Lecture Presentation Slides 1-2!1!2011

Engineering Mechanics 100 DYNAMICS

Basic Concepts and Rectilinear Motion of Particles

Dr Yu(Roger) Dong Department of Mechanical Engineering Email: [email protected]

L1-2

Statics

VS.

Dynamics

Bodies at REST Bodies in MOTION

Dynamics of Liquids

Dynamics of Solid Bodies

e.g. Robotics

Why Dynamics So Important? e.g. Hydrodynamics

Dynamics of Gasses/Air e.g. Aerodynamics

1.0 Introduction Mechanics:

The action and effects of forces on bodies Bodies at rest, or in equilibrium

Statics Mechanics

Chaos Pendulum Video

Bodies in motion, or out of equilibrium

Dynamics

In Equilibrium

v=0

Out of Equilibrium

Be static or move with constant velocity v=0.2 m/s

m

m

Static

Move with v=constant

Accelerate with the change of velocity m

a=0.5 m/s2 θ

Study of motion without reference to the Kinematics forces producing motion: Relations applied only between position, velocity, acceleration and time

Dynamics

Kinetics

Relation between unbalanced forces and the change in motion they produce

v a


Kinematics: how fast, how far and how long

B

the motion takes A

Kinematics: e.g. Motion of rocket from position A to B


Kinetics: What forces were involved to produce the motion? θ

F Kinetics: e.g. Motion of pendulum ball applied by F

- Weight - Friction - Tension - Spring Force - Support Force How about the resulting acceleration?

1.1 Basic Concepts 1) Vectors (magnitude +direction) and Scalars (magnitude only) position – “s” (or “s”)

distance –“s”

velocity – “v” (or “v”)

speed – “v”

force – “F” (or “F”)

moment – “M”

2) Space: Geometric region occupied by bodies which is used to determine the position relative to the reference system. Time: Measure of the succession of events with an absolute quantity in Newtonian mechanics Mass: Quantitative measure of the inertia Force: Vector action of one body to another 3) SI Units

Basic units

1.1 Basic Concepts (cont’) 4) Particles:

- a body of negligible dimensions - a body with dimensions irrelevant to the motion or the action of forces upon it B

A 5) Rigid Body:

=

Equivalent Particle

B

A

- important overall dimensions of the body or changes in position of the body - negligible deformation (change in shape) of the body Negligible spring deformation

=

Rigid body

6) Flexible Body: - deformed body under loads - beyond the scope of this course and about to cover in Year 2 Strength of Materials 232

1.2 Newton’s Laws of Motion These are fundamental laws relating forces and motion. Law I. A particle remains at rest or continues to move in a straight line with a constant velocity if there is no unbalanced force acting on it. ∑F=0

Sir Isaac Newton (1643-1727)

In equilibrium

Law II. The acceleration of a particle is proportional to the resultant force acting on it and is in the direction of this force. ∑F=ma

Out of equilibrium

Law III. The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear. F

F'

F= F'

Laws I and II are strictly true only in an absolute frame of reference (i.e. A particle does not accelerate for Law I and does not rotate for Law II)

Law II (Most commonly used in dynamics) ∑F=ma ∑F: resultant force acting on a body (vector)

Where

m: mass of the body (scalar)

a: the resulting acceleration of the body (vector)

This equation relates applied forces (∑F) to the motion of a body (a).

F1

Fn

∑F

F2

m Fn-1 .. …

F4

F3

=

m

∑F=F1+F2+F3+F4+…..Fn-1+Fn=ma

1.3 Topics to Cover Section 1. Basic Concepts and Rectilinear Motion of Particles

Topic Covered • • •

2. Plane Curvilinear Motion of Particles

•

3. Relative Motion of Particles

Introduction to dynamics Kinematic equations, derivation and application Linear motion with variable acceleration

Num. Lectures

2

Projectile motion in x-y rectangular coordinate system Curvilinear motion in n-t coordinate system

2

•

Relative motion analysis of two particles using translating axes

1

4. Kinetics of Particles: Force and Acceleration

•

Application of equation of motion (∑ ∑F=ma) using x-y coordinate system Application of equation of motion (∑ ∑F=ma) using n-t coordinate system

5. Impulse and Momentum

• •

Principle of linear impulse and momentum Conservation of linear momentum

• • •

Work down by forces, power Principle of work and kinetic energy Kinetic energy, potential energy, gravitational/elastic potential energy Work and energy equation Energy conservation

6. Principle of Work and Energy

•

•

• • 7. Plane Kinematics and kinetics of Rigid Bodies

• • •

Kinematics of angular motion Kinetics of rigid bodies, translation motion of rigid bodies Fixed axis motion of rigid bodies

2

1 2

2

1.4 Reference Books and Online Materials J.L.

Meriam, L.G. Kraige (2006), Engineering Mechanics: Dynamics, SI Version, 6th ed., Wiley. R.C.

Hibbeler (2010), Engineering Mechanics: Dynamics,12th ed., Pearson.

Lecture Notes Lecture Slides Tutorials Problems /Solutions Lab Sheets iLectures

Login from OASIS (www.oasis.curtin.edu.au)

2. Rectilinear Kinematics: Continuous Motion 2.1 Introduction Kinematics ─ “Geometry of motion” to describe the motion without reference to the forces producing it. ─ Only relations between position, velocity, acceleration and time.

Kinematics Examples

Motion of Gears

Flight Trajectory of Rocket

Car Route Planning

Motion can be constrained (forced to follow a specific path: e.g. car trip, train on tracks) or unconstrained (can move in any direction: e.g. aircraft flight path, trajectory of a ball after it is thrown)

Tennis Ball Bouncing Unconstrained Motion

Train Running on Tracks Constrained Motion

y

P

2-D Coordinate Systems to Describe Motion: • Rectangular coordinate (x, y)

r

t θ

• Polar coordinate (r,θ)

n

• Normal (perpendicular) and Tangent (along the path) coordinates

x

2.2 General Notions for Velocity and Acceleration Consider a particle moving along a straight line. Let point O be the origin, or reference point. -s

o

P

P'

+s

∆s

s

t

t+∆t Displacement of the Particle

s p ' − s p = (s + ∆ s ) − s = ∆ s

Average Velocity of the Particle

v ave =

s p' − s p t p' − t p

=

( s + ∆s ) − s = ∆ s (t + ∆t ) − t ∆t

Limit of vave as ∆t→0 Instantaneous Velocity of the Particle

 ∆s  ds v = lim  = ∆t →0 ∆t   dt

or

v=

ds = s& dt

1

o

-s

P

P' v

Average Acceleration of the Particle

aave =

v p' − v p t p' − t p

=

+s v+∆v

(v + ∆v ) − v = ∆v (t + ∆t ) − t ∆t

Limit of aave as ∆t→0 Instantaneous Acceleration of the Particle

 ds  d  2 2 or a = dv =  dt  = d s = &s& dt dt dt 2

 ∆v  dv a = lim  = = v& ∆t →0 ∆t   dt

Alternatively, acceleration and velocity can be related based on the chain rule

a=

dv dv ds dv = =v dt ds dt ds

3

or

vdv = ads

or

s&ds& = &s&ds

2.3 Rectilinear Motion with Constant Acceleration

1

ds = v dt

(v-s-t)

2

dv =a dt

(a-v-t)

3

vdv = ads

(a-v-s)

Integration Limits Initial Conditions (Lower Limit)

&

Final Conditions (Upper Limit)

Time=

t0

t

Position=

s0

s

Velocity=

v0

v

(a) Velocity as a Function of Time (v=v(t)) dv =a dt

2

∫

v

v0

t

t

t0

t0

[v ]vv

dv = ∫ adt = a ∫ dt

= a [t ]t 0 t

0

v − v0 = a(t − t0 ) (b) Position as a Function of Time (s=s(t)) 1

ds =v dt

∫

s

s0

ds =

∫

t

t0

vdt =

∫ [v t

t0

0

+ a ( t − t 0 ) ]dt

1 s − s0 = v 0 ( t − t 0 ) + a ( t − t 0 ) 2 2

(c) Velocity as a Function of Position (v=v(s)) 3 vdv = ads

∫

v

v0

s

vdv = ∫ ads s0

v 2 − v02 = 2a(s − s0 )

In special cases t0=0 t0=0, s0=0 and v0=0

v − v0 = at

v = at

1 s − s0 = v0 t + at 2 2 1 s = at 2 2

v 2 − v02 = 2a(s − s0 )

v 2 = 2as

Graphical Interpretation

a

Assuming t0=0

a =Const. a –t Curve

O

t v v=v0+at at

v –t Curve v0

v v0

O

t s 1 s = s0 + v0 t + at 2 2

s –t Curve

s0 O

1 v 0 t + at 2 2

s0 t

s t

WORKED EXAMPLE # 1.1 A high speed train is travelling along a straight level road bed at a speed of 240 km/hr. Determine its stopping distance if the deceleration is constant and equal to 7.0 m/s2. How much time elapsed during which the brakes were applied to stop this train? The motion of train is subjected to constant deceleration until it stops. Using v 2 − v 02 = 2a∆s , the stopping distance can be calculated as 2

 240 × 10 3   0 −  3600 v 2 − v 02   = 317.5m ∆s = = 2 × (− 7.0 ) 2a For the elapsing time, using a =

v − v0 , thus ∆t

 240 × 10 3 0 −  v − v0  3600 ∆t = = a −7

   = 9 .52 s

WORKED EXAMPLE # 1.2 A car passes you at point 1 travelling at an initial velocity of 6 m/s, and then accelerates at a constant rate to reach a velocity of 30 m/s at point 2. This occurs over an 8 second period.

1

2

(a) What is the required constant acceleration during the initial 8 sec period? (b) Calculate the distance covered by the car in this 8 sec period. (c) Once the car passes point 2 at t= 8 s, the acceleration becomes a function of time

1 8

given by a(t ) = − t + 4. Determine an equation for the velocity of the car as a function of time v(t) for t>8 s.

a=

(a) Using v 2 − v 1 = a ( t 2 − t 1 )

v 2 − v1 30 − 6 = = 3m / s 2 t 2 − t1 8

1 (b) Using ∆s = s2 − s1 = v 0 ( t 2 − t1 ) + a ( t 2 − t1 ) 2 2

∆s = 6 × 8 +

1 × 3 × 8 2 = 144 m 2

(c) When t>8s from point 2

a=

dv 1 = − t+4 dt 8

1 t2 v = ∫ adt = ∫ (− t + 4)dt = − + 4t + C 8 16

(C is a constant)

To determine C, using the initial condition @ point 2 (i.e. t=8 s, v=30 m/s)

82 30 = − + 4 × 8 + C 16

v = −

1 2 t + 4t + 2 16

C=2

3. Linear Motion with Variable Acceleration Depending on the nature of a problem, acceleration (a) may also be known in different forms including (a) a is a given function of time

“a=a(t)”

(b) a is a given function of velocity

“a=a(v)”

(c) a is a given function of displacement

“a=a(s)”

(a) Given a=a(t), develop v-t and s-t relationships

dv = a (t ) 2 dt

∫

v

v0

dv =

∫

t

t0

t

Tip: Velocity v(t) as a function of

t0

time can be found by integrating a(t)

v − v 0 = ∫ a ( t )dt 1

ds = v(t ) dt

s − s0 =

a ( t )dt

∫

s

s0

∫

t

t0

t

ds = ∫ v ( t )dt

v ( t )dt

t0

Tip: Distance s(t) as a function of time can be found by integrating v(t)

(b) Given a=a(v), develop v-t and s-v relationships 2

dv = a (v ) dt

∫

∫

v v0

v

v0

dv = a (v )

∫

t t0

dt

dv = t − t0 a (v )

This gives a relationship between velocity v and time taken t.

3

vdv = a(v)ds

s v ∫v0 a(v )dv = ∫s0 ds v

v ∫v0 a(v )dv = s − s0 v

This gives the distance travelled s before the velocity v is reached.

(c) Given a=a(s), develop v-s and s-t relationships 3 vdv = a( s )ds

∫

v

v0

s

vdv = ∫ a( s )ds s0

s

v 2 − v02 = 2∫ a( s )ds s0

This gives velocity v(s) as a function of distance s.

1

ds = v (s ) dt

∫

t

t0

ds s0 v ( s )

dt = ∫

s

∫

s

t − t0 =

s0

ds v(s)

This gives a relationship between distance s and time taken t.

Graphical Interpretation a=slope of v-t curve

v=slope of s-t curve

v= s

ds dt

a=

v

a

a=

ds v = = s& dt

t

dv = v& dt

1

a

v

1

t1

dv dt

t

t

t1

t2 s2

t2

s1

t1

s2 − s1 = ∫ ds = ∫ vdt

s=area under v-t curve

t

t dt

t1

t2 v2

t2

v1

t1

v 2 − v1 = ∫ dv = ∫ adt

v=area under a-t curve

t dt

t2

Graphical Interpretation (Cont’d) area under a-s curve

∫

v2

v1

vdv =

∫

s2

s1

a ( s )ds

or

(

)

s2 1 2 v 2 − v12 = ∫ a ( s )ds s1 2

v

a

1

dv ds

a s s1

s ds

s2

s1

 dv  a = v   ds 

a = v x slope of v-s curve

s

s2

s

WORKED EXAMPLE #1.3 A motorcycle starts from rest and travels on a straight road with a constant acceleration of 5 m/s2 for 8 sec, after which it maintains a constant speed for 2 sec. Finally it decelerates at 7 m/s2 until it stops. Plot a-t, v-t diagrams for the entire motion.Determine the total distance travelled. a (m/s2)

Sketch a-t diagram from the known accelerations, thus

 5 (0 ≤ t < 8 s )  a =  0 (8 ≤ t < 10s) − 7 (10 ≤ t ≤ t ' ) 

5

(segment I) (segment II)

8

(segment III)

v

∫

0

t

∫

dv =

0

5 dt

v = 5t

When t =8 s, v =5× ×8= 40m/s. Using this as the initial condition for segment II, thus

8 ≤ t < 10 s

∫

v

40

dv =

∫

t

t

( − 7 ) dt

8

0 dt

v = 40m / s

Similarly, for segment III

10 ≤ t ≤ t '

∫

v

40

dv =

∫

10

10

-7

Since dv=adt, the v-t diagram is determined by integrating the straight line segments of a-t diagram. Using the initial condition t=0, v=0 for segment I, we have

0 ≤ t < 8s

t (s)

v = −7t + 110

a-t Diagram

t' (=15.71)

When v=0 (i.e. motorcycle stops)

0 = −7 t '+110

t ' = 15 .71 s

v (m/s) 40

Thus, the velocity as the function of time can be expressed as

s1

s2 8

5t (0 ≤ t < 8 s )   v= 40 (8 ≤ t < 10s ) − 7t + 110 (10 ≤ t ≤ 15.71s ) 

s3 10

t (s) 15.71

v-t Diagram

The total distance travelled (using the area under v-t diagram)

1   1 s = s1 + s 2 + s 3 =  × 8 × 40  + (2 × 40 ) +  × 5 .71 × 40  = 354 .2 m 2  2 

WORKED EXAMPLE # 1.4 A test projectile is fired horizontally into a viscous liquid with a velocity v0.The retarding force is proportional to the square of the velocity, so that the acceleration becomes a=-kv2. Derive expressions for distance D travelling in the liquid and the corresponding time t required to reduce the velocity to v0/2.Neglect any vertical motion. (2/40 in M+K) Note the acceleration a is non-constant. Using

∫

D

0

dx =

v0 2 v0

∫

v0 2

 ln v  D = −   k  v0 Using

x

vdv = adx = − kv 2dx

a= v0 2 v0

∫

v

0 vdv dv 2 = − ∫v0 kv − kv 2

v0

v0 1 ln 2 0 .693 = − ln 2 = = k v0 k k

dv = − kv 2 dt

dv = − kv 2

∫

t

0

dt

t=

1 k

v0 2

1 1 = v  kv 0   v0

v

WORKED EXAMPLE #1.5 ax (m/s2) The acceleration of a particle which moves in the positive x-direction varies with its position as 0.4 shown. If the velocity of the particle is 0.8 m/s when x=0, determine the velocities v of the 0.2 particle when x=0.6 and 1.4 m. (adapted from 2/23

x (m)

in M+K) Using

∫

x

0

adx =

v

v  v −v vdv = =   ∫v 0 2  2  v0 2

v

2

2 0

0.4 0.6 0.8

1.2 1.4

For x=0.6m

v=

v + 2∫ 2 0

0 .6

0

adx =

1   0 .8 2 + 2 ×  ( 0 .4 × 0 .4 ) + ( 0 .3 + 0 .4 ) × 0 .2  = 1 .05 m / s 2  

Area under ax-x curve (0≤x ≤ 0.6) For x=1.4m 1.4 1   v = v 02 + 2 ∫ adx = 0.8 2 + 2 × ( 0.4 × 0.4) + ( 0.2 + 0.4) × 0.4 + 0.4 × 0.2 + 0  = 1.17 m / s 0 2  

Area under ax-x curve (0≤x ≤ 1.4)

Where v0=0.8 m/s

WORKED EXAMPLE #1.6 The v-s diagram for a testing vehicle travelling on a v (m/s) straight road is shown. Determine the acceleration of the vehicle at s=50 m and s=150 m. Draw the 8 a-s diagram. s (m) Since the equations for segments of v-s diagram are given, we can use ads=vdv to determine a-s diagram.

0 ≤ s < 100 m a=v

100

v = 0.08 s

dv d = ( 0 .08 s ) ( 0 .08 s ) = 0 . 0064 s ds ds

200

a (m/s2) 0.64

100 ≤ s ≤ 200 m v = −0.08 s + 16 d a = ( − 0 . 08 s + 16 ) ( − 0 . 08 s + 16 ) = 0 .0064 s − 1 .28 ds -0.64 When s=50 m, then

100

a = 0.0064× 50 = 0.32m / s 2 (acceleration in segment I)

2 When s=150 m, then a = 0.0064× 150 − 1.28 = −0.32m / s (deceleration in segment II)

200 s (m)