ENGINEERING DYNAMICS

1. P1X Dynamics & Relativity:
Newton & Einstein
Chris Parkes
October 2007
Dynamics
Motion
Forces – Newton’s Laws
Simple Harmonic Motion
Circular Motion
http://ppewww.ph.gla.ac.uk/~parkes/teaching/Dynamics/Dynamics.html
Part I - “I frame no hypotheses;
for whatever is not deduced from
the phenomena is to be called a
hypothesis; and hypotheses,
whether metaphysical or physical,
whether of occult qualities or
mechanical, have no place in
experimental philosophy.”
READ the
textbook!
section numbers
in syllabus

2. Motion
• Position [m]
• Velocity [ms-1]
– Rate of change of position
• Acceleration [ms-2]
– Rate of change of velocity
t
x
v
t
dx
dt
dt
dx
v 
2
2
dt
x
d
dt
dv
a 

e.g
0
a
0
0

3. Equations of motion in 1D
– Initially (t=0) at x0
– Initial velocity u,
– acceleration a,
2
2
1
0 at
ut
x
x 


at
u
v
dt
dx



a
a
dt
x
d


2
2
s=ut+1/2 at2,
where s is displacement from
initial position
v=u+at
)
(
2
2
)
(
2
2
1
2
2
2
2
2
2
2
at
ut
a
u
v
t
a
uat
u
at
u
v








Differentiate w.r.t. time:
v2=u2+2 as

4. 2D motion: vector quantities
• Position is a vector
– r, (x,y) or (r,  )
– Cartesian or
cylindrical polar co-
ordinates
– For 3D would specify
z also
• Right angle triangle
x=r cos , y=r sin 
r2=x2+y2, tan  = y/x
Scalar: 1 number
Vector: magnitude & direction,
>1 number
0 X
Y
x
y
r


5. vector addition
• c=a+b
cx= ax +bx
cy= ay +by
scalar product
x
y
a
b
c
can use unit vectors i,j
i vector length 1 in x direction
j vector length 1 in y direction
finding the angle between two vectors
2
2
2
2
cos
y
x
y
x
y
y
x
x
b
b
a
a
b
a
b
a
ab
b
a








a,b, lengths of a,b
Result is a scalar
y
y
x
x b
a
b
a
ab
b
a 


 
cos
a
b


6. Vector product
e.g. Find a vector perpendicular to two vectors

sin
b
a
c
b
a
c




















x
y
y
x
z
x
x
z
y
z
z
y
z
y
x
z
y
x
b
a
b
a
b
a
b
a
b
a
b
a
b
b
b
a
a
a
k
j
i
b
a
c
ˆ
ˆ
ˆ
a
b
c

Right-handed
Co-ordinate system

7. dt
dv
a
dt
dv
a
y
y
x
x 
 ,
Velocity and acceleration vectors
• Position changes with time
• Rate of change of r is
velocity
– How much is the change in a
very small amount of time t
0
X
Y
x
r(t)
r(t+t)
t
t
r
t
t
r
dt
r
d
v

 )
(
)
( 


 Limit at  t0
2
2
)
(
)
(
dt
r
d
t
t
v
t
t
v
dt
v
d
a 






dt
dy
v
dt
dx
v y
x 
 ,

8. Projectiles
Motion of a thrown / fired
object mass m under gravity
x
y
x,y,t
v

Velocity components:
vx=v cos 
vy=v sin 
x direction y direction
a:
v=u+at:
s=ut+0.5at2:
ax=0
ay=-g
vx=vcos  + axt = vcos  vy=vsin  - gt
This describes the motion, now we can use it to solve problems
x=(vcos )t y= vtsin  -0.5gt2
Force: -mg in y direction
acceleration: -g in y direction

9. Relative Velocity 2D
V boat 2m/s
V Alice 1m/s
V
relative to shore






27
,
2
/
1
tan
/
5
2
1 2
2


s
m
V
Relative Velocity 1D
e.g. Alice walks forwards along a boat at 1m/s and the boat moves at 2m/s.
What is Alice’s velocity as seen by Bob ?
If Bob is on the boat it is just 1 m/s
If Bob is on the shore it is 1+2=3m/s
If Bob is on a boat passing in the opposite direction….. and the earth is
spinning…
Velocity relative to an observer
e.g. Alice walks across the boat at 1m/s.
As seen on the shore:
θ

10. Changing co-ordinate system
vt
Frame S
(shore)
Frame S’
(boat) v boat w.r.t shore
(x’,y’)
Define the frame of reference – the co-ordinate system –
in which you are measuring the relative motion.
x
x’
Equations for (stationary) Alice’s position on boat w.r.t shore
i.e. the co-ordinate transformation from frame S to S’
Assuming S and S’ coincide at t=0 :
'
'
y
y
vt
x
x


 Known as Gallilean transformations
As we will see, these simple relations do not hold in
special relativity
y

11. • First Law
– A body continues in a state of rest or uniform
motion unless there are forces acting on it.
• No external force means no change in velocity
• Second Law
– A net force F acting on a body of mass m [kg]
produces an acceleration a = F /m [ms-2]
• Relates motion to its cause
F = ma units of F: kg.m.s-2, called Newtons [N]
Newton’s laws
We described the motion, position, velocity, acceleration,
now look at the underlying causes

12. • Third Law
– The force exerted by A on B is equal and opposite to
the force exerted by B on A
Block on table
Weight
(a Force)
Fb
Fa
•Force exerted by
block on table is Fa
•Force exerted by
table on block is Fb
Fa=-Fb
(Both equal to weight)
Examples of Forces
weight of body from gravity (mg),
- remember m is the mass, mg is the force (weight)
tension, compression
Friction,

13. Force Components
2
1 F
F
R 

1
F
2
F
R


sin
cos
F
F
F
F
y
x


x
F
y
F F
i
F
F x
x
ˆ

j
F
F y
y
ˆ


•Force is a Vector
•Resultant from vector sum
•Resolve into perpendicular components

14. Free Body Diagram
• Apply Newton’s laws to particular body
• Only forces acting on the body matter
– Net Force
• Separate problem into each body
Body 1
Tension
In rope
Block weight
Friction
Body 2
Tension in rope
Block Weight
e.g.
F
Supporting Force
from plane
(normal
force)

15. Tension & Compression
• Tension
– Pulling force - flexible or rigid
• String, rope, chain and bars
• Compression
– Pushing force
• Bars
• Tension & compression act in BOTH
directions.
– Imagine string cut
– Two equal & opposite forces – the tension
mg
mg
mg

16. • A contact force resisting sliding
– Origin is chemical forces between atoms in the two
surfaces.
• Static Friction (fs)
– Must be overcome before an objects starts to move
• Kinetic Friction (fk)
– The resisting force once sliding has started
• does not depend on speed
Friction
mg
N
F
fs or fk
N
f
N
f
k
k
s
s





17. Simple Harmonic Motion
• Occurs for any system with Linear restoring Force
» Same form as Hooke’s law
– Hence Newton’s 2nd
– Satisfied by sinusoidal expression
– Substitute in to find 
Oscillating system that can be described by sinusoidal function
Pendulum, mass on a spring, electromagnetic waves (E&B fields)…
x
k
F 

x
m
k
dt
x
d
a
m
F 


 2
2
t
A
x 
sin
 or t
A
x 
cos
 A is the oscillation amplitude
 is the angular frequency
t
A
dt
x
d
t
A
dt
dx
t
A
x 



 sin
cos
sin 2
2
2






m
k
m
k


 
2
 in radians/sec 

2

f
f
T
1

Period
Sec for 1 cycle
Frequency
Hz, cycles/sec

18. SHM General Form
)
sin( 
 
 t
A
x
A is the oscillation amplitude
- Maximum displacement
Displacement
Oscillation frequency
f

 2

Phase
(offset of sine wave
in time)
T
f /
1



19. SHM Examples
1) Mass on a spring
• Let weight hang on spring
• Pull down by distance x
– Let go!
In equilibrium
F=-kL’=mg
L’
x
Restoring Force F=-kx
m
k


Energy: 2
2
1
.
. mv
E
K  (assuming spring has negligible mass)
2
2
1
kx
U  potential energy of spring
But total energy conserved
At maximum of oscillation, when x=A and v=0
2
2
1
kA
E 
Total Similarly, for all SHM (Q. : pendulum energy?)

20. SHM Examples 2) Simple Pendulum
l
x



sin
but if  is small
Working along swing: 
 sin
mg
F 

x
l
g
dt
x
d


2
2
Hence, Newton 2:
c.f. this with F=-kx on previous slide
and
l
g


Angular frequency for
simple pendulum,
small deflection
x
mg sin

mg
L
L
x
mg
mg
F 


 
 sin
Not actually SHM, proportional to sin, not 
•Mass on a string

21. Circular Motion
x
y
=t
R
t=0
s
360o = 2 radians
180o =  radians
90o = /2 radians
t
R
t
R
dt
d
v
t
R
t
R
dt
d
v
y
x






cos
)
sin
(
sin
)
cos
(





t
R
wt
R
dt
d
v
dt
d
a
t
R
t
R
dt
d
v
dt
d
a
y
y
x
x







sin
)
cos
(
)
(
cos
)
sin
(
)
(
2
2









•Acceleration
• Rotate in circle with constant angular speed 
R – radius of circle
s – distance moved along circumference
=t, angle  (radians) = s/R
• Co-ordinates
x= R cos  = R cos t
y= R sin  = R sin t
• Velocity
N.B. similarity
with S.H.M eqn
1D projection of a
circle is SHM

22. Magnitude and direction of motion
2
2
2
2
2
2
2
2
2
2
2
cos
sin R
t
R
t
w
R
v
v
v y
x 


 




And direction of velocity vector v
Is tangential to the circle
o
x
y
t
t
v
v
90
tan
1
sin
cos
tan















v

2
4
2
4
2
2
4
2
2
2
2
sin
cos R
t
R
t
w
R
a
a
a y
x



 




And direction of acceleration vector a
a
y
a
x
a
y
x
2
2






•Velocity
v=R
•Acceleration
a= 2R=(R)2/R=v2/R
a= -2r Acceleration is towards centre of circle

23. Force towards centre of circle
• Particle is accelerating
– So must be a Force
• Accelerating towards centre of circle
– So force is towards centre of circle
F=ma= mv2/R in direction –r
or using unit vector
• Examples of central Force
1. Tension in a rope
2. Banked Corner
3. Gravity acting on a satellite
r
r
v
m
F ˆ
2



24. Gravitational Force
Myth of Newton & apple.
He realised gravity is universal
same for planets and apples
2
2
1
r
m
m
G
F 
Newton’s law of Gravity
Inverse square law 1/r2, r distance between masses
The gravitational constant G = 6.67 x 10-11 Nm2/kg2
F
F
m1
m2
r
Gravity on
earth’s surface
m
R
Gm
R
m
m
G
F
E
E
E
E









 2
2
Or mg
F  Hence,
2
2
81
.
9 

 ms
R
Gm
g
E
E
mE=5.97x1024kg,
RE=6378km
Mass, radius of earth
•Explains motion of planets, moons and tides
•Any two masses m1,m2 attract each other
with a gravitational force:

25. Satellites
N.B. general solution is an ellipse not a circle - planets travel in ellipses around sun
M
m
R
R
mv
R
Mm
G
F
2
2


R
M
G
v 
2
R
M
G
v 
Distance in one revolution s = 2R, in time period T, v=s/T
GM
R
R
v
R
T 
 2
/
2 
 T2R3 , Kepler’s 3rd Law
•Special case of satellites – Geostationary orbit
•Stay above same point on earth T=24 hours
km
R
GM
R
E
000
,
42
2
60
60
24
2
3



 
•Centripetal Force provided by Gravity

26. Dynamics I – Key Points
1. 1D motion, 2D motion as vectors
– s=ut+1/2 at2 v=u+at v2=u2+2 as
– Projectiles, 2D motion analysed in
components
2. Newton’s laws
– F = ma
– Action & reaction
3. SHM
4. Circular motion (R,)
)
sin( 
 
 t
A
x
r
r
v
m
F ˆ
2


Oscillating system that can be described by sinusoidal function
Force towards centre of circle
x
k
F 
