LEC 02 (Constant Acc 051)

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King Fahd University of Petroleum King Fahd University of Petroleum & Minerals & Minerals Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 2

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Engineering Mechanics lectures

Transcript of LEC 02 (Constant Acc 051)

Page 1: LEC 02 (Constant Acc 051)

King Fahd University of King Fahd University of Petroleum & MineralsPetroleum & Minerals

Mechanical EngineeringDynamics ME 201

BYDr. Meyassar N. Al-Haddad

Lecture # 2

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ObjectiveObjective

To introduce the concepts of position, displacement, velocity, and acceleration.

To study particle motion along a straight line.

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Rectilinear KinematicsRectilinear KinematicsSection 12.2Section 12.2

Rectilinear : Straight line motion Kinematics : Study the geometry of the motion dealing

with s, v, a. Rectilinear Kinematics : To identify at any given instant, the

particle’s position, velocity, and acceleration.

(All objects such as rockets, projectiles, or vehicles will be considered as particles “has negligible size and shape”

particles : has mass but negligible size and shape

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PositionPosition

Position : Location of a particle at any given instant with respect to the origin

r : Displacement ( Vector )s : Distance ( Scalar )

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Distance & DisplacementDistance & Displacement

Displacement : defined as the change in position.

r : Displacement ( 3 km ) s : Distance ( 8 km )

Total length

For straight-line Distance = Displacement s = r

s r

Vector is direction oriented r positive (left )r negative (right)

QUT

City

My PlaceX

3km

River

8 km

N

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Velocity & SpeedVelocity & Speed

Velocity : Displacement per unit time Average velocity : V = rt

Speed : Distance per unit time Average speed : spsTt (Always positive scalar )

Speed refers to the magnitude of velocity Average velocity :

avg = s / t

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Velocity (con.)Velocity (con.)

Instantaneous velocity :

For straight-line r = s

dtdr

trV

t

lim0

dtdsv

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ProblemProblem

A particle moves along a straight line such that its position is defined by s = (t3 – 3 t2 + 2 ) m. Determine the velocity of the particle when t = 4 s.

dtdsv

At t = 4 s, the velocity = 3 (4)2 – 6(4) = 24 m/s

tt 63 2

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AccelerationAcceleration

Acceleration : The rate of change in velocity {(m/s)/s}

Average acceleration :

Instantaneous acceleration :

If v ‘ > v “ Acceleration “ If v ‘ < v “ Deceleration”

VVV

tVaavg

2

2

0lim

dtsd

dtdv

tva

t

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ProblemProblem

A particle moves along a straight line such that its position is defined by s = (t3 – 3 t2 + 2 ) m. Determine the acceleration of the particle when t = 4 s.

At t = 4

ttdtdsv 63 2

66 tdtdva

a(4) = 6(4) - 6 = 18 m/s2

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ProblemProblem A particle moves along a straight line such that its position is defined

by s = (t3 – 12 t2 + 36 t -20 ) cm. Describe the motion of P during the time interval [0,9]

)6)(2(336243 2 ttttdtdsv

)4(6246 ttdtdva

t 0 2 4 6 9s -20 12 -4 -20 61v 36 0 -12 0 63a -24 -12 0 12 30

Total time = 9 secondsTotal distance = (32+32+81)= 145 meterDisplacement = form -20 to 61 = 81 meterAverage Velocity = 81/9= 9 m/s to the rightSpeed = 9 m/sAverage speed = 145/9 = 16.1 m/sAverage acceleration = 27/9= 3 m/s2 to the right

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Relation involving s, v, and aRelation involving s, v, and aNo time tNo time t

dtdva

vdsdt

dtdsv

advdt

adv

vds

dvvdsa

Acceleration

Velocity

Position s

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Problem 12.18Problem 12.18 A car starts from rest and moves along a straight line with

an acceleration of a = ( 3 s -1/3 ) m/s2. where s is in meters. Determine the car’s acceleration when t = 4 s.

Rest t = 0 , v = 0

31

3sv

dvvdsa

vs

dvvdss00

31

3

232

21)3(

23 vs

31

3sdtdsv

dtdss 331

ts

dtdss00

31

3

ts 323 3

2

23

)2( ts

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For constant accelerationFor constant accelerationa = aa = acc

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Velocity as a Function of TimeVelocity as a Function of Time

dtdvac

dtadv c

dtadvt

c

v

vo 0

tavv c 0

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Position as a Function of TimePosition as a Function of Time

tavdtdsv c 0

dttavdst

c

s

so 0

0 )(

200 2

1 tatvss c

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velocity as a Function of Positionvelocity as a Function of Position

s

sc

v

v

dsadvv00

dsadvv c

)(2 020

2 ssavv c

)(21

21

020

2 ssavv c

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Free Fall Free Fall Ali and Omar are standing at the top of a cliff of heightAli and Omar are standing at the top of a cliff of height HH. Both throw a ball with initial speed. Both throw a ball with initial speed vv00, Ali straight, Ali straight downdown and Omar straightand Omar straight upup. The speed of the balls . The speed of the balls

when they hit the ground arewhen they hit the ground are vvAA andand vvOO respectively.respectively. Which of the following is true:Which of the following is true:

(a)(a) vvAA < < vvOO (b) (b) vvAA = = vvOO (c) (c) vvAA > > vvOO

vv00

vv00

OmarOmarAliAli

HHvvAA vvOO

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Free fall…Free fall…

Since the motion up and back down is symmetric, intuition should tell you that v = v0

We can prove that your intuition is correct:

vv00

OmarOmar

HH

vv = v= v00

0HHg2vv 20

2 )(Equation:Equation:

This looks just like Omar threw This looks just like Omar threw the ball down with speed the ball down with speed vv00, so, sothe speed at the bottom shouldthe speed at the bottom shouldbe the same as Ali’s ball.be the same as Ali’s ball.

y = 0y = 0

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Free fall…Free fall…

We can also just use the equation directly:We can also just use the equation directly:

H0)g(2vv 20

2 Ali :Ali :

vv00

vv00

AliAli OmarOmar

y = 0y = 0

H0g2vv 20

2 )(Omar:Omar:same !!same !!

y = Hy = H

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SummarySummary

Time dependent acceleration Constant acceleration

dtdsv

)(ts

2

2

dtsd

dtdva

dvvdsa

tavv c 0

200 2

1 tatvss c

)(2 020

2 ssavv c

This applies to a freely falling object:

22 /2.32/81.9ga sftsm

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