P1X Dynamics & Relativity:

Newton & Einstein

Chris ParkesOctober 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.”

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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

dtdt

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 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

Force towards centre of circle

xkF