Universiti Malaysia Pahang Universiti Malaysia Pahang Manufacturing Engineering Dynamics Lecture # 2
Universiti Malaysia PahangUniversiti Malaysia Pahang
Manufacturing EngineeringDynamics
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
dtdr
trV
t
lim0
dtdsv
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
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
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
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
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
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
For constant accelerationFor constant accelerationa = aa = acc
Velocity as a Function of TimeVelocity as a Function of Time
dtdvac
dtadv c
dtadvt
c
v
vo 0
tavv c 0
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
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
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
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
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