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

pdF

)(

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