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)
