Problems Chap 4 45, 53, Chap 5: 13, 19, 27, 45, 50, 53Problems Chap 4 45, 53, Chap 5: 13, 19, 27, 45, 50, 53

Summary Lecture 4Summary Lecture 4

Dynamics4.84.8 Relative motion in 1DRelative motion in 1D4.94.9 Relative motion in 3-DRelative motion in 3-D4.74.7 Circular motion (study at home) Circular motion (study at home) 5.25.2 Newton 1Newton 15.3-5.55.3-5.5 Force, mass, and Newton 2Force, mass, and Newton 25.75.7 Newton 3 Newton 3 5.85.8 Some examples Some examples 6.1-26.1-2 FrictionFriction6.46.4 Drag forceDrag force

Summary Lecture 4Summary Lecture 4

Dynamics4.84.8 Relative motion in 1DRelative motion in 1D4.94.9 Relative motion in 3-DRelative motion in 3-D4.74.7 Circular motion (study at home) Circular motion (study at home) 5.25.2 Newton 1Newton 15.3-5.55.3-5.5 Force, mass, and Newton 2Force, mass, and Newton 25.75.7 Newton 3 Newton 3 5.85.8 Some examples Some examples 6.1-26.1-2 FrictionFriction6.46.4 Drag forceDrag force

Labs and Tutes start this week.

Check times and places on notice board: level 2

physics podium.

Thursday 12 – 2 pm

“Extension” lecture.

Room 211 podium level

Turn up any time

If it is moving with a constant velocity, IT WILL CONTINUE TO DO SO

In the absence of a FORCE

a body is at rest

1660 AD

A body only moves if it is driven.

In the absence of a FORCE

A body at rest WILL REMAIN AT REST

350 BC

DynamicsDynamicsAristotle

For an object to MOVE

we need a force.

Newton

For an object to CHANGE its motion

we need a force

Newtons mechanics applies for motion in

an inertial frame of reference! ???????

He believed that there existed an absolute (not accelerating) reference frame, and an absolute time.

The laws of physics are always the same in any inertial reference frame.

His laws applied only when measurements were made in this reference frame…..

Newton clarified the mechanics of motion in the “real world”.

…or in any other reference frame that was at rest or moving at a constant velocity relative to this absolute frame.

Inertial

reference

frame

Inertial

reference

frame

Inertial reference

frame

Newton believed that there existed an absolute (not accelerating) reference frame, and an absolute time.

The laws of physics are always the same in any inertial reference frame.

Einstein recognised that all measurements of position and velocity (and time) are relative.

There is no absolute reference frame.

Frames of Reference

The reference frames may have a constant relative velocity

We need to be able to relate one set of measurements to the other

The connection between inertial reference frames is the “Gallilean transformation”.

In mechanics we need to specify position, velocity etc. of an object or event.

This requires a frame of reference

x

y

z

o

p

The reference frames for the same object may be different

x’

y’

z’

o

Ref. Frame P (my seat in plane)

T (Lunch Trolley)xTP

Ref. Frame G (ground)

xPG

xTG

xTG = xTP + xPG

Vel. = d/dt(xTG) vTG = vTP + vPG

Accel = d/dt (vTG) aTG = aTP + aPG

In any inertial frame the laws of physics are the same

VPG(const

)

Gallilean transformations

0

N

E

GROUND

N’

E’

AIR

r PG

r AG

r PA

aPG = (vPG) = aPA + aAGdtd

vAG

rPG = rPA + rAG

dtd

vPG = (rPG) = vPA + vAG

PLooking

from above

0

In any inertial frame the laws of physics are the same

AG

PA

PA AG

VPG = VPA + VAG VPA = 215 km/h to East

VAG = 65 km/h to North

hkmv

v

vvv

PG

PG

agpaPG

/225

422546225

22

Tan = 65/215

= 16.8o

Ground Speed of Plane

AG

PA

PA AG

Ground Speed of Plane

In order to travel due East, in what direction relative to the air must the plane travel? (i.e. in what direction is the plane pointing?), and what is the plane’s speed rel. to the ground?

Do this at home and then do sample problem 4-11 (p. 74)

VPA = 215 km/h to East

VAG = 65 km/h to North

Isaac Newton

1642-1727

Newton’s 1st Law

If a body is at rest, and no force acts on it, IT WILL REMAIN AT REST.

If it is moving with a constant velocity, IT WILL CONTINUE TO DO SO

If a body is moving at constant velocity, we can always find a reference frame where it is AT REST.

At rest moving at constant velocity

An applied force changes the velocity of the body

a F

ForceForceIf things do not need pushing to move at constant velocity, what is the role of FORCE???

Inertial mass

The more massive a body is, the less it is accelerated by a given force.

a = F 1m

F = am

The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line

in which that force is impressed.

a = F/m

?

How to measure the Inertial Mass

m

kfreq

mgweightextensionF

Balances the gravitational weight

Which would work in a space ship away from gravity?

mmeasures inertial mass

Newton’s 2nd Law

a = F/m

aa = = vector SUM of ALL EXTERNAL forces acting on the body

vector SUM of ALL EXTERNAL forces acting on the body

mm______________

F = ma

vector SUM of ALL EXTERNAL forces acting on the body

vector SUM of ALL EXTERNAL forces acting on the body

= ma= ma

Newton’s 2nd Law

F=maF = iFx + jFy + kFz a = iax + jay + kaz

Fx=max

Fy=may

Fz=maz

Applies to each component of the vectors

FA+ FB+ FC= 0 since a= F/m and a = 0

F = 0

Fx = 0 Fy = 0

FCcos - FAcos47= 0

= 220 sin 47 +170 sin 28

= 160 +80 240 N

FA sin 47 + FCsin – FB = 0

FB = FA sin 47 +FCsin

220N

? N

170N

Fy = 0

cos= cos47 =280220170

170 cos - 220 cos47= 0

(280)

A (M2 – M1)g

B (M1 – M2)g

C M1g

D Unknown

m2g

N

TM2

m1g

TM1

Apply F = ma to each body

i.e Fx = max and Fy = may

For m2

Vertically

Fy = -N + m2g = m2ay = 0

N = m2g

Horizontally

Fx = T = m2a

21

1

mmgm

a

For mass m1

Vertical (only)

m1g - T = m1a

Analyse the equation!

m1g - m2a = m1a

m1g = m2a + m1a = a(m2 + m1)

41%

26%

33%

2.5 x 103 N 2.5 x 103 N 3.9 x 104 N

2.5 x 103 N 2.5 x 103 NT

3.9 x 104 N

F = ma m is mass of road train

5.0 x 103 N

3.9 x 104 N

F = ma

a = (39 – 5) x 103/(4 x 104)

a = F/m

a

a = 0.85 m s-2

2.5 x 103 N 2.5 x 103 N Don’t care!T

F = ma m is mass of trailer!

F is net force on TRAILER

2.5 x 103 N

Ta = 0 So F= 0

T - 2.5 x 103 = 0

T = 2.5 x 103 N

a = 0

2.5 x 103 N 2.5 x 103 N

3.9 x 104 N

F = ma m is mass of roadtrain

No driving force so F = -5.0 x 103 N a = - 5.0 x 103/4.0 x 104

= - 0.125 m s -2

Use v2 = vo2 + 2a(x – x0)

x – x0 = 400 mv = 0 u = 20 m s-1,

Newton’s 3rd Law

Forces come in pairs

To every action there is an equal and opposite reaction.

The action-reaction pairs

ALWAYS act on DIFFERENTDIFFERENT bodies

NAG

mg

What is reaction pair to the weight force mg?

Is it N (normal reaction)?

N

The reaction pair is the force of the apple on the Earth

N is the force of the ground on the apple NGA

The reaction pair to NGA (or N) is the force of the apple on the ground, NAG

NGA

NO! . N and mg act on the SAME body

According to stationary observer

R

mgF = ma

Taking “up” as +ve

R - mg = ma

R = m(g + a)

If a = 0 ==> R = mg normal weight

If a is +ve ==> R = m(g + a) weight increase

If a is -ve ==> R = m(g - a) weight decrease

R is reaction force

= reading on scales

Measured weight in an accelerating Reference Frame

accel a

Spring scales

Man in lift

According to traveller

F = ma

R - mg = ma

BUT in his ref. frame a = 0!

so R = mg!!

How come he still sees R changing as lift accelerates?

R

mg

R is reaction force

= reading on scales

Only if it is an inertial frame of reference! The accelerating lift is NOT!

Didn’t we say the laws of physics do not depend on the frame of reference?

Here endeth

the lessonlecture

No. IV

Summary Lecture 5Summary Lecture 5

6.1-2 Friction

6.4 Drag force

Terminal velocity

6.5 Taking a curve in the road

Summary Lecture 5Summary Lecture 5

6.1-2 Friction

6.4 Drag force

Terminal velocity

6.5 Taking a curve in the road

Problems Chap 6: 5, 14, 29 , 32, 33, Problems Chap 6: 5, 14, 29 , 32, 33,

Staff-Student Liaison Committee

This committee meets 3 times each semester, to consider reports from lectures, labs. and tutorials. The meetings are after 5 pm, and last ~1 hour.

We need a rep. from this class to report comments from this lecture group.

mgmg

Why doesn’t Mick Doohan fall over?Why doesn’t Mick Doohan fall over?

Friction provides the central force

Friction provides the central force

In the rest reference frameIn the rest reference frame

What is Friction•Surfaces between two materials are not even

•Microscopically the force is atomic

Smooth surfaces have high friction

•Causes wear between surfaces

Bits break off

•Lubrication separates the surfaces

The Source of Friction between two surfaces

fF

mg

Static Friction

As F increases friction f increases in the opposite direction. Therefore Total Force on Block = zero does not move

F is now greater than f

and slipping begins

If no force F

No friction force fSurface with friction

As F continues to increase, at a critical point most of the (“velcro”) bonds break and f decreases rapidly.

fF

f depends on surface properties.

Combine these properties into a coefficient of friction

f N

is usually < 1

Static f < or = s N

Surface with friction

Kinetic f = k N

f

F

f < fmax (= kN )

Static friction

Kinetic friction

Coefficient of Kinetic friction < Coefficient of Static friction

Slipping begins (fmax = sN )fmax

mg

Ff

At crit

F = f

mg sin crit = f = S N

Independent of m, or g.

Property of surfaces only

S = tan crit

mg sin

mg cos

= S mg coscrit

crit

crit

cosmgsinmg

thus S =

F1

F2

Making the most of Friction

A F1 > F2

B F1 = F2

C F1 < F2

mg

NN

mg

f1 f2

f = S N

Friction force does not depend on area!

f = S mg

So why do Petrol Heads use fat tyres?

To reduce wear

Tyres get hot and sticky which effectively increases .

The wider the tyre the greater the effect.

To reduce wear

Tyres get hot and sticky which effectively increases .

The wider the tyre the greater the effect.

The tru

th!

Frictio

n is not a

s sim

ple as P

hysics 1

41 says!

The tru

th!

Frictio

n is not a

s sim

ple as P

hysics 1

41 says!tribophysics

Force of Tyre on road

Force of road on Tyre

acceleration

What force drives the car?

Driving Torque

Braking force

Friction road/tyres

v

d

f

v2 =vo2 + 2a(x-xo)

0 = vo2 + 2ad a

vd

2

20

F = ma

= smgMax value of a is when f is max.

Stopping Distance depends on friction

amax = - sg

-fmax = mamax -smg = mamax

vo

N

mg

fmax = sN

Thus since a

vd

2

20

maxmin a

vd

2

20

gv

ds

min

2

20

dmin depends on v2!! Take care!!

If v0 = 90 kph (24 m s-1) and = 0.6 ==> d = 50 m!!

r

v Fcent

mg

N

rmv

F2

cent

Fcent is provided by friction.

If no slipping the limit is when

Fcent = fs(limit)= sN = smg

grμv

rmv

mgμ

s

2

s

So that

Does not depend on m

So for a given s (tyre quality) and given r there is a maximum vel. for safety.

If s halves, safe v drops to 70%….take care!

Taking a curve on Flat surface

Lateral Acceleration of 4.5 g

The lateral acceleration experienced by a Formula-1 driver on a GP circuit can be as high as 4.5 g

This is equivalent to that experienced by a jet-fighter pilot in fast-turn manoeuvres.

Albert Park GP circuitCentral force provided Central force provided by friction.by friction.Central force provided Central force provided by friction.by friction.

mg

N

= v= v22/Rg/Rg

= 4.3= 4.3

= v= v22/Rg/Rg

= 4.3= 4.3

mvmv22/R = /R = N = N = mgmgmvmv22/R = /R = N = N = mgmg

R = 70 m mv2/R

V=

55 m

s-1

for racing tyres is ~ 1 (not 4!).

How can the car stay on the road?

for racing tyres is ~ 1 (not 4!).

How can the car stay on the road?

Soft rubber

Grooved tread

Are these just for show, or advertising?

200 km/h

Another version of Newton #2

amF p= mv =momentum

F is a measure of how much momentum is transferred in time t

t

pF

dt

vdm

dt

)vm(d

dt

pd

Momentum p transferred over a time t gives a force:-

Distance travelled in 1 sec @ velocity v Distance travelled in 1 sec @ velocity v

Volume of air hitting each spoiler (area A) in 1 secVolume of air hitting each spoiler (area A) in 1 sec

Area A m2

mass of air (density ) hitting each spoiler in 1 secmass of air (density ) hitting each spoiler in 1 sec

Momentum of air hitting each spoiler in 1 secMomentum of air hitting each spoiler in 1 sec

If deflected by 900, mom change in 1 secIf deflected by 900, mom change in 1 sec

Newton says this is the resulting forceNewton says this is the resulting force

@ 200 kph v = 55 m s-1

A ~ 0.5 m2

~ 1 kg m-3

F ~ 3 x 104 N

~ 3 Tonne!

= v m

= v x A m3

= x v x A kg

= x v2 x A kg m s-1

mvmv22/R = /R = N = N = mgmgmvmv22/R = /R = N = N = mgmg

mvmv22/R = /R = N = N = (m + 3000) g(m + 3000) gmvmv22/R = /R = N = N = (m + 3000) g(m + 3000) g

VISCOUS DRAG FORCEVISCOUS DRAG FORCEDRAG

VISCOUS DRAG FORCE

Assumptions

low viscosity (like air)

turbulent flow

What is it?

like fluid friction

a force opposing motion as fluid flows past object

What does the drag force depend on?D D velocity (v velocity (v22))

D D effective area (A) effective area (A)

D D fluid density ( fluid density (

D D A vA v22

D= ½ C A v2

D D velocity (v velocity (v22))

D D effective area (A) effective area (A)

D D fluid density ( fluid density (

D D A vA v22

D= ½ C A v2

C is the Drag coefficient.

It incorporates specifics like

shape, surface texture etc.

v

Fluid of density

V m

Volume hitting object in 1 sec. =AV

Mass hitting object in 1 sec. = AV

momentum (p) transferred to object in 1 sec. = ( AV)V

Force on object = const AV2

t

pF

Area A

In 1 sec a length of V metres hits the object

V

mg

mg

D

mg

D

V

V=0

F = mg - D

F = mg -1/2CAv2

D increases as v2

until F=0

i.e. mg= 1/2CAv2

AC

mg2v

AC

mg2v

term

term2

0mgAv1/2Cdt

dvm 2

F = mg –DD

mg

ma = mg -D

D- mgdt

dvm

2/1Ac

m2

)]e1(Ac

gm2[v

t

2/1]Ac

gm2[v

2/1Ac

m2

)]e1(Ac

gm2[v

t

AC

mg2vterm

When entertainment defies reality

D= ½ CAv2

Assume C = 1

v = 700 km h-1

Calculate:

Drag force on presidents wife

Compare with weight force

Could they slide down the wire?

D= ½ CAv2

Assume C = 1

v = 700 km h-1

Calculate:

The angle of the cable relative to horizontal.

Compare this with the angle in the film (~30o)

In working out this problem you will prove the expression for the viscous drag force

2AvC2

1F