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

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

Mechanical EngineeringDynamics ME 201

BYDr. Meyassar N. Al-Haddad

Lecture # 2

ObjectiveObjective

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

To study particle motion along a straight line.

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

PositionPosition

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

r : Displacement ( Vector )

s : Distance ( Scalar )

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

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

Velocity (con.)Velocity (con.)

Instantaneous velocity :

For straight-line r = s

dt

dr

t

rV

t

lim0

dt

dsv

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.

dt

dsv

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

tt 63 2

AccelerationAcceleration

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

Average acceleration :

Instantaneous acceleration :

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

VVV

t

Vaavg

2

2

0lim

dt

sd

dt

dv

t

va

t

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

ttdt

dsv 63 2

66 tdt

dva

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

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 ttttdt

dsv

)4(6246 ttdt

dva

t 0 2 4 6 9

s -20 12 -4 -20 61

v 36 0 -12 0 63

a -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

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

dt

dva

v

dsdt

dt

dsv

a

dvdt

a

dv

v

ds

dvvdsa

Acceleration

Velocity

Position s

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

3

1

3sv

dvvdsa

vs

dvvdss00

3

1

3

23

2

2

1)3(

2

3vs

3

1

3sdt

dsv

dtdss 33

1

ts

dtdss00

3

1

3

ts 32

3 3

2

2

3

)2( ts

For constant accelerationFor constant accelerationa = aa = acc

Velocity as a Function of TimeVelocity as a Function of Time

dt

dvac

dtadv c

dtadvt

c

v

vo

0

tavv c 0

Position as a Function of TimePosition as a Function of Time

tavdt

dsv c 0

dttavdst

c

s

so

0

0 )(

200 2

1tatvss c

velocity as a Function of Positionvelocity as a Function of Position

s

s

c

v

v

dsadvv00

dsadvv c

)(2 020

2 ssavv c

)(2

1

2

10

20

2 ssavv c

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 . The speed of the

balls when they hit the ground areballs when 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

HH

vvAA vvOO

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

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

SummarySummary

Time dependent acceleration Constant acceleration

dt

dsv

)(ts

2

2

dt

sd

dt

dva

dvvdsa

tavv c 0

200 2

1tatvss c

)(2 020

2 ssavv c

This applies to a freely falling object:

22 /2.32/81.9ga sftsm