Rigid Body Dynamics (MENG233) Instructor: Dr. Mostafa Ranjbar
Jan 03, 2016
Rigid Body Dynamics(MENG233)
Instructor:
Dr. Mostafa Ranjbar
Textbook
Engineering Mechanics, Dynamics
R. C. Hibbeler, 10th Edition
Grading system
• 25% Attendance, Quizzes, Homework Assignments
• 30% Midterm Exam
• 45% 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 - 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
0
0
0
x
y
F
F
M
x x
y y
F ma
F ma
M I
Review of Vectors and Scalars
• A Vector quantity has both magnitude and direction.
• A Scalar quantity has magnitude only.
• Scalars (e.g)Scalars (e.g)– distance– speed– mass– temperature– pure numbers– time– pressure– area, volume– charge– energy
• Vectors (e.g.)Vectors (e.g.)– displacement– velocity– acceleration– force– weight (force)– momentum
Vectors
• Can be represented by an arrow (called the “vector”).
• Length of a vector represents its magnitude.
• Symbols for vectors: – (e.g. force) F , or F (bold type), or
F
F 2 F
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
• 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
• DistanceDistance
– Total length of path travelled
– Must be greater than (or equal to) magnitude of displacement
– Only equal if path is straight
– Symbol d
• DisplacementDisplacement
– 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
distanceAverage Speed =
timedt
displacementAverage Velocity =
timeo
st
x x xv
t t
Velocity & 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.)
• Instantaneous velocity :
• For straight-line r = s
dt
dr
t
rV
t
lim0
dt
dsv
Average Speed and Average Velocity
t
d
time
distance= Speed Average
t
xx
t
xv
t
s
o
time
ntdisplaceme =Velocity Average
# 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.
dt
dsv
At t = 4 s, the velocity = 3 (4)2 – 6(4) = 24 m/s
tt 63 2
Acceleration• 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
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.
• At t = 4
ttdt
dsv 63 2
66 tdt
dva
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]
)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 meter
Displacement = form -20 to 61 = 81 meterAverage Velocity = 81/9= 9 m/s to the right
Speed = 9 m/sAverage speed = 145/9 = 16.1 m/s
Average acceleration = 27/9= 3 m/s2 to the right
Relation involving s, v, and aNo time t
dt
dva
v
dsdt
dt
dsv
a
dvdt
a
dv
v
ds
dvvdsa
Acceleration
Velocity
Position s
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)
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 accelerationa = ac
Velocity as a Function of Time
dt
dvac
dtadv c
dtadvt
c
v
vo
0
tavv c 0
Position as a Function of Time
tavdt
dsv c 0
dttavdst
c
s
so
0
0 )(
200 2
1tatvss c
Velocity 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
Summary• 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
Thank you