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P1X*Dynamics & Relativity:
Newton & Einstein
Chris ParkesOctober 2004
Dynamics
Motion
Forces
Energy & Momentum Conservation
Simple Harmonic Motion
Circular Motionhttp://ppewww.ph.gla.ac.uk/~parkes/teaching/DynRel/DynRel.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 numbersin syllabus
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
dxv
2
2
dt
xd
dt
dva
e.g
0
a
0
0
Equations of motion in 1D
– Initially (t=0) at x0
– Initial velocity u,
– acceleration a,
221
0 atutxx
atuvdt
dx
aadt
xd
2
2
s=ut+1/2 at2,
where s is displacement from initial position
v=u+at
)(2
2)(2
2122
22222
atutauv
tauatuatuv
Differentiate w.r.t. time:
v2=u2+2 as
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 product
e.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
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
dt
dyv
dt
dxv yx ,
2
2)()(
dt
rd
t
tvttv
dt
vda
dt
dva
dt
dva y
yx
x ,
Projectiles
Force: -mg in y directionacceleration: -g in y direction
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 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
Relative Velocity 1De.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 observerRelative Velocity 2D
e.g. Alice walks across the boat at 1m/s.
As seen on the shore: V boat 1m/sV Alice 2m/s
V relative to shore
4.63,1/2tan
/521 22
smV
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,
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
Linear Momentum Conservation• Define momentum p=mv
• Newton’s 2nd law actually
• So, with no external forces, momentum is conserved.
• e.g. two body collision on frictionless surface in 1D
amdt
vdm
dt
vmd
dt
)(
before
after
m1 m2
m1 m2
v0 0 ms-1
v1v2
For 2D remember momentum is a VECTOR, must apply conservation, separately for x and y velocity components
Initial momentum: m1 v0 = m1v1+ m2v2 : final momentum
constpdt
pdF ,0,0 Also true for net forces
on groups of particlesIf
then constpp
FF
ii
ii
,0
Energy Conservation
• 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 can neither be created nor destroyed
•Energy can be converted from one form to another
Energy measured in Joules [J]
Collision revisited• We identify two types of collisions
– Elastic: momentum and kinetic energy conserved
– Inelastic: momentum is conserved, kinetic energy is not• Kinetic energy is transformed into other forms of energy
Initial K.E.: ½m1 v02
= ½ m1v12+ ½ m2v2
2 : final K.E.
m1 v1
m2 v2
See lecture example for cases of elastic solution 1. m1>m2
2. m1<m2
3. m1=m2
Newton’s cradle
Impulse• Change in momentum from a force acting
for a short amount of time (dt)
• NB: Just Newton 2nd law rewritten
dtFppJ 12
Impulse Where, p1 initial momentum p2 final momentum
amdt
vdm
dt
pd
dt
ppF
12
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 = 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 Power [Js-1Watts]
• Energy can be converted into work– Electrical, chemical,Or letting the
– weight fall (gravitational)• Hydro-electric power station
Work is the change in energy that results from applying a force
Fs
x
F
mgh of water
dt
dWP So, for constant Force vF
dt
xdF
dt
xFdP
This stored energy has the potential to do work Potential EnergyWe are dealing with changes in energy
0h
• choose an arbitrary 0, and look at p.e.
This was gravitational p.e., another example :
Stored energy in a SpringDo 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
hmgxFW )(
Work done by a variable forceConsider small distance dx over which force is constant
F(x)
dx
Work W=Fx dx
So, total work is sum
0 X
X
dxxFdxFW0
)(
Graph of F vs x,
integral is area under graph
work done = area
F
XdxFor spring,F(x)=-kx:
Fx
X
221
02
21
00
][)( kXkxkxdxdxxFW XXX
Stretched spring stores P.E. ½kX2
Work - Energy
• For a system conserving K.E. + P.E., then– Conservative forces
• 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
FdxUUW ordx
dUF
e.g. spring kxdx
kxdFkxU
][ 221
221
Conservative & Dissipative Forces
dx
dUF
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 Examples
• Mass on a string
1) Simple Pendulum
l
xsinIf is small
Working Horizontally: LxmgmgF sin
xl
g
dt
xd
2
2
Hence, Newton 2:
c.f. this with F=-kx on previous slide x
mg sin
mg andl
g Angular frequency for
simple pendulum,small deflection
SHM Examples2) 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?)
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
• For a body moving in a circle of radius r at speed v, the angular momentum isL=r (mv) = mr2= I
The rate of change of angular momentum is
– The product rF is called the torque of the Force
• Work done by force is Fs =(Fr)(s/r)= Torque angle in radians
Power = rate of doing work
= Torque Angular velocity
Angular Momentum
(using v=R)I is called moment of inertia
Framr
armdt
vdrmvmr
dt
Lddtd
)(
Torque
dt
dTorque
s
r
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,1
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
Moment of Inertia• Have seen corresponding angular quantities
for linear quantities– x; v; pL– Mass also has an equivalent: moment of Inertia, I– Linear K.E.:– Rotating body v, mI:– Or p=mv becomes:Conservation of ang. mom.:e.g. frisbee solid sphere hula-hoop
pc hard disk neutron star space station
221.. mvEK
221.. IEK
IL
2211 II
R
R
R1
R2
221 MRI
252 MRI
)( 22
212
1 RRMI
masses m distance from
rotation axis r dmrrmI
iii
22
Dynamics Top Five1. 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• Energy (P.E., K.E….) and momentum• Elastic/Inelastic collisions
4. SHM, Circular motion
5. Angular momentum• L=r (mv) = mr2= I • Moment of inertia
rr
vmF ˆ
2
tAx sin