Dr Yu(Roger) Dong

Department of Mechanical Engineering

Email: Y.Dong@curtin.edu.au

Engineering Mechanics 100

DYNAMICSDYNAMICSDYNAMICSDYNAMICS

L1-2

Basic Concepts and Rectilinear Motion

of Particles

Why Dynamics So Important?

Statics VS. Dynamics

Bodies at REST Bodies in MOTION

Dynamics of Solid Bodies

Dynamics of Gasses/Air

Dynamics of Liquids

e.g. Hydrodynamicse.g. Robotics

e.g. Aerodynamics

1.0 Introduction

Mechanics

Mechanics: The action and effects of forces on bodies

Statics

Dynamics

Bodies at rest , or in equilibrium

Bodies in motion , or out of equilibrium

In Equilibrium Be static or move with constant velocity

v=0 mv=0.2 m/s

m

Static Move with v=constant

Out of Equilibrium Accelerate with the change of velocity

θ

a=0.5 m/s2m

Chaos Pendulum Video

Dynamics

Kinematics

Kinetics

Study of motion without reference to the forces producing motion: Relations applied only between position , velocity , acceleration and time

Relation between unbalanced forces and the change in motion they produce

A

B

va

Kinematics : e.g. Motion of rocket from position A to B

Kinetics :e.g. Motion of pendulum

ball applied by F

F

θ

�Kinematics : how fast, how far and how long

the motion takes

�Kinetics : What forces were involved to

produce the motion?

- Weight- Friction- Tension- Spring Force- Support Force

How about the resulting acceleration?

1.1 Basic Concepts

1) Vectors (magnitude +direction) and Scalars (magnitude only)

position – “ s” (or “ s”) distance –“ s”

velocity – “ v” (or “ v”) speed – “ v”

force – “ F” (or “ F”) moment – “ M”

2) Space: Geometric region occupied by bodies which is used to

determine the position relative to the reference system.

Time: Measure of the succession of events with an absolute

quantity in Newtonian mechanics

Mass: Quantitative measure of the inertia

Force : Vector action of one body to another

3) SI Units

Basic units

1.1 Basic Concepts (cont’)

4) Particles : - a body of negligible dimensions- a body with dimensions irrelevant to the motion or

the action of forces upon it

A

B

A

BEquivalent Particle

=5) Rigid Body : - important overall dimensions of the body or changes

in position of the body - negligible deformation (change in shape) of the body

6) Flexible Body : - deformed body under loads - beyond the scope of this course and about to cover in Year 2 Strength of Materials 232

Negligible spring deformation

=Rigid body

1.2 Newton’s Laws of Motion

These are fundamental laws relating forces and motion .

Law I. A particle remains at rest or continues to move in a straight line with a constant velocity if there is no unbalanced force acting on it.

Sir Isaac Newton(1643-1727)

Law II. The acceleration of a particle is proportional to the resultant force acting on it and is in the direction of this force.

∑F=0 In equilibrium

∑F=ma Out of equilibrium

Law III. The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear.

F= F'F F'

Laws I and II are strictly true only in an absolute frame of refe rence (i.e. A particle does not accelerate for Law I and does not rotate for Law II )

Law II (Most commonly used in dynamics)

∑F=ma

Where ∑F: resultant force acting on a body (vector)

m: mass of the body (scalar)

a: the resulting acceleration of the body (vector)

∑F

m=∑F=F1+F2+F3+F4+…..Fn-1+Fn=ma

F1Fn

F2

F3F4

Fn-1

m

…..

This equation relates applied forces (∑F) to the motion of a body (a).

1.3 Topics to Cover

2• Kinematics of angular motion• Kinetics of rigid bodies, translation motion of

rigid bodies• Fixed axis motion of rigid bodies

7. Plane Kinematics and kinetics of Rigid Bodies

2• Work down by forces, power• Principle of work and kinetic energy• Kinetic energy, potential energy,

gravitational/elastic potential energy• Work and energy equation• Energy conservation

6. Principle of Work and Energy

1• Principle of linear impulse and momentum• Conservation of linear momentum

5. Impulse and Momentum

2• Application of equation of motion ( ∑∑∑∑F=ma)

using x-y coordinate system• Application of equation of motion ( ∑∑∑∑F=ma)

using n-t coordinate system

4. Kinetics of Particles: Force and Acceleration

1• Relative motion analysis of two particles using

translating axes3. Relative Motion of Particles

2• Projectile motion in x-y rectangular coordinate

system• Curvilinear motion in n-t coordinate system

2. Plane Curvilinear Motion of Particles

2• Introduction to dynamics• Kinematic equations, derivation and

application• Linear motion with variable acceleration

1. Basic Concepts and RectilinearMotion of Particles

Num. LecturesTopic CoveredSection

1.4 Reference Books and Online Materials

��J.L. Meriam, L.G. Kraige (2006), J.L. Meriam, L.G. Kraige (2006), Engineering Mechanics: Dynamics, Engineering Mechanics: Dynamics, SI Version, 6th ed., Wiley.SI Version, 6th ed., Wiley.

��R.C. Hibbeler (2010), Engineering R.C. Hibbeler (2010), Engineering Mechanics: Dynamics,12th ed., Mechanics: Dynamics,12th ed., Pearson.Pearson.

Lecture Notes

Lecture Slides

Tutorials Problems/Solutions

Lab Sheets

iLecturesLogin from OASIS

(www.oasis.curtin.edu.au)

2. Rectilinear Kinematics: Continuous Motion

2.1 Introduction

Kinematics Examples

Motion of Gears Flight Trajectory of Rocket Car Route Planning

Kinematics ─ “Geometry of motion” to describe the motion without reference

to the forces producing it.

─ Only relations between position, velocity, accelera tion and time.

Motion can be constrained (forced to follow a specific path: e.g. car trip, t rain

on tracks)

or unconstrained (can move in any direction: e.g. aircraft flight pa th, trajectory

of a ball after it is thrown)

Tennis Ball BouncingUnconstrained Motion

Train Running on TracksConstrained Motion

2-D Coordinate Systems to Describe Motion:

• Rectangular coordinate (x, y)

• Polar coordinate (r, θ)

• Normal (perpendicular) and Tangent (along the path) coordinates

y P

x

r

θ

t

n

2.2 General Notions for Velocity and Acceleration

Consider a particle moving along a straight line. L et point O be the origin, or reference point.

-s +so P'P

s ∆s

t t+∆t

Displacement of the Particle

(((( )))) ssssss pp∆∆∆∆====−−−−∆∆∆∆++++====−−−−'

(((( ))))(((( )))) t

stttsss

tt

ssv

pp

ppave ∆∆∆∆

∆∆∆∆====−−−−∆∆∆∆++++−−−−∆∆∆∆++++====

−−−−

−−−−====

'

'

Average Velocity of the Particle

Instantaneous Velocity of the

Particle

Limit of vave as ∆t→0

dtds

ts

vt

====

∆∆∆∆∆∆∆∆====

→→→→∆∆∆∆ 0lim or s

dtds

v &======== 1

-s +so P'P

v v+∆v

(((( ))))(((( )))) t

vtttvvv

tt

vva

pp

ppave ∆∆∆∆

∆∆∆∆====−−−−∆∆∆∆++++−−−−∆∆∆∆++++====

−−−−

−−−−====

'

'

vdtdv

tv

at

&========

∆∆∆∆∆∆∆∆====

→→→→∆∆∆∆ 0lim

dsdv

vdtds

dsdv

dtdv

a ============

Average Acceleration of the Particle

Instantaneous Acceleration of the

Particle

Limit of aave as ∆t→0

Alternatively, acceleration and velocity can be related based on the chain rule

or sdt

sddtdtds

d

dtdv

a &&========

======== 2

2

3 or adsvdv==== or dsssds &&&& ====

2

2.3 Rectilinear Motion with Constant Acceleration

vdtds ====

adtdv ====

adsvdv====

1

2

3

(v-s-t)

(a-v-t)

(a-v-s)

Integration Limits

Initial Conditions (Lower Limit) & Final Con ditions (Upper Limit)

Time= t 0 t

Position= s 0 s

Velocity= v 0 v

(a) Velocity as a Function of Time (v=v(t))

adtdv ====2 ∫∫∫∫∫∫∫∫∫∫∫∫ ========

t

t

t

t

v

vdtaadtdv

000

[[[[ ]]]] [[[[ ]]]]tt

vv tav

00====

)( 00 ttavv −−−−====−−−−

(b) Position as a Function of Time (s=s(t))

vdtds ====1 [[[[ ]]]]dtttavvdtds

t

t

t

t

s

s ∫∫∫∫∫∫∫∫∫∫∫∫ −−−−++++========000

)( 00

20000 )(

21

)( ttattvss −−−−++++−−−−====−−−−

(c) Velocity as a Function of Position (v=v(s))

adsvdv====3 ∫∫∫∫∫∫∫∫ ====s

s

v

vadsvdv

00

)(2 020

2 ssavv −−−−====−−−−

In special cases

atv ==== 2

21

ats ==== asv 22 ====t0=0, s0=0 and v 0=0

t0=0 atvv ====−−−− 02

00 21

attvss ++++====−−−− )(2 020

2 ssavv −−−−====−−−−

Graphical Interpretation

Assuming t 0=0

a =Const.

a

tO

a –t Curve

v=v0+atv

tO

v –t Curve

tO

s –t Curve

s2

00 21

attvss ++++++++====

t

s0

v0

ss0

20 2

1attv ++++

at

v0

v

WORKED EXAMPLE # 1.1

A high speed train is travelling along a straight level road bed at a speed of 240 km/hr. Determine its stopping distance if the deceleration is constant and equal to 7.0 m/s2. How much time elapsed during which the brakes were applied to stop this train?

The motion of train is subjected to constant deceleration until it stops. Using , the stopping distance can be calculated as savv ∆∆∆∆====−−−− 22

02

(((( )))) mavv

s 5.3170.72

360010240

0

2

23

20

2

====−−−−××××

××××−−−−====

−−−−====∆∆∆∆

For the elapsing time, using , thus tvv

a∆∆∆∆−−−−==== 0

sa

vvt 52.9

7

360010240

03

0 ====−−−−

××××−−−−====

−−−−====∆∆∆∆

WORKED EXAMPLE # 1.2

A car passes you at point 1 travelling at an initial velocity of 6 m/s, and then accelerates at a constant rate to reach a velocity of 30 m/s at point 2. This occurs over an 8second period.

(a) What is the required constant acceleration during the initial 8 sec period?

(b) Calculate the distance covered by the car in this 8 sec period.

(c) Once the car passes point 2 at t= 8 s, the acceleration becomes a function of time

given by . Determine an equation for the velocity of the car as a function of

time v(t) for t>8 s.

1 2

481

)( ++++−−−−==== tta

(a) Using )( 1212 ttavv −−−−====−−−− 2

12

12 /38

630sm

ttvv

a ====−−−−====

−−−−−−−−====

(b) Using 21212012 )(

21

)( ttattvsss −−−−++++−−−−====−−−−====∆∆∆∆ ms 1448321

86 2 ====××××××××++++××××====∆∆∆∆

(c) When t>8s from point 2

481 ++++−−−−======== t

dtdv

a

Ctt

dttadtv ++++++++−−−−====++++−−−−======== ∫∫∫∫∫∫∫∫ 416

)481

(2

(C is a constant )

To determine C, using the initial condition @ point 2 (i.e. t=8 s, v=30 m/s )

C++++××××++++−−−−==== 84168

302

2====C

24161 2 ++++++++−−−−==== ttv

3. Linear Motion with Variable Acceleration

Depending on the nature of a problem, acceleration (a) may also be known in different forms including

(a) a is a given function of time “a=a(t)”

(b) a is a given function of velocity “a=a(v)”

(c) a is a given function of displacement “a=a(s)”

(a) Given a=a(t), develop v-t and s-t relationships

)( tadtdv ====2 ∫∫∫∫∫∫∫∫ ====

t

t

v

vdttadv

00

)(

∫∫∫∫====−−−−t

tdttavv

0

)(0Tip : Velocity v(t) as a function of time can be found by integrating a(t)

)(tvdtds====1

∫∫∫∫====−−−−t

tdttvss

0

)(0Tip : Distance s(t) as a function oftime can be found by integrating v(t)

∫∫∫∫∫∫∫∫ ====t

t

s

sdttvds

00

)(

(b) Given a=a(v), develop v-t and s-v relationships

)(vadtdv ====2 ∫∫∫∫∫∫∫∫ ====

t

t

v

vdt

vadv

00 )(

00 )(

ttva

dvv

v−−−−====∫∫∫∫

dsvavdv )(====3 ∫∫∫∫∫∫∫∫ ====s

s

v

vdsdv

vav

00 )(

00 )(

ssdvvavv

v−−−−====∫∫∫∫

This gives a relationship between velocity v and time taken t.

This gives the distance travelled s before the velocity v is reached.

(c) Given a=a(s), develop v-s and s-t relationships

dssavdv )(====3 ∫∫∫∫∫∫∫∫ ====s

s

v

vdssadvv

00

)(

∫∫∫∫====−−−−s

sdssavv

0

)(220

2

This gives velocity v(s) as a function of distance s.

)(svdtds ====1 ∫∫∫∫∫∫∫∫ ====

s

s

t

t svds

dt00 )(

∫∫∫∫====−−−−s

s svds

tt0 )(0

This gives a relationship between distance s and time taken t.

Graphical Interpretation

sdtds

v &========

1

vdtdv

a &========

1

dt

v

t

dt

a

v=slope of s-t curve a=slope of v-t curve

v=area under a-t curves=area under v-t curve

dtds

v ====dtdv

a ====

∫∫∫∫∫∫∫∫ ========−−−− 2

1

2

112

t

t

s

svdtdsss ∫∫∫∫∫∫∫∫ ========−−−− 2

1

2

112

t

t

v

vadtdvvv

tt

t1 t2

s

tt1

t

t2

v

tt1 t2

a

Graphical Interpretation (Cont’d)

ds

a

====dsdv

va

(((( )))) ∫∫∫∫====−−−− 2

1

)(21 2

122

s

sdssavv∫∫∫∫∫∫∫∫ ==== 2

1

2

1

)(s

s

v

vdssadvv or

dsdv

1

ss1 s s2

v

area under a-s curve

a = v x slope of v-s curve

ss1 s2

a

s

WORKED EXAMPLE #1.3

A motorcycle starts from rest and travels on a straight road with a constant acceleration of 5 m/s2 for 8 sec, after which it maintains a constant speed for 2 sec. Finally it decelerates at 7 m/s2 until it stops. Plot a-t, v-t diagrams for the entire motion.Determine the total distance travelled.

Sketch a-t diagram from the known accelerations, th us

5

-7

≤≤≤≤≤≤≤≤−−−−<<<<≤≤≤≤<<<<≤≤≤≤

====)'10(7

)108(0

)80(5

tt

st

st

a

(segment I )

(segment II )

(segment III )

Since dv=adt, the v-t diagram is determined by integrating the straight line segments of a-t diagram. Using th e initial condition t=0, v=0 for segment I , we have

st 80 <<<<≤≤≤≤ ∫∫∫∫∫∫∫∫ ====tv

dtdv00

5 tv 5====

When t =8 s, v =5××××8= 40m/s . Using this as the initial condition for segment II, thus

st 108 <<<<≤≤≤≤ ∫∫∫∫∫∫∫∫ ====tv

dtdv840

0 smv /40====

Similarly, for segment III

'10 tt ≤≤≤≤≤≤≤≤ ∫∫∫∫∫∫∫∫ −−−−====tv

dtdv1040

)7( 1107 ++++−−−−==== tv

a (m/s2)

t (s)

8 10 t' (=15.71)

a-t Diagram

When v=0 (i.e. motorcycle stops)

110'70 ++++−−−−==== t st 71.15' ====

Thus, the velocity as the function of time can be expressed as

≤≤≤≤≤≤≤≤++++−−−−<<<<≤≤≤≤<<<<≤≤≤≤

====)71.1510(1107

)108(40

)80(5

stt

st

stt

v

The total distance travelled ( using the area under v-t diagram )

(((( )))) mssss 2.3544071.521

40240821

321 ====

××××××××++++××××++++

××××××××====++++++++====

v (m/s)

t (s)

40

8 10 15.71

v-t Diagram

s1 s2 s3

WORKED EXAMPLE # 1.4

A test projectile is fired horizontally into a viscous liquid with a velocity v0.The retarding force is proportional to the square of the velocity, so that the acceleration becomes a=-kv2. Derive expressions for distance D travelling in the liquid and the corresponding time t required to reduce the velocity to v0/2.Neglect any vertical motion. (2/40 in M+K)

v0v

xNote the acceleration a is non-constant .

Using dxkvadxvdv 2−−−−========

∫∫∫∫∫∫∫∫∫∫∫∫ −−−−====−−−−

==== 2220

0

0

0

0

v

v

v

v

D

kvdv

kvvdv

dx

kkv

v

kkv

D

v

v

693.02ln2ln1ln

0

020

0

========−−−−====

−−−−====

Using2kv

dtdv

a −−−−========

∫∫∫∫∫∫∫∫ ====−−−−

tv

vdt

kvdv

0

22

0

00

2 1110

0kvvk

t

v

v

====

====

WORKED EXAMPLE #1.5

The acceleration of a particle which moves in the positive x-direction varies with its position as shown. If the velocity of the particle is 0.8 m/swhen x=0, determine the velocities v of the particle when x=0.6 and 1.4 m. (adapted from 2/23 in M+K )

ax (m/s2)

x (m)

0.4

0.2

0.4 0.8 1.2Using

22

20

22

00

0

vvvvdvadx

v

v

v

v

x −−−−====

======== ∫∫∫∫∫∫∫∫

smadxvv /17.102.04.04.0)4.02.0(21

)4.04.0(28.02 24.1

0

20 ====

++++××××++++××××++++++++××××××××++++====++++==== ∫∫∫∫

For x=1.4m

Where v0=0.8 m/sArea under a x-x curve(0≤x ≤ 1.4)

For x=0.6m

1.40.6

smadxvv /05.12.0)4.03.0(21

)4.04.0(28.02 26.0

0

20 ====

××××++++++++××××××××++++====++++==== ∫∫∫∫

Area under a x-x curve(0≤x ≤ 0.6)

WORKED EXAMPLE #1.6

The v-s diagram for a testing vehicle travelling on a straight road is shown. Determine the acceleration of the vehicle at s=50 m and s=150 m. Draw the a-s diagram.

v (m/s)

s (m)100 200

8

Since the equations for segments of v-s diagram are given,we can use ads=vdvto determine a-s diagram.

ms 1000 <<<<≤≤≤≤

ssdsd

sdsdv

va 0064.0)08.0()08.0( ============

ms 200100 ≤≤≤≤≤≤≤≤

28.10064.0)1608.0()1608.0( −−−−====++++−−−−++++−−−−==== ssdsd

sa

1608.0 ++++−−−−==== sv

sv 08.0====

When s=50 m, then (acceleration in segment I ) 2/32.0500064.0 sma ====××××====

When s=150 m, then (deceleration in segment II )2/32.028.11500064.0 sma −−−−====−−−−××××====

a (m/s2)

100 200 s (m)

0.64

-0.64