Lecture 1 - Introduction and Rectilinear Motion

DYNAMICSDYNAMICSBDA 2013BDA 2013

Dr. Waluyo Adi Siswanto

Lecture 1 – Introduction, Rectilinear MotionLecture 1 – Introduction, Rectilinear Motion

Lecture Module 1 Dr. Waluyo A. Siswanto 2

Course ObjectivesCourse Objectives

To understand basic concepts in dynamics (kinematics and kinetics) of particles and rigid bodies.

To predict the effects of force and motion while carrying out the creative design functions of engineering.

To apply problem-solving procedures that will be applicable in succeeding courses and throughout the engineering careers.


Learning OutcomesLearning Outcomes

At the end of this course, students will be able to: Determine the position, velocity and acceleration of

particles and rigid bodies using kinematics approach. Use law of motion to relate forces and acceleration of

particles and rigid bodies. Apply the principle of work and energy to solve

kinetics problem on particles or rigid bodies. Solve the kinetics problems using the principles of

impulse and momentum. Apply all methods that have studied in real situation. Communicate and work effectively in a group through

given assignments and presentations.


What you can do…What you can do…


ReferencesReferences Siswanto W.A., 2008, Principles of Engineering

Dynamics, Penerbit UTHM Meriam J.L. and Kraige L. G., 2007. Engineering

mechanics – Dynamics, 6th Edition, John Wiley & Sons, Inc.

Bear F.P. and Johnson E. R., 2007. Vector Mechanics for Engineers – Dynamics, 8th S.I. Edition, Mc Graw Hill

Hibbeler R.C., 2004. Engineering Mechanics –Dynamics, 3rd S.I. Edition, Prentice Hall.

Bedford A. & Wallace T. Fowler, 2005. Engineering Mechanics-Dynamics, S.I. Edition, Prentice Hall.

Mohd Imran Ghazali, Izzuddin Zaman, et.all., 2007, Koleksi Soalan Dinamik, Penerbit UTHM.

Saifulnizam Jamian; Izzuddin Zaman, et.all., 2007, Dynamics, Penerbit UTHM.


Subject Delivery (S4 and S6) Subject Delivery (S4 and S6)

LecturingExplaining theories, methods and understanding related to the topics

TutoringDiscussing sample problems related to the topics

All course materials, announcements, and students marks will be published in E-Learning

S4: Tuesday (BS A6, 20.00 – 21.50) Friday (BS A6, 08.00 – 09.50)

S6: Monday (BS C2, 08.00 – 08.50) Wednesday (DK D, 10.00 – 12.50)


Assessment SchemeAssessment Scheme

Students attendance * Individual assignment (3) (10%) Group Assignment (10%) Test 1 (15%) Test 2 (15%) Final Examination (50%)

* Attendance must not less than 80%


Time Tables Time Tables BDA 2013 Dynamics, S4BDA 2013 Dynamics, S4

Part 1Ass 1

07/07 10/07 14/07 17/07 21/07 24/07 28/07 31/07

1 2 3 4

Ass 1: First assignmentAss 2: Second assignmentAss 3: Third assignment

Ass 1 Due: First assignment deadline submissionAss 2 Due: Second assignment deadline submissionAss 3 Due: Third assignment deadline submission

Proj: Project commencing Test 1: First Test Project Presentation: Presentation of the project Test 2: Second Test

Part 2Ass 2 Ass 2 Due

Test 2: 17/9 or 18/9

25/08 28/08 01/09 04/09 08/09 11/09 15/09 18/09

7 8 9 10


29/09 02/10 06/10 09/10 13/10 16/10 20/10 23/10

11 12 13 14

04/08 07/08 11/08 14/08

5 6 Ass 1 Due

Test 1: 13/8 or 14/8

ProjProject Presentation


Time Tables Time Tables BDA 2013 Dynamics, S6BDA 2013 Dynamics, S6

Part 1Ass 1

06/07 08/07 13/07 15/07 20/07 22/07 27/07 29/07

1 2 3 4

Ass 1: First assignmentAss 2: Second assignmentAss 3: Third assignment

Ass 1 Due: First assignment deadline submissionAss 2 Due: Second assignment deadline submissionAss 3 Due: Third assignment deadline submission

Proj: Project commencing Test 1: First Test Project Presentation: Presentation of the project Test 2: Second Test

Part 2Ass 2

24/08 26/08 31/08 02/09 07/09 09/09 14/09 16/09

7 8 9 10


28/09 30/09 05/10 07/10 12/10 14/10 19/10 21/10

11 12 13 14

03/08 05/08 10/08 12/08

5 6

ProjProject Presentation

Ass 1 DueTest 1: 13/8 or 14/8

Ass 2 DueTest 2: 17/9 or 18/9


Topics (Part 1)Topics (Part 1)

KINEMATICS OF PARTICLES Rectilinear motion Curvilinear motion Rectangular components of velocity and

acceleration Tangential and normal components Radial and transverse components Relative motion of moving particle

KINETICS OF PARTICLES Force Mass and Acceleration Work and Energy

TEST 1TEST 1



KINETICS OF PARTICLES Impulse and Momentum Direct Impact Oblique Impacts

KINEMATICS OF RIGID BODIES Translation, rotation and general plane motion Absolute motion Relative velocity and acceleration Zero velocity instantaneous center

TEST 2TEST 2



KINEMATICS OF RIGID BODIES Relative motion of rotating frame

KINETICS OF RIGID BODIES Mass Moment of Inertia Moment, Inertia and Acceleration Translation and Rotation and General

Plane Motion Work and Energy in Rigid Bodies

FINAL EXAMFINAL EXAM


INTRODUCTIONINTRODUCTION


DynamicsDynamics

Dynamics consists two distinct parts: kinematics and kinetics.

Kinematics deals with the study of motion without reference to the force which cause motions

Kinetics relates the action of forces on bodies to their resulting motions


Kinematics: Kinematics: Motion onlyMotion only


Kinetics: Kinetics: Interaction force - motionInteraction force - motion

Force

Path of motion because of the force


Vector Analysis Review: Vector Analysis Review: NotationNotation

A

length of arrow denotes magnitude of F

extension of arrow denotes line of action

arrow head denotesthe direction of F

321

F

3,2,1F

3 2 3F i j k

Vectors are defined as mathematical expression possessing magnitude and direction, which add according to the parallelogram law.

(note: scalar has only magnitude)


Vector Analysis Review Vector Analysis Review MagnitudeMagnitude

2 2 22 1 2 1 2 1magnitude A x x y y z z = A


Analysis Vector Review Analysis Vector Review Parallelogram Law in Two-Dimensional VectorsParallelogram Law in Two-Dimensional Vectors

A

B

A+B

P

cossin

x

y

A

A AA A

cossin

x

y

B

B BB B

cos cossin sin

A + BA + B

A Bβ

α


Analysis Vector Review Analysis Vector Review VectorVector DirectionDirection

z

Ax

Ay

Az

ij

k

A=Ayi+ Ayj + Ayk

αβ

γ α : angle between x coord and A

β : angle between y coord and A

γ : angle between z coord and A

y

x


Analysis Vector Review Analysis Vector Review VectorVector Direction – Unit VectorDirection – Unit Vector

cos xAA

cos yAA

cos zAA

yx zA

AA AA A A

e i j k

(cos ) (cos ) (cos )A e i j k

AAA e

2 2 2cos cos cos 1

β

yAy

A

γ

Az

A

α

Ax

A

z

x


Particle and rigid bodyParticle and rigid body

A particle is a point mass. This means the mass is concentrated at a single point and the particle has neither dimensions (height, width, etc) nor orientation (angular position)

Under certain conditions a physical body can be modeled as a particle; for example, when considering translation of a body, or when all forces acting on a body pass through the centre of mass, or when the dimensions of a body are very much smaller than those of its path of motion


When a body can be replaced by a When a body can be replaced by a particleparticle

Replace by a particle (no rotation involved)

The body motion cannot be replaced by a particle(rotation involved)


Translation - RotationTranslation - Rotation

Translation Combination Translation and Rotation


Acceleration due to gravityAcceleration due to gravity

2

2( )og ge

e

RR h

2 49,780327 1 0,005279sin 0,000023sin og

In general applications, the standard values for g of 9,81 m/s2 or 32,2 ft/s2

will be sufficiently accurate.

Re = 6,371 . 106 m h : altitude


KINEMATICS OF PARTICLE KINEMATICS OF PARTICLE TRANSLATIONTRANSLATION


Translation of a particle: Translation of a particle: straight, curve and circular pathsstraight, curve and circular paths

Straight path

Curve path Circular path

Straight path

Curve path Circular path

There is no orientation of the body

Can be simplified as particle


Average and InstantanneousAverage and Instantanneous

0lim sv

t

t

dsvdt

v stime (s)

Posi

tion

(m)

s

t

Instantaneous velocity

Average Velocity avg

svt

0lim

t

t

va

Instantaneous acceleration

Average Acceleration avg

vat

2

2

va d d sdt dt

a v s

Δt

Δs


Calculating distance and velocity, Calculating distance and velocity, graphicallygraphically

v

tt0 t

v 0 0

s t

s t

ds dt

0

v

v

v a 0

t

t

d dt

The area under v(t) is the DISTANCE TRAVEL

The area under a(t) is the NET CHANGE IN VELOCITY

a

tt0 t1


Differential equation of motionDifferential equation of motion

v ad dt

dsd dt dt dsdt

v v v a a a v v ad ds

s ds s ds

Multiply by v

or


Rectilinear motion Rectilinear motion at constant velocityat constant velocity

v 0 0

s t

s t

ds dt

0 0( )v s s t t

0 0( )+ v s s t t


Velocity at constant accelerationVelocity at constant acceleration

0

v

v

v a 0

t

t

d dt 0 0v v a t t

0 0v v + a t t

0

0

v

v

v v a s

s

d ds 2 20 0

12v v a s s

2 20 0v v + a 2 s s


Distance at constant accelerationDistance at constant acceleration

v 0 0

s t

s t

ds dt 0 0v + a t t 0 0

s t

s t

ds ( )dt

0 0

0 0 0

0 0 0

0 0

v a

v a a

t t

t t

t t t

t t t

s s dt t t dt

dt tdt t dt

2 20 0 0 0 0 0

1( )2

s s v t t a t t at t t


Practical situationPractical situation

Usually a particle start moving when the time is set to 0 and the distance goes from 0

0 00 and 0t s at and distance t s

0 0v v + a t t

2 20 0v v + a 2 s s

2 20 0 0 0 0 0

1( )2

s s v t t a t t at t t

0 tv v + a

2 20 2 sv v + a

20

1( )2

s v t a t

0

0

0 0 0 0


Example Problem 1-1Example Problem 1-1

The displacement of a mechanical component follows a ruled path as a function of time. The function dis(t) = 2t3 – 24t + 6 meter.

• Derive the velocity and acceleration based on the given displacement function dis(t).

• Calculate the time to reach velocity of 72 m/s. Calculate the corresponding acceleration at that time.

• Plot the displacement, velocity and acceleration versus time for the first 4 seconds every 1 second.



A particle starts from rest at x = -2 m and moves along x-axis with the velocity history shown below. •Derive the velocity equation in every stage of t (0 to 0,5), (0,5 to 1,0) and (1,0 to 2,0)•Calculate the corresponding acceleration and displacement at t= 0; 0,5; 1,0; and 2,0

v (m

/s)

0

3

0.5 1.0 1.5 2.0

-1.0

t (s)



A train is traveling at a constant speed of 108 km/h when the driver sees a stationary obstacle on the track 135m ahead. He immediately applies the brakes at full capacity. If the train’s constant deceleration on this section of track is

4,5 m/s2, determine whether the train will stop before colliding with the obstacle.

Calculate the time interval between the driver applying the brakes and the train coming to rest.




See you again …….See you again …….

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Lecture 1 - Introduction and Rectilinear Motion