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P1X*Dynamics & Relativity:
Newton & Einstein
Part I - “I frame no hypotheses;
Dynamics
Motion
READ the
textbook!
section numbers
in syllabus
Forces
Energy & Momentum
Conservation
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.”
Simple Harmonic Motion
Circular Motion
http://ppewww.ph.gla.ac.uk/~parkes/teaching/DynRel/DynRel.html
October 2004
Chris Parkes
Motion
• Position [m]
• Velocity [ms-1]
x
e.g
dx
dx
v
dt
– Rate of change of position 0
2
• Acceleration [ms-2] a  dv  d
dt
– Rate of change of velocity
dt
x
v
2
0
a
0
dt
t
t
Equations of motion in 1D
– Initially (t=0) at x0
– Initial velocity u,
– acceleration a,
x  x0  ut  at
1
2
Differentiate w.r.t. time:
dx
 v  u  at
dt
2
d x
aa
2
dt
s=ut+1/2 at2,
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 x by  a y bx 
c
Right-handed
Co-ordinate system
b

a
Velocity and acceleration vectors
• Position changes with time
Y
• Rate of change of r is
velocity
– How much is the change in a
very small amount of time t
d r r (t  t )  r (t ) Limit at  t0
v

dt
t
0
yd
xd
 yv ,
 xv
td
td
d v v(t  t )  v(t ) d 2 r
a

 2
dt
t
dt
vd
y vd
 ya , x
 xa
td
td
r(t)
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
x
a:
v=u+at:
x direction
ax=0
vx=vcos  + axt = vcos 
s=ut+0.5at2: x=(vcos )t
vy=v sin 
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 2 m/s. what is Alices’ 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 1m/s
As seen on the shore:
V  1  2  5m / s
2
2
tan   2 / 1,  63.4
V
relative to shore
V Alice 2m/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 forceFexerted by B on A
b
•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,
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
Linear Momentum Conservation
• Define momentum p=mv
d p d (mv)
nd
• Newton’s 2 law actually F  dt  dt  m ddtv  ma
• So, with no external forces, momentum is
dp
conserved. F  0, dt  0, p  const
Also true for net forces
on groups of particles
• e.g. two body collision on frictionless If F   F 0,
surface in 1D
then p   p const
i
i
i
i
before
m1
m2
0 ms-1
Initial momentum: m1 v0 = m1v1+ m2v2 : final momentum
after
v0
m1
m2
v2
v1
For 2D remember momentum is a VECTOR, must apply
conservation, separately for x and y velocity components
Energy Conservation
•Energy can neither be created nor destroyed
•Energy can be converted from one form to another
• Need to consider all possible forms of energy in a
system e.g:
–
–
–
–
–
Kinetic energy (1/2 mv2)
Potential energy (gravitational mgh, electrostatic)
Electromagnetic energy
Work done on the system
Heat (1st law of thermodynamics of Lord Kelvin)
• Friction  Heat
Energy measured in Joules [J]
Collision revisited
m1
v1
m2
• We identify two types of collisions
– Elastic: momentum and kinetic energy conserved
Initial K.E.: ½m1 v02 = ½ m1v12+ ½ m2v22 : final K.E.
– Inelastic: momentum is conserved, kinetic energy is not
• Kinetic energy is transformed into other forms of energy
See lecture example for cases of elastic solution
Newton’s cradle
1. m1>m2
2. m1<m2
3. m1=m2
v2
Impulse
• Change in momentum from a force acting
for a short amount of time (dt)
Impulse J  p 2  p1  Fdt
Where, p1 initial momentum
p2 final momentum
• NB: Just Newton 2nd law rewritten
F
p 2  p1
dt

dp
dt
m
dv
 ma
dt
Approximating
derivative
Impulse is measured in Ns.
change in momentum is measured in kg m/s.
since a Newton is a kg m/s2 these are equivalent
Q) Estimate
the impulse
For Greg
Rusedski’s
serve
[150 mph]?
Work & Energy
Work is the change in energy that results from applying a force
• Work = Force F times Distance s, units of Joules[J]
– More precisely W=F.x
– F,x Vectors so W=F x cos
• e.g. raise a 10kg weight 2m
• F=mg=10*9.8 N,
• W=Fx=98*2=196 Nm=196J
• The rate of doing work is the
P
dW
dt
So, for constant Force
Power
P
F

x
F
[Js-1Watts]
dF x
dx
F
 F v
dt
dt
• Energy can be converted into work
– Electrical, chemical,Or letting the
mgh of
– weight fall (gravitational)
water
• Hydro-electric power station
s
This stored energy has the potential to do work Potential
We are dealing with changes in energy
• choose an arbitrary 0, and look at  p.e.
This was gravitational p.e., another example :
Energy
h
0
W  F  x  (mg)h
Stored energy in a Spring
Do work on a spring to compress it or expand it
Hooke’s law
BUT, Force depends on extension x
Work done by a variable force
Work done by a variable force
Consider small distance dx over which force is constant
F(x)
Work W=Fx dx
X
dx
So, total work is sum W   F  dx   F ( x)dx
0
X
0
Graph of F vs x,
F
integral is area under graph
work done = area
For spring,F(x)=-kx:
x
F
X
dx
X
X
0
0
X
W   F ( x)dx    kxdx  [ 12 kx2 ]0X   12 kX 2
Stretched spring stores P.E. ½kX2
Work - Energy
dU
W  U  U    Fdx or F  
dx
2
1
d
[
2
2 kx ]
1
 kx
e.g. spring U  2 kx  F  
dx
Conservative & Dissipative Forces
• For a system conserving K.E. + P.E., then
– Conservative forces
dU
F 
dx
• But if a system changes energy in some other way
(“dissipative forces”)
– Friction changes energy to heat
Then the relation no longer holds –
the amount of work done will depend on the path taken
against the frictional force
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 Examples
1) Simple Pendulum
• Mass on a string
If  is small

x
mg sin
mg
Working Horizontally:
sin  
x
l
F  mg sin   mg Lx
c.f. this with F=-kx on previous slide
Hence, Newton 2:
and

g
l
d 2x
g
 x
2
dt
l
Angular frequency for
simple pendulum,
small deflection
SHM Examples
2) 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?)
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 
    90o
•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
Angular Momentum
• For a body moving in a circle of radius r at speed
v, the angular momentum is
L=r (mv) = mr2= I 
(using v=R)
I is called moment of inertia
The rate of change of angular momentum is
dL d
dv
 dt (r  mv)  mr 
 mr  a
dt
dt
 r  ma  r  F

r
s
– The product rF is called the torque of the Force
• Work done by force is Fs =(Fr)(s/r)
= Torque  angle in radians
Power
d
 Torque 
 Torque  
dt
= rate of doing work
= Torque  Angular velocity
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
=5.97x10


mE m
GmE
E
Gravity on
m
F G
 
RE=6378km
2
2

earth’s surface
RE
RE 
Mass, radius of earth

GmE
1
 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
Moment of Inertia
• Have seen corresponding angular quantities
masses m
for linear quantities I   mi ri 2  r 2dm distance
from

i
rotation axis r
– x; v; pL
– Mass also has an equivalent: moment of Inertia, I
– Linear K.E.: K .E.  mv
– Rotating body v, mI: K .E.  I
– Or p=mv becomes: L  I 
Conservation of ang. mom.: I11  I 2  2
e.g.
frisbee
solid sphere
hula-hoop
2
2
I  1 M (R  R )
1
2
2
1
2
pc hard disk
I  MR
1
2
2
neutron star
I  52 MR 2

R

2
space station

1
2
R1
R2
R
2
Dynamics Top Five
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
3. Conservation Laws
•
•
4.
Energy (P.E., K.E….) and momentum
Elastic/Inelastic collisions
SHM, Circular motion
v2
F  m rˆ
x  A sin t
r
5. Angular momentum
•
L=r (mv) = mr2= I 
•
Moment of inertia