1PHYS 1121 and 1131.www.phys.unsw.edu.au/~jw/1131.html www.physclips.unsw.edu.auLecture timetable. Lectures are shared for 1121 and 1131. However, there are three different lecturesets.Group A (1121&1131) Group B (1121&1131) Group C (overflow – probably temporary)L1 Mon 1-2 Mon 12-1 Tue 11-12L2 Wed 5-6 Wed 11-12 Wed 3-4L3 Fri 11-12 Thur 11-12 Fri 2-3No tuts or labs in week 1.Course pack. (i) Buy the lab book at the bookshop (ii) then present this at the First Year Lab to getthe rest of the material before your first tut or lab class.1121 vs 1131. Material covered is the same. 1131 has harder tutorials, harder exam.Text. Serway and Jewett: Physics for Scientists and EngineersAssumed mathematical knowledge:Plotting and reading graphsAppropriate use of significant figuresSolving quadratic equationsTrigonometric functions and some identitiesExponential and logSolving simultaneous equationsDifferentiation and integrationSolving simple differential equationsThe following sections will be very rapid

Introduction to vector addition and subtractionVector components and resolving vectors

Assumed physics knowledge:Officially none. However, we'll go quickly through parts of mechanics (eg projectiles).

Tutorial noticesTutorial books are part of the course pack. Buy the lab book at the bookshop, then show it to staff atthe FY lab to get the rest of the course pack.

Note for Physics Majors, Advanced Science program or Combined degree withPhysics MajorYour tutorial class is ..... TBA

Assessment: PHYS1131 PHYS1121Lab 20% 20%

Test 1 OR Test 1R;whichever is greater 35% 35%

Test 2 35% 35%Tutorial quizzes 10% 10%

Test 1 is mid-session (90 mins + 10 mins reading)Test 2 and 1R end of session (180 mins + 10 mins reading). You may spend it all on Test 2Labs. Run independently from lecture syllabus.

2Outline of physics in first yearPHYS 1121-1131 PHYS 1221-1231

(session 1) (session 2)

Mechanics WavesThermal; Intro toElectromagnetism Quantum and Solid State_________________________________PHYS 1121-1131 Mechanics (weeks 1-5)Kinematics describing motion

(vectors)

Dynamics forces Newton's

explanation of motion

work and energy~Hamilton'sexplanation

Gravity & planetary mechanicsMomentum and collisions_____________________________________

www.phys.unsw.edu.au/~jw/1131.html

Physics (from the Greek, φυσικος, "natural", and φυσις ,"nature")"The science treating of the properties of matter and energy,or of the action of the different forms of energy on matter ingeneral (excluding chemistry and biology)" (OED)

"Physics is the science of Nature in the broadest sense."(wordnet.princeton.edu/perl/webwn)

Classical mechanics:Motion, vectors, dynamics, work and energy, momentum &collisions, gravitation etc, rotation

Kinematics• constant acceleration• relative velocities• circular motion• simple harmonic motion (brief for now)• projectiles

3Kinematics - study of motion

s

tso

not so straightforwardMeasure lengths to get (relative) positionsMeasure durations to get (relative) timesHow?Rulers and counting, then indirect methodsCycles " " " " " "

What are time and space?Move the clocks and rulers around?

Does this change them?

Have identical ones? How to set them?

Can you (always) give an object a position or an event a time? Can youdo so more than once?

Can you make continuous measurements?

What is a point? A particle?

Motion with constant acceleration

ay = constant ≡ dvydt

vy = ∫ ay dt subscript y for y direction

= ayt + constvy = vy0 + ayt (i) subscript 0 for t = 0

y = ∫ vy dt

= ∫ vy0 + ayt

= vy0t + 12 ayt2 + constant

y = y0 + vy0t + 12 ayt2. (ii)

Eliminate t from (i) and (ii)

y - y0 = vy0t + 12 ayt2 = .... ⇒

2ay(y - y0) = vy2 − vy02 (iii)

do we need this subscript notation?

4Example. Joe runs at constant speed 6 ms-1 towards astationary bus. When he is 30 m from it, the bus acceleratesaway at 2 m.s-2. Can he overtake it?

Given: vJ = 6 ms-1, vb0 = 0, ab = 2 m.s-2, aJ = 0,xb0 - xJ0 = 30 mTranslate the question: Does xJ = xb at any t?

xb = xb0 + vb0t + 12 abt2 (i)

xJ = xJ0 + vJ0t + 12 aJt2 (ii)

Eliminate xJo by choice of origin, xbo = 30 m

Draw a diagram or two

xb = xJ substitute (i) and (ii)12 abt2 - vJot + xbo = 0

t = vJo ± √v2Jo - 2abxbo

ab are the units correct?

Put in numbers

√-ve ∴ no solutions, ∴ no overtaking.

Example. Ball 1 thrown vertically up at 5 ms-1 from 20 mabove ground. Simultaneously, ball 2 thrown vertically downat 5 ms-1 from 20 m above ground. What are their speedswhen they hit the ground, and the interval between collisions?

vo1 = 5 ms-1 vo2 = - 5 ms-1 yo= 20 m

When y = 0, t = ?Use (ii) to get t1 and t2. Use (i) to get vy1 and vy2.

5How much do you lose if you miss the gun?Two runners at different times (∆t apart). During the(constant) acceleration phase, when are they ∆x apart ?

x1 = 12 at2

x2 = 12 a(t')2

t' = t − ∆t

x1 − x2

= 12 at2 −

12 a(t − ∆t)2

{ solve for t}

_______________________________________________

• Vectors and components• Unit vectors

Vectorshave direction and magnitudee.g. displacement, velocity, acceleration, force, spin, electricfield...

2 m towards door; 3.7 ms-1 at 31° E. of N., -9.8 ms-2 up, 4 N in +ve x direction etc(cf Scalars: mass, length, heat, temperature...)

Notation: a in Serway and Jewetta˜ when hand writing

a here to remind you

a in some books, handwriting

6Addition

y

z

y'

x'c = a + b

a

b

put them head to tail to add.

Subtraction

y

za

b

d = a + (−b) = a − b

to subtract, rewrite the equation:a_ − b_ = a_ + (− b_)

or a_ − b_ = d_ → a_ = d_ + b_

_______________________________________________

More kinematics• Circular motion• Simple harmonic motion (brief: more in S2)• Projectile motion

7Uniform circular motionWrite θ = ωt. where ω = const

ω is the angular velocity

As ∆t and ∆θ → 0, ∆v_ → right angles to v_

∴ a_ ≡ lim∆t→0

∆v_

∆t ( )// − r_ i.e. centripital

acceleration

As ∆θ → 0, ∆s → r∆θ (arc becomes straight line)

v = dsdt = r

dθdt = rω

Again, use arc ≅ str line of triangle, here for ∆v_ :

|∆v_| ≅ |v∆θ| (n.b.: |∆v_| =/ ∆|v_| )

lim

∆t→0 |dv_ | = |vdθ|

|a_| ≡ |dv_|dt

= v dθdt = vω

a = v2

r = ω2r but a_ // − r_

so a_ = − ω2r_

Example. Car travelling a v goes over hill with vertical radiusr = 30 m (>> height of car) at summit. Assume it doesn't slowdown. How high must v be for the car to lose contact with theground?

r

v

a

If gravity is only downwards force, then if acentrip > g, g is notenough acceleration for circular motion.

a = v2

r > g

v > √rg = 17 m.s-1 = 62 kph

8ExampleWhat is the acceleration of this theatre?Due to rotation of the Earth:

arot = ω2r

=

2π

T2 r (TEarth; rSydney to Earth's axis)

=

2π

23.9*3600 s2 (5.3 106 m)

= 28 mm.s-2

Due to Earth's orbit:

aorb =

2π

T2 r (Torbit, rorbit)

=

2π

365.24*24*3600 s2 (1.5 1011 m)

= 7 mm.s-2 Aristotle?

Due to Sun's orbit:

Harmonic motion: repeats after period Tfrequency f = 1/TSimple Harmonic Motion (SHM)

Important for vibrations, waves, circuits.....We do it quantitatively in S2.Here qualitative only (but see notes on web)

9SHM as a projection of circular motionω is angular velocity Uniform → θ = ωt

θ = ωt x

y

R

R cos ωt

R sin ωt

projection on x axis: x = R cos ωt

y = R sin ωt

Circular motion SHMr constant, θ = ωt x = r cos ωt

v = rω vx = −rω sin ωt

a = rω2 ax = −rω2 cos ωt

10Double the frequency

Other motion examples:Example An object with initial speed vo slows to a stop withv = vo e−t/τ. What is τ (Greek letter t)? Because the exponent must have no units, itmust have the units of time. Called Characteristic time

(Incidentally, this could come from

a resistive force F = − Cv: prove it.)i) How long does it take to stop?ii) How far does it go before stopping?

i) ?ii) ds = v.dt

s = ∫t = 0

∞ v.dt = ∫

t = 0

∞ vo e−t/τ.dt

= −voτ [e−t/τ] 0

∞

= voτ.

11Vector components and unit vectors

x

y

ˆ

j

i

ax

ay

ay j

ax i

ax i ay ja = +

ax = a cos θ, ay = a sin θax is the component of a in the x direction - scalarî is unit vector: magnitude of 1 in x direction.

a = ax î + ay j

a = √ax2 + ay2 θ = tan-1 ayax

Addition by componentsc_ = a_ + b_ = (ax î+ ay j ) + (bx î+ by j )cx î + cy j_ = (ax + bx) î + (ay + by) jcx = ax + bx and cy = ay + by

vector eqn inn dimensions →

n independentalgebraic eqns.

important case: 0 = 0 î + 0 j

∴ if a_ + b_ = 0, ( )e.g. mechanicalequilibrium

ax + bx = 0 and ay + by = 0

Resolving vectors is often useful.Choose convenient axes:

Component of W_ in dirn of plane = - Wsinθ

Total force in y' dirn = - Wsinθ

= m d2y'dt2

Components in normal direction add to zero:N - Wcosθ = 0

12Problem. What is force

F exerted by air jet?

θ

air jet

reaction

W

F

T

If stationary, ΣF = 0.In direction of string:T − W cos θ = 0In direction of jet:

W sin θ − F = 0 More about this after Newton

In three dimensions:

x (East)

z (up)

ˆ

k

i

ax

j

y (North)

r_ = rx î + ry j + rz k (sometimes just i, j, k )

right hand convention:

î j k in dirn ofthumb index middle fingers of right hand

i

j

k

13Pythagoras' theorem in three dimensions

x

y

z

a

ax

ay

az

h

What is magnitude of a?Hypotenuse h:

h2 = ax2 + ay2.Now look at triangle h,az,a:

a2 = h2 + az2

= ax2 + ay2 + az2.

a = √ax2 + ay2 + az2

Exampler = (A.sin ωt) î + (A.cos ωt) j + (Bt) kwhere A and B are constants.What shape is r_ ? What is a_ ?

a = d2

dt2 r = -(Aω2.sin ωt) î - (Aω2.cos ωt) j

a is in xy plane.

a = √ax2 + ay2 = ... = Aω2 = constant

14Relative velocities

( )Gallilean/Newtonian relativity watch for hiddenassumptions

origin of frame (x',y') is at r f_ and moves with vf_ w.r.t (x,y)frame. No rotation.

r'_ = r_ - r f_

v'_ = ddt r'_

= ddt r_ -

ddt r f_

= v_ - vf_

Example A boat heads East at 8 km.hr-1. The current flowsSouth at 6 km.hr-1. What is the boat's velocity relative to theEarth?

vb

vc

v = vboat + vcurrentTo add the vectors, draw them head-to-tail.

vb

vc

v = v + b vc

θ

magnitude: v = √vb2 + vc2

= √(8 km.hr-1)2 + (6 km.hr-1)2

= √(82 + 62) (km.hr-1)2

= √(64 + 36) √(km.hr-1)2

= 10 km.hr-1

direction: θ = tan-1 6 km.hr-1

8 km.hr-1 = tan-1 0.75

= 37°

Answer: 10 km.hr-1 at 40° South of East

15Vector subtractionDraw the vectors head to head to subtract them.

y

za

b

d = a + (−b) = a − b

to subtract, rewrite the equation:a_ − b_ = a_ + (− b_)

or a_ − b_ = d_ → a_ = d_ + b_

In other words, you can rearrange vector subtractionso that it becomes vector addition

Example A sailor wants to travel East at 8 km.hr-1. Thecurrent flows South at 6 km.hr-1. What direction must shehead, and what speed should she make relative to the water?

v = vboat + vcurrentvboat = v − vcurrentTo subtract the vectors, draw them head-to-head.

vvc

v

vcθ

vb

magnitude: vb = √v2 + vc2

= √(8 km.hr-1)2 + (6 km.hr-1)2

= 10 km.hr-1

direction: θ = tan-1 vcv = tan-1

6 km.hr-1

8 km.hr-1

= 37°

She must head 37° North of East and travel at 10 km.hr-1with respect to the water.

16Puzzle River flows East at 10 km/hr. Sailing boat travelsEast down the river.Can the boat travel fastera) with no wind?b) with 10 km/hr wind from West?

Consider motion relative to the water.

b) v'w = 0

a) 10 km/hr headwind.

How does sailboat move against wind? www.phys.unsw.edu.au/~jw/sailing.html

Example A cockroach on a turntable crawls in the radialdirection (initially the x direction) at speed v. The turntablerotates at ω anticlockwise. Describe his path in i, jcoordinates.

v

ω

θr

ωtvt

We can specify his position as (r,θ), where r = vt, and θ = ωt.

x = r cos θ, y = r sin θpath is r (t) = vt cos ωt i + vt sin ωt jWhat is this shape?

17A SE wind blows at 30 km/hr. If you are travelling North,how fast must you travel before the wind is coming exactlyfrom your right? From 30° E of N?From the ground: From your frame:

w

E or x

Nor y

v

w

x'

y'v

or

w

x'

y'v

30°

Relative motion: (add vel wrt me to my vel to get vel wrt ground)vw = vyou + v'w so v'w = vw − vyou

i) Do it algebraically. Given:vw = − vw cos 45° i + vw sin 45° jvyou = 0 i + vyou jv'w = v'wx i + 0 j

y direction: vw sin 45° = vyou + 0vyou = vw sin 45° = 21 km/hr

x direction: − vw cos 45° = 0 + v'wx

v'wx = − vw cos 45° = 21 km/hr from E

ii) Do it with vector diagrams:

wv

wv

youv

'

45°

Same answers, but directly.

Second case: v'w is 30° E of N

wv

wv

youv

'

45°

60°

x direction: vw cos 45° = v'w cos 60°.y direction: vyou = vw sin 45° + v'w sin 60°2 equations, 2 unknowns, first gives

v'w = vw cos 45°

cos 60° = 42 km/hr

Substitute:vyou = 60 km/hr

18Projectiles

Question. A man fires a gun horiztonally. At the same timehe drops a bullet. Which hits the ground first? Explain yourreasoning.(To simplify, let's say it happens on the moon.)

F = ma. So a force in y direction does not cause anacceleration in x direction.

Independence of x and y motion.

Question. A man shoots at a coconut. At the instant that hefires, the coconut falls. What happens

h

Lθ

obvious method (c for coconut, b for bullet)h = L tan θ

yc = h − 12gt2 yb = vyot − 1

2gt2

yc − yb = h − vyot = h − vosinθ.t

L = t vocos θ → T = L

vocos θ

yc − yb = h − vo sinθ.Lvo cos θ

= h − L tan θ

= 0

Alternatively : consider a frame of reference falling with g.

g

19ProjectilesWithout air, ay = −g = constant. ax = 0

(Galileo: independence of horiz. & vert motion)

(ii) → y = yo + vyot − 12 gt2.

(ii) → x = xo + vxt.

Choose axes so that xo = yo = 0 and eliminate t:

y = vyo

x

vx − 1

2 g

x

vx

2 (*)

y = 0 when x = 0 or R

(*) ⇒ vyo

R

vx = 1

2 g

R

vx

2 (**)

vyo = vo sin θ, vx = vo cos θ → R = R(vo,θ)

Set ∂R∂θ = 0 to obtain θ for maximum range.

vyo

R

vx = 1

2 g

R

vx

2 (**)

solve (**) for R:

R = 2vxvyo

g (check units)

= 2v0 sin θ.v0 cos θ

g

= v02 sin 2θ

g

∂R∂θ =

2v02 cos 2θg

∂R∂θ = 0 when 2θ = 90°

20Example The human cannon of Circus Oz has a muzzlevelocity vo. For their next trick, they will fire the humancanonball into a horizontal teflon tube at height h above thecanon mouth. To avoid damage to the canonball, he mustarrive with purely horizontal velocity. Calculate the postion ofthe canon and its angle to the vertical.i) draw a diagram

ii) put in symbols for quantities

iii) translate question

h

L

vθ

o

v

given h, vo and final vy = 0

Find L and θRelate h, vy and vyo. vyo depends on θ.

During flight, the acceleration is −g upwards. The desired vy iszero, so

0 = vy2 = vyo2 + 2ay(∆y) = vo2 cos2 θ − 2gh

∴ vo2 cos2 θ = 2gh

∴ cos θ = √2ghvo

θ = cos −1

√2gh

vo

Find time of flight. vxt = L.

0 = vy = vyo + ayt

∴ t = vo cos θ

g

L = vo t sin θ.

= vo sin θ vo cos θ

g

<optional> = vo

2

2g sin 2 θ

where θ = cos −1

√2gh

vo simplify optional

21Projectile in 3D

Example. Throw object from origin at 20 ms-1, 30° East andat 45° to horizontal. Take x = North, y = East, describe itsmotion as a function of t in i, j, k notation.

x (East)

z (up)

ˆ

k

i

ax

j

y (North)

vzo = vo sin 45° = 14 ms-1

vhoriz = vo cos 45° = 14 ms-1 (= const)

vx = vhoriz cos 60° = 7 ms-1 (= const)

vy = vhoriz cos 30° = 12 ms-1 (= const)

r_ = rx î + ry j + rz k

= (7 ms-1).t î + (12 ms-1).t j

+ ( )(14 ms-1)t - 12(9.8 ms-2)t2 k

Spin a laser pointer steadily: full turn takes 1 second. Howfast does the spot move on the wall?

θ

x

D

ω c ≅ infinite

Steady circular motion:

θ = ωt where ω = 2π1 s = 2π rad/s

Geometry: x = D tan (90−θ) = D cot θ

vspot = dxdt =

d(D cot θ)dθ

dθdt

ddθ

cos θsin θ = −

sin θsin θ −

cos2 θsin2 θ = −

sin2 θ + cos2 θsin2 θ

vspot = − D 1

sin2 θ ω = −

ωDsin2 θ

Does it make sense?

What happens when θ → 0 or when D → ∞?

Consider a laser at the centre of a circular room. In 1 s, the spot travels 2πD.What if D = 108 m.

22Example. Electron enters a uniform electric field at (x,y,t)=(0,0,0), with velocity vxi. The field extends from x = 0 tox = L, but is zero for x < 0 and x > L. In the field, the electron is accelerated at ayj . Write an equation for theposition of the electron for x > L.

y

y

vx i

vx i vyf j+

ay j

L,yf

x ≤ 0 (t ≤ 0) v = vxi = const

0 ≤ x ≤ L (0 ≤ t ≤ L/vx)

x = x0 + vx0t + 12 axt2 vx = vx0 + axt

y = y0 + vy0t + 12 ayt2 vy = vy0 + ayt

yf = 12 aytf2 = 1

2 ay

L

vx

2 vyf = ay

L

vx

x ≥ L x = vxt (still no acceleration)

y = yf + vyf

t −

L

vx

position is vxti +

1

2 ay

L

vx

2 + ay

L

vx

t −

L

vxj

23How to make a bullet-proof vest. (Out of order in syllabus - deliberate)

A gun fires a 5 g bullet at v = 400 m.s-1.

M

m

vk

Fired at a plate (mass of 500 g), such a bullet penetrates 5 mminto it, but not through. Will such a plate protect me from aspeeding bullet?Estimate only. Why?If bullet stops with constant acceleration:

vf2 − vi2 = 2a ∆x

Force exerted by bullet

F = mbullet|a| ~ m vi2

2∆x = 100 kN !

Collision takes distancemean v = 25 µs

Collision with say 500 g. If momentum conserved, then plate

moves at 400 m.s-1 5g500g = 4 ms-1, so moves 4 ms-1*25 µs

= 0.1 mm during collision, so yes, external forces negligible(so far!) so momentum ~ conserved.I get hit by a 500 g mass travelling at 4 m/s.Next steps:make block lighter.What if it is flexible?What is pressure distribution?What are elastic properties of chest? Etc

How did we solve it?Examine the problem in the worldIdentify the important parameters and variables.Where necessary, make measurementsMake appropriate assumptions and approximationsMake a mathematical model of the situation(s)(Solve the maths)Interpret the solution quantitatively, in terms of the real worldproblem

This course involves a lot of thinking and problem solving. It will be challenging but these are important skills.

We'll come back to this problem

24Particle dynamicsNewton's laws:

force, mass, acceleration also weightFriction - coefficients of frictionHooke's LawDynamics of circular motion

Aristotle: v_ = 0 is "natural" stateGalileo & Newton: a_ = 0 is "natural" state

Newton's LawsFirst "zero (total) force ⇒ zero acceleration"more formally:

If Σ F_ = 0, ∃ reference frames in which a_ = 0∃ ≡ "there exists"

called Inertial framesobservation: w.r.t. these frames, distant stars don't accelerate

In inertial frames:

Second Σ F_ = m a_ Σ is important

(Σ Fx = max Σ Fy = may Σ Fz = maz) 3D →

3 equations

1st law is special case of 2nd

What does the 2nd law mean?Σ F_ = mi a_ and W_ = mg g_

are mi and mg necessarily the same?

called inertial and gravitational masses

F_ = m a_ a_ is already defined.i) Does this equation define m?ii) Does this law define F_ ?iii) Is it a physical law?iv) All of the above?v) How?

25Newton 1: "Every body persists in its state of rest or of uniformmotion in a straight line unless it is compelled to change that state byforces impressed on it."

postulateAn inertial frame of reference is one in which Newton's 1st law istrue. definitionSuch frames exist (and w.r.t. these frames, distant stars don'taccelerate) observation:∴ if Σ F_ = 0, _a = 0 w.r.t. distant stars.

Force causes acceleration. F_ // _a , F_ ∝ _adefinition

Newton 2: To any body may be ascribed a scalar constant, mass,such that the acceleration produced in two bodies by a given force isinversely proportional to their masses,

i.e. for same F, m2m1

= a1a2

Already have metre, second, choose a standard body for kg, thenchoose units of F (Newtons) such that

F_ = m _a(this eqn. is laws 1&2, definition of mass and units of force)

Newton 3: "To every action there is always opposed an equalreaction; or the mutual actions of two bodies upon each other arealways equal and directed to contrary parts"OrForces always occur in pairs, F_ and -F_ , one acting on each of a pairof interacting bodies.

Important conclusion: internal forces in a system add to zero.

Third F_AB = - F_BA

Why so?

26Example Where is centre of earth-moon orbit?

Fe = Fm = Fg equal and opposite NB signconventions

each makes a circle about common centre of mass

Fg = mmam = mmω2rm

Fg = meae = meω2re

∴ rmre

= memm

= 5.98 1024 kg7.36 1022 kg

= 81.3 (i)

earth-moon distance re + rm = 3.85 108 m (ii)

solve → rm = 3.80 108 m, re = 4.7 106 m∴ centre of both orbits is inside earth (later: c. of mass)

Example. As the bus takes a steady turn with radius 8 m at constantspeed, you notice that a mass on a string hangs at 30° to the vertical.How fast is the bus going?Draw a diagram with physics

string

T

W

a

30°

m

We know:tension in direction of string,weight down,acceleration horizontalcircular motion

mass in circular motion with bus, so force on m:

Fhoriz = ma = m v2r

Only the tension has a horizontal component, so

T sin 30° = m v2r

Need one more eqn: mass is not falling down, ie

vertical acceleration = 0, soT cos 30° = mg

Eliminate T: tan 30° = m v2r

1mg

→ v = √gr tan 30° = 6.7 m/s = 24 kph.Check dimensions. Reasonable? And so...?

27Problem. Horse and cart. Wheels roll freely.

Why should I pull?The force of the carton me equals my forceon it, but opposes it.Σ F = 0. We'll neveraccelerate.

Let's go!

Fc Fc

Fc Fc

FgFg

Horizontal forces on cart (mass mc)

Fc

Fc = mcac = mcaHorizontal forces on horse (mass mh)

Fc

Fg Fg − Fc = mhaHorizontal forces on Earth (mass mE)

Fg mE >> mh + mc

28"light" ropes etc.Here, light means m << other masses

Truck (mt) pulls wagon (mw) with rope (mr).All have same a_.

i) wagon: − F2 = mwa.ii) rope: F1 - F2 = mraiii) truck: − F1 + Fext = mta

(ii)/(i) →F1 − F2

− F2 =

mramwa

∴ if mr << mw, F1= F2.Forces at opposite ends of light ropes etc are equal and opposite.

Example. Train. Wheels roll freely. Loco exerts horizontal force Fon the track. What are the tensions T1 and T2 in the two couplings?

m m

lococar 2 car 1

FF

T1m

T2

Whole train accelerates together with a.Look at the external forces acting on the train (horiz. only).

m m m

FF = (m+m+m)a → a = F/3m

Look at horiz forces on car 2:

mT2

T2 = ma = F/3and on cars 2 and 1 together

mT1

m T1 = 2ma

29Hooke's Law.No applied force(x = 0)

Under tensionx > 0

_Fspring in opposite direction to x.Experimentally, |_Fs| ∝ |x| over small range of x

F = − kx Hooke's Law.linear elastic behaviour - more in S2.

Why linear elasticity?Intermolecular forces F and energies U:

r

Frepulsive

attractive

total

U

r

Hooke's law

See tut problem on interatomic forces

30Mass and weight(inertial) mass m defined by F = maobservation:near earth's surface and without air,all (?) bodies fall with same a ( = −g)weight W = − mg

Warning: do not confuse mass and weight, or their units

kg → mass N → force (kg.m.s-2)kg wt = weight of 1 kg = mg = 9.8 N

What is your weight?

Why is W ∝ m? Why is mg ∝ mi?ma = F = W = (Grav field).(grav. property of body)• Mach's Principle• Principle of General Relativity• Interactions with vacuum field

31Example Grav. field on moon gm= 1.7 ms-2. An astronaut weighs800 N on Earth, and, while jumping, exerts 2kN while body moves0.3 m. What is his weight on moon? How high does he jump onearth and on moon?

mgE = WE → m = 800 N

9.8 ms-2 = 82 kg

Wm = mgm = 82kg 1.7ms-2 = 140 N

Vertical (y) motion with constaccel. While feet are on ground,Σ F = 2 kN - WE

= 1.2 kN (Earth)Moon:Σ F = 2 kN - mgm = 1.9 kN

Jump has two parts:

feet on ground

a = Σ Fm vi = 0, vf = vj

feet off ground a = - g vi = vj, vf = 0

vv

y

y∆

h

y

feet leave ground

v j

ymeasure y of centreof mass.

While on ground:

vj2 − vo2 = 2aj∆y = 2 Σ Fm ∆y

Earth → vj = 3.0 ms−1 Moon → vj = 3.7 ms−1

While above ground:

v2 − vj2 = − 2gh → h = vj2

2ghE = 0.5 m. hm = 4 m

32ExampleLight pulley, light string. What isacceleration of the masses?

Let a be accel (down) of m1 = accel (up) of m2.Newton 2 for m1: T − m1g = − m1a

" " " m2: T − m2g = + m2asubtract: − m1g + m2g = − m1a − m2a

a = m1 − m2m1 + m2

g

(Check: if m1 = m2, a = 0. If m2 = 0, a = g.)

________________________________________________

F

F

Ncontact

friction

Contact forces

The normal component of a contact force is calledthe normal force N. The component in the planeof contact is called the friction force Ff.

Normal force: at right angles to surface, is provided by deformation.• If ∃ relative motion, kinetic friction (whose direction

opposes relative motion)• If ∃ no relative motion, static friction (whose direction

opposes applied force)Define coefficients of kinetic (k) and static (s) friction:

|Ff| = µkN |Ff| ≤ µsNFriction follows this approximate empirical law

µs and µk are approx. independent of N and of contact area.

Often µk < µs.(It takes less force to keep sliding than to start sliding.)

33

Example. θ is graduallyincreased to θc whensliding begins. What isθc? What is a at θc?

Newton 2 in normal dirn:N - mg cos θ = 0 (i)

Newton 2 in dirn down plane:mg sin θ − Ff = ma. (ii)

No sliding: a = 0∴ (ii) ⇒ mg sin θ = Ff

≤ µs N(i) ⇒ = µ s mg cosθ

mg sin θ ≤ µs mg cosθtan θ ≤ µs, θc = tan−1µs useful technique for µs

Sliding at θ = θc: a > 0

∴ (ii) ⇒ a = g sin θc − Ffm

= g sin θc − µkNm

(i) ⇒ = g sin θc − µkg cosθc

= = g cosθc (µs − µk)

34Example. Rear wheel drivecar, 300 kg wt on each frontwheel, 200 kg wt on rear.Rubber-road:µs = 1.5, µk = 1.1

Neglect rotation of car during accelerations. Assume that brakesproduce 1.8 times as much force on front wheels as on back. (i)What is max forward accel without skidding? What is maximumdeceleration (ii) not skidding? (iii) 4 wheel skid?(i) FfRs ≤ µsNR = µsWR

= 1.5x200x9.8 N = 2.9 kN

amax = Fm =

2 x 2.9 kN (2x300 + 2x200)kg

= 5.8 ms−2

Stopping.For all wheels, Ffs ≤ µsN = µsWFfF = 1.8 FfR. µsWF = 1.5 µsWR,∴ front wheels skid first, when FfF > µsWF.

max total friction = (front + rear) =

2 + 2

1.8 .µsWF

= 1400 kg.wt = 14 kN

ii) amax = Fmax

m = 14 kN

1000 kg = 14 ms-2

iii) a = Σ Fkm =

Σ µkWm = ... = 11 ms-2

Questions:Does area of rubber-road contact make a difference?Does the size of the tire make a difference?

35ExampleConical pendulum(Uniform circular motion.)What is the frequency?

Apply Newton 2 in two directions:Vertical: ay = 0 ∴ Σ Fy = 0∴ T cos θ − W = 0

T = mg

cos θHorizontal:

mv2

r = mac = T sin θ

= mg sin θ

cos θ

∴ v2

r = g tan θ

∴ v = √rg tan θ

∴ 2πrT = √rg tan θ

∴ f = 1T =

12π √g tan θ

r

36Example.Apply force F at θ to horizontal.Mass m on floor, coefficients µsand µk. For any given θ, what F isrequired to make the mass move?Eliminate 2 unknowns N and Ff → F(θ,µs,m,g)

Stationary if Ff ≤ µsN (1)Newton 2 vertical: N = mg + F sin θ (2)Newton 2 horizontal: F cos θ = Ff (3)(1,3) → stationary if F cos θ ≤ µsN

F cos θ ≤ µs(mg + F sin θ) (using (2))

F (cos θ − µs sin θ) ≤ µsmg (*)

note importance of sign of (.....) if (cos θ − µs sin θ) = 0,

θ = θcrit = tan−1(1/µs).

If θ < θc, then (cos θ − µs sin θ) > 0

stationary if F ≤ µsmg

cos θ − µs sin θ

i.e. moves when F > Fcrit = µsmg

cos θ − µs sin θ

If θ > θc, then (....) > 0

(*) ⇒ stationary if F ≥ µsmg

cos θ − µs sin θi.e. stationary no matter how large F becomes.

37Centripetal acceleration and forceCircular motion with ω = const. and v const. eg bus going round acorner

F = ma centrip

Orconsider hammer thrower

T

Wresultant

Resultant force produces acceleration in the horizontal direction,towards the centre of the motionCentripetal force, centripetal acceleration

Example Plane travels in horizontal circle, speed v, radius r. Forgiven v, what is the r for which the normal force exerted by the planeon the pilot = twice her weight? What is the direction of this force?

Centripetal force F = m v2

r = N cos θ

Vertical forces: N sin θ = mg

eliminate θ: N2 = m2

v4

r2 + g2

N2

m2 − g2 = v4

r2→ r =

v2

√N2

m2 − g2

sin θ = mgN =

12

∴ 30° above horizontal, towards axis of rotation

38

r

Example. Foolhardy lecturer swings abucket bricks in a vertical circle. Howfast should he swing so that the bricksstay in contact with the bucket at thetop of the trajectory?

Draw diagram & identify important variables

pose question mathematically.

N

Wa centrip

m

W and N provide centripetal force.mg + N = macFor contact, we need

N ≥ 0so mac ≥ mg how to express ac?

ac = v2

r = rω2 = r

2π

T

2

T is easy to measure

T = 2π√ rac

≤ 2π√ rg

r ~ 1m → T ≤ 2 s. please check for errors!

Question. Three identical bricks.What is the minimum force youmust apply to hold them still likethis?

Vertical forces on middle brick add to zero:

2 Ff = mg

Define µs

Ff ≤ µs N

∴ N ≥ Ffµs

= mg2µs

F f

Ff

N N

gm

Bricks not accelerating horizontally, so normal force from hands =normal force between bricks.

∴ (each) hand must provide ≥ mg2µs

horizontally.

Vertically, two hands together provide 3mg.

3mg/2

mg/2µs

minimumforce fromhands

39PHYS 1131 Work and Energy Joe Wolfe, UNSWThe scalar product. dot productWhy? e.g. Work: scalar, related to F_, ds_ and θ.

dW = |F_| |ds__| cos θ

(also dV = |E_| |ds__| cos θ etc)

∴ define

a_.b_ ≡ ab cos θ (= b_.a_)

Apply to unit vectors: i_.i_ = 1 . 1 cos 0° = 1 = j_.j_ = k_.k_ i_.j_ = 1 . 1 cos 90° = 0 = j_.k_ = k_.i_

Scalar product by components

a_ . b_ = (ax i_ + ay j_ + az k_).(bx i_ + by j_ + bz k_)= (axbx) i_ .i_ + (ayby) j_.j_ + (azbz) k_.k_ + (axby + aybx) i_.j_ + (..) j_.k_ + (..) k_.i_

a_ . b_ = axbx + ayby + azbz

Problem. Find the angle betweena_ = 4 i_ − 3 j_ + 7 k_b_ = 2 i_ + 5 j_ −3 k_ab cos θ = a_ . b_ = axbx + ayby + azbz

cos θ = axbx + ayby + azbz

√ax2 + ay2 + az2 √bx2 + by2 + bz2

→ θ = 122°

Definition of work

F F

dsθ

When force varies, use differential displacement ds__

dW = F ds cos θ(F) (ds cosθ) → F X component of ds // F, or(F cosθ) (ds) → ds X component of F // ds

W = ∫0

L F cos θ ds

if F and θ are constant, we get W = FLcos θSI Unit: 1 Newton x 1 metre = 1 Joule

40Example. How much work is done by lifting 100 kg vertically by1.8 m very slowly?

Slow ∴ Fapplied ≅ mgW = mg d cos 0° = 1.8 kJ.

Not a lot, yet it is hard to do, because the force is inconveniently large.Consider:

TT T

mg

mg

If the rope and pulleys are light, and if the acclerations arenegligible, thenForce on LH pulleyma ≅ 0 = 2T − mg∴ T = mg/2If mass rises by D, word done = mgD.But rope shortens on both sides of rising pulley,if mass rises by D, rope must be pulled 2D,so work done = T * 2D = mgD

Example. What is the work done by gravity in a circular orbit?

W = ∫ F ds cos θ = 0

Example. Fgrav ∝ 1/r2. How much work is done to move m = 1tonne from earth's surface (r = 6500 km) to r = ∞?

W = ∫ F ds cos θ= ∫ F dr

F dsθdr

g

F = - Fgrav = Cmr2

more later

On surface F/m = 9.8 ms-2

∴ C = (9.8 ms-2)(6.5 106 m)2 = 4.1 1014 m3s-2

W = ∫6 5 0 0 k m

∞

Cmr2

dr

= - Cm

1

∞ - 1

6.5 106

= 6.3 1010 J = 63 GJ.

41Work to deform spring

F springF spring F applied

x

No applied force (x = 0)Hooke's law: F = − kx

Work done by spring = ∫ Fspring.dx

= ∫ -kx.dx = − 12 kx2

Work done on spring = ∫ Fapplied.dx

= ∫ kx.dx = + 12 kx2 (= work stored in spring)

The work-energy theorem(Total) force F acts on mass m in x direction.

vi vfvF

Work done by F = ∫i

f Fdx (use F = ma)

= ∫i

f m

dvdt dx = ∫

i

f m

dxdt dv

= ∫i

f mv.dv = [ 1

2 mv2 ]f

i

Work done by F = 12 mvf2 -

12 mvi2 ≡ ∆K

Define kinetic energy K ≡ 12 mv2

Increase in kinetic energy of body = work done by total forceacting on it.

This is a theorem, ie a tautologybecause it is only true by definition of KE and by Newton 2.

∴ restatement of Newton 2 in terms of energy. Not a new law

42Power. is the rate of doing work

Average power P-

≡ W∆t

Instantaneous power P = dWdt

SI unit: 1 Joule per second ≡ 1 Watt (1 W)Example Jill (m = 60 kg) climbs the stairs in Matthews Bldgand rises 50 m in 1 minute. How much work does she do againstgravity? What is her average output power? (neglectaccelerations)W = ∫ F_ . ds__ = ∫ Fy dyFy ≅ mg

W = mg ∫ dy = mg ∆y

= 29 kJ (cf K = 12mv2 ~ 20 - 40 J)

P-

≡ W∆t =

mg ∆y∆t = 490 W

(humans can produce 100s of W, car engines several tens of kW)(1 horsepower ≡ 550 ft.lb.s-1 = 0.76 kW)

Potential energy.e.g. Compress spring, do W on it, but get no K. Yet can getenergy out: spring can expand and give K to a mass. → Idea ofstored energy.

e.g. Gravity : lift object (slowly), do work but get no K. Yet objectcan fall back down and get K.Recall Wagainst grav = mg ∆y i.e. W = W(y)

But: Slide mass slowly along a surface. Do work againstfriction , but can't recover this energy mechanically. Not all forces"store" energy

Conservative and non-conservative forces(same examples)

Wagainst grav = − ∫i

f Fgdr cos θ

= − ∫i

f Fg dz

= mg ∫i

f dz

= mg (zf - zi) in uniform field

W is uniquely defined at all r_, i.e. W = W(r_)If zf - zi are the same, W = 0.∴ Work done against gravity round a closed path = 0Gravity is a conservative force

43Compare spring

Wagainst spring = − ∫i

f Fspring.dx = − ∫

i

f -kx.dx

= 12 k(xf2 − xi2)

W is uniquely defined at all x, i.e. W = W(x)xf = xi ⇒ W = 0.∴ Work done round a closed path = 0Spring force is a conservative force

so it has stored or potential energy: symbol U.

with frictiondWagainst fric = − Ff ds cos θbut Ff always has a component opposite ds∴ dW always ≥ 0. (we never get work back)

∴ cannot be zero round closed path, ∴ W =/ W(r_)∴ friction is a non-conservative forceNote that direction of friction (dissipative force) is always against motion.

Potential energyFor a conservative force F_ (i.e. one where work done against it,W = W(r_)) we can define potential energy U by ∆U = Wagainst.i.e.

∆U = − ∫i

f F dr cos θ

Same example: spring

∆Uspring = − ∫i

f Fspring.dx

= 12 k(xf2 − xi2)

Choice of zero for U is arbitrary.Here U = 0 at x = 0 is obvious, so

Uspring = 12 kx2

44From energy to force:U = − ∫ F ds where ds is in the direction // F

F = − dUds

in fact Fx = − dUdx

, Fy = − dUdy

, Fz = − dUdz

Spring: Uspring = 12 kx2 ∴ Fspring = − kx

Gravity: Ug = mgz ∴ Fg = − dUdz = − mg

Energy of interaction:

r

Frepulsive

attractive

total

U

r

Hooke's law

Conservation of mechanical energyRecall: Increase in K of body = work done by total force actingon it. (restatement of Newton 2)

But, if all forces are conservative, work done by these forces =− ∆U (definition of U)

∴ if only conservative forces act, ∆K = − ∆UWe define mechanical energy

Ε ≡ K + Uso, if only conservative forces act, ∆E = 0.

we can make this stronger.Work done by non-conservative forcesDefine internal energy Uint where

∆Uint = − Work done by n-c forces (= + Work done against n-c forces)

Recall defn of K: ∆K = work done by Σ force∴ ∆K = − ∆U − ∆Uint

∴ ∆K + ∆U + ∆Uint = 0If n-c forces do no work, then ∆Uint = 0, so:

If non-conservative forces do no work,∆E ≡ ∆K + ∆U = 0

or: mechanical energy E is conservedEquivalent to Newton 2, but useful for many mechanics problemswhere integration is difficult.State the principle carefully!Never, ever write: "kinetic energy = potential energy"

Classic problem. Child pushes off with vi. How fast is the s/hegoing at the bottom of the slide? Neglect friction (a non-conservative force).

45

W

N

W

N

Σ Fh

v

i) By Newton 2 directly:

v = ∫top

bottom a dt = ∫

top

bottom

Fm

= ∫top

bottom g cos θ dt = .....

ii) Using work energy theorem (Newton 2 indirectly):

Non-conservative forces do no work, ∴ mechanical energy isconserved, i.e.∆E = ∆K + ∆U = 0K f - K i + Uf - Ui = 0

or Ef = Ei

K f + Uf = Ki + Ui

12 mvf2 −

12 mvi2 + mgyf − mgyi = 0

rearrange → vf = √vi2 + 2g(yi - yf)Conservation of energyobservation: for many forces, W = W(r_) ∴ useful to define U = U(r_).observation: for all systems yet studied, Uint is a state function,i.e. Uint = Uint(measured variables)Hence idea of internal energy. e.g.:Friction, (− Uint) = heat produced when work is done against friction.

Air resistance (− Uint) is sound and heat.

Combustion engines and animals: +Uint comes from chemical energy

∆K + ∆U + ∆Uint = 0is statement of Newton 2 plus definitions of K, U, Uint.

The statement that ∆Uint is a state function is the first law ofthermodynamics. It is a law, ie falsifiable. More on this in Heat.

Example. Freda (m = 60 kg) rides pogo stick (m << 60 kg) withspring constant k = 100 kN.m-1. Neglecting friction, how far doesspring compress if jumps are 50 cm high?

b

t

yb

ytxb

Non-conservative forces do no work, ∴ mechanical energy isconserved, i.e.Ebottom = Etop Kb + Ub = Kt + Ut (U = Ugrav + Uspring)

12 mvhoriz2 + (mgyb +

12 kxb2)

≅ 12 mvhoriz2 + (mgyt +

12 kxt2)

mg(yt − yb) ≅ 12 kxb2

∴ xb ≅ √2mg(yt − yb)k ≅ 80 mm.

46

h

v

r1

h2

Example. Slide starts at height h1. Later there is a hump withheight h2 and (vertical) radius r. What is the minimum value ofh2 − h1 if slider is to become airborne? Neglect friction, airresistance.

Over hump, ac = v2

r (down). Airborne if g < ac, i.e. if v22 > gr.

No non-conservative forces act soE2 = E1U2 + K2 = U1 + K1

mgy2 + 12 mv2

2 = mgy1 + 12 mv1

2

12 mv2

2 = mg(y1 − y2)

(y1 − y2) = v2

2

2g > gr2g =

r2

Example Bicycle and rider (80 kg) travelling at 20 m.s-1 stopwithout skidding. µs = 1.1. What is minimum stopping distance?How much work done by friction between tire and road? Betweenbrake pad and rim? Wheel rim is ~300 g with specific heat ~ 1kj.kg-1, how hot does it get?friction → declleration → stopping distance

|a| = Ffm ≤

µsNm = µsg

|a| ≤ µsg

vf2 − vi2 = 2as → s = vf2 − vi2

2a

s ≥ |− vi2

2µsg| = 19 m

Work done by friction between tire and road?No skidding, ∴ no relative motion, ∴ W = 0.Between pad and rim? Here ∃ relative motion.All K of bike & rider → heat in rim and pad

W = ∆K = Kf − Ki = − 16 kJQ = mC∆T ...... ∆T ~ 50 °C

47Example Which way is it easier to drag an object?

mF

mF

m

F'

θ

F

θ

mg

N

Ff

Suppose we move at steady speed, a = 0. Which requires less F? Whichrequires less work?

mechanicalequilibrium →

horizontal F cos θ = Ffvertical N + F sin θ = mg

sliding → Ff = µkNF cos θ = µkN

eliminate N → F cos θ = µk(mg − F sin θ)

F = µkmg

cos θ + µk sin θwhen θ = 0, F' = µkmg

F < F' if cos θ + µk sin θ > 1, i.e. if µk large

& θ smallWork done = Fs cos θ = Ffs = µkNs = µks(mg − F sin θ) decreases with θ

Question

Rh

r

v

rr'

90°

How high should h be so that it can loop the loop?

48Example. A hydroelectric dam is 100 m tall. Assuming that theturbines and generators are 100% efficient, and neglecting friction,calculate the flow of water required to produce 10 MW of power.The output pipes have a cross section of 5 m2.

v

generatorh

dm

water

Nett effect: ~ stationary water lost from top of dam, water appearswith speed v at bottom.

Let flow be dmdt .

dW ≡ work doneby water

= − Work done

on water = − energy increase

of waterdW = − dE = − dK− dU

= − ( )12 dmv2 − 0 − ( )0 − dm.gh = dm

gh − v2

2

P = dWdt =

dmdt

gh − v2

2

Problem: v depends on dmdt

v

Av.dt

dVdt =

A.(v.dt)dt = Av

Density: ρ ≡ mass

volume = mV so m = ρV

dmdt = ρdV

dt = ρAv

P = ρAv

gh − v2

2

v3 − (2gh)v + 2PρA = 0 can solve cubit, but messy

Neglect v3 → v = P

ghρA = 2 m/s

and indeed we see that v3<< other terms

49Gravity Notes for PHYS 1121-1131. Joe Wolfe, UNSW

Gravity: where does it fit in?Gravity[general

relativity]

Electricforce*

Weaknuclearforce

Strongnuclearforce

Colour

force

gravitons photons intermediate

vector bosons pions gluons

electro-weak

Grand Unified Theories

Some triesfor classical

gravity

Theories Of Everything* Electromagnetism "unified" by Maxwell, and also by Einstein: Magnetismcan be considered as the relativistic correction to electric interactions whichapplies when charges move.

• Only gravity and electric force have macroscopic ("infinite")range.

mgraviton? = mphoton = 0

• Gravity weakest, but dominates on large scales because it isalways attractive

Greeks to Galileo:i) things fall to the ground ('natural' places)ii) planets etc move (variety of reasons)but no connection (in fact, natural vs supernatural)

Newton's calculation: acceln of moon

= rmωm2

= (3.8 108 m)( )2π27.3 24 3600

2

= 2.7 10-3 m.s-2

acceln of "apple" = 9.8 mms-2

aappleamoon

= 3600; rmRe

= 385000 km

6370km = 60;

rm

Re

2 = 3600

Newton's brilliant idea: What if the apple and the moon accelerate according to the same law? →What if every body in the universe attracts every other, inverse square law?

50

Newton's law of gravity:

F = − G m1m2

r2

Negative sign means F_ // − r_

Why is it inverse square? Wait for Gauss' law in electricity.

F_12 = − F_21 Newton 3

Newton already knew Kepler's empirical law:

For planets, r3 ∝ Τ 2 orbit radius and period

Now if F ∝ ac ∝ 1r2

then constant = acr2 = rω2r2 = r3ω2

Planet r from sun T ω rω2 r3ω2

million km Ms rad.s-1 ms-2

Mercury 58 7.62 8.25 10-7 3.95 10-5 1.31 1020 m3s-2

Venus 108 19.4 3.23 10-7 1.13 10-5 1.32 1020 m3s-2

Earth 150 31.6 1.99 10-7 5.94 10-6 1.33 1020 m3s-2

etc

51How big is G? Cavendish's experiment (1798)

F = − G m1m2

r2

From deflection and spring constant, calculate F, know m1 and m2, ∴ cancalculate G. G = 6.67 10-11 Nm2kg-2 ( or m3kg-1s-2

Now also weight of m: |W| = mg ≅ G m.MeRe2

∴ Cavendish first calculated mass of the earth:

Me = gre2

G = 9.8 m.s-2 x (6.37 106 m)2

6.67 10-11 Nm2kg-2 = 6.0 1024 kg

see http://www.physicscentral.com/action/action-01-5-print.htmland http://physics.usask.ca/~kolb/p404/cavendish/

Some numbersWhat is force between two oil tankers at 100 m?

F = − G m1m2

r2

What happens when more there are ≥ 3 bodies?Superposition principle.

F_all objects together = Σ F_individual

or F_1 = Σi F_1i

force on m1due to masses mi

continuousbody

F_1 = ∫body

d F_

52Shell theoremA uniform shell of mass M causes the same gravitational

force on a body outside is as does a point mass M located at thecentre of the shell, and zero force on a body inside it.

dθ

dm

R

Fg

M

mF=0

r

Fg = GMm

R2

Example. If ρearth were uniform (it isn't), how long would it takefor a mass to fall through a hole through the earth to the other side?

Mr = ρ.43 πr3

∴ Fr = − G mρ.4

3 πr3

r2

F = − Kr where K = Gmρ.43 π

∴ motion is SHM with ω = √ Km

Τ = 2πω =

2π

√Gρ.43 π

= 2π

√GM/R3 = .... = 84 minutes

∴ falls through (one half cycle) in 42 minutes (actually faster for real density profile)

53Gravity near Earth's surface

W = |Fg| = G MmRe2

W = mgo = G Mem

r2

go is acceln in an inertial (non-rotating) frame

go = G Mer2

Usually, r ≅ Re, but

go = G Me

(Re + h)2 = gs

Re

Re + h2

= gs

1

1 + h/Re

2 where gs isgo at surface

Other complications:i) Earth is not uniform (especially the crust) useful for prospecting

ii) Earth is not sphericaliii) Earth rotates (see Foucault pendulum)

(Weight) = − (the force exerted by scales)At poles, F_ − N_ = 0At latitude θ, F_ − N_ = ma_

where a = rω2 = (Re cos θ)ω2

= .... = 0.03 ms-2 at equator = 0 at poles

We define g_ = N_ m =

F_ − ma_ m

So g_ is greatest at the poles, least at the equator, and does not (quite)point towards centre.

horizontal _| g_Earth is flattened at poles

54Puzzle: How far from the earth is the point at which thegravitational attractions towards the earth and that towards the sunare equal and opposite? Compare with distance earth-moon(380,000 km)

Fe = FsGMem

d2 = GMsm(re - d)2

Me(re − d)2 = Msd2

re2 − 2red + d2 = MsMe

d2

Ms

Me − 1 d2 + 2red − re2 = 0

d = .... = ?

Gravitational field. A field is ratio of force on a particle to someproperty of the particle. For gravity, (gravitational) mass is theproperty:

F_grav

m = g_ = g_(r_) is a vector field

cf electric field F_elec

q = E_(r_) later in syllabus

Gravitational potential energy. Revision:Potential energyFor a conservative force F_ (i.e. one where work done against it, W = W(r_))we can define potential energy U by ∆U = Wagainst. i.e.

∆U = − ∫i

f F_ . dr__

near Earth's surface, F_ g = mg_ ≅ constant

= − ∫i

f (-mgk_) . (dxi_ + dyj_ + dzk_)

= mg k_ . k_ ∫i

f dz = mg (zf - zi)

choose reference at zi = 0, so U = mgz

Gravitational potential energy of m and M.

M mr

F dsg

∆Ug ≡ − ∫i

f F_g . ds__ = ∫

i

f Fgdr = ∫

i

f G

Mmr2

dr = − GMm[1rf

− 1ri]

Convention: take ri = ∞ as reference: U(r) = − GMm

rU = work to move one mass from ∞ to r in the field of the other. Always negative.Usually one mass >> other, we talk of U of one in the field of the other, but it is U of both.

55Escape "velocity".Escape "velocity" is minimum speed ve required to escape, i.e. toget to a large distance (r → ∞).

M

mrR

v

Projectile in space: no non-conservative forces soconservation of mechancial energy

Ki + Ui = Kf + Uf

12 mve2 −

GMmR = 0 + 0

vesc = √2GMR

For Earth: vesc = √2 6.67 10-11 m3kg-1s-2 5.98 1024 kg6.37 106 m

= 11.2 km.s-1 = 40,000 k.p.h.

Put launch sites near equator: veq = Reωe = 0.47 km.s-1

Question In Jules Verne's "From the Earth to the Moon", theheros' spaceship is fired from a cannon*. If the barrel were 100 mlong, what would be the average acceleration in the barrel?

vf2 − vi2 = 2as

a = ve2 - 0

2s = (1.12 104 ms-2)2

2 x 100 m

= 630,000 ms-2 = 64,000 g

* why? If you burn all the fuel on the ground, you don't have toaccelerate and to lift it. Much more efficient.

Planetary motion"The music of the spheres" - Plato

Leucippus & Democritus C5 B.C.heliocentric universe

Hipparchus (C2 BC) & Ptolemy (C2 AD) geocentric universeTycho Brahe (1546-1601) - very many, very careful, naked eyeobservations.Johannes Kepler joined him. He fitted the data to these empiricallaws:Kepler's laws:1 All planets move in elliptical orbits, with the sun at one focus.

Except for Pluto and Oort cloud objects, these ellipses are ≅ circles.

Msun >> mplanet, so sun is ≅ c.m.

2 A line joining the planet to the sun sweeps out equal areas inequal time.

Slow at apogee (distant), fast at perigee (close)

3 The square of the period ∝ the cube of the semi-major axisSlow for distant, fast for close

56Newton's explanations:Law of areas:

Area = 12 r.rδθ

i.e. for same δt, 12 r2δθ = constant

Conservation of angular momentum L_. Sun at c.m.

∴ |L_| = |r_ x p_| = |r_ x mv_| momentum= p. see later

= mrvtangential

= mr.rω = mr2δθδt

= mδt r

2δθ = constant.

Conservation of L_ ⇒ Kepler 2.Law of periods: (we consider only circular orbits)

Kepler 3: T2 ∝ r3

Newton 2 → F = ma = m rω2

G Mmr2

= mr

2π

T2

T2 =

4π2

GM r3 → Kepler 3

(works for elipses with semi-major axis a instead of r)

Newton 2 & Newton's gravity ⇒ Kepler 3

Newton 2 & Newton's gravity also ⇒ Kepler 1

Example What is the period of the smallest earth orbit? (r ≅ Re)What is period of the moon? (rmoon = 3.82 108 m)

T1 = √ 4π2

GM r3 = √ 4 π2

6.67 10-11 5.98 1024 (6.37 106)3 s

= 84 min

Kepler 3: T2 ∝ r3

T2T1

=

r2

r1

3/2 =

3.82 108

6.37 1063/2

= 464

T2 = 464 T1 = 27.2 days

For other planets: most have moons, so the mass of the planet canbe calculated from

T2 =

4π2

GM r3

57Orbits and energyNo non-conservative forces do work, so mechanical energy isconstant:

E = K + U

= 12 mv2 −

GMmr

Let's remove v. Consider circular orbit:

v2

r = ac = Fm =

GMmr2m

∴ 12 mv2 = 1

2 GMm

r E = K + U

= 12 GMm

r − GMm

r

= − GMm

2r

i.e. E = 12 U, or K = − 1

2 U, or K = − E.

Small r ⇒ U very negative, K large. (inner planets fast, outer slow)

Example A spacecraft in orbit fires rockets while pointingforward. Is its new orbit faster or slower?

F_ // ds__ ∴ Work done on craft

W = ∫ F_ . ds__ > 0.

∴ E = − GMm

2r increases, i.e. it becomes less negative. (R is

larger). K = - E, ∴ K smaller, so it travels more slowly.called "Speeding down"

Quantitatively:Ki = − Ei Kf = − Ef = − (Ei + ∆E)Kf = Ki − ∆E12 mvf2 = 1

2 mvf2 − W

Looks odd, but need lots of work to get to a high, slow orbit.

58Manœuvring in orbit.

To catch up, vessel 1 fires engines backwards, and loses energy. Itthus falls to a lower orbit where it travels faster, until it catches up. Itthen fires its engines forwards in order to slow down (it climbsback to the original, slower orbit).

Example: In what orbit does a satellite remain above the samepoint on the equator?

Called the Clarke Geosynchronous Orbit

i) Period of orbit = period of earth's rotationii) Must be circular so that ω constant

T = 23.9 hours

T2 =

4π2

GM r3

r = √3

GMT2

4π2 = .....

= 42,000 km popular orbit!

59PHYS 1121-1131. Systems of particles and extended bodies,Centre of mass, Momentum, Collisions Joe Wolfe, UNSW

Centre of massIn a finite body, not all parts have the same acceleration. Not even if it is rigid.How to apply F_ = m a_?

mi

mj

r ir j

r cm

origin

Total mass M = Σ miDefine the centre of mass as thepoint with displacement

r_ cm = Σ mi r_i

M

Why? n particles, mi at positions r_i, F_i acts on each. Total forceacting on all particles:

m i

m jF i

F j

r ir j

origin

Σ F_i = Σ mi a_i

= Σ mi d2

dt2 r_i

if masses constant:

= Σ d2

dt2 mi r_i

= d2

dt2 Σ mi r_i

= Md2

dt2

Σ

mi r_iM

But we defined r_ cm = Σ mi r_i

M

Then Σ F_i = Md2

dt2 r_ cm = M a_ cm

(total force) = (total mass)*(acceleration of centre of mass)

Look at forces in detail:

m i

mj

Fi

F j

F i,ext

F j,ext

F i,int

F j,int

Each F_i is the sum ofinternal forces (from otherparticles in the body/system) and external forces(from outside the system)

Σ F_i = Σ F_i, internal + Σ F_i, external

Newton 3: All internal forces F_ij between ith and jth particles arereaction pairs F_ji = − F_ij

∴ Σ internal forces = 0∴ Σ F_i = Σ F_i, external = F_ external

∴ F_ external = M a_ cm

( )totalexternal force = ( )total

mass *( )acceleration of centre of mass

60For n discrete particles, centre of mass at

r_ cm = Σ mi r_iΣ mi

= Σ mi r_i

M (i)

For a continuous body, elements of mass dm at r_

r_ cm =

∫body

r_ dm

∫body

dm =

∫body

r_ dm

M (ii)

Can rearrange (i):

0 = Σ mi r_i − mi r_cm

M → Σ mi (r_i − r_cm) = 0

(ii) → ∫body

(r_i − r_ cm)dm = 0

Later, when doing rotation, we'll consider

W

cmN

which is a useful way to find c.m. experimentally.

Example. Where is the c.m. of the earth moon system?

r_ cm = Σ mi r_iΣ mi

Take origin at centre of earth.

x cm = mexe + mmxm

me + mm =

mmdme + mm

= 4,600 km i.e. inside the earth.recall: we derived the centre of rotation of the system

Example

r

O

m1m 2

cm

On a square plate (mass mp), weplace m1 and m2 as indicated.mp = 80 g, m1 = 200 g and m2 =100 gWhere is the cm of the system?

r_ cm = Σ mi r_iΣ mi

= mp(1.5i + 1.5j ) + m1(2.0j ) + m2(2.0i + 2.0j )

mp + m1 + m2

= (320g)i + (720g)j

380g = 0.8 i + 1.9j check that Σ mi ( r_i − r_cm) = 0

61Example. Rod, cross-section A, made of length a of material withdensity ρ1 and length b of material with density ρ2. Where is c.m.?If ρ1/ = 2ρ2, and a = 2b, where is cm?

r_ cm = ∫ r_i dm∫ dm

dm = ρdV = ρAdx

x cm = ∫ x dm∫ dm

= ∫

-b

0 ρ1Ax dx + ∫

0

a ρ2Ax dx

∫-b

0 ρ1A dx + ∫

0

a ρ2A dx

= − 1

2 ρ1b2 + 1

2 ρ2a2

ρ1b + ρ2a

= a2 − rb2

2(a + rb) where r = ρ1ρ2

62MomentumDefinition: p_ ≡ mv_

Later we'll see that this is a low velocity approximation to

p_ = mov_

√1 − v2/c2

Generalised form of

Newton 2: Σ F_ = ddt p_

Σ F_ = m ddt v_ + v_

ddt m

If m constant, Σ F_ = m a_

System of particles: What is system? - you choose.

P_ = Σ p_ i and M = Σ mi

P_ = Σ mi v_ i = Σ mi ddt r_ i

= ddt Σ mi r_ i = M

ddt

Σ mi r_ i

M P_ = Mv_ cm

If M constant: ddt P_ = Ma_ cm

Σ F_ i = Σ ddt p_ i = ddt P_

All internal forces are in pairs F_ji = − F_ij

∴ F_ext = ddt P_ conclude:

i) Motion of cm is like that of particle mass M at r_ cmsubjected to F_ext.

ii) If F_ext = 0, momentum is conserved

Internal vs external work.Problem. Skateboarder pushes away from a wall

v

F F

Point of application of F_ does not move, ∴ normal force does nowork, but K changes. Where does energy come from?

Fext = Macm

Fext dx = Macm dxcm = Mdvcm dxcm

dt = Mvcmdvcm

"Centre of mass work"

Wcm = ∫i

f Fext dx = ( )1

2 Mvcm2f − ( )1

2 Mvcm2i

Work done = that which would have been done if Fext had acted on cm.

63Example 90 kg man jumps (vj = 5 ms-1) into a (stationary) 30 kgdinghy. What is their final speed? (Neglect friction.)

v i

v fm m m d

No external forces act in horizontal direction so Px is conserved. Pi = Pf

man dinghy man dinghy mmvj + 0 = (mm + md)vf

vf = mm

mm + md vj

Example Rain falls into an open trailer(area 10 m2) at 10 litres.min-1.m-2.

Neglecting friction, what F required to maintain constant speed of10 ms-1?

10 litres has mass 10 kg

Fx = ddt (mvx) = m

ddt vx + vx

ddt m

= 10 ms-1 x

10 kg.m-2

60 s 10 m2

= 17 N.

Example. Rocket has mass m = m(t), which decreases as it ejects

exhaust at rate r = − dmdt and at relative velocity u. What is the

acceleration of the rocket? ( )dmdt

= rate of increase of

mass of rocket < 0

No external forces act so momentum conserved. In the frame of therocket, forwards direction:

dprocket + dpexhaust = 0m.dv + (−dm).(−u) = 0

dv = − udmm

a = dvdt = −

um .

dmdt

a = urm 1st rocket equation

dv = − u dmm = dv = − u d(ln m)

∫i

f dv = vf - vi = u ln

mimf

2nd rocket equation

need high exhaust velocity u (c?), else require mi >> mf

64Collisions Definition: in a collision, "large" forces act betweenbodies over a "short" time.In comparison, we shall often neglect the momentum change due toexternal forces.Example 1:

Fab = Fba = 0

v a v b

Fab = Fba = large!

F ab F ba

Fab = Fba = 0forces that crumple cars during (brief) collision are much larger than friction force(tires - road), ∴ neglect Fext.

In previous example: car decelerates from 30 kph to rest in a 60 cm 'crumplezone'. Average a = 58 ms-2, so force on car during collision ~ ma ~ 58 kN,compared with friction at ~ 10 kN

Example 2Jupiter

spacecraft

doesn't "hit"

Here, start and finish of collision not well definedAt large separation before and after, Fab = Fba ≅ 0During collision (fly-by), forces are considerable.However, Fgrav ∝ 1/r2, so much smaller at large distances.

Impulse (J_) and momentumNewton 2 ⇒ dp_ = F_ dt

∴ ∫i

f dp_ = ∫

i

f F_ dt so

Definition: J_ ≡ p_f − p_i = ∫i

f F_ dt

Usual case: external forces small, act for small time, therefore ∫i

f F_ext dt is small.

m1 m 2

F F1 2

∆ p_1 = ∫i

f F_1 dt ≡

-F_1 ∆t ∆ p_2 = ∫

i

f F_2 dt = − ∫

i

f F_1 dt

∴ ∆ p_1 = − ∆ p_2∴ ∆ P_ = ∆ p_1 − ∆ p_2 = 0

If external forces make negligible change in the momentum of asystem, then the momentum of the system is conserved.

65Example. Cricket ball, m = 156 g, travels at 70 ms-1. What impulseis required to catch it? If the force applied were constant, whataverage force would be required to stop it in 1 ms? in 10 ms? Whatstopping distances in these cases?

vivf = 0

m = 0.156 kg, vi = 70 m.s-1 vf = 0.J_ ≡ p_f − p_i

= m(vf - vi) to right= .... = 10.9 kgms-1 to left.

J_ = ∫i

f F_ dt if F_ constant, J = F∆t.

∴ Fav = J/∆t.

Const F ⇒ const a. s = vav∆t. vav = 35 m.s-1

∆t 1 ms 10 msF 11 kN 1.1 kN ouch!

s 3.5 cm 35 cm

Example. (Common method to measure speed of bullet.) Bullet (m) withvb fired into stationary block (M) on string. (i) What is their(combined) velocity after the collision? (ii) What is the kinetic energyof the bullet? (iii) of the combination? (iv) How high does theblock then swing?

a-b): collision, no horizontal external forces ∴ momentumconserved. Friction does work, so mechanical energy is lost, notconserved

b-c): during this phase, external forces do act, so momentum islost, not conserved. However, there are no non-conservative forces,so mechanical energy conserved.

Mm

bvM+m

tv

a) b) c)

h

v=0

Note the different stages:

66Analyse a) to b)

Mm

bvM+m

tv

a) b)

No horizontal ext forces during collision ∴ momentum conservedi) Pxi = Pxf

mvb = (m + M)vt

vt = m

m + M vb

ii) K b = 12 mvb2

iii) K t = 12 (m + M)vt2 = 1

2 (m + M)

m

m + M vb2

= 12

m2

m + M vb2 < Kb.

Conclusion: Ui = Uf, Ki =/ Kf.Mechanical energy is not conserved - deformation of block is notelastic; heat is produced.During a collision with negligible external forces,

P_ = Mv_cm is conserved

M constant ∴ v_cm is constant ∴ 12 Mv_cm

2 constant

K of c.m. is not lost. But the K of components with respect to c.m.can be lost.Greatest possible loss of K: if all final velocities = v_cm, i.e. if allobjects stick together after collision.Called completely inelastic collision.

v1i v2i

m1 m 2

v

m1 m 2+

Completely elastic collision is one in which no non-conservativeforces do work, so mechanical energy is conserved.

Part (iv) of previous example (b-c):

M+m

tv

b) c)

h

v=0

here the external forces (gravity and tension) do do work and changemomentum. But there is no non-conservative force and so in thispart of the process conservation of mechanical energy applies:

∆U + ∆K = 0(M + m)g (∆h - 0) + (Kt - 0) = 0

∆h = ... = 12

m2

g(m + M)2 vb2

67Puzzle. Fly travelling West at 5m/s meets train travelling East at 30m/s.

?

v

t

initally, vfly = − 5 ms-1

finally, vfly = + 30 ms-1

At some time t, the fly travels at 0 ms-1

i) does this occur before or after the fly first touches thewindscreen.ii) how fast was the windscreen going when the fly wasgoing 0 ms-1?

68Elastic collision in one dimensionv 1i v 2i

m1 m 2

(bang)

v1f v2f

m1 m 2

Collision: neglect external forces ⇒pi = pfm1v1i + m2v2i = m1v1f + m2v2f (i)

elastic ⇒ Ki = Kf12 m1v1i2 + 1

2 m2v2i2 = 1

2 m1v1f2 + 1

2 m2v2f2 (ii)

usually know m1, m2, v1i, v2i. Two unknowns (v1f, v2f), ∴ we canalways solve.

Or: transform to frame where (e.g.) v1 = 0Or: transform to centre of mass frame.

Example. Take m1 = m2, v2i = 0, v1i = v.v v2i

m m

= 0

m m

v 1f v 2f

neglect external forces ⇒ pi = pfmv + 0 = mv1f + mv2f (i)

12 mv2 + 0 = 1

2 mv1f2 + 1

2 mv2f2 (ii)

(i) → v2f = v − v1f (iii)substitute in (ii) →

12 mv2 + 0 = 1

2 mv1f2 + 1

2 m(v2 + v1f2 - 2vv1f)

∴ 0 = v1f2 − vv1f0 = v1f(v1f − v) 2 solutions

Either: v1f = 0 and (iii)→ v2f = vi.e. 1st stops dead, all p and K transferred to m2

or: v1f = v and (iii)→ v2f = 0i.e. missed it.

Example Show that, for an elastic collision in one dimension, therelative velocity is unchanged.

i.e. show v1i - v2i = v2f - v1fp and K conservation gave:(i) m1(v1i - v1f) = m2(v2i - v2f)

(ii) 12 m1(v1i2 - v1f2) = 1

2 m2(v2i2 - v2f2)

If they hit, (v1i - v1f) /= 0, (v2i - v2f) /= 0(ii)/(i) ⇒ v1i + v1f = v2i + v2f

∴ v1i - v2i = v2f - v1f

i.e. relative velocity the same before and afterSolve → v2f = 2m1

m1 + m2 v1i +

m2 - m1m2 + m1

v2i

69Example Two similar objects, mass m, collide completelyinelastically.case 1: v1i = v, v2i = 0.case 2: v1i = v, v2i = − v.What energy is lost in each case?p conserved → mv1i + mv2i = 2mvf

vf = v1i + v2i

2

∆Κ = Kf − Ki = 12 (2m)vf2 − 1

2 mv1i2 − 1

2 mv2i2

case 1: ∆K = 12 (2m)

v + 0

22 - 1

2 mv2

= − 14 mv2

case 2: ∆K = 12 (2m)

0 + 0

22 − 1

2 mv2 − 1

2 mv2

= − mv2 4 times as much energy lost

Elastic collisions in 2 (& 3) dimensions

x

y

v 1i

b v2fθ2

θ1

v2f

Choose frame in which m2 stationary, v1i in x dirn

b is called impact parameter (distance "off centre")px conserved m1v1i = mv1f cos θ1 + mv2f cos θ2py conserved 0 = mv2f sin θ2 − mv1f sin θ1K conserved

12 m1v1i2 = 1

2 mv1f2 + 1

2 mv2f2 + ∆K (iii)

where ∆K = 0 for elastic case3 equations in v1f, v2f, θ1 and θ2: need more info (often given θ1 or θ2)

Incidentally: for billiard balls, neglecting rotation and friction (reasonable during collision, but not after)

θ2b

(R + R) sin θ2 = b θ2 = sin-1 b

2R

i) Note that as θ → 90°, small error in b gives large error in θ2. ii) Does b = R give θ2 = 30°?

70

θ

φ

s1

s2m1

m1

m2

C

vv1

v2

Example. Police report of roadaccident. Car 1, mass m1 strikesstationary car m2 at point C. They thenslide to rest in positions shown. Givenµk (assumed same for both) find theinitial speed v of m1. Can you checkassumption? (real example)

After collision, a for both = Ffm = −

Wm = − µg

vf2 − vi2 = 2as = − 2µgs

0 − v12 = − 2µgs1

v1 = √2µgs1 v2 = √2µgs2Neglect external forces during collision: ∆P = 0Px: m1v = m1v1 cos θ + m2v2 cos φ (i)Py: 0 = m1v1 sin θ − m2v2 sin φ (ii)

(i) ⇒ v = √2µgs1 cos θ + m2m1

√2µgs2 cos φ

Note the "spare" equation—we can use it to check the model or assumptions:

(ii) ⇒ m1√2µ1gs1 sin θ = m2√2µ2gs2 sin φµ2µ1

= s1m12 sin2 θs2m22 sin2 φ

(The µ may not be the same for the two: surfaces different etc)

1

2

n

S

G

h

Example A building has S stories,each of height h. An explosiondestroys strength of nth floor. Howlong before (n+1)th floor hitsground, falling vertically?Assume inelastic collisions betweenfloors. To obtain a lower estimate,assume negligible strength betweenfloors.

71Example

v z

v b

v f

θ

x

y

Balloonist Albert writes message on a bottle (1 kg) and drops it overthe side. It is falling vertically at 40 m.s-1 when caught by parachutistZelda (m = 50 kg), travelling at 1 ms-1 at 45° to vertical. Collision(bottle—Zelda's hand) lasts 10 ms.

i) If only gravity acted, what is ∆p for Zelda over 10 ms ?ii) Neglecting ext forces during collision, what is thevelocity of (Zelda+bottle) after collision?iii) What impulse is applied to bottle during collision?iv) What is the impulse applied to Zelda?v) What is the average force during collision?

i) due to W_ , ∆p_ = W_ ∆t = .. = 5 kgm.s-1 downii) Neglect ext forces ⇒ momentum conserved.

mb v_bi + mZ v_Zi = m(Z+b) v_(Z+b)f1(-40 j_) + 50(1 cos 45° i_ − 1 cos 45° j_) = 51(vx i_ + vy j_)

i_ dirn: vx = cos 45° .5051 x1 = 0.7 ms-1

j_ dirn: vy = − 1x40 − 50 cos 45°

51 = −1.5 ms-1

∴ |vf| = √vx2 + vy2 = 1.6 ms-1

θf = tan-1 vyvx

⇒ 67° to horizontal

iii) J _b ≡ p_bf − p_bi = 1x(vx i_ + vy j_) - 1(-40 j_)

= (1.6 i_ + 38 j_) kgm.s-1

iv) J _Z = − J _b = − (1.6 i_ + 38 j_) kgm.s-1

| J _Z| = √1.62 + 382 = 38 kgm.s-1

v) -F_Z =

∆ pZ_∆t =

| J _Z|∆t = .. = 380 N