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