MAE 242: DYNAMICS
Instructor
Dr. Samir N. Shoukry
Professor, MAE & CEE
assistants
K. Mc Bride
Dhananjy Rao
Dony Oomen
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© Samir N. Shoukry, 2004, Dynamics MAE 242
Key to a grade of A in MAE242
• Attend all classes and pay full attention in each class and ask questions.
BEFORE YOU COME TO CLASS YOU MUST READ THE RELATED
MATERIAL.
•Solve all the book example problems following each class. YOU MUST do
using pencil and paper; JUST REDING IS A WASTE OF TIME.
•SOLVE ALL HOMEWORK PROBLEMS: IF YOU CANNOT SOLVE A
PROBLEM GO AND ASK QUESTIONS UNTILL YOU SOLVE THE
PROBLEM ON YOUR OWN.
•KEEP TIDY RECORD OF THE SOLUTIONS.
•ALWAYS VISIT ESB115 and Ask me, William or McBride.
•REMEMBER: NO ONE WILL EXPLAIN THIS MATERIAL BETTER
THAN DR. Shoukry , Dr. William, or Mr. McBride
•YOU HAVE A GREAT OPPORTUNITY …. USE IT.
© Samir N. Shoukry, 2004, Dynamics MAE 242
Syllabus
D a te
L ec.
No.
T o p ic
R e a d in g s
H W P ro b le m s
HW Due on
8 /2 3 - 8 /2 7
2
R e c tilin e a r K in e m a tic s
1 2 .1 - 1 2 .3 1 2 : 1 4 , 3 3 , 4 3 , 5 0
3 1 -A u g
8 /3 0 – 9 /3
4
G e n e ra l C u rv ilin e a r M o tio n
1 2 .4 - 1 2 .6 1 2 : 6 7 , 7 1 , 8 5 , 9 1
7 -S e p
9 /0 6 - 9 /1 0
6
C u rv ilin e a r M o tio n
9 /1 3 - 9 /1 7
8
R e la tiv e M o tio n
9 /2 0 -9 /2 4
10
1 2 .7 -1 2 .8
In -p la n e M o tio n o f R ig id B o d ie s
12: 118, 120, 164, 169
1 4 -S e p
1 2 .9 - 1 2 .1 0 1 2 : 1 7 3 , 1 7 8 , 1 9 6 , 2 0 1
2 1 -S e p
1 6 .1 - 1 6 .4 1 6 . 7 , 1 2 , 3 8 , 4 6
2 8 -S e p
T est # 1 o n T h u rsd a y 9 /3 0 /2 0 0 4 fro m 7 :0 0 – 8 :3 0 P M
9 /2 7 -1 0 /0 1
12
R e la tiv e M o tio n A n a ly sis
1 6 .5 - 1 6 .7 1 6 : 1 0 7 , 1 2 0 ,
1 0 /0 4 -1 0 /8
14
N e w to n 's L a w s
1 3 .1 - 1 3 .6 1 3 : 1 0 , 2 2 , 7 5 , 9 7 , 1 0 2
1 0 /1 1 -1 0 /1 5
16
P la n e r K in e tic s o f R ig id B o d ie s
1 0 /1 9
1 7 .1 -1 7 .5
17: 33, 44, 50, 54
5 -O c t
1 2 -O c t
1 9 -O c t
T e s t # 2 o n T u e s d a y 1 0 /1 9 @ 7 :0 0 p m - 8 :3 0 p m
1 0 /1 8 -1 0 /2 2
18
P la n e r K in e tic s o f R ig id B o d ie s
17
17: 66, 75, 97,
2 6 -O c t
1 0 /2 5 -1 0 /2 9
20
G e n e ra l P la n e M o tio n
17
17: 103, 106, 108
2 -N o v
***
1 0 /2 9
E le c tio n D a y R e c e s s
1 1 /2
1 1 /1 -1 0 /5
21
W o rk , E n e rg y , P o w e r, E ffic ie n c y
1 4 .1 - 1 4 .4 1 4 : 7 , 1 3 , 2 3 , 4 8
9 -N o v
1 1 /0 8 -1 1 /1 2
23
C o n se rv a tio n o f E n e rg y
1 4 .5 - 1 4 .6 1 4 : 7 0 , 7 9 , 8 9
1 6 -N o v
1 1 /1 5 -1 1 /1 9
25
Im p u lse , M o m e n tu m & Im p a c t
1 5 .1 - 1 5 .4 1 5 : 1 3 , 2 0 , 3 8 , 5 2
3 0 -N o v
T h a n k s g iv in g B re a k fro m N o v e m b e r 2 0 th to N o v e m b e r 2 8 th
1 1 /2 9 -1 2 /3
1 2 /6 - 1 2 /1 0
1 2 /1 3 - 1 2 /1 8
27
W o rk a n d E n e rg y
1 8 .1 - 1 8 .5 1 8 : 2 , 1 7 , 2 3 , 4 8
7 -D e c
R e vie w
th
F IN A L E X A M O N M o n d a y D e c e m b e r 1 3 fro m 3 :0 0 to 5 :0 0 P M
© Samir N. Shoukry, 2004, Dynamics MAE 242
How motion is described?
Kinematics is the science of characterizing motion
Kinetics looks into the forces that cause the motion
DISTANCE?
POSITION?
VELOCITY?
ACCELERATION?
© Samir N. Shoukry, 2004, Dynamics MAE 242
DISTANCE
To describe the distance of travel, we
need to define a reference point.
P
S
The ball traveled a distance S from a reference point P.
If the time of travel is t , the average speed is:
S
t
Is distance a vector quantity?
How the position of P may be defined?
© Samir N. Shoukry, 2004, Dynamics MAE 242
Position
M o rg a n to w n
N ew Y o rk C ity
r = P osition vector of M organ tow n
N o rth
r = P osition vector of N ew Y ork C ity
E a st
W est
P o in t o f referen ce th at m ay b e p laced
an y w h ere … , say W
C hash
arlesto
in g ton
n, W
DV
C.
So u th
Position is defined by a vector that originates from an
origin. The vector is called POSITION VECTOR.
The origin may be selected at any convenient point.
© Samir N. Shoukry, 2004, Dynamics MAE 242
Speed
Example: Your average speed from Morgantown to New
York was 70 miles/hour. On your way you took a speeding
ticket for speeding at a velocity of 92.5 miles/hour.
Morgantown
S  455 miles
S  455 miles
t  6 . 5 hours
Averge Speed 
455
 70 MPH
6 .5
So, What is Velocity?
© Samir N. Shoukry, 2004, Dynamics MAE 242
New York
Velocity
While your average speed on the trip was 70 MPH, your speed
varied from 5 MPH driving out of your garage, to 92.5 MPH
when you where caught speeding at mile post 36 on I68.
Velocity is an instantaneous vector that
describes the rate of change in position
with respect to time.
Velocity =
instantaneous change in position
Change in time
Velocity is a VECTOR quantity
© Samir N. Shoukry, 2004, Dynamics MAE 242
Δs
r
dr

dt
r’
Position, Speed, and Velocity
P
A particle moves from P to P’ in Δt seconds
Δs
Δr
Δr is the change in position vector r in Δt seconds
P’
r
r’
r
t 0  t
Velocity = v  lim
v  dr
dt
P
O
P’
r
The magnitude v of the VECTOR v is the speed
of the particle. Since Δr=PP’, as t0 the
distance PP’  Δs. Thus, the average speed:
Δr
Δt
r’
P
P’
O
v  lim
t 0
PP '
t
 lim
t 0
s
t

r
r’
ds
dt
O
The VELOCITY VECTOR is always TANGENT to the path
© Samir N. Shoukry, 2004, Dynamics MAE 242
v
Acceleration and Deceleration
We enter I79 at speed of 45 MPH. Since the allowed speed limit is 70
MPH; one increases the velocity of his car. While a Cadillac can
accelerate to 79 MPH in 6 seconds a Subaru reaches the same velocity in
15 seconds. Driving at 85 MPH, a police car spotted; the driver reduce
his speed to 65 MPH in 16 seconds. He was spotted speeding at 82 MPH.
Acceleration is a vector that describes the rate of change
in velocity with respect to time.
Acceleration =
instantaneous change in velocity
Change in time
Acceleration is a VECTOR quantity
© Samir N. Shoukry, 2004, Dynamics MAE 242
dv

dt
Position, Velocity, & Acceleration
dr
=r
dt
v dt = dr
dr
dt =
v
dv r
=
a=
dt
a dt = dv
dt =
v=
dv
a
dr = dv
v a
© Samir N. Shoukry, 2004, Dynamics MAE 242
a = v dv
dr
Inverse Problem
1. Acceleration is a time function i.e: a=f(t)
dv
f(t) =
dt
Separate variables
f(t) dt = dv
Integrate
v = f(t) dt + C

r = ( f(t) dt + C ) dt + C’
 
2. Acceleration is a time position i.e: a=f(r)
dv
f(r) = v
dr
Separate variables
v=f(r) = dr
dt
f(r)dr =vdv
dt = dr
f(r)
Integrate
v2 
= f(r)dr + C
2 
t=  1 dr+ C’
 f(r)
© Samir N. Shoukry, 2004, Dynamics MAE 242
500

U se : a ds  v dv 
80

a ds 
0
a 
6400
 6.4 ft / s
v dv  a [ S ]
500
0
[
v
2
0
2
2 * 500
U se : a dt  dv 
t 
80
t
80
 6.4 dt 

0
0
 12.5 ft / s
2
dv  6.4 t  [ v ] 0
80
2
6.4
© Samir N. Shoukry, 2004, Dynamics MAE 242
80
]0
T he avargae velocity is found from the b eginning and
ending positions and the total travel ti m e.
V av 
2.5  0.5
C
 0.333 m / s
6
-1.5
0
B
T he average speed is found from the tota l distance of
travel devided by the total tr avel tim e.
A verage speed=
{(0.5  (  1.5))  ( 2.5  (  1.5))}
1 m / s
6
© Samir N. Shoukry, 2004, Dynamics MAE 242
0.5
A
2.5
A t t= 6 seconds, S  (6  9 * 6  15 * 6 )   18 ft
3
2
T o determ ine the total distance traveled in 6 s, w e
m ust find w hen the particle changes its direction. T his
occurs w hen the drivative of the positio n is zero.
ds
 3 t  18 t  15
2

3( t  5)( t  1)  0
st
at t  1, s  7 ft and t  5, s   25 ft .
-25 ft
7 ft
-18 ft
Total distance=7+(7-(-25))-(-25-(-18))=46 ft
© Samir N. Shoukry, 2004, Dynamics MAE 242