Transcript Transparancies for Dynamics

P1X Dynamics & Relativity:
Newton & Einstein
Part I - “I frame no hypotheses;
Dynamics
READ the
textbook!
section numbers
in syllabus
Motion
Forces – Newton’s Laws
Simple Harmonic Motion
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.”
Circular Motion
http://ppewww.ph.gla.ac.uk/~parkes/teaching/Dynamics/Dynamics.html
October 2007
Chris Parkes
Motion
• Position [m]
• Velocity [ms-1]
– Rate of change of position
dx
v
dt
x
e.g
dx
0
v
dt
t
• Acceleration [ms-2]
– Rate of change of velocity
dv d 2 x
a
 2
dt dt
0
a
0
t
Equations of motion in 1D
– Initially (t=0) at x0
– Initial velocity u,
– acceleration a,
s=ut+1/2 at2,
x  x0  ut  at
1
2
Differentiate w.r.t. time:
dx
 v  u  at
dt
2
d x
aa
2
dt
2
where s is displacement from
initial position
v=u+at
v 2  (u  at ) 2  u 2  2uat  a 2t 2
v 2  u 2  2a(ut  12 at 2 )
v2=u2+2 as
2D motion: vector quantities
• Position is a vector
– r, (x,y) or (r,  )
– Cartesian or
cylindrical polar coordinates
– For 3D would specify
z also
Scalar: 1 number
Vector: magnitude & direction,
>1 number
Y
• Right angle triangle
x=r cos , y=r sin 
r2=x2+y2, tan  = y/x
r
0

x
y
X
vector addition
• c=a+b
y
cx= ax +bx
cy= ay +by
b
can use unit vectors i,j
a
c
i vector length 1 in x direction
x
j vector length 1 in y direction
scalar product
finding the angle between two vectors
a  b  ab cos   a x bx  a y by a,b, lengths of a,b
Result is a scalar
a xbx  a y by
a b
cos 

2
2
2
2
ab
a x  a y  bx  by
a

b
Vector product
e.g. Find a vector perpendicular to two vectors
c  ab
c  a b sin 
iˆ
c  a  b  ax
bx
ˆj
ay
by
kˆ  a y bz  a z by 


a z   a z bx  a x bz 
bz  a xby  a y bx 
c
Right-handed
Co-ordinate system
b

a
Velocity and acceleration vectors
• Position changes with time
• Rate of change of r is
velocity
Y
– How much is the change in a
very small amount of time t
v
d r r (t  t )  r (t )
Limit at  t0

dt
t
vx 
dx
dy
, vy 
dt
dt
d v v(t  t )  v(t ) d 2 r
a

 2
dt
t
dt
ax 
dv y
dv x
, ay 
dt
dt
r(t)
0
x
r(t+t)
X
Projectiles
Motion of a thrown / fired
object mass m under gravity
y
Velocity components:
v
vx=v cos 

x,y,t
vy=v sin 
x
a:
v=u+at:
x direction
ax=0
vx=vcos  + axt = vcos 
s=ut+0.5at2: x=(vcos )t
Force: -mg in y direction
acceleration: -g in y direction
y direction
ay=-g
vy=vsin  - gt
y= vtsin  -0.5gt2
This describes the motion, now we can use it to solve problems
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
Relative Velocity 2D
e.g. Alice walks across the boat at 1m/s.
V boat 2m/s
As seen on the shore:
θ
V
2
2
V  1  2  5m / s relative to shore
tan   1 / 2,  27
V Alice 1m/s
Changing co-ordinate system
Define the frame of reference – the co-ordinate system –
in which you are measuring the relative motion.
(x’,y’)
y
Frame S
(shore)
vt
Frame S’
(boat) v boat w.r.t shore
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 :
x  x'vt Known as Gallilean transformations
As we will see, these simple relations do not hold in
y  y'
special relativity
We described the motion, position, velocity, acceleration,
now look at the underlying causes
Newton’s laws
• 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]
• Third Law
– The force exerted by A on B is equal and opposite to
the force exerted by B on A
Fb
•Force exerted by
block on table is Fa
Block on table
Fa=-Fb
Weight
(a Force)
Fa
•Force exerted by
table on block is 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,
Force Components
•Force is a Vector
•Resultant from vector sum
R  F1  F2
F1
R
F2
•Resolve into perpendicular components
Fx  F cos 
Fy
F
Fy  F sin 

F x  Fxiˆ
F y  Fy ˆj
Fx
Free Body Diagram
• Apply Newton’s laws to particular body
• Only forces acting on the body matter
– Net Force
F
• Separate problem into each body
e.g.
Body 1
Supporting Force
from plane
(normal
force)
Friction
Tension
In rope
Block weight
Body 2
Tension in rope
Block Weight
Tension & Compression
• Tension
– Pulling force - flexible or rigid
• String, rope, chain and bars
mg
• Compression
– Pushing force
• Bars
mg
mg
• Tension & compression act in BOTH
directions.
– Imagine string cut
– Two equal & opposite forces – the tension
Friction
• 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
N
fs or fk
F
mg
fs  s N
fk  k N
Simple Harmonic Motion
Oscillating system that can be described by sinusoidal function
Pendulum, mass on a spring, electromagnetic waves (E&B fields)…
• Occurs for any system with
F  k x
Linear restoring Force
» Same form as Hooke’s law
d2x
k
F  ma  2   x
dt
m
– Hence Newton’s 2nd
– Satisfied by sinusoidal expression
x  A sin t or x  A cos t A is the oscillation amplitude
 is the angular frequency
– Substitute in to find 
dx
d 2x
x  A sin t 
 A cos t  2   A 2 sin t
dt
k
k
   
m
m
2
dt
 in radians/sec
Frequencyf   Period T  1
Hz, cycles/sec
2 Sec for 1 cycle
f
SHM General Form

x  A sin( t   )
Displacement
A is the oscillation amplitude
- Maximum displacement
Phase
(offset of sine wave
in time)
Oscillation frequency
  2f
f  1/ T
SHM Examples
1) Mass on a spring
• Let weight hang on spring
• Pull down by distance x
– Let go!
L’
In equilibrium
F=-kL’=mg
Energy:
x
K .E.  12 mv2
U  12 kx2
Restoring Force F=-kx
k

m
(assuming spring has negligible mass)
potential energy of spring
But total energy conserved
At maximum 2of oscillation, when x=A and v=0
Total E  12 kA Similarly, for all SHM (Q. : pendulum energy?)
SHM Examples 2) Simple Pendulum
•Mass on a string

Working along swing: F  mg sin 
Not actually SHM, proportional to sin, not 
but if  is small sin     x
L
l
x
F  mg sin   mg Lx
c.f. this with F=-kx on previous slide
mg sin
mg
Hence, Newton 2:
d 2x
g


x
2
dt
l
and

g
l
Angular frequency for
simple pendulum,
small deflection
Circular Motion
360o = 2 radians
180o =  radians
90o = /2 radians
• 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
•Acceleration
d
v x  ( R cos t )   R sin t
dt
d
v y  ( R sin t )  R cos t
dt
d
d
a x  (v x )  ( R sin t )   R 2 cos t
dt
dt
d
d
a y  (v y )  ( R cos wt )   R 2 sin t
dt
dt
y
R
s
=t
x
t=0
N.B. similarity
with S.H.M eqn
1D projection of a
circle is SHM
Magnitude and direction of motion
•Velocity
v 2  vx  v y  R 2 w2 sin 2 t  R 2 2 cos 2 t   2 R 2
2
2
v=R
tan  
And direction of velocity vector v
Is tangential to the circle
vy
vx

cos t
1

 sin t
tan 
    90 o
•Acceleration
2

a
a  ax  a y 
2
v
2

R 2 w4 cos 2 t  R 2 4 sin 2 t   4 R 2
a= 2R=(R)2/R=v2/R
And direction of acceleration vector a
a= -2r
a x   2 x
a y   2 y
Acceleration is towards centre of circle
Force towards centre of circle
•
Particle is accelerating
–
•
•
1.
2.
3.
So must be a Force
Accelerating towards centre of circle
– So force is towards centre of circle
F=ma= mv2/R in direction –r 2
v
or using unit vector F  m rˆ
r
Examples of central Force
Tension in a rope
Banked Corner
Gravity acting on a satellite
Gravitational Force
Myth of Newton & apple.
He realised gravity is universal
same for planets and apples
•Any two masses m1,m2 attract each other
with a gravitational force:
F
F
m1m2
F G 2
r
r
m2
m1
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
•Explains motion of planets, moons and tides
24kg,


m
m
Gm
m
=5.97x10
E
E 
Gravity on
F  G E 2  
m
2 
RE=6378km
earth’s surface
RE
 RE 
Mass, radius of earth
GmE
2
 9.81ms
Or F  mg Hence, g 
2
RE
N.B. general solution is an ellipse not a circle - planets travel in ellipses around sun
Satellites
•Centripetal Force provided by Gravity
Mm mv2
F G 2 
R
R
M
2
M
v G
v

G
R
R
m
R
M
Distance in one revolution s = 2R, in time period T, v=s/T
R
T  2R / v  2R
GM
T2R3 , Kepler’s 3rd Law
•Special case of satellites – Geostationary orbit
•Stay above same point on earth T=24 hours
3
24  60  60  2
R  42,000km
R2
GM E
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
Oscillating system that can be described by sinusoidal function
F  k x
4.
x  A sin( t   )
Circular motion (R,)
2
v
F  m rˆ Force towards centre of circle
r