01 MENG212 Introductionf

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MENG 212 Fall 2014 Dr. Jong B. Lee, ME @NYIT 1

MENG 212

Engineering Mechanics: Dynamics ( 1)

Sept. 03 2014

2014 Jong B. Lee PhD

ME @NYIT

Introduction toDynamics

MENG 212 Engineering Mechanics II: Dynamics. Jong B. Lee, PhD, All rights reserved.

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MENG 212

Engineering Mechanics: Dynamics Fall2014

Class Hour

Wednesday 5:45PM ~ 8:25PM

Class room

HSH #212

Office Address: HSH Room 224A

Office Phone: (516) 686 7955

Course web site: http://iris.nyit.edu/~jlee26

Email:jongblee@nyi t.edu

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MENG 212

Office Hours

I adopt an open door policy

You are encouraged to come to my office and

ask questions, consult, provide feedback, or

give suggestions at anytime during the day

However, I may not be available all the time

Set times for offices hours are the office thissemester are:

Mon. ~ Wed. 02:30 PM 03:30 PM

or by appointment via email or phone

can be changed without pre-notification

3 MENG 212 Engineering Mechanics II: Dynamics. Jong B. Lee, PhD, All rights reserved.

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MENG 212

Description of Engineering Mechanics:

Dynamics

This course teaches students how to apply Newtonian

physics to relatively simple physical situations. It followson from the Statics course, but considers systems that

are not in equilibrium i.e. with velocity and acceleration.Some of the topics covered are pure kinematics (a

mathematical description of motion only), while others

are kinetic (determine motion in problems involving theconcepts of force and energy). The course restricts

itself to 2-D (planar) mechanisms.

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MENG 212 Objectives

The student should understand the basic physical

concepts of dynamics.

The student should understand and be able to relate the

kinematics of particles and rigid bodies to the solution ofdynamics problems in straight line and curvilinear

motion.

The student should understand and be able to apply

Newtons Laws to particles and rigid bodies to solve

problems related to dynamic behavior.

The student should be able to apply the methods of

work, momentum and energy to particles and rigid

bodies associated with dynamic behavior.


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MENG 212 Course outcomes:After successful completion of this course, you will

have

1. Understand basic kinematics concepts displacement, velocity

and acceleration (and their angular counterparts).

2. Understand basic dynamics concepts force, momentum, work

and energy.

3. Understand and be able to apply Newtons laws of motion.

4. Understand and be able to apply other basic dynamics concepts- the Work-Energy principle, Impulse-Momentum principle and

the coefficient of restitution.

5. Learn to solve dynamics problems. Appraise given information

and determine which concepts apply, and choose an appropriate

solution strategy.

6. Gain an introduction to basic machine parts such as pulleys andmass-spring systems.

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MENG 212

Class Discussion

Communication is very important in achieving our

collective goals and objectives

Feel free to voice your opinions and ask questionsanytime during a class period

Remember you are here to learn and I am here to

teach and that teaching and learning are

intertwined

So you can help me teach you as much as I can

help you learn

I urge you to be an active participant in the learning

process and recognize that it takes a team effort to

realize meaningful things in life


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MENG 212

Major Courses in mechanical engineering

Engineering mechanics: The study of how

bodies react to forces acting on them

Statics: The study of bodies in equilibrium

Dynamics

Strength and Materials

Vibration

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Solid Mechanics

Thermodynamics

Fluid Mechanics

3 Major

Mechanics


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MENG 212

Engineering Mechanics

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EngineeringMechanics

SolidMechanics

FluidMechanics

Rigid BodyMechanics

DeformableBody

Mechanics

Statics:F=ma,

Dynamics:

v = 0

a = 0F = 0

F 0

F = ma

a = ?

v = ?

Mechanics

of Materials

Elastics

Plastics

External

Load

Stress

Strain


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Adv. Mach. Designby Numerical Method

Statics &Dynamics

AppliedMach. Design

Stress Analysis

MENG 212

Mechanical Course Flow

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Applied SolidMechanics

Strength andMaterials

ElementMach. Design

Elastics

SeniorMach. Design


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MENG 212 Topics covered

Kinematics of Particle

Kinetics of a Particle

Force and Acceleration

Work and Energy

Impulse and Momentum

Planar Kinematics ofa Rigid Body Force and Acceleration

Work and Energy

Impulse and Momentum

Three-Dimensional Kinematics of a Rigid Body

Three-Dimensional Kinetics of a Rigid Body

Vibrations


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MENG 212 Text Book

Engineering Mechanics: Dynamics, 13th edition,

R. C. Hibbeler

ISBN-10: 0132911272

Pearson

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References Engineering Mechanics: Dynamics, 7th ed., by by J. L.

Meriam and L. G. Kraige, Willey

Vector Mechanics for Engineers Dynamics, 10thEdition by BEER

Engineering Mechanics: Statics and Dynamic, Google

eBook, C. L. Rao, J. LAKSHINARASHIMAN, R.

SETHURAMAN, S. M. SIVAKUMAR

Engineering Dynamics: A Comprehensive Introduction

N. Jeremy Kasdin & Derek A. Paley, PrincetonUniversity Press.

Engineering Mechanics: Dynamics, 5th ed. A. M.

Bedford and W. Fowler, Pearson

Engineering Mechanics: Dynamics, 4th ed., I. H.

Shames, Pearson


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MENG 212

Complete class syllabus and rule will begiven today

Class syllabus can be changed without pre-

notification

Please check at course website

PLEASE READ SYLLABUS CAREFULLY,

and let me know if you have any questions

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Exams and Quizzes Ideally, all quizzes and exams are closed book

There will be two midterm exams which constitute20% and 20%

each of the grade

Mid term Exam I: Oct. 08 (20%)

Mid term Exam II: Nov. 12 (20%)

There will be one final exam which constitutes20% of the grade

Final Exam: Dec. 17

There will be number of quizzes which constitute35% of grade

Eight to Ten Quizzes: (Mostly every week)

Participation (Homework and attendance, etc): 5% Total: 100%

This schedule and constitution of the grade can be changed


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Grade

Grading Policy

The following straight scale will be used:

Grade I, IF, W and WF: Please check onUniversity Catalog

Remember final drop day or add/drop from theUniversity Academic Calendar

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A: 94-100

A-: 90-93

B+: 88-89

B: 83-87

B-: 80-82

C+: 78-79

C: 73-77

C-: 70-72

D+: 68-69D: 61-67

F: 0-60


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Grade How to curve an exam and assign grades

We have all given exams where the grades end up

lower than we hoped.

If the class does significantly lower than I think they

should have, I will consider curving the exam. How do Ido it?

Whats the goal of the curve?

How do I curve an exam?

Flat scale

Least squares regression

Linear scale


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Exams Grading Policy

Only neatly wri t ten problem s wil l be graded

A correct answer without a correct outline of the workwill not carry any grade

All incorrect work must be clearly crossed out on thepage

In cases where more than one solution is presentedfor a problem, the solution with the most errors will begraded

Each solution must have proper units

No units or inappro priate units: 0 credit

Class attendance and participation in discussions arenot strongly recommended, it is mandatory

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MENG 212 Quizzes

Will cover theoretical aspects (definitions andderivations) and problem solving skills

Will be closed book, closed notes, with no crib sheet

Will be announced couple of days before the exam Will contain one to four problems No formula sheet

No makeup quizzes wil l be given

Homework

Due of the homework will be after they are assigned Problems will be graded only if they are written neatly

Late assignment is no credit


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MENG 370 Midterm and Final Exams

Will contain four to eight problems

Will be comprehensive Closed book and closed notes No formula sheet

No makeup exams wil l be given

Remember, M ake-up exams wil l not be avail able

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MENG 212

Class Rules

Cheating will be dealt with

according to the rules of the

University

Materials to be covered in an exam

will be announced at least oneweek prior to the exam

Cell phone is not allowed, only

calculator


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MENG 212

Student Code of Conduct

It is the responsibility of each student to adhere to

the principles of academic integrity

Academic integrity means that a student is honest

with him/herself, fellow students, instructors, and

the University in matters concerning his or her

educational endeavors

Thus, a student should not falsely claim the work of

another as one's own, or misrepresent him/herselfso that the measures of one's academic

performance do not reflect his/her own work or

personal knowledge

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MENG 212 Student Code of Conduct

In this regard, cheating will not be tolerated

Cheating includes (but is not limited to) any

communication (written or oral) during

examinations and sharing of work, such as

using the same models or computer

programs or copying work

All homework and projects must be an

individual effort unless specifically noted


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MENG 212 Student Code of Conduct

Students who cheat on any assignment or during

any examination will be assigned a failing grade for

the course

Therefore, avoid all appearance of improperbehavior!

Students who witness cheating should report the

incident to the instructor as soon as possible.

Students are also welcome to discuss any concernsrelated to cheating with Dr. Lu, Chair of Mechanical

Engineering

Dropping: Find the last day to drop this course form

the university academic calendar

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MENG 212

Attendance Sheet

Please sign if your name, student

id number is corrected on theattendance sheet, otherwise make

correction

Please fill in the entire line on the

form if your name is missed


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MENG 212

Questions

Please raise your hand or stop me at any

time when you have a question.

Please do not talk () to your classmates

during the lecture. If you absolutely need

to speak with someone, please feel freeto go out the classroom

Please shut down your cell phone!!!

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MENG 212 Class Materials

Lecture notes will be provided prior to or afterthe class time through course website

User Name:students

Password:

Any changes will be mentioned in class

Please visit and check here frequently forupdates? Even though you are not able to

attend class, please download lecture notes tocatch up class.

Detailed homework, quizzes and examsinformation will be noticed via course websiteso that problems caused by not vis i t ingcourse websi te is your responsib i l i ty


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Course Schedule & Outline

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Class schedule, outlineand office hours can be changed withoutpre-notification

Week # Date Course Contents

1 Sept. 03 Introduction to Dynamics

2 Sept. 10 Kinematics of Particle

3 Sep t. 17 K inet ics o f a Par tic le: Fo rce and Accele ra tion

4 Sept. 24 Kinetics of a Particle: W ork and Energy

5 Oc t. 01 Kinetics of a Partic le: Impulse and Momentum

6 Oct. 08 Mid Term Exam I

7 Oct. 15 Planar Kinematics of a Rigid Body

8 Oct. 22 P lanar Kinemat ics o f a Rig id Body: Fo rce and Accelerat ion

9 Oc t. 29 Planar Kinematic s of a Rig id Bo dy: Wo rk and En ergy

10 Nov. 05 P lanar Kinemat ics of a R ig id Body: Impu lse and Momen tum11 Nov. 12 Mid Term Exam II

12 Nov. 19 Three -D imensional K inemat ics o f a R ig id Body

13 Nov. 26 No Class - Thanksg iv ing Holiday

14 D ec. 03 Three-Dimen sion al Kinetic s of a Rigid Bo dy

15 Dec. 10 Vibrations

16 Dec. 17 Final Exam


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Wrap Up

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Read the syl labusin detail!

Please visit course web site frequently

http://iris.nyit.edu/~jlee26

Course information is subject to change,

so always check here for the latest info

Off ice hours w i l l not be held this week.

They begin next week.

Welcome, good luck, and enjoy!


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Dynamics Dynamics is that branch of mechanics which

deals with the motion of bodies under the

action of forces

Kinematics

study of motion w/o reference to the forces causing motions

study of the geometry of motion. Kinematics is used to relatedisplacement, velocity, acceleration, and time without reference

to the cause of motion.

Kinetics

relates the action of forces on bodies to their resulting

motions

study of the relations existing between the forces acting on a

body, the mass of the body, and the motion of the body.Kinetics is used to predict the motion caused by given forces orto determine the forces required to produce a given motion.

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Dynamics

Dynamics includes

Rectilinear motion: position, velocity, and

acceleration of a particle as it moves along a

straight line.

Curvilinear motion: position, velocity, and

acceleration of a particle as it moves along a

curved line in two or three dimensions.


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Dynamics

Basic Concepts

Space: geometric region occupied by bodies

Time: a measure of the succession of events

and is considered as absolute quantity in

Newtonian mechanics

Mass: quantitative measure of the inertia or

resistance to change in motion of a body

Force: vector action of one body on another

Particle: a body of negligible dimensions

Rigid body: a body whose changes in shape

are negligible compared w/ the changes in

position of a body as a whole

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Dynamics

Newtons Laws

Law 1: A particle remains at rest or continues to

move w/ uniform velocity (in a straight line w/ a

constant speed) if there is no unbalanced force

acting on it

Law 2: The acceleration of a particle is

proportional to the resultant force acting on it

and is in the direction of this force (F=ma)

Law 3: The forces of action and reaction b/w

interacting bodies are equal in magnitude,

opposite in direction, and collinear


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Dynamics

Units

SI Units (US Customary Units)

Mass: kg (slug)

Length: m (ft)

Time: sec. (sec.)

Gravitation:

F: the mutual force of attraction between two particles

G: a universal constant called the constant of gravitation

m1,m2: the masses of the two particles

r: the distance b/w the centers of the particles

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2

21

r

mmGF


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Dynamics Gravitation

Gravitational acceleration

Mis the mass of the larger body, is a unit vector

directed from the large mass to the smaller mass.

Negative sign means the force is an attractive force

In the same way,

g=9.80665 m/s2 (32.1740ft/s2)

Variation of g with altitude

go: gravitational acceleration at the sea level, h:altitude,R: the radius of the earth, me: the mass of

the earth35

2

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r

mmGF

rr

GMg

2

r

22

hR

Rgg o

2R

mGg e


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Kinematics of a Particle Rectilinear Kinematics: Continuous

Motion

Find the kinematic quantities (position,

displacement, velocity, and acceleration) of a

particle traveling along a straight path.

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Applications

Relations between s(t),

v(t), and a(t) for general

rectilinear motion.

Relations between s(t),

v(t), and a(t) when

acceleration is constant.


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Kinematics of a Particle

Applications

The motion of large objects,

such as rockets, airplanes,

or cars, can often beanalyzed as if they were

particles.

Why?

If we measure the altitude

of this rocket as a function

of time, how can we

determine its velocity and

acceleration?


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Applications

A sports car travels along a straight road.

Can we treat the car as a particle?

If the car accelerates at a constant rate, how

can we determine its position and velocity at

some instant?

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Overview of Mechanics

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Statics: The study of

bodies in equilibrium.Dynamics:

1. Kinematicsconcerned with

the geometric aspects of motion

2. Kinetics - concerned with

the forces causing the motion

Mechanics: The study of how bodies

react to forces acting on them.


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Rectilinear Kinematics: Continuous Motion

A particle travels along a straight-

line path defined by the coordinate

axis s.

The position of the particle at any

instant, relative to the origin, O, is

defined by the position vector r, or

the scalar s. Scalar s can be positive

or negative. Typical units for rand s

are meters (m) or feet (ft).

The displacement of the particle isdefined as its change in position.

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Vector form:

r = r - r Scalar form: s = s - s

The total distance traveled by the particle, sT, is a

positive scalar that represents the total length of the

path over which the particle travels.


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Rectilinear Kinematics: Continuous Motion Velocity

Velocity is a measure of the rate of change in the position of aparticle. It is a vector quantity (it has both magnitude and direction).

The magnitude of the velocity is called speed, with units of m/s or

ft/s.

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The average velocity of a particle during a

time interval tis

vav g= r / t

The instantaneous velocity is the time-derivative of position.

v= dr/ dt

Speed is the magnitude of velocity: v=ds/dt

Average speed is the total distance traveled divided by elapsed

time: (vsp)avg = sT / tMENG 212 Engineering Mechanics II: Dynamics. Jong B. Lee, PhD, All rights reserved.

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Rectilinear Kinematics: Continuous Motion Acceleration

Acceleration is the rate of change in the velocity of a particle. It is avector quantity. Typical units are m/s2 or ft/s2.

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As the text indicates, the derivative equations for velocity and

acceleration can be manipulated to get a ds = v dv

The instantaneous acceleration is the time

derivative of velocity.

Vector form: a = dv / dt

Scalar form: a = dv / dt = d2s / dt2

Acceleration can be positive (speed

increasing) or negative (speed decreasing).


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Summary

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Differentiate position to get velocity and acceleration.

v = ds/dt ; a = dv/dt or a = v dv/ds

Integrate acceleration for velocity and position.

Note that so and vo represent the initial position and

velocity of the particle at t = 0.

Velocity:

t

o

v

vo

dtadv s

s

v

v oo

dsadvvor t

o

s

so

dtvds

Position:


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Constant Acceleration

The three kinematic equations can be integrated for the

special case when acceleration is constant (a = ac) to

obtain very useful equations. A common example ofconstant acceleration is gravity; i.e., a body freely falling

toward earth. In this case, ac= g = 9.81 m/s2= 32.2 ft/s2

downward. These equations are:

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tavv co yields t

o

c

v

v

dtadvo

2coo

s

t(1/2) atvss yields t

os

dtvdso

)s-(s2a)(vv oc2

o

2 yields s

s

c

v

v oo

dsadvv


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Example

A particle travels along a straight line to

the right with a velocity of v = (4t3t2)

m/s where t is in seconds. Also, s = 0

when t = 0.

The position and acceleration of the

particle when t = 4s.


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Example: Solution

Take a derivative of the velocity to

determine the acceleration

(or in the direction) when t = 4s

Calculate the distance traveled in 4s by

integrating the velocity using so = 0:

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mttss

dtttdsvdtdsdt

dsv

o

s

so

322

34

4

0

32

4

0

2

22/2064

34smat

dt

ttd

dt

dva


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Example A particle is moving along a straight line

such that its velocity is defined as v = (-

4s2) m/s, where sis in meters.

The velocity and acceleration as

functions of time if s = 2 m when t = 0.

Since the velocity is given as a functionof distance, use the equation v=ds/dt.

Express the distance in terms of time.

Take a derivative of it to calculate the velocity

and acceleration.


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Example: Solution Since v=-4s2

Determine the distance by integrating using so=2

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2

244

s

dsdts

dt

dsv

18

21

2

14

2

11

2

114

14

4

22

1

0

2

2

2

0

ts

st

sst

sst

dsss

dsdt

sst

ss

s

t

o


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Rectilinear Kinematics: Erratic Motion v-tGraph

Plots of velocity vs. time can be

used to find acceleration vs. time

curves. Finding the slope of the

line tangent to the velocity curve atany point is the acceleration at that

point (or a = dv/dt).

Therefore, the acceleration vs.

time (or a-t) graph can be

constructed by finding the slope at

various points along the v-t graph.

Also, the distance moved

(displacement) of the particle is

the area under the v-t graph during

time t.


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Rectilinear Kinematics: Erratic Motion

a-tGraph

Given the acceleration vs.

time or a-t curve, the

change in velocity (v)during a time period is the

area under the a-t curve.

So we can construct a v-t

graph from an a-tgraph if

we know the initial velocity

of the particle.

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Rectilinear Kinematics: Erratic Motion a-sGraph

A more complex case is presented bythe acceleration versus position or a-

s graph. The area under the a-scurve represents the change invelocity

This equation can be solved for v1,allowing you to solve for the velocity

at a point. By doing this repeatedly,

you can create a plot of velocityversus distance.

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)vdvads(Recall

graphsaunder theArea

2

1 2

1

2

0

2

1

s

s

adsvv


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Rectilinear Kinematics: Erratic Motion v-sGraph

Another complex case is presented bythe velocity vs. distance or v-s graph.

By reading the velocity v at a point onthe curve and multiplying it by theslope of the curve (dv/ds) at this same

point, we can obtain the acceleration

at that point. Recall the formula

Thus, we can obtain an a-s plot from

the v-s curve

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ds

dvva


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Example The s-t graph for a sports car moving

along a straight road

The v-t graph and a-t graph over the time

interval shown


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Example: Solution The v-t graph can be constructed by

finding the slope of the s-t graph at key

points. What are those?

When 0


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Example: Solution Similarly, the a-t graph can be constructed by finding

the slope at various points along the v-t graph. Using

the results of the first part where the velocity was

found:

When 0


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Example: Solution

Now find the distance traveled:

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sm

v

m

sss

mtdttvdts

mdttvdts

savg

/348

144

time

distancetotal

1445490

54482

1

3

148

3

1

90302

1

5

1

5

1

)480(

4830300480

48

30

248

30

4830

230

0

300


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Curvilinear motion: General and rectangular

components

Describe the motion of a particle traveling along a

curved path. Relate kinematic quantities in terms of the rectangular

components of the vectors.

Applications

General Curvilinear Motion

Rectangular Components of Kinematic Vectors

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Curvilinear motion: General and rectangular components

Applications

The path of motion of a

plane can be tracked

with radar and its x, y,

and z coordinates

(relative to a point on

earth) recorded as a

function of time. How can we determine

the velocity or

acceleration of the plane

at any instant?


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Curvilinear motion: General and rectangular components

Applications

A roller coaster car

travels down a fixed,

helical path at a constant

speed.

How can we determine its

position or acceleration at

any instant?

If you are designing the

track, why is it important

to be able to predict the

acceleration of the car?

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General Curvilinear Motion A particle moving along a curved

path undergoes curvilinear motion.

Since the motion is often three-

dimensional, vectors are used todescribe the motion.

A particle moves along a curvedefined by the path function, s.

The position of the particle at anyinstant is designated by the vector r =

r(t). Both the magnitude and

direction of r may vary with time.

If the particle moves a distance s

along the curve during time interval

t, the displacement is determined byvector subtraction: r = r - r


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General Curvilinear Motion Velocity

Velocity represents the rate of changein the position of a particle.

The average velocity of the particle

during the time increment t is, vavg =r/t .

The instantaneous velocity is the time-derivative of position, v = dr/dt .

The velocity vector, v, is alwaystangent to the path of motion.

The magnitude of v is called the speed.Since the arc length s approaches the

magnitude of r as t0, the speed can

be obtained by differentiating the pathfunction (v = ds/dt). Note that this is

not a vector!

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General Curvilinear Motion Acceleration

Acceleration represents the rate ofchange in the velocity of a particle.

If a particles velocity changes fromv to v

over a time increment t, the averageacceleration during that increment is:aavg= v/t = (v - v)/t

The instantaneous acceleration is the

time-derivative of velocity: a = dv/dt =d2r/dt2

A plot of the locus of points defined by thearrowhead of the velocity vector is calleda hodograph. The acceleration vector is

tangent to the hodograph, but not, in

general, tangent to the path function


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General Curvilinear Motion Rectangular components

It is often convenient to describe themotion of a particle in terms of its x, y,

z or rectangular components, relative

to a fixed frame of reference. The position of the particle can be

defined at any instant by the positionvector, r=xi+yj+zk.

The x, y, z components may all befunctions of time, i.e.,x=x(t), y=y(t),

and z=z(t).

The magnitude of the position vectoris: r=(x2+y2+z2)0.5

The direction of ris defined by the unit

vector: ur= (1/r)r

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Curvilinear Motion: Rectangular Components

Velocity

The velocity vector is the time derivative of the position vector :v=dr/dt=d(xi)/dt+d(yj)/dt +d(zk)/dt

Since the unit vectors i, j, kare constant in magnitude anddirection, this equation reduces to v=vxi+vyj+vzk, where

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dt

dzzv

dt

dyyv

dt

dxxv zyx ,,

The magnitude of the velocity

vector is

The direction of v is tangent to the

path of motion.

222zyx

vvvv


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Curvilinear Motion: Rectangular Components

Acceleration

The acceleration vector is the time derivative of the

velocity vector (second derivative of the position vector):a=dv/dt=d2r/dt2=axi+ayj+azk

where

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dt

dvzva

dt

dvyva

dt

dvxva zzz

y

yyx

xx ,,

The magnitude of the acceleration

vector is

The direction of a is usually not

tangent to the path of the particle.

222zyx

aaaa


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Example The box slides down the slope described

by the equation y=(0.05x2)m, where xis in

meters. vx=-3m/s, ax=-1.5m/s2at x=5m.

The ycomponents of the velocity and the

acceleration of the box at x=5m.


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Example: Solution Find the y-component of velocity by

taking a time derivative of the position y

= (0.05x2)

Find the acceleration component by

taking a time derivative of the velocity

Substituting the x-component of the

acceleration, velocity at x=5into

78

xxxxxxdt

d

dt

dvy

xxxxyvxy

y

y

1.01.01.0

1.0205.005.0 2

yy and


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Example: Solution

Since

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2

22

2

/15.0

/5.1/5.1v

5at

/15.05.151.031.0

1.01.0/5.1351.01.0

5/5.1,/3

sma

smsm

mx

sm

xxxxy

smxxy

mxatsmaxsmvx

y

y

xx


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Example

The particle travels along the pathy=0.5x2. When t=0, x=y=z=0.

The particles distance and themagnitude of its acceleration when t=1s,

if vx=(5t)ft/s, where t is in seconds.

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Example: Solution

x-components

y-components

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statsfttdt

dxa

stfttxdttdt

stsftsfttx

x

t

x

x

1/55:onAccelerati

1at5.25.25v:position

1at/5/5v:asknowVelocity

2

2

0

stsftxxxxya

stsftxxxxy

stftxy

y 1at/5.37:onAccelerati

1at/5.1225.0:Velocity

1at125.35.0:asknowposition

2

2


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Example: Solution

The position vector and the acceleration

vector are

Position vector: r= [ x i+ yj] ft

where x= 2.5 ft, y= 3.125 ft

Magnitude: r = (2.52 + 3.1252)0.5 = 4.00 ft

Acceleration vector: a = [ ax i+ ayj] ft/s2

where ax = 5 ft/s2, ay = 37.5 ft/s

2

Magnitude: a = (52 + 37.52)0.5 = 37.8 ft/s2

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Motion of a Projectile Analyze the free-flight motion of a

projectile

Applications

Kinematic Equations for Projectile

Motion


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Motion of a Projectile Applications

A good kicker instinctively knows at what angle, q, and

initial velocity, vA, he must kick the ball to make a field

goal.

For a given kick strength, at what angle should the ballbe kicked to get the maximum distance?

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Motion of a Projectile

Applications

A basketball is shot at a certain angle. What parameters

should the shooter consider in order for the basketball

to pass through the basket? Distance, speed, the basket location, anything else?


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Applications

A firefighter needs to know the maximum height on the

wall she can project water from the hose. What

parameters would you program into a wrist computer tofind the angle, , that she should use to hold the hose?

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Projectile motion can be treated as two

rectilinear motions, one in the horizontal

direction experiencing zero acceleration

and the other in the vertical direction

experiencing constant acceleration (i.e.,

from gravity).


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Motion of a Projectile For illustration, consider the two

balls on the left. The red ball falls

from rest, whereas the yellow ballis given a horizontal velocity. Each

picture in this sequence is takenafter the same time interval. Notice

both balls are subjected to the

same downward acceleration since

they remain at the same elevation

at any instant. Also, note that the

horizontal distance betweensuccessive photos of the yellow

ball is constant since the velocity inthe horizontal direction is constant.

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Motion of a Projectile Kinetic equations: Horizontal motion

Since ax=0, the velocity in the horizontal direction

remains constant (vx=vox) and the position in thex

direction can be determined by:

Why is ax equal to zero (what assumption must be

made if the movement is through the air)?

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txxoxo

v


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Motion of a Projectile Kinetic equations: Vertical motion

Since the positivey-axis is directed upward, ay=

g. Application of the constant acceleration

equations yields:

For any given problem, only two of these three

equations can be used. Why?

90

oyoy

oyo

oyy

yyg

gttyy

gt

2vv

2

1v

vv

22

2


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Example

Given: vA and

Find: Horizontal distance it travels and

vC. Apply the kinematic relations in x- and

y-directions.


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Example: Solution

Using vax=10cos30 and vay=10sin30

We can write

vx = 10 cos 30 vy = 10 sin 30 (9.81) t

x = (10 cos 30) t

y = (10 sin 30) t (9.81)t2

Since y=0 at C

0=(10sin30)t- (9.81)t2t=0, 1.019s

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Example: Solution

Velocity components at C are

vCx=10cos30=8.66 m/s

vCy=10sin 30(9.81)(1.019)= -5 m/s=5m/s

Horizontal distance the ball travels is;

x = (10cos30)t

x = (10cos30)1.019=8.83 m

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smvC

/10566.8 22


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Example Projectile is fired with vA=150m/sat point A.

The horizontal distance it travels (R) and the time in

the air.

Establish a fixed x, y coordinate system (in this

solution, the origin of the coordinate system is placedat A). Apply the kinematic relations in x- and y-

directions.

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Example: Solution Place the coordinate system at point A. Then, write

the equation for horizontal motion.

+ xB = xA + vAx tAB

where xB = R, xA = 0, vAx = 150 (4/5) m/s

Range, R, will be R = 120 tAB

Now write a vertical motion equation. Use the distance

equation.

+ yB = yA + vAy tAB 0.5 g tAB2

where yB = 150, yA = 0, and vAy = 150(3/5) m/s

We get the following equation: 150=90tAB+0.5(9.81) tAB2

Solving for tAB first, tAB = 19.89 s.

Then, R = 120 tAB = 120 (19.89) = 2387 m


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Example A skier leaves the ski jump ramp at A = 25

o and hitsthe slope at B.

The skiers initial speed vA

Establish a fixed x,y coordinate system (in thissolution, the origin of the coordinate system is placedat A). Apply the kinematic relations in x- and y-

directions.

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Example: Solution

Motion in x-direction

Motion in y-direction

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AA

AB

ABAABoxAB

vvt

tvtvxx

27.88

25cos

80

25cos01005

4

smv

vvv

tgtvyy

A

AA

A

ABABoyAB

/42.19

27.8881.9

2

127.8825sin064

2

1

2

2


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Example

The golf ball is struck with a velocity of 80 ft/s

as shown.

Find distance dto where it will land.

Establish a fixed x, y coordinate system (in

this solution, the origin of the coordinate

system is placed at A). Apply the kinematic

relations in x- and y-directions.

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Example: Solution

Motion in x-direction

Motion in y-direction

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ftd

dd

ddd

tgtvyy ABABoyAB

166,0

007415.0233.10

02146.02.322102146.055sin80010sin

2

1

2

2

2

dt

tdtvxx

AB

ABABoxAB

02146.0

55cos80010cos


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01 MENG212 Introductionf

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