P1X Dynamics & Relativity: Newton & Einstein Chris Parkes ctober 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
P1X Dynamics & Relativity : Newton & Einstein. 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.”. - PowerPoint PPT Presentation
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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 trianglex=r cos , y=r sin r2=x2+y2, tan = y/x
Scalar: 1 number
Vector: magnitude & direction, >1 number
0 X
Y
x
yr
vector addition• c=a+b
cx= ax +bx
cy= ay +by
scalar product
x
y
a
b
ccan 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
2222cos
yxyx
yyxx
bbaa
baba
ab
ba
a,b, lengths of a,b
Result is a scalaryyxx babaabba cos
a
b
Vector producte.g. Find a vector perpendicular to two vectors
sinbac
bac
xyyx
zxxz
yzzy
zyx
zyx
baba
baba
baba
bbb
aaa
kji
bac
ˆˆˆ
a
b
c
Right-handed Co-ordinate system
dt
dva
dt
dva y
yx
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
trttr
dt
rdv
)()(
Limit at t0
2
2)()(
dt
rd
t
tvttv
dt
vda
dt
dyv
dt
dxv yx ,
ProjectilesMotion 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 directiona:
v=u+at:
s=ut+0.5at2:
ax=0 ay=-gvx=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 directionacceleration: -g in y direction
Relative Velocity 2D
V boat 2m/sV Alice 1m/s
V relative to shore
27,2/1tan
/521 22
smV
Relative Velocity 1De.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
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 shorei.e. the co-ordinate transformation from frame S to S’Assuming S and S’ coincide at t=0 :
'
'
yy
vtxx
Known as Gallilean transformations
As we will see, these simple relations do not hold in special relativity
y
• 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
• 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,
Force Components
21 FFR
1F
2F
R
sin
cos
FF
FF
y
x
xF
yF F
iFF xxˆ
jFF yyˆ
•Force is a Vector•Resultant from vector sum
•Resolve into perpendicular components
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
TensionIn rope
Block weightFriction
Body 2
Tension in rope
Block Weight
e.g.
F
Supporting Force from plane(normal force)
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
mgmg
mg
• 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
Ffs or fk
Nf
Nf
kk
ss
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 functionPendulum, mass on a spring, electromagnetic waves (E&B fields)…
xkF x
m
k
dt
xdamF
2
2
tAx sin or tAx cos A is the oscillation amplitude is the angular frequency
tAdt
xdtA
dt
dxtAx sincossin 2
2
2
m
k
m
k 2 in radians/sec
2
ff
T1
PeriodSec for 1 cycle
FrequencyHz, cycles/sec
SHM General Form
)sin( tAx
A is the oscillation amplitude- Maximum displacement
DisplacementOscillation frequency
f 2
Phase(offset of sine wave
in time)
Tf /1
SHM Examples1) Mass on a spring
• Let weight hang on spring
• Pull down by distance x– Let go!
In equilibriumF=-kL’=mg
L’
xRestoring Force F=-kx
m
k
Energy: 221.. mvEK (assuming spring has negligible mass)
221 kxU potential energy of spring
But total energy conservedAt maximum of oscillation, when x=A and v=0
221 kAE Total Similarly, for all SHM (Q. : pendulum energy?)
SHM Examples 2) Simple Pendulum
l
xsinbut if is small
Working along swing: sinmgF
xl
g
dt
xd
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
LxmgmgF sin
Not actually SHM, proportional to sin, not
•Mass on a string
Circular Motion
x
y=t
R
t=0
s
360o = 2 radians180o = radians90o = /2 radians
tRtRdt
dv
tRtRdt
dv
y
x
cos)sin(
sin)cos(
tRwtRdt
dv
dt
da
tRtRdt
dv
dt
da
yy
xx
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-ordinatesx= 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
Magnitude and direction of motion22222222222 cossin RtRtwRvvv yx
And direction of velocity vector vIs tangential to the circle o
x
y
t
t
v
v
90
tan
1
sin
costan
v
24242242
222
sincos RtRtwR
aaa yx
And direction of acceleration vector a
a
ya
xa
y
x
2
2
•Velocity
v=R
•Acceleration
a= 2R=(R)2/R=v2/R
a= -2r Acceleration is towards centre of circle
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
rr
vmF ˆ
2
Gravitational ForceMyth of Newton & apple.
He realised gravity is universal same for planets and apples
221
r
mmGF
Newton’s law of GravityInverse square law 1/r2, r distance between massesThe gravitational constant G = 6.67 x 10-11 Nm2/kg2
FF
m1
m2r
Gravity on earth’s surface
mR
Gm
R
mmGF
E
E
E
E
22
Or mgF Hence,2
2 81.9 msR
Gmg
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:
SatellitesN.B. general solution is an ellipse not a circle - planets travel in ellipses around sun
M
m
RR
mv
R
MmGF
2
2
R
MGv 2
R
MGv
Distance in one revolution s = 2R, in time period T, v=s/T
GM
RRvRT 2/2 T2R3 , Kepler’s 3rd Law
•Special case of satellites – Geostationary orbit•Stay above same point on earth T=24 hours
kmR
GM
R
E
000,42
26060242
3
•Centripetal Force provided by Gravity
Dynamics I – Key Points1. 1D motion, 2D motion as vectors
– s=ut+1/2 at2v=u+at v2=u2+2 as– Projectiles, 2D motion analysed in
components2. Newton’s laws
– F = ma– Action & reaction
3. SHM
4. Circular motion (R,)
)sin( tAx
rr
vmF ˆ
2
Oscillating system that can be described by sinusoidal function