Rigid Body Dynamics
(MENG233)
Instructor:
Dr. Mostafa Ranjbar
Textbook
Engineering Mechanics, Dynamics
R. C. Hibbeler, 10th Edition
Grading system
• 25%
• 30%
• 45%
Attendance, Quizzes, Homework Assignments
Midterm Exam
Final Exam
So, You Must be Active!
Homework
• Homework problems are pre-assigned in syllabus
every week.
• All homework problems assigned during a given
week are due in class on the following week unless
stated otherwise.
• Late Homework will not be accepted
• Attendance will be checked during each lecture.
Course coverage
•
•
•
•
Kinematics of a Particle. (Ch. 12)
Kinetics of a Particle: Force and Acceleration. (Ch. 13)
Kinetics of a Particle: Work and Energy. (Ch. 14)
Kinetics of a Particle: Impulse and Momentum. (Ch. 15)
• Planar Kinematics of a Rigid Body. (Ch. 16)
• Planar Kinematics of a Rigid Body: Force and
Acceleration. (Ch. 17)
• Planar Kinematics of a Rigid Body: Work and Energy.
(Ch. 18)
• Planar Kinematics of a Rigid Body: Impulse and
Momentum. (Ch. 19)
Areas of mechanics (section12.1)
1) Statics
-
Concerned with body at rest.
2) Dynamics
-
F
F
M
x
0
y
0
0
Concerned with body in motion
1. Kinematics: is a study the
geometric aspect of the motion.
2. Kinetics: Analysis of forces that
causing the motion
F
F
M
x
 ma x
y
 ma y
 I
Review of Vectors and Scalars
• A Vector quantity has both magnitude
and direction.
• A Scalar quantity has magnitude only.
• Scalars (e.g)
• Vectors (e.g.)
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
distance
speed
mass
temperature
pure numbers
time
pressure
area, volume
charge
energy
displacement
velocity
acceleration
force
weight (force)
momentum
Vectors
• Can be represented by an arrow (called the
“vector”).

F
• Length of a vector represents its
magnitude.
• Symbols for vectors:
–
(e.g. force) F , or F (bold type), or
F
2F
F
Position
•
Position : Location of a particle
at any given instant with respect
to the origin
r : Displacement ( Vector )
s : Distance ( Scalar )
Distance & Displacement
•
•
•
Displacement : defined as the
change in position.
r : Displacement ( 3 km )
s : Distance
( 8 km )
Total length
N
River
City
My Place
X
3km
QUT
•
For straight-line
Distance = Displacement
s
=
r
Ds 
Dr
Vector is direction oriented
Dr positive (left )
Dr negative (right)
8 km
• Distance
• Displacement
– Total length of path
travelled
– Must be greater
than (or equal to)
magnitude of
displacement
– Only equal if path is
straight
– Symbol d
– Refers to the change
in particle’s position
vector
– Direct distance
– Shortest distance
between two points
– Distance between
Start and End points
– “as the crow flies”
– Can be describe with
only one direction
– Symbol S
– S = X final - X initial
Average Speed and Average
Velocity
distance d
Average Speed =

time
t
displacement s
Average Velocity =

time
t
Dx x  x o
v

t
t
Velocity & Speed
•
•
•
Velocity : Displacement per unit time
Average velocity :
V = Dr / Dt
•
•
Speed : Distance per unit time
Average speed :

•
•
usp  sT / Dt (Always positive scalar )
Speed refers to the magnitude of velocity
Average velocity :
uavg = Ds / Dt
Velocity (con.)
•
Instantaneous velocity :
Dr
V  lim
Dt 0 Dt
•
For straight-line
Dr = Ds
ds
v
dt
dr

dt
Average Speed and Average Velocity
distance d
Average Speed =

time
t
displaceme nt s
Average Velocity =

time
t
Dx x  xo
v

t
t
# Problem
• 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.
ds
v
 3t 2  6t
dt
At t = 4 s, the velocity = 3 (4)2 – 6(4) = 24 m/s
Acceleration
•
•
Acceleration : The rate of change in
velocity {(m/s)/s}
 V
D
V

V
Average acceleration :
aavg 
•
Instantaneous acceleration :
DV
Dt
Dv dv d 2 s
a  lim

 2
Dt 0 Dt
dt dt
•
•
If v ‘ > v
If v ‘ < v
“ Acceleration “
“ Deceleration”
Problem
•
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.
v
ds
 3t 2  6t
dt
a
dv
 6t  6
dt
• At t = 4
a(4) = 6(4) - 6 = 18 m/s2
Problem
•
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]
ds
 3t 2  24t  36  3(t  2)(t  6)
dt
dv
a
 6t  24  6(t  4)
dt
v
t
0
2
4
6
9
s
v
a
-20
36
-24
12
0
-12
-4
-12
0
-20
0
12
61
63
30
Total time = 9 seconds
Total distance = (32+32+81)= 145 meter
Displacement = form -20 to 61 = 81 meter
Average Velocity = 81/9= 9 m/s to the right
Speed = 9 m/s
Average speed = 145/9 = 16.1 m/s
Average acceleration = 27/9= 3 m/s2 to the right
Relation involving s, v, and a
No time t
Position s
Velocity
ds
v
dt
Acceleration a 
dv
dt
dt 
ds
v
dt 
dv
a
a ds  v dv
ds dv

v
a
Problem 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)

a ds  v dv
s

s ds  3 dt
v
1
3
 3s ds   vdv
0
0
2
s

1
3
t
ds   3 dt
0
2
3
3
s  3t
2
1
3
ds
v
 3s
dt
s
0
3
1 2
3
(3) s  v
2
2
v  3s
1
3
1
3
s  (2t )
3
2
For constant acceleration
a = ac
Velocity as a Function of Time
dv
ac 
dt
dv  ac dt
v
t
 dv   a
c
vo
dt
0
v  v0  act
Position as a Function of Time
ds
v
 v0  ac t
dt
s
t
 ds   (v
0
so
 act ) dt
0
1 2
s  s0  v0t  ac t
2
Velocity as a Function of Position
v dv  ac ds
v
s
 v dv   a
c
v0
ds
s0
1 2 1 2
v  v0  ac ( s  s0 )
2
2
v  v  2ac ( s  s0 )
2
2
0
Summary
•
Time dependent acceleration
s (t )
ds
v
dt
2
dv d s
a
 2
dt dt
a ds  v dv
•
Constant acceleration
v  v0  act
1 2
s  s0  v0t  ac t
2
v  v  2ac ( s  s0 )
2
2
0
This applies to a freely falling object:
a  g  9.81 m / s 2  32.2 ft / s 2
Thank you