg 2s

m

= N


26/78


ZOL BAHRI RAZALI

26 |

13.2 The equation of motion

The resultant force FR = FF = am.

The resultant of these forces produce the vector am. , its magnitude and direction can be

represented on the kinetic diagram.

F2 F2

FR = FP ma

F1 PF1

a

F.B.D. Kinetic diagram


27/78


ZOL BAHRI RAZALI

27 |

13.3 Equation of motion for a system of particles

F i + F i = ii am ( F = ma )

where F i - resultant ext. force.

fi - resultant int. force.

ama = ii am F = ... gam

If r a is position vector which locates at the center of mass G.

SpringIf the particle is connected to an elastic spring.

F s = ks

s = oll

where : l deformed length

ol = unreformed length.

KinematicsIf the velocity are position of the particle is to be found, apply the

kinematics equation.

acceleration is a function of time.

dtdva ,

dtdsv

acceleration is a function of displacement : integrate ... dvvdsa

acceleration is constant , to determined the velocity or position of the particle.

- v = vo + act

- s = so + vot + 21 a ct

2

- v 2 = vo

2 + 2 ca ( s-s o )

In all cases, make sure the positive inertial coordinates directions.

miai

Kinetic diagram

Fi

fi

F.B.D.


28/78


ZOL BAHRI RAZALI

28 |

13.4 Equation of motion : rectangular coordinates

When the particle moving relative to a inertial x, y, z frame, the force actingmaybe expressed in term of their i, j, k. components.

F = am. F ix + F yj + F z k= )( kajaiam zyx

F x = xam. F

y= yam ..

F z = zam.

Procedure for analysis.

1) FBD :- Draw the FBD, it provides a graphical representation that accounts for all

the forces (F) which act on the particle.

2) Equation of motion :- use scalar or Cartesian vector analysis (3-D)for the solution.

a) Friction :

If the particle contacts the rough surface, it maybe necessary to use the frictional

Equation.

F f = .Nk where :

[ Ff

always act on the FBD such k

- coefficient of kinetic friction.

that it opposes the motion of the F f - magnitude of the friction

particle relative to the surface it N - Normal forces.

contacts ]

See Example:

- 13.1

- 13.2

- 13.3

- 13.4


29/78


ZOL BAHRI RAZALI

29 |

Exercise : 13.1

- 13.2

- 13.3

- 13.4

- 13.5

13.5 Equation of motion : Normal and tangential coordinates

When the particle moves over a curved path, the equation of motion for the

particle maybe written in the tangential, normal and binormal directions.

F = ma

F tu t + F n u n + F b u b = nt amam ..

where Ft,F

n, F

b- force components acting on the particle in the tangential, normal and

binormal.

There is no motion of the particle in the binormal direction, since the particle is

constrained to move along the path.

F tt am. dt

dvat ( time rate of change in the

magnitude of velocity )

F nn am.

F .ob

F t acts in the direction of motion, the particles speed will be increase. Opposite direction, the speed will slow down.

g

van

2

( time rate of change in the velocity direction )

vector always acts in the positiven direction ( toward the paths centre of

curvature), F n which cause na , also acts in this direction.


30/78


ZOL BAHRI RAZALI

30 |

Procedure for Analysis

The motion of a particle along a known curved path, normal and tangential coordinates

should be considered.

FBDDraw the FBD for the particle

Equation of motionas mention

Kinematics g

vands

dvvat

dtdvat

2

,.,

If the path is defined as y = f (x), the radius of curvature at thepoint where the particle located can be obtained:

= [1 + (dx

dy ) 2 ] 3/2 /d2y / dx2

See Example:

- 13.6

- 13.7

- 13.8

- 13.9

Exercise : 13.48

- 13.49

- 13.50

- 13.51

- 13.52


31/78


ZOL BAHRI RAZALI

31 |

CHAPTER 14

KINEMATICS OF A PARTICLE

(WORK AND ENERGY)

Chapter ObjectivesTo develop the principle of work and energy and apply it solve problems that

involve force, velocity and displacementTo study problems that involves power and efficiencyTo introduce the concept of a conservative force and apply the theorem of

conservation of energy to solve kinetics problems

14.1 The work of a force

A force F does work when the particle undergoes a displacement in the directionof the force.

Eg : particles moves along the paths, from

position r to r, dr = r r, where dr = ds. If

the angle , the work dU is done by F is ascalar quantity

dU = F.ds cos

Definition of the dot product dU = F.dr

dU = F.ds cos = F.dr

dU = 0 when F perpendiculars.

dr = ds cos

Work of a variable force.

If the particle undergoes finite displacement

along its path, s1

to s2

, work is determined

by integration.

u21 =

2

1

.s

rdrF

2

1

coss

sF .


32/78


ZOL BAHRI RAZALI

32 |

If ( F cos vs s ) is plotted, workingcomponents as the area under the curve,

from position s 1 to s 2 .

Work of constant force moving along a straight line

If the Fc

has a constant magnitude and acts at a constant angle from its straight line

path, so work done F c is

u21 = F c cos

2

1

s

sd s

= F c cos ( s 2 - s 1 ).

Work F c represents the area of the rectangle.

Work of weight

The particle moves along the path s, from s1

to s2

.At the immediate point, the

displacement dr = dx i + dy j + dz k.

Since w = -w j

u21 = drf.

= 2

1

)).((r

rdzkdyjdxiwj


33/78


ZOL BAHRI RAZALI

33 |

= 2

1

11 12

)( yywwdy

u21 = -w (y 2 - y1 )

This, work done is equal to the magnitude of

the particles weight times its vertical displacement.

Work of spring

The magnitude of force developed in a linear elastic spring when the spring is displaced a

distance s from its unstreched position is Fs = ks , where k is the spring stiffness. If the

spring is elongated or compressed from a position s1 to a further position s2 , the work

done on the spring by Fs is positive, since in each case, the force and displacement are inthe same direction.

U1-2 = 2

1

s

sFsds =

2

1

s

sksds

= .2

12

1 21

2

2ksks

When the particle is attach to a spring, then the force Fs exerted on the particle isopposite to that exerted on the spring.

The force will do negative work on the particle when the particle is moving so as afurther elongate ( or compress ) the spring.

U1-2 = - ( ).21

21 2

1

2

2ksks


34/78


ZOL BAHRI RAZALI

34 |

If both same directionpositive work .opposite directionnegative work.


35/78


ZOL BAHRI RAZALI

35 |

14.2 Principle work and energy

If the particle has a mass, m and is subjected to a system of external forces, represented

by the resultant FR = Ft = ma t.

Applying the kinematics equation:

dsdvvat .

2

1

s

stdsF =

2

1

.v

vmvdv

=2

1

2

2 21

21 mvmv

U21 = the sum of the work done by all the forces acting as the particle

moves from point 1 -2.

T = 22

1 mv T = particle final kinetic energy

mv2

1 2 = particle initial kinetic energy

T1 + U1-2 = T2

Note : i ) Ft = tam. , to obtain ta , integrate ta = .. dsdvv

ii ) F n = nam. cannot be used, since these force do no work on the particle onthe forces directed normal to the path.


36/78


ZOL BAHRI RAZALI

36 |

Procedure for analysis

The principle of work and energy is used to solve the kinetic problems that involve

velocity , force and displacement.

FBD : Draw FBD in order to account for all the forces.

Principle of w E :

T1 + U1-2 = T2

Kinetic energy at the initial / final point always positive T = 22

1 mv

A forces does work when it moves through its displacement in the direction of the

force.( +ve same direction )

Force that are functions of displacement must be integrate to obtain the work.

The work of a weightweight magnitude and the vertical displacement.

Uw = wy ( + ).

The work of spring , s = 21 ks

2.


37/78


ZOL BAHRI RAZALI

37 |

14.3 Principle of Work and Energy for a system of particles

thi particles

m i massF i resultant external forcefi resultant internal force

Which applies in the tangential direction, the principle of work and energy:

2

1

2

1

2

2

2

21)()(

21 i

i

i

i

s

s

s

siittii vmdsfidsFivm

For all of the thi particles.

2

1

2

1

2

2

2

1 21)()(

21 i

i

i

i

s

s

i

siittii vmdsfidsFivm

T1+ U 21 = T 2 systems finalkinetic energy

work done by all the

external or internal forces.

systems initial kinetic energy.


38/78


ZOL BAHRI RAZALI

38 |

Work of Friction Caused by Sliding.

The cases where a body is sliding over thesurface on another body in the presence of

friction.

Applied force P just balance the resultant

friction force ..Nk

22

21

21 mvNPsmv sk

is satisfied when P = k N

The sliding motion will generate heat, a form of energy which seems not to be accounted

for in the W&E equation.

See Example:

- 14.2

- 14.3

- 14.4

- 14.5

- 14.6

Quiz:- 14.114.41 (each student has to answer one question)

v v

P P

s

W

P

F = kN

N


39/78


ZOL BAHRI RAZALI

39 |

14.4 Power and Efficiency

Defined as the amount of work performance per unit of time.

P =dtdu where du = F.dr

=dt

drF.

= F.dt

dror P = F.V.

Power is a scalar, where in the formulation; V represents the velocity of the point which

is acted upon by the force, F.

I W a H = I sJ = I N. sm = I w

Note:

The term power provides the useful basis for determining the type of motor ormachine which is required to do a certain amount of work in a given time. For example,

two pumps may each be able to empty a reservoir if given enough time, however thepump having the larger power will be complete the job sooner.

The mechanical efficiency of a machine is defined as the ratioof the output of useful power produce by the machine to the

input of power supplied to the machine.

inputpower

outputpowerE

If the energy applied,inputenergy

outputenergyE

Since machines consist of a series of moving parts, frictional forces will always bedeveloped within the machine, and as result, extra energy or power is needed to

overcome these forces.

The efficiency of a machine is always less than 1

Power

Efficiency


40/78


ZOL BAHRI RAZALI

40 |


o Determine the external force F

o If accelerating , amF . o One F and V have been found, P = F.V = Fr cos

o In some problem F per unit time, P = dtdu

See Example:

- 14.7

- 14.8

Exercise

- 14.42

- 14.43

- 14.44

- 14.45

- 14.46


41/78


ZOL BAHRI RAZALI

41 |

14.5 Conservative forces and potential energy

Conservative force

When the work done by a force in moving a particle from one point to another isindependent of the path followed by the particle, the force is called a conservativeforce.

e.g. : - weight of the particle

: - the force of an elastic spring

weightdepends on the particles vertical displacement

spring depends only on the extension/compression.

in contrastforce of friction exerted on a moving objectsdepends on the

path/neoconservative.

Potential Energy

Energy - capacity for doing works.- from the motion of particlekinetic energy.

- from the position of particle ( fixed datum / reference ) - potential

energy

(potential energy due to gravity ( weight ) and elastic spring is important.)

Gravitational potential energy: Vg = Wy


42/78


ZOL BAHRI RAZALI

42 |

Elastic potential energy: V e =2

21

sk

[ V e always + ve, the spring has the

capacity for always doing positivework ] when the spring back tooutstretch position.

Potential Function

If a particle is subjected to both gravitational and elastic forces, - potential function.

V = V g + Ve

U21 = V 1 - V 2 work done by a conservative force.

E.g. : potential function for a particle ofweight W, suspended from a spring can be

expressed in term of its position, s ,

measured from a datum.

V = Vg + Ve

= - W(s) + 22

1 ks

If the particle moves from s1 to lower s2,

U )2

1()2

1(2

22

2

112121ksWsksWsVV

= )2

12

1()(2

1

2

212ksksssW


43/78


ZOL BAHRI RAZALI

43 |

14.6 Conservation of Energy

When the particle acted upon by a system of both conservative and neoconservative,

work done by conservative position.

222111 )( VTnonconsUVT

If noncons is zero =>2211

VTVT

Note : conservative force (not follow the path )weight / springneoconservative force (follow exactly the path )friction.

2211VTVT conservation of mechanical energy, conservation energy.

Note : during the motion, sum of potential and kinetic energies remains constant.

For this occur, kinetic energy must be transformed into potential energy andvice versa.

E.g. : The ball of weight w is dropped from a height, h above the ground (datum).

at initial position, mechanical energy

E = T1

+ V1

= )(

2

1 21 ge vvmv

= whmv 212

1

= 0 + wh

= .wh

When the ball has fallen a distance ,2

h its speed can be determined by using:

)(22

0

2

oc yyavv Therefore:

=)(2

oc yya

= )

2(2 hg 22 TVE

v = )2

(2 hg = 2)(2

1)2

( ghmhw

= gh2 = 2)(2

1)2

( ghg

whw

= .wh


44/78


ZOL BAHRI RAZALI

44 |

When the ball strikes the ground, P.E. (0) and

)(222

oco yyavv Total energy:

= ).(.2 hg

= gh2 33 TVE

= 2)2(2

10 gh

g

w

= .wh


45/78


ZOL BAHRI RAZALI

45 |

CHAPTER 15

KINETICS OF A PARTICLE:

IMPULSE & MOMENTUM

Chapter ObjectivesTo develop the principle of linear impulse and momentum for a particleTo study the conservation of linear momentum for particlesTo analyze the mechanics of impactTo introduce the concept of angular impulse and momentumTo solved problems involving steady fluid streams and propulsion with variable

mass

15.1 Principle of Linear Impulse & Momentum

Equation of motion for a particle of mass m:

dt

dvmmaF

Rearranging:

2

1

1

2

t

t

v

v dvmdtF

or

2

1

12

t

t

mvmvdtF Eq. 1

Linear Momentum

principle of linear impulse &

momentum(time integration of the

equation of motion)

linear particlesimpulse linear

momentum


46/78


ZOL BAHRI RAZALI

46 |

Vector of L = mv in Eq. 1 is referred to as the particles linear momentum Magnitude of mv units of mass-velocity

Integral I = F dt in Eq. 1 is referred to as the linear impulse Measures the effect of a force during the time the force acts. Magnitude: force-time If forceexpressed as a function of time, impulse = direct evaluation of the integral If forceconstant direction during the time period t1 to t2, impulse = area under the

curve of force vs. time:

2

1

t

t

dtFI

If forceconstant in magnitude & direction:

)( 122

1

ttFdtFI c

t

t

c shaded rectangular area

21

2

1

mvdtFmv

t

t

Initial momentum of the particle at t1 plus the vector sumof all the impulses applied to the particle during the time

interval t1 to t2 is equivalent to the final momentum of the

particle at t2.

Linear Impulse

Principle of Linear Impulse & Momentum


47/78


ZOL BAHRI RAZALI

47 |

The scalar x, y, z components of the previous equation are:

21

21

21

)()(

)()(

)()(

2

1

2

1

2

1

z

t

t

zz

y

t

t

yy

x

t

t

xx

vmdtFvm

vmdtFvm

vmdtFvm

To solve a linear impulse and momentum:

a) FBD - establish the x, y, z inertial frame of reference

- draw the particles FBD account for all the forces that produce impulses on

the particle

- establishdirection & sense of the particles initial & final velocities

b) Principle of Impulse & Momentum

- apply the principle:21

2

1

mvdtFmv

t

t

- if motion occurs on x-y planethe 2 scalar component equation can be formulated

by:

1. resolving the vector components of F from FBD

2. using the data on the impulse and momentum diagrams

Scalar Equation


See Example:

15.1 15.2 15.3


48/78


ZOL BAHRI RAZALI

48 |

15.2 Principle of Linear Impulse & Momentum for a System of Particles

dt

dvmF iii

Multiplying both sides by dt and integrating between the limits t = t 1, vi = (vi)1 &t = t2, vi = (vi)2,

21 )()(2

ii

t

t

iii vmdtFvm

The initial linear momenta of the system

added vectorially to the impulses of allthe external forces acting on the system

during the time period t1 to t2 are equalto the systems final linear momenta.

Location of mass centre G of the system:

iiiG mmrmrm ,

Taking the time derivatives:

iiG vmvm

Total linear momentum of the system of particles plus (vectorially)

the external impulses acting on the system of particles during the

time interval t1 to t2 equal to the aggregate particles final linear

momentum.

Sum of

external

forces


49/78


ZOL BAHRI RAZALI

49 |

15.3 Conservation of Linear Momentum for a System of Particles

When the sum of the external impulses acting on a system of particles is zero:

21 )()( iiii vmvm

In other form:

21)()( GG vv

Applied when particles collide or interact.

Nonimpulsive forces: - causing negligible impulses- including any force that is very small compared to otherlarger (impulsive) forces

Impulsive forces: - forces which are very large & act for a very short periodof time; produce a significant change in momentum

- normally occur due to an explosion or the striking of one

body against another

conservation of linear momentum

See Example:

15.4 15.5 15.6 15.7 15.8

Exercise:

15.32 15.33 15.34 15.35 15.36


50/78


ZOL BAHRI RAZALI

50 |

15.4 Impact

Impact occurs when two bodies collide with each other during a very short periodof time causing relatively large (impulsive) forces to be exerted between twobodies.

E.g.: striking of a hammer on a rail golf club on a ball

Central impactthe direction of motion ofthe mass centers of the two collidingparticles is along a line passing through the

mass centers of the particles

Oblique impactwhen the motion of one orboth of the particles is at angle with the lineof impact

Particle have the initial momenta, (vA)1 > (vB)1 During the collisionthe particles will undergo a period of deformationequal but

opposite deformation impulse P dt At maximum deformationboth particles move with common velocity v Then a period of restitution occurs (particles will either return to original shape or

remain deform)restitution impulse R dt pushes particles apart from another,where P dt > R dt

Just after separation, particles final momenta = (vB)2 > (vA)2

Analysis of Central Impact


51/78


ZOL BAHRI RAZALI

51 |

To determine the velocities, apply conservation of momentum for the system of

particles by referring to the figure above:

+ mA (vA)1 + mB (vB)1 = mA (vA)2 + mB (vB)2

To solve for final velocities (initial values of particles will be known in mostcases), consider deformation phase:

+ mA (vA)1 - P dt = mA v

For restitution phase:

+ mA (vA)1 - R dt = mA ( vA)2

Ratio of the restitution impulse to the deformation impulse = coefficient ofrestitution, e:

vv

vv

dtP

dtRe

A

A

1

2

)(

)(

For particle B:

1

2

)(

)(

B

B

vvvv

dtP

dtR

e

If remove unknown v:

11

22

)()(

)()(

BA

AB

vv

vve

+


52/78


ZOL BAHRI RAZALI

52 |

In general, value of e = between zero and one:

1. Elastic Impact (e = 1): If the collision between the two particles is perfectlyelastic, deformation impulse P dt = equal and opposite to restitution impulseR dt

2. Plastic Impact (e = 0): No restitution impulse given to the particles (R dt = 0);after collision, both particles couple or stick together and move with common

velocity.

Use the following two equations:1. The conservation of momentum applies to the system of particles,

mv1 = mv22. The coefficient of restitution, e = [(vB)2(vA)2] / [(vA)1(vB)1], relates the

relative velocities of the particles along the line of impact, just before andjust after collision.

In determining (vAx)2, (vAy)2, (vBx)2 and (vBy)2,consider these four equations:

1. Momentum of the system is conserved along

the line of impact, x axis, so that m(vx)1 =m(vx)2

2. The coefficient of restitution, e = [(vBx)2(vAx)2] / [(vAx)1(vBx)1], relates the relative

velocity components of the particles along theline of impact (x axis)

3. Momentum of particle A is conserved alongthe y axis, perpendicular to the line of impact,

since no impulse acts on particle A in thisdirection.

4. Momentum of particle B is conserved alongthe y axis, perpendicular to the line of impact,

since no impulse acts on particle B in thisdirection.

Coefficient of Restitution


(Central Impact)


(Oblique Impact)


53/78


ZOL BAHRI RAZALI

53 |

See Example:

15.9 15.10 15.11

15.5 Angular Momentum

Angular momentum of a particle about point O = moment of the particleslinear momentum about O.

Also being referred to as the moment of momentum.

If a particle is moving along a curve lying in thex-y plane, magnitude ofHo:

(Ho)z = (d)(mv)

Common unit = kg m2/ s Directionright-hand rule

If the particle is moving along a space curve, angular momentum Ho:

Ho = rx mv

In Cartesian components:

zyx

zyx

mvmvmv

rrr

kji

Ho

Scalar Formulation

Vector Formulation


54/78


ZOL BAHRI RAZALI

54 |

CHAPTER 16

PLANAR KINEMATICS OF

A RIGID BODY

Chapter ObjectivesTo classify the various types of rigid-body planar motionTo investigate rigid-body translation and show how to analyze motion about a

fixed axisTo study planar motion using an absolute motion analysisTo provide relative motion analysis of velocity and acceleration using a translating

frame of referenceTo show how to find the instantaneous center of zero velocity and determine the

velocity of a point on a body using this methodTo provide a relative motion analysis of velocity and acceleration using a rotating

frame of reference

16.1 Rigid-Body Motion

Particles of a rigid body move along paths equidistant from a fixed plane Has 3 types:

1. Translation- every line segment on the body remains parallel to its original direction during the

motion- rectilinear translation: path of motionalong equidistant straight lines- curvilinear translation: path of motionalong curved lines which are equidistant

Planar motion


55/78


ZOL BAHRI RAZALI

55 |

2. Rotation about a fixed axis- all particles of the body (except those lie on the

axis of rotation) move along circular paths

3. General plane motion- undergoes a combination of translation

and rotation

16.2 Translation

Position:- location of pointsA andBdefined from fixedx, y reference frameusing

position vectors rA and rB

- x,ycoordinate systemfixed in the body where origin =A (base point)- position ofB with respect toA = relative position vector rB/A (r ofB with respect to

A)


56/78


ZOL BAHRI RAZALI

56 |

- vector addition: rB = rA + rB/A

Velocity:- relationship between instantaneous velocities of A and Bobtained by taking the

time derivative of the position equation:

vB = vA + drB/A/dt

Since drB/A/dt= 0 due to the magnitude ofrB/A = constant, and vB = vA = absolute

velocities,

vB = vA

Acceleration:- time derivative of velocity equation:

aB = aA

- velocity and acceleration equation indicates that all points in a rigid bodysubjected to either rectilinear or curvilinear translation move with the same velocityand acceleration.

16.3 Rotation About a Fixed Axis

Angular motion

- only lines or bodies undergo angular motion- angular motion of a radial line r located within the shaded plane and directed from

point O on the axis of rotation to point P

1. Angular position

- angular position or r = defined by angle - measured between a fixed reference line and r

2. Angular displacement

- defined by the change in the angular position,

measured as a differential d- has a magnitude of d, measured in degrees, radians

or revolutions, where 1 rev = 2 rad.


57/78


ZOL BAHRI RAZALI

57 |

- Since motion is about a fixed axis, direction of dwhich always along the axis

- Directiondetermine by the right hand rule

3. Angular velocity

- defined as the time rate of change in angular position,

, where = d / dt +

- has a magnitude measured in rad/s- directionalways along the axis of rotation where

the sense of rotation being referred as clockwise or

counterclockwise- arbitrarily chosen counterclockwise as positive

4. Angular acceleration

- measures the time rate of change of the angular velocity

- magnitude: = d / dt or = d2 / dt2 +

- directiondepends on whether is increasing or decreasing- e.g.: if is decreasing, = angular deceleration, directionopposite to

- by eliminating dt from the above equation,

d = d +

5. Constant angular acceleration

- when angular acceleration of the body is constant,

= c

+ = o + c t

+ = o + o t + c t2

+ 2 = o2 + 2c ( - o)

where o = initial angular positiono = initial angular velocity


58/78


ZOL BAHRI RAZALI

58 |

Motion of point P- as rigid body rotates, point P travels along a circular

path of radius r and center at point O.

1. Position- defined by the position vector r, which extends from

O to P

2. Velocity

- has a magnitude of rvrvr ,

- since r= constant, rvvrvr ,0

- since rv ,

- direction ofv = tangent to the circular path- magnitude and direction of vaccounted from:

prv

where rp: directed from any point on the axis of rotation

to point P

- to establish the direction of vright hand rule- by referring to the figure,

since

rv

rvrr

rv

p

p

,sin

sin

3. Acceleration

- can be expressed in terms of its normal and tangential components:

where

rara

dtdrvr

vadtdva

nt

nt

2

2

,

,/,,

/,/

- tangential componentsrepresents the time rate of change in the velocitysmagnitude

- normal componenttime rate of change in the velocitys direction


59/78


ZOL BAHRI RAZALI

59 |

- acceleration in terms of vector cross product:

since

)(

&

::

pp

p

p

p

p

rra

rvdt

dr

dt

d

dt

drr

dt

d

dt

dva

- by referring to the next figure, rra pt sin

- applying right hand rule yields pr in the direction

ofat

- hence obtainrr

aaa nt2

- magnitude:22

tn aaa

Procedure for Analysis:

To determine velocity and acceleration of a point located on a rigid body that is rotating

about a fixed axis:

a) Angular Motion1. Establish positive sense of direction along the axis of rotation and show it

alongside each kinematics equation as it is applied.2. If a relationship is known between any two of the 4 variables , , and t, then a

third variable can be obtained by using one of the following kinematics equation

which relates all 3 variables:

dddt

d

dt

d ,,

3. For constant angular acceleration, use:

)(20

2

0

2

2

2

100

0

c

c

c

tt

t

4. ,, - determine from algebraic signs of numerical quantities.

b) Motion of P1. Velocity of P and components of acceleration can be determine from:

ra

ra

rv

n

t

2


60/78


ZOL BAHRI RAZALI

60 |

2. If geometry of problem is different to visualize, use:

rra

rra

rrv

pn

pt

p

2)(

Note:- rpdirected from any point on the axis of rotation to point P- rlies in the plane of motion P- vectorsexpressed in terms of its i, j, k components.

See Example 16.1 and 16.2.

16.5 Relative-Motion Analysis: Velocity

General motion: combination of translation androtation

To view motions separatelyuse relative-motion analysis, involving 2 sets of coordinateaxes

Fixed referencemeasures the absolute positionof 2 points A & B on the body

Translating referencedo not rotate with thebody; only allowed to translate with respect to

the fixed frame; originattached to the selected

base point A

Position vector rAspecifies the location of base point A Relative position rB/Alocates point B with respect to point A by vector addition, position of B: rB = rA + rB/A

Points A & Bundergo displacements drA & drB during an instant of time dt Consider general plane motion by its component parts:

- entire bodytranslates by drAA moves to its final position and B to B

Position

Displacement


61/78


ZOL BAHRI RAZALI

61 |

- rotated about A by d - B moves to its final position (relative displacementdrB/A)

- displacement of B:

ABAB drdrdr /

to determine the relationship between the velocities of points A and B take thetime derivative (divide displacement equation by dt):

dt

dr

dt

dr

dt

dr ABAB /

ABAB vvv /

absolute

velocities of

points A & B

Velocity

due to rotation about A

due to translation of A

due to translation & rotation

relative

velocity

vB/A

relative velocity of B with

respect to A

velocity of base point A

velocity of point B


62/78


ZOL BAHRI RAZALI

62 |

since vB/A also representing the effect of circular motion about A:

ABAB

ABAB

rvv

rv

/

//

A) Vector Analysis

1. Kinematics Diagram

Establish the directions of the fixed x,y coordinates and draw a kinematicsdiagram of the body

Indicate vA, vB, , rB/A If magnitudes of vA, vB or are unknown, the sense of direction can be

assumed

2. Velocity Equation

To apply ABAB rvv / , express the vectors in Cartesian vector form and

substitute them into the equation.

Evaluate the cross product and then equate the i andj components to obtaintwo scalar equations.

If negative answer obtained for an unknown magnitude, direction of vectoropposite to that shown on the kinematics diagram.

B) Scalar Analysis

1. Kinematics Diagram

Draw a kinematics diagram to show the relative motion Consider body to be pinned momentarily at base point A, magnitude: vB/A =

rB/A Direction of vB/Aestablished from the diagram

relative-position vector drawn

from A to B

angular velocity of the body

velocity of base point A

velocity of point B



63/78


ZOL BAHRI RAZALI

63 |

2. Velocity Equation

From equation vB = vA + vB/A, represent each vectors graphically by showingmagnitudes and directions.

Scalar equationdetermine from x & y components of these vectors.

See Example:

Exercise

16.4 16.37 16.58 16.63

16.6 16.7 16.8 16.9


64/78


ZOL BAHRI RAZALI

64 |

CHAPTER 17

KINETICS OF A RIGID BODY

(FORCE AND ACCELERATION)

Chapter ObjectivesTo introduce the methods used to determine the mass moment of inertia of a bodyTo develop the planar kinetic equation of motion for a symmetric rigid bodyTo discuss applications of these equations to bodies undergoing translation,

rotation about a fixed axis and general plane motion

17.1 Moment of Inertia

A body has a definite size and shape.F = m.a (mass is a measure of the body resistance to acceleration)

Rotational aspect, caused by moment, M

M =I where Imoment inertia- moment inertia is a measure of the resistance of a body to

angular acceleration (M =I)

Moment inertiaas the integral of the second moment about an axis of all the element ofmass, dm which compose the body.

dmrIm

2

moment arm, r perpendicular distance from z axis Value of I, different for each axis If material having variable density, , which dm = dV,

dVrIV

2

When being a constant,

dVrIV

2


65/78


ZOL BAHRI RAZALI

65 |

Procedure for analysis.

For integration, consider only symmetric bodies having surface which are generated by

revolving a curve about an axis.

Shell element Dish element

- height, z, radius, r = y radius, y

- thickness, dy - thickness, dz- dV = (2y) (z) dy - dv = (y2)dz

See Example:

- 17.1- 17.2

ParallelAxis Theorem

If the moment of inertia of the body about an axis passing through the bodys mess centre

is known, then moment of inertia about any other parallel axis may be determined byparallel axis theorem.


66/78


ZOL BAHRI RAZALI

66 |

Using Phythagorean theorem,

r2 = (d+x)

2 +y2

Hence, moment of inertia,

m m m

mm

dmddmxddmyxI

dmxddmrI

222

22

'2)''(

])'[(

WhereIGmoment inertia about z axis passing through the mass center, G

mmass of the body

dperpendicular distance.

Radius of Gyration

Moment of inertia of a body about a specified axis, using the radius of gyration, k.

m

IkmkI ,2

Similarity betweenk

&r, from

dI=

r2

dm, moment of inertia of an elemental mass, dm ofthe body about an axis.

Composite bodies

The body of constructed of a number of simple shape such as disk, spheres and rods, the

moment of inertia of the body about any axis, z can be determined by adding

algebraically the moment of inertia of all the composite shape.

)( 2mdII G

See Example:

- 17.3- 17.4

IG zero through total mass, m

(since r2

= x2+y

2) mass

center


67/78


ZOL BAHRI RAZALI

67 |

17.2 Planar Kinetic Equations of Motion

Limit of studyplanar kinetic to rigid bodies, which along with their loadings, are

considered to be symmetrical.

The inertial frame of reference x, y, z has its

origin coincident with the arbitrary point P inthe body. By definition, these axes do not rotate

and are either fixed or translate with constant

velocity.

Equation of translation motion.

The external forces represent the effect of gravitation electrical, magnetic or contact forcebetween adjacent bodies.

The analysis of a system of particles:

GmaF

(the translation equation of motion far the mass centre of a rigid body)

The sum of all the external forces acting on the body is equal to the bodys mass timesthe acceleration of its mass centre, G

For x-y plane,

yGy

xGx

amF

amF

)(

)(

Equation of rotational motion

Particle FBD Particle kinetic diagram


68/78


ZOL BAHRI RAZALI

68 |

where : Firesultant external force

firesultant internal force far I particle cause by interactions with adjacentparticle

mimass of particle

aiinstant acceleration

If moments of the force acting on the particle are summed about point P:

iiii amrfrFr

or iiiP amrM )(

The moment about P can be expressed in term of acceleration of point P.

If the body has an angular acceleration and angular velocity w,

)()(2rrarmM piiP

)()([

2

rrrrarm pi

Cross product operation with Cartesian component,

)]}([)(])()[(){()( yjxikyjxijaiayjximkM yPxPiiP

kyxaxaym yPxPi ])()([22

])()([)(2raxaymM yPxPiip

Letting mi dm,

))()())((2

dmraxdmaydmMm

yP

m

xP

m

p

MPrepresent only the moment of the external forces acting on the body about P.


69/78


ZOL BAHRI RAZALI

69 |

Note: the resultant of moment of the internal force is zero, since for the entire body, these

forces occur in equal and opposite collinear, thus the moment of each pairs of forcesabout P cancels.

The integral of first and second term are wed to locate the bodys centre mass G

Since y dm = m and x dm =x m, and if point P coincides with the mass centre a forthe body,x == 0,

Gp IM

This rotational equation of motion state that the sum of the moment of all the external

forces computed about the bodys mass center G, is equal to the product of moment of

inertia of the body about an axis passing through G and the bodys angular acceleration.

PkpM )(

where kkinetics momentWhen moments of the external force shown, on the free body diagram are summed about

point P, they are equivalent to the sum of the `kinetic-moments of the component ofmaG

about P plus the kinetic moment ofIG.

Equation of motion:

GG

yGy

xGx

IM

amF

amF

)(

)(

or PkpM )(


70/78


ZOL BAHRI RAZALI

70 |

17.3 Equation of Motion :Translation

When the rigid body undergoes a translation, all the particles of the body have the sameacceleration,

aG = a

= 0

Rotational equation of motion applied at point G,

MG = 0

Rectilinear translation:All the particles of the body travel along parallel straight line path.

SinceIG = 0, only maG on the kinetic diagram:

0

)(

)(

G

yGy

xGx

M

amF

amF


71/78


ZOL BAHRI RAZALI

71 |

CHAPTER 18

PLANAR KINEMATICS OF

A RIGID BODY:

WORK AND ENERGY

Chapter ObjectivesTo develop formulation for the kinetic energy of a body and define the various

ways a force and couple do workTo apply the principle of work and energy to solve rigid-body planar kinetics

problems that involve force, velocity and displacementTo show how the conservation of energy can be used to solve rigid-body planar

kinetic problems

18.1 Kinetic Energy

Consider the rigid body shownwith an arbitraryith particle of the body, having a mass dm, is

located at r from the arbitrary point P.

If at the instant shown the particle has a velocity vi,then the particles kinetic energy, Ti = dm vi

2

Kinetic energy of the entire body:

m

ivdmT2

2

1

In terms of velocity of point P,

jxvijv

yjxikjviv

vvv

ypxp

ypxp

pipi

])[(])[(

)()()(

/

Square of magnitude of vi:

222

222222

222

)(2)(2

)(2)()(2)(

])[(])[(

rxvyvv

xxvvyyvv

xvyvvvv

ypxpp

ypypxpxp

ypxpiii


72/78


ZOL BAHRI RAZALI

72 |

Substitute the equation of K.E.:

mm m

ypxp

m

p dmrdmxvdmyvvdmT )()()()()()(22

212

21

2

212

21 )()( Pypxpp ImxvmyvmvT

For ,0 yx 2

2

12

2

1

GG ImvT

For a body which having either rectilinear orcurvilinear translation, = 0,

2

21

GmvT

Body has both translation & rotational kinetic energy:

2

2

12

2

1

GG ImvT

Note that vG = rG,22

21 )( GG ImrT

entire

mass m

of the

body

bodys center of mass G with

respect to P

bodys

moment of

inertia IP

Translation

Rotation about a

fixed axis

moment of

inertia, Io


73/78


ZOL BAHRI RAZALI

73 |

2

021 IT

Kinetic Energy: 2212

21 GG ImvT

18.2 The Work of a Force

a) Work of a Variable Force

s

F dsFU cos

where:

UFwork done by force

Fexternal force

spath

- angle between the tails ofthe forcevector & the differential

displacement

b) Work of a Constant Force

sFU CFC )cos(

where:

UFCwork done by forceFCexternal force

Stranslation

- angle of direction

General Plane

Motion


74/78


ZOL BAHRI RAZALI

74 |

c) Work of a Weight

- undergoes a vertical displacement y

- upward directionnegative work(weight and displacementoppositedirections)

yWUW

d) Work of a Spring Force

)(2

1212

221 skskUS

for12

ss

where:ks = Fs = spring force

s1 = initial compression position

s2 = further position

e) Forces That Do No Work

- act at a fixed points on the body orhaving direction perpendicular to the

displacement- refer to figurerolling resistance force

Frdoes no work since acting on a

round body as it rolls without slipping

over a rough surface

- due to Fracts at a point which has zerovelocity (instantaneous center, IC)during any instant of time dt.

- work of Fr = 0

See Example 18.1.


75/78


ZOL BAHRI RAZALI

75 |

18.3 The Work of a Couple

when body translatespositive work of one force

cancels the negative work of the other body undergoes different rotation d about an axis,

each force undergoes displacement ds = (r/2) d total work done:

dM

dFr

dFdFdU rrM

)(

)()(22

when body rotates through finite angle (rad), work

of a couple:

2

1

dMUM

for constant magnitude,

UM= M (2 - 1)

work is positive for M and (2 - 1) are having thesame direction.

See Example 18.2.

18.4 Principle of Work and Energy

2211 TUT

bodys initial translational & work done by all the external bodys final

rotational kinetic energy forces & couple moments translational &

rotational kinetic energy


76/78


ZOL BAHRI RAZALI

76 |

Work of a weight Magnitude of work Vertical displacement

stretch / compression of

spring

spring stiffness


Kinetic Energy (Kinematic Diagrams)

- translation -2

21 GmvT - rotation - 2

21 GIT

- special case (rotation about a fixed axis) - 202

1 IT

- use kinematic diagramdetermine vG, and relationship between them

Work (Free-Body Diagram - FBD)

- draw FBDcount for all forces and couple moments- integrate forcesobtain work

- graphicallywork = area under force-displacement curve- work of a weight:

UW = W y

- work of spring:

US = k s2

- work of coupleproduct of couple moment and angle (rad) through whichit rotates

- workpositive when force (couple moment)same direction asdisplacement (rotation)

Principle of Work & Energy

2211 TUT

bodys initial translational & work done by all the external bodys final

rotational kinetic energy forces & couple moments translational &

rotational kinetic energy


77/78


ZOL BAHRI RAZALI

77 |

See Example 18.2.

See Example:

- 18.3

- 18.4

- 18.5

- 18.6

18.5 Conservation of Energy

Determine by knowing the height ofthe bodys center of gravity

Vg = W yG P.E.positive when yG = positive

Ve = + k s2

In deformed position, the spring forceacting on the body always has thecapacity for doing positive work when

spring is returned back to its original

undeformed position

GravitationalPotential Energy

Elastic

Potential Energy


78/78


ZOL BAHRI RAZALI

For body subjected to both gravitational and elastic forces,

Total P.E., V = Vg +Ve

or

Principle of work energy:

222111 )( VTUVT noncons

For ,0)( 21 nonconsU

2211VTVT conservation of mechanical energy


Potential Energy

- draw initial and final position diagram

- if center of gravity, G, performing a vertical displacement, establish a

horizontal datum to measure Vg.

- use V = Vg +Ve, where Vg = W yG (+/-),

Ve = + k s2

Kinetic Energy

- translation -2

21

GmvT

- rotation - 221 GIT

Conservation of

Energy

Documents

Eng. Dynamics as Easy as 123